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Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements

July 20, 2024July 19, 2024 by Prasanna

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements

Question 1.
What is a measurement? How is measured quantity expressed?
Answer:

A measurement is a comparison with internationally accepted standard measuring unit.
The measured quantity (M) is expressed in terms of a number (n) followed by a corresponding unit (u) i.e., M = nu.

Example:
Length of a wire when expressed as 2 m, it means value of length is 2 in the unit of m (metre).

Question 2.
State true or false. If false correct the statement and rewrite. Different quantities are measured in different units.
Answer: True.
[Note: Choice of unit depends upon its suitability for measuring the magnitude of a physical quantity under consideration. Hence, we choose different scales for same physical quantity.]

Question 3.
Describe briefly different types of systems of units.
Answer:
System of units are classified mainly into four types:

C.G.S. system:
It stands for Centimetre-Gram-Second system. In this system, length, mass and time are measured in centimetre, gram and second respectively.
M.K.S. system:
It stands for Metre-Kilogram-Second system. In this system, length, mass and time are measured in metre, kilogram and second respectively.
F.P.S. system:
It stands for Foot-Pound-Second system. In this system, length, mass and time are measured in foot, pound and second respectively.
S.I. system:
It stands for System International. This system has replaced all other systems mentioned above. It has been internationally accepted and is being used all over world. As the SI units use decimal system, conversion within the system is very simple and convenient.

Question 4.
What are fundamental quantities? State two examples of fundamental quantities. Write their S.J. and C.G.S. units.
Answer:
Fundamental quantities:
The physical quantities which do not depend on any other physical quantity for their measurements i.e., they can be directly measured are called fundamental quantities.
Examples: mass, length etc.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 1

Question 5.
What are fundamental units? State the S.l. units of seven fundamental quantities.
Answer:
Fundamental units:
The units used to measure fundamental quantities are called fundamental units.
S.I. Units of fundamental quantities:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 2

Question 6.
State and describe the two supplementary units.
Answer:
The two supplementary units are:
i) Plane angle (dθ):
a. The ratio of kngth of arc (ds) of an circle to the radius (r) of the circle is called as Plane angle (dθ)
i.e., dθ = \(\frac{\mathrm{ds}}{\mathrm{r}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 3
b. Thus, dθ is angle subtended by the arc at the centre of the circle.
c. Unit: radian (rad)
d. Denoted as θ^c
e. Length of arc of circle = Circumference of circle = 2πr.
∴ plane angle subtended by entire circle at its centre is θ = \(\frac{2 \pi \mathrm{r}}{\mathrm{r}}\) = 2π^c

ii) Solid angle (dΩ):
a. solid angle is 3-dimensional analogue of plane angle.
b. Solid angle is defined as area of a portion of surface of a sphere to the square of radius of the sphere.
i.e., dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 4
c. Unit: Steradian (sr)
d. Denoted as (Ω)
e. Surface area of sphere = 4πr²
∴ solid angle subtended by entire sphere at its centre is Ω = \(\frac{4 \pi r^{2}}{r^{2}}\) = 4π sr

Question 7.
Derive the relation between radian and degree. Also find out 1” and 1’ in terms of their respective values in radian. (Take π = 3.1416)
Answer:
We know that, 2 π^c = 360°
∴ π^c = 180°
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 5

Question 8.
What are derived quantities and derived units? State two examples. State the corresponding S.L. and C.G.S. units of the examples.
Answer:

Derived quantities: Physical qUantities other than fundamental quantities which depend on one or more fundamental quantities for their measurements are called derived quantities.
Derived units: The units of derived quantities which are expressed in terms of fundamental units for their measurements are called derived units.
Examples and units:

Question 9.
Classify the following quantities into fundamental and derived quantities: Length, Velocity, Area, Electric current, Acceleration, Time, Force, Momentum, Energy, Temperature, Mass, Pressure, Magnetic induction, Density.
Answer:
Fundamental Quantities: Length, Electric current, Time, Temperature, Mass.

Derived Quantities: Velocity, Area, Acceleration, Force, Momentum, Energy. Pressure, Magnetic induction, Density

Question 10.
Classify the following units into fundamental, supplementary and derived units:
newton, metre, candela, radian, hertz. square metre, tesla, ampere, kelvin, volt, mol, coulomb, farad, steradian.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 7

Question 11.
List the conventions followed while using SI units.
Answer:
Following conventions should be followed while writing S.I. units of physical quantities:

Unit of every physical quantity should be represented by its symbol.
Full name of a unit always starts with smaller letter even if it is named after a person, eg.: 1 newton, 1 joule, etc. But symbol for unit named after a person should be in capital letter, eg.: N after scientist Newton, J after scientist Joule, etc.
Symbols for units do not take plural form.
Symbols for units do not contain any full stops at the end of recommended letter.
The units of physical quantities in numerator and denominator should be written as one ratio. For example the SI unit of acceleration is m/s² or m s^-2 but not m/s/s.
Use of combination of units and symbols for units is avoided when physical quantity is expressed by combination of two. For example, The unit J/kg K is correct while joule/kg K is not correct.
A prefix symbol is used before the symbol of the unit.
- a. Prefix symbol and symbol of unit constitute a new symbol for the unit which can be raised to a positive or negative power of 10.
  For example,
  1 ms = 1 millisecond = 10^-3 s
  1 μs = 1 microsecond = 10^-6 s
  1 ns = 1 nanosecond = 10^-9 s
- b. Use of double prefixes is avoided when single prefix is available
  10^-6 s = 1 μs and not 1 mms
  10^-9 s = 1 ns and not 1 mμs
Space or hyphen must be introduced while indicating multiplication of two units e.g., m/s should be written as m s^-1 or m-s^-1.

Solved Examples

Question 12.
What is the solid angle subtended by the moon at any point of the Earth, given the diameter of the moon is 3474 km and its distance from the Earth 3.84 × 10⁸ m?
Solution:
Given: Diameter (D) = 3474 km
∴ Radius of moon (R) = 1737 km
= 1.737 × 10⁶ m
Distance from Earth r = 3.84 × 10⁸ m
To find: Solid angle (dΩ)
Formula: dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)

Calculation:
From formula,
dΩ = \(\frac{\pi \mathrm{R}^{2}}{\mathrm{r}^{2}}\) ……..( cross-sectional area of disc of moon = πR²)
dΩ = \(\frac{\pi \times\left(1.737 \times 10^{5}\right)^{2}}{\left(3.84 \times 10^{8}\right)^{2}}\)
= \(\frac{3.412 \times(1.737)^{2} \times 10^{10}}{(3.84)^{2} \times 10^{16}}\)
= antilog{log(3.142) + 2log(1.737) – 2log(3.84)} × 10^-6
= antilog {0.4972 + 2(0.2397) – 2(0.5843)} × 10^-6
= antilog{0.4972 + 0.4794 – 1.1686} × 10^-6
= antilog{\(\overline{1}\) .8080} × 10^-6
= 6.428 × 10^-1 × 10^-6
= 6.43 × 10^-5 sr
Solid angle subtended by moon at Earth is 6.43 × 10^-5 sr
[Note: Above answer is obtained substituting value of r as 3.142]

Question 13.
Pluto has mean diameter of 2,300 km and very eccentric orbit (oval shaped) around the Sun, with a perihelion (nearest) distance of 4.4 × 10⁹ km and an aphelion (farthest) distance of 7.3 × 10⁹ km. What are the respective solid angles subtended by Pluto from Earth’s perspective? Assume that distance from the Sun can be neglected.
Solution:
Given: Radius of Pluto. R = \(\frac{2300}{2}\) km
= 1150km
Perihelion distance r_p = 4.4 × 10⁹ km
Aphelion distance r_a = 7.3 × 10⁹ km
To find: Solid angles (dΩ_p and dΩ_a)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 8
Solid angle at perihelion distance is 2.146 × 10^-13 sr and at aphelion distance is 7.798 × 10^-14 sr.

Question 14.
Define a metre.
Answer:
The metre is the length of the path travelled by light in vacuum during a time interval of 1/299, 792, 458 of a second.
Answer:

Question 15.
What ¡s parallax?
Answer:

Parallax is defined as the apparent change in position of an object due to a change in position of an observer.
Explanation: When a pencil is held in front of our eyes and we look at it once with our left eye closed and then with our right eye closed, pencil appears to move against the background. This effect is called parallax effect.

Question 16.
What is parallax angle?
Answer:
i) Angle between the two directions along which a star or planet is viewed at the two points of observation is called parallax angle (parallactic angle).
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 9
ii) It is given by θ = \(\frac{b}{D}\)
where, b = Separation between two points of observation.
D = Distance of source from any point of observation.

Question 17.
Explain the method to determine distance of a planet from the Earth.
Answer:

Parallax method is used to determine distance of different planets from the Earth.
To measure the distance ‘D’ of a far distant planet S, select two different observatories (E₁ and E₂).
The planet should be visible from E₁ and E₂ observatories simultaneously i.e. at the same time.
E₁ and E₂ are separated by distance ‘b’ shown in figure.
∴ E₁E₃ = b
The angle between the two directions along which the planet is viewed, can be measured. It is parallax angle, which in this case is L ∠E₁E₂ = θ
The planet is far away from the (Earth) observers, hence
b < <D
∴ \(\frac{b}{D}\) < < 1 and ‘θ’ is also very small.
Hence, E₁E₂ can be considered as arc of length b of circle with S as centre and D as radius.
:. E₁S = E₂S = D
∴ θ = \(\frac{b}{D}\) . . . .(θ is taken in radian)
∴ D = \(\frac{b}{\theta}\)
Thus, the distance ‘D’ of a far away planet ‘S’ can be determined using the parallax method.

Question 18.
Explain how parallax method is used to measure distance of a star from Earth.
Answer:

The parallax measured from two farthest distance points on Earth for stars will be too small and hence cannot be measured.
Instead, parallax between two farthest points (i.e., 2 ΔU apart) along the orbit of Earth around the Sun (s) is measured.

Question 19.
Explain how size of a planet or star is measured.
Answer:

To determine the diameter (d) of a planet or star, two diametrically opposite points of the planet are viewed from the same observatory.
If d is diameter of planet or star, angle subtended by it at any single point on the Earth is called angular diameter of planet.
Let angle α be angle between these two directions.
If distance between the Earth and planet or star (D) is known, α = \(\frac{\mathrm{d}}{\mathrm{D}}\)
This relation gives, d = α D
Thus, diameter (d) of planet or star can be determined.

Question 20.
Name the devices used to measure very small distances such as atomic size.
Answer:
Devices used are:
Electron microscope, tunnelling electron microscope.

Question 21.
Just as large distances are measured in AU, parsec or light year, atomic or nuclear distances are measured with the help of microscopic units. Match the units given in column A with their corresponding SI unit given in column B.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 39
Answer:
i. – (b)
ii. – (a)

Solved Examples

Question 22.
A star is 5.5 light years away from the Earth. How much parallax in arcsec will it subtend when viewed from two opposite points along the orbit of the Earth?
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 12
Solution:
Two opposite points-A and B along the orbit of the Earth are 2 AU apart. The angle subtended by AB at the position of the star is

= antilog{log(2.992) – log(5.5) – log(9.46)} × 10^-4
= antilog {0.4761 – 0.7404 – 0.9759} × 10^-4
= antilog {\(\overline{2}\).7598} × 10^-4
= 5.751 × 10^-2 × 10^-4
= 5.75 × 10^-6
= 5.75 × 10^-6 rad
= 5.75 × 10^-6 × 57.297 × 60 × 60 arcsec
…. (converting radian into arcsecond)
= 1.186 arcsec
Parallax is 1.186 arcsec

Question 23.
The moon is at a distance of 3.84 × 10⁸ m from the Earth. If viewed from two diametrically opposite points on the Earth, the angle subtended at the moon is 1° 54′. What is the diameter of the Earth?
Solution:
Given
Distance (D) = 3.84 × 10⁸ m
Subtended angle (α)
= 1° 54′ = (60’+ 54′)= 114′
= 114 × 2.91 × 10^-4 rad
= 3.317 × 10^-2 rad
To find: Diameter of Earth (d)
Formula: d = αD
Calculation: From formula,
d = 3.317 × 10^-2 × 3.84 × 10⁸
= 1.274 × 10⁷ m
Diameter of Earth is 1.274 × 10⁷ m.

Question 24.
Explain the method to measure mass.
Answer:
Method for measurement of mass:

Mass, until recently, was measured with a standard mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Serves, near Paris, France.
As platinum – iridium piece was seen to pick up microparticles and found to be affected by atmosphere, its mass could no longer be treated as constant.
Hence, a new definition of mass was introduced in terms of electric current on 20th May 2019.
Now, one kilogram mass is described in terms of amount of current which has to be passed through electromagnet to pull one side of extremely sensitive balance to balance the other side which holds one standard kg mass.
To measure mass of small entities such as atoms and nucleus, atomic mass unit (amu) is used.
It is defined as (\(\frac{1}{12}\))^th mass of an unexcited atom of carbon -12(C¹²).
1 amu ≈ 1.66 × 10^-27 kg.

Question 25.
That can he the reason for choosing Carbon-12 to define atomic mass unit?
Answer:

Unlike oxygen and hydrogen, which exhibit various isotopes in higher proportions, carbon- 12 is the single most abundant (98% of available carbon) isotope of carbon.
it is also very stable.
Hence, it makes more accurate unit of measuring mass and is used to define atomic mass unit.

Question 26.
Define mean solar day. Explain the method for measurement of time.
Answer:

A mean solar day is the average time interval from one noon to the next noon.
Method for measurement of time:
The unit of time, the second, was considered to be \(\frac{1}{86400}\) of the mean solar day, where a mean solar day = 24 hours
= 24 × 60 × 60
= 86400 s
However, this definition proved to be unsatisfactory to define the unit of time precisely because solar day varies gradually due to gradual slowing down of the Earth’s rotation. Hence, the definition of second was replaced by one based on atomic standard of time.
Atomic standard of time is now used for the measurement of time. In atomic standard of time, periodic vibrations of caesium atom is used.
One second is time required for 9,192.631,770 vibrations of the radiation corresponding to transition between two hyperfine energy states of caesium-133 (Cs- 133) atom.

Question 27.
Define dimensions and dimensional formula of physical quantities. Give few examples of dimensional formula.
Answer:

Dimensions:
The dimensions of a physical quantity are the powers to which the fundamental units must be raised in order to obtain the unit of a given physical quantity.
Dimensional formula:
When any derived quantity is represented with appropriate powers of symbols of the fundamental quantities, SUCh an expression is called dimensional formula.
It is expressed by square bracket with no comma in between the symbols.
Examples of dimensional formula:
a. Speed = \(\frac{\text { Distance }}{\text { time }}\)
∴ Dimensions of speed = \(\frac{[\mathrm{L}]}{[\mathrm{T}]}\) = [L¹M⁰T^-1]
[Note: As power of M is zero, it can be omitted from dimensional formula. Therefore, dimensions of speed can be written as [L¹T¹]

Question 28.
A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic function:
i) y = a sin \(\frac{2 \pi t}{T}\)
ii) y = a sin v t
iii) y = \(\frac{\mathrm{a}}{\mathbf{T}} \sin \frac{\mathrm{t}}{\mathrm{a}}\)
iv) y = \(\frac{a}{\sqrt{2}}\left[\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right]\)
Here, a is maximum displacement of particle, y ¡s speed of particle, T is time period of motion. Rule out the wrong formulae on dimensional grounds.
Answer:
The argument of trigonometrical function, i.e., angle is dimensionless. Now,
i) The argument, \(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L⁰M⁰T⁰]
which is a dimensionless quantity.
Hence, formula (i) is correct.

ii) The argument,
[vt] = [LT^-1] [T] = [L] = [L¹M⁰T⁰]
which is not a dimensionless quantity.
Hence, formula (ii) is incorrect.

iii) The argument,
\(\left[\frac{\mathrm{t}}{\mathrm{a}}\right]=\frac{[\mathrm{T}]}{[\mathrm{L}]}\) = [L^-1M⁰T¹]
which is not a dimensionless quantity.
Hence, formula (iii) is incorrect.

iv) The argument,
\(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L⁰M⁰T⁰]
which is a dimensionless quantity.
Hence, formula (iv) is correct.

Question 29.
State principle of homogeneity of dimensions.
Answer:
Principle of homogeneity of dimensions: The dimensions of all the terms on the two sides of a physical equation relating different physical quantities must be same.

Question 30.
State the uses of dimensional analysis.
Answer:
Uses of dimensional analysis:

To check the correctness of a physical equation.
Correctness of a physical equation by dimensional analysis:

A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
For example, consider the equation of motion.
v = u + at ……………. (1)
Writing the dimensional formula of every term, we get
Dimensions of LH.S. [v] [L¹M⁰T^-1],
Dimensions of R.H.S. = [u] + [at]
= [L¹M⁰T^-1] + [L¹M⁰T^-2] [L¹M⁰T^-1]
= [L¹M⁰T^-1] + [L¹M⁰T^-1]
⇒ [L.HS.] = [R.H.S.]
As dimensions of both side of equation is same, physical equation is dimensionally correct.

To derive the relationship between related physical quantities.
Expression for time period of a simple pendulum by dimensional analysis:

Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
T ∝ l^a g^b
i.e., T = k l^a g^b ………… (1)
where, k = proportionality constant, which is dimensionless.
The dimensions of T = [L⁰M⁰T¹)
The dimensions of l = [L¹M⁰T⁰]
The dimensions of g = [L¹M⁰T²]
Taking dimensions on both sides of equation (1),
[L⁰M⁰T¹] = [L¹M⁰T⁰]^a [L¹M⁰T^-2]^b[L⁰M⁰T¹] = [L^a+bM⁰T^-2b]
Equating corresponding power of L, M and T on both sides, we get
a + b = 0 …………. (2)
and -2b = 1
∴ b = –\(\frac{1}{2}\)
Substituting ‘b’ in equation (2), we get
a = \(\frac{1}{2}\)
Substituting values of a and b in equation (1),
we have,
Experimentally, it ¡s found that k = 2π
∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
This is the required expression for time period of a simple pendulum.

To find the conversion factor between the units of the same physical quantity in two different systems of units.
Conversion factor between units of same physical quantity:

let ‘n’ be the conversion factor between the units of work.
∴ 1 J = n erg ………….. (1)
Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
From (1),
\(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)

n= 10⁴ × 10³ × 1 = 10⁷
Hence, the conversion factor, n = 10⁷
There fore, from equation (1), we have,
∴ 1 J = 10⁷ erg.

Question 31.
Explain the use of dimensional analysis to check the correctness of a physical equation.
Answer:
Correctness of a physical equation by dimensional analysis:

A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
For example, consider the equation of motion.
v = u + at ……………. (1)
Writing the dimensional formula of every term, we get
Dimensions of LH.S. [v] [L¹M⁰T^-1],
Dimensions of R.H.S. = [u] + [at]
= [L¹M⁰T^-1] + [L¹M⁰T^-2] [L¹M⁰T^-1]
= [L¹M⁰T^-1] + [L¹M⁰T^-1]
⇒ [L.HS.] = [R.H.S.]
As dimensions of both side of equation is same, physical equation is dimensionally correct.

Question 32.
Time period of a simple pendulum depends upon the length of pendulum (l) and acceleration due to gravity (g). Using dimensional analysis, obtain an expression for time period of a simple pendulum.
Answer:
Expression for time period of a simple pendulum by dimensional analysis:
i) Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
T ∝ l^a g^b
i.e., T = k l^a g^b ………… (1)
where, k = proportionality constant, which is dimensionless.

ii) The dimensions of T = [L⁰M⁰T¹)
The dimensions of l = [L¹M⁰T⁰]
The dimensions of g = [L¹M⁰T²]
Taking dimensions on both sides of equation (1),
[L⁰M⁰T¹] = [L¹M⁰T⁰]^a [L¹M⁰T^-2]^b
[L⁰M⁰T¹] = [L^a+bM⁰T^-2b]

iii) Equating corresponding power of L, M and T
on both sides, we get
a + b = 0 …………. (2)
and -2b = 1
∴ b = –\(\frac{1}{2}\)

iv) Substituting ‘b’ in equation (2), we get
a = \(\frac{1}{2}\)

v) Substituting values of a and b in equation (1),
we have,
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 15

vi) Experimentally, it ¡s found that k = 2π
∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
This is the required expression for time period of a simple pendulum.

Question 33.
Find the conversion factor between the S.I. and the C.OES. units of work using dimensional analysis.
Answer:
Conversion factor between units of same physical quantity:

let ‘n’ be the conversion factor between the units of work.
∴ 1 J = n erg ………….. (1)
Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
From (1),
\(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)

n= 10⁴ × 10³ × 1 = 10⁷
Hence, the conversion factor, n = 10⁷
There fore, from equation (1), we have,
∴ 1 J = 10⁷ erg.

Question 34.
State the limitations of dimensional analysis.
Answer:
Limitations of dimensional analysis:

The value of dimensionless constant can be obtained with the help of experiments only.
Dimensional analysis cannot be used to derive relations involving trigonometric (sin θ, cos θ, etc.), exponential (e^x, e^x², etc.), and logarithmic functions (log x, log x³, etc) as these quantities are dimensionless.
This method is not useful if constant of proportionality is not a dimensionless quantity.
If the correct equation contains some more terms of the same dimension, it is not possible to know about their presence using dimensional equation.

Question 35.
If two quantities have same dimensions, do they always represent the same physical content?
Answer:
When dimensions of two quantities are same, they do not always represent the same physical content.
Example:
Force and momentum both have same dimensions but they represent different physical content.

Question 36.
A dimensionally correct equation need not actually be a correct equation but dimensionally incorrect equation is necessarily wrong. Justify.
Answer:
i) To justify a dimensionally correct equation need not be actually a correct equation, consider equation, v² = 2as
Dimensions of L.H.S. = [v²] = [L²M⁰T²]
Dimensions of R.H.S. = [as]= [L²M⁰T²]
⇒ [L.H.S.] = [R.H.S.]
This implies equation v² = 2as is dimensionally correct.
But actual equation is, v² = u² + 2as
This confirms a dimensionally correct equation need not be actually a correct equation.

ii) To justify dimensionally incorrect equation is necessarily wrong, consider the formula,
\(\frac{1}{2}\) mv = mgh
Dimensions of L.H.S. = [mv] = [L¹M¹T^-1]
Dimensions of R.H.S. = [mgh] = [L²M¹T^-2]
Since the dimensions of R.H.S. and L.H.S. are not equal, the formula given by equation must be incorrect.
This confirms dimensionally incorrect equation is necessarily wrong.

Question 37.
State, whether all constants are dimensionless or unitless.
Answer:
All constants need not be dimensionless or unitless.
Planck’s constant, gravitational constant etc., possess dimensions and units. They are dimensional constants.

Solved Examples

Question 38.
If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?
Solution:
i) Force = Mass × Acceleration Force

∴ Dimensional equation of mass
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 18
= [F¹L^-4T²]

i) The dimensional equation of mass is [F¹L^-1T²].
ii) The dimensional equation of density is [F¹L^-4T²].

Question 39.
A calorie is a unit of heat and it equals 4.2 J, where 1 J = kg m² s^-2. A distant civilisation employs a system of units in which the units of mass, length and time are α kg, β m and δ s. Also J’ is their unit of energy. What will be the magnitude of calorie in their units?
Solution:
1 cal = 4.2 kg m² s^-2
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 19
New unit of energy is J’
Dimensional formula of energy is [L²M¹T^-2] According to the question,

Question 40.
Assume that the speed (v) of sound in air depends upon the pressure (P) and density (ρ) of air, then use dimensional analysis to obtain an expression for the speed of sound.
Solution:
It is given that speed (v) of sound in air depends upon the pressure (P) and density (ρ) of the air.
Hence, we can write, v = k P^a ρ^b ……….. (1)
where, k is a dimensionless constant and a and b are powers to be determined.
Dimensions of y = [L¹M⁰T^-1]
Dimensions of P = [L^-1M¹T^-2]
Dimensions of ρ = [L^-3M¹T⁰]
Substituting the dimensions of the quantities on both sides of equation (1),
∴ [L¹M⁰T^-1] = [L^-1M¹T^-2]^a [L^-3M¹T⁰]^b
∴ [L¹M⁰T^-1] = [L^-aM^aT^-2a] [L^-3bM^bT⁰]
∴ [L¹M⁰T^-1] = [L^-a-3bM^a+bT^-2a]
Comparing the powers of L, M and T on both sides, we get,
-2a = -1
∴ a = \(\frac{1}{2}\)
Also, a + b = O
∴ \(\frac{1}{2}\) + b = 0 b = – \(\frac{1}{2}\)
Substituting values of a and b in equation (1), we get
y = k P^{\(\frac{1}{2}\)} ρ^{–\(\frac{1}{2}\)}
∴ v = k \(\sqrt{\frac{\mathrm{p}}{\rho}}\)

Question 41.
Density of oil is 0.8 g cm³ in C.G.S. unit. Find its value in S.I. units.
Solution:
Dimensions of density is [L^-3M¹T⁰]
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 21
= 0.8 [10^-3] [10^-2]^-3
= 0.8 [10^-3] [10]⁶
n = 0.8 × 10³
Substituting the value of ‘n’ in equation (1).
we get, 0.8 g cm3 = 0.8 × 10³ kg m^-3.
Density of oil in S.l unit is 0.8 × 10³ kg m^-3.

Question 42.
The value of G in C.G.S system is 6.67 × 10^-8 dyne cm² g^-2. Calculate its value in S.l. system.
Solution:
Dimensional formula of gravitational constant
[L³M^-1T^-2]
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 22
n = 6.67 × 10^-8 × 10^-6 × 10³
n = 6.67 × 10^-11
From equation (1),
6.67 × 10^-8 dyne cm² g^-2
= 6.67 × 10^-11 N-m² kg^-2
Value ofG in S.l. system is 6.67 × 10^-11 N-m² kg^-2.

Question 43.
What is accuracy?
Answer:
Accuracy is how close a measurement is to the actual value of that quantity.

Question 44.
What is precision?
Answer:
Precision is a measure of how consistently a device records nearly identical values i.e., reproducible results.

Question 45.
A scale in a lab measures the mass of object consistently more by 500 g than their actual mass. How would you describe the scale in terms of accuracy and precision?
Answer:
The scale is precise but not accurate.
Explanation: Precision measures how consistently a device records the same answer; even though it displays the wrong value. Hence, the scale is precise.

Accuracy is how well a device measures something against its accepted value. As scale in the lab is always off by 500 g, it is not accurate.
[Note: The goal of the observer should be to get accurate as well as precise measurements.]

Question 46.
List reasons that may introduce possible uncertainties in an observation.
Answer:
Possible uncertainties in an observation may arise due to following reasons:

Quality of instrument used,
Skill of the person doing the experiment,
The method used for measurement,
External or internal factors affecting the result of the experiment.

Question 47.
What is systematic error? Classify errors into different categories.
Answer:

Systematic errors are errors that are not determined by chance but are introduced by an inaccuracy (involving either the observation or measurement process) inherent to the system.
Classification of errors:
Errors are classified into following two groups:
Systematic errors:
- Instrumental error (constant error),
- Error due to imperfection in experimental technique,
- Personal error (human error).
Random error (accidental error)

Question 48.
What is instrumental (constant) error?
Answer:
Instrumental error:

It arises due to defective calibration of an instrument.
Example: If a thermometer is not graduated properly, i.e., one degree on the thermometer actually corresponds to 0.99°, the temperature measured by such a thermometer will differ from its value by a constant amount.

Question 49.
What is error due to imperfection in experimental technique?
Answer:
Error due to imperfection in experimental technique:

The errors which occur due to defective setting of an instrument is called error due to imperfection in experimental technique.
For example the measured volume of a liquid in a graduated tube will be inaccurate if the tube is not held vertical.

Question 50.
What is personal error?
Answer:
Personal error (Human error):

The errors introduced due to fault of an observer taking readings are called personal errors.
For example, while measuring the length of an object with a ruler, it is necessary to look at the ruler from directly above. If the observer looks at it from an angle, the measured length will be wrong due to parallax.

Question 51.
What is random error (accidental)?
Answer:

Random error (accidental):
The errors which are caused due to minute change in experimental conditions like temperature, pressure change in gas or fluctuation in voltage, while the experiment is being performed are called random errors.
They can be positive or negative.
Random error cannot be eliminated completely but can be minimized by taking multiple observations and calculating their mean.

Question 52.
State general methods to minimise effect of systematic errors.
Answer:
Methods to minimise effect of systematic errors:

By using correct instrument.
Following proper experimental procedure.
Removing personal error.

Question 53.
Define the term:
Arithmetic mean
Answer:
Arithmetic mean:
a. The most probable value of a large number of readings of a quantity is called the arithmetic mean value of the quantity. This value can be considered to be true value of the quantity.

b. If a₁, a₂, a₃, …………… a_n are ‘n’ number of readings taken for measurement of a quantity, then their mean value is given by,
a_mean = \(\frac{a_{1}+a_{2}+\ldots \ldots .+a_{n}}{n}\)
∴ a_mean = \(\frac{1}{n} \sum_{i=1}^{n} a_{i}\)

Question 54.
What does a = a_mean ± ∆a_mean signify?
Answer:
a = a_mean ± ∆ a_mean signifies that the actual value of a lies between (a_mean – ∆ a_mean) and (a_mean + ∆ a_mean).

Question 55.
What is meant by the term combination of errors?
Answer:
Derived quantities may get errors due to individual errors of fundamental quantities, such type of errors are called as combined errors.

Question 56.
Explain errors in sum and in difference of measured quantity.
Answer:
Errors in sum and in difference:
i) Suppose two physical quantities A and B have measured values A ± ∆A and B ± ∆B. respectively, where ∆A and ∆B are their mean absolute errors.

ii) Then, the absolute error ∆Z in their sum.
Z = A + B
Z ± ∆Z = (A ± ∆A) + (B ± ∆B)
= (A + B) ± ∆A ± ∆B
∴ ± ∆Z = ± ∆A ± ∆B.

iii) For difference. i.e.. if Z = A – B.
Z ± ∆Z = A ± ∆A) – (B ± ∆B)
= (A – B) ± ∆A ∓ ∆B
∴ ± ∆Z = ± ∆A ∓ ∆B,

iv) There are four possible values for ∆Z. namely (+∆A – ∆B), (+∆A + ∆B), (-∆A -∆B), (-∆A + ∆B). Hence, maximLim value of absolute error is ∆Z = (∆A+ ∆B) in both the cases.

v) Thus. when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.

Question 57.
Explain errors in product of measured quantity.
Answer:
Errors in product:
i) Suppose Z = AB and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= AB ± A∆B ± B∆A ± ∆A∆B
Dividing L.H.S by Z and R.H.S. by AB we get
\(\left(1 \pm \frac{\Delta Z}{Z}\right)=\left[1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm\left(\frac{\Delta A}{A}\right)\left(\frac{\Delta B}{B}\right)\right]\)
Since ∆A/A and ∆B/B are very small, product is neglected. Hence, maximum relative error in Z is \(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)

ii) Thus, when two quantities are multiplied, the maximum relative error in the result is the sum of relative errors in each quantity.

Question 58.
Explain errors due to power (index) of measured quantity.
Answer:
Errors due to the power (index) of measured quantity:

Suppose
Z = A³ = A × A × A
then, \(\)
Hence the relative error in Z = A³ is three times the relative error in A.
This means if Z = Aⁿ
This implies, the quantity in the formula which has large power is responsible for maximum error.

Question 59.
The radius of a sphere measured repeatedly yields values 5.63 m, 5.54 m, 5.44 m, 5.40 m and 5.35 m. Determine the most probable value of radius and the mean absolute, relative and percentage errors.
Solution:
Given: a₁ = 5.63 m, a₂ = 5.54 m, a₃ = 5.44 m
a₄ = 5.40 m, a₅ = 5.35 m,
To find:
i) Most probable value (Mean value)
ii) Mean absolute error
iii) Relative error
iv) Percentage error
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 24
From formula (ii),
Absolute errors:
∆a₁ = |a_mean – a₁| = |5.472 – 5.63| = 0.158
∆a₂ = |a_mean – a₂| = |5.472 – 5.54| = 0.068
∆a₃ = |a_mean – a₃| = |5.472 – 5.44| = 0.032
∆a₄ = |a_mean – a₄| = |5.472 – 5.40| = 0.072
∆a₅ = |a_mean – a₅| = |5.472 – 5.35| = 0.122

From formula (ii),
∆a_mean = \(\frac{0.158+0.068+0.032+0.072+0.122}{5}\)
= \(\frac{0.452}{5}\)
= 0.0904 m
From formula (iii),
Relative error = \(\frac{0.0904}{5.472}\)
= 1.652 × 10^-2
(after rounding off to correct significant digits)
= 1.66 × 10^-2
= 0.0166
∴ Percentage error = 1.66 × 10^-2 × 100 = 1.66%
i) The mean value is 5.472 m.
ii) The mean absolute error is 0.0904 m.
iii) The relative error is 0.0166.
iv) The percentage error is 1.66%
[Note: Answer to relative error is rounded off using rules of significant figures and of rounding off]

Question 60.
Lin an experiment to determine the volume of an object, mass and density are recorded as m = (5 ± 0.15) kg and p = (5 ± 0.2) kg m³ respectively. Calculate percentage error in the measurement of volume.
Solulion:
Given: M = 5kg, ∆M = 0.15 kg, ρ = 5 kg/m³,
∆ρ = 0.2 kg/m³
To find: Percentage error in volume (V)
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 25
The percentage error in the determination of volume is 7%.

Question 61.
The acceleration due to gravity is determined by using a simple pendulum of length l = (100 ± 0.1) cm. If its time period is T = (2 ± 0.01) s, find the maximum percentage error in the measurement of g.
Solution:
Given: ∆l = 0.1 cm, l = 100 cm, ∆T = 0.01 s,
T = 2s
To find: Percentage error
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 26
Percentage error in measurement of g is 1.1 %.

Question 62.
Find the number of significant figures in the following numbers,
i. 25.42
ii. 0.004567
iii. 35.320
iv. 91.000
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 27

Solved Examples

Question 63.
Add 7.21, 12.141 and 0.0028 and express the result to an appropriate number of significant figures.
Solution:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 28
In the given problem, minimum number of digits after decimal is 2.
∴ Result will be rounded off upto two places of decimal.
Corrected rounded off sum is 19.35.

Question 64.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (i) the total mass of the box? (ii) the difference in the masses of the pieces to correct significant figures?
Solution:
i) Total mass of the box
= (2.3+ 0.02017 + 0.02015) kg
= 2.34032 kg
Since, the last number of significant figure is 2, therefore, the total mass of the box = 2.3 kg

ii) Difference of mass = (20.17 – 20.15) = 0.02g Since, there are two significant figures so the difference in masses to the correct significant figures is 0.02 g.

i) The total mass of the box to correct significant figures is 2.3 kg.
ii) The difference in the masses to correct significant figures is 0.02 g.

Apply Your Knowledge

Question 65.
Write the dimensions of a and b in the relation
E = \(\frac{b-x^{2}}{a}\)
Where E is energy, x ¡s distance and t is time.
Answer:
The given relation is E = \(\frac{b-x^{2}}{a}\)
As x is subtracted from b,
∴ dimensions of b are x²;
i.e., b = [L²]
∴ We can write equation as E = \(\frac{\mathrm{L}^{2}}{\mathrm{a}}\)
Or a = \(\frac{\mathrm{L}^{2}}{\mathrm{E}}=\frac{\mathrm{L}^{2}}{\left[\mathrm{~L}^{2} \mathrm{MT}^{-2}\right]}\) = [L⁰M^-1T²]

Question 66.
What is the difference between 6.0 and 6.00? which Is more accurate?
Answer:
6.0 indicates the measurement is correct up to first decimal place, whereas 6.00 indicates that the measurement is correct up to second decimal place. Thus, 6.00 is a more accurate value than 6.0.

Question 67.
A child walking on a footpath notices that the width of the footpath is uneven. He reported this to his school principal and the complaint was forwarded to the municipal officer.
i. What is the possible error encountered?
ii. What is the relative error in width of footpath if width of footpath in 10 m length are noted as 5 m, 5.5 m, 5 m, 6 m and 4.5 m?
Answer:
i) The error encountered is personal error.

ii) Mean value of widths
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 29
The relative error in width of footpath is 0.084.

Question 68.
A factory owner kept five identical spheres between two wooden blocks on a ruler as shown in figure. He called all his workers and told them to take reading, to check their efficiency and knowledge.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 30
i. What is the area of central sphere?
ii. What is the absolute error in reading of diameter of second sphere?
Answer:
i) From above diagram radius of central sphere is
r = 1 cm
∴ Area = πr² = 3.142 × (1)²= 3.142 cm²
The area of central sphere is 3.142 cm².

ii) Mean value of all reading of diameters
d_mean = \(\frac{\mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}+\mathrm{d}_{4}+\mathrm{d}_{5}}{5}=\frac{2+2+2+2+2}{5}\)
= \(\frac{10}{5}\) = 2 cm
Absolute error in reading of second sphere.
∆d₂ = |d_mean – d₂| = 2 – 2 = 0
The absolute error in reading of diameter of second sphere is zero.

Question 69.
A potential difference of V = 100 ± 2 volt, when applied across a resistance R gives a current of 10 ± 0.5 ampere. Calculate percentage error in R given by R V/I.
Answer:
Here. V = 100 ± 2 volt and I = 10 ± 0.5 ampere
Expressing limits of error as percentage error,
We have
V = 100 volt ± \(\frac{2}{100}\) × 100% = 10 volt ± 2%
and I = 10 ampere ± \(\frac{0.5}{10}\) × 100%
= 10 ampere ± 5%
∴ R = \(\frac{V}{I}\)
∴ %error in R = %error in V + %error in I
= 2% + 5% = 7%

Quick Review

Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 31

Multiple Choice Questions

Question 1.
A physical quantity may be defined as
(A) the one having dimension.
(B) that which is immeasurable.
(C) that which has weight.
(D) that which has mass.
Answer:
(A) the one having dimension.

Question 2.
Which of the following is the fundamental unit?
(A) Length, force, time
(B) Length, mass, time
(C) Mass, volume, height
(D) Mass, velocity, pressure
Answer:
(B) Length, mass, time

Question 3.
Which of the following is NOT a fundamental quantity?
(A) Temperature
(B) Electric charge
(C) Mass
(D) Electric current
Answer:
(B) Electric charge

Question 4.
The distance of the planet from the earth is measured by __________.
(A) direct method
(B) directly by metre scale
(C) spherometer method
(D) parallax method
Answer:
(D) parallax method

Question 5.
The two stars S₁ and S₂ are located at distances d₁ and d₂ respectively. Also if d₁ > d2₂ then following statement is true.
(A) The parallax of S₁ and S₂ are same.
(B) The parallax of S₁ is twice as that of S₂
(C) The parallax of S₁ is greater than parallax of S₂
(D) The parallax of S₂ is greater than parallax of S₁
Answer:
(D) The parallax of S₂ is greater than parallax of S₁

Question 6.
Which of the following is NOT a unit of time?
(A) Hour
(B) Nano second
(C) Microsecond
(D) parsec
Answer:
(D) parsec

Question 7.
An atomic clock makes use of _________.
(A) cesium-133 atom
(B) cesium-132 atom
(C) cesium-123 atom
(D) cesium-131 atom
Answer:
(A) cesium-133 atom

Question 8.
S.I. unit of energy is joule and it is equivalent to
(A) 10⁶ erg
(B) 10^-7 erg
(C) 10⁷ erg
(D) 10⁵ erg
Answer:
(C) 10⁷ erg

Question 9.
[L¹M¹T^-1] is an expression for __________.
(A) force
(B) energy
(C) pressure
(D) momentum
Answer:
(D) momentum

Question 10.
Dimensions of sin θ is
(A) [L²]
(B) [M]
(C) [ML]
(D) [M⁰L⁰T⁰]
Answer:
(D) [M⁰L⁰T⁰]

Question 11.
Accuracy of measurement is determined by
(A) absolute error
(B) percentage error
(C) human error
(D) personal error
Answer:
(B) percentage error

Question 12.
Zero error of an instrument introduces .
(A) systematic error
(B) random error
(C) personal error
(D) decimal error
Answer:
(A) systematic error

Question 13.
The diameter of the paper pin is measured accurately by using ________.
(A) Vernier callipers
(B) micrometer screw gauge
(C) metre scale
(D) a measuring tape
Answer:
(B) micrometer screw gauge

Question 14.
The number of significant figures in 11.118 × 10^-6 is
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(C) 5

Question 15.
0.00849 contains ___________ significant figures.
(A) 6
(B) 5
(C) 3
(D) 2
Answer:
(C) 3

Question 16.
3.310 × 10² has ___________ significant figures.
(A) 6
(B) 4
(C) 2
(D) 1
Answer:
(B) 4

Question 17.
The Earth’s radius is 6371 km. The order of magnitude of the Earth’s radius is
(A) 10³ m
(B) 10⁹ m
(C) 10⁷ m
(D) 10² m
Answer:
(C) 10⁷ m

Question 18.
__________ is the smallest measurement that can be made using the given instrument
(A) Significant number
(B) Least count
(C) Order of magnitude
(D) Relative error
Answer:
(B) Least count

Competitive Corner

Question 1.
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X,
where X = \(\frac{A^{2} \frac{1}{B^{2}}}{C^{\frac{1}{3}} D^{3}}\), will be:
(A) -10 %
(B) 10 %
(C) \(\left(\frac{3}{13}\right) \%\)
(D) 16 %
Answer:
(D) 16 %
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 33

∴ Percentage error in x is given as,
\(\frac{\Delta x}{x}\) × 100 – (error contributed by A) – (error contributed by B) + (error contributed by C) + (error contributed by D)
= 2% + 1% + 1% + 12%
= 16%

Question 2.
The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n – 1) divisions of main scale. The least count of the vernier callipers is,
(A) \(\frac{1}{n(n+1)}\) cm
(B) \(\frac{1}{(n+1)(n-1)}\) cm
(C) \(\frac{1}{n}\) cm
(D) \(\frac{1}{n^{2}}\) cm
Answer:
(D) \(\frac{1}{n^{2}}\) cm
Hint:
1 V.S.D. = \(\frac{(n-1)}{n}\) M.S.D.
LC. = 1 M.S.D. – 1 V.S.D.
= 1 M.S.D. – \(\frac{(n-1)}{n}\) M.S.D.
= \(\frac{1}{n}\) M.S.D.
= \(\frac{1}{n}\) × \(\frac{1}{n}\) cm
∴ L.C. = \(\frac{1}{n^{2}}\) cm

Question 3.
A student measures time for 20 oscillations of a simple pendulum as 30 s. 32 s, 35 s and 31 s. 1f the minimum division in the measuring clock is I s, then correct mean time in second is
(A) 32 ± 3
(B) 32 ± 1
(C) 32 ± 2
(D) 32 ± 5
Answer:
(C) 32 ± 2
Hint:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 35
Hence rounding off,
∆t = ± 2 s
∴ t ± ∆t = 32 ± 2 s

Question 4.
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference leveL If screw gauge has a zero error of— 0.004 cm, the correct diameter of the ball is
(A) 0.521 cm
(B) 0.525 cm
(C) 0.053 cm
(D) 0.29 cm
Answer:
(D) 0.29 cm

Hint:
Least count of screw gauge = 0.001 cm = 0.01mm
Main scale reading = 5 mm.
Zero error = – 0.004 cm = -0.04 mm
Zero correction = +0.04 mm
Observed reading = Mainscale reading + (Division × least count)
Observed reading = 5 + (25 × 0.01) = 5.25 mm
Corrected reading = Observed reading + Zero correction
Corrected reading = 5.25 + 0.04
= 5.29 mm = 0.529 cm

Question 5.
The density of the material in the shape of a cube is determined by measuring three sides of the cube and its mass. 1f the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
(A) 4.5%
(B) 6%
(C) 2.5°
(D) 3.5%
Answer:
(A) 4.5%
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 36

Question 6.
Let x = \(\left[\frac{a^{2} b^{2}}{c}\right]\) be the physical quantity. If the percentage error in the measurement of physical quantities a, b and c is 2, 3 and 4 percent respectively then percentage en-or in the measurement of x is
(A) 7%
(B) 14%
(C) 21%
(D) 28%
Answer:
(B) 14%
Hint:
Given: x = \(\frac{a^{2} b^{2}}{c}\)
Percentage error is given by.
\(\frac{\Delta x}{x}=\frac{2 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{\Delta c}{c}\)
= (2 × 2) + (2 × 3) + 4
= 4 + 6 + 4 = 14
∴ \(\frac{\Delta \mathrm{x}}{\mathrm{x}} \%\) = 14%

Question 7.
A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\) is [c is velocity of light, G is universal constant of gravitation and e is charge]:
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 37
Answer:
(A) \(\frac{1}{\mathrm{c}^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Hint:
Let the physical quantity formed of the dimensions of length be given as.
[L] = [c]^x [G]^y \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) …………….. (i)
Now,
Dimensions of velocity of light [c]^x = [LT^-1]^x
Dimensions of universal gravitational constant
[G]^y = [L³T²M^-1]^y
Dimensions of \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) = [ML³T^-2]^z
Substitrning these in equation (i)
[L] [LT^-1]^x [M^-1L³T^-2]^y [ML³T^-2]^z
= L^x+3y+3z M^-y+z T^-x-2y-2z
Solving for x, y, z
x + 3y + 3z = 1
-y + z = 0
x + 2y + 2z = O
Solving the above equation,
x = -2, y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
∴ L = \(\frac{1}{c^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Question 8.
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 × 10^-2 m
rise of water, h = 1.45 × 10^-2 m
Using g = 9.80 m/s² and the simplified relation
T = \(\frac{\mathrm{rhg}}{2}\) × 10³ N/m, the possible error in surface tension is closest to:
(A) 0.15%
(B) 1.5%
(C) 2.4%
(D) 10%
Answer:
(B) 1.5%
Hint:
D = 1.25 × 10^-2 m; h = 1.45 × 10^-2 m
The maximum permissible error in D
= ∆D = 0.01 × 10^-2 m
The maximum permissible error in h
= ∆h = 0.01 × 10^-2 m
g is given as a constant and is errorless.
Maharashtra Board Class 11 Physics Important Questions Chapter 1 Units and Measurements 38

Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination

July 19, 2024July 18, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 9 Control and Co-ordination Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 9 Control and Co-ordination

1. Multiple choice questions

Question 1.
The nervous system of mammals uses both electrical and chemical means to send signals via neurons. Which part of the neuron receives impulse?
(a) Axon
(b) Dendron
(c) Nodes of Ranvier
(d) Neurilemma
Answer:
(b) Dendron

Question 2.
……………. is a neurotransmitter.
(a) ADH
(b) Acetyl CoA
(c) Acetyl choline
(d) Inositol
Answer:
(c) Acetyl choline

Question 3.
The supporting cells that produce myelin sheath in the PNS are …………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(d) Schwann cells

Question 4.
A collection of neuron cell bodies located outside the CNS is called …………….
(a) tract
(b) nucleus
(c) nerve
(d) ganglion
Answer:
(d) ganglion

Question 5.
Receptors for protein hormones are located …………….
(a) in cytoplasm
(b) on cell surface
(c) in nucleus
(d) on Golgi complex
Answer:
(b) on cell surface

Question 6.
If parathyroid gland of man Eire removed, the specific result will be …………….
(a) onset of aging
(b) disturbance of Ca⁺⁺
(c) onset of myxoedema
(d) elevation of blood pressure
Answer:
(b) disturbance of Ca⁺⁺

Question 7.
Hormone thyroxine, adrenaline and non¬adrenaline are formed from ……………
(a) Glycine
(b) Arginine
(c) Ornithine
(d) Tyrosine
Answer:
(d) Tyrosine

Question 8.
Pheromones are chemical messengers produced by animals and released outside the body. The odour of these substance affects …………….
(a) skin colour
(b) excretion
(c) digestion
(d) behaviour
Answer:
(d) behaviour

Question 9.
Which one of the following is a set of discrete endocrine gland?
(a) Salivary glands, thyroid, adrenal, ovary
(b) Adrenal, testis, ovary, liver
(c) Pituitary, thyroid, adrenal, thymus
(d) Pituitary, pancreas, adrenal, thymus
Answer:
(c) Pituitary, thyroid, adrenal, thymus

Question 10.
After ovulation, Graafian follicle changes into …………….
(a) corpus luteum
(b) corpus albicans
(c) corpus spongiosum
(d) corpus callosum
Answer:
(a) corpus luteum

Question 11.
Which one of the following pairs correctly matches a hormone with a disease resulting from its deficiency?
(a) Parathyroid hormone – Diabetes insipidus
(b) Luteinising hormone – Diabetes mellitus
(c) Insulin – Hyperglycaemia
(d) Thyroxine – Tetany
Answer:
(c) Insulin – Hyperglycaemia

Question 12.
……………. is in direct contact of brain in humans.
(a) Cranium
(b) Dura mater
(c) Arachnoid
(d) Pia mater
Answer:
(d) Pia mater

2. Very very short answer questions.

Question 1.
What is the function of red nucleus?
Answer:
Red nucleus plays an important role in controlling posture and muscle tone, modifying some motor activities and motor coordination.

Question 2.
What is the importance of corpora quadrigemina?
Answer:
Corpora quadrigemina consists of 4 solid rounded structures, viz. superior and inferior colliculi. Superior colliculi control visual reflexes while inferior colliculi control auditory reflexes.

Question 3.
What does the cerebellum of brain control?
Answer:
Cerebellum of brain is an important centre which maintains equilibrium of body, posture, balancing orientation, moderation of voluntary movements and maintenance of muscle tone.

Question 4.
Name the three ear ossicles.
Answer:
Malleus [hammer], incus [anvil] and stapes [stirrup].

Question 5.
Name the anti abortion hormone.
Answer:
Progesterone.

Question 6.
Name an organ which acts as temporary endocrine gland.
Answer:
Placenta. Corpus luteum in ovary.

Question 7.
Name the type of hormones which bind to the DNA and alter the gene expression.
Answer:
Steroid hormones.

Question 8.
What is the cause of abnormal elongation of long bones of arms and legs and of lower jaw.
Answer:
Hypersecretion of growth hormones in adults causes abnormal elongation of long bones of arms and legs and of lower jaw i.e. acromegaly.

Question 9.
Name the hormone secreted by the pineal gland.
Answer:
Melatonin.

Question 10.
Which endocrine gland plays important, role in improving immunity?
Answer:
The endocrine gland, thymus plays an important role in improving immunity.

3. Match the organism with the type of nervous system found in them.

Column A	Column B
(1) Neurons	(a) Earthworm
(2) Ladder type	(b) Hydra
(3) Ganglion	(c) Flatworm
(4) Nerve net	(d) Human

Answer:

Column A	Column B
(1) Neurons	(d) Human
(2) Ladder type	(c) Flatworm
(3) Ganglion	(a) Earthworm
(4) Nerve net	(b) Hydra

4. Very short answer questions.

Question 1.
Describe the endocrine role of islets of Langerhans.
OR
Islets of Langerhans.
Answer:
Endocrine cells of pancreas form groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

Alpha (α) cells : They are 20% and secrete glucagon. Glucagon is a hyperglycemic hormone. It stimulates liver for glucogenolysis and increases the blood glucose level.
Beta (β) cells : They are 70% and secrete insulin. Insulin is a hypoglycemic hormone. It stimulates liver and muscles for glycogenesis. This lowers blood glucose level.
Delta (δ) cells : They are 5% and secrete somatostatin. Somatostatin inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract. In general it is a growth inhibiting factor.
PP cells or F cells : They form 5%. They secrete pancreatic polypeptide (PP) which inhibits the release of pancreatic juice.

Question 2.
Mention the function of testosterone?
Answer:
Testosterone is a steroid sex hormone secreted by testes and cortex of adrenal glands. It controls the secondary sexual characters in males.

Question 3.
Give symptoms of the disease caused by hyposecretion of ADH.
Answer:
Polydipsia, i.e. frequent thirst and polyuria, i.e. frequent urination are the symptoms of the disease caused by hyposecretion of ADH.

5. Short answer questions

Question 1.
Rakesh got hurt on his head when he fell down from his motorbike. Which inner membranes must have protected his brain? What other roles do they have to play
Answer:

When Rakesh fell down from his motorbike, the inner membranes that protected his brain were meninges, viz. dura mater, arachnoid membrane and pia mater. Morevover, CSF must have also acted as a shock absorber.
Dura mater : It is the outer tough membrane protective in function.
Arachnoid membrane : It is the middle web-like membrane which communicates with fluids of upper sub dural space and lower sub arachnoid space.
Pia mater : It is the innermost highly vascularised nutritive membrane in close contact with brain and spinal cord.

Question 2.
Injury to medulla oblongata may prove fatal.
OR
Injury to medulla oblongata causes sudden death. Explain.
Answer:

Medulla oblongata is the region of the brain that controls all the involuntary activities.
Vital activities such as heartbeats, respiration, vasomotor activities, peristalsis, etc. are under the control of medulla oblongata.
When medulla oblongata is injured, all these vital functions are instantly stopped.
Therefore, injury to medulla oblongata causes sudden death.

Question 3.
Distinguish between the sympathetic and parasympathetic nervous system on the basis of the effect they have on:
Heartbeat andUrinary Bladder.
Answer:

Sympathetic Nervous System	Parasympathetic	Nervous System
(1) Heartbeat	Increases	Decreases
(2) Urinary bladder	Relaxes and stores urine	Contracts causing micturition

Question 4.
While holding a tea cup Mr. Kothari’s hands rattle. Which disorder he may be suffering from and what is the reason for this?
Answer:

This condition is due to Parkinson’s disease.
It is due to degeneration of dopamine- producing neurons in the CNS.
80% of the patients develop this condition along with stiffness, difficulty in walking, balance and coordination.

Question 5.
List the properties of the nerve fibres.
Answer:

Excitability / irritability
Conductivity
Stimulus
Summation
All or none
Refractory period
Synaptic delay
Synaptic fatigue
Velocity.

Question 6.
How does tongue detect the sensation of taste?
Answer:

The surface of tongue is with gustatoreceptors.
These receptors are sensitive to the chemicals [sweet, salt, sour, bitter and umami (savory)] present in the food.
The receptor cells get stimulated, generate the impulse which is given to the sensory neuron.

Question 7.
State the site of production and function of Secretin, Gastrin and Cholecystokinin.
Answer:

Hormone	Site of production	Functions
1. Secretin	Duodenal mucosa	Stimulates secretion of pancreatic juice and bile from pancreas and liver respectively.
2. Gastrin	Gastric mucosa	Stimulates gastric glands to secrete gastric juice.
3. Cholecystokinin	Duodenal mucosa	Stimulates pancreas and gall bladder to release pancreatic enzymes and bile respectively.

Question 8.
An adult patient suffers from low heart rate, low metabolic rate and low body temperature. He also lacks alertness, intelligence and initiative. What can be this disease? What can be its cause and cure ?
Answer:

The above symptoms indicate that the person is suffering from Myxoedema.
Myxoedema is condition caused due to hypothyroidism.
Hypothyroidism causes deficiency of thyroid hormones like T₃ and T₄ (thyroxine). This results in low BMR.
This condition can be cured by giving injections of thyroxine or tablets containing hormone preparation.

Question 9.
Where is the pituitary gland located? Enlist the hormones secreted by anterior pituitary.
Answer:
The pituitary gland is attached to hypothalamus on the ventral surface of brain. It is lodged in a bony depression called sella turcica of sphenoid bone.
For names of hormones:

GH : [Growth Hormone/STH : Somatotropic Hormone]
TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
ACTH – [adrenocorticotropic hormone]
PRL – [prolactin]

Gonadotropins-

FSH [follicle stimulating hormone]
LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Explain how the adrenal medulla and sympathetic nervous system function as a closely integrated system.
Answer:

Adrenal medulla originates from embryonic neuro – ectoderm.
It consists of rounded group of large granular cells called chromaffin cells. They are modified post-ganglionic cells of sympathetic nervous system which have lost normal processes and acquired glandular function.
These cells are connected with pre-ganglionic fibres of sympathetic nervous system.
Hence adrenal medulla is an extension of sympathetic nervous system.
Thus adrenal medulla and sympathetic nervous system functions as a closely integrated system.

Question 11.
Name the secretion of alpha, beta and delta cells of islets of Langerhans. Explain their role.
Answer:

Pancreatic islet cells	Secretion	Functions
1. Alpha cells	Glucagon	Stimulates glycogenolysis in the liver
2. Beta cells	Insulin	Stimulates glycogenesis in the liver and muscles
3. Delta cells	Somatostatin	Inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract.

Question 12.
Which are the two types of goitre? What are their causes?
Answer:
(1) Goitre is the enlargement of thyroid gland. It is easily visible at the base of neck when a person is suffering from it.

(2) Goitre is of two types.

Simple goitre : It is also called endemic goitre. This is due to iodine deficiency in the food. This causes iodine deficit in blood. In an attempt to take more iodine from blood, the blood supply to the gland increases. This results in swelling on the thyroid.
Exophthalmic goitre : It is also called toxic goitre. This is due to hyperactive thyroid gland. This can happen if there is overstimulation of thyroid due to excess of ACTH. This disorder is also called Grave’s disease or hyperthyroidism.

Question 13.
Name the ovarian hormone and give their functions.
Answer:

Hormone	Functions
Oestrogen	It is responsible for secondary sexual characters in female.
Progesterone	Essential for thickening of uterine endometrium, thus preparing the uterus for implantation of fertilized ovum. It is responsible for development of mammary glands during pregnancy. It inhibits uterine contractions during pregnancy.
Relaxin	It relaxes the cervix of the pregnant female and ligaments of pelvic girdle during parturition.
Inhibin	It inhibits the FSH and GnRH production.

6. Answer the following.

Complete the table.

Location	Cell type	Function
PNS	————-	Produce myelin sheath.
PNS	Satellite cells	————-
————	Oligodendrocytes	Form myelin sheath around central axon.
CNS	————	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	————	Form the epithelial lining of brain cavities and central canal.

Answer:

Location	Cell type	Function
PNS	Schwann cells	Produce myelin sheath.
PNS	Satellite cells	Supply nutrients to surrounding neurons, protect and cushion nearby neurons.
CNS	Oligodendrocytes	Form myelin sheath around central axon.
CNS	Microglia	Pathogens are destroyed by phagocytosis. (Phagocytose)
CNS	Ependyma	Form the epithelial lining of brain cavities and central canal.

7. Long answer questions.

Question 1.
Explain the process of conduction of nerve impulses up to development of action potential.
Answer:

The origin and maintenance of resting potential depends on the original state of no stimulation.
Any stimulus or disturbance to the membrane will make the membrane permeable to Na⁺ ions. This causes rapid influx of Na⁺ ions.
The voltage gated Na⁺/K⁺ channels are unique. They can change the potential difference of the membrane as per the stimulus received and also the gates operate separately and are self closing.
During resting potential, both gates are closed and resting potential is maintained.
However during depolarization, the Na⁺ channels open but not the K⁺ channels. This causes Na⁺ to rush into the axon and bring about a depolarisation. This condition is called action potential.
Extra cellular fluid (ECF) becomes electronegative with respect to the inner membrane which becomes electropositive.

Question 2.
Draw the neat labelled diagrams.
a. Human ear.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 1

b. Sectional view of human eye.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 2

c. Draw the neat labelled diagram of sagittal section or L.S. of human brain
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 3

d. Draw the neat labelled diagram of Multipolar Neuron.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 4

Question 3.
Answer the questions after observing the diagram given below.
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 5
a. What do the synaptic vesicles contain?
Answer:
Synaptic vesicles contain a neurotransmitter – acetyl choline.

b. What process is used to release the neurotransmitter ?
Answer:
Exocytosis.

c. What should be the reason for the next impulse to be conducted?
Answer:
Removal of neurotransmitter by the action of acetyl cholinesterase.

d. Will the impulse be carried by post synaptic membrane even if one pre-synaptic neuron is there?
Answer:
As far as impulse is transmitted by pre-synaptic neuron, it will be received by post-synaptic neuron.

e. Can you name the channel responsible for their transmission?
Answer:
Ca⁺⁺ channel

Question 4.
Explain the Reflex Pathway with the help of a neat labelled diagram.
OR
With the help of a neat and labelled diagram, describe reflex arc.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 9 Control and Co-ordination 6
I. Reflex action : Reflex action Is defined as a quick, automatic involuntary and often unconscious action brought about when the receptors are stimulated by external or internal stimuli.

II. Reflex arc : Reflex actions are controlled by CNS. Reflex arc is the structural or functional unit of reflex action. Simple reflex arc is formed of the following five components.
(1) Receptor organ : The sensory part that receives the stimulus is called receptor organ. It can be any sense organ that receives the stimulus and converts it into the impulse, e.g. skin, eye, ear, tongue, nasal epithelium, etc.

(2) Sensory neuron or afferent neuron:
Sensory part carrying impulse from receptor organ to CNS is called sensory neuron. Its cyton is located in dorsal root ganglion. Its dendron is long and connected to receptor while the axon enters in the grey matter of spinal cord to form a synapse.

(3) Association, adjustor or intermediate neuron : It is present in the grey matter of spinal cord. Receiving impulse from sensory neuron, interpreting it and generating motor impulse are done by association neuron.
(4) Motor neuron (effector) : The cyton of motor neuron is present in the ventral horn of grey matter and axon travels through ventral root. It conducts motor impulse from spinal cord to effector organ.

(5) Effector organ : Effector organ is a specialized part of the body which is excited by receiving the motor impulse. It gives proper response to the stimulus, e.g. muscles or glands. The path of reflex action is followed by the unidirectional impulse. It originates in the receptor organ and ends in effector organ through CNS.

Question 5.
Krishna was going to school and on the way he saw a major bus accident. His heartbeat increased and hands and feet become cold. Name the part of the nervous system that had a role to play in this reaction.
Answer:

The symptoms observed in Krishna were due to sympathetic nervous system. Emergency conditions trigger sympathetic nervous system to stimulate adrenal medulla.
The cells of adrenal medulla secrete catecholamines like adrenaline and nor¬adrenaline.
These hormones have direct effect on the pacemaker of the heart which causes increase in the heart rate and other associated symptoms.
This is a typical fright reaction caused by intervention of sympathetic nervous system.

Question 6.
What will be the effect of thyroid gland atrophy on the human body?
Answer:

Atrophy means degeneration. Atrophy of thyroid gland will result in deficient secretion of thyroid hormones leading to hypothyroidism. Deficiency of thyroid hormones [T₃ and T₄] and thyrocalcitonin will cause following effects on the body.
Decrease in BMR i.e. basal metabolic rate, decrease in the blood pressure, heart beat, body temperature, etc.
Occurrence of myxoedema in which there is abnormal deposition of fats under the skin giving puffy appearance in adults.
Irregularities in menstrual cycle in case of female patients.
Hair become brittle and fall.
Calcium metabolism also disturbs due to lack of thyrocalcitonin.

Question 7.
Write the names of hormones and the glands secreting them for the regulation of following functions
(a) Growth of thyroid and secretion of thyroxine.
Answer:
TSH by adenohypophysis.

(b) Helps in relaxing pubic ligaments to facilitate easy birth of young ones.
Answer:
Relaxin by degenerating corpus luteum of the ovary.

(d) Controls calcium level in the blood.
Answer:
Calcitonin [hypocalcemic hormone] by thyroid and parathormone [ hypercalcemic hormone] by parathyroid glands.

(e) Controls tubular absorption of water in kidneys.
Answer:
ADH by hypothalamus.

(f) Urinary elimination of water.
Answer:
Atrial natriuretic factor by atria of heart.

(g) Sodium and potassium ion metabolism.
Answer:
Aldosterone by adrenal cortex.

(h) Basal Metabolic rate.
Answer:
T₃ and T₄ by thyroid gland.

(I) Uterine contraction.
Answer:
Oxytocin by hypothalamus.

(j) Heartbeat and blood pressure.
Answer:
Adrenaline, non-adrenaline [stimulation] and acetylcholine [inhibition] by adrenal medulla.

(k) Secretion of growth hormone.
Answer:
GHRF by hypothalamus.

(l) Maturation of Graafian follicle.
Answer:
FSH by anterior pituitary.

Question 8.
Explain the role of hypothalamus and pituitary as a coordinated unit in maintaining homeostasis.
Answer:

Homeostasis is maintenance of constant internal environment of the body.
When certain hormones from any endocrine glands are secreted in excess quantity, the : inhibiting factors from hypothalamus, automatically exert negative feedback and stop the production of stimulating hormones from pituitary.
Similarly, if any hormone is in deficit, then j the concerned gland is given message through releasing factor. This way the hormone production remains in a balanced state or homeostasis.
E.g. If thyroxine from thyroid gland is secreted in excess, the secretion of TSH from pituitary is stopped by stopping the production of TRF from hypothalamus.
Though most of the endocrine glands are under the influence of pituitary gland, it is in turn controlled by hypothalamus.
Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
There is negative feedback mechanism in controlling the secretions of the endocrine glands.
Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.

Following are the releasing and inhibiting factors produced by hypothalamus:

Somatotropin/GHRF : It stimulates release of growth hormone.
Somatostatin/GHRIF : It inhibits the release of growth hormone.
Adrenocorticotropin Releasing Hormone / CRF : It stimulates the release of ACTH by the anterior pituitary gland.
Thyrotropin Releasing Hormone /TRF : It stimulates the release of TSH by anterior pituitary gland.
Gonadotropin Releasing Hormone (GnRH) : It stimulates pituitary to secrete gonadotropins.
Prolactin Inhibiting Hormone (Prolactostatin) : It inhibits prolactin released by anterior pituitary gland.
Gastrin Releasing Peptide (GRP).
Gastric Inhibitory Polypeptide (GIP).

Question 9.
What is adenohypophysis ? Name the hormones secreted by it.
Answer:

Adenohypophysis is the large anterior lobe of pituitary gland.
It is derived from embryonic ectoderm in the form of Rathke’s pouch which is a small outgrowth from the roof of embryonic stomodaeum.
It is made up of epitheloid secretory cells.

It secretes following hormones:

GH : [Growth Hormone/STH : Somatotropic Hormone]
TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
ACTH – [adrenocorticotropic hormone]
PRL – [prolactin]

Gonadotropins-

FSH [follicle stimulating hormone]
LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Describe, in brief, an account of disorders of adrenal gland.
Answer:
(1) Disorders of adrenal cortical secretions are caused due to hyposecretion and hypersecretion of adrenal corcoid hormones.

(2) Hyposecretion of corticosteroids causes Addison’s disease.

(3) The symptoms of Addison’s disease are low blood sugar, low body temperature, feeble heart action, low BR acidosis, low Na⁺ and K⁺ concentration in plasma, excessive loss of Na⁺ and water in urine, impaired kidney functioning and kidney failure, etc. it leads to weight loss, general weakness, nausea, vomiting and diarrhoea.

(4) Hypersecretion of corticoids causes Cushing’s disease.

(5) The symptoms of Cushing’s disease are high blood sugar level, glucosuria, alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, obesity, wasting of limb muscles, etc.

Question 11.
Explain action of steroid hormones and proteinous hormones.
OR
Explain the mode of action of steroid hormones.
Answer:
The hormones always act on their target organs or tissues to induce their effects. The target tissues have specific binding sites or receptor sites which contain hormone receptors.
I. Steroid hormones:

The steroid hormones are lipid soluble and can easily cross the lipoproteinous plasma membrane.
The hormone receptors for steroid hormones are present in cytoplasm or in nucleus.
Hormone-receptor complex formed in cytoplasm enters the nucleus and regulate the gene expression or chromosome function.
In some cases the receptors are present inside the nucleus where hormone receptor complex is formed.
These complexes interact with the genome to evoke biochemical changes that result in physiological and developmental functions.

II. Protein hormones:

The hormone receptors for protein hormones are present on the cell membrane (i.e. membrane bound receptors).
When the hormone binds to its receptor, it forms hormone-receptor complex. Each receptor is specific to a specific hormone.
The hormones which interact with membrane bound receptors normally do not enter the target cell but generate second messengers. Such as cyclic AMP Ca⁺⁺ or IP (Inositol triphosphate), etc.
This leads to certain biochemical changes : in the target tissue.
Thus, the tissue metabolism and consequently the physiological functions are regulated by hormones.

Question 12.
Describe in brief an account of disorders of the thyroid.
OR
What are the functional disorders of thyroid gland? Describe in brief.
Answer:
Disorders of thyroid gland are of three types, viz. hypothyroidism, hyperthyroidism and simple goitre.
(1) Hypothyroidism : Hypothyroidism is deficient secretion of thyroxine. This hyposecretion causes two types of disorders, viz. cretinism in children and myoxedema in adults.
(i) Cretinism : Hyposecretion of thyroxine in childhood causes cretinism. The symptoms of cretinism are retardation of physical and mental growth.

(ii) Myxoedema : Deficiency of thyroxine in adults causes this disorder. It is also referred to as Gull’s disease. Symptoms are thickening and puffiness of the skin and subcutaneous tissue particularly of face and extremities. Patients with low BMR. It also causes mental dullness, loss of memory, slow action.

(2) Hyperthyroidism : Excessive secretion of thyroxine causes exophthalmic goitre or Grave’s disease. There is slight enlargement of thyroid gland. It increases BMR, heart rate, pulse rate and BE Reduction in body weight due to rapid oxidation, nervousness, irritability. Peculiar symptom is exophthalmos, i.e. bulging of eyeballs with staring look and less blinking. This is caused by deposition of fats behind the eye balls in eye sockets. There is muscular weakness and loss of weight.

(3) Simple goitre (Iodine deficiency goitre) : Simple goitre occurs due to deficiency of iodine in diet or drinking water. Simple goitre causes enlargement of thyroid gland. Thyroid gland in an attempt to get more iodine from the blood, swells due to increased blood supply. Prevention of goitre can be done by administering iodized table salt. It is also called endemic goitre as it is common in hilly areas.

Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance

July 19, 2024July 18, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 4 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 4 Molecular Basis of Inheritance

1. Multiple Choice Questions

Question 1.
Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer:
(d) Streptococci

Question 2.
The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer:
(c) Endonucleases

Question 3.
Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer:
(b) Cytoplasm

Question 4.
The enzyme required for transcription is ………………..
(a) DNA polymerase
(b) RNApolymerase
(c) Restriction enzyme
(d) RNase
Answer:
(b) RNA polymerase

Question 5.
Transcription is the transfer of genetic information from ………………..
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer:
(a) DNA to RNA

Question 6.
Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of these
Answer:
(a) Replication

Question 7.
In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Question 8.
How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(a) 3

Question 9.
Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer:
(d) Tryptophan operon

Question 10.
Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer:
(b) i, iv, iii, ii

2. Very Short Answer Questions

Question 1.
What is the function of an RNA primer during protein synthesis?
Answer:
During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as template.
[Note : RNA primer has no direct role in protein synthesis.]

Question 2.
Why is the genetic code considered as commaless?
Answer:
The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless.

Question 3
Genome
Answer:
Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

Question 4.
Which enzyme does remove supercoils from replicating DNA?
Answer:
Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA.

Question 5.
Why are Okazaki fragments formed on lagging strand only?
Answer:
Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time.

Question 6.
When does DNA replication take place?
Answer:
In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division.

Question 7.
Define term Codogen and Codon
Answer:
Codogen is a triplet of nucleotides present on the DNA which specifies one particular amino acid.
Codon is a triplet of nucleotides present on the m-RNA which specifies one particular amino acid.

Question 8.
What is degeneracy of genetic code?
Answer:
Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons.

Question 9.
Which are the nucleosomal ‘core’ histones?
Answer:
Two molecules each of histone proteins, viz. H₂A. H₂B, H₃ and H₄ are the nucleosomal ‘core’ histones.

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10^-16 m).
DNA is associated with histone and non-histone proteins.
Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. H₂A, H₂B, H₃ and H₄ forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by 1 3/4 turns.
H₁ protein binds the DNA thread where it enters and leaves the nucleosome.
Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
Each nucleosome contains 200 bp of DNA.
Packaging involves formation of – Beads on string (10 nm diameter), Solenoid fibre (looks like coiled telephone wire, 30 nm diameter/300Å), Chromatin fibre and Chromosome.
Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.

Question 2.
Enlist the characteristics of genetic code.
Answer:
The characteristics of genetic code are

Genetic code is triplet, commaless and non-overlapping.
It is degenerate and non-ambiguous.
It is universal
It has polarity.

Question 3.
Applications of DNA fingerprinting.
Answer:
Applications of DNA fingerprinting are as follows:

In forensic science to solve rape and murder cases.
Finds out the biological father or mother or both, of the child, in case of disputed parentage.
Used in pedigree analysis in cats, dogs, horses and humans.

Question 4.
Explain the role of lactose in ‘Lac Operon’.
Answer:

A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
Lactose binds to repressor and inactivates it.
Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
Thus, lactose acts as an inducer.
When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.

4. Short Answer Questions

Question 1.
Human genome project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed r in 2003.

2. The main aims:

To sequence 3 billion base pairs of DNA in human genome and to map an estimated 33000 genes.
To store the information collected from the project in databases.
To develop tools and techniques for analysis of the data.
Transfer of the related technologies to the private sectors, such as industries.
Taking care of the legal, ethical and social issues which may arise from project.
To sequence the genomes of several other organisms such as bacteria e.g. E.coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophil, rice, Arabidopsis), Mus musculus, etc.

3. Significance:

HGP has a major impact in the fields like Medicine, Biotechnology, Bioinformatics and the Life sciences.
More understanding of functions of genes, proteins and human evolution.

Question 2.
Describe the structure of operon.
Answer:

An operon is a unit of gene expression and regulation.
It includes the structural genes and their control elements. Control elements are promoters and operators.
The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
Operators are present between the promoters and structural genes.
There is repressor protein that binds to the operator region of the operon.
There are regulatory genes which are responsible for the formation of repressors which interact with operators.

Question 3.
In the figure below A, B and C are three types of
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 1
Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 2
Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

Question 5.
Match the entries in Column I with those of Column II and choose the correct answer.

Column I	Column II
A. Alkali treatment	i. Separation of DNA fragments on gel slab
B. Southern blotting	ii. Splits DNA fragments into single strands
C. Electrophoresis	iii. DNA transferred to nitrocellulose sheet
D. PCR	iv. X-ray photography
E. Autoradiography	v. Produce fragments different sizes
F. DNA treated with REN	vi. DNA amplification

Answer:

Column I	Column II
A. Alkali treatment	ii. Splits DNA fragments into single strands
B. Southern blotting	iii. DNA transferred to nitrocellulose sheet
C. Electrophoresis	i. Separation of DNA fragments on gel slab
D. PCR	vi. DNA amplification
E. Autoradiography	iv. X-ray photography
F. DNA treated with REN	v. Produce fragments different sizes

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 3
DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

Nucleotides (dAMP dGMR dCMP and dTMP) present in the nucleoplasm, are activated by ATP in presence of an enzyme phosphorylase.
This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATE dGTR dCTP and dTTE

Point of Origin or Initiation point:

Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
At the point ‘O’, enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:

Enzyme DNA helices breaks weak hydrogen bonds in the vicinity of ‘O’.
The strands of DNA separate and unwind. This unwinding is bidirectional.
SSBP (Single strand binding proteins) remains attached to both the separated strands and prevent them from recoiling (rejoining).

Replicating fork:

Y-shape replication fork is formed due to unwinding and separation of two strands.
The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

Synthesis of new strands:

Each separated strand acts as a template for the synthesis of new complementary strand.
A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

The template strand with free 3′ is called the leading template.
The template strand with free 5′ end is called the lagging template.
The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
New strands are always formed in 5′ → 3′ direction.
The new strand which develops continuously towards replicating fork is called the leading strand.
The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.
Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

Question 2.
Describe the process of transcription in protein synthesis.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 4
Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

RNA polymerase binds to promoter site.
It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
Only antisense strand functions as template.

(2) Elongation:

The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes m-RNA molecule free.
As the m-RNA grows, the transcribed region of DNA molecule becomes spirally coiled and regains double helical form.

(3) Termination:
When RNA polymerase reaches the terminator site on the DNA, both enzyme and newly formed m-RNA (primary transcript) gets released.

Question 3.
Describe the process of translation in protein synthesis.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 5
Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

Small subunit of ribosome binds to the m-RNA at 5’ end.
Start codon is positioned properly at P-site.
Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
The large subunit of ribosome joins with the smaller subunit in the presence of Mg⁺⁺.
Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

3. Elongations of polypeptide chain:
Addition of amino acid occurs in 3 Step cycle-
i. Codon recognition.
Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide bond formation.

Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
It takes less than 0.1 second for formation of peptide bond.
Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation.

Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
As ribosome move over the m-RNA, all the codons on m-RNA are exposed oiie by one for translation.

4. Termination and release of polypeptide:
When stop codon (UAA, UAG, UGA) gets exposed at the A-site, the release factor binds to the stop codon, thereby terminating the translation process
The polypeptide gets released in the cytoplasm.
Two subunits of ribosome dissociate and last t-RNA and m-RNA are released in the cytoplasm.
m-RNA gets denatured by nucleases immediately.

Question 4.
Describe Lac ‘Operon’.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 4 Molecular Basis of Inheritance 6
Lac operon consists of the following components:
(1) Regulator gene:

Regulator gene precedes the promoter gene.
It may not be present immediately adjacent to operator gene.
Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:

It precedes the operator gene.
It is present adjacent to operator gene.
RNA Polymerase enzyme binds at promoter site.
Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator gene:

It precedes the structural genes.
When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural gene:

There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

Question 5.
Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
Answer:

DNA as the genetic material has to be chemically and structurally stable.
It should be able to generate its replica.
Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) The process of translation occurs at the ribosome.
Answer:

Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
Ribosome has three binding sites for t-RNA : P-site (peptidyl t-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
In eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
Thus ribosome plays an important role in translation.

(iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer:
The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfer specific amino acid to m-RNA Ribosome complex. It binds with amino acid at its 3′ end.

(iv) Transcription must occur before translation may occur.
Answer:
In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm.

Question 6.
Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	—————-
————–	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	—————–
Bite mark	—————–	Saliva
————-	Surface area	Hair, semen, sweat, urine

Answer:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	Saliva
Door	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	Saliva
Bite mark	Teeth impression	Saliva
Clothes	Surface area	Hair, semen, sweat, urine

Maharashtra Board Class 12 Biology Solutions Chapter 8 Respiration and Circulation

July 17, 2024July 16, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 8 Respiration and Circulation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 8 Respiration and Circulation

1. Multiple choice questions

Question 1.
The muscular structure that separates the thoracic and abdominal cavity is …………………..
(a) pleura
(b) diaphragm
(c) trachea
(d) epithelium
Answer:
(b) diaphragm

Question 2.
What is the minimum number of plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a R.B.C.?
(a) two
(b) three
(c) four
(d) five
Answer:
(a) two

Question 3.
…………………. is a sound producing organ.
(a) Larynx
(b) Pharynx
(c) Tonsils
(d) Trachea
Answer:
(a) Larynx

Question 4.
The maximum volume of gas that is inhaled during breathing in addition to T.V. is …………………..
(a) residual volume
(b) IRV
(c) GRV.
(d) vital capacity
Answer:
(b) IRV

Question 5.
………………….. muscles contract when the external intercostals muscles contract.
(a) Internal abdominal
(b) Jaw
(c) Muscles in bronchial walls
(d) Diaphragm
Answer:
(d) Diaphragm

Question 6.
Movement of cytoplasm in unicellular organisms is called …………………..
(a) diffusion
(b) cyclosis
(c) circulation
(d) thrombosis
Answer:
(b) cyclosis

Question 7.
Which of the following animals do not have closed circulation?
(a) Earthworm
(b) Rabbit
(c) Butterfly
(d) Shark
Answer:
(c) Butterfly

Question 8.
Diapedesis is performed by …………………..
(a) erythrocytes
(b) thrombocytes
(c) adipocytes
(d) leucocytes
Answer:
(d) leucocytes

Question 9.
Pacemaker of heart is …………………..
(a) SA node
(b) AV node
(c) His bundle
(d) Purkinje fibers
Answer:
(a) SA node

Question 10.
Which of the following is without nucleus?
(a) Red blood corpuscle
(b) Neutrophil
(c) Basophil
(d) Lymphocyte
Answer:
(a) Red blood corpuscle

Question 11.
Cockroach shows which kind of circulatory system?
(a) Open
(b) Closed
(c) Lymphatic
(d) Double
Answer:
(a) Open

Question 12.
Diapedesis can be seen in …………………..
(a) RBC
(b) WBC
(c) Platelet
(d) neuron
Answer:
(b) WBC

Question 13.
Opening of inferior vena cava is guarded by …………………..
(a) bicuspid valve
(b) tricuspid valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 14.
…………………. wave in ECG represent atrial depolarization.
(a) P
(b) QRS complex
(c) Q
(d) T
Answer:
(a) P

Question 15.
The fluid seen in the intercellular spaces in Human is …………………..
(a) blood
(b) lymph
(c) interstitial fluid
(d) water
Answer:
(b) lymph

2. Match the columns

Question 1.
Respiratory surface Organism

Respiratory surface	Organism
(1) Plasma membrane	(a) Insect
(2) Lungs	(b) Salamander
(3) External gills	(c) Bird
(4) Internal gills	(d) Amoeba
(5) Trachea	(e) Fish

Answer:

Respiratory surface	Organism
(1) Plasma membrane	(d) Amoeba
(2) Lungs	(c) Bird
(3) External gills	(b) Salamander
(4) Internal gills	(e) Fish
(5) Trachea	(a) Insect

3. Very Short Answer Questions

Question 1.
Why does trachea have ‘C’-shaped rings of cartilage?
Answer:
Trachea is supported by ‘C’-shaped rings of J cartilage which prevent it from collapsing and always keep it open.

Question 2.
Why is respiration in insect called direct respiration?
Answer:
Respiration in insect is called direct because tracheal tubes exchange O₂ and CO₂ directly with the haemocoel which then exchange them with tissues.

Question 3.
Why is gas exchange very rapid at alveolar level?
OR
Why does gas exchange in the alveolar region very rapid?
Answer:
Gas exchange is very rapid at alveolar level because numerous alveoli (about 700 millions) in the lungs provide large surface area for gaseous exchange.

Question 4.
Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the entry of food into trachea while eating.

4. Short Answer Questions

Question 1.
Why is it advantageous to breathe through the nose than through the mouth?
Answer:
Breathing through nose is better than breathing through the mouth because of the following reasons:

The nostrils are smaller than the mouth so air exhaled through the nose creates a backflow of air into the lungs.
As we exhale more slowly through the nose than we do through the mouth, the lungs have more time to extract oxygen from the air that we have already taken in.
The hairs inside nostrils filter any dust particles and microbes in the air and it only lets the clean air pass through.
The air gets warm and humidified in nostrils as it passes into our bodies.
Moreover breathing through the mouth can dry the oral cavity and lead to bad breath, gum disease and tooth decay.

Question 2.
Identity the incorrect statement and correct it.
(a) A respiratory surface area should have a. large surface area.
(b) A respiratory surface area should be kept dry.
(c) A respiratory surface area should be thin, may be 1 mm or less.
Answer:
Statement (a) and statement (c) are correct whereas statement (b) is incorrect. A respiratory surface area should be kept moist, is the correct statement.

Question 3.
Given below are the characteristics of some modified respiratory movement. Identify them.
a. Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.
Answer:
Sneezing

b. An inspiration followed by many short convulsive expiration accompanied by facial expression.
Answer:
Laughing, Crying.

Question 4.
Blood plasma.
Answer:

Plasma is a straw coloured, slightly alkaline viscous fluid part of the blood, having 90-92% water and 8-10% soluble proteins.
Serum albumin, serum globulin, heparin, fibrinogen and prothrombin are the plasma proteins which form 7% of the plasma.
Glucose, amino acids, fatty acids and glycerol are the nutrients dissolved in plasma.
Nitrogenous wastes (urea, uric acid, . ammonia and creatinine) and respiratory gases (oxygen and carbon dioxide) is present in plasma.
Enzymes and hormones too are transported Ada plasma.
Inorganic minerals are also present in plasma such as bicarbonates, chlorides, phosphates and sulphates of sodium, potassium, calcium and magnesium.

Question 5.
Blood clotting/Coagulation of blood.
OR
Explain blood clotting in short.
Answer:

The process of converting the liquid blood into a semisolid form is called blood clotting or coagulation.
The process of clotting may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
Intrinsic and extrinsic processes involve interaction of various substances called clotting factors by a step wise or cascade mechanism.
There are in all twelve clotting factors numbered as I to XII (factor VI is not in active use).
Interaction of these factors in a cascade manner leads to formation of enzyme, Thromboplastin which helps in the formation of enzyme prothrombinase.
Prothrombinase inactivates heparin and also converts inactive prothrombin into active thrombin.
Thrombin converts soluble blood protein- fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.
These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 6.
Describe pericardium.
Answer:

Pericardium is the double layered peritoneum that encloses the heart. It consists of two layers, viz. fibrous pericardium and serous pericardium.
Fibrous pericardium is the outer layer having tough, inelastic fibrous connective tissue whereas serous pericardium is the v inner double layered membrane. It has in turn an outer parietal layer and inner visceral layer.
Parietal layer of serous pericardium lies on the inner side of fibrous pericardium.
Visceral layer also known as epicardium adheres to heart and thus forms outer covering over the heart.
There is a pericardial fluid in the pericardial space which is present in between the parietal and visceral layers of serous pericardium.

Question 7.
Describe valves in the human heart.
Answer:
Human heart has following main valves:

Tricuspid valve : Tricuspid valve is present between the right atrium and right ventricle. It has three cusps or flaps. It prevents the backflow of blood into right atrium.
Bicuspid valve : Bicuspid valve, also called mitral valve is present between the left atrium and left ventricle. It has two flaps. It prevents the backflow of blood in left atrium. Both tricuspid and bicuspid valves are attached to papillary muscles with tendinous chords or chordate tendinae to prevent valves from turning back into atria at the time of systole.
Semilunar valve : These are present at the opening of pulmonary artery and systemic aorta. They prevent the back flow of blood when ventricles undergo systole.
Thebesian valve : Thebesian valve is present at the opening of coronary sinus.
Eustachian valve : Eustachian valve is present at the opening of inferior vena cava.

Question 8.
What is the role of papillary muscles and chordae tendinae in human heart?
Answer:

Papillary muscles are large and well- developed muscular ridges present along the inner surface of the ventricles.
Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae.
Chordae tendinae are inelastic fibres present in the lumen of ventricles.
The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles and regulate the opening and closing of bicuspid and tricuspid valves.

Question 9.
Explain in brief the factors affecting blood pressure.
Answer:

Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

5. Give Scientific Reason

Question 1.
Closed circulation is more efficient than open circulation.
Answer:

Closed circulation considerably enhances the speed, precision and efficiency of circulation.
The blood flows more rapidly, it takes less time to circulate through the closed system and return to the heart.
This fastens the supply and removed of materials to and from the tissues by the blood as compared to open circulation.
In open circulation, there are no blood vessels such as arteries or veins, to pump the blood.
Therefore, the blood pressure is very low.
Organisms with an open circulatory system typically have a relatively high volume of hemolymph and low blood pressure. Closed circulation is thus more efficient than open circulation.

Question 2.
Human heart is called as myogenic and autorhythmic?
Answer:

The heart shows auto rhythmicity because the impulse for its rhythmic movement develops inside the heart. Such heart is called myogenic.
Some of the cardiac muscle fibres become auto rhythmic (self-excitable) and start generating impulse during development.
These autorhythmic fibres perform two important function, viz. acting as a pacemaker and setting the rhythm for heart.
They also form conducting system for conduction of nerve impulses throughout the heart muscles.

Question 3.
In human heart, the blood flows only in one direction.
Answer:

In veins there are valves, which prevent the back flow of the blood.
In arteries, blood flows with unidirectional pressure.
Hence the circulation takes place only in one direction.

Question 4.
Arteries are thicker than veins.
Answer:

Arteries have relatively thick walls to enable them to withstand the high pressure of blood ejected from the heart.
Arteries expand when the pressure increases as the heart pushes blood out but then recoil (shrink) Wn the pressure decreases when the heart relaxes between heartbeats.
This expansion and recoiling occurs to maintain a smooth blood flow.
Veins, on the other hand, have thinner walls and larger lumen veins have no need for thick walls as then need not have to withstand high pressure like arteries.
Moreover, as veins transport relatively low pressure blood, they are commonly equipped with valves to promote the unidirectional flow of blood towards the heart.

Question 5.
Left ventricle is thick than all other chambers of heart.
OR
Left ventricle has thicker wall than the right ventricle.
Answer:

Left ventricle pumps oxygenated blood to all parts of the body. Therefore, there is greater pressure from the blood in left ventricle.
Right ventricle sends deoxygenated blood to lungs for oxygenation. This does not put more pressure and lungs are in vicinity of the heart.
Due to these functional differences between the two ventricles, left ventricle has thicker wall than that of the right ventricle.

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

Open circulation	Closed circulation
1. In open circulation, blood flows through large open spaces and channels called lacunae and haemocoels among the tissues.	1. In closed circulation, blood flows through a network of blood vessels all over the body.
2. Tissues are in direct contact with the blood.	2. Blood does not come in direct contact with tissue.
3. Blood flows with low pressure and usually does not contain any respiratory pigment like haemoglobin.	3. Blood flows with high pressure and contains respiratory pigment like haemoglobin.
4. Exchange of material takes place directly between blood and cells or tissues of the body.	4. Exchange of material takes place between blood and body tissues through an intermediate fluid called lymph.
5. Volume of blood flowing through a tissue cannot be controlled as blood flows out in open space.	5. Volume of blood can be regulated by the contraction and relaxation of the smooth muscles of the blood vessels.
6. Open circulatory system is found in arthropods and some molluscs.	6. Closed circulatory system is found in annelids, echinoderms and all vertebrates.

Question 2.
Arteries and veins.
Answer:

Arteries	Veins
1. The blood vessels that arise from the heart and carry blood away from heart are called arteries.	1. The blood vessels that bring blood to the heart are called veins.
2. Arteries are thick walled blood vessels, situated in deep layers in the body.	2. Veins cure thin walled blood vessels, situated superficially in the body.
3. Arteries do not have valves.	3. Veins have valves.
4. Tunica adventitia, the outermost layer of arteries is thick and elastic.	4. Tunica externa, the outermost layer of veins is thin.
5. Tunica media is very thick and contain elastic fibres.	5. Tunica media is thin layer and contain involuntary muscle fibres.
6. The lumen of arteries is small.	6. The lumen of the veins is very spacious.
7. With the exception of pulmonary arteries, all other arteries carry oxygenated blood.	7. With the exception of pulmonary veins, all other veins carry deoxygenated blood.
8. Blood in the arteries show high blood pressure.	8. Blood in the veins show lesser blood pressure.

Question 3.
Blood and Lymph.
Answer:

Blood	Lymph
1. Contains blood plasma with proteins and all three types of blood cells namely RBCs, WBCs and blood platelets.	1. Contains blood plasma without blood proteins, RBCs and platelets and contains lymphocytes.
2. Red in colour due to presence of RBCs.	2. Light yellow in colour and does not contain RBCs.
3. Carries oxygen in the body.	3. Does not carry oxygen.
4. The flow of blood in blood vessels is fast.	4. The flow of lymph in lymph capillaries is slow.
5. Lymphocytes are present.	5. Lymphocytes are present, more in number than those present in the blood.

Question 4.
Blood capillary and Lymph capillary.
Answer:

Blood capillary	Lymph capillary
1. Reddish, easy to observe.	1. Colourless, difficult to observe.
2. Joined to arterioles at one end and to venules at another end.	2. Blind (closed at the tip).
3. Narrower than lymph capillaries.	3. Wider than blood capillaries.
4. Wall consists of normal endothelium and basement membrane.	4. Wall consists of thin endothelium and poorly developed basement membrane.
5. Contains red blood.	5. Contains colourless lymph.
6. Have relatively high pressure.	6. Have relatively low pressure.

Question 5.
Intrinsic and Extrinsic process of clotting.
Answer:

Intrinsic process	Extrinsic process
1. The intrinsic pathway requires only clotting factors found within the blood itself – in particular, clotting factor XII (Hageman factor) from the platelets.	1. The extrinsic pathway is initiated by factors external to the blood, in the tissues adjacent to damaged blood vessel – in particular, it is initiated by clotting factor III, thromoboplastin from the damaged tissues.
2. It is a longer, multistep process and it takes a little longer for the blood to clot by this mechanism.	2. It involves fewer chemical reaction steps and produce a clot a little more quickly than the intrinsic pathway.

7. Long Answer Questions

Question 1.
Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated
OR
While working with the car engine in a closed garage, John suddenly felt dizzy and fainted what is the possible reason?
Answer:

As Smita and John were working with the car engine running in a closed garage, they must be suffering from carbon monoxide poisoning.
Carbon monoxide (CO) is a highly toxic gas produced when fuels burn incompletely from automobile engines.
Because of strong affinity of haemoglobin with carbon monoxide, it readily combines with carbon monoxide to from a stable compound, carboxyhaemoglobin. Thus, less haemoglobin is available for oxygen transport depriving the cells of oxygen.
Exposure to carbon monoxide can usually leads to throbbing headache, drowsiness, breathlessness and often person gets fainted. In extreme cases carbon monoxide poisoning usually leads to unconsciousness, convulsions, cardiovascular failure, coma and eventually death.

The breathless persons can be treated by following method:

Oxygen treatment : The best way to treat carbon monoxide poisoning is to breathe in pure oxygen (high-dose oxygen treatment)
Oxygen chamber : Doctor may temporarily place her in a pressurized oxygen chamber (also known as a hyperbaric oxygen chamber)

Question 2.
Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and stated wheezing. What could be the possible condition and how can he be treated?
Answer:
(1) It indicates that Shreyas might be suffering from allergic reactions. He may have come in contact with allergens such as pollen, dust, pet dander or other environmental substances on his way in the garden. Or Shreyas may be already a patient of Asthma and his symptoms may have aggrevated due to wintry climate.

(2) If a person is allergic to a substance, such as pollen, his immune system reacts to the substance as if it was foreign and harmful, and tries to destroy it.

(3) The body reacts to these allergens by making and releasing substances known as IgE antibodies. These IgE antibodies attach to most cells in the body which release histamine. Histamine is the main substance responsible for pollen allergy symptoms such as difficulty in breathing, wheezing, sneezing, itchy throat, etc.

(4) Treatment : There are several drugs to treat the allergic reactions:

Antihistamines such as cetirizine or diphenhydramine.
Decongestants, such as pseudoephedrine or oxymetazoline.
Medications that combine an antihistamine and decongestant such as Actifed and Claritin-D.

Question 3.
Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?
Answer:
(1) When the heart contracts, it creates pressure that pushes blood out of heart. This pressure acts like a wave. This “wave” of pressure is the pulse you feel. But this pressure is not constant.

(2) When the heart pumps the blood out of it at the time of systole, there is maximum pressure in the arteries. This pressure weakens considerably when it reaches capillaries, and so the veins which are away from the heart are under less pressure. Due to low pressure veins have valves to prevent backflow of blood.

(3) The pressure in the arteries can be felt every time the heart beats, especially in arteries which come to surface of the body like that of the wrist and neck but not in veins.

(4) The pressure in veins is always weaker than in arteries, resulting in a weaker pulse to the point that it is undetectable by touch
alone.

(5) Owing to this, when we keep finger on the arteries of wrist or neck, we feel a pulse but not when we keep it on a vein.

Question 4.
A man’s pulse rate is 68 and cardiac output is 5500 cm3. Find the stroke volume.
Answer:
Cardiac output is the volume of blood pumped out per min for a normal adult human being it is calculated as follows:
Cardiac output = Heart rate × Stroke volume
Given : Cardiac output = 5500 cm³
Pulse rate = Heart rate = 68
By using these values stroke volume of is calculated as follows:
∴ Cardiac output = Heart rate × Stroke volume
∴ Stroke volume = Cardiac output/Heart rate
= 5500/68
= Approx. 80. ∴ Stroke volume is 80 ml.

Question 5.
Which blood vessel leaving from the heart will have the maximum content of oxygen and why?
Answer:

The Aorta leaving the heart from left ventricle carry the maximum content of oxygen.
Deoxygenated blood becomes oxygenated in the pulmonary capillaries surrounding the alveoli of lungs. The oxygenated blood from lungs is collected by the four pulmonary veins.
These pulmonary veins carry that oxygenated blood to left atrium of heart. During atrial systole that blood is carried to left ventricle.
Left ventricle then pumps that oxygenated blood to Aorta during ventricular systole. Therefore, aorta has the maximum content of oxygen.

Question 6.
If the duration of the atrial ‘systole is 0.1 second and that of complete diastole is 0.4 second, then how does one cardiac cycle complete in 0.8 second?
Answer:

The time duration required to complete one cardiac cycle is 0.8 second.
Cardiac cycle is divided into three important phases, viz, atrial systole, ventricular systole and joint diastole.
Atrial systole in normal condition lasts for 0.1 second, ventricular systole follows atrial systole and lasts for 0.3 second whereas joint diastole or complete diastole lasts for about 0.4 second.
In this way one cardiac cycle is completed in 0.8 second.

Question 7.
How is blood kept moving in the large veins of the legs?
Answer:
1. When heart undergoes systole, it pushes the blood with pressure in aorta. This pressure moves the entire circulation of the blood throughout the body. Aorta gives rise to dorsal aorta after supplying to upper parts of body. Then it divides into two arteries which enter two legs. The blood is forced to move in the legs due to blood pressure and also aided by gravity.

2. In addition, the muscles in legs help transport blood back to our heart. As the muscles of our body contract and relax to move our limbs, they squeeze the blood in veins and the blood is then pushed towards the heart.

3. The veins in legs also have valves to keep this process going and prevent blood from flowing back down towards the feet.

4. In this way blood is kept moving in the large veins of the legs.

Question 8.
Describe histological structure of artery, vein and capillary.
Maharashtra Board Class 12 Biology Solutions Chapter 8 Respiration and Circulation 1
Answer:
Histological structure of artery and vein.

Artery is a thick walled blood vessel that carries oxygenated blood. (Exception is pulmonary artery which carries deoxygenated blood from heart to lungs for oxygenation.)
All the arteries arise from heart and carry blood away from the heart.
Each artery is made up of three layers, viz. tunica externa, tunica media and tunica interna.
Tunica externa or adventitia is the thickest layer of all. It is the outermost coat made up of connective tissue with elastic and collagen fibres.
Tunica media is the middle coat made up of smooth muscle fibres and elastic fibres. It withstands high blood pressure during ventricular systole. It is also thick.
Tunica interna or intima is the innermost coat made of endothelium and elastic layer.

Histology of Capillaries:

Capillaries are the smallest and thinnest blood vessels. Capillaries are formed by the division and re-division of the arterioles.
The wall of the capillary is made up of endothelium or squamous epithelium.
The capillary wall is permeable to water and dissolved substances.
Exchange of respiratory gases, nutrients, excretory products, etc. takes place through the capillary wall.
Capillaries unite to form venules.

Question 9.
What is blood pressure? How is it measured? Explain factors affecting blood pressure.
Answer:
1. Blood pressure:

The pressure exerted by blood on the wall of the blood vessels is called blood pressure. Pressure exerted by blood on the wall of arterial wall is arterial blood pressure. Blood pressure is described in two terms viz. systolic blood pressure and diastolic blood pressure.
Systolic blood pressure is the pressure exerted on arterial wall during ventricular contraction (systole). For a normal healthy adult the average value is 120 mmHg.
Diastolic blood pressure is the pressure on arterial wall during ventricular relaxation (diastole). For a normal healthy adult it is 80 mmHg.
B. E = SP/DP = 120/80 mmHg. Blood pressure is normally written as 120/80 mmHg. Difference between systolic and diastolic pressure is called pulse pressure normally, it is 40 mmHg.

2. Measurement of blood pressure:

Blood pressure is measured with the help of an instrument called sphygmomanometer.
The instrument consists of inflatable rubber bag cuff covered by a cotton cloth. It is connected with the help of tubes to a mercury manometer on one side and a rubber bulb on the other side.
During measurement, the person is asked to lie in a sleeping position. The instrument is placed at the level of heart and the cuff is tightly wrapped around upper arm.
The cuff is inflated till the brachial artery is blocked due to external pressure. Then pressure in the cuff is slowly lowered till the first pulsatile sound is produced. At this moment, pressure indicated in manometer is systolic pressure. Sounds heard during this measurement of blood pressure are called as Korotkoff sounds.
Pressure in the cuff is further lowered till any pulsatile sound cannot be heard due to smooth blood flow. At this moment, pressure indicated in manometer is diastolic pressure an optimal blood pressure (normal) level reads 120/80 mmHg.

3. Factors affecting blood pressure:

Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

Question 10.
Describe human blood and give its functions.
Answer:
Blood Composition:

Blood is a red coloured fluid connective tissue derived from embryonic mesoderm.
It has two components – the fluid plasma (55%) and the formed elements i.e. blood cells (44%).
Plasma is a straw coloured, slightly alkaline and viscous fluid having 90% water and 10% solutes such as proteins, nutrients, nitrogenous wastes, salts, hormones, etc.
Blood corpuscles are of three types, viz. erythrocytes (RBCs), white blood corpuscles (WBCs) and thrombocytes (platelets).

(5) Red blood corpuscles or Erythrocytes:

Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
The RBC size : 7 pm in diameter and 2.5 pm in thickness.
The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
The average life span of RBC : 120 days.
RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
The old and worn out RBCs are destroyed in liver and spleen.
Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
Maintenance of blood pH as haemoglobin acts as a buffer.
Maintenance of the viscosity of blood.

(6) White blood corpuscles / Leucocytes:
Maharashtra Board Class 12 Biology Solutions Chapter 8 Respiration and Circulation 3
1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.

3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.

4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.

5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

In neutrophils, the cytoplasmic granules take up neutral stains.
Their nucleus is three to five lobed.
It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
Neutrophils are about 70% of total WBCs.
They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
Eosinophils are about 3% of total WBCs.
They are non-phagocytic in nature.
Their number increases (i.e. eosinophilia) during allergic conditions.
They have antihistamine property.

(III) Basophils:

The cytoplasmic granules of basophils take up basic stains such as methylene blue.
They have twisted nucleus.
In size, they are smallest and constitute about 0.5% of total WBCs.
They too are non-phagocytic.
Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

6. Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

Agranulocytes with a large round nucleus are called lymphocyte.
They are about 30% of total WBCs.
Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
They are phagocytic in function.
They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
At the site of infections they are seen in more enlarged form.

(7) Thrombocytes/Platelets:

Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
Their life span is about 5 to 10 days.
Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
Thrombocytes possess thromboplastin which helps in clotting of blood.
Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

(8) Functions of blood:

Transport of oxygen and carbon dioxide
Transport of food
Transport of waste product
Transport of hormones
Maintenance of pH
Water balance
Transport of heat
Defence against infection
Temperature regulation
Blood clotting/coagulation
Helps in healing

Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation

July 17, 2024July 16, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation

Multiple-choice Questions

Question 1.
The nasal cavity is divided into right and left nasal chambers by a …………………..
(a) sphenoid
(b) palatine
(c) mesethmoid
(d) zygomatic
Answer:
(c) mesethmoid

Question 2.
The right lung is divided into …………………..
(a) 3 lobes
(b) 2 lobes
(c) 4 lobes
(d) 6 lobes
Answer:
(a) 3 lobes

Question 3.
Carbon dioxide is carried in the blood mainly as …………………..
(a) sodium carbonate
(b) sodium bicarbonate
(c) carbaminohaemoglobin
(d) carbonic acid
Answer:
(b) sodium bicarbonate

Question 4.
Transport of oxygen is carried out by …………………..
(a) plasma
(b) lungs
(c) RBCs
(d) nostrils
Answer:
(c) RBCs

Question 5.
Respiration taking place at the alveoli of lungs is called …………………..
(a) internal respiration
(b) external respiration
(c) cellular respiration
(d) tissue respiration
Answer:
(b) external respiration

Question 6.
The volume of air inspired or expired during normal breathing is …………………..
(a) ERV
(b) IRV
(c) TV
(d) VC
Answer:
(c) TV

Question 7.
What is the partial pressure of oxygen and carbon dioxide respectively in the atmospheric air?
(a) PPO₂ 159 mmHg, PPCO₂2 0.3 mmHg
(b) PPO₂ 104 mmHg, PPCO₂ 40 mmHg
(c) PPO₂ 40 mmHg, PPCO₂ 45 mmHg
(d) PPO₂ 95 mmHg, PPCO₂ 40 mmHg
Answer:
(b) PPO₂ 104 mmHg, PPCO₂ 40 mmHg

Question 8.
The vital capacity of human lung is equal to …………………..
(a) 3500 ml
(b) 4600 ml
(c) 500 ml
(d) 1200 ml
Answer:
(b) 4600 ml

Question 9.
The exchange of gases between alveolar air and alveolar capillaries occurs by …………………..
(a) osmosis
(b) active transport
(c) absorption
(d) diffusion
Answer:
(d) diffusion

Question 10.
Oxygen dissociation curve will shift to right on the decrease of …………………..
(a) acidity
(b) carbon dioxide concentration
(c) temperature
(d) pH
Answer:
(d) pH

Question 11.
Respiratory organs in scorpion are …………………..
(a) gills
(b) book lungs
(c) skin
(d) book gills
Answer:
(b) book lungs

Question 12.
Breakdown of alveoli of lungs resulting in reducing surface area for gas exchange is known as …………………..
(a) emphysema
(b) sneezing
(c) pneumonia
(d) tuberculosis
Answer:
(a) emphysema

Question 13.
During inspiration, the diaphragm …………………..
(a) relaxes
(b) contracts
(c) expands
(d) shows no change
Answer:
(b) contracts

Question 14.
Over inflation of the lungs is prevented due to …………………..
(a) Bohr’s effect
(b) Conditioned reflex
(c) Hering-Breuer reflex
(d) Haldane effect
Answer:
(c) Hering-Breuer reflex

Question 15.
Which of the following prevents collapsing of trachea?
(a) Muscles
(b) Diaphragm
(c) Ribs
(d) Cartilaginous rings
Answer:
(d) cartilaginous rings

Question 16.
Which one of the following produces antibodies ?
(a) Monocytes
(b) Erythrocytes
(c) Lymphocytes
(d) Monocytes
Answer:
(c) Lymphocytes

Question 17.
Plasma protein which initiate blood coagulation is …………………..
(a) prothrombin
(b) fibrinogen
(c) thrombin
(d) fibrin
Answer:
(a) prothrombin

Question 18.
The covering of heart is …………………..
(a) perichondrium
(b) pericardium
(c) periosteum
(d) peritoneum
Answer:
(b) pericardium

Question 19.
Left atrioventricular aperture is guarded by …………………..
(a) tricuspid valve
(b) Eustachian valve
(c) bicuspid valve
(d) semilunar valve
Answer:
(c) bicuspid valve

Question 20.
The pulmonary trunk and systemic aorta are joined by …………………..
(a) chordae tendinae
(b) columnae carnae
(c) ligamentum arteriosum
(d) Purkinje fibres
Answer:
(c) ligamentum arteriosum

Question 21.
Atrioventricular node is located in …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 22.
…………………. is most commonly used to feel pulse.
(a) Radial vein
(b) Brachial artery
(c) Brachial vein
(d) Radial artery
Answer:
(d) Radial artery

Question 23.
QRS is related to …………………..
(a) atrial contraction
(b) ventricular contraction
(c) atrial relaxation
(d) ventricular relaxation
Answer:
(b) ventricular contraction

Question 24.
Blood is a fluid connective tissue derived from …………………..
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) epithelium
Answer:
(b) mesoderm

Question 25.
What is the increase in number of RBCs called?
(a) Erythropoiesis
(b) Polycythaemia
(c) Erythrocytopenia
(d) Erythroblastosis
Answer:
(b) Polycythaemia

Question 26.
What is the increase in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(c) Leucocytosis

Question 27.
In which of the following diseases there is uncontrolled increase in number of WBCs ?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(d) Leukaemia

Question 28.
What is the decrease in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(b) Leukopenia

Question 29.
Which is the correct arrangement of types of WBCs with respect to their number in blood?
(Consider Neutrophil = N, Eosinophil = E, Basophil = B, Monocyte = M and Lymphocyte = L)
(a) NLMEB
(b) BEMLN
(c) NEBLM
(d) MEBLN
Answer:
(a) NLMEB

Question 30.
Which is the correct order in which the proteins participate in clotting of blood?
(a) Prothrombinase → Prothrombin → Thromboplastin → Thrombin
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin
(c) Prothrombin → Thromboplastin → Thrombin → Prothrombinase
(d) Thrombin → Prothrombin → Thromboplastin → Prothrombinase
Answer:
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin

Question 31.
Decrease in platelet count is called …………………..
(a) thrombocytopenia
(b) thrombocytosis
(c) thrombokinase
(d) thromboplastin
Answer:
(a) thrombocytopenia

Question 32.
Atrioventricular groove is also called a …………………..
(a) foramen ovale
(b) ligamentum arteriosum
(c) coronary sulcus
(d) ductus arteriosus
Answer:
(c) coronary sulcus

Question 33.
The coronary sinus opens into the …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 34.
Name the valve from the following that guards the opening of inferior vena cava.
(a) Tricuspid valve
(b) Semilunar valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 35.
Name the valve from the following guarding the opening of coronary sinus …………………..
(a) Thebesian valve
(b) Eustachian valve
(c) Tricuspid valve
(d) Semilunar valve
Answer:
(a) Thebesian valve

Question 36.
What is an oval aperture in the interatrial septum of the foetus called?
(a) Fossa ovalis
(b) Foramen ovalis
(c) Ligamentum arteriosum
(d) Ductus arteriosus
Answer:
(b) Foramen ovalis

Question 37.
What is the meaning of stroke volume?
(a) Amount of blood in the body
(b) Pressure of contraction of heart
(c) Amount of blood put out of the ventricles in one minute
(d) Amount of blood put out of the ventricles in one beat
Answer:
(d) Amount of blood put out of the ventricles in one beat

Question 38.
How much amount of blood is put out of the heart during one minute?
(a) Equal to cardiac output
(b) Equal to stroke volume
(c) Equal to half of blood volume
(d) Equal to quarter of blood volume
Answer:
(a) Equal to cardiac output

Question 39.
What is the time taken for one cardiac cycle of normal human being?
(a) 0.1 second
(b) 0.3 second
(c) 0.4 second
(d) 0.8 second
Answer:
(d) 0 .8 second

Question 40.
Deposition of fatty substances in the lining of arteries results in …………………..
(a) arteriosclerosis
(b) atherosclerosis
(c) hyperglycemia
(d) hypotension
Answer:
(b) atherosclerosis

Question 41.
Largest number of white blood corpuscles are …………………..
(a) eosinophils
(b) basophils
(c) neutrophils
(d) monocytes
Answer:
(c) neutrophils

Question 42.
Which of the following animal have open circulatory system?
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Rabbit
Answer:
(b) Cockroach

Question 43.
Which of the following leucocytes have unlobed nucleus?
(a) lymphocyte
(b) eosinophils
(c) neutrophils
(d) basophils
Answer:
(a) lymphocyte

Question 44.
Carbonic anhydrase is found in …………………..
(a) WBC
(b) RBCs
(c) thrombocytes
(d) blood plasma
Answer:
(b) RBCs

Question 45.
The typical Lubb – Dup sounds heard in the heart of a healthy person are due to …………………..
(a) closing of cuspid valves followed by the closing of the semilunar valves
(b) closing of semilunar valves
(c) closing of tricuspid valves
(d) closing of bicuspid valves
Answer:
(a) closing of cuspid valves followed by the closing of the semilunar valves

Match the columns

Question 1.

Animal	Respiratory organ
(1) Fishes	(a) Trachea
(2) Birds/Reptiles	(b) Moist cuticle
(3) Insects	(c) Gills
(4) Earthworm	(d) Lungs

Answer:

Animal	Respiratory organ
(1) Fishes	(c) Gills
(2) Birds/Reptiles	(d) Lungs
(3) Insects	(a) Trachea
(4) Earthworm	(b) Moist cuticle

Question 2.

Respiratory organs	Alternative name
(1) Larynx	(a) Lid of larynx
(2) Trachea	(b) Air sacs
(3) Alveoli	(c) Sound box
(4) Epiglottis	(d) Windpipe

Answer:

Respiratory organs	Alternative name
(1) Larynx	(c) Sound box
(2) Trachea	(d) Windpipe
(3) Alveoli	(b) Air sacs
(4) Epiglottis	(a) Lid of larynx

Question 3.

Respiratory capacities	Respiratory volumes
(1) Residual volume (RV)	(a) 500 ml
(2) Vital capacity (VC)	(b) 2000 – 3000 ml
(3) Tidal volume (TV)	(c) 1100 – 1200 ml
(4) Inspiratory reserve volume (IRV)	(d) 4100 – 4600 ml

Answer:

Respiratory capacities	Respiratory volumes
(1) Residual volume (RV)	(c) 1100 – 1200 ml
(2) Vital capacity (VC)	(d) 4100 – 4600 ml
(3) Tidal volume (TV)	(a) 500 ml
(4) Inspiratory reserve volume (IRV)	(b) 2000 – 3000 ml

Question 4.

Disease	Symptoms
(1) Asthma	(a) Fully blown out alveoli
(2) Bronchitis	(b) Inflammation of lungs with cough and fever
(3) Emphysema	(c) Spasm of Bronchial muscles
(4) Pneumonia	(d) Inflammation of bronchi

Answer:

Disease	Symptoms
(1) Asthma	(c) Spasm of Bronchial muscles
(2) Bronchitis	(d) Inflammation of bronchi
(3) Emphysema	(a) Fully blown out alveoli
(4) Pneumonia	(b) Inflammation of lungs with cough and fever

Question 5.

Valves in heart	Location
(1) Bicuspid/Mitral valve	(a) Opening of inferior vena cava
(2) Tricuspid valve	(b) Opening of coronary sinus
(3) Eustachian valve	(c) Left atrioventricular aperture
(4) Thebesian valve	(d) Right atrioventricular aperture

Answer:

Valves in heart	Location
(1) Bicuspid/Mitral valve	(c) Left atrioventricular aperture
(2) Tricuspid valve	(d) Right atrioventricular aperture
(3) Eustachian valve	(a) Opening of inferior vena cava
(4) Thebesian valve	(b) Opening of coronary sinus

Question 6.

Blood vessel	Functions
(1) Pulmonary aorta	(a) Carries oxygenated blood to left atrium
(2) Superior vena cava	(b) Carries oxygenated blood to all body parts
(3) Pulmonary vein	(c) Carries deoxygenated blood from upper parts of body to right atrium
(4) Aorta	(d) Carries deoxygenated blood to lungs

Answer:

Blood vessel	Functions
(1) Pulmonary aorta	(d) Carries deoxygenated blood to lungs
(2) Superior vena cava	(c) Carries deoxygenated blood from upper parts of body to right atrium
(3) Pulmonary vein	(a) Carries oxygenated blood to left atrium
(4) Aorta	(b) Carries oxygenated blood to all body parts

Question 7.

Cells	Functions
(1) T-lymphocytes	(a) Phagocytic in function
(2) Neutrophils	(b) Responsible for Humoral immunity
(3) Eosinophils/Acidophils	(c) Responsible for cell-medicated immunity
(4) B-lymphocytes	(d) Anti-allergic [Antihistamine] in function

Answer:

Cells	Functions
(1) T-lymphocytes	(c) Responsible for cell-medicated immunity
(2) Neutrophils	(a) Phagocytic in function
(3) Eosinophils/Acidophils	(d) Anti-allergic [Antihistamine] in function
(4) B-lymphocytes	(b) Responsible for Humoral immunity

Question 8.

Waves recorded in ECG	Heart activity
(1) P wave	(a) Ventricular repolarization
(2) QRS complex	(b) Atrial depolarization
(3) T wave	(c) Isoelectric segment
(4) ST segment	(d) Ventricular depolarization

Answer:

Waves recorded in ECG	Heart activity
(1) P wave	(b) Atrial depolarization
(2) QRS complex	(d) Ventricular depolarization
(3) T wave	(a) Ventricular repolarization
(4) ST segment	(c) Isoelectric segment

Question 9.

Events in cardiac cycle	Time duration
(1) Atrial systole	(a) 0.3 second
(2) Atrial diastole	(b) 0.5 second
(3) Ventricular systole	(c) 0.1 second
(4) Ventricular diastole	(d) 0.7 second

Answer:

Events in cardiac cycle	Time duration
(1) Atrial systole	(c) 0.1 second
(2) Atrial diastole	(d) 0.7 second
(3) Ventricular systole	(a) 0.3 second
(4) Ventricular diastole	(b) 0.5 second

Classify the following to form Column B as per the category given in Column A

Question 1.
Classify the following composition of blood plasma given below as per Column ‘A’ and complete Column ‘B’. Select from the given options
(i) Serum albumin
(ii) Bicarbonates
(iii) Urea
(iv) Sulphates of sodium
(v) Fibrinogen
(vi) Uric acid

Column A	Column B
(1) Plasma proteins	————
(2) Nitrogenous waste	————
(3) Inorganic salts	————

Answer:

Column A	Column B
(1) Plasma proteins	Serum albumin Fibrinogen
(2) Nitrogenous waste	Urea, Uric acid
(3) Inorganic salts	Bicarbonates, Sulphates of sodium

Question 2.
Classify the following animals having different respiratory organs given below as per Column ‘A’ and complete Column ‘B’.
Select from the given options:
(i) Scorpion
(ii) Reptiles
(iii) Amphibian tadpoles of frog
(iv) Spiders
(vi) Salamanders
(v) Birds

Column A	Column B
(1) External gills	————
(2) Book lungs	————
(3) Lungs	————

Answer:

Column A	Column B
(1) External gills	Amphibian tadpoles of frog, Salamanders
(2) Book lungs	Scorpion, Spiders
(3) Lungs	Reptiles, Birds

Question 3.
Classify the following disorders of respiratory system given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Pneumonia
(ii) Asbestosis
(iii) Emphysema
(iv) Laryngitis
(v) Chronic bronchitis
(vi) Silicosis

Column A	Column B
(1) Occupational disorders	————
(2) Disorders due to smoking and air pollution	————
(3) Disorders due to viruses and bacteria	————

Answer:

Column A	Column B
(1) Occupational disorders	Asbestosis, Silicosis
(2) Disorders due to smoking and air pollution	Emphysema, Chronic bronchitis
(3) Disorders due to viruses and bacteria	Pneumonia, Laryngitis

Question 4.
Classify the following white blood corpuscles given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Eosinophils
(ii) T-lymphocytes
(iii) Neutrophils
(iv) Basophils
(v) B-lymphocytes
(vi) Monocytes

Column A	Column B
(1) Phagocytic cells	————
(2) Cells involved in giving immune response	————
(3) Cells that increase during allergic and anti-allergic responses	————

Answer:

Column A	Column B
(1) Phagocytic cells	Neutrophils Monocytes
(2) Cells involved in giving immune response	T-lymphocytes-B-lymphocytes
(3) Cells that increase during allergic and anti-allergic responses	Eosinophils, Basophils

Very Short Answer Questions

Question 1.
How many molecules of ATP are formed when one molecule of glucose is oxidized?
Answer:
38 molecules of ATP are formed when one molecule of glucose is oxidized.

Question 2.
What are the three regions of nasal chamber?
Answer:
Vestibule, respiratory part and sensory part are the three regions of nasal chamber.

Question 3.
What is meant by respiratory cycle?
Answer:
Alternate inspiration and expiration together make one respiratory cycle.

Question 4.
Why is it dangerous to sleep in a garage where automobiles have running engines?
Answer:
It is dangerous to sleep in a garage where automobiles have running engines because it may cause carbon monoxide poisoning.

Question 5.
In which form major part of CO₂ is transported in the blood?
Answer:
CO₂ is transported in the blood in the form of sodium and potassium bicarbonates.

Question 6.
Which are the parts of plant that help in the process of gaseous exchange?
Answer:
The parts of plants that help in the process of gaseous exchange are stomata, lenticels, etc.

Question 7.
Which respiratory membranes help in gaseous exchange between the alveolar air and the blood?
Answer:
The layer of squamous epithelium lining the alveolus, basement membrane and a layer of squamous epithelium lining the capillary wall help in gaseous exchange between the alveolar air and the blood.

Question 8.
When will the oxygen dissociation curve shift towards the right?
Answer:
The oxygen dissociation curve will shift towards the right due to increase in H⁺ concentration, increase in _PPCO₂ rise in temperature and rise in DPG (2, 3 diphosphoglycerate), formed in RBCs during glycolysis.

Question 9.
What is the action of carbonic anhydrase in the RBCs of blood?
Answer:
In the RBCs, CO₂ combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. In the presence of carbonic anhydrase carbonic acid immediately dissociates into HCO₃^–and H⁺ ions leading to large accumulation of HCO₃^– inside the RBCs.

Question 10.
How much energy is required for the formation of single molecule of ATP ?
Answer:
For the formation of a single molecule of ATP about 7.3 Kcal of energy is required.

Question 11.
What is Hamburger’s phenomena?
Answer:
The diffusion of Chloride ions into the RBCs to main the ionic balance between RBCs and the plasma is called Hamburger’s phenomena or chloride shift.

Question 12.
What is the role of Hering-Breuer reflex in respiration?
Answer:
The Hering-Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

Question 13.
How much blood is present in the human body and from which embryonic germ layer is it derived?
Answer:
An average adult has about 4 to 6 litres of blood, which is red coloured fluid connective tissue derived from embryonic mesoderm.

Question 14.
What is the percentage of plasma in the blood and how much water does it contain?
Answer:
There is 55% of plasma in the blood and it contains 90 to 92% water.

Question 15.
What is the average life span of RBCs?
Answer:
RBCs have a life span of about 120 days.

Question 16.
What is normal RBC count and total WBC count?
Answer:
Average RBC count in adult human is 5.1 to 5.8 million per cubic mm and average total WBC count in adult human is 5000 to 9000 per cubic mm.

Question 17.
What is erythropoiesis?
Answer:
The process of formation of Red Blood Cells is called erythropoiesis.

Question 18.
What is increase in the RBC number called?
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
What is leucopenia and erythrocytopenia ?
Answer:
The decrease in the number of white blood cells is called leucopenia whereas decrease in the number of red blood cells is called erythrocytopenia.

Question 20.
Where are Eustachian valve and Thebesian valve located?
Answer:
Eustachian valve is present at the opening of inferior vena cava while Thebesian valve is present near the opening of coronary sinus.

Question 21.
What is foramen ovale and how is it related to fossa ovalis?
Answer:
Foramen ovale is an oval opening in the interatrial septum of the foetal heart representing the fossa ovalis which lies as a depression on the right side of interatrial septum.

Question 22.
When is a person described as having hypertension?
Answer:
When the blood pressure values Eire more than 140 mm Hg systolic pressure and more than 90 mm Hg diastolic pressure, then the person is described as having hypertension.

Question 23.
What are the effects of excessive hypertension?
Answer:
Excessive hypertension of values about 220/120 mm Hg can cause blindness, nephritis, stroke or paralysis.

Question 24.
What is the difference between anemia and leukemia?
Answer:
Anemia is disorder caused due to the deficiency of heaemoglobin while leukemia is blood cancer in which there is abnormal increase in the number of white blood cells.

Question 25.
What is the difference between tachycardia and bradycardia?
Answer:
The faster heart rate over 100 beats per minute is called tachycardia, while the slower heart rate below 60 beats per minute is called bradycardia.

Question 26.
What is the difference between chordae tendinae and columnae carnae?
Answer:
Chordae tendinae are chords that connect bicuspid and tricuspid valves with the papillary muscles in ventricles while columnae carnae are series of irregular muscular ridges present on the inner surface of the ventricles.

Question 27.
Which valves prevent the backward flow of blood at the time of ventricular systole?
Answer:
Semilunar valves located at the base of pulmonary artery and systemic aorta prevent the backward flow of blood at the time of ventricular systole.

Question 28.
What are the time intervals for atrial systole, ventricular systole and joint diastole?
Answer:
Atrial systole is for 0.1 second, ventricular systole is for 0.3 second and joint diastole is for 0.4 second.

Question 29.
In the electrocardiogram shown below, which wave represents ventricular diastole?
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 1
Answer:
‘T’ wave represents ventricular diastole.

Question 30.
Mention the role of pacemaker in human heart.
Answer:
Pacemaker can generate wave of contraction or cardiac impulse for rhythmic contraction of heart.

Question 31.
Which structure in the heart is called pacemaker?
Answer:
Sinuatrial node [S. A. node] in the heart wall is called a pacemaker.

Question 32.
What is electrocardiograph?
Answer:
The instrument which is used to record action potentials generated in the heart muscles is called an electrocardiograph or ECG machine.

Question 33.
What is angina pectoris?
Answer:
Angina pectoris is the pain in the chest caused due to reduction in blood supply to cardiac muscle caused due to narrowed and hardened coronary arteries.

Question 34.
What is pulse pressure?
Answer:
Difference between systolic and diastolic pressure is called pulse pressure which is normally 40 mm Hg.

Question 38.
What would happen if respiration takes place in one single step?
Answer:
If respiration takes place in one single step, then the chemical energy released at once during that step might result in a brief blast of light and heat and may lead to death of the cell. Hence respiration is a step-wise process.

Question 39.
Why do the veins have valves?
Answer:
The veins have valves at regular intervals to prevent backflow of blood as blood flows through veins with low pressure.

Question 40.
What is Bohr effect?
Answer:
Bohr effect is the shift of oxyhaemoglobin dissociation curve due to change in partial pressure of CO in blood.

Question 41.
What is Haldane effect?
Answer:
Decrease of pH of blood, due to increase in the number of H⁺ ions, HCO₃ changes into H₂O and CO₂ by the presence of oxyhaemoglobin is called Haldane effect.

Give definitions of the following

Question 1.
Respiration
Answer:
It is a biochemical process of oxidation of organic compounds in an orderly manner for the liberation of chemical energy in the form of ATP.

Question 2.
Breathing
Answer:
It is a physical process by which gaseous exchange takes place between the atmosphere and the lungs. It involves inspiration and expiration.

Question 3.
Tidal Volume (TV)
Answer:
It is the volume of un¬ inspired or expired during normal breathing. It is 500 ml.

Question 4.
Inspiratory reserve volume (IRV)
Answer:
The maximum or the extra volume of air that is inspired during forced breathing in addition to TV (2000 to 3000 ml).

Question 5.
Expiratory reserve volume (ERV)
Answer:
The maximum volume of air that is expired during forced breathing after normal expiration. (1000 to 1100 ml).

Question 6.
Dead space (DS)
Answer:
The volume of air that is present in the respiratory tract (from nose to the terminal bronchioles), but not involved in gaseous exchange (150 ml).

Question 7.
Residual volume (RV)
Answer:
The volume of air that remains in the lungs and the dead space even after maximum expiration (1100 to 1200 ml).

Question 8.
Total lung capacity
Answer:
The maximum amount of air that the lungs can hold after a maximum forceful inspiration (5200 to 5900 ml).

Question 9.
Vital capacity (VC)
Answer:
The maximum amount of air that can be breathed out after of maximum inspiration. It is the sum total of TV, IRV and ERV and is 4100 to 4600 ml.

Question 10.
Oxygen dissociation curve
Answer:
The relationship between HbO₂ saturation and oxygen tension (PPO₂) is called oxygen dissociation curve.

Question 11.
Phosphorylation
Answer:
The process that involves trapping the heat energy in the form of high energy bond of ATP molecule is called phosphorylation.

Question 12.
Artificial ventilation
Answer:
It is the method of inducing breathing in a person when natural respiration has ceased or is faltering.

Question 13.
Ventilator
Answer:
A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 14.
Cyclosis
Answer:
Cyclosis is the streaming movement of the cytoplasm shown by almost all living organisms. E.g. Paramoecium, Amoeba, etc.

Question 15.
Single circulation
Answer:
The movement of blood once through the heart during each circulation cycle is called single circulation.

Question 16.
Double circulation
Answer:
The movement of blood twice through the heart during one circulation cycle is called double circulation.

Question 17.
Erythropoiesis
Answer:
The process of formation of RBCs is called erythropoiesis.

Question 18.
Polycythemia
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
Erythrocytopenia
Answer:
The decrease in the number of RBCs is called Erythrocytopenia.

Question 20.
Hematocrit
Answer:
The hematocrit is ratio of the volume of RBCs to total blood volume of blood.

Question 21.
Diapedesis
Answer:
Leucocytes perform amoeboid movement. Due to this kind of movement they can move out of the capillary walls. This is called diapedesis.

Question 22.
Leucocytosis
Answer:
Increase in the number of leucocytes or WBCs is called leucocytosis.

Question 23.
Leucopenia
Answer:
The decrease in the number of white blood cells is called leucopenia.

Question 24.
Leukaemia
Answer:
Pathological Increase in the number WBCs is called leukaemia or blood cancer.

Question 25.
Thrombocytopenia
Answer:
Decrease in the number of blood platelets is called thrombocytopenia.

Question 26.
Blood Coagulation
Answer:
Conversion of liquid blood into semisolid jelly is called blood coagulation or blood clotting.

Question 27.
Pericardium
Answer:
Double layered peritoneum that covers the heart from outside is called pericardium.

Question 28.
Pacemaker
Answer:
Pacemaker is the region that has power of generation of wave of contraction. In heart, sinoatrial node is called pacemaker.

Question 29.
Heartbeat
Answer:
The rhythmic contraction and relaxation of the heart is called heartbeat.

Question 30.
Pulse
Answer:
A pressure wave that travels through the arteries after each ventricular systole is called pulse.

Question 31.
Heart rate
Answer:
The rate with which the heart beats per minute is called the heart rate.

Question 32.
Stroke volume
Answer:
The amount of blood thrown out of the ventricles during one systole is called the stroke volume.

Question 33.
Cardiac output
Answer:
The amount of blood thrown out of the ventricles during one minute is called cardiac output.

Question 34.
Tachycardia
Answer:
Higher heart rate over 100 beats per minute is called tachycardia.

Question 35.
Bradycardia
Answer:
Lower heart rate which is lesser than 60 per minute is called bradycardia.

Question 36.
Myogenic
Answer:
When the initiation and further regulation of heartbeats take place in the muscles then such a heart is called myogenic.

Question 37.
Cardiac cycle
Answer:
Consecutive systole and diastole constitutes a single heartbeat or cardiac cycle.

Question 38.
Arterial blood pressure
Answer:
The pressure exerted by blood on the wall of artery is called arterial blood pressure.

Question 39.
Angiology
Answer:
Study of blood vessels is called angiology.

Question 40.
Angiography
Answer:
X-ray or imaging of the cardiac blood vessels to locate the position of blockages is called angiography.

Question 41.
Heart transplant
Answer:
Replacement of severely damaged heart by normal heart from brain- dead or recently dead donor is called heart transplant.

Question 42.
Silent Heart Attack
Answer:
Silent heart attack, also known as silent myocardial infarction, is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.

Question 43.
Electrocardiogram
Answer:
Graphical recording of electrical variations detected at the surface of body during their propagation through the wall of heart is electrocardiogram (ECG).

Question 44.
Lymph
Answer:
It is a fluid connective tissue with almost similar composition to the blood except RBCs, platelets and some proteins.

Give functions of the following

Question 1.
Epiglottis.
Answer:
The epiglottis prevents the entry of food into the trachea by closing the glottis temporarily.

Question 2.
Carbonic anhydrase.
Answer:
Carbonic anhydrase enzyme is found inside the RBCs only to accelerate the rate of formation of carbonic acid from CO₂ and H₂O.

Question 3.
Ventilators.
Answer:
Ventilators used in hospitals are part of life supporting system, which help in breathing by

Pushing oxygen into the lungs
Removing carbon dioxide from the lungs

Question 4.
Erythrocytes.
Answer:
Erythrocytes carry oxygen to all cells of the body from the lungs and bringing carbon dioxide from all the cells back to lungs.

Question 5.
Neutrophils.
Answer:
Neutrophils are responsible for destroying pathogens by the process of phagocytosis.

Question 6.
Thrombocytes/Platelets.
Answer:
Platelets secrete platelet factors which are essential in blood clotting. They also seal v the ruptured blood vessels by formation of platelet plug/thrombus. They secrete serotonin, a local vasoconstrictor.

Question 7.
Pericardial fluid.
Answer:
Pericardial fluid acts as a shock absorber and protects the heart from mechanical injuries. It also keeps the heart moist and acts as lubricant.

Question 8.
Heart walls.
Answer:
The epicardium and endocardium are protective in function whereas myocardium is responsible for contraction and relaxation of heart.

Question 9.
Valves in heart.
Answer:
Valves in the heart prevent the backflow of the blood at the time of systole and help in maintaining a unidirectional flow of blood.

Question 10.
Chordae tendinae.
Answer:
Chordae tendinae attach the bicuspid and tricuspid valves to the ventricular wall (papillary muscles) and regulate their opening and closing.

Question 11.
Semilunar valves.
Answer:
Semilunar valves prevent the backward flow of blood from pulmonary aorta and the aorta into the respective ventricles.

Question 12.
Sinoatrial node [SA] or Pacemaker.
Answer:
SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.

Question 13.
Electrocardiogram (ECG).
Answer:
ECG helps to diagnose the abnormality in conducting pathway, enlargement of heart chambers, damage to cardiac muscles, reduced blood supply to cardiac muscles and causes of chest pain.

Question 14.
Blood.
Answer:
Functions of blood:

Transport of oxygen and carbon dioxide
Transport of food
Transport of waste product
Transport of hormones
Maintenance of pH
Water balance
Transport of heat
Defence against infection
Temperature regulation
Blood clotting/coagulation
Helps in healing

Name the following

Question 1.
Name two animals in which moist skin acts as a respiratory surface.
Answer:
Earthworm, Frog

Question 2.
Name the respiratory organs in insects and fish.
Answer:
Insects – Tracheal tubes and spiracles
Fish – Internal gills

Question 3.
Name any two disorders of respiratory system.
Answer:
Asthma and pneumonia are the two disorders of respiratory system.

Question 4.
Name the structural and functional unit of lungs.
Answer:
Alveolus is the structural and functional unit of lungs.

Question 5.
Name the energy currency of cell.
Answer:
ATP is the energy currency of cell.

Question 6.
Name the site where actual exchange of O₂ and CO₂ takes place between air and blood in the body of man.
Answer:
Alveolus of lung.

Question 7.
Name any two respiratory centres required for regulation of breathing.
Answer:
Inspiratory centre, Expiratory centre, Pneumotaxic centre and Apneustic centre.

Question 8.
Name the muscles which move ribs up and down.
Answer:
External intercostal muscles.

Question 9.
Name two phyla where haemocoel is present.
Answer:
Phylum-Arthropoda and Phylum-Mollusca.

Question 10.
Name the animal-group which show single circulation.
Answer:
Fishes

Question 11.
Name the cells which produce thrombocytes.
Answer:
Megakaryocytes produce thrombocytes.

Question 12.
Name the process of formation of red blood corpuscles.
Answer:
Erythropoiesis

Question 13.
Name the space in which human heart is located.
Answer:
Mediastinum is the space in which human heart is located.

Question 14.
Name the types of lymphocytes depending upon functions.
Answer:
B-lymphocytes and T-lymphocytes.

Question 15.
Name the layers of peritoneum that surrounds the heart sequentially from outside to inside.
Answer:
Fibrous pericardium, parietal layer of serous pericardium and visceral layer of serous pericardium.

Question 16.
Name the connection between the pulmonary trunk and systemic aorta.
Answer:
Ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 17.
Name the valve between left atrium and left ventricle and give its significance.
Answer:
Between left atrium and left ventricle is mitral or bicuspid valve which maintains the unidirectional flow of blood by preventing hs backflow.

Question 18.
Name the walls of an artery.
Answer:
Outer tunica externa, middle tunica media and inner tunica interna.

Question 19.
Name the instrument used to measure blood pressure.
Answer:
Sphygmomanometer is used to measure blood pressure.

Question 20.
Name the plasma proteins involved in the process of blood clotting.
Answer:
Prothrombin and fibrinogen.

Question 21.
Name the various components of conducting system of the heart.
Answer:
Conducting system of the heart consists of SA node, AV node, bundle of His and Purkinje fibers.

Question 22.
Name the neurotransmitters that decrease and increase the heart rate in human beings respectively.
Answer:
Acetylcholine decreases heart rate and adrenaline or epinephrine increases the heart rate in human.

Question 23.
Who discovered ECG?
Answer:
Willem Einthoven discovered ECG.

Distinguish between the following

Question 1.
Pharynx and Larynx.
Answer:

Pharynx	Laryix
1. Pharynx is a short, vertical tube.	1. Larynx is a sound producing organ located at the end of pharynx.
2. Mouth leads to the pharynx.	2. Larynx leads to the oesophagus.
3. Vocal cords are absent.	3. Vocal cords are present.
4. Pharynx does not increase in size at the time of puberty.	4. Larynx increases in size at the time of puberty.
5. Pharynx does not show Adam’s apple.	5. Larynx shows Adam’s apple in adult males.

Question 2.
Inspiration and Expiration.
Answer:

Inspiration	Expiration
1. Inspiration is an active process.	1. Expiration is a passive process.
2. During inspiration diaphragm contracts and becomes flattened.	2. During expiration diaphragm relaxes and becomes dome shaped.
3. During inspiration intercostal muscles contract.	3. During expiration intercostal muscles relax.
4. During inspiration ribs are pulled outwards and sternum is raised.	4. During expiration ribs are pulled inwards and sternum is lowered.
5. During inspiration the space in the thoracic cavity increases.	5. During expiration the space in the thoracic cavity decreases.
6. During inspiration pressure in the lungs decreases.	6. During expiration pressure in the lungs increases.
7. During inspiration the volume of the lungs increase.	7. During expiration the volume of the lungs decreases.
8. During inspiration air comes inside the body.	8. During expiration air goes out of the body.

Question 3.
External respiration and Internal respiration.
Answer:

External respiration	Internal respiration
1. The respiratory processes occurring in lungs is called external respiration.	1. The respiratory processes that occur in tissues is called internal respiration.
2. During external respiration O₂ from the lungs enters into the lung capillaries by diffusion.	2. During internal respiration O₂ from the blood enters the tissue cells.
3. During external respiration CO₂ from the lung capillaries diffuse into the lungs.	3. During internal respiration CO₂ from the tissues enters into the blood.
4. During external respiration exchange of gases takes place between the air and the lungs.	4. During internal respiration exchange of gases take place between the blood and the tissue.
5. Formation of oxyhaemoglobin takes place during external respiration.	5. Oxyhaemoglobin dissociates into oxygen and haemoglobin during internal respiration.
6. During external respiration CO₂ is released.	6. During internal respiration carbamino haemoglobin is formed which is carried to the lungs.

Question 4.
Transport of O₂ and Transport of CO₂.
Answer:

Transport of O₂	Transport of CO₂
1. Transport of O₂ takes place from lungs to the tissues and cells.	1. Transport of CO₂ takes place from tissues and cells to the lungs.
2. Oxygen is carried as oxyhaemoglobin to the tissues with the help of RBCs.	2. Carbon dioxide is carried as carbaminohaemoglobin from the tissues with the help of plasma and RBCs.
3. Oxygen does not form oxides or other products during its transport.	3. CO₂ forms bicarbonates with sodium and potassium during its transport.
4. O₂ does not form acids during its transport.	4. CO₂ dissolves in water to form carbonic acid.

Question 5.
Vital Capacity of Lung and Total Lung Capacity.
Answer:

Vital Capacity of Lung	Total Lung Capacity
1. It is the maximum amount of air a person can expire and inspire to their maximum extent.	1. It is the maximum amount of air that the lungs can hold after a maximum forceful inspiration.
2. It is the sum total of inspiratory reserve volume, tidal volume and expiratory reserve volume.	2. It is the sum total of vital capacity and residual volume.
3. It ranges from 4100 to 4600 ml.	3. It ranges from 5200 to 5800 ml.

Question 6.
Inspiratory Reserve Volume (IRV) and Expiratory Reserve Volume (ERV).
Answer:

Inspiratory Reserve Volume (IRV)	Expiratory Reserve Volume (ERV).
1. It is the maximum volume of air, or the extra volume of air, that is inspired during forced breathing.	1. It is the maximum volume of air that is expired during forced breathing.
2. Its value is 2000/3000 ml.	2. Its value is 1000/1100 ml.

Question 7.
T. S. of artery and T.S. of vein.
Answer:

T. S. of artery	T.S. of vein
1. Histologically in transverse section of artery there are three walls, tunica externa, tunica media and tunica interna or endothelium.	1. Histologically in transverse section of vein there are three walls, tunica externa, tunica media and tunica interna or endothelium.
2. Tunica media is thick and muscular.	2. Tunica media is thinner as compared to artery.
3. Lumen of artery is narrow.	3. Lumen of vein is broad.

Question 8.
Erythrocytes and Leucocytes.
Answer:

Erythrocytes	Leucocytes
1. Erythrocytes have a definite shape which is elliptical or oval.	1. Leucocytes do not have definite shape as they are amoeboid.
2. They are enucleated.	2. They are nucleated.
3. Erythrocytes contain haemoglobin and hence appear red.	3. Leucocytes are devoid of any respiratory pigment and hence appear colourless.
4. The normal erythrocyte count is 4.3 to 5.8 million per cubic mm of blood.	4. The normal leucocyte count is 4000 to 11000 per cubic mm of blood.
5. The life span of erythrocytes is 100 to 120 days.	5. The life span of leucocytes is 3 to 4 days.
6. The diameter of erythrocytes is 7.2 m and thickness is about 2 to 2.2 m.	6. The size of leucocytes varies with its subtypes and is of average size of 8 to 15 m.
7. Erythrocytes are formed by the process of erythropoiesis in red bone marrow.	7. Leucocytes are formed by the process of leucopoiesis in bone marrow, tonsils, lymph nodes, spleen, thymus, etc.
8. Erythrocytes transport the respiratory gases.	8. Leucocytes help in the formation of antibodies besides fighting against foreign antigens by phagocytic activity.

Question 9.
Eosinophils and Basophils.
Answer:

Eosinophils	Basophils
1. Cytoplasmic granules present in eosinophils are stained with acidic stains.	1. Cytoplasmic granules present in basophils are stained with basic stains.
2. Nucleus is bilobed.	2. Nucleus is twisted.
3. Eosinophils constitute 3% of total WBCs.	3. Basophils constitute 0.5% of total WBCs.

Question 10.
Neutrophils and Eosionophils.
Answer:

Neutrophils	Eosinophils
1. Cytoplasmic granules present in neutrophils are stained with neutral stains.	1. Cytoplasmic granules present in eosinophils are stained with acidic stains.
2. Nucleus is three to five lobes showing polymorphic form.	2. Nucleus is bilobed.
3. Neutrophils constitute 62% of total WBCs.	3. Eosinophils constitute 3% of total WBCs.

Question 11.
Lymphocytes and Monocytes.
Answer:

Lymphocytes	Monocytes
1. Large round nucleus but size of the cell is smaller.	1. Large kidney shaped nucleus and largest size among WBCs.
2. Lymphocytes form 25-33% of WBCs.	2. Monocytes form 3-9% of WBCs.

Question 12.
Granulocytes and Agranulocytes.
Answer:

Granulocytes	Agranulocytes
1. WBCs with granular cytoplasm are called granulocytes. Thus, cytoplasmic granules are present.	1. WBCs with agranular cytoplasm are called agranulocytes. Thus, cytoplasmic granules are absent.
2. Nuclei of granulocytes are variously lobed.	2. Nuclei of agranulocytes are not lobed.

Question 13.
Single circulation and Double circulation.
Answer:

Single circulation	Double circulation
1. Blood flows only once through the heart in a complete cycle.	1. Blood flows twice through the heart during one complete circulation. Systemic – to and fro ‘ from heart to body and pulmonary – to and fro from heart to lungs.
2. Heart pumps deoxygenated blood only.	2. Heart pumps both deoxygenated and oxygenated blood to lungs and body respectively.
3. Blood is oxygenated in gills.	3. Blood is oxygenated in lungs.
4. Occurs only in fishes.	4. Occurs in amphibians, reptiles, birds and mammals.

Question 14.
Systolic blood circulation and Diastolic blood circulation.
Answer:

Systolic blood circulation	Diastolic blood circulation
1. Blood is passed from right ventricle to lungs by pulmonary artery during systolic circulation. Similarly, from left ventricle the oxygenated blood is given to the entire body through systemic aorta during systolic circulation.	1. Blood is passed to left atrium from lungs by pulmonary veins during diastolic circulation. Similarly, deoxygenated blood from entire body is brought back to heart through vena cava during diastolic circulation.
2. Systolic blood circulation is under maximum pressure as heart is forcing the blood to come out of heart.	2. Diastolic blood circulation is under minimum blood pressure as heart is relaxed during diastole.

Question 15.
Atria and Ventricles.
Answer:

Atria	Ventricles
1. Atria are upper chambers of the heart.	1. Ventricles are lower chambers of the heart.
2. Atria are thin walled.	2. Ventricles are thick walled.
3. Atria are receiving chambers.	3. Ventricles are distributing chambers.
4. Interatrial septum divides the two auricles (atria).	4. Interventricular septum divides the two ventricles.
5. Right atrium is larger in size than left atrium.	5. Left ventricle is larger in size than the right ventricle.

Question 16.
S.A. Node and A.V. Node.
Answer:

S.A. Node	A.V. Node
1. Sinoatrial node is present in the right ventricle near the opening near the opening of the superior vena cava.	1. Atrioventricular node is present in the right ventricle near the opening of the coronary sinus.
2. S.A. node is the pacemaker of the heart and it starts atrial systole.	2. A.V. node starts ventricular systole through bundles of His and Purkinje’s fibre system.

Question 17.
Pulmonary circulation and Systemic circulation.
Answer:

Pulmonary circulation	Systemic circulation
1. The course of blood from the right ventricle to the left atrium of the heart through the lungs is called pulmonary circulation.	1. The course of blood from the left ventricle to the right atrium of the heart through the body is called systemic circulation.
2. Pulmonary circulation is mainly for sending the blood for oxygenation in the lungs from the heart and bringing it back to the heart after oxygenation.	2. Systemic circulation is for sending the deoxygenated blood from the body to the heart and sending oxygenated blood from the heart to the body.

Question 18.
Atrio ventricular valves and Semilunar valves.
Answer:

Atrio ventricular valves	Semilunar valves
1. Atrio ventricular valves Eire present between the atria and ventricles. On the right side there is tricuspid valve whereas on the left side there is bicuspid valve.	1. Semilunar valves are present at the opening of pulmonary artery and systemic aorta.
2. Atrio ventricular valves prevent the back flow of blood from ventricles to atria at the time of systole.	2. Semilunar valves prevent the back flow of blood from pulmonary artery and systemic aorta back to the heart.

Question 19.
Hypertension and Hypotension.
Answer:

Hypertension	Hypotension
1. Blood pressure values more than 140 mm Hg SP and 90 mm HG DP is called hypertension.	1. Blood pressure values less them 120 mm Hg SP and 70 mm HG DP is called hypotension.
2. Excessive hypertension can result into lethal complications such as stroke or paralysis.	2. Hypotension may not be lethal if immediate measures are taken to raise the blood pressure.

Give reasons

Question 1.
ATP is called energy currency of the cell.
Answer:

During cellular respiration, the oxidation of food (glucose) takes place.
This happens in the mitochondria using the oxygen present in the blood.
ATP molecules are formed during this oxidation.
ATP is used for various vital body processes and also for maintaining the body temperature to constancy.
Since energy is stored in the form of ATP it is called an energy currency of the cell.

Question 2.
The vestibule of nasal chamber has fine hair.
Answer:

Vestibule is the anterior most part of the nasal chamber.
The hairs present in this region trap the dust particles and prevent them from entering into the interiors of the respiratory passage.
Therefore, the vestibule of nasal chambers has fine hair.

Question 3.
Glottis is guarded by a flap called epiglottis.
Answer:

The oesophagus and trachea lie side by side.
There is possibility that food particles may enter respiratory passage at the time of gulping.
However, the epiglottis prevents the entry of food into the respiratory passage by closing it temporarily.
Thus, for preventing the entry of food particles into respiratory passage, the glottis is guarded by a flap called epiglottis.

Question 4.
Alveoli are very flexible.
Answer:

Alveoli are made up of collagen and elastin fibres.
They are very thin (0.0001 mm) and lined by non-ciliated squamous epithelium.
All the above structural components make the alveoli very flexible.

Question 5.
Expiration is called a passive process.
Answer:

During expiration, intercostal muscles relax. This results in pulling the ribs inwards.
Diaphragm also relaxes and returns to its normal dome shape.
The collective contraction of ribs and diaphragm results in the reduction of
thoracic cavity and hence automatically air is pushed out of the lungs.
Since the pressure on the lungs increase rushing the air to outside, expiration is called a passive process.

Question 6.
Pericardium acts as a defence wall for the heart.
Answer:

Pericardium protects the heart. It is double layered peritoneum, having outer fibrous and inner serous pericardium layers.
Fibrous pericardium being tough gives protection to the heart.
Serous pericardium has two layers, parietal and visceral layer or epicardium.
In between these two layers, there is pericardial fluid, which helps to absorb shocks and provide nourishment.
In this way pericardium acts as a defence wall.

Question 7.
Valves are present in veins.
Answer:

Veins carry blood to the heart.
At that time the backward flow of the blood should be prevented.
Therefore, valves are present in veins.

Question 8.
Atria are thin walled than ventricles.
Answer:

Atria are receiving chambers, while ventricles are distributing chambers.
The blood is driven out from ventricles.
Ventricles are therefore, strong and with thicker walls.
Atria are thin walled as compared to ventricles.

Question 9.
Heart is called a pump.
Answer:

The heart acts as a pumping organ. It shows continuous pumping action.
The rhythmic contraction or systole and relaxation or diastole of heart forms one heartbeat.
Such heartbeats occur about 72 times per minute.
The heart efficiently pumps about 5 litres of blood per minute. Therefore, the heart is called a pump.

Question 10.
In normal human heart, there is no mixing of oxygenated and deoxygenated blood.
Answer:

In normal human heart, there is completely formed atrioventricular septum.
This septum keeps the deoxygenated and oxygenated blood separate.
Hence there is no mixing of the two types of blood.

Question 11.
Blood pressure is inversely related to the elasticity of the blood vessels.
Answer:

When the blood gushes through the blood vessels, the walls of blood vessels -can expand a little due to their elasticity.
But as the age advances, the elasticity is reduced and then the blood vessels do not expand.
Hence the flowing blood gets more resistance and the blood pressure can rise.
Lesser the elasticity more will be the blood pressure, whereas more the elasticity of the vessel wall, then the pressure will not rise.
In this way, the blood pressure is inversely related to the elasticity of the blood vessels.

Write short notes

Question 1.
Chloride shift or Hamburger’s phenomenon.
Answer:

About 70% of CO₂ is transported in the form of sodium bicarbonates/potassium bicarbonates from tissue cells to lungs.
In the RBCs, CO₂ combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. This action is rapid in RBCs as compared to that in the plasma.
Carbonic acid being unstable, immediately dissociates into HCO₃^– and H⁺in the presence of same enzyme, leading to large accumulation of HCO₃^– inside the RBCs. It thus moves out of RBCs. This can bring about imbalance of the charge inside the RBCs.
To maintain the ionic balance between the RBCs and the plasma, Cl^– diffuses into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
HCO₃^– that comes in the plasma joins to Na⁺/K⁺ forming NaHCO₃/KHCO₃ which can maintain pH of blood. The remaining H⁺ ions in the RBCs are buffered by haemoglobin by the formation of oxyhaemoglobin.
At the level of lungs, due to the low partial pressure of carbon dioxide of the alveolar air, hydrogen ion and bicarbonate ions combine to form carbonic acid and under the influence of carbonic anhydrase again yields carbon dioxide and water.

Question 2.
Regulation of breathing.
Answer:
(1) Respiration is under dual control, i.e. nervous and chemical. Normal breathing is an involuntary process. Steady state of respiration is controlled by neurons located in the pons and medulla and are known as the respiratory centres. They regulate the rate and depth of breathing.

(2) These centres are divided into three groups : dorsal group of neurons in the medulla (inspiratory centre), ventro-lateral group of neurons in medulla (inspiratory and expiratory centre) and pneumotaxic centre located in the pons and apneustic centre which is antagonistic in action to pneumotaxic centre.

(3) During inspiration, when the lungs expand to a critical point, the stretch receptors are stimulated and impulses are sent along the vagus nerves to the expiratory centre. It then sends out inhibitory impulses to the inspiratory centre.

(4) The inspiratory muscles relax and expiration follows. As the air leaves but, the lungs are deflated and the stretch receptors are no longer stimulated. Thus, the inspiratory centre is no longer inhibited and a new respiration begins. These events are called the Hering – Breuer reflex. The Hering – Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

(5) The respiratory centre has connections with the cerebral cortex that means we can voluntarily change our pattern of breathing. Voluntary control is protective because it enables us to prevent water or irritating gases from entering the lungs.

Question 3.
Carbon monoxide poisoning.
Answer:

Carbon monoxide poisoning is caused when carbon monoxide is combined with haemoglobin.
Haemoglobin is said to have 250 times more affinity for carbon monoxide than that for the oxygen.
Therefore, haemoglobin with carbon monoxide forms a stable compound, the carboxyhemoglobin.
Due to the formation of carboxyhaemoglobin, the haemoglobin no longer carries oxygen to the cells and tissues. Tissues then suffer from oxygen starvation. This leads to asphyxiation and in extreme cases it leads to death.
Carbon monoxide poisoning occurs in closed rooms with incompletely burning substances such as stove burners or furnaces and garages having running automobile engines.
Person suffering from carbon monoxide poisoning has to be administered with oxygen-carbon dioxide mixture, so that high levels of CO₂ makes carbon monoxide dissociated from haemoglobin.

Question 4.
Artificial ventilation.
Answer:
(1) Artificial ventilation is the artificial respiration. It is the method of inducing breathing in a person when natural respiration has ceased or is faltering. If used properly and quickly, it can prevent death due to drowning, choking, suffocation, electric shock, etc.

(2) The process involves two main steps:
a. Establishing and maintaining an open air passage from the upper respiratory tract to the lungs.
b. Force inspiration and expiration as in mouth to mouth respiration or by mechanical means like ventilator.

(3) A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 5.
Erythrocytes.
Answer:

Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
The RBC size : 7 pm in diameter and 2.5 pm in thickness.
The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
The average life span of RBC : 120 days.
RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
The old and worn out RBCs are destroyed in liver and spleen.
Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
Maintenance of blood pH as haemoglobin acts as a buffer.
Maintenance of the viscosity of blood.

Question 6.
Heartbeat.
Answer:

The rhythmic contraction and relaxation of the heart is called heartbeat.
Each heartbeat includes one systole and one diastole. During systole the heart contracts and during diastole it relaxes.
The rate with which the heart beats is called heart rate. The normal heart rate is 72 beats per minute.
Tachycardia means faster heart rate of about more than 100 beats per minute.
Bradycardia means slower heart rate that is below 60 beats per minute.

Question 7.
Pulse.
Answer:

A pressure wave that travels through the arteries after each ventricular systole is called a pulse.
The pulse can be felt in any artery that lies near the surface of the body.
The radial artery at the wrist is most commonly used to feel the pulse.
The pulse rate per minute indicates the heart rate. Since each heartbeat generates one pulse in the arteries, the pulse rate is same as that of heart rate, i.e. 72 times per minute.

Question 8.
Peacemaker.
Answer:

Pacemaker is the region in tile heart which initiates the beating.
The natural pacemaker of the heart is sinoatrial node (SA node).
The pacemaker is autorhythmic, it is able to repeatedly and rhythmically generate impulses.
SA node is responsible for initiation of cardiac excitation. Therefore, it is called a pacemaker.

Question 9.
Blood pressure.
Answer:

Blood pressure is the pressure exerted by the flowing blood on the walls of arteries.
Blood pressure described in two terms viz. systolic blood pressure and diastolic blood pressure. Systolic blood pressure is the maximum pressure of blood when heart undergoes ventricular systole. It is responsible for flow of blood in the arteries. Normal systolic pressure is 120 mm Hg.
Diastolic blood pressure is the minimum pressure of blood when heart undergoes diastole. Normal diastolic pressure is 80 mm Hg.
Blood pressure is represented as 120/80 mm Hg for a normal human being.

Question 10.
Hypertension.
Answer:

In a normal healthy person the blood pressure values are 120 mm Hg (systolic)/ 80 mm Hg (diastolic).
When the blood pressure is persistently more than 140 mm Hg systolic pressure and 90 mm Hg diastolic arterial blood pressure then it is said to be hypertension or high blood pressure.
Excessively high blood pressure is very dangerous as high blood pressure of about 220/120 mm Hg may cause rupturing of blood vessels.
Rupture of eye blood vessels can lead to blindness.
If blood vessels of kidney are affected then nephritis is caused.
Hemorrhage occurring in the brain can lead to stroke or paralysis. Therefore, hypertension is commonly called silent killer. It may be present for years with no distinct symptoms.
The factors causing hypertension are arteriosclerosis (reduction of elasticity of blood vessels), atherosclerosis (deposition, of cholesterol inside the blood vessels wall), obesity, physical or emotional stress, alcoholism, smoking, cholesterol rich diet, increased secretion of renin, epinephrine or aldosterone, etc.

Question 11.
Coronary artery disease (CAD).
Answer:

Coronary artery disease is a condition caused due to problems like atherosclerosis.
In this disease, coronary arteries are narrowed due to deposition of fatty substances.
Due to this the blood flow to the heart is reduced.
In coronary heart disease, the heart muscle is damaged because of an inadequate amount of blood due to obstruction of its blood supply.
The symptoms of CAD depend upon the degree of obstruction.
Symptoms are mild chest pain or angina pectoris.
In severe cases it results in heart attack known as myocardial infarction.

Question 12.
Angina pectoris.
Answer:

Angina pectoris is the pain in the chest. It results from a reduction in blood supply to cardiac muscle due to narrowed and hardened coronary arteries.
Atherosclerosis and arteriosclerosis can cause this problem. Basically, the coronary arteries are affected during angina pectoris.
It causes heaviness and severe pain in the chest. The pain can also be felt at the neck, lower jaw, left arm and left shoulder.
Angina pectoris often occurs during exertion, when the heart demands more oxygen and narrowed blood vessels cannot supply. It disappears with rest.

Question 13.
Heart failure.
Answer:

Heart failure is caused due to progressive weakening of the heart muscle. This results in the failure of the heart to pump the blood effectively.
Hypertension increase the after load on the heart leading to significant enlargement of the heart.
This finally results in heart failure.
Factors responsible for heart failure are advanced age, malnutrition, chronic infections, toxins, severe anaemia or hyperthyroidism, etc.
Any problem leading to degeneration of heart muscle, may result in heart failure.

Question 14.
Atherosclerosis.
Answer:

Atherosclerosis is the deposition of fatty substances and cholesterol on the inner lining of eateries.
This deposition results in the formation of atherosclerotic plaque.
It results in the decrease of the lumen of the blood vessels causing increasing resistance for the blood to flow which in turn results in the hypertension.
Atherosclerosis of the coronary arteries results in decrease in the blood flow to the heart muscles.
Due to such condition, coronary heart disease is caused.

Question 15.
ECG.
Answer:

Electrocardiogram or ECG is the graphic v record of electrical variations produced by the heart during one heartbeat or cardiac cycle.
ECG is taken with the help of an instrument called electrocardiograph or ECG machine. Electrocardiograph records the action potentials generated by the heart muscles.
The electrical activity of heart is represented in the form of a graph plotted with time on X-axis against voltage displacement on Y-axis.
A normal ECG is a graph having series of ridges and furrows. There are waves such as P-wave, QRS complex and T-wave.
P-wave is a small upwards wave representing impulse generated by SA node. P-wave is caused by atrial depolarization that results in atrial contraction.
QRS-complex wave begins as a downward deflection, continues as a large upright triangular wave and ends’ as a downward wave.
QRS-complex wave represents spreading of impulse from SA node to AV node, then to bundle of His and Purkinje fibres. It causes ventricular depolarization resulting in ventricular contraction.
T-wave is a broad upward wave which represents ventricular repolarization resulting in ventricular relaxation.
Functions of ECG are mainly for diagnosis and also for prognosis. It is useful to detect abnormal functioning of heart as in coronary artery diseases, heart block, angina pectoris, tachycardia, ischemic heart disease, myocardial infarction, cardiac arrest, etc.

Question 16.
Angiography.
Answer:

Angiography is an X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Depending upon the degree of blockage, remedial procedures like angioplasty or by¬pass surgery are performed.
In angioplasty a stent is inserted at the site of blockage to restore the blood supply while in by-pass surgery, the atherosclerotic region is by-passed with part of vein or artery taken from any other suitable part of the body, like hands or legs.

Question 17.
Silent Heart Attack or silent myocardial infarction.
Answer:

Silent heart attack is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.
Symptoms of silent heart attack are so mild that a person often confuses it for regular « discomfort and thereby ignores it.
Men are more affected by silent heart attack than women.

Question 18.
Heart Transplant.
Answer:

Heart transplant is the replacement of severely damaged heart by normal heart from brain-dead or recently dead donor,
Heart transplant is necessary in case of patients with end-stage heart disease and severe coronary arterial disease.

Short Answer Questions

Question 1.
What is meant by respiration? How is it useful in the production of energy?
Answer:

Respiration is the biochemical process in which organic compound such as glucose are oxidized to liberate chemical energy.
During respiration energy is released in gradual and step wise process. The released energy is in the form of bonds of ATP (Adenosine Tri Phosphate) molecules are shown below:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 38 ATP
ATP is the biologically useful energy. ATP drives most of the life process.
When cell requires the energy, ATP is hydrolyzed and is converted into ADP with subsequent release of energy.
The respiratory system, blood and the body cells play an important role in the process of respiration.

Question 2.
How does exchange of gases take place at the alveolar level?
Answer:
1. Exchange of gases between the alveolar air and the blood is known as external respiration.

2. Simple squamous epithelial layer of alveolus is intimately associated with a similar layer lining the capillary wall. Both of these layers are thin walled and together they make up the respiratory membrane through which gaseous exchange occurs between the alveolar air and the blood.

3. Diffusion of gases will take place from an area of higher partial pressure to an area of lower partial pressure until the partial pressure in the two regions reaches equilibrium.

4. The partial pressure of carbon dioxide of blood entering the pulmonary capillaries is 45 mmHg while partial pressure of carbon dioxide in alveolar air is 40 mmHg. Due to this difference, carbon dioxide diffuses from the capillaries into the alveolus.

5. Similarly, partial pressure of oxygen of blood in pulmonary capillaries is 40 mmHg while in alveolar blood it is 104 mmHg. Due to this difference oxygen diffuses from alveoli to the capillaries.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 2

Question 3.
What is the role of haemoglobin in the transport of oxygen in the blood?
Answer:

Haemoglobin is a respiratory pigment present in cytoplasm of RBCs. About 97% of oxygen is transported by these haemoglobin molecules from lungs to tissues.
Haemoglobin has a high affinity for Oa and combines with it to form oxyhaemoglobin. One molecule of Hb has four FeT, each of which can pick up a molecule of oxygen (O₂). Hb + 4O₂ → Hb (4O₂)
Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
Hb (4O₂) → Hb + 4O₂
In the alveoli where PPOa is high and PPCO₂ is low, oxygen binds with haemoglobin, but in tissues, where PPO₂ is lower and PPCO₂ is high, Oxyhaemoglobin dissociates and releases O₂ for diffusion into the tissue cells.

Question 4.
What is blood? What is the normal quantity of blood in an adult human being?
Answer:

Blood is the fluid connective tissue that circulates in the body.
Blood is derived from mesoderm.
It is bright red, slightly alkaline fluid having pH about 7.4. It is salty, viscous fluid heavier than water.
The average sized adult has about 5 litres of blood in his/her body which constitutes about 8% of the total body weight.

Question 5.
Describe the structure and the function of thrombocytes.
Answer:

Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
Their life span is about 5 to 10 days.
Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
Thrombocytes possess thromboplastin which helps in clotting of blood.
Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

Question 6.
Describe the structure of the heart wall.
Answer:

The heart wall is composed of three layers, viz. outer epicardium, middle myocardium and inner endocardium.
Epicardium is composed of single layer of mesothelium having flat epithelial cells.
Myocardium is composed of cardiac muscle fibres. These muscle fibres perform the function of systole and diastole by showing contraction and relaxation of muscle wall of the heart.
Endocardium is composed of single layer of flat epithelial cells called endothelium.

Question 7.
Name the two heart sounds. How and when are they produced?
Answer:

In one normal heartbeat the heart sounds like lubb and dup are produced once each.
The rhythmic contraction (Systole) and relaxation (diastole) forms are heartbeat. The heart sounds are due to closure of valves.
Lubb sound is produced during ventricular systole when the cuspid valves close both the atrioventricular apertures preventing blood flow into atria.
Dub sound is produced during ventricular diastole when semilunar valves are closed, preventing backflow of blood from pulmonary trunk and systemic aorta into ventricles.

Question 8.
What is double circulation? What is its significance?
Answer:
(1) Double circulation : Movement of blood twice through the heart during one circulation cycle is called double circulation. Body → heart → lungs → heart → body is the course of double circulation.

(2) Significance of double circulation:
a. Double circulation is more effective type of circulation in which oxygenated and deoxygenated type of blood do not intermix.
b. The systemic circulation i.e. from body to heart and back to body while the pulmonary circulation i.e. from heart to lungs and back to heart circulate the blood uniformly.

(3) Coronary and hepatic portal circulation is also achieved due to double circulation.

Question 9.
Describe pulmonary and systemic circulation.
Answer:

In human beings, there is double circulation because blood passes twice through the heart during one cardiac cycle.
The blood follows two routes viz. pulmonary and systemic.
Pulmonary circulation is circulation between heart and lungs. Systemic circulation is the circulation between the heart and body organs (except lungs).
During pulmonary circulation, the blood passes from the right ventricle to the left atrium of the heart through lungs.
The right ventricle pumps deoxygenated blood into the pulmonary trunk which carries it to lungs for oxygenation. The oxygenated blood from the lungs is brought to left atrium by two pairs of pulmonary veins.
During systemic circulation, the blood from the left ventricle passes to the right atrium of heart through body organs.
The left ventricle pumps oxygenated blood into the systemic aorta which carries it to all body organs except lungs. The deoxygenated blood from the body organs is brought to right atrium by superior and inferior venae cavae.

Question 10.
How is cardiac activity regulated?
Answer:

Normal activities of the heart are auto regulated. The specialized muscles help in this regulation.
The heart is said to be myogenic due to this ability.
In the medulla oblongata of brain, there is cardiovascular centre,
From this centre, sympathetic and parasympathetic nerves innervate the sinoatrial node.
Sympathetic nerves secrete adrenaline and it stimulates and increases the heartbeat.
Parasympathetic nerves secrete acetylcholine and it decreases the heart rate.
Rate of heartbeat is controlled in response to inputs from various receptors like proprio-receptors.
These receptors monitor the position of limbs and muscles. There are chemoreceptors which monitor chemical change in blood and baroreceptors that monitor the stretching of main arteries and veins.

Chemical control on heart rate:

Hypoxia, acidosis, alkalosis cause decrease in cardiac activity.
Hormones like epinephrine and nor epinephrine enhance the cardiac activity.
Elevated blood level of K+ and Na+ decreases the cardiac activity.

Question 11.
What are the main features of respiratory surface?
Answer:
The respiratory surface, for the efficient gaseous exchange should have the following features:

It should have a large surface area.
It should be thin, highly vascular and permeable to allow exchange of gases.
It should be moist.

Question 12.
What is the co-relationship between activeness of organism and complexity of transport system?
Answer:

As the size of an organism increases, its surface area to volume ratio decreases. This means it has relatively less surface area available for substances to diffuse through.
Large multicellular organisms therefore cannot rely on diffusion alone to supply their cells with Substances such as food and oxygen and to remove waste products. Large multicellular organisms require specialized transport system.
In short for the organisms to become active, they must be having complex transport system to bring about their vital functions rapidly.

Question 13.
What are the granules in granulocytes?
Answer:

Neutrophils : Granules . of neutrophils contain cationic proteins and other proteins that are used to kill bacteria, some enzymes to breakdown bacterial proteins, lysozymes to breakdown bacterial cell wall. etc.
Eosinophils : Granules of eosinophil contains a unique toxic basic protein receptors that bind to IgE used to help in killing parasites.
Basophils : Granules of basophils contain abundant histamine, heparin and platelet activating factor.

Question 14.
What is haemoglobin count in normal human beings? What is the function of haemoglobin?
Answer:

The normal haemoglobin in adult male is 13-18 mg/100 ml of blood.
In a normal adult female, it is about 11.5-16.5 mg/100 ml of blood.
In anaemic individuals there is lesser amount of haemoglobin.
Functions of haemoglobin is to transport oxygen from lungs to tissues and carbon dioxide from tissues of lungs.
Haemoglobin acts as a buffer and maintains the blood pH.

Question 15.
Why has the heart-recipient to rely upon life-time supply of immunosuppressants?
Answer:
Person who has undergone heart transplant needs lifetime supply of immunosuppressants because in these persons organ rejection is a constant threat. Keeping away the immune system from attacking the transplanted organ requires constant supply of immunosuppressant drugs. These drugs help prevent immune system from attacking (“rejecting”) the donor organ. Typically, these drugs are taken for the life-time for maintaining transplanted organ.

Question 16.
Why is it difficult to hold one’s breath beyond a limit?
Answer:

It is difficult to hold one’s breath beyond a limit because the pressure of oxygen and carbon dioxide in blood changes as one holds his breath.
When the breath is held beyond a limit, the urge to breath becomes irresistible.
When the breath is held forever, body becomes starved of oxygen and person may fall unconscious and the instinct to breath would take over.

Question 17.
Why and when do the leucocytes perform diapedesis?
Answer:

Diapedesis is the movement of leucocytes through the wall of blood capillaries into the tissue space.
Leucocytes perform diapedesis as an important part of their reaction to tissue injury or infection.
This process forms the part of the innate immune response, involving recruitment of non-specific leucocytes.
Monocytes also use this process during their development into macrophages.
Diapedesis helps leucocytes to perform their functions like phagocytosis, production of antibodies, secretion of inflammatory response chemicals, etc.

Question 18.
Why are obese persons prone to hypertension?
Answer:
(1) Being overweight or obese is a major cause of hypertension, accounting for 65% to 75% of the risk for human primary hypertension.

(2) Following factors play an important role in initiating obesity hypertension:

Physical compression of the kidneys by fat.
Activation of the renin-angiotensin – aldosterone system
Increased sympathetic nervous system activity.
Obesity means more body-surface area. In order to supply blood to these parts. Heart and blood vessels work more resulting into hypertensions.

(3) Blood pressure rises as body weight increases and therefore obese persons are prone to hypertension.

Question 19.
Why does the transplanted heart beats at higher rate than normal?
Answer:

The transplanted heart beats at higher rate than normal (about 100 to 110 beats per minute) because the nerves leading to the heart are cut during the operation. These nerves stimulate the pacemaker i.e. Sinoatrial node.
The new heart also responds more slowly to exercise and does not increase its rate as quickly as before.

Question 20.
Why do large animals cannot carry out respiration without the help of circulatory system?
Answer:

Large animals have various organ systems which always work in a coordinated manner.
These animals provide large respiratory surfaces (numerous alveoli) for the exchange of gases. But these respiratory gases must be carried to the cells of tissues which are away from the respiratory surfaces.
To carry these gases to tissues, there is need of transport system. These gasses are transported from respiratory surfaces to the cells of tissues through blood as a transporting medium.
Therefore, large animals cannot carry out respiration without the help of circulatory system.

Question 21.
What is immunity? Name its types.
Answer:

Immunity is the general ability of a body to recognize, neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.
There are two basic types of immunity, viz. innate immunity and acquired immunity, Acquired immunity is further divided into four types, i.e. Natural acquired active immunity, Natural acquired passive immunity, Artificial acquired active passive immunity and Artificial acquired passive immunity.

Question 22.
Why does the platelet count decrease in dengue patient?
Answer:

The causative pathogen of dengue is dengue virus which induces bone marrow suppression. Since in bone marrow blood cells are formed its suppression causes the deficiency of blood cells leading to low platelet count.
The dengue virus also links with platelets in the blood when there is a virus-specific antibody present in the human body.
When vascular endothelial cell which are infected with dengue virus gets combined with platelets, they tend to destroy platelets. This is one of the major causes of low platelet count in dengue fever.
Even the antibodies that are produced after infection of the dengue virus also cause the destruction of platelets, thus lowering the platelet count.

Question 23.
Why does our immune system fail against pathogens like Trypanosoma cruzi and Plasmodium?
Answer:

Microbes have evolved a diverse range of strategies to destroy the host immune system. The protozoan parasite Trypanosoma cruzi and Plasmodium show similar such adaptations to disturb host defence mechanism.
This parasite attacks host tissues including both peripheral and central lymphoid tissues.
This causes systemic acute response in host body which the parasite tries to overcome. The parasite in fact weakens both innate and acquired immunity.
It interferes with the antigen presenting function of dendritic cells via an action on hosts like lectin receptors. These receptors also induce suppression of CD₄⁺ T cells responses. Therefore, our immune system fail against such pathogens.

Question 24.
What is the relation between immunity and organ transplantation?
Answer:

Those who undergo an organ transplant face the possibility that their immune system will reject their new organ and that they will always be at a higher risk for infections.
The immune system is able to recognize the difference between cells that belong to our body and those that do not by learning to identity protein markers (antigens) that are found on cell and infection surfaces.
In people, the antigens or markers that identity their immune system are referred to as the human leukocyte antigen (HLA).
Antigens that are recognized as unfriendly invaders stimulate an immune response to destroy them.
Therefore, when organ transplantation is done, the immune responses are temporarily stalled. This helps in acceptance of the graft in the recipient’s body.

Question 25.
How do monocytes perform amoeboid movement and phagocytosis?
Answer:
Monocytes can perform phagocytosis. They do this by using intermediary or opsonising proteins such as antibodies or complement that coat the pathogen. They also bind to the microbe directly via pattern-recognition receptors that recognize pathogens. In this way they perform amoeboid movement and indulge in phagocytosis.

Question 26.
How do monocytes modify into macrophages?
Answer:
Monocytes upon having inflammation, selectively travel to the sites of inflammation. Here they produce inflammatory cytokines and contribute to local and systemic inflammation. They are highly infiltrative. They differentiate into inflammatory macrophages, which then remove PAMPs or pathogen-associated molecular patterns and cell debris.

Chart based/Table based questions

Question 1.
Complete the following:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	—————
2. Amphibian tadpoles of frog, salamanders	—————-	—————-
3. Fish	Aquatic	————–
4. Reptiles, Birds and Mammals	—————-	—————-

Answer:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	Tracheal tubes and spiracles
2. Amphibian tadpoles of frog, salamanders	Aquatic	External gills
3. Fish	Aquatic	Internal gills
4. Reptiles, Birds and Mammals	Terrestrial	Lungs

Question 2.

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO₂	—————	—————
PPCO₂	—————-	—————-

Answer:

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO₂	104 mm Hg	40 mm Hg
PPCO₂	40 mm Hg	45 mm Hg

Question 3.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 3
Answer:

Question 4.
Complete the following flow chart
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 5
Answer:

Question 5.
Complete the following Table

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	—————-
QRS wave	—————–	Ventricular depolarization
T wave	—————–	Ventricular repolarization

Answer:

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	Atrial depolarization
QRS wave	Ventricular contraction	Ventricular depolarization
T wave	Ventricular contraction	Ventricular repolarization

Question 6.

Cardiovascular disorders	Symptom
Coronary Artery Diseases (CAD)	Deposition of calcium, fat, cholesterol and ———————
—————-	Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack	Myocardial infarction without ———————-

Answer:

Cardiovascular disorders	Symptom
Coronary Artery Diseases (CAD)	Deposition of calcium, fat, cholesterol and fibrous tissues in blood vessels.
Angina pectoris	Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack	Myocardial infarction without showing symptoms of classical heart attack.

Question 7.

Instrument / Technique	Purpose of use
Sphygmomanometer	———————
—————-	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
—————	To measure ECG.

Answer:

Instrument / Technique	Purpose of use
Sphygmomanometer	To measure blood pressure.
Angiography	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Electrocardiograph	To measure ECG.

Diagram based questions

Question 1.
Give the name and function of A and ‘B’ from the diagram given below
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 7
Answer:

Name	Function
(A) Epiglottis	Epiglottis prevents the entry of food into trachea.
(B) Tracheal cartilage	Tracheal cartilage prevents collapse of trachea and always keeps it open.

Question 2.
Label parts A’ and ‘B’ from the following diagram and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 8
(a) What is the partial pressure of O₂ in part ‘B’?
(b) What is the partial pressure of CO₂ in part A’?
(c) How many alveoli are present in the lungs?
Answer:
Label A → Blood from pulmonary artery Label B → Alveolus
(a) The partial pressure of O₂ in alveolar air is 104 mm Hg.
(b) The partial pressure of CO₂ in pulmonary capillaries is 45 mmHg.
(c) There are about 700 million alveoli in the lungs.

Question 3.
Label parts A’ and ‘B’ from the given diagram and give their functions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 9
Answer: Part A → Sinoatrial [SA] node
Function : SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.
Part B → Atrioventricular [AV] node
Function : Atrioventricular [AV] node acts as pace-setter of heart.

Question 4.
Sketch and label the dorsal (posterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 10

Question 5.
Sketch and label the ventral (anterior) view of human heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 11

Question 6.
Sketch and label – Electrocardiogram or ECG.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 12

Question 7.
Sketch and label – T.S. of Artery, Vein and Capillary.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 13

Question 8.
Observe the diagrams of blood cells and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 14
(a) Which of the above is agranulocyte ?
(b) Describe its origin and structure.
(c) Mention its types.
(d) Explain its function.
Answer:
(a) The figure ‘D’ is agranulocyte.
(b) Structure : Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed.
(c) Agranulocytes are of two types, viz. lymphocytes and monocytes.
(d) Functions of agranulocytes : Agranulocytes are responsible for immune response of body by producing antibodies and monocytes are phagocytic in function.

Question 9.
Observe the diagrammatic representation of double circulation and answer the given questions.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 15

Why the circulation shown in the above diagram is called double circulation?
What are the two main routes of double circulation?
Which blood vessels carry oxygented blood to heart and deoxygenated blood to lungs?
Which blood vessels carry deoxygented blood to heart and oxygenated blood to body organs?

Answer:

During circulation, blood passes twice through the heart, therefore it is called double circulation.
(1) Pulmonary circulation which is from heart to lungs and back from lungs to heart.
(2) Systemic circulation which is from heart to body and back from all body organs to the heart.
Oxygenated blood is carried to the heart by pulmonary veins. Dexoygenated blood is carried to the lungs by pulmonary artery.
Deoxygenated blood is carried to heart by superior and inferior vena cavae. Oxygenated blood is carried to the body organs by systemic or dorsal aorta.

Question 10.
Observe the cardiac cycle given below and answer the following questions
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 16

Which phases of cardiac cycle are shown in the above diagrammatic representation ?
How much time is taken for entire heart to be in diastole?
How much longer is ventricular systole as compared to atrial systole?

Answer:

There are four main phases of cardiac cycle shown in the above diagram. They are
(1) AS : Arterial systole. (2) AD : Atrial diastole. (3) VS : Ventricular systole.
(4) VD : Ventricular diastole which is along with joint diastole.
Diastole of entire heart is called joint diastole, which is for about 0.4 second.
Ventricular systole is almost for the double time than the atrial systole. Atrial systole is for 0.15 second whereas ventricular systole is for 0.3 second.

Long Answer Questions

Question 1.
Describe the respiratory system of human.
Answer:
Respiratory system of human : Human respiratory system consists of nostrils, nasal chambers, pharynx, larynx, trachea, bronchi, bronchioles, lungs, diaphragm and intercostal muscles.
1. Nostrils and nasal chambers:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 17

Oxygen rich air is taken in the body through the nostrils or external nares. They are external opening of the nose. Carbon dioxide and water vapour are also released out of the body through the same passage i.e. the nostrils.
Internal nares open into the pharynx. The space between external and internal nares is knows as nasal chamber which is lined internally by mucous membrane and ciliated epithelium.
Nasal chamber is divided into two parts by a cartilage called mesethmoid. Each part of these halves is further divided into three regions, viz. vestibule, respiratory part and sensory part.
Vestibule is the anteriormost part of nasal chamber. In the vestibule fine hairs are present. They filter out the dust particles and prevent them from going inside.
Respiratory part is the second region which is richly supplied with the capillaries. Air is made warm and moist in this region.
Sensory part is lined by sensory epithelium. It is concerned with the detection of smell.

2. Pharynx:

Pharynx is a short and vertical tube measuring about 12 cm in length. In pharynx the respiratory and food passages cross each other.
The upper part of pharynx is known as naso-pharynx which conducts the air. The lower part is called laryngo-pharynx or oro¬pharynx which conducts food to the oesophagus.
Tonsils that are made up of lymphatic tissue are present in the pharynx. They kill the bacteria by trapping them in the mucus.

3. Larynx:

Larynx produces sound. In males it increases in size at puberty. This is termed as Adam’s apple. It is clearly seen in the neck region.
From pharynx air enters the larynx. The opening through which it enters is called glottis. Glottis is guarded by a flap called epiglottis.
Epiglottis prevents the entry of food particles into the trachea.
TWo folds of elastic tissue called vocal cords are seen along the side of glottis. When they vibrate the sound is produced.

4. Trachea:

The trachea or wind pipe is about 12 cm long and 2.5 cm wide.
It is situated in front of the oesophagus and runs downwards in the thorax through the neck.
The trachea is made up of fibrous muscular tissue wall which is supported by ‘C’-shaped cartilages. These cartilaginous rings are 16 to 20 in number.
Internally the tracheal wall bears ciliated epithelium and mucous glands.
When any foreign particle enters the trachea inadvertently. It is thrown out by coughing action.
Mucous and ciliary action remove the dust particles and push them upwards to the larynx. These particles are then gulped and taken into the oesophagus.

5. Bronchi and bronchioles:

At the distal end, the trachea divides into two bronchi (Singular – bronchus). Bronchi lie below the sternum or breast bone.
Each bronchus has a complete ring of cartilage for support. The two bronchi enter into the lungs on either side.
After entering into the lungs each bronchus divides into secondary and tertiary bronchi. The tertiary bronchi divide and re-divide to form minute bronchioles.
Bronchioles do not have cartilages in their walls. Each bronchiole ends into a balloon like alveolus.
Owing to the presence of alveoli the lungs become spongy and elastic.

6. Lungs:

Lungs are principal respiratory organs located in the thoracic cavity.
They are pinkish, soft, hollow, paired, elastic and distensible organs.
Each lung is enclosed in a pleural sac which consists of two membranes, viz. an outer parietal and inner visceral.
The parietal and visceral membranes enclose pleural cavity which is filled with pleural fluid. The pleural fluid lubricates and prevents friction when pleural membranes slide on each other.
Lungs are highly vascular as they are richly supplied with blood capillaries.
The left lung has two lobes while the right lung has three lobes. Each lobe has many bronchioles and alveolar sacs. The alveolar sacs are spherical and thin walled.
Each alveolar sac contains about 20 alveoli. The alveoli appear as a bunch of grapes. The lobule in the lung thus consists of alveolar ducts, alveolar sacs and alveoli.
Each alveolus has thin and elastic walls. It is about 0.1 mm in diameter. Alveoli are covered by network of capillaries from pulmonary artery and pulmonary vein. A network of pulmonary capillaries supply the alveolus.
The alveolar wall is 0.0001 mm thick and made up of simple, non-ciliated, squamous epithelium. It has collagen and elastin fibres.
Every lung has about 700 million alveoli. They increase the surface area of the lungs for exchange of gases.

Question 2.
Describe the process of respiration in man.
OR
Describe the mechanism of respiration in human beings.
Answer:
Respiration includes breathing, external respiration, internal respiration and cellular respiration.
A. Breathing : During breathing air comes in and goes out of the lungs. The rate of gaseous exchange is speeded up by breathing. Breathing involves two processes, viz. inspiration and expiration
1. Inspiration:

Inspiration is the process in which the air containing oxygen is taken inside the lung.
Inspiration is the active process which is possible due to intercostal muscles, sternum and diaphragm.
During inspiration, intercostal muscles contract, ribs are pulled outward as a result of which the space in the thoracic cavity is increased.
At the same time the lower part of the breast bone is raised and diaphragm flattens by contraction.
The volume of thoracic cavity is thus increased.
Pressure in the lungs decreases as the lungs expand and their volume is increased. Owing to this the atmospheric air enters inside the body through respiratory passage and reaches the lungs.

2. Expiration:

Expiration is the process in which air containing carbon dioxide and water vapour is expelled out of the lungs.
Expiration is a passive process.
During expiration, intercostal muscles relax and ribs are pulled inwards.
The diaphragm relaxes and becomes dome shaped. Intercostal muscles contract simultaneously and due to these events, the volume of the thoracic cavity is reduced.
The pressure on the lungs is thus increased as a result of which they are compressed.
Due to this, air rushes out of the lungs and is expelled out through the nose.

B. External respiration : The process of external respiration takes place in the lungs where oxygen from the lungs diffuses into the blood capillaries present in the lung tissue. Similarly carbon dioxide from the blood capillaries diffuses out and enters in the alveoli in the lungs.

C. Internal respiration : Internal respiration takes place in the cells of the body. Oxygen brought by the blood is given to the cells and the tissues during internal respiration. Similarly carbon dioxide passes into the blood cells from the cells and tissues.

D. Cellular respiration : Cellular respiration takes place in the mitochondria of the cell, where oxygen is utilised to liberate energy in the form of ATP molecules.

Question 3.
Describe the respiratory disorders.
Answer:

Following are some respiratory disorders : Emphysema is caused due to alveolar abnormalities. Chronic bronchitis results into coughing and shortness of breath.
Viral and bacterial respiratory diseases. Acute bronchitis, sinusitis, laryngitis and pneumonia are some of the inflammatory diseases caused either due to virus or due to bacteria.
Allergens like pollen or pet dander can cause asthma. In asthma constriction of bronchioles takes place causing periodic wheezing and difficulty in breathing.
Occupational hazards cause respiratory diseases like silicosis or asbestosis. In these disorders there is inflammation fibrosis leading to lung damage.

Treatments of respiratory diseases:

Bacterial diseases can be completely cured by specific antibiotics.
Viral diseases need to be taken care of by using vaporizers and decongestants.
Asthma needs treatment by inhalers and nebulizers.
For occupational disorders proper mask and other protective gear is a must.
Lethal diseases like pneumonia should be controlled by medication and rest.

Question 4.
How does transport of O₂ and CO₂ take place in man?
Answer:
1. Transport of O₂:

Only 3% of the total oxygen is transported in a dissolved state by the plasma.
The remaining 97% is transported in the form of oxyhaemoglobin in the RBCs.
Hemoglobin present is RBCs combines with oxygen to form oxyhaemoglobin.
Hb + 4O₂ → Hb (4O₂)
Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
Binding of oxygen with haemoglobin in the alveoli and release of oxygen into the tissue cells depends upon the difference in partial pressure of O₂ and CO₂.

2. Transport of CO₂: Carbon dioxide is transported by RBCs and plasma in three different forms.

By plasma in solution form (7%) : About 7% of CO₂ is transported in a dissolved form as carbonic acid (which can be broken down into CO₂ and H₂O).
CO₂ + H₂O = H₂CO₃.
By bicarbonate ions (70%) : Nearly 70% of carbon dioxide is transported in the form of sodium bicarbonate/potassium bicarbonate in the plasma.
RBCs contains an enzyme, carbonic anhydrase. In the presence of this enzyme CO₂ combines with water to form carbonic acid.
Carbonic anhydrase also brings about dissociation of carbonic acid immediately tending to large accumulation of HCO^3- ions inside the RBCs.
CO₂ + H₂O Carbonic anhydrese H₂CO₂ Carbonic anhydrase H₊ + HCO^3-
The bicarbonate ions moves out of RBCs and this would bring about imbalance of the charge inside the RBCs.
To maintain the ionic balance, Cl^– ions diffuse from plasma into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
HCO^3- ions from the plasma then joins to Na⁺/K⁺ forming NaHCO₃/KHCO₃ (to maintain PH of blood).
HCO^3- + Na⁺ → NaHCO₃ Sodium bicarbonate
H⁺ is taken up by haemoglobin to form Reduced Hb (HHb).
At the level of the lungs due to the low partial pressure of the alveolar air, hydrogen ion and bicarbonate ions recombine to form carbonic acid and in presence of carbonic anhydrase it again yields carbon dioxide and water.
H⁺ + HCO_3- Carbonic anhydrase H₂CO₃ Carbonic anhydrase CO₂ + H₂O.

3. By red blood cells (23%):

Carbon dioxide binds with the amino group of the haemoglobin and form a loosely bound compound carbaminohaemoglobin Hb + CO₂ = HbCO₂
Due to low partial pressure of CO₂ at alveolus carbaminohaemoglobin decomposes releasing the carbon dioxide.

Question 5.
Describe different types of leucocytes.
OR
Describe five types of leucocytes, with the help of diagrams. Add a note on their functions.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 19

Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.
Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.
They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.
Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.
Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

In neutrophils, the cytoplasmic granules take up neutral stains.
Their nucleus is three to five lobed.
It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
Neutrophils are about 70% of total WBCs.
They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
Eosinophils are about 3% of total WBCs.
They are non-phagocytic in nature.
Their number increases (i.e. eosinophilia) during allergic conditions.
They have antihistamine property.

(III) Basophils:

The cytoplasmic granules of basophils take up basic stains such as methylene blue.
They have twisted nucleus.
In size, they are smallest and constitute about 0.5% of total WBCs.
They too are non-phagocytic.
Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

Agranulocytes with a large round nucleus are called lymphocyte.
They are about 30% of total WBCs.
Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
They are phagocytic in function.
They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
At the site of infections they are seen in more enlarged form.

Question 6.
Give an account of external features of the human heart.
Answer:
(1) The heart is hollow, muscular, conical organ about the size of one’s fist with broad base and narrow apex tilted towards left measuring about 12 cm in length. 9 cm in breadth and weighing about 250 to 300 grams.

(2) The human heart has four chambers, two atria which are superior, small, thin walled receiving chambers and two ventricles which are inferior, large, thick walled, distributing chambers.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 20

(3) Externally there is a transverse groove between the atria and the ventricles which is known as atrioventricular groove or coronary sulcus.

(4) Between the right and left ventricles there is interventricular sulcus (pi. sulci). In these sulci the coronary arteries and coronary veins are present.

(5) Oxygenated blood to the heart is supplied by coronary arteries while coronary veins collect deoxygenated blood from the heart. The coronary veins join to form coronary sinus which opens into the right atrium.

(6) Right atrium is larger in size than the left atrium. Deoxygenated blood from all over the body is brought through superior vena cava and inferior vena cava and poured into right atrium. Oxygenated blood from lungs is brought to heart by two pairs of pulmonary veins which carry it to the left atrium.

(7) Pulmonary trunk is seen arising from the right ventricle, which carries deoxygenated blood to lungs. While systemic aorta arises from the left ventricle and carries oxygenated blood to all parts of the body.

(8) The pulmonary trunk and systemic aorta are connected by ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 7.
With the help of well labelled diagram describe the internal structure of human heart.
OR
Sketch and label internal view of heart.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 21
The heart shows four chambers with two atria and two ventricles.

I. Atria:
1. Right atrium:

There are two atria which are separated from each other by interatrial septum. They are thin walled receiving chambers on the upper side.
The right atrium receives deoxygenated blood from upper part of body through superior vena cava and from the lower part of the body by inferior vena cava. In the right atrium opens the coronary sinus.
Eustachian valve guards the opening of inferior vena cava while opening of coronary sinus is guarded by Thebesian valve.
On the right side of interatrial septum is seen an oval depression called the fossa ovalis. In the interatrial septum of the foetus there is an oval opening called foramen ovale. Fossa ovalis is remnant of this foramen ovale.
Right atrium opens into the right ventricle.

2. Left atrium:

The oxygenated blood from the lungs is brought into left atrium through four openings of pulmonary veins.
Left atrium opens into the left ventricle.

II. Ventricles:

There are two ventricles which are separated from each other by interventricular septum. They are two thick walled distributing chambers situated on the lower side of the heart.
Left ventricle has thickest wall as it pumps blood to all parts of the body.
The inner surface of the ventricle is thrown into a series of irregular muscular ridges called columnae carnae or trabeculae carnae.
Each atrium opens into the ventricle of its side through atrioventricular aperture. These apertures are guarded by valves made up of connective tissue. The right atrioventricular valve has three flaps hence called tricuspid valve. Left atrioventricular valve has two flaps hence called bicuspid valve or mitral valve.
Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles.
From the right ventricle arises pulmonary trunk which carries deoxygenated blood to lungs for oxygenation.
From the left ventricle arise systemic aorta which distributes oxygenated blood to all parts of the body.
Pulmonary aorta and systemic aorta has three semilunar valves at the base which prevent backward flow of blood during ventricular diastole.

Question 8.
With the help of suitable diagram, describe the conducting system of human heart.
Answer:
(1) The human heart is myogenic.

(2) Conducting system of the heart consists of sinoatrial node, atrioventricular node, bundles of His and Purkinje’s fibre system.

(3) The pacemaker of the heart is sinoatrial node because here the heartbeat originates. Pacemaker has power of generation of wave of contraction. This is modified cardiac tissue, also called a nodal tissue.

(4) SA node is situated in the wall of right atrium near the opening of superior vena cava. The wave of contraction generated by SA node is conducted by cardiac muscle fibres to both the atria. This results in contraction resulting into atrial systole.
Maharashtra Board Class 12 Biology Important Questions Chapter 8 Respiration and Circulation 22

(5) The atrioventricular node (AV node) is located in the wall of right atrium near the opening of coronary sinus. AV node receives the wave of contraction generated by SA node through intermodal pathways.

(6) Bundle of His arises from AV node and divides into right and left bundle branches. These are located in the interventricular septum.

(7) The bundle branches further form Purkinje fibres which penetrate into myocardium of ventricles.

(8) The bundle of His and Purkinje fibres conduct the wave of contraction from AV node to myocardium of ventricles causing ventricular systole.

Question 9.
Describe the detail cardiac cycle.
OR
Explain the working of heart.
Answer:
(1) The working of heart or cardiac cycle is formed by atrial systole, ventricular systole and joint diastole. It takes place in 0.8 second.

(2) During atrial systole from right atrium, the deoxygenated blood is poured into right ventricle through atrioventricular aperture. Similarly, from left atrium, the oxygenated blood enters the left ventricle through atrioventricular aperture. This entire atrial systole lasts for 0.1 second.

(3) After auricular systole follows the ventricular systole. During ventricular systole, the deoxygenated blood from the right ventricle enters the pulmonary trunk, which carries blood to lungs for oxygenation. At the same time, the oxygenated blood from the left ventricle enters the aorta which is then supplied to all parts of the body. The ventricular systole lasts for 0.3 second.

(4) Joint diastole or complete cardiac diastole is the phase taking place after the systole, when the entire heart undergoes relaxation, for 0.4 second.

(5) During joint diastole, the right atrium receives deoxygenated blood from all parts of the body through superior vena cava, inferior vena cava and coronary sinus. The left atrium receives oxygenated blood from the lungs through two pairs of pulmonary veins.

Question 10.
What is blood clotting? How and when does it occur?
Answer:

Blood clotting is coagulation of blood in order to stop the blood flow and resuting blood loss at the time of injury.
When the blood vessel is intact, blood does not clot due to the presence of active anticoagulants like heparin and antithrombin. But when there is an injury causing rupture of a blood vessel, bleeding starts.
This bleeding is stopped by the process of blood clotting during which liquid blood is converted into semisolid jelly.

The events occurring during blood clotting are as follows:

Release of thromboplastin from thrombocytes and injured tissue.
Formation of enzyme prothrombinase in the blood due to initiation of thromboplastin.
Conversion of inactive prothrombin into active thrombin by prothrombinase in the presence of Ca ions.
Conversion of soluble fibrinogen into insoluble fibrin by thrombin.
Formation of a clot by enmeshing platelets, other blood cells and plasma in the fibrin fibres enmesh.

These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 11.
What is repolarization and depolarization ?
Answer:
Repolarization is a stage of an action potential in which the cell experiences a reduction of voltage due to the efflux of potassium (K⁺) ions along its electrochemical gradient. This phase occurs after the cell reaches its highest voltage from depolarization.

Depolarization occurs in the four chambers of the heart : both atria first and then both ventricles. The SA node sends the depolarization wave to the atrioventricular (AV) node which-with about a 100 minutes delay to let the atria finish contracting-then causes contraction in both ventricles, seen in the QRS wave.

Question 12.
What is the correlation between depolarization and repolarization as well as contraction and relaxation of the heart?
Answer:
Depolarization and Repolarization:

When cardiac cells are at rest, they are polarized, meaning no electrical activity takes place.
The cell membrane of the cardiac muscle cell separates different concentrations of ions, such as sodium, potassium, and calcium. This is called the resting potential.
Electrical impulses are generated by specialized cardiac cells automatically.
Once an electrical cell generates an electrical impulse, this electrical impulse causes the ions to cross the cell membrane and causes the action potential, also called depolarization.
The movement of ions across the cell membrane through sodium, potassium and calcium channels, is the drive that causes contraction of the cardiac cells/muscle.
Depolarization with corresponding contraction of myocardial muscle moves as a wave through the heart. Depolarization thus corresponds with contraction of heart.
Repolarization is the return of the ions to their previous resting state, which corresponds with relaxation of the myocardial muscle. Repolarization thus corresponds with relaxation of heart.
Depolarization and repolarization are electrical activities which cause muscular activity.
The electrical changes in the myocardial cell during the depolarization – repolarization cycle is detected on ECG.

Question 13.
How are the signals detected and amplified by electrocardiograph?
Answer:

The action potential created by contractions of the heart wall spreads electrical currents from the heart throughout the body.
The spreading electrical currents create different potentials at points in the body, which can be sensed by electrodes placed on the skin.
The electrodes are made of metals and salts and they act as biological transducers.
Ten electrodes are attached to different points on the body while taking ECG.
There Eire three main leads responsible for measuring the electrical potential difference between arms and legs.
Electrical potential difference between electrodes is recorded.
As in all ECG lead measurements, the electrode connected to the right leg is considered the ground node.
These ECG signals are acquired using a biopotential amplifier and then displayed using instrumentation software. This is recorded on ECG machine or electrocardiograph. The recorded ECG is anailysed by an expert.

Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

July 16, 2024July 15, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition

Multiple choice questions

Question 1.
For the following, which is the aspect of growth? Example – Increase in length, size and number of cells.
(a) Quantitative
(b) Qualitative
(c) Both (a) and (b)
(d) Three dimensional
Answer:
(a) Quantitative

Question 2.
In vascular plants, growth takes place due to ………………..
(a) conducting tissues
(b) embryo
(c) meristems
(d) stem cell
Answer:
(c) meristems

Question 3.
What is growth?
(a) Temporary, irreversible increase in an organism.
(b) Permanent, reversible increase in an organism.
(c) Permanent, irreversible decrease in an organism.
(d) Permanent, irreversible increase in an organism.
Answer:
(d) Permanent, irreversible, increase in an organism

Question 4.
Based on location, plants have these three types of meristems.
(a) Basal, Intercalary and Lateral
(b) Apical, Basal and Lateral
(c) Apical, Interfascicular and Lateral
(d) Apical, Intercalary, Lateral
Answer:
(d) Apical, Intercalary, Lateral

Question 5.
Meristem cells are ………………..
(a) thin walled, vacuolated with prominent nucleus
(b) thick walled, non-vacuolated without nucleus
(c) thin walled, non-vacuolated with prominent nucleus
(d) thick walled, vacuolated with prominent nucleus
Answer:
(c) thin walled, non-vacuolated with prominent nucleus

Question 6.
Select correct sequence of phases of growth.
(a) Phase of formation, phase of elongation, phase of maturation
(b) Phase of cell division, phase of enlargement, phase of elongation
(c) Phase of formation, phase of maturation, phase of elongation
(d) Phase of enlargement, phase of cell division, phase of maturation
Answer:
(a) Phase of formation, phase of elongation, phase of maturation

Question 7.
In which phase of growth, growth rate is at accelerated pace?
(a) Lag phase
(b) Log phase
(c) Steady phase
(d) Stationary phase
Answer:
(b) Log phase

Question 8.
Water is essential for growth because it is necessary ………………..
(a) for turgidity
(b) as nutrient
(c) as raw material
(d) for gravity
Answer:
(a) for turgidity

Question 9.
Which equipment is suitable for measuring linear growth of shoot?
(a) Horizontal microscope
(b) Spectroscope
(c) Crescograph
(d) Auxanometer
Answer:
(d) Auxanometer

Question 10.
Which is correct expression of absolute growth rate (AGR)?
(a) AGR = \(\frac { dn }{ dt }\)
(b) AGR = \(\frac { dt }{ dn}\)
(c) AGR = \(\frac { RGR }{ n }\)
(d) AGR = \(\frac { n }{ RGR }\)
Answer:
(a) AGR = \(\frac { dn }{ dt }\)

Question 11.
Arithmetic growth in plants shows ……………….. graph.
(a) Sigmoid
(b) J-shaped
(c) linear
(d) elliptical
Answer:
(c) linear

Question 12.
What is grand period of growth?
(a) The total time required for all phases to occur
(b) The total time required for exponential phase
(c) The total time required for Lag and Log phase together
(d) The toted time required for stationary phase
Answer:
(a) The total time required for all phases to occur

Question 13.
Which tissue is formed by process of de-differentiation?
(a) Intrafascicular cambium
(b) Secondary phloem
(c) Interfascicular cambium
(d) Secondary xylem
Answer:
(c) Interfascicular cambium

Question 14.
The example of environmental plasticity- heterophylly observed is ………………..
(a) Cotton
(b) Coriander
(c) Larkspur
(d) Buttercup
Answer:
(d) Buttercup

Question 15.
Synthesis of IAA takes place from amino acid ………………..
(a) Methionine
(b) Tryptophan
(c) Valine
(d) Aspartic acid
Answer:
(b) Tryptophan

Question 16.
Find the odd one out.
(a) IAA
(b) 2, 4-D
(c) NAA
(d) IBA
Answer:
(a) IAA

Question 17.
The selective herbicide is ………………..
(a) IBA
(b) GA₃
(c) 2, 4 D
(d) NAA
Answer:
(c) 2, 4 D

Question 18.
This hormone promotes rooting in artificial method of cutting ………………..
(a) Gibberellin
(b) Auxin
(c) Cytokinin
(d) Dormin
Answer:
(b) Auxin

Question 19.
Chemically the peculiar structure of gibberellins is ……………….. ring.
la) pyrole ring
(b) purine ring
(c) gibbeane ring
(d) pyrimidine
Answer:
(c) gibbeane ring

Question 20.
First natural cytokinin was obtained from ……………….. by Letham.
(a) Maize grains
(b) Coconut milk
(c) Rice seedling
(d) Tomato
Answer:
(a) Maize grains

Question 21.
A low ratio of cytokinin to auxin induces ……………….. in plants.
(a) rooting
(b) shooting
(c) bud formation
(d) flowering
Answer:
(a) rooting

Question 22.
Apical dominance : Auxin : : Fruit ripening : ?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Answer:
(c) Ethylene

Question 23.
Which hormone is known as stress hormone ?
(a) Auxin
(b) Gibberellin
(c) Ethylene
(d) Abscissic acid
Answer:
(d) Abscissic acid

Question 24.
Abscissic acid is synthesised from ………………..
(a) Methionine
(b) Malic acid
(c) Mevalonic acid
(d) Mucin
Answer:
(c) Mevalonic acid

Question 25.
Photoperiodic response is because of pigment ………………..
(a) Cytochrome
(b) Phytochrome
(c) Anthocyanin
(d) Phycobilin
Answer:
(b) Phytochrome

Question 26.
The favourable temperature for vernalization is ………………..
(a) 1 to 6 °C
(b) 11 to 16 °C
(c) 10 to 16 °C
(d) – 1 to 1 °C
Answer:
(a) 1 to 6 °C

Question 27.
Identify the group of non-mineral elements needed by plants.
(a) PO₄, CO₃, SO₄
(b) C, H, O
(c) N, P K
(d) C, H, N
Answer:
(b) C, H, O

Question 28.
The deficiency symptoms of these minerals are visible in young leaves ………………..
(a) Ca, CO
(b) S, P
(c) Ca, S
(d) Zn, Mg
Answer:
(c) Ca, S

Question 29.
The symptom chlorosis is observed as ………………..
(a) yellowing of leaf
(b) premature leaf fall
(c) malformation of leaf
(d) localized death of tissue
Answer:
(a) yellowing of leaf

Question 30.
…………….. is constituent of chlorophyll.
(a) Mn
(b) Mg
(c) Mo
(d) Fe
Answer:
(b) Mg

Question 31.
This is essential for O₂ evolution in photosynthesis and for proper solute concentration
(a) Ca
(b) Cu
(c) Cl
(d) Co
Answer:
(c) Cl

Question 32.
For active absorption of mineral uptake, energy is supplied by ………………..
(a) respiration
(b) chemosynthesis
(c) photosynthesis
(d) transpiration
Answer:
(a) respiration

Question 33.
Cyanobacteria fix nitrogen in specialised cells called ………………..
(a) Velamen
(b) Haustoria
(c) Heteocysts
(d) Hormogonia
Answer:
(c) Heteocysts

Question 34.
In root nodule, symbiotic nitrogen fixer organism is ………………..
(a) Rhizopus
(b) Pseudomonas
(c) Rhizobium
(d) Nitrosomonas
Answer:
(c) Rhizobium

Question 35.
In plant body, amides are transported through ………………..
(a) sieve tubes
(b) xylem vessels
(c) phloem parenchyma
(d) plasmodesmata
Answer:
(b) xylem vessels

Question 36.
Building blocks of proteins are ………………..
(a) amides
(b) amino acids
(c) carboxylic acid
(d) nitrates
Answer:
(b) amino acids

Question 37.
Flowering plants Aster, Dahlia and Chrysanthemum are ………………..
(a) SDP
(b) LDP
(c) DNP
(d) SDP or LDP
Answer:
(a) SDP

Question 38.
Experimental material of Garner and Allard for discovery of photoperiodism was ………………..
(a) Cucumber and Tomato
(b) Dahlia and Aster
(c) Soybean and Tobacco
(d) Cabbage and Spinach
Answer:
(c) Soybean and Tobacco

Question 39.
Which of the following is used for the production of long seedless grapes ?
(a) Auxin
(b) Cytokinin
(c) Ethylene
(d) Gibberellin
Answer:
(d) Gibberellin

Question 40.
In vernalization, the cold stimulus is perceived by ………………..
(a) axillary bud
(b) floral bud
(c) leaves
(d) apical bud (shoot apex)
Answer:
(d) apical bud (shoot apex)

Question 41.
Xanthium is ………………..
(a) SDP
(b) LDP
(c) DNP
(d) not a flowering plant
Answer:
(a) SDP

Question 42.
Growth starts slowly during the ………………..
(a) lag phase
(b) exponential phase
(c) maturation phase
(d) stationary phase
Answer:
(a) lag phase

Question 43.
Cytokinins induce the formation of ………………..
(a) shoot apex
(b) intrafascicular cambium
(c) cork cambium
(d) interfascicular cambium
Answer:
(d) interfascicular cambium

Question 44.
Bakane disease in rice is associated with the discovery of ………………..
(a) cytokinins
(b) gibberellins
(c) auxins
(d) ethylene
Answer:
(b) gibberellins

Question 45.
ABA is also known as ………………..
(a) antitoxin
(b) antivirulent
(c) antioxidant
(d) antigibber ellin
Answer:
(d) antigibberellin

Question 46.
Gibberellins were first discovered from ………………..
(a) bacteria
(b) fungi
(c) algae
(d) gymnosperms
Answer:
(b) fungi

Question 47.
Which of the following is trace element?
(a) Mg
(b) Nitrogen
(c) Sulphur
(d) Mn
Answer:
(d) Mn

Question 48.
Which of the following is a macronutrient?
(a) Ca
(b) Mn
(c) Zn
(d) Mo
Answer:
(a) Ca

Question 49.
Nitrogen is an important constituent of ………………..
(a) carbohydrates
(b) sugars
(c) proteins
(d) polyphosphates
Answer:
(c) proteins

Question 50.
Deficiency of phosphorus causes ………………..
(a) stunted growth
(b) inward rolling of leaf margin
(c) brittle cell walls
(d) necrotic spots
Answer:
(a) stunted growth

Question 51.
Which of the following is not an essential element for plant?
(a) Sulphur
(b) Boron
(c) Iron
(d) Cadmium
Answer:
(d) Cadmium

Question 52.
…………….. is a constituent of middle lamella.
(a) Mg
(b) K
(c) Ca
(d) P
Answer:
(c) Ca

Match the columns

Question 1.

Column A (Phase of growth)	Column B (Condition)
(1) Lag phase	(a) Growth rate faster
(2) Log phase	(b) Growth rate steady state
(3) Stationary phase	(c) Growth rate slow

Answer:

Column A (Phase of growth)	Column B (Condition)
(1) Lag phase	(c) Growth rate slow
(2) Log phase	(a) Growth rate faster
(3) Stationary phase	(b) Growth rate steady state

Question 2.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 1
Answer:
(1) -(c) Arithmetic growth
(2) – (a) Rate of growth against time
(3) -(b) Geometric growth curve

Question 3.

Column A	Column B
(1) Use of measuring scale	(a) Record of primary growth
(2) Crescograph	(b) Increase in height of plant
(3) Auxanometer	(c) Measure growth in field
(4) Horizontal microscope	(d) Measurement of linear growth of shoot

Answer:

Column A	Column B
(1) Use of measuring scale	(b) Increase in height of plant
(2) Crescograph	(a) Record of primary growth
(3) Auxanometer	(d) Measurement of linear growth of shoot
(4) Horizontal microscope	(c) Measure growth in field

Question 4.

Column A	Column B
(1) Auxin	(a) Bolting in rosette plants
(2) Cytokinin	(b) Stimulate flowering in SDP
(3) Gibberellins	(c) Promotion of growth of lateral buds
(4) Abscissic acid	(d) Apical dominance

Answer:

Column A	Column B
(1) Auxin	(d) Apical dominance
(2) Cytokinin	(c) Promotion of growth of lateral buds
(3) Gibberellins	(a) Bolting in rosette plants
(4) Abscissic acid	(b) Stimulate flowering in SDP

Question 5.

Column A	Column B (Symptoms observed)
(1) Deficiency of Cu	(a) Malformed leaves
(2) Deficiency of Bo	(b) Leaves with yellow edges
(3) Deficiency of Zn	(c) Brown heart disease
(4) Deficiency of K	(d) Die back of shoot

Answer:

Column A	Column B (Symptoms observed)
(1) Deficiency of Cu	(d) Die back of shoot
(2) Deficiency of Bo	(c) Brown heart disease
(3) Deficiency of Zn	(a) Malformed leaves
(4) Deficiency of K	(b) Leaves with yellow edges

Question 6.

Column A	Column B (Organisms)
1. Symbiotic nitrogen fixation	a. Nitrobacter
2. Denitrification	b. Cyanobacteria
3. Free living nitrogen fixers	c. Rhizobium
4. Nitrification	d. Paracoccus

Answer:

Column A	Column B (Organisms)
1. Symbiotic nitrogen fixation	c. Rhizobium
2. Denitrification	d. Paracoccus
3. Free living nitrogen fixers	b. Cyanobacteria
4. Nitrification	a. Nitrobacter

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the given plant growth regulators as per their specific control of event in plant life cycle in Column A and complete Column B.
(IAA, GA, Cytokinin, Abscissic acid)

Column A	Column B
(1) Shedding of leaves	————-
(2) Induce flowering in LDP	————
(3) Apical dominance	————-
(4) Induce RNA synthesis	————-

Answer:

Column A	Column B
(1) Shedding of leaves	Abscissic acid
(2) Induce flowering in LDP	GA
(3) Apical dominance	IAA
(4) Induce RNA synthesis	Cytokinin

Question 2.
Classify the given specific effects of different types of auxins in Column A and complete Column B with the examples.
(IAA, NAA, 2, 4-D, IBA, 2, 4, 5-T)

Column A	Column B
(1) Selective synthetic herbicide	————-
(2) Seedless fruits	————
(3) Flowering in pineapple	————-
(4) Synthetic auxin	————-
(5) Agent orange	————

Answer:

Column A	Column B
(1) Selective synthetic herbicide	2, 4-D
(2) Seedless fruits	IAA
(3) Flowering in pineapple	NAA
(4) Synthetic auxin	IBA
(5) Agent orange	2, 4, 5 – T

Question 3.
Classify the given plant hormones as their specific effect observed in plants in Column A and complete Column B.
(ABA, GA, Ethylene, Kinetin)

Column A	Column B
(1) Bolting of rosette plants	————-
(2) Epinasty	————
(3) Closure of stomata	————-
(4) Proliferation of callus	————-

Answer:

Column A	Column B
(1) Bolting of rosette plants	GA
(2) Epinasty	Ethylene
(3) Closure of stomata	ABA
(4) Proliferation of callus	Kinetin

Question 4.
Classify the given disease to their cause given in Column B.
(Ethylene, GA, Deficiency of BO, Deficiency of Cu)

Column A	Column B
(1) Brown heart disease	————-
(2) Bakane disease of Rice	————
(3) Die back of shoot	————-
(4) Degreening of Banana	————-

Answer:

Column A	Column B
(1) Brown heart disease	Deficiency of BO
(2) Bakane disease of Rice	GA
(3) Die back of shoot	Deficiency of Cu
(4) Degreening of Banana	Ethylene

Question 5.
Classify the given organisms related to Nitrogen cycle in Column B.
[Nitrobacter, Rhizobium, Pseudomonas, Nitrosococcus)

Column A	Column B
(1) Symbiont in root nodule	————-
(2) Conversion of nitrite to nitrate	————
(3) Denitrification process	————-
(4) Conversion of ammonia to nitrite	————-

Answer:

Column A	Column B
(1) Symbiont in root nodule	Rhizobium
(2) Conversion of nitrite to nitrate	Nitrobacter
(3) Denitrification process	Pseudomonas
(4) Conversion of ammonia to nitrite	Nitrosococcus

Very short answer questions

Question 1.
Enlist the types of meristems that we observe in plants.
Answer:
In plants, there are apical, intercalary and lateral meristems.

Question 2.
Give characteristic features of meristematic cells.
Answer:
Meristematic cells are thin walled, non- vacuolated with prominent nuclei having granular cytoplasm and are capable of cell division.

Question 3.
What is the role of oxygen in growth?
Answer:
Oxygen is required for respiration of cells and release of energy for the process of growth.

Question 4.
What is the role of water for growth?
Answer:
Water maintains turgidity of the cell and is chief component of protoplasm as well as it is a medium for various biochemical reactions.

Question 5.
Mention the mathematical formula for rate of absolute growth.
Answer:
Absolute Growth Rate = AGR = \(\frac { dn }{ dt }\) where dn is cell number and dt is time interval.

Question 6.
What is exponential phase of growth curve?
Answer:
In exponential phase or log phase, growth rate is faster. It accelerates and reaches its maximum.

Question 7.
Is there any relation between phases of growth and regions of growth curve?
Answer:
Yes, there is relation between the two as initially growth is slow, which accelerates and ultimately it slows down and becomes steady which is observed in growth curve.

Question 8.
Which plant organ does show both arithmetic and geometric growth?
Answer:
Embryo that develops from zygote inside the seed shows initially the growth which is geometric and later on arithmetic.

Question 9.
Describe the example of plasticity related to internal stimuli.
Answer:
In plants like cotton, coriander and larkspur heterophylly is observed where leaves in juvenile stage and adult stage show different forms.

Question 10.
What is the role of growth hormones in plants?
Answer:
Growth hormones inhibit, promote or modify the growth in plants.

Question 11.
What is the peculiarity of growth hormones ?
Answer:
Growth hormones are needed in very small amount to evoke the response and they act at a site away from their place of production.

Question 12.
Which is the first hormone to be discovered in plants?
Answer:
Auxin IAA i.e. Indole acetic acid is the first hormone to be discovered in plants.

Question 13.
Give the full form of 2, 4 – D.
Answer:
It is synthetic auxin 2, 4-dichlorophenoxy acetic acid.

Question 14.
What is the effect of NAA and 2, 4-D foliar spray?
Answer:
Foliar spray of synthetic hormones induce flowering in plants litchi and pineapple and prevent premature fruit drop in apples, pear and oranges.

Question 15.
How cytokinins control apical dominance ?
Answer:
Cytokinins promote growth of lateral buds by cell division and thus control apical dominance.

Question 16.
What was the discovery of Richmond and Lang?
Answer:
Richmond and Lang discovered that cytokinins delay the process of ageing and senescence, abscission in plant organs.

Question 17.
What is ‘epinasty’?
Answer:
It is an effect of ethylene where it causes drooping of leaves and flowers.

Question 18.
Why is auxin called a growth regulator?
Answer:
Auxin is synthesized at meristematic region of plants and it controls cell enlargement, cell elongation and stimulates growth of stem and root, apical dominance. Hence it is growth promoting hormone.

Question 19.
What is effect of gibberellin application on apple?
Answer:
Gibberellin causes parthenocarpy in apple.

Question 20.
How can we overcome apical dominance?
Answer:
By application of cytokinin we can overcome apical dominance effect.

Question 21.
Which is standard bioassay method for auxins?
Answer:
Avena curvature test/Avena coleoptile test is a standard bioassay method for auxins.

Question 22.
ABA is called as stress hormone why?
Answer:
Answer: ABA induces dormancy in seeds by inhibiting growth. Thus plants can tide over adverse environmental conditions. Hence it is called as stress hormone.

Question 23.
What was the plant material for study of photoperiodism by Garner and Allard?
Answer:
The flowering response of Soybean and Maryland mammoth variety of tobacco was studied by Garner and Allard.

Question 24.
What is photomorphogenesis?
Answer:
Control of morphogenesis by light and phytochrome pigment is called photomorphogenesis.

Question 25.
What is critical concentration of minerals ?
Answer:
The concentration of the essential elements below which plant growth is retarded is termed as critical concentration.

Question 26.
What is a role of Sulphur in plants?
Answer:
Sulphur is constituent of amino acids, proteins, vitamins (mainly thaimine, biotin CoA) and Ferredoxin.

Question 27.
What is a role of nitrogen in plants?
Answer:
Nitrogen is constituent of amino acids, proteins, nucleic acid, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.

Question 28.
Which is the process by which mainly we get nitrogen in human tissues?
Answer:
Industrial nitrogen fixation by Haber – Bosch Nitrate process is responsible for nitrogen found in human tissues.

Question 29.
Give equation of Haber – Bosch process.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 2

Question 30.
What is nitrogen assimilation?
Answer:
Nitrogen present in the soil as nitrates, nitrites and ammonia is absorbed by plants and converted into nitrogenous organic compounds is nitrogen assimilation.

Give definitions of the following

Question 1.
Growth
Answer:
Growth can be defined as permanent, irreversible increase in the bulk of an organism with change in form.

Question 2.
Efficiency index
Answer:
The increased growth per unit time is called efficiency index.

Question 3.
Absolute growth rate (AGR)
Answer:
The measurement and comparison of total growth per unit time is called absolute growth rate.

Question 4.
Relative growth rate (RGR)
Answer:
The growth of a particular system per unit time expressed on a common basis or alternately it is the ratio of growth in the given time per initial growth.

Question 5.
Differentiation
Answer:
A permanent change in structure and function of cells that leads to their maturation is called differentiation.

Question 6.
Redifferentiation
Answer:
When cells produced from de-differentiation lose their capacity to divide and mature for specific function it is known as re-differentiation.

Question 7.
Development
Answer:
The progressive changes in shape, form and degree of complexity which includes growth, maturation and morphogenesis is referred as development.

Question 8.
Growth Hormone or Growth regulators
Answer:
The internal factors that influence growth by inhibiting, promoting or modifying it are called growth hormones or regulators.

Question 9.
Apical dominance
Answer:
In higher plants growing apical bud inhibits the growth of lateral buds. This is known as apical dominance.

Question 10.
Critical photoperiod
Answer:
That length of photoperiod above or below which the plant shows flowering is critical photoperiod.

Question 11.
Photomorphogenesis
Answer:
The control of morphogenesis of plants by light and phytochrome is photomorphogenesis.

Question 12.
Symptom or hunger sign
Answer:
Any visible deviation from the normal structure and function of the plant is called symptom or hunger sign.

Question 13.
Active absorption of minerals
Answer:
Uptake of mineral ions against concentration gradient, with expenditure of energy (ATP) is called active absorption.

Question 14.
Nitrogen fixation
Answer:
Free nitrogen of air N₂ is converted to nitrogenous salts so that it is made available to plants is called nitrogen fixation.

Question 15.
Nitrification
Answer:
Soil microbes, mainly : chemoautotrophs convert ammonia into j nitrate, the form of nitrogen which can be used by plants, this process is nitrification.

Question 16.
Amidation
Answer:
Ammonia may be absorbed by amino acids to produce amides. This process is called amidation.

Name the following

Question 1.
Instrument developed by Indian physiologist to measure growth.
Answer:
Crescograph developed by Sir J.C. Bose.

Question 2.
Instrument to measure linear growth of shoot.
Answer:
Auxanometer.

Question 3.
Type of growth curve for geometric growth.
Answer:
J-shaped or exponential growth curve.

Question 4.
The process by which cork cambium is formed.
Answer:
Dedifferentiation.

Question 5.
Ability of plant to form different kinds of structures.
Answer:
Plasticity.

Question 6.
Condition where plant exhibits different types of leaves on same plant.
Answer:
Heterophylly.

Question 7.
Growth promoter hormones.
Answer:
Auxins, gibberellins and cytokinins.

Question 8.
Growth inhibitors of plants.
Answer:
Ethylene and abscissic acid.

Question 9.
First hormone that is discovered in plant and its precursor.
Answer:
Amino acid tryptophan is precursor of Indole acetic acid (IAA).

Question 10.
A synthetic hormone that acts as selective herbicide.
Answer:
2, 4 – D (dichlorophenoxy acetic acid).

Question 11.
A plant hormone discovered from fungus.
Answer:
Gibberellic acid (GA).

Question 12.
A plant hormone also present in fish (Herring) sperm DNA.
Answer:
Kinetin (cytokinin).

Question 13.
A plant hormone also present in urine of person suffering from Pellagra.
Answer:
Auxin.

Question 14.
Most widely used source of ethylene for fruit ripening.
Answer:
Ethephon – 2 chloroethyl phosphoric acid.

Question 15.
A precursor from which GA and ABA are synthesized.
Answer:
Mevalonic acid.

Question 16.
Plant antitranspirant.
Answer:
Abscissic acid.

Question 17.
Chemical stimulant of low temperature effect on flowering of plants.
Answer:
Vernalin.

Question 18.
Methods of synthesis of amino acids.
Answer:
Reductive animation and Transamination.

Question 19.
Common amides present in plants.
Answer:
Asparagine and Glutamine.

Question 20.
Nitrogen fixing prokaryotic organisms.
Answer:
Diazotrophs or Nitrogen fixers.

Question 21.
Special structures of cyanobacteria where N₂ fixation occurs.
Answer:
Heterocysts.

Question 22.
A synthetic cytokinin hormone.
Answer:
6 – benzyl adenine.

Give Functions/Significance/Importance of the following

Question 1.
Meristems
Answer:
In plants meristems are situated at specific regions where growth takes place by constant and continuous addition of new cells. Meristems have capacity to divide (Mitotic divisions).

Question 2.
Synthetic auxin 2, 4-D
Answer:
It is selective herbicide which kills dicot weeds and foliar spray of 2, 4-D induces flowering in litchi and pineapple, prevents premature fruit drop in apples, pear, oranges.

Question 3.
Coconut milk
Answer:
Coconut milk contains natural cytokinin substance kinetin which is used as nutritional supplement for callus tissue culture where proliferation is noticed due to promotion of cell division.

Question 4.
Ethylene/Ethephon
Answer:
It is used for fruit ripening and as it causes degreening effect by increasing activity of chlorophyllase enzyme for banana and citrus fruits.

Question 5.
Abscissic acid/ABA
Answer:
It is natural growth inhibiting substance in plants and it acts as plant anti Iranspirant causing closure of stomata. It is stress hormone that induces plant to bear the adverse environmental conditions like drought.

Question 6.
Phytochrome
Answer:
It is a proteinaceous pigment present in leaves which perceives stimulus of light for flowering. As it is interconvertible in two forms, it promotes flowering in SDP and inhibits flowering in LDE.

Question 7.
Vernalization
Answer:
It is response of flowering to low temperature treatment which helps in cultivation of crops in regions where they do not occur naturally and also crops can be produced earlier.

Question 8.
Nitrogen cycle
Answer:
Nitrogen is essential macronutrient for plant growth, it is constituent of amino acids, proteins, nucleic acids, vitamins, hormones, ATP coenzymes, chlorophyll molecule. It is limiting nutrient for plant productivity and agricultural ecosystem. Through food chain it moves to consumers, i.e. animals and decomposers.

Distinguish between the following

Question 1.
Phase of cell division and Phase of cell enlargement
Answer:

Phase of cell division	Phase of cell enlargement
1. In this phase cells of meristems divide by mitotic division.	1. In this phase newly formed cells become vacuolated and turgid, osmotically active.
2. Rate of growth at slow pace.	2. Rate of growth at accelerated pace.
3. This is described as lag phase.	3. This is described as log phase.
4. There is no synthesis of any new material.	4. Synthesis of new wall materials and other materials takes place.

Question 2.
Long day plants and Short day plants
Answer:

Long day plants	Short day plants
1. Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants (LDP).	1. Plants that flower only when they are exposed to light period shorter than the critical photoperiod are called short day plants (SDP).
2. The plants usually flower in summer.	2. The plants usually flower in winter or late summer.
3. These plants require short night period for flowering. Hence, they are also known as short night plants.	3. These plants require long night period for flowering. Hence, they are also known as long night plants.
4. Plants such as Pea, Radish, Sugar, Beet, Cabbage, Spinach, Wheat, poppy are LDP	4. Plants such as Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) Cocklebur (Xanthium) are SDP

Question 3.
Passive absorption of Minerals and Active absorption of minerals
Answer:

Passive absorption of minerals	Active absorption of minerals
1. The movement of mineral ions into root cells due to diffusion, without expenditure of energy is called passive absorption.	1. The uptake of mineral ions by root cells that requires expenditure of ATP energy is known as active absorption.
2. The movement is according to concentration gradient.	2. The movement of mineral ions is against concentration gradient.
3. It takes place by direct ion exchange, diffusion, indirect ion exchange – mass flow and Donnan equilibrium.	3. It takes place by mineral ion accumulation in root hair and carrier concept.

Given reasons

Question 1.
Water is essential for growth in plants.
Answer:

Meristematic cells divide and form new cells.
Absorption of water is necessary for maintaining turgidity in the newly formed cells.
Turgidity results in enlargement of cells in phase of cell elongation.
Water is essential component of protoplasm of cells.
biochemical reactions. Therefore it is essential for growth in plants.

Question 2.
2, 4-D is used as herbicide.
Answer:

2, 4-D is a synthetic auxin which kills dicot weeds.
Our most of the food crops are cereals, i.e. monocot plants.
Weeds are unwanted plants which otherwise lower the productivity hence to kill them. Selective herbicide is used.

Question 3.
In morphogenesis of plants cytokinin auxin ratio is important.
Answer:

Auxins and cytokinins are growth promoting substances which stimulate cell division and cell enlargement.
A high cytokinin promotes shooting in plants.
A low ratio of cytokinin to auxin induces root development.
A high ratio of cytokinin to auxin induces growth of buds and shoot development.
Thus cytokinin and auxin ratio and their interactions control morphogenesis in plants.

Question 4.
ABA is described as an antitranspirant.
Answer:

ABA is a growth inhibiting hormone.
ABA is responsible for causing efflux of K⁺ ions from guard cells of stomata.
As a result of this, osmotic changes occur and guard cells become flaccid resulting in closure of stomata.
Transpiration mainly occurs through open stomata and due to closure the activity is checked. Hence it is described as antitranspirant.

Question 5.
Some deficiency symptoms of mineral are visible in young leaves while some appear in older leaves.
Answer:

When mineral element is present below a certain critical concentration it is said to be deficient.
Symptoms are indicated in the form of certain morphological changes on the mobility of element.
These symptoms depend on the mobility of element inside the plant body.
When the element is relatively immobile like S, Ca then the symptoms appear first in young leaves.
When the elements are actively mobilised inside plant body, they are transported to young tissues then the symptoms are visible in older, i.e. senescent leaves e.g. N, Mg, K.

Question 6.
In Donnan equilibrium of passive absorption of minerals concentration of cations increases inside the cell.
Answer:

Minerals exist in soil in the form of charged particles.
Certain negatively charged (anions) get accumulated on the inner side of cell membrane after their entry inside cell.
These anions cannot diffuse out through semipermeable cell membrane.
Thus additional mobile cautions are needed inside the cell to balance these fixed anions.
Hence, the concentration of cations increases inside the cell.

Question 7.
A sudden drop in active absorption of minerals is noticed if roots are deprived of oxygen supply.
Answer:

Absorption of mineral ions from soil against concentration gradient is known as active absorption.
It requires energy (ATP) for absorption by absorbing root cell.
The source of energy is respiration of cells for supply of ATE
When the roots are deprived of oxygen, their respiration process is affected and thus energy is not supplied in required amount. Hence, a sudden drop in absorption of minerals is noticed.

Question 8.
Nitrogen is a limiting nutrient in the agricultural system.
Answer:

Nitrogen is a major nutrient for plant growth.
Proper carbon/nitrogen ratio in soil is necessary for plant growth.
It is component of proteins in the form of amino acids.
Proteins are synthesised from photosynthetic products sugars.
Nitrogen exists in atmosphere but it is inert, non-reactive.
Plants need nitrogen in a reactive form usually nitrate in soil.
This supply need to be maintained through biological and physical nitrogen fixation.
Otherwise productivity is affected hence it is limiting nutrient in the agricultural ecosystem.

Question 9.
Cucumber and sunflower are regarded as photoneutral plants.
Answer:

In cucumber and sunflower, the flower is not controlled by light period.
Both these plants flower in all light periods.
Cucumber and sunflower, therefore, are regarded as photoneutral plants.

Write short notes on the following

Question 1.
Meristems
Answer:

In vascular plants, growth is indeterminate and occurs at specific regions where meristems are located.
Meristems are of three types based on their location – apical, intercalary and lateral.
Meristems are thin walled cells with prominent nucleus with granular cytoplasm, non-vacuolated.
Mitotic divisions take place in meristematic cells.

Question 2.
Phase of cell formation
Answer:

Formative phase is the first phase of growth.
During this phase, the meristematic cells undergo mitosis to form new cells.
During formative phase, the rate of growth is slow.
The phase of cell formation is also called lag phase.
This phase is also known as phases of cell division.

Question 3.
Development
Answer:

Development is progressive changes taking place in shape, form and degree of complexity in an organism.
In plants, it includes all the changes taking place in sequence from seed germination to senescence or death of plant.
Development is an orderly process.
It includes growth, morphogenesis, maturation and senescence.

Question 4.
Plasticity
Answer:

Plasticity is the capacity of plant being molded or formed.
It is ability of plant to develop different kinds of structures in response to environmental factors or stimuli.
Different kinds of structures can be developed in plants due to internal stimuli in different phases, i.e. juvenile and adult.
Heterophylly is shown in plant in different phases or in different environmental conditions.
In coriander and cotton plants, two different kinds of leaves are observed in young (juvenile) and mature (adult) plant.
In buttercup, two different kinds of leaves are observed in terrestrial (on land) and aquatic (in water) habitat.

Question 5.
Phytohormones/Plant Growth Regulators
Answer:

Phytohormones or plant growth regulators are internal factor that influence growth.
They inhibit, promote or modify the plant growth.
Plant hormones are organic substances produced naturally in plants and required in small amount.
Their place of production and site of the activity are different.
Auxins, gibberellins, cytokinins are growth promoters and ethylene, abscissic acid are growth inhibitors.

Question 6.
Phytochrome
Answer:

These are proteinaceous pigments present
Phytochrome exists in two interconvertible forms. P_r and P_fr.
Phytochromes are located in cell membranes of chlorophyllous cells.
When P_fr absorbs red light it is converted to P_r.
P_fr is accumulated in plants during daytime and inhibits flowering in SDP but initiates flowering in LDP

Question 7.
Venralization
Answer:

Effect of temperature on flowering of plants is known as vernalization.
For inducing early flowering pretreatment of seeds or seedlings is done at 1 to 6 °C for about a months duration.
Shoot apical meristem is believed to be site of vernalization stimulus.
Vernalization stimulus is in a form of chemical named vernalin.
Vernalization is effective in ereals (wheat) and crucifers.

Question 8.
Apical dominance
Answer:

The presence of apical bud inhibits the growth of lateral buds. This phenomenon in which the apical (terminal) bud is active and lateral buds remain inactive is called apical dominance.
It is believed that apical dominance is controlled by an auxin which is synthesized in the apical bud.
From the apical bud, the auxin migrates to the lateral buds and inhibits their growth.
When apical bud is removed, the lateral buds grow and form branches.
For producing more branches therefore, the apical buds are removed.
Cytokinins reverse apical dominance effect by promoting growth of lateral buds by cell division.

Question 9.
Toxicity of micronutrients or mineral toxicity
Answer:

Micronutrients are required in minute quantities by plants.
Their moderate decrease causes deficiency symptoms while their moderate increase causes toxicity.
The reduction in dry weight of a tissue by 10% by any mineral is known as toxicity.
It is not easy to identify toxicity symptoms.
Most of the time, the excess of an element inhibits the uptake of another element resulting in causing the deficiency symptom of that element.
Manganese inhibits calcium translocation towards apex of stem and exhibits symptoms of chlorosis with grey spots appearing on leaves.
This is because manganese competes with iron and magnesium for uptake.
Therefore what we see as symptoms of manganese toxicity, may be the deficiency symptoms of Fe, Mg and Ca.

Question 10.
Day neutral plants (DNP)
Answer:

Day neutral plants are those in which flowering is not affected by the duration of the light period.
Day neutral plants flower in all photoperiods.
Day neutral plants are also known as photoneutral or intermediate plants.
Plants such as cucumber, shoe-flower, sunflower, tomato, maize, balsam, etc. are day neutral plants.

Short Answer Questions

Question 1.
What is de-differentiation?
Answer:

The differentiated cells which are formed may again gain the capacity to divide as per need.
Permanent cells (mature cells) undergo de-differentiation and become meristematic.
This acquired feature of living permanent cells is known as de-differentiation.
e.g. Formation of interfascicular cambium and cork cambium from parenchyma cells of medullary rays and outer cortical cells respectively.

Question 2.
Discuss about natural Auxin.
Answer:

F.W. Went named the growth promoting substance as Auxin.
Auxin was also isolated from urine of patient of pellagra.
Indole 3 acetic acid was first hormone discovered in plants.
IAA is most common and natural hormone synthesized at growing tips and responsible for cell elongation.
It is synthesized from tryptophan and shows polar transport – Basipetal transport.

Question 3.
Discuss about discovery of Gibberellins.
Answer:

Gibberellins are growth promoting hormones and were isolated form fungus Gibberella Jujikuroi by Kurasawa.
Rice plants when infected with this fungus show stem elongation i.e. Bakane disease.
Yabuta and Sumuki isolated gibberellins in crystalline form, from fungal culture and named it gibberellins.
Gibberellins are synthesized from mevalonic acid at young leaves, seeds, root and stem tips and show non-polar transport.

Question 4.
Discuss about discovery of cytokinin.
Answer:

Cytokinins are growth promoting hormones that stimulate cell division.
Skoog and Miller discovered first cytokinin when they were investigating nutritional requirements of tobacco callus culture.
It was observed that the callus proliferated when there was addition of coconut milk as supplement.
The degraded sample of herring (fish) sperm DNA also showed similar growth of tobacco callus. They named the substance as kinetin.

Question 5.
Discuss about discovery of abscissic acid.
Answer:

Abscissic acid (ABA) is a natural growth inhibiting hormone.
It was observed by Carns and Addicott that shedding of cotton balls occur due to chemical substance abscission I and II.
From the buds of Acer, Wareing isolated substance that causes bud dormancy and named it as dormin.
These two chemical substances were identical and now known as abscissic acid.
ABA is synthesized in leaves, fruits and seeds from mevalonic acid.

Question 6.
What is day neutral plant (DNP)? Give any two examples.
Answer:

The plants which do not require specific duration of light period or dark period flowering are day neutral plants (DNP).
They flower throughout the year, as they do not need specific photoperiod.
The flowering in these plants is independent of photoperiod.
Examples – cucumber, tomato, cotton, sunflower, maize and balsam.

Question 7.
Discuss about discovery of phytochromes.
Answer:

Phytochromes are proteinaceous pigments present in cell membrane of green cells.
Phytochromes receive photoperiodic stimulus and induce flowering in plants in response to light duration.
Hendrick and Borthwick observed that in SD plants flowering is inhibited if dark period is interrupted by flash of red light (660 nm).
If flash of far red light (730 nm) is given then again flowering is observed in SD plants.
Pigment system of plant receives photoperiodic stimulus and these pigments exist in two interconvertible forms P^fr and P^r

Question 8.
What is mineral nutrition of plants?
Answer:

Plants require inorganic materials for the synthesis of food.
These elements are obtained by plants in the form of minerals mainly form soil.
Chemical analysis of plant ash reveals that about 40 different minerals are needed by plants which are taken from surroundings, (air, soil and water)
These minerals are absorbed in dissolved form, i.e. ionic form through root system mainly root hairs.
Some minerals are required in large amounts (major) while some are needed only in traces or small amounts (minor).

Question 9.
What are symptoms of mineral deficiency in plants?
Answer:
Any visible deviation from the normal structure and function of plants due to lack or unavailability of particular element below its critical concentration is deficiency symptom of that mineral element.

The common symptoms observed in plants are as follows:

Stunting : Retarded growth and thus stem appears short and condensed.
Chlorosis : This is loss or lack of chlorophyll that result in yellowing of leaf.
Necrosis : It is localized death of tissue. Mottling : This is appearance of green or non-green patches or spots on leaves. Abscission : This is premature fall of leaves, buds, fruits and flowers.

Question 10.
Enlist the role of following minerals and the symptoms caused due to their deficiency : (a) Calcium (b) Boron and (c) Chlorine.
Answer:
(a) Calcium:
Role : Involved in selective permeability of cell membranes, activator of certain enzymes, required as calcium pectate in middle lamella of cell wall at root and stem apex (for cell division).
Deficiency symptom : stunted growth.

(b) Boron:
Role : Required for uptake and utilization of Calcium (Ca²⁺), pollen germination, cell differentiation, carbohydrate translocation. Deficiency symptom : Brown heart disease

(c) Chlorine:
Role : Na⁺ and K⁺ help to determine solute concentration and anion – cation balance in cells, necessary for oxygen evolution in photosynthesis.
Deficiency symptom : Poor growth of plant.

Question 11.
What is Donnan Equilibrium?
Answer:

Donnan equilibrium is a process of passive absorption of minerals in plants which is without any expenditure of energy.
It is assumed that certain anions after their entry by diffusion into the cell get fixed on inner side of cell membrane.
Additional mobile cations are needed to balance this fixed anions as they cannot diffuse outside.
Concentration of cations thus increases due to accumulation.
This passive absorption of anions or cations from exterior against their concentration gradients so as to neutralize the effect of cations or anion is known as Donnan equilibrium.

Question 12.
Explain physical nitrogen fixation.
Answer:

Conversion of free nitrogen of air into nitrogenous compounds that are made available to plants for uptake is known as : nitrogen fixation.
Physical nitrogen fixation occurs in step- : wise manner and it takes place in atmosphere j and soil.
Under the influence of electric discharge, lightning and thunder, atmospheric nitrogen combines with oxygen to form nitric oxide.
Nitric oxide is then oxidized to nitrogen peroxide in presence of oxygen.
Nitrogen peroxide combines with rainwater to form nitrous and nitric acid which come on ground as acid rains.
On ground, alkali radicals (mainly of Ca, K) react with nitric acid to produce nitrites and nitrates which are absorbable forms for plants.

Question 13.
Give equations of physical nitrogen fixation.
Answer:
(1) Physical nitrogen fixation occurs in stepwise manner in atmosphere and on land (soil)
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 3

Question 14.
Give the equations of amino acid synthesis.
Answer:
1. Macromolecule proteins are made up of building blocks of amino acids.
2. Amino acids are synthesized by two methods – Reductive animation and transamination.
3. Reductive amination – Ammonia reacts with alpha keto glutaric acid to form glutamic acid.

4. Transamination – Amino group of one amino acid is transferred to other carboxylic acid at keto position.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 5

Question 15.
Explain lag phase, log phase and steady phase of growth.
Answer:
1. In plants, growth curve is always sigmoid, i.e. S-shaped. This is because growth starts slowly during formative phase, becomes rapid during elongation phase and finally slows down to a steady state during the maturation phase.

2. The standard growth curve shows three phases, viz. lag phase, log phase and stationary phase.
(i) Lag phase or initial growth phase : This is the initial phase of growth. During this phase of growth, the rate of growth is slow. It corresponds to formative phase of growth where new cells are formed due to cell division.

(ii) Log phase or exponential phase : This is the second phase of growth. During this phase, the growth is rapid and maximum. It corresponds to the phase of cell elongation.

(iii) Stationary phase or steady phase : The stationary phase is the third and last phase of growth. In this phase, growth slows down and becomes steady. The cells undergo differentiation during stationary phase.

Chart based or Table based questions

Question 1.
Complete the chart of plasticity.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 6
Answer:

Question 2.
Complete the flow chart of development.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 8
Answer:

Question 3.
Complete the table of growth hormones.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 10
Answer:

Question 4.
Complete the table of mineral nutrition of plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 12
Answer:

Diagram based questions

Question 1.
Draw diagram of photoperiodic response of SDP and LDP.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 14

Question 2.
With the help of diagram show arithmetic growth and geometric growth.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 15

Question 3.
Draw the sigmoid growth curve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 16

Question 4.
Observe the diagram and answer the questions related to it.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 17

From which cells newly formed cells are added? How?
How is rate of growth in zone of cell elongation?
What is peculiarity of zone of differentiation? How is rate of growth in this region?

Answer:

Meristematlc cells add new cells by mitotic division.
Rate of growth is at accelerated pace.
The rate of growth is at steady state and cells become specialised to perform specific function become mature.

Question 5.
Observe the adjacent graph indicating growth. What is correct labelling of A, B and C respectively?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 18
Answer:
A. Stationary Phase
B. Exponential Phase
C. Lag Phase

Question 6.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 19
(1) Observe the above graph indicating increase in height of plants. Which type of growth indicates this pattern of graph?
(2) Give its mathematical expression.
Answer:
(1) Arithmetic growth

(2) Lt = Lo + rt where
Lt = Length at time t,
Lo = Length at time zero r = growth rate,
t = time of growth

Question 7.
Observe the figure given below of two different leaves ‘A’ and ‘B’ Which leaf shows much higher relative growth rate ?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 20
Answer:
Relative Growth Rate (RGR) = Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 10 – 5 = 5, \(\frac { 5 }{ 5 }\) × 100 = 100%
Leaf B = 55 – 50 = 5, \(\frac { 5 }{ 50 }\) × 100 = 10%
Hence fig. A shows more relative growth.

Question 8.
Observe the diagrams A and ‘B’ showing growth in two leaves. Which diagram shows more relative growth?
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 21
Answer:
Relative Growth Rate (RGR)
= Growth per unit time as % of intial size.
RGR = \(\frac { Growth per unit time }{ intial size }\) × 100
Leaf A = 20 – 10 = 10, \(\frac { 10 }{ 10 }\) × 100 = 100%
Leaf B = 60 – 50 = 10, \(\frac { 10 }{ 50 }\) × 100 = 20%
Hence Diagram A shows more relative growth.

Question 9.
Identify the type of growth curves observed in plants.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 22
Answer:
Constant linear growth curve for the arithmetic growth.
Exponential (J shaped) growth curve for the geometric growth.
Sigmoid growth curve related to distinct phases of growth.

Question 10.
Fill in the blanks in the given nitrogen cycle.
Maharashtra Board Class 12 Biology Important Questions Chapter 7 Plant Growth and Mineral Nutrition 23
Answer:

Long answer questions

Question 1.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:

Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]
It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.
A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.
Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.
The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.
While preparing the required nutrient medium particular nutrient can be totally avoided and then the effect of lack of that nntr’cnt can be studied in variation of plant growth.
Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.
For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chlorosis is noticed if Magnesium is lacking as it is a structural component of chlorophyll pigment.

Question 2.
Explain the active absorption of minerals.
Answer:

Plants absorb minerals from the soil with their root system.
Minerals are absorbed from the soil in the form of charged particles, positively charged cations and negatively charged anions.
The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
The ATP energy derived from respiration in root cells is utilized for active absorption.
Ions get accumulated in the root hair against the concentration gradient.
These ions pass into cortical cells and finally reach xylem of roots.
Along with the water these minerals are carried to other parts of plant.

Question 3.
What is growth? What are its characteristics ?
Answer:
Growth : Growth is a “vital process which brings about an irreversible increase in an organism or its part with respect to its size, weight, form and volume.”

Characteristics of growth:

Growth is a permanent increase in size, weight, shape, volume and dry weight of a plant.
The change occurring due to growth is permanent and irreversible.
Growth is an intrinsic process caused due to internal activities.
Growth occurs by cell division and cell elongation followed by cell maturation which lead to the formation of different types of tissues.
Growth in plants is mostly localized, i.e. restricted to some regions of plants possessing meristematic tissues or meris terns.
Growth has a qualitative aspect where development takes place in an orderly manner and differentiation leads to higher and more complex state.

Question 4.
Describe the phases of growth.
Answer:
There are three phases of growth, viz, formation phase, elongation phase and maturation phase.
(1) Formative phase (Phase of cell division) : This is the first phase of growth. In this phase, the meristematic cells undergo mitosis to produce new cells. Owing to the formation of new cells, there occurs a slight increase in the size of the organ.

(2) Elongation phase (Phase of cell enlargement) : This is the second phase of growth. In this phase, the new cells that are formed, undergo enlargement as a result of which the size and volume of the cells increase. Enlargement of cells occur mostly in linear direction as a result of which the elongation of the root and stem takes place. The enlargement of cells causes a considerable increase in size and weight of an organ and a plant as a whole.

(3) Maturation phase (Phase of cell maturation and differentiation) : Maturation phase is the third and last phase of growth. In this phase, the elongated cells undergo maturation and differentiation to form various types of plant tissues like parenchyma, sclerenchyma, xylem and phloem.

Question 5.
Explain the various conditions of growth that are essential.
Answer:

For a proper growth of plant various environmental and physiological conditions are necessary.
Carbon/Nitrogen ratio in soil is having effect on growth as both carbon and nitrogen are structural elements in carbohydrates, proteins and other biomolecules.
Water is essential component of protoplasm and required for turgidity of cells during cell enlargement phase. It is a medium in which various biochemical reactions ocfcur.
Nutrients are necessary for proper growth. Macronutrients and micronutrients have their specific role.
Temperature of 25 – 35 °C is optimum for growth.
Light is essential for seed germination and photosynthesis.
Oxygen is necessary for respiration and supply of energy.
Gravitational force decides direction of growth for root system and shoot system.
Growth hormones are organic compounds that are involved in various physiological aspects and control of growth.

Question 6.
What are plant growth regulators? Give the characteristics of plant growth regulators.
Answer:
Plant growth regulators : Plant growth regulators or phytohormones or plant hormones as they are also called are organic compounds synthesised by the plants which promote, inhibit or control the growth and other physiological processes.

Characteristics of plant growth regulators:

Plant growth regulators are organic compounds other than nutrients.
They are synthesised at the apices of root, stem and leaves from where they are transported to other parts of plants where they produce their effects.
They are required in minute quantities.
A single plant growth regulator can control or regulate the various aspects of growth.

Question 7.
Enlist the five main types of growth regulators and state the role of abscissic acid in plants.
Answer:
Five main types of growth regulators:

Auxins
Gibberellins
Cytokinins
Ethylene and
Abscissic acid.

Role of abscissic acid (ABA) in plants:

Abscissic acid influences abscission and dormancy.
ABA accelerates senescence of leaves, flowers and fruits.
It is a stress hormone as it is produced during drought and other unfavourable climatic conditions.
ABA induces dormancy in seeds, buds and tubers.
It acts as growth inhibitor as it retards growth.
ABA plays an important role in closing of stomata to check transpiration.
It inhibits and delays cell division and suppresses cambial activity by inhibiting mitosis in vascular cambium.
ABA inhibits flowering in LDP and stimulates flowering in short day plants (SDP).

Question 8.
Write an account of auxins as growth regulators.
Answer:

Auxins are plant growth regulators produced naturally by plants.
They are weak organic acids capable of promoting cell elongation during the growth of stem and root.
Auxins are synthesized in shoot and root apices besides young leaf primordia.
Auxins may be natural or synthetic.
Naturally occurring auxins are indole-3- acetic acid (IAA) and its derivatives.
NAA, 2, 4-D and 2,4, 5-T are synthetic auxins.
Auxins in higher concentration promote the growth of stem.
Auxins play an important role in initiation and promotion of cell division.
Auxins help in the formation of adventitious roots from cuttings when applied in lower concentration.
Auxins play an important role in apical dominance.
Auxin prevents abscission by preventing the action of hydrolytic enzymes in abscission layer.
Auxins are used to produce parthenocarpic fruits in plants like orange, apple, tomato and grapes.

Question 9.
Give an account of physiological effects and application of auxin with examples.
Answer:

Auxins are growth promoting substances synthesized at meristematic regions of plants.
The primary effect of auxin is cell enlargement and it stimulates growth of stem and root.
Apical dominance – The phenomenon where growing apical bud inhibits the growth of lateral bud is apical dominance which is controlled by auxin synthesized at apical bud.
Owing to activity of inducing multiplication of cells it is used in plant tissue culture to produce callus.
Auxin stimulates formation of lateral and adventitious roots hence used for rooting propagation of cuttings.
2, 4-D is a synthetic herbicide which kills dicot weeds.
Induced parthenocarpy in fruits-oranges, banana, grapes, lemons is by application of auxin.
Foliar spray of NAA and 2, 4-D induces flowering in litchi and pineapple.
Premature fruit drop of apples, pear and oranges is prevented.
Auxins break seed dormancy and promote germination.
Auxins promote early differentiation of xylem and phloem, cell elongation, increase rate of respiration, prevent formation of abscission layer.

Question 10.
Explain the application of gibberellins.
Answer:

Gibberellins are growth promoting hormone and it is present in root tips, stem tips and seeds.
Gibberellins break dormancy of bud, dormancy of seed.
They promote seed germination in cereals by activating or synthesising enzyme amylase to produce sugar.
Gibberellins induce elongation of the cells in stem hence increase in internode length is noticed.
In rosette plants like cabbage it causes ‘bolting’ that is increase in internode length before flowering.
Gibberellins are more effective in inducing parthenocarpy than auxins in plants like tomato, apple and pear.
It is also used to increase fruit size and length of bunches in grapes.
By its application genetically dwarf plants can be converted to phenotypically tall e.g. Maize.
It overcomes requirement of vernalization, delays senescence and prevents abscission.
Application of gibberellins causes production of male flowers on female plants.

Question 11.
Describe about the physiological effects and applications of cytokinin.
Answer:

Cytokinins are growth promoting hormone that promotes cell division. Kinetin, zeatin are examples of cytokinin.
They promote cell division as well as cell enlargement.
High cytokinin promotes shoot development.
Growth of lateral buds is promoted by cytokinins. Thus it controls apical dominance.
Process of ageing and senescence, abscission of plant organs is delayed by their application.
It promotes formation of interfascicular cambium.
It has a role in breaking seed dormancy and promotes seed germination.
Cytokinins induce RNA synthesis.
Cytokinin and auxin ratio and their interactions control morphogenesis and cell differentiation.

Question 12.
Discuss about physiological effects and applications of ethylene.
Answer:

Ethylene is a gaseous growth inhibitor hormone.
It promotes ripening of fruits like bananas, apples and mangoes. The commercial application of ethephon is done.
It initiates growth of lateral roots.
Dormancy of buds and seeds is broken by its application.
It accelerates formation of abscission layer and thus abscission of leaves, flowers and fruits is observed.
Ethylene is responsible for checking growth of lateral buds thus causes apical dominance and retards flowering.
Process of senescence of plant organs is enhanced.
Epinasty, i.e. drooping of leaves and flowers results due to its application in some plants.
It increases activity of chlorophyllase enzyme causing degreening effect in banana and Citrus fruits.

Question 13.
Discuss about experiment of Hendricks and Borthwick for discovery of phytochromes.
Answer:

Phytochrome pigments receive photoperiodic stimulus and control flowering in plants.
Hendricks and Borthwick observed that in SDP flowering is inhibited if continuous dark period is interrupted even by a short duration or flash of red light of wavelength 660 nm.
If this interruption is again exposed to flash of far red light of wavelength 730 nm, then these plants flower.
From this they concluded that some pigment system in plant receives the photoperiodic stimulus.
These pigment proteins are called phytochromes and it exists in two interconvertible forms – P_r and P_fr.
These pigments are located in cell membranes of green cells.
P_fr is biologically active form and during daytime it gets accumulated in the plant.

Question 14.
Give schematic representation of nitrogen cycle and enlist important steps of this cycle.
Answer:

Atmospheric nitrogen is a source of nitrogen cycle.
The important steps of the cycle are Nitrogen fixation, Nitrification, Ammonification, Nitrogen assimilation by plants. Amino acid synthesis and amidation, Denitrification and sedimentation.
Amino acid are building blocks of proteins. Amides are amino acid with two amino groups.
Schematic representation of Nitrogen

Question 15.
What is nitrogen cycle? Describe it briefly.
Answer:

The cyclic movement of nitrogen between the atmosphere, biosphere and geosphere in different forms is called nitrogen cycle.
Nitrogen cycle is one of the most important biogeochemical cycles.
The nitrogen cycle involves many processes such as cycling of nitrogen through the biosphere, atmosphere and geosphere, nitrogen fixation, nitrogen uptake, formation of biomass, ammonification, nitrification and denitrification, etc.
Bacteria such as Nitrosomonas, Nitrosococcus and Nitrobacter are the nitrifying bacteria which play an important role in nitrification.
Denitrification is carried out by the bacteria Pseudomonas denitrifficans. From this it is obvious that bacteria play a major role in nitrogen cycle.
Nitrogen fixation occurs by physical, industrial and biological methods. Prokaryotic organism play an important role in biological nitrogen fixation.

Maharashtra Board Class 12 Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

July 15, 2024July 14, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization ?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

It is a process of maturation of cells derived from apical meristems.
Differentiation is a permanent change in structure and function of cells that leads to its maturation.
Cell undergoes major anatomical and physiological change during differentiation process.
In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
The cells mature to perform specific function.
Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Arithmetic growth	Geometric growth
1. In arithmetic growth only one daughter cell continues to divide, while the other undergoes differentiation and maturation.	1. In geometric growth both the daughter cells continue to divide and redivide again and again.
2. Rate of growth is constant.	2 Rate growth is initially slow but later on rapid rate.
3. Linear curve is obtained.	3. Exponential curve is obtained.
4. Mathematical expression is L_t = Lo + rt whereL_t = length of time ‘t’ L_o = Length at time zero r_t = growth rate, t = time of growth	4. Mathematical expression is W_t = Wo^{e rt} where, W_t = final size, Wo = initial size, r = growth rate, t = time of growth E = base of natural logarithm
5. e.g. Elongation of root	5. e.g. Divisions of zygote during embryo development.

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

Due to chilling treatment crops can be produced earlier.
Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

When growth occurs in plants three distinct phases of growth are noticed.
Phase of cell formation is first phase where meristematic cells divide and new cells added.
In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
In phase of cell maturation cells get differentiated.
When we compare the growth rate it differs in these three phases.
In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
In stationary phase of maturation growth rate slows down and comes to steady state.
When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

Effect of light duration on flowering of plants is known as photoperiodism.
Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
These bacteria are chemoautotrophic and utilize these processes for their metabolism.
Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

Plants absorb mineral nutrients from their surroundings.
For a proper growth of plants about 35 to 40 different elements are required.
Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–, Sulphur as SO₄^2- etc.
Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
C, H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.

Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation

July 6, 2024July 5, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 6 Plant Water Relation

Multiple Choice Questions

Question 1.
What is the reason behind various properties of water?
(a) Physical state
(b) Hydrogen bonding
(c) Colour
(d) Fluidity
Answer:
(b) Hydrogen bonding

Question 2.
Which special type of tissue is present in epiphytic roots?
(a) Velamen
(b) Lenticel
(c) Aerenchyma
(d) Haustoria
Answer:
(a) Velamen

Question 3.
In which zone of root we come across root hairs?
(a) meristematic region
(b) zone of elongation
(c) zone of absorption
(d) maturation zone
Answer:
(c) zone of absorption

Question 4.
In double layered cell wall of root hair, outer layer is of ……………..
(a) cellulose
(b) pectin
(c) cutin
(d) suberin
Answer:
(b) pectin

Question 5.
What is rhizosphere?
(a) Root ball formed by growth of roots
(b) Covering of root tip
(c) Microenvironment around root
(d) region of root hairs
Answer:
(c) Microenvironment around root

Question 6.
A root hair is derived from …………….. cell.
(a) epidermal
(b) cortical
(c) endodermal
(d) pericycle
Answer:
(a) epiderinal

Question 7.
When we keep raisins in water they swell up, due to ……………..
(a) exosmosis
(b) plasmolysis
(c) imbibition
(d) diffusion
Answer:
(c) imbibition

Question 8.
This phenomenon is associated with exchange of gases ……………..
(a) osmosis
(b) diffusion
(c) imbibition
(d) plasmolysis
Answer:
(b) diffusion

Question 9.
What is correct expression for diffusion pressure deficit (D.PD.)?
(a) D.PD. = O.R – S.E
(b) D.PD. = T.E – O.E
(c) D.PD. = W.E – T.R
(d) D.PD. = O.R – T.R
Answer:
(d) D.PD. = O.P – T.P

Question 10.
What is true for a turgid cell?
(a) T.R is zero
(b) T.R = S.R
(c) D.P.D. is zero
(d) W.R = S.R
Answer:
(c) D.P.D. is zero

Question 11.
The particles which easily diffuse through cell membrane are ……………..
(a) lipid soluble
(b) water soluble
(c) hydrophilic
(d) lipophobic
Answer:
(a) lipid soluble

Question 12.
Which proteins help in facilitated diffusion process?
(a) cutin
(b) mucin
(c) aquaporins
(d) lipoproteins
Answer:
(c) aquaporins

Question 13.
D.PD. is now known as ……………..
(a) water potential
(b) solute potential
(c) pressure potential
(d) osmotic potential
Answer:
(a) water potential

Question 14.
Water always flows from ……………..
(a) more negative water potential to less negative water potential
(b) high water potential area to low water potential area
(c) low water potential area to high water potential area
(d) negative water potential area to area of zero water potential
Answer:
(b) high water potential area to low water potential area

Question 15.
Plasmolysed cell becoming turgid is process of ……………..
(a) replasmolysis
(b) incipient plasmolysis
(c) deplasmolysis
(d) exosmosis
Answer:
(c) deplasmolysis

Question 16.
In a fully turgid cell ……………..
(a) T.P = O.P
(b) T.P = S.P
(c) O.P = S.P
(d) T.RP = 0 (zero)
Answer:
(a) T.P = O.P

Question 17.
Casparian strip of endodermis has material ……………..
(a) pectin
(b) suberin
(c) cutin
(d) porin
Answer:
(b) suberin

Question 18.
Water from pericycle is forced into xylem due to ……………..
(a) aquaporin
(b) plasmodesmata
(c) root pressure
(d) ion-channels
Answer:
(c) root pressure

Question 19.
Water absorbed from root hair when passes through intercellular spaces and cell wall it is …………….. pathway.
(a) apoplast
(b) symplast
(c) transmembrane
(d) plasmodesmata
Answer:
(a) apoplast

Question 20.
In root system, secondary roots arise from ……………..
(a) cortical cells
(b) endodermis
(c) pericycle
(d) passage cell
Answer:
(c) pericycle

Question 21.
Active absorption of water occurs during ……………..
(a) early morning
(b) daytime
(c) bright sunlight
(d) night time
Answer:
(d) night-time

Question 22.
The value of root pressure is usually about ……………..
(a) +1 to +12 bars
(b) -1 to +1 bars
(c) +1 to +2 bars
(d) +1 to +21 bars
Answer:
(c) +1 to +2 bars

Question 23.
The ascent of sap in plants takes place through ……………..
(a) xylem
(b) phloem
(c) parenchyma
(d) endodermis
Answer:
(a) xylem

Question 24.
From the following, which water will show greatest water potential (ψ)
(a) pure water
(b) salt water
(c) sugar water
(d) salt + sugar + water
Answer:
(a) pure water

Question 25.
Due to entry of water in a cell, the pressure potential ……………..
(a) increases
(b) decreases
(c) remains unaffected
(d) becomes zero
Answer:
(a) increases

Question 26.
Which mineral element is not remobilized in plants?
(a) P
(b) K
(c) S
(d) Ca
Answer:
(d) Ca

Question 27.
Phloem sap analysis is done using isotope ……………..
(a) ¹⁶O
(b) ¹⁴C
(c) ¹⁵N
(d) ³⁵S
Answer:
(b) ¹⁴C

Question 28.
In plants, food is mainly translocated in form of ……………..
(a) starch
(b) glucose
(c) sucrose
(d) amino acids
Answer:
(c) sucrose

Question 29.
Transport of food through phloem is ……………..
(a) unidirectional
(b) bidirectional
(c) three dimensional
(d) absent
Answer:
(b) bidirectional

Question 30.
Guttation occurs through ……………..
(a) stomata
(b) lenticels
(c) hydathodes
(d) velamen
Answer:
(c) hydathodes

Question 31.
Amount of cuticular transpiration is about ……………..
(a) 0.1-1%
(b) 8 – 10%
(c) 90 – 93%
(d) 2 – 8%
Answer:
(b) 8 – 10%

Question 32.
Epistomatic leaf shows ……………..
(a) stomata on upper side
(b) stomata on lower side
(c) stomata on both surfaces
(d) absence of stomata
Answer:
(a) stomata on upper side

Question 33.
Dumbbell shaped guard cells are found in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(b) grasses

Question 34.
Stomatal transpiration occurs during night time in ……………..
(a) most dicots
(b) grasses
(c) gymnosperms
(d) desert plants
Answer:
(d) desert plants

Question 35.
Reservoir of K⁺ ions is ……………..
(a) guard cells
(b) epidermal cells
(c) subsidiary cells
(d) mesophyll
Answer:
(c) subsidiary cells

Question 36.
Guard cells are surrounded by ……………..
(a) epidermal hairs
(b) mesophyll cells
(c) accessory cells
(d) lenticels
Answer:
(c) accessory cells

Question 37.
When guard cells close stomata at night which acid prevents uptake of K⁺ and C^– ions?
(a) Abscissic acid
(b) Pyruvic acid
(c) Malic acid
(d) Acetic acid
Answer:
(a) Abscissic acid

Question 38.
Which type of injury is noticed in plants if there is excessive transpiration?
(a) Burning
(b) Chlorosis
(c) Necrosis
(d) Wilting
Answer:
(d) Wilting

Question 39.
Maximum transpiration occurs through ……………..
(a) cuticle
(b) lenticels
(c) stomata
(d) bark
Answer:
(c) stomata

Question 40.
What will be the condition of guard cells during night-time?
(a) Show increased turgor pressure
(b) Become flaccid
(c) Increase uptake of K⁺ and Cl^– ions
(d) Starch converted to sugar
Answer:
(b) Become flaccid

Question 41.
Cell A has water potential – 10 bars and cell B has – 5 bars, the movement of water will occur from ……………..
(a) A to B
(b) B to A
(c) No movement
(d) Either A to B or B to A
Answer:
(b) B to A

Question 42.
What is the correct symbol and unit of water potential?
(a) ψ and ha
(b) ω and ha
(c) ω and Pa
(d) ψ and Pa
Answer:
(d) ψ and Pa

Question 43.
When root system absorbs water, which is the first physical process concerned with this activity?
(a) Osmosis
(b) Imbibition
(c) Facilitated diffusion
(d) Diffusion
Answer:
(b) Imbibition

Match the columns

Question 1.

Column A (Scientist)	Column B (Theory)
(1) Munch	(a) Proton transport theory
(2) Bohem	(b) Pressure flow theory
(3) J. Pristley	(c) Capillary theory
(4) Levitt	(d) Root Pressure theory

Answer:

Column A (Scientist)	Column B (Theory)
(1) Munch	(b) Pressure flow theory
(2) Bohem	(c) Capillary theory
(3) J. Pristley	(d) Root Pressure theory
(4) Levitt	(a) Proton transport theory

Question 2.

Column A (Scientist)	Column B (Theory)
(1) Dixon and Joly	(a) Starch-sugar inter conversion
(2) Steward	(b) Osmotic absorption theory
(3) Atkins and Pristley	(c) Non-osmotic absorption theory
(4) Kramer and Thimann	(d) Cohesion tension theory

Answer:

Column A (Scientist)	Column B (Theory)
(1) Dixon and Joly	(d) Cohesion tension theory
(2) Steward	(a) Starch-sugar inter conversion
(3) Atkins and Pristley	(b) Osmotic absorption theory
(4) Kramer and Thimann	(c) Non-osmotic absorption theory

Question 3.

Column A	Column B (New Terminology)
(1) D.P.D.	(a) Osmotic potential
(2) O.E	(b) Pressure potential
(3) T.P	(c) Water potential

Answer:

Column A	Column B (New Terminology)
(1) D.P.D.	(c) Water potential
(2) O.E	(a) Osmotic potential
(3) T.P	(b) Pressure potential

Classify the following to form Column B as per the category given in Column A

Question 1.
B, Co, Mn, P, N, S

Column A	Column B
Macro elements	——–, ———-, ——–
Micro elements	——–, ———-, ——–

Answer:

Column A	Column B
Macro elements	P, N, S
Micro elements	B, Co, Mn

Question 2.
Capillary water, Combined water, Hygroscopic water, Gravitational water.

Column A	Column B
(1) Water adsorbed on soil particles	—————
(2) Water penetrated deep in soil	————–
(3) Water present as hydrated oxides	————-
(4) Water present between soil particles	————-

Answer:

Column A	Column B
(1) Water adsorbed on soil particles	Hygroscopic water
(2) Water penetrated deep in soil	Gravitational water
(3) Water present as hydrated oxides	Combined water
(4) Water present between soil particles	Capillary water

Very short answer question

Question 1.
Why water acts as a thermal buffer?
Answer:
The water has high specific heat, high heat of vaporization and high heat of fusion hence it acts as a thermal buffer.

Question 2.
Why water can easily rise in capillaries?
Answer:
Water has high surface tension and high adhesive and cohesive force hence it can easily rise in capillaries.

Question 3.
Enlist the kind of water available in soil environment.
Answer:
Soil environment has gravitational water, hygroscopic water, combined water and capillary water.

Question 4.
Mention examples of imbibition process.
Answer:
Soaking of seeds, swelling up of raisins, kneading of flour and warping of wooden doors in rainy season.

Question 5.
What is the mechanism of imbibition?
Answer:
When imbibition takes place water molecules (imbibate) get tightly adsorbed on imbibant without formation of solution.

Question 6.
Why water enters plant cell by process of diffusion?
Answer:
Water present around cell wall has more diffusion pressure than inner cell sap hence water moves in the cell through freely permeable cell wall by diffusion.

Question 7.
What is isotonic condition in osmotic system?
Answer:
A condition where concentration of solution has neither gain nor loss of water in an osmotic system is isotonic.

Question 8.
What is effect on protoplasm when cell is plasmolysed ?
Answer:
When cell is plasmolysed, protoplast of cell shrinks and recedes from the cell wall thus gap is observed between cell wall and protoplast.

Question 9.
What is apoplast pathway?
Answer:
When water absorbed by root hair passes across the roots through cell wall and intercellular spaces of root cortex it is called apoplast pathway.

Question 10.
Which experiment is a proof for existence of root pressure?
Answer:
When a stem of potted plant is cut above the soil, xylem sap is seen oozing through cut end, that proves presence of root pressure.

Question 11.
Which analysis indicates that minerals are absorbed by plants?
Answer:
Analysis of plant ash contents is indication of absorbed minerals.

Question 12.
Which ions are readily remobilized in plants?
Answer:
Ions of phosphorus, sulphur and nitrogen are remobilized from older plants, parts (leaves) to younger parts.

Question 13.
What is radial and tangential translocation?
Answer:
When lateral translocation of food occurs in root or stem, transport from phloem to pith is radial translocation while that from phloem to cortex is tangential translocation.

Question 14.
What is loading of vein?
Answer:
In turgid cell, due to increased turgor pressure of photosynthetic cell, sugar is forced into the sieve tube of the vein, which is known as loading of vein.

Question 15.
What is unloading of vein?
Answer:
At the sink end, turgor pressure is lowered and hence turgor pressure gradient is developed from sieve-tube which translocates food passively along concentration gradient, this is vein unloading.

Question 16.
What is peculiarity of wall of guard cells?
Answer:
The inner wall of guard cell that faces opening is thick and inelastic while its outer wall or lateral wall is thin and elastic.

Question 17.
Give reaction of starch-sugar interconversion theory.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 1

Question 18.
How do land plants absorb water and mineral salts from the soil?
Answer:
Land plants absorb water and mineral salts from soil with the help of their roots.

Question 19.
In which form the water is lost from leaves to the atmosphere?
Answer:
The water is lost in the form of vapour from leaves to the atmosphere.

Question 20.
How does plants lift the water from soil up to canopy without any pump?
Answer:
Plants have vascular tissue system of xylem, mainly through vessels and tracheids. Water is conducted upwards due to pull created in this continuous channel of water.

Question 21.
What is a peculiarity of epiphytic plants like orchids?
Answer:
The epiphytic plants like orchids have epiphytic roots with special water vapour absorbing layer of velamen tissue that absorbs water vapour from air.

Question 22.
What are root hairs?
Answer:
Root hairs are the extensions of epiblema or epidermal cells in the region of absorption.

Question 23.
Why do the wooden doors become very hard to close and open in rainy season?
Answer:
During rainy season wooden doors swell up due to imbibition of water. The humidity or moisture content of air increases and wood which is hydrophilic takes up this moisture.

Question 24
How does the water come out through the surface of porous earthen pot?
Answer:
Earthen pot has tiny pores through which water diffuses out. Due to evaporation of this water from surface water kept inside becomes cool.

Question 25.
Describe external structure of root hair.
Answer:
A root hair is a cytoplasmic extension of epiblema cell which is colourless, unbranched, ephemeral, very delicate tubular structure of about 1 to 10 mm long.

Question 26.
What is gravitational water?
Answer:
Water present in the soil that percolates deep inside due to gravitational force is called gravitational water.

Question 27.
How can we describe osmotic movement of water related to energy?
Answer:
The osmotic movement of water is on the basis of free energy which is used to do work.

Question 28.
Which process is responsible for transport of minerals and to make them available to cells?
Answer:
The minerals absorbed by roots are transported upwards through sap and from veins by process of diffusion cells uptake them.

Question 29.
Give examples of vertical translocation of food in downward direction and upward direction.
Answer:
Food is translocated in downward direction from leaves (source) to root, while during germination of seed, bulbil, corm it is in upward direction.

Question 30.
What is translocation of food?
Answer:
The transport of food from one part of plant to the other part, i.e. source to sink is called translocation of food.

Question 31.
Which plant tissues are involved in transport of water, minerals and food?
Answer:
A complex plant tissue xylem, mainly with its tracheids and vessels is involved in transport of water and minerals while complex tissue phloem with sieve tubes and companion cells is involved in transport of food.

Question 32.
Which is a direct pathway of transport available in plants through root system?
Answer:
Secondary roots that originate from pericycle that is outer layer of vascular cylinder, bypass endodermis with Casparian strip allow direct apoplast pathway to enter xylem and phloem.

Question 33.
For which type of plants root pressure theory of ascent of sap is applicable?
Answer:
Root pressure theory is applicable to plants having height up to 10 to 20 metres.

Question 34.
What is hydrogen bond?
Answer:
Hydrogen bond results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom like O or N.

Question 35.
When you burn an incense stick in one corner of room, its fragrance spreads all over the room in a short time. How does it happen?
Answer:
The process of diffusion is responsible for spreading the fragrance of an incense stick in one corner of room.

Give definitions of the following

Question 1.
Facilitated diffusion
Answer:
The passive absorption of solutes when mediated by a carrier is called facilitated diffusion.

Question 2.
Osmotic pressure / Osmotic potential
Answer:
The pressure exerted due to osmosis is osmotic pressure now termed as osmotic potential.

Question 3.
Deplasmolysis
Answer:
When plasmolysed cell is placed in hypotonic solution, endo-osmosis occurs making cell turgid again then this is called deplasmolysis.

Question 4.
Translocation of food
Answer:
The movement of food from one part of plant to the other part is called translocation of food.

Name the following

Question 1.
Condition of cell wall and cell membrane : based on permeability.
Answer:
Cell wall freely permeable and Cell membrane semipermeable.

Question 2.
Weak solution having low osmotic concentration.
Answer:
Hypotonic

Question 3.
Strong solution having high osmotic concentration.
Answer:
Hypertonic

Question 4.
A suberised layer on endodermis.
Answer:
Casparian strip

Question 5.
One example of each epistomatic, hypostomatic and amphistomatic leaf.
Answer:

Epistomatic leaf – Lotus
Hypostomatic – Nerium.
Amphistomatic – Grass

Question 6.
Pathway for entry of water into xylem from endodermis.
Answer:
Symplast pathway

Question 7.
A material deposited on endodermis which forms barrier.
Answer:
Suberin

Question 8.
Water imbibed or adsorbed on soil particles.
Answer:
Hygroscopic water

Question 9.
Water potential was previously known as.
Answer:
D.ED. or Diffusion pressure deficit

Question 10.
Free energy per molecule in chemical system.
Answer:
Chemical potential

Question 11.
The substance synthesized in cell responsible for increasing osmotic concentration as per Munch theory.
Answer:
Glucose

Question 12.
A waxy substance present in layer on outer surface of epidermis.
Answer:
Cutin in layer cuticle

Question 13.
Anatomical structure through which guttation occurs.
Answer:
Hydathode

Question 14.
Type of leaves in hydrophytes based on distribution of stomata.
Answer:
Epistomatic

Question 15.
Type of leaves in xerophytes based on distribution of stomata.
Answer:
Hypostomatic

Question 16.
The form in which food is transported in plants and then stored in plants.
Answer:
Transported in the form of sugar, sucrose and stored as starch.

Question 17.
Chief vascular element concerned with transport of food in plant.
Answer:
Sieve tubes of phloem.

Give functions/significance of the following

Question 1.
Zone of absorption of root.
Answer:
This zone has unicellular root hairs which absorb water available in rhizosphere.

Question 2.
Diffusion.
Answer:

Necessary for absorption of water by root hairs
For absorption of minerals
Conduction of water against gravity
Exchange of gases
Transport and distribution of food.

Question 3.
Turgor pressure.
Answer:

Keeps cells and cell organelles stretched
Provides support
During growth essential for cell enlargement
Maintains shape of cell
Facilitates opening and closing of stomata

Question 4.
Osmosis.
Answer:

Absorption of water into root
Maintains turgidity of cell
Facilitates cell to cell movement of absorbed water
Resistance to drought or frost injury
Helps in drooping movement of leaflet.
E.g. Touch me not plant

Question 5.
Root pressure.
Answer:
Hydrostatic pressure developed in living cells of root helps in forcing water from pericycle into xylem
Helps in upward conduction of water against gravity

Question 6.
Transpiration.
Answer:

Removal of excess of water
Helps in passive absorption of water and minerals
Helps in ascent of sap – transpiration pull
Maintains turgor of cells
Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 7.
Guard cells.
Answer:

They contain few chloroplasts so capable of photosynthesis.
They can change their size hence opening and closing of stomatal aperture.

Distinguish between the following

Question 1.
Diffusion and Osmosis.
Answer:

Diffusion	Osmosis
1. Diffusion is the movement of ions, atoms or molecules of solute, liquids or gases.	1. Osmosis Is the special type of diffusion of water or solvent.
2. Diffusion involves the flow of diffusing particles in both the directions.	2. Osmosis involves the unidirectional flow of solvent molecules.
3. Diffusion does not occur through a semi- permeable membrance.	3. Osmosis occurs through a semi-permeable membrance.
4. Diffusion occurs from a place of higher concentration of diffusing particles to a place of lower concentration.	4. Osmosis occurs from a solution of lower concentration to a solution of higher concentration.

Question 2.
Active absorption and Passive absorption.
Answer:

Active absorption	Passive absorption
1. Active absorption is a physiological process.	1. Passive absorption is a physical process.
2. Active absorption takes place by the process of osmosis.	2. Passive absorption takes place by suction force.
3. It involves the expenditure of energy on the part of absorbing cells.	3. It does not involve the expenditure of energy on the part of absorbing cells.
4. Active absorption takes place against the concentration gradient.	4. Passive absorption takes place along the concentration gradient.

Question 3.
Cuticular transpiration and Stomatal transpiration.
Answer:

Cuticular transpiration	Stomatal transpiration
1. Cuticular traspiration takes place through the cuticle.	1. Stomatal traspiration takes place through the stomata.
2. Cuticular transpiration accounts for 8 to 10% of total loss of water from plants.	2. Stomatal transpiration accounts for 80 to 90% of total loss of water from plants.
3. Cuticular transpiration depends upon the thickness of the cuticle.	3. Stomatal traspiration depends upon the number and size of stomata.

Give reasons or explain the statements

Question 1.
Water is considered as ‘elixir of life’.
Answer:

Water plays an important role in living organisms.
About 90 – 95% water is present in cell which is functional and structural unit of living organisms.
It helps in maintaining turgidity and shape of cells and cell organelles.
Due to its various properties, it is medium of biochemical reactions, transporting medium and thermal buffer also.
Therefore it is absolutely necessary for life i.e. ‘elixir of life’.

Question 2.
Water is a best transporting medium.
Answer:

Water is in liquid state at room temperature.
It is best solvent for most of the solutes. Thus called universal solvent.
It is inert inorganic compound with neutral pH i.e. pH 7 when in pure form.
Hence it is best medium for dissolved minerals.

Question 3.
Water is significant molecule that connects physical world with biological processes.
Answer:

Water is an important constituent of cell. About 90 – 95% of protoplasm is water.
Water in liquid state is best solvent in which various minerals and food molecules are dissolved and transported.
Water acts as the thermal buffer has high specific heat.
Water molecules have high adhesive and cohesive forces of attraction.
It can rise in capillaries due to high surface tension and adhesive forces, e.g. Ascent of sap in plants.
Due to all these important factors it is a significant molecule connecting physical world with biological processes.

Question 4.
Absorption of water by roots from soil is with physical processes inhibition, diffusion and osmosis.
Answer:

Water is absorbed from rhizosphere with the help of unicellular root hairs.
Root hairs have plasma membrane and thin, double layer cell wall of pectin and cellulose.
During imbibition water molecules get tightly adsorbed to the wall of hydrophilic colloids.
Cell wall is freely permeable membrane hence through diffusion water passes into the cell.
Osmosis is a special kind of diffusion of solvent through a semipermeable membrane and as plasma membrane is semipermeable, water enters cell by osmotic mechanism.
Hence all these three physical processes occur sequentially when water is absorbed by roots.

Question 5.
Additional apoplastic pathway through secondary roots is beneficial to plants.
Answer:

Secondary roots develop from the pericycle which is inside endodermis.
Protoxylem is situated close to pericycle in root.
Endodermis have suberized layer Casparian strip which forces water to move in the symplast so that it can enter the vascular xylem.
Since secondary roots originate form pericycle, they bypass the Casparian strip.
Therefore, a direct pathway to xylem and phloem is available without moving into symplast.

Write shorts notes

Question 1.
Role of water or Biological importance of water.
Answer:

Water is absolutely necessary for life.
It is a major constituent of protoplasm.
If provides aqueous medium for various metabolic reactions that take place in plant.
It is raw material for photosynthesis.
It helps in maintaining the turgidity of cells.
It is excellent solvent for various organic materials.
It is transporting medium for dissolved minerals.
It is thermal buffer.

Question 2.
Properties of water.
Answer:

Water is a compound and it is in liquid state at room temperature.
It is an inert inorganic compound with neutral pH.
It has high specific heat, high heat of vapourization, high heat of fusion.
It has high surface tension.
Water molecule has good adhesive and cohesive forces of attraction.
The various properties of water are result of weak hydrogen bonding between the water molecules.

Question 3.
Lenticular transpiration.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 2

Small, raised structures, which are composed of loose cells with large intercellular spaces situated on bark are lenticels.
They are observed on bark of old stem, root and on woody pericarp of fruits.
Lenticular transpiration is of very small amount about 0.1% to 1% of total transpiration.
It is very slow process but occurs throughout the day.

Question 4.
Structure of stomatal apparatus.
Answer:

Stomata are minute openings mainly located in the epidermal surfaces of young stem and leaves.
It is composed of two guard cells and accessory cells which form the opening stomatal pore for transpiration.
Guard cells are kidney shaped in dicot plants and dumbbell shaped in grasses.
Guard cells have unevenly thickened wall, inner wall thick, inelastic while outer wall is thin and elastic.
Guard cells are nucleated cells with few chloroplasts and hence can perform photosynthesis.
Accessory cells are specialized cells that surround the guard cells and are reservoirs of K⁺ ions.

Question 5.
Apoplast pathway.
Answer:

The movement of water across the root cells occur by two pathways viz. apoplast and symplast.
Apoplast movement of water in plants occurs exclusively through cell walls and intercellular spaces in living cells of cortex.
Apoplast pathway is up to the endodermis as the suberized layer of Casparian strip obstructs the movement.
Additional apoplastic route giving direct contact to xylem and phloem is through secondary roots which originate at pericycle and bypass endodermis.
The cellulosic walls of root hair, cortical cells are hydrophilic and permeable to water.

Question 6.
Water potential.
Answer:

Chemical potential of water is called water potential.
The Greek letter (ψ) psi represents water potential.
The unit of measurement is in bars/pascals (Pa) or atmosphere.
Water potential of pure water is always zero.
Water potential of protoplasm is equal to D.PD. but it has a negative value.
Water always moves from less negative potential to more negative water potential.

Short answer questions

Question 1.
Enlist macronutrients and micronutrients required for plant growth.
Answer:

Some minerals which are required in large amounts for plant metabolism are macronutrients. E.g. C, H, O, E N, S, Mg, Ca, K. etc.
Some minerals which are required in small amounts for plant metabolism are micronutrients. E.g. Cu, Co, Mn, B, Zn, Cl, etc.

Question 2.
Explain about the factors that affect water absorption.
Answer:

Water is absorbed by unicellular root hairs from soil.
Presence of capillary water in soil is needed as this water from soil is absorbed by root hairs.
Soil temperature of 20-30°C favours water absorption.
Rate of absorption is decreased by high concentration of solutes in soil.
Soil should be properly aerated, poorly aerated soil shows poor absorption rate.
Increased transpiration accelerates rate of absorption of water.

Question 3.
Explain mechanism of sugar transport through phloem.
Answer:

The part of plant where food is synthesized is known as source and where it is utilized is sink.
Food is translocated through phloem tissue in the form of sucrose along the concentration gradient from source to sink.
Munch’s pressure flow theory or mass flow hypothesis is widely accepted theory for sugar transport.
Glucose is synthesized in photosynthetic cells hence endo-osmosis occurs due to increased osmotic concentration.
With this turgor pressure increases and sugar is forced in sieve tube of vein from photosynthetic cell – Vein loading.
At the sink, sugar is utilized and excess amount converted to starch, hence osmotic concentration is lowered and exosmosis takes place.
Turgor pressure is lowered hence turgor pressure gradient is set and food is translocated passively against concentration gradient – Vein unloading.

Question 4.
Explain the principles of cohesion tension theory and its limitations?
Answer:

Cohesion tension theory is widely accepted theory for translocation of water proposed by Dixon and Joly.
It is based on principles cohesion and adhesion of water molecules along with transpiration pull.
Strong attractive force between water molecules is cohesive force and the strong force of attraction between water molecules and wall of lumen of xylem vessels is called adhesive force.
Owing to these cohesive and adhesive forces a continuous column of water is maintained in xylem elements from root to aerial parts, tips of leaves.
Transpiration pull is developed in the xylem vessel in leaves.
This pull is transmitted downwards and due to suction pressure, water is pulled passively against gravity, i.e. ascent of sap.

Limitations of theory-

For activity of transpiration pull, water column should be maintained constantly and continuously. Owing to changes in temperature during day and night, gas bubbles may be formed in the water channel. This will break the continuity.
According to this theory, tracheids are more efficient than vessels because of their tapering end walls which support water column. Vessels are tubular structure with open ends.

Question 5.
What is the meaning of specific heat, heat of vaporization and heat of fusion?
Answer:

Specific heat : The specific heat is the amount of heat per unit mass required to raise the temperature by one degree. The specific heat of water is 1 calorie/gram degree C.
Heat of vapourization : It is the amount of energy that must be added to a liquid substance to transform a quantity of that substance to gas. It is also known as heat of evaporation.
Heat of fusion : It is amount of energy typically heat, provided to a specific quantity of the substance to change its state from solid to liquid at constant pressure.

Question 6.
What are adhesive and cohesive forces?
Answer:

Adhesive and cohesive forces are attractive forces between molecules of same substance.
Cohesive forces exist between molecules of the same type. E.g. between water-water molecule.
Adhesive forces exist between dissimilar molecules. E.g. water molecule and lignin deposited wall of xylem element.

Chart based/Table based questions

Question 1.
Complete the table based on types of solution.
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 3
Answer:

Question 2.
Complete the table.

Event	Physical process
1. Soaking of Seeds	—————-
2. Water entering guard cells	—————-
3. Exchange of gases	—————
4. Loss of water in liquid form	—————
5. Water coming out through earthen pot	—————
6. Loss of water in vapour form	—————
7. Absorption of solutes passively by carrier	—————
8. Spreading of fragrance of incense stick	—————

Answer:

Event	Physical process
1. Soaking of Seeds	Imbibition
2. Water entering guard cells	Osmosis
3. Exchange of gases	Diffusion
4. Loss of water in liquid form	Guttation
5. Water coming out through earthen pot	Diffusion
6. Loss of water in vapour form	Transpiration
7. Absorption of solutes passively by carrier	Facilitated Diffusion
8. Spreading of fragrance of incense stick	Diffusion

Diagram based questions

Question 1.
Zones of root
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 5

Question 2.
Structure of root hair
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 6

Question 3.
Diffusion of water
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 7

Question 4.
Pathway for water uptake
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 8

Question 5.
L. S. of sieve tube (Transport of food)
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 9

Question 6.
Structure of stomata and Types of guard cells
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 6 Plant Water Relation 10

Long answer questions

Question 1.
Write a note on macronutrients and micronutrients required for plant growth.
Answer:

Plants absorb mincral nutrients from their surroundings.
For a proper growth of plants about 35 to 40 different elements are required.
Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–. Sulphur as SO₄^2- etc.
Based on their requirement in quantity. they are classified as major nutrients or macronutrients and those needed In small amounts are minor or micronutrients.
Macroclements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O. R Mg. N, K. S and Ca. -Ca pectate cell wall component. Mg component of chlorophyll.
C. H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
Microelements arc required In traces as they mainly have catalytic role as co-factors or activators of enzymes.
Microelernents may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
The Important micronutrients for plant growth are Mn. B. Cu, Zn, Cl.

Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation

July 2, 2024July 1, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 6 Plant Water Relation

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
Passive absorption is the chief method of absorption (98%).
There is no expenditure of energy in passive absorption.
Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
It occurs during daytime when there is active transpiration.
Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
Active absorption may be osmotic or non- osmotic type.
For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
The movement is against the gravity and described as ascent of sap.
The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

The loss of water in the form of vapour is called transpiration.
Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
Opening and closing of stomata is controlled by turgidity of guard cells.
When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

Removal of excess of water
Helps in passive absorption of water and minerals
Helps in ascent of sap – transpiration pull
Maintains turgor of cells
Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

Root pressure theory is proposed by J. Pristley.
For translocation of water, activity of living cells of root is responsible.
Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
Root pressure is an osmotic phenomenon.

Limitation of this theory:

Not applicable to tall plants above 20 metres.
Even in absence of root pressure ascent of sap is noticed.
In actively transpiring plants, root pressure is not developed.
In taller gymnosperms, root pressure is zero.
Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

Capillarity theory of water translocation is proposed by Bohem.
Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
Xylem vessels and tracheids are tubular elements having their lumen.
In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

The loss of water in the form of water vapour is called transpiration.
About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
The process is necessary evil because water which is important for plant is lost in the process.
At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
Root hair cell becomes flaccid and ready to absorb soil water.
Water is passed on similarly in inner cortical cells.
Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R)	Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure.	1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis.	2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane.	3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

Soil is the chief source of minerals for the plants.
Minerals get dissolved in the soil water.
Minerals are absorbed by the plants in the ionic form mainly through roots.
Absorption of minerals is independent of water.
Absorbed minerals are pulled upwards along with xylem sap.
Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:
Maharashtra Board Class 12 Biology Solutions Chapter 6 Plant Water Relation 1

Water from soil is absorbed by plants with the help of root hairs.
Root hairs are present in zone of absorption.
Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
Root hairs are nothing but cytoplasmic extensions of epiblema cell.
Root hairs are long tube like structures of about 1 to 10 mm.
They are colourless, unbranched and very delicate structures.
A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
Water molecules get adsorbed on cell wall of root hair (imbibition).
They enter the root hair cell by diffusion through cell wall which is freely permeable.
By process of osmosis they enter through plasma membrane which is semipermeable.
The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
The pathway of water is by apoplast and symplast.
When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
Ascent of sap occurs through lumen of xylem elements.
Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
When guard cells are flaccid that results in closure of stomata.
According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

Plants absorb minerals from the soil with their root system.
MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
Ions get accumulated in the root hair against the concentration gradient.
These ions pass into cortical cells and finally reach xylem of roots.
Along with the water these minerals are carried to other parts of plant.

Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life

June 22, 2024June 21, 2024 by Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 5 Origin and Evolution of Life

Multiple Choice Questions

Question 1.
Which statement is true for the theory of spontaneous generation?
(a) Life came from outer space.
(b) Life can arise from dead matter.
(c) Life can arise from non-living things only
(d) Life arises spontaneously by miracle
Answer:
(c) Life can arise from non-living things only

Question 2.
When man makes the selection during animal husbandry and plant breeding programmes, then it is an example of ……………….
(a) reverse evolution
(b) artificial selection
(c) mutation
(d) natural selection
Answer:
(b) artificial selection

Question 3.
Palaeontological evidences for evolution are ……………….
obtained in the form of
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs
Answer:
(c) fossils

Question 4.
The homologous organs such as bones of forelimbs of whale, bat, cheetah and man are similar in structure, because ……………….
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function
(d) they have biochemical similarities
Answer:
(b) they share a common ancestor

Question 5.
Which type of evolution gives rise to analogous organs?
(a) divergent evolution
(b) parallel evolution
(c) genetic drift
(d) convergent evolution
Answer:
(d) convergent evolution

Question 6.
Hardy Weinberg’s equation of genetic equilibrium i.e. (p + q)² = p² + 2pq + q² = 1 is used in ……………….
(a) population genetics
(b) Mendelian genetics
(c) biometrics
(d) molecular genetics
Answer:
(a) population genetics

Question 7.
Which type of rocks show maximum fossils ?
(a) Sedimentary rocks
(b) Igneous rocks
(c) Metamorphic rocks
(d) Any type of rock
Answer:
(a) Sedimentary rocks

Question 8.
Industrial melanism observed in moth, Biston bitularia shows ………………. type of natural selection.
(a) stabilising
(b) directional
(c) disruptive
(d) artificial
Answer:
(b) directional

Question 9.
Variations during mutations of meiotic recombinations are ……………….
(a) random and directionless
(b) random and directional
(c) small and directional
(d) random, small and directional
Answer:
(a) random and directionless

Question 10.
Theory of special creation is based on ………………. beliefs.
(a) scientific
(b) religious
(c) traditional
(d) mythological
Answer:
(b) religious

Question 11.
Which sentence holds true for theory of biogenesis?
(a) Living organisms arise from non-living things.
(b) Theory of biogenesis can explain origin of life.
(c) Continuity of life can be explained by this theory.
(d) This theory was disproved by Louis Pasteur.
Answer:
(c) Continuity of life can be explained by this theory

Question 12.
Who gave the Big-Bang theory that explained the origin of life ?
(a) Georges Lemaitre
(b) Oparin and Haldane
(c) Charles Darwin
(d) J.B.S. Haldane
Answer:
(a) Georges Lemaitre

Question 13.
Primitive atmosphere of the earth was of type ……………….
la) oxidizing
(b) reducing
(c) aerobic
(d) oxido-reducing
Answer:
(b) reducing

Question 14.
What is the meaning of protobiogenesis?
(a) The origin of life on the earth.
(b) The origin of protozoans on the earth.
(c) The origin of protists on the earth.
(d) The origin of protons of the earth.
Answer:
(a) The origin of life on the earth

Question 15.
What are the first form of life on the earth called ?
(a) Pre-cells or Protobionts
(b) Protoproteins
(c) Coacervates
(d) Chromophores
Answer:
(a) Pre-cells or Protobionts

Question 16.
Which of the following is a landmark in the origin of life ?
(a) Formation of oxygen
(b) Formation of carbohydrates
(c) Formation of proteins
(d) Formation of water
Answer:
(c) Formation of proteins

Question 17.
The first chemicals formed on the earth were ………………. etc.
(a) oxygen, CFC, ozone
(b) DNA. RNA and nucleotides
(c) salt, sugar, proteins
(d) water, ammonia, methane
Answer:
(d) water, ammonia, methane

Question 18.
The unique feature of hot dilute soup is that there was no free ………………. in it.
(a) nitrogen
(b) oxygen
(c) carbon
(d) sulphur
Answer:
(b) oxygen

Question 19.
On which plant did Hugo de Vries work during his experimentations?
(a) Hibiscus rosa sinensis
(b) Oenothera lamarkiana
(c) Mirabilis jalapa
(d) Pisum sativum
Answer:
(b) Oenothera lamarkiana

Question 20.
Which one out of the following is a connecting link between fish and amphibian ?
(a) Archaeopteryx
(b) Seymouria
(c) Ichthyostega
(d) Dinosaurus
Answer:
(c) Ichthyostega

Question 21.
Homologous organs always lead to ………………. evolution.
(a) convergent
(b) divergent
(c) parallel
(d) radiating
Answer:
(b) divergent

Question 22.
Find the odd one out
(a) Caecum
(b) Nictitating membrane
(c) Coccyx
(d) Sacrum
Answer:
(d) Sacrum

Question 23.
The most common types of fossils are ……………….
(a) moulds
(b) casts
(c) actual remains
(d) model
Answer:
(c) actual remains

Question 24.
In geological time scale which period showed dominance of reptiles ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(b) Jurassic

Question 25.
Which was the period of beginning of modern birds ?
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 26.
Which epoch showed mammals at height of evolution ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(c) Mlocene

Question 27.
When did Holocene began ? (mya = million years ago)
(a) 2 mya
(b) 1 mya
(c) 0.5 mya
(d) 0.01 mya
Answer:
(d) 0.01 mya

Question 28.
Cichlid fishes in Lake Victoria are representatives of ………………. type of speciation.
(a) allopatric
(b) sympatric
(c) isolation
(d) random
Answer:
(b) sympatric

Question 29.
Which is the offspring of male horse and female donkey?
(a) Mule
(b) Hinny
(c) Marino
(d) Baroque
Answer:
(b) Hinny

Question 30.
Which one out of the following is a living fossil ?
(a) Coelacanth
(b) Lung fish
(c) Shark
(d) Rays
Answer:
(a) Coelacanth

Question 31.
Giant cephalopods like Nautilus were present in ………………. period.
(a) Silurian
(b) Devonian
(c) Ordovician
(d) Permian
Answer:
(c) Ordovician

Question 32.
First modern birds were formed during ………………. period.
(a) Triassic
(b) Jurassic
(c) Cretaceous
(d) Eocene
Answer:
(c) Cretaceous

Question 33.
When did man emerge during evolutionary time period ?
(a) Eocene
(b) Oligocene
(c) Miocene
(d) Pliocene
Answer:
(d) Pliocene

Question 34.
Find the odd monkey out:
(a) Baboons
(b) Gibbons
(c) Macaques
(d) Langurs
Answer:
(b) Gibbons

Question 35.
Who was man with ape-brain ?
(a) Pithecanthropus
(b) Dryopithecus
(c) Australopithecus
(d) Neanderthal man
Answer:
(c) Australopithecus

Question 36.
Who was the first true man ?
(a) Pithecanthropus
(b) Cro-magnon
(c) Australopithecus
(d) Neanderthal man
Answer:
(a) Pithecanthropus

Question 37.
Which one of the following is not present in human beings ?
(a) S curves in vertebral column
(b) Orthognathus face
(c) Simian gap
(d) Chin
Answer:
(c) Simian gap

Question 38.
Which is the correct sequence of human evolution ?
(a) Australopithecus → Ramapithecus → Homo sapiens → Homo habilis
(b) Homo erectus → Homo habilis → Homo sapiens
(c) Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
(d) Australopithecus → Ramapithecus → Homo erectus → Homo habilis → Homo sapiens.
Answer:
(c) Ramapithecus – Homo hablUs -‘ Homo erectus Homo sapiens

Question 39.
Fossils of Homo erectus were obtained in ………………. and ……………….
(a) Kenya, Shivalik hills
(b) Java, Peking
(c) Africa, Asia
(d) Neanderthal valley, Taung
Answer:
(b) Java, Peking

Question 40.
Dentition more like that of the modern man was seen for the first time in ……………….
(a) Dryopithecus
(b) Ramapithecus
(c) Australopithecus
(d) Homo habilis
Answer:
(d) Homo habilis

Question 41.
Lemur and Tarsier belongs to ……………….
(a) Prosimi
(b) Hyalobatidae
(c) Pongidae
(d) Homonidae
Answer:
(a) Prosimi

Match the columns

Question 1.

Geological time	Animal life
(1) Cambrian	(a) Amphibians
(2) Ordovician	(b) First terrestrial animals
(3) Silurian	(c) Jawless fishes
(4) Devonian	(d) Trilobite

Answer:

Geological time	Animal life
(1) Cambrian	(d) Trilobite
(2) Ordovician	(c) Jawless fishes
(3) Silurian	(b) First terrestrial animals
(4) Devonian	(a) Amphibians

Question 2.

Human stage	Cranial capacity in CC
(1) Homo sapiens	(a) 650-800
(2) Homo neanderthalensis	(b) 900
(3) Homo habilis	(c) 1400
(4) Homo erectus	(d) 1450

Answer:

Human stage	Cranial capacity in CC
(1) Homo sapiens	(d) 1450
(2) Homo neanderthalensis	(c) 1400
(3) Homo habilis	(a) 650-800
(4) Homo erectus	(b) 900

Classify the following to form Column B as per the category given in Column A

Question 1.
Nictitating membrane, Seymouria, Lung fish, Flipper of whale and wing of bird, Wing of insect and wing of bird, wisdom
tooth, Eye of octopus, an eye of mammal, vertebrate heart and brain.

Column I	Column II
(1) Homologous organs	————-
(2) Analogous organs	————-
(3) Vestigial organs	————-
(4) Connecting links	————-

Answer:

Column I	Column II
(1) Homologous organs	Flipper of whale and wing of bird, Vertebrate heart and brain
(2) Analogous organs	Wing of insect and wing of bird. Eye of octopus, an eye of mammal
(3) Vestigial organs	Nictitating membrane, wisdom tooth
(4) Connecting links	Seymouria, Lung fish

Question 2.
Origin of conifers, Abundance of trilobites, Diversification of fishes, All types of marine algae, Formation of forests, Rise of dinosaurs, Extinction

Geological time period	Major events
Cambrian	————-
Devonian	————-
Permian	————-
Triassic	————-

Answer:

Geological time period	Major events
Cambrian	Abundance of trilobites, All types of marine algae
Devonian	Diversification of fishes, Formation of forests
Permian	Origin of conifers. Rise of modern insects.
Triassic	Rise of dinosaurs, Extinction of seed ferns

Very short answer questions

Question 1.
What is protobiogenesis?
Answer:
The origin of life on the earth is called protobiogenesis.

Question 2.
What are panspermia?
Answer:
Panspermia or cosmozoa are considered to be spores through which life came on the earth from distant planets.

Question 3.
What are protobionts?
Answer:
Protobionts were first form of life which were formed by nucleic acids along with other inorganic and organic molecules. They have some properties of living form.

Question 4.
What are eobionts?
Answer:
Eobionts or protocells are the first primitive living system which are formed by colloidal aggregations of lipids and proteinoids.

Question 5.
When did universe originate? How?
Answer:
Universe originated about 20 billion years ago by huge titanic explosion called big- bang.

Question 6.
What were the different energy sources during primitive times when earth was cooling?
Answer:
The different available energy sources during primitive times were ultra-violet rays, radiations, lightning and volcanic activities.

Question 7.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 8.
How was first cell formed on the primitive earth?
Answer:
In the protocell, when RNA or DNA system developed and they started regulating various metabolic activities, then it was called a first cell.

Question 9.
How was first cell on the earth in its metabolism ?
Answer:
First cell was anaerobic, heterotrophic and obtained energy by chemoheterotrophic processes.

Question 10.
What are ribozymes?
Answer:
Ribozymes are catalytic RNA which act as biocatalyst.

Question 11.
Who disproved Lamarck’s theory?
Answer:
August Weismann disproved Lamarck’s theory.

Question 12.
Write theory of germplasm as suggested by Wallace.
Answer:
The theory of germplasm says that variations produced in germ cells or germplasm are inherited to next generations, but the somatic variations present in somatoplasm or somatic cells are not inherited.

Question 13.
Who are the main contributors of modem : synthetic theory of evolution?
Answer:
R. Fischer, J. B. S. Haldane, T. Dobzhansky, Huxley, E. Mayr, Simpson, Stebbins, Fisher, Sewall Wright, Medel, T. H. Morgan, etc. are the main contributors of modern theory of evolution.

Question 14.
What is variation?
Answer:
The variations are the differences that occur in morphology, physiology, nutrition, habit, behavioural patterns, etc.

Question 15.
What is mutation?
Answer:
Sudden, large, inheritable and drastic change occurring in the genetic constitution is called mutation.

Question 16.
What is gene frequency?
Answer:
Gene frequency is the relative frequency of an allele or a gene at a particular locus in a population, as compared to other genes, expressed as a fraction or percentage.

Question 17.
What is the other name for genetic drift?
Answer:
Sewall wright effect is the other name for genetic drift.

Question 18.
Why is genetic drift called founder’s effect?
Answer:
The allelic frequency of new population which undergoes genetic drift becomes different from the original one, thus the original drifted population becomes different and are called ‘founders’. Since founders are formed therefore genetic drift is called founder’s effect.

Question 19.
What are fossils?
Answer:
Fossils are the dead remains of plants and animals from prehistoric times, which are found in different forms such as moulds, casts, actual remains or compression seen in various geological layers.

Question 20.
What is embryology?
Answer:
Embryology is the branch of biology and medicine which deals with study of embryos and their development.

Give definition of the following

Question 1.
Gene flow
Answer:
The transfer of genes between two genetically different populations among themselves is called gene flow.

Question 2.
Genetic drift
Answer:
Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

Question 3.
Chromosomal aberrations
Answer:
The structural, morphological change in chromosome due to rearrangement, is called chromosomal aberrations.

Question 4.
Mendelian population
Answer:
Small interbreeding group of a population is defined as Mendelian population.

Question 5.
Crossing over
Answer:
Exchange of genetic material between non-sister chromatids of homologous chromosomes in sexually reproducing organisms, during gamete formation is called crossing over.

Question 6.
Saltation
Answer:
Saltation is defined as single step large mutation.

Name the following

Question 1.
The book written by Charles Darwin after returning from voyage.
Answer:
Origin of species by natural selection.

Question 2.
Five main postulates of Darwinism.
Answer:

Overproduction
Struggle for existence
Organic variations
Natural selection
Origin of next species.

Question 3.
Five key factors of evolution as suggested by Stebbins.
Answer:

Gene mutations
Mutations in the chromosome structure and number
Genetic recombinations
Natural selection
Reproductive isolation.

Question 4.
Name the four types of chromosomal aberrations.
Answer:

Deletion
Translocation
Duplication
Inversion

Question 5.
Name of the insect that displayed industrial melanism and the name of the scientist who studied them.
Answer:
Kettlewell reported that Biston betularia or peppered moth displayed industrial melanism.

Question 6.
Example of homologous organs in plants.
Answer:
Thorns of Bougainvillea and tendrils of Cucurbita.

Question 7.
Example of analogous organs in plants.
Answer:
Sweet potato and potato.

Question 8.
Pre-mating isolating mechanisms.
Answer:

Habitat or ecological isolating mechanism
Seasonal or temporal isolating mechanism
Ethological isolating mechanism
Mechanical isolating mechanism

Question 9.
Post-mating isolating mechanisms.
Answer:

Gamete mortality
Zygote mortality
Hybrid sterility

Question 10.
Name of connecting link between reptiles and birds.
Answer:
Archaeopteryx is the connecting link between reptiles and birds.

Question 11.
Name of connecting link between amphibians and reptiles.
Answer:
Seymouria is the connecting link between amphibians and reptiles.

Question 12.
Name of connecting link between fish and amphibians.
Answer:
Ichthyostegia is a missing link between fish and amphibians.

Question 13.
Name the period that was dominant for Amphibia.
Answer:
Carboniferous is dominant period for Amphibia.

Question 14.
Name the epoch when mammals were at the height of evolution.
Answer:
Miocene epoch.

Question 15.
Name the era when birds began to origin.
Answer:
Mesozoic era.

Question 16.
Name the three subclasses of Class Mammalia.
Answer:

Marsupials
Monotremes
Eutheria

Question 17.
Name the three subfamilies of family Hominoidea.
Answer:

Hyalobatidae
Pongidae
Hominidae

Give the significance of the following

Question 1.
Significance of Natural selection
Answer:

Natural selection is the main driving force behind the evolution.
Natural selection favours those genetic variations which have better fitness value.
Such organisms are at selective advantage and they produce more offspring than the rest. Such organisms have greater survival and reproductive capacity.
In this way natural selection helps in the evolution of new species.
Natural selection favours differential reproduction of gene and brings about changes in the gene frequency.
Natural selection brings about evolutionary changes.
Natural selection also eliminates the genes carrying harmful mutations. This is called mutation balance in which allele frequency of harmful recessives remain constant generation after generation.

Distinguish between the following

Question 1.
Gene flow and Genetic drift.
Answer:

Gene flow	Genetic drift
1. Gene flow is the alteration in the gene frequency due to migrations.	1. Genetic drift is alteration in the gene frequency by pure chance.
2. Gene flow occurs due to exchange of genes in the adjacent populations through interbreeding.	2. Genetic drift occurs due to accidental and sudden elimination of a particular gene.
3. Larger populations tend to show more migrations and hence more gene flow.	3. Smaller populations have greater chances of genetic drift.
4. Gene flow occurs due to emigration and immigration.	4. Genetic drift occurs only within the specified population.

Question 2.
Directional, Stabilizing and Disruptive selection.
Answer:

Directional selection	Stabilizing selection	Disruptive selection
1. Natural selection operating in a linear direction is called directional selection.	1. Natural selection operating to balance or stabilize the population is called stabilizing selection.	1. Natural selection which disrupts the mean characteristics of a population is called disruptive selection.
2. In directional selection, more individuals acquire characters other than the mean character value.	2. In stabilizing selection, more individuals of a population acquire a mean character value.	2. In disruptive selection, more number of individuals acquire extreme or peripheral character value.
3. Directional selection eliminates one of the extremes of the phenotypic range and favour the other.	3. Stabilizing selection tends to favour the intermediate forms and eliminate both the phenotypic extremes.	3. Disruptive selection favours extreme phenotypes and eliminate intermediate.
4. It streamlines variations.	4. It reduces variations.	4. It increases variations.
5. This kind of selection is the most common.	5. This kind of selection is common.	5. This kind of selection is rare.
6. Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction. E.g. Industrial melanism, DDT resistance in mosquito, etc.	6. This selection leads to evolutionary change but tend to maintain phenotypic stability within population. E.g. All the populations which have adapted to their environment.	6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems. E.g. African seed cracker finches with different sized beaks

Question 3.
Homologous organs and Analogous organs.
Answer:

Homologous organs	Analogous organs
1. Homologous organs are structurally similar to one another.	1. Analogous organs are structurally dissimilar to one another.
2. Homologous organs are functionally dissimilar from one another.	2. Analogous organs are functionally similar to one another.
3. Homologous organs help in tracing the evolutionary relationships.	3. Analogous organs do not help in tracing the evolutionary relationships.
4. Homologous organs lead to divergent evolution.	4. Analogous organs lead to convergent evolution.
5. Animals residing in different habitats but having closer evolutionary relationship show homologous organs. E.g. Forelimbs of frog, lizard, bird, bat, whale, man, etc.	5. Diverse animals residing in the similar habitat show analogous organs. E.g. Wing of a bird and wing of an insect.

Question 4.
Allopatric and Sympatric speciation.
Answer:

Allopatric speciation	Sympatric speciation
1. Allopatric speciation is the formation of a new species due to separation of a segment of population from the original population.	1. Sympatric speciation is the formation of species within single population.
2. There is geographical barrier cutting across the species range during such speciation.	2. There is no geographical isolation during sympatric speciation.
3. Allopatric speciation does not have physiological barrier.	3. Sympatric speciation is due to physiological or reproductive isolating barrier.
4. Migration of individual are also helpful in allopatric speciation. E.g. African elephant and Indian elephant.	4. Mutations are helpful in sympatric speciation. E.g. Cichlid fishes in Lake Victoria.

Give scientific reasons

Question 1.
Simple organic molecules did not show decomposition in primitive oceans.
Answer:
Simple organic molecules which were formed during chemical evolution, accumulated at the bottom of water bodies. At that time there was no free oxygen and enzymes. Therefore, simple organic molecules did not show decomposition in primitive oceans.

Question 2.
Archaeopteryx is called connecting link between reptiles and birds.
Answer:
(1) Archaeopteryx shows reptilian as well as avian characters.

(2) Reptilian characters are as follows:

Jaws with homodont (all similar) teeth. Bones are nonpneumatic i.e. solid.
Ribs have a single head. Sternum without keel.
Abdominal ribs present which are like the crocodilian ribs.
Forearms have three digits ending in distinct claws while the hind limb has four digits ending in clawed digits.

(3) Avian characters shown by it are as follows:

Forearms modified into wings.
Feathery exoskeleton.
Skull bones are completely fused.
Cranium is rounded with large orbits and a single condyle.
Jaws are modified into beak.
Limb bones have first toe in opposable manner. Foot present with clawed digits. Since it showed characters of both the classes, it is considered as the connecting link between the two.

Question 3.
Birds are glorified reptiles.
Answer:
Huxley, the evolutionary biologist gave this statement after studying the characters of birds and reptiles. The fossil bird, Archaeopteryx was discovered which showed characters of both Reptilia and Aves. It showed transformation of reptilian characters into bird characters. Hence, birds are said to be glorified reptiles with feathery exo-skeleton and other glorious characteristics.

Question 4.
Analogous organs do not have significant role in evolution.
Answer:
Analogous organs lead to convergent evolution, i.e. different organisms show same superficial structural similarities due to similar functions or habitat. But anatomically and structurally they are different. These organs do not help to trace the common ancestry. Therefore, they are said to have no significant role in evolution.

Question 5.
Australopithecus is described as a man with ape brain.
Answer:
(1) Australopithecus can be considered as a connecting link between ape and man due to the following ape-like and man like characteristics shown by it.

(2) The ape-like characteristics of Australopithecus:

The jaws and teeth were larger than those of modern man.
The face was prognathous, i.e. it had a muzzle like slope
The chin was absent
The eye-brow ridges projected over the eyes
Their cranial capacity ranged from 450-600 c.c.

(3) The man-like characteristics of Australopithecus:

It walked nearly or completely straight due to erect posture.
The vertebral column had a distinct lumbar curve with broad basin-like pelvic girdle.
Dentition was man-like with the smoothly rounded parabolic dental arch.
A simian gap was absent. Australopithecus is therefore, rightly described as a man with ape brain.

Write short notes on the following

Question 1.
Evidences of Darwinism.
Answer:
(1) Height of neck of Giraffe : Long-necked Giraffe came into existence in the following way. Long-necked Giraffe could pluck and eat more leaves from tall trees and woody climbers. So it was well adapted to the environment. Short-necked one could not get food and thus perished in the struggle. This adaptation was transmitted to their offspring.

(2) Black colour peppered moths : The example of industrial melanism seen in U.K. is an excellent example of natural selection in action. Black coloured moths evolved gradually as new species from the previous white coloured forms.

(3) DDT resistance in mosquitoes : Intensive DDT spraying destroyed all types of mosquitoes. Some mosquitoes developed resistance to DDT and survived in spite of DDT spray. They reproduced more and were thus selected naturally.

Question 2.
Drawbacks and Objections to Darwinism.
Answer:

Darwin took into consideration minute fluctuating variation as principal factors. But these are neither heritable nor are part of evolution.
Darwin did not distinguish somatic and germinal variation and considered all variations are heritable.
‘Arrival of the fittest’ was not explained by him.
Darwin was unable to explain the cause, origin and inheritance of variations and of vestigial organs.
He also could not explain extinction of species.
Gradual accumulation of useful variations forms the new species, but their intermediate forms were not recognised.
Darwin could not explain existence of neutral flowers and the sterility of hybrids.

Question 3.
The main features of mutation theory.
Answer:

Mutations are large, sudden and discontinuous variations in a population.
Changes caused due to mutations are inheritable.
The raw material for organic evolution is provided by mutations.
Mutation can be useful or harmful. Useful mutations are at evolutionary advantage as they are selected by nature.
Accumulation of the useful mutations over a period of time leads to the origin and establishment of new species.
Harmful or non-adaptive mutation may persist or get eliminated by nature.

Question 4.
Directional selection.
Answer:

Natural selection bringing about directional change without disrupting the balance is called directional selection.
In a population when more individuals acquire characters which are other than the mean character value, then it is called directional selection.
Natural selection usually acts to eliminate one of the extremes of the phenotypic range and favour the other. E.g. systematic elimination of homozygous recessives.
Directional selection operates for many generations, it results in an evolutionary trend within a population and shifting a peak in one direction.
E.g. Industrial melanism, DDT resistant mosquito, etc.

Question 5.
Stabilizing selection/Balancing selection.
Answer:

Stabilizing selection is the type of natural selection which balances the population, hence it is also known as balancing selection.
In such population more individuals acquire a mean character value.
Such selection tends to favour the intermediate forms and eliminate both the phenotypic extremes.
E.g. More number of infants with intermediate weight survive better as compared to overweight or underweight infants.
Stabilizing selection reduces variations.
It tends to maintain phenotypic stability within population, and does not bring about drastic evolutionary changes.
A population showing stabilizing selection is well-adapted to its environment.

Question 6.
Homologous organs.
Answer:

The structural similarities between the homologous organs indicate that they have a common ancestry.
Different homologous organs indicate divergent evolution or adaptive radiation.
Homologous organs help in tracing the phylogenetic relationships.
Homologous organs are those organs which are structurally similar but functionally dissimilar.

E.g.
(i) Forelimbs of frog, lizard, bird, bat, whale and man are homologous to each other. All the limbs are morphologically similar in construction such as similar limb bones but are dissimilar in function. Frog limbs are meant for hopping, lizard limbs help in crawling, birds and bats fly with the . help of forelimbs while whale uses it for swimming and man for handling the objects.

(ii) Vertebrate heart and brain. & In plants, thorns of Bougainvillea and tendrils of Cucurbita represent homology.

Question 7.
Analogous organs.
Answer:

Analogous organs are similar in function but dissimilar in structural details.
They do not help to trace the relationship in the evolution but help to understand the convergent evolution.
Structural modifications in the organs are due to similar habitat.

E.g.
(i) Wing of an insect and wing of a bird, both are useful in flight so they are functionally similar but are structurally different. Insect wing is formed by exoskeleton expansion while bird wing is the modified forelimb.

(ii) Eye of the Molluscan octopus and of eye of mammals. They differ in their retinal position, structure of lens and origin of different eye parts, but both perform function of vision.

(iii) The flippers of penguins (birds) and dolphins (mammals).

(iv) Sweet potato which is a root modification and potato which is a stem modification, both perform similar function of storing starchy food.

Question 8.
Vestigial organ.
Answer:

Vestigial organs are rudimentary organs which are imperfectly developed and non¬functional, degenerate structures.
These organs in animals become functionless thus their presence in the body is not required.
But they are simply present as they descend down during evolution and continue to exist.
In the process of evolution, they may disappear totally.
They indicate evolutionary line as they were once functional in the ancestors.

Examples of vestigial organs in human beings:

Caecum and vermiform appendix : These are functional in herbivorous animals where they help in cellulose digestion. In humans they are functionless.
Nictitating membrane situated in the eyes of humans. It is a remnant of third eyelid.
Coccyx or tail vertebrae which shows remnant of tail, Wisdom teeth or 3rd molars. These organs indicate that human beings descended from ape like ancestors.

Question 9.
Types of fossils.
Answer:
There are four main types of fossils : actual remains, moulds, casts and compressions.
1. Actual remains : The most common type of fossil is actual remains in which the plants, animals and human bodies are seen embedded in permafrost of arctic or alpine snow. Due to severe cold temperature, the bodies remain preserved in the actual state, E.g., Fossil of Woolly Mammoth in Siberia. Many insects and smaller arthropods remained embedded and thus preserved in amber or hardened resin.

2. Moulds : Hardened encasements formed in the outer parts of organic remains of animals or plants form moulds. The organisms later decays leaving cavities or the impression in permanent form. E.g. Footprints.

3. Casts : Casts are hardened pieces of mineral matter which is deposited in the cavities of moulds.

4. Compressions : A thin carbon film indicates the outline of external features of ancient organism, but other structural details are not seen.

Question 10.
Major changes that occurred in human evolution.
Answer:

Major changes that took place in evolution of man are as follows :
Increase in size and complexity of brain and enhanced intelligence.
Increase in cranial capacity.
Bipedal locomotion.
Opposable thumb.
Erect posture.
Shortening of forelimbs and lengthening of hind limbs.
Development of chin. Orthognathous face.
Broadening of pelvic girdle and development of lumbar curvature.
Social and cultural development such as articulated speech, art, development of tools, etc.

Question 11.
Dryopithecus.
Answer:

Dryopithecus is also called Proconsul. Leakey discovered the fossils of Dryopithecus, on an island in Lake Victoria of Africa. Also the fossil was found in Haritalyanga in Bilaspur district of Himachal Pradesh.
It was a group of apes that lived in Miocene epoch about 20 to 25 million years ago.’
Several species of Dryopithecus are available, the important among these is African fossil D. africanus.
Dryopithecus has a close similarity to chimpanzee and also walked like a modern chimpanzee.
The structure of its limbs and wrists show that knuckle walking was lesser in it. It used the flat of its hands like a monkey.
It had arms and legs of the same length and had a semi-erect posture.

Question 12.
Ramapithecus.
Answer:

Ramapithecus was on direct line of evolution of man.
It was called an ape-man like primate.
Its fossils were obtained in the form of teeth and jaw bones in the rocks of Siwalik Hills in India by Lewis and also in Kenya.
It existed during late Miocene and early Pliocene epoch about 14 to 12 million years ago.
It walked erect on its hind limbs.
It had close similarity with chimpanzee.
Some scientists believe that Dryopithecus evolved into Ramapithecus.

Question 13.
Australopithecus.
Answer:

Australopithecus is considered as connecting link between ape and man.
Its fossils were obtained from Toung valley in South Africa, from Ethiopia and Tanzania.
It was in late Pliocene or early Pleistocene epoch about 4 to 1.8 million years ago.
It was about 4 feet tall. It had prognathus face, with larger jaws. Chin was absent. Lumbar curvature was present.
It walked upright.
The cranial capacity was about 450 to 600 CC. Therefore, it was called man with ape brain.

Question 14.
Homo habilis.
Answer:

Homo habilis is described as Handy man. His fossils were obtained from Olduvai Gorge in Tanzania, Africa.
He existed in late Pliocene or early Pleistocene about 2.5 to 1.4 million years ago.
He was lightly built.
Fossil of lower jaw was obtained which showed that his dentition was more like modern man with small molars.
He walked erect. His cranial capacity was 640 to 800 cc.
He did not eat meat and made stone tools.

Question 15.
Homo erectus.
Answer:

Homo erectus was also known as Java man or Peking Man due to his fossils obtained from these areas.
He was also called ape man.
He lived in the middle Pleistocene epoch about 1.5 million years ago.
He was 5 feet in height with prognathous face, massive jaws, huge teeth and bony eye brow ridges.
Chin was absent.
He walked erect.
The cranial capacity was 900 cc.
He was omnivorous and probably used fire and ate meat.

Question 16.
Neanderthal man.
Answer:

The scientific name of Neanderthal man is Homo neanderthalensis. He is described as advanced prehistoric man.
It was called Neanderthal man because its first fossil was collected from Neanderthal valley in Germany by Fuhlrott (1856).
It was heavily built and short and had outwardly curved thigh bones.
The facial features were as follows : prominent brow ridges, thick skull bones, low and slanting forehead, deep jaw without a chin, etc.
Neanderthal man existed in late Pleistocene epoch about 1,00,000 to 40,000 years ago. It was widely spread in Europe, Asia and North America. It became extinct about 25,000 years ago.
The cranial capacity of Neanderthal man was about 1400 cc, which was roughly equal to that of modern man. He used hide for dressing.
It showed intellectual development in constructing and using flint tools and fire.
The Neanderthal men used to bury their dead bodies along with their tools and perform ceremonies.

Short answer questions

Question 1.
Enlist the steps in the process of chemical evolution.
Answer:

Origin of Earth and Primitive atmosphere.
Formation of ammonia, water and methane.
Formation of simple organic molecules.
Formation of complex organic molecules.
Formation of Nucleic acids.
Formation of Protobionts or Procells.
Formation of first cell.

Question 2.
When did Earth originate? Which transformations took place later?
Answer:

Earth originated about 4.6 billion years ago as a part of the solar system.
When it was formed, it was a rotating cloud of hot gases and cosmic dust. It was then appearing like a nebula.
Later the condensation and cooling started which resulted in stratification.
Heavier elements like nickel and iron settled to the core. Lighter elements like helium, hydrogen, nitrogen, oxygen, carbon, etc. remained on the surface and they formed the primitive atmosphere.
This atmosphere of the earth was of a reducing type, devoid of free oxygen and very hot.

Question 3.
How were simple organic molecules formed on the earth?
Answer:
1. Initially earth’s temperature was very high but as the cooling process started, lighter elements reacted chemically with each other.

2. The early atmosphere was rich in hydrogen, carbon, nitrogen and sulphur. Hydrogen was most active and hence it reacted with other elements to form chemicals on earth like CH4, NH3, H20 and H2S.

3. With decreasing temperature of the earth, steam condensed into water that resulted in heavy rainfall. This constantiy falling rainwater got accumulated on the land to form different water bodies and especially oceans. It also cooled down the earth.

4. The early molecules of hydrocarbons, ammonia, methane and water underwent reactions like condensation, polymerisation, oxidation and reduction due to different energy sources such as ultra-violet rays, radiations, lightning and volcanic activities.

5. These reactions resulted in formation of simple organic molecules like monosaccharides, amino acids, purines, pyrimidines, fatty acids, glycerol, etc.

Question 4.
How were complex organic molecules formed during chemical evolution?
Answer:

The primitive broth in which simple organic molecules were suspended, was neutral and free from oxygen.
In this broth polymerisation took place and simple organic molecules aggregated to form new complex organic molecules like polysaccharides, fats, proteins, nucleosides and nucleotides.
Protoproteins were formed by polymerisation of amino acids. These protoproteins later formed proteins.
Formation of protein molecules is considered as landmark in the origin of life. Later the enzymes were formed which accelerated the rate of other chemical reactions.

Question 5.
How were protobionts formed with the help of nucleic acids during chemical evolution?
Answer:

By the reaction between phosphoric acid, sugar and nitrogenous bases (purines and pyrimidines), nucleotides may have been formed.
These nucleotides joined together to form nucleic acids such as RNA and DNA.
Nucleic acids acquired self-replicating ability which is a fundamental property of living form.
They later formed protobionts. They were the first form of life formed by nucleic acids along with inorganic and organic molecules.
Protobionts were the prebiotic chemical aggregates having some properties of living system. Aggregation of organic molecules due to coacervation formed these protobionts.

Question 6.
Why variations are seen in population?
Answer:
Variations are seen in population due to gene flow, genetic drift, genetic recombinations that occur at the time of gamete formation, crossing over and sudden drastic changes like gene mutations or chromosomal aberrations. All the above factors are constantly operating over every population. Due to these evolutionary processes, variation take place in a population.

Question 7.
In which conditions the gene frequency of a population will remain constant?
Answer:
In the condition of no migrations of the organisms, no mutations, no sexually reproduction consisting of crossing over, no genetic drift, no recombinations and variation, the gene frequency of a population will remain constant. Such hypothetical conditions will never exist because even one set of sexually reproducing organisms forms an offspring which is slightly different from its parents. This means that there is constant change of gene frequency.

Question 8.
What is carbon dating and how does it work?
Answer:
Carbon dating is the method to find out the age of the fossil or any other organic matter. In carbon dating, the relative proportions of the carbon isotopes, carbon-12 and carbon-14 which are present in the organic matter, is estimated. The ratio between them changes as radioactive carbon-14 decays and is not replaced by exchange with the atmosphere. From these findings the age of that organic matter can be concluded.

Question 9.
What is a connecting link? Give suitable examples of connecting links.
Answer:

A connecting link is an intermediate or transitional state between two systematic groups of organisms.
It bears characters common to both these groups on either side of its position. Thus it represents an evolutionary line.
Connecting links are also called a missing link.
E.g. Archaeopteryx, the extinct bird is a connecting link between Reptiles and Aves.
Seymouria is a connecting link between Amphibia and Reptilia.
Ichthyostega is a connecting link between Pisces and Amphibia.

Question 10.
What is geological time scale? How is it divided ?
Answer:

Geological time scale is the arrangement of major divisions of geological time into eras, periods and epochs on the time scale.
This division is based on the study of fossilized organisms obtained from the different strata of the earth.
The characteristic significant events that occurred in the organization of organisms helped the geologists to understand the geological time scale.
The major divisions of geological time are called eras.
The eras are divided into periods and the periods into epochs.
By studying fossils in the earth crust, the evolutionary changes in the organisms have been traced out.

Question 11.
What is meant by palaeontological evidences ?
Answer:

Palaeontology means the study of fossils. Palaeontological evidences are the fossilized forms of various organisms which are obtained from different strata of the earth. They represent the dead remains of plants and animals that lived in the past in various geological layers.
The older and more primitive forms of life are excavated from the lower strata of the soil whereas the recent ones are situated on the upper layers of the soil.
Fossils are formed in variety of materials such as sedimentary rocks, amber, volcanic gas, ice, peat bogs, soil, etc.
They provide the true, direct and reliable evidences of evolution.

Question 12.
What are the molecular evidences that show the evolution?
Answer:

Different organisms have basic similarities in their molecules and the cellular constituents.
All living organisms have the same basic structural and functional unit, i.e. cell.
Cell organelles such as endoplasmic reticulum, Golgi bodies, mitochondria, etc. are present in different types of organisms.
Proteins and gene performing different functions have the same basic pattern which shows a common ancestry.
Catabolic activities of liberating energy, synthesis of macromolecules such as proteins, carbohydrates, nucleic acids, etc. are similar in different organisms.
ATP is the common energy currency of all the organisms.
All the above facts are called molecular evidences in favour of evolution.

Question 13.
Arrange the following stages of the human evolution in the order of their increasing cranial capacity, (a) Neanderthal man (b) Cro-Magnon man (c) Homo erectus (d) Homo habilis.
Answer:

Homo habilis (650-800 cc)
Homo erectus (850-1200 cc)
Neanderthal man (1400 cc)
Cro-Magnon man (1450 cc)

Question 14.
Since your earlier school days you have been solving mysteries/puzzles labelled as use your brain power. Did you ever wonder why human brain has such a capacity? Why and how we evolved along these lines? What is the extent of similarity between humans, chimpanzees and monkeys?
Answer:
Human beings have extremely well- developed brain. Especially the cerebral hemispheres are very large constituting 85% of the brain weight. Due to such cerebrum, there are many, neurons. They are responsible for faculties such as speech, memory, emotions, thought process. The mind or psyche is well developed due to over developed cerebral hemispheres.

That’s why human brain has tremendous capacity of Chimpanzees are also comparatively more intelligent than the monkeys. However, the development of speech and language is lacking in them. The cranial capacity of chimpanzee and monkey is 275-500 cc and 45-50 cc respectively, whereas humans have 1450-1500 cc. Larger the brain, more is the intelligence and all other mental faculties which only humans show.

Question 15.
Even though the cranium of elephant is larger than that of man, humans are considered more intelligent than elephant. Why is it so?
Answer:
Elephant’s brain is large weighing about 5 kg. But elephant’s body weight too is very high. The proportion of body brain weight is highest in human beings. Therefore, humans are considered more intelligent than elephant. Moreover, the cerebral cortex of elephant is not very well developed, instead they have well developed cerebellum which helps in locomotion and balancing their huge bodies. Human brain has very well-developed cerebrum which brings about cognitive behaviour and intelligence.

Chart based/Table based questions

Question 1.
Give the graphical representation of Hardy-Weinberg’s principle in the form of Punnet square.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 1
Genotypes = AA + 2 Aa + aa
Gene frequencies = p² + 2pq + q²

Question 2.
Make a chart showing the types of isolating mechanisms.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 2

Diagram based questions

Question 1.
Give diagrammatic representation to show RNA world.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 3

Question 2.
Sketch and label four types of chromosomal aberrations
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 4

Question 3.
Sketch the graphs to show directional and stabilizing selection.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 5

Long Answer Questions

Question 1.
Write about four old theories which suggested about how did life originate on the earth.
Answer:
1. Theory of special creation : Theory of special creation is the oldest theory which is based on religious beliefs. According to this theory, all the living organisms were created by supernatural power. However, since there are no scientific proofs to this theory, it is not accepted.

2. Cosmozoic theory/Theory of Panspermia : This theory says that life did not originate on the earth but it was exported from the other planets in the form of biological spores or microorganisms which were named as cosmozoa or panspermia. They may have descended to the earth from other planets. Recently, NASA has reported fossils of bacteria-like organisms on a piece of Martian rock recovered from Antarctica. Such facts may throw some light on the cosmozoic theory.

3. Theory of spontaneous generation or Abiogenesis : There was a belief that life originated from non-living material spontaneously. This theory was later disproved by Louis Pasteur.

4. Theory of biogenesis : This theory says that living organisms can originate only from pre-existing living beings. It is same as reproduction. But this theory of biogenesis was unable to explain origin of life on earth. It explains only the continuity of life.

Question 2.
Haldane described ‘Hot dilute soup’ in his theory. Describe how this soup led to formation of some important molecules.
Answer:
(1) The primitive sea containing molecules of organic substances without free oxygen was described as ‘hot dilute soup or primitive broth’ by Haldane. He proposed the theory of chemical evolution.

(2) According to this theory, the chemical evolution took place in the following steps : (a) Origin of earth and its primitive atmosphere, (b) Formation of ammonia, water and methane. These molecules dissolved in rainwater and formed the seas, (c) Then synthesis of simple organic compounds took place, followed by formation of complex organic compounds such as nucleic acids.

(3) The early molecules underwent chemical reactions such as condensation, polymerization, oxidation and reduction.

(4) The biologically important molecules such as monosaccharides, amino acids, purine, pyrimidine, fatty acids and glycerol were formed due to these reactions, utilizing the sources of energy on the primitive earth.

(5) Since oxygen was lacking, there was no degradation. Enzymes were also absent and hence there was formation of complex molecules in the hot dilute soup.

(6) This further led to the formation of pre-cells or protobiont. These aggregates were called coacervates by Oparin or microspheres by Sidney Fox. This further gave rise to first cells on the earth.

Question 3.
Explain the process of formation of eobionts.
Answer:

Protobionts or coacervates were colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).
They grew in size by taking up material from surrounding aqueous medium.
During their growth they became thermodynamically unstable and split into smaller units. These were called microspheres.
They were proteinoids formed from colloidal hydrophilic complexes surrounded by water molecules.
These bodies were like primitive cells having outer double-membrane. Across this membrane diffusion and osmosis may have occurred. They were more stable than coacervates.
Coacervates and microspheres were non-living colloidal aggregations of lipids and proteinoids respectively.
But they showed growth and division like living cells.
These colloidal aggregations turned into first primitive living system called eobionts or protocell.

Question 4.
Describe RNA World hypothesis.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 6
(1) RNA world hypothesis is based on discovery of catalytic RNA or ribozymes. It was proposed by Carl Woese, Francis Crick and Leslie Orgel in 1960 whereas Ribozymes were discovered by Sidney Altman and Thomas Cech in 1980.

(2) According to this hypothesis, early life must have been based most probably on RNA.

(3) Factors supporting this hypothesis are:

RNA is found abundantly in all living cells.
It is structurally related to DNA.
Chains of RNA can evolve or undergo mutations, replicate and catalyse reactions.
Biomolecules like Acetyl-Co-A have a nucleotide in their molecular structure.
Ribosome acts as a protein assembly unit in the cell and is seen in many types of cells.
In ribosomes, translation process is catalysed by RNA.

(4) The primitive molecules underwent repeated replication and mutation forming varieties of RNA molecules with varying sizes and catalytic properties.

(5) They later developed their own protein coats and machinery to survive the assembly of primitive cell.

(6) From them DNA was developed which was double stranded stable structure.

(7) It further kept on evolving giving rise to rich biodiversity on earth.

Question 5.
Explain in brief Darwinism and its five main postulates.
Answer:
Darwinism means theories of natural selection and speciation as put forth by Charles Darwin. The five main postulates of his theories are as follows: Overproduction or prodigality, Struggle for existence, Organic variations, Natural selection, Origin of new species (speciation).
1. Overproduction (Prodigality of nature) : There is a natural tendency to produce more number of progeny in geometric ratio for continuing the species. E.g. Salmon fish produces about 28 lakh eggs in a single season. Single pair of elephants would produce 19,000,000 elephants. But the size of given species in a given area remains relatively constant because of fluctuations that occur seasonally.

2. Struggle for existence Due to over¬production there is struggle for existence between the members of population for limited supply of food or to overcome adverse environmental conditions or for a space or to escape from enemies, etc.

3. Organic variations : There are differences in morphology, physiology, nutrition, habit, behavioural patterns, etc., among the members of same species or members of different species. These variations act as raw material for evolution.

4. Natural selection : Some organisms possess better variations to get adapted and survive under existing environmental conditions, while some do not have. Better adapted organisms are selected by the nature while those with unfavourable variations perish. The principle by which useful variations are preserved by nature, is called ‘Natural Selection’. It is also called ‘survival of fittest’ by H. Spencer.

5. Origin of new species (speciation) : Favourable variations are transmitted from generation to generation, resulting into better adapted generations. Gradually these adaptations with few new modifications become fixed in the life cycle, forming a new species.

Question 6.
Explain modern Synthetic Theory of Evolution in brief.
Answer:
(1) Modern synthetic theory of evolution is the result of modification of Darwinism and theory of mutations by taking into consideration studies of genetics, ecology, anatomy, geography and palaeontology.

(2) Five key factors of modern synthetic theory are gene mutations, mutations in the chromosome structure and number, genetic recombinations, natural selection and reproductive isolation. All these finally contribute in the evolution of new species or process of speciation.

(3) Population or Mendelian population is the small group of ‘interbreeding populations’. For every Mendelian population there is a gene pool which is constituted by total number of genotypes in it. The genotype of an organism in a population is constant, but the gene pool constantly undergoes change due to different factors such as mutations, recombination, gene flow, genetic drift, etc.

(4) Every gene has two alleles. The proportion of a particular allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency. Thus any change in the gene frequency in the gene pool affects population.

(5) The five main factors are broadly divided into three main concepts as follows:
(i) Genetic variations caused due to various aspects of mutation, recombination and migration. Such variations cause change in the gene frequency. Gene mutations or point mutation change the phenotype of the organism, leading to variation. Recombination is caused due to crossing over in which new genetic combinations are produced. Sexual reproduction due to fertilization of gametes also cause recombinations. All these lead to variations, Gene flow is movement of genes into or out of the population, either due to migrations or dispersal of gametes.

Gene flow therefore change the gene frequencies of the population. Genetic drift is a random change which occurs by pure chance. It occurs in small populations but change the gene frequency. Chromosomal aberrations are structural or morphological changes in the chromosomes causing rearrangement of the sequence of genes.

(ii) Natural selection is said to be the main driving force in evolution. It brings about evolutionary changes by selecting favourable gene combinations by differential reproduction of genes. This brings about changes in gene frequency from one generation to next generation.

(iii) Isolation means the separation of the population of a particular species into smaller units which prevents interbreeding between them. This over a long time period leads to speciation or formation of new species.

Question 7.
What are different types of chromosomal aberrations?
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 5 Origin and Evolution of Life 7
Chromosomal aberrations:
(1) The structural, morphological change, which take place in chromosome due to rearrangement, is called chromosomal aberrations.

(2) The aberrations change the sequence of the genes. This causes variations. Chromosomal aberrations are mainly of following four types:

Deletion : Loss of genes from chromosome.
Duplication : Genes are repeated or doubled in number on chromosome.
Inversion : A particular segment of chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twist. There is no loss or gain of gene complement of the chromosome.
Translocation : Transfer or transposition of a part of chromosome or a set of genes to a non-homologous chromosome is called translocation. It is effected naturally by the transposons present in the cell.

Question 8.
What are the different pre-zygotic isolating mechanisms?
Answer:
(1) Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.

(2) By various mechanisms the two groups remain isolated. These mechanisms are of following types:
(i) Habitat isolation : Habitat isolation is the phenomenon in which members of a population living in the same region occupy different habitats. Hence the potential mates do not interbreed among themselves.

(ii) Seasonal isolation : In seasonal isolation, members of a population share the same region but attaining sexual maturity at the different times of the year. They thus remain isolated reproductively preventing interbreeding among themselves.

(iii) Ethological isolation : Ethological isolation is seen when members of two populations have different mating behaviours. This prevents interbreeding.

(iv) Mechanical isolation : Mechanical isolation is seen when the members of two populations have differences in the structure of reproductive organs. Due to such differences interbreeding is not possible.

Question 9.
What are the different post-zygotic isolating mechanisms?
Answer:

In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
Thus the populations remain isolated without the actual genetic exchange.

Post-mating isolating mechanisms are divided into the following categories:

Gamete mortality : In gamete mortality, there is death of gametes. Sperm transfer may take place but the egg is not fertilized due to gamete mortality.
Zygote mortality : In zygote mortality, the zygote is formed but it fails to thrive. Though the egg is fertilized the zygote does not survive.
Hybrid sterility : In this isolation, there is the formation of hybrid as the gametes or zygotes do not die but the hybrid formed is sterile. Sterile hybrid cannot contribute genetically to further generations.

Question 10.
What is Hardy-Weinberg equilibrium? Explain it in brief.
Answer:

Hardy and Weinberg were two scientists who proposed a concept of genetic equilibrium popularly known as Hardy- Weinberg principle or equilibrium.
This principle states that gene, allele or genotype frequencies remain the same from generation to generation unless disturbed by factors like mutation, non-random mating, genetic drift, etc.
For explaining the concept of equilibrium they assumed that there are two alleles located at a single locus (A and a).
Their respective frequencies are p and q.
The frequency of genotype AA is p, for 2Aa is 2pq and for aa is q.
The equilibrium equation is p² + 2pq + q² = 1
It says that if sum total of gene frequencies is 1, then sum total of genotype frequencies is also equal to 1. When the equilibrium is disturbed then only evolution occurs.

Question 11.
Human being is said to be most evolved, intelligent living being. Yet we are not self-sufficient. Think of various aspects for which we depend on other living beings for our survival.
Answer:
Human brain is evolved and super- specialised but yet in many aspects human beings are much dependent on other natural factors. Human body is not with any protective exoskeleton, or organs of offence and defence.

Unless well dressed, he cannot cope up with severe cold temperatures as he lacks natural protective fur. He cannot run fast as the other animals can. Neither he can digest uncooked food. He has overcome all his shortcomings by using his brain power.

He has managed to take fur and feathers from other animals by killing them. He also uses other natural fibres from plants to cover his body. He has also finished fish from the oceans by over-exploitation and polluting the natural habitats of these creatures.

He takes meat from other animals by killing them and for the purpose he domesticates them to satisfy his hunger. Before technical age, animals were used as beasts of burden and as means of transport. Thus human history has shown excessive use of horses, camels, elephants, etc.

Man snatches milk from other animals like cows and buffaloes which is for their young ones. But the entire diary industry and human needs for dairy products have been taken care of by these herbivores.

Apart from all such uses, man also uses other animals for experimentations and pharmaceutical industries. In this way, man has mastered other animal kingdom due to his intelligence.