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Maharashtra State Board HSC 11th Commerce Book Keeping & Accountancy Important Questions and Answers

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Maharashtra Board 11th HSC Important Questions

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 3 Ionic Equilibria

Question 1.
What are electrolytes?
Answer:
Electrolytes: The substances which in their aqueous solutions (or in any polar solvents) dissociate or ionize forming positively charged ions (cations) and negatively charged ions (anions) are called electrolytes. For example, NaCl, HCl, etc.

Question 2.
What is an ionic equilibrium?
Answer:
Ionic equilibrium: The equilibrium between ions and unionized molecules of an electrolyte in solution is called an ionic equilibrium.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)

Question 3.
What are the types of electrolytes?
Answer:
There are two types of electrolytes as follows :
(a) Strong electrolyte The electrolytes which ionise completely or almost completely are called strong electrolytes. For example, NaCl, HCl, H₂SO₄, etc.
(b) Weak electrolytes: The electrolytes which dissociate to a less extent are called weak electrolytes. For example, CH₃COOH, NH₄OH, etc.

Question 4.
Define degree of dissociation.
Answer:
Degree of dissocsavIt is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium.
It is denoted by a and represented by,

OR α = \(\frac{\text { Per cent dissociation }}{100}\)
∴ Per cent dissociation = α × 100

Question 5.
Define acid and base. Give examples.
Answer:
Acid : A hydrogen containing substance which gives H⁺ ions in aqueous solution is called an acid. For example, HCl, CH₃COOH, etc.
\(\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Base : A substance that contains OH group and produces hydroxide ions (OH^–) in aqueous solution is called a base.
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

Question 6.
What are the limitations of Arrhenius acid-base theory ?
Answer:
Limitations of Arrhenius theory :

1. This theory is applicable only for aqueous solutions and not for non-aqueous solutions.
2. It fails to explain the acidic nature of non-hydrogen compounds like BF₃, AlCl₂, FeCl₃, etc.
3. It fails to explain the basic nature of non-hydroxy compounds like NH₃, amines, Na₂CO₃, KCN, aniline, etc. in their aqueous solutions.
4. It does not explain role of solvent or existence of H₃O⁺ in an aqueous solution of an acid.

Question 7.
Explain neutralisation reaction according to Arrhenius theory.
Answer:
Neutralisation reaction : According to Arrhenius theory neutralisation is a reaction between an acid and a base in their aqueous solutions produciijg salt and unionised water.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{NaOH}_{(\mathrm{aq})} \rightarrow \mathrm{NaCl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
Since strong acid, strong base and salt dissociate completely, the above reaction is represented as,

Hence, according to Arrhenius theory neutralisation reaction is defined as a reaction between H⁺ ions and OH^– ions forming unionised water molecules.

Question 8.
Explain Bronsted-Lowry theory of acids and bases.
Answer:
Acid : According to Bronsted-Lowry theory acid is a substance that donates a proton (H⁺) to another substance.
Base : According to this theory base is a substance that accepts a proton (H⁺) from another substance. For example,

Since HCl donates a proton it is an acid while NH₃ accepts a proton it is a base.

Question 9.
For each of the following reactions, identify the Lowry-Bronsted conjugate acid-base pairs :

Answer:

In this Acid-1 (CH₃COOH) and Base-1 (CH₃COO^–) is one acid base conjugate pair while Base-2 (NH₃) and Acid-2 \(\mathrm{NH}_{4}^{+}\) is another conjugate pair.

Question 10.
Mention a conjugate acid and a conjugate base for each of the following :
(a) H₂O (b) \(\mathrm{HSO}_{4}^{-}\) (C) Br^– (d) H₂CO₃ (e) \(\mathbf{H}_{2} \mathbf{P O}_{4}^{-}\) (f) \(\mathbf{N H}_{4}^{+}\)
Answer:

Substance	Conjugate acid	Conjugate base
(a) H2O	H3O+	OH–
(b) \(\mathrm{HSO}_{4}^{-}\)	H2SO4	\(\mathrm{SO}_{4}^{-2}\)
(c) Br–	HBr	–
(d) H2CO3	–	\(\mathrm{HCO}_{3}^{-}\)
(e) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)	H3PO4	\(\mathrm{HPO}_{4}^{-2}\)
(f) \(\mathrm{NH}_{4}^{+}\)	–	NH3

Question 11.
Identify conjugate acid-base pairs in the following :


Answer:

Question 12.
Define acids and bases on the basis of Lewis concept. Give examples.
Answer:
Lewis concept of an acid and a base is based on the electronic theory.
Acid : It is defined as any species (molecule or ion) that can accept a pair of electrons. E.g. BF₃, AlCl₃ and all electron deficient species like cations (K⁺, Ag⁺) and molecules having incomplete octet, like BeF₂, BF₃.

Base : It is defined as any species (molecule or ion) that can donate a pair of electrons. E.g. NH₃, C₂H₅NH₂ and all electron rich species like anions (Cl^–, OH^–) and all molecules with lone pair of electrons.

Question 13.
Explain : (A) BF₃ is a Lewis acid, (B) NH₃ is a Lewis base.
Answer:
(A) According to Lewis theory, an acid is a substance which can accept a pair of electrons.
In BF₃ molecule, the octet of B is incomplete, hence it needs two electrons or a pair of electrons to complete its octet. Hence BF₃ acts as a Lewis acid.

(B) According to Lewis theory, a base is a substance which can donate a pair of electrons.
In NH₃ molecule, nitrogen atom has one lone pair of elctrons to donate.

The reaction between BF3 and NH3 can be represented as,

Question 14.
Classify the following into Lewis acids and bases :
CN^–, Cl^–, S^2-, Cu⁺⁺, H₂O, OH^–, BF₃, Ag⁺.
Answer:
(1) Lewis acid : Cu⁺⁺, BF₃, Ag⁺
(2) Lewis bases : CN^–, Cl^–, S^2-, OH^–.

Question 15.
Explain amphoteric nature of water.
Answer:
(1) Since water acts as an acid as well as a base, it is amphoteric in nature.
(2) H₂O has a tendency to donate a proton forming OH^– as well as has a tendency to accept a proton forming H₃O⁺.


Therefore H₂O is amphoteric in nature.

Question 16.
How are acids and bases classified on the basis of extent of their dissociation?
Answer:
On the basis of extent of dissociation acids and bases are classified as follows :
(1) Strong acids and strong bases : The acids and bases which dissociate to a greater extent or almost completely are called strong acids and strong bases.
\(\mathrm{HCl}_{\text {(aq) }} \longrightarrow \mathrm{H}_{\text {(aq) }}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-}\)
\(\mathrm{NaOH}_{(\mathrm{aq})} \longrightarrow \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

(2) Weak acids and weak bases : The acids and bases which dissociate partially are called weak acids and weak bases. There exists an equilibrium between undissociated molecules and ions in solution.

Question 17.
Give examples of (a) strong acids and strong bases, (b) weak acids and bases.
Answer:
(a) Strong acids : HCl, H₂SO₄
Strong bases : NaOH, KOH
(b) Weak acids : HCOOH, CH₃COOH
Weak bases : NH₄OH, C₂H₅NH₂

Question 18.
Define and explain dissociation constant of a weak acid.
Answer:
Dissociation constant of a weak acid : It is defined as the equilibrium constant for dissociation equilibrium of a weak acid and denoted by K_a.
Explanation : Consider an aqueous solution of a weak acid HA.
\(\mathrm{HA}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-}\)
The equilibrium constant called dissociation constant Ka is represented as,
K_a = \(\frac{\left[\mathrm{H}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)

Question 19.
Derive the expression of Ostwald’s dilution law in case of a weak acid (HA).
OR
Derive the relationship between degree of dissociation and dissociation constant of a weak acid.
Answer:
Expression of Ostwald’s dilution law in case of a weak acid : Consider the dissociation of a weak acid HA. Let V dm³ of a solution contain one mole of weak acid HA. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^3-3. Let α be the degree of dissociation of HA.

Applying the law of mass action to this dissociation equilibrium, we have,

As the acid is weak, a is very small as compared to unity,
∴ (1 – α) ≈ 1.

This is an expression for Ostwald’s dilution law. This shows that the degree of dissociation of a weak acid is directly proportional to the volume of solution containing one mole of acid or inversely proportional to square root of its concentration.

Question 20.
Derive the relationship between degree of dissociation and dissociation constant of a weak base.
OR
Derive the expression of Ostwald’s dilution law in case of a weak base.
Answer:
Expression of Ostwald’s dilution law in case of a weak base : Consider the dissociation of a weak base BOH. Let V dm³ of a solution contain one mole of weak base BOH. Then the concentration of a solution is, C = \(\frac{1}{V}\) mol dm^-3. Let α be the degree of dissociation of BOH.

Consentration at equilibrium (mol dm^-3) \(\frac{(1-\alpha)}{V} \quad \frac{\alpha}{V} \quad \frac{\alpha}{V}\)
Applying the law of mass action to this dissociation equilibrium, we have,

This is an expression of Ostwald’s dilution law. Thus, the degree of dissociation of a weak base is directly proportional to the square root of the volume of the solution containing one mole of a base or inversely proportional to the square root of its concentration.

Solved Examples 3.4

Question 21.
Solve the following:

(1) At 298 K, 0.01 M formic acid solution is 1.2% dissociated. Calculate the dissociation constant of formic acid.
Solution :
Given : Concentration of HCOOH = C = 0.01 M
Percent dissociation = 1.2
Dissociation constant = K_a = ?
∴ Degree of dissociation = \(\frac{\text { Percent dissociation }}{100}\)
α = \(\frac{1.2}{100}\)
= 1.2 × 10^-2
K_a = Cα²
= 0.01 × (1.2 × 10^-2)²
= 1.44 × 10^-6
Ans. Dissociation constant = K_a = 1.44 × 10^-6

(2) The degree of dissociation of ammonium hydroxide is 0.0232 in 0.5 M solution. What will be the dissociation constant of ammonium hydroxide ?
Solution :
Given : Degree of dissociation = α = 0.0232
Concentration of NH₄OH = C = 0.5 M
Dissociation constant = K_b = ?
K_b = Cα²
= 0.5 × (0.0232)²
= 2.692 × 10^-4
Ans. Dissociation constant of NH₄OH
= K_b = 2.692 × 10^-4.

(3) Calculate the hydrogen ion concentration in 0.1 M acetic acid solution when the acetic acid is 2% dissociated in the solution.
Solution : The dissociation of acetic acid is represented below :
CH₃COOH ⇌ CH₃ – COO^– + H⁺
Given : Dissociation = 2%, C = 0.1 M
[H₃O⁺] = ?
The concentration of hydrogen ion, [H⁺], is given by the following formula :
[H₃O⁺] = αC
α = Degree of dissociation = \(\frac{\text { Per cent dissociation }}{100}\)
= \(\frac{2}{100}\) = 0.02
[H₃O⁺] = Hydrogen ion concentration = ?
C = Molar concentration of acetic acid
= 0.1 M = 0.1 mol dm^-3
∴ [H₃O⁺] = C × α = 0.1 × 0.02
= 0.002 = 2.0 × 10^-3 mol dm^-3
Ans. Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3.

(4) Calculate the percentage dissociation of 0.01 M NH₄OH solution. The K_b for NH₄OH is 1.75 × 10^-5.
Solution :
Given : Concentration of NH₄OH = C = 0.01 M
Dissociation constant of NH₄OH
= K_b= 1.75 × 10^-5
Percentage dissociation = ?
K_b = Cα²

∴ Per cent dissociation = α × 100
= 4.183 × 10^-2 × 100
= 4.183
Ans. Dissociation of NH₄OH = 4.183%

(5) 0.01 mole of a weak base is dissolved in 8 dm³ of water. The dissociation constant of the base is 4.0 × 10^-10. Calculate the degree of dissociation of the base in the solution.
Solution :
Given : V = 8 dm³; n = 0.01 mole; K_b = 4 × 10^-10
α = ?
The degree of dissociation and dissociation constant of a weak base are related to each other by the following formula :
K_b = α²C OR α = \(\sqrt{\frac{K_{b}}{C}}\)
K_b = Dissociation constant of the base = 4.0 × 10^-10
α = Degree of dissociation of the base = ?
C = Molar concentration of the base
= 0.01 mole in 8 dm³

Ans. Degree of dissociation of the base = 5.656 × 10^-4

(6) The dissociation constant of benzoic acid (C₆H₅COOH) is 6.6 × 10^-5. Calculate the hydrogen ion concentration of a solution containing 1.22 g of benzoic acid in 2000 mL of water.
Solution :
Given : K_a = 6.6 × 10^-5; V = 2000 mL;
W = 1.22 g; [H⁺] = ?
Molar mass of benzoic acid (C₆H₅COOH) = 122
The concentration of the solution is 1.22 g benzoic acid in 2000 ml (2 dm³) of solution.
1.22 g = \(\frac{1.22}{122}\) = 0.01 mol
∴ Molar concentration of benzoic acid = \(\frac{0.01}{2}\)
= 0.005 mol dm^-3
The dissociation constant and degree of dissociation of a weak acid are,
K_a = α²C OR α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
α = Degree of dissociation of benzoic acid = ?
C = 0.005 mol dm^-3

Since benzoic acid is a monobasic acid, [H⁺] = aC
∴ [H⁺] = 0.1149 × 0.005
= 5.745 × 10^-4 mol dm^-3
Ans. Hydrogen ion concentration
= 5.745 × 10^-4 mol dm^-3

(7) The degree of dissociation of acetic acid in its 0.1 M solution is 0.0132 at 25 °C. Calculate the degree of dissociation in its 0.01 M solution.
Solution :
Given : C = 0.1 M, α = 0.0132, C’ = 0.01 M,
α = 0.0132, α’= ?
K_a = α²C
C = Molar concentration of acetic acid = 0.1 M
α = Degree of dissociation in 0.1 M solution = 0.0132
∴ α = \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
If the concentration is C’, α’ = \(\sqrt{\frac{K_{\mathrm{a}}}{C^{\prime}}}\)

Ans. Degree of dissociation in 0.01 M solution = 4.175 × 10^-2

Question 22.
Explain autoionisation of water. Derive a relation for ionic product of water.
Answer:
Pure water ionises to a very less extent. The ionisation equilibrium is represented as follows,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The equilibrium constant K for the above equilibrium is represented as,
K = \(\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}}\)
∴ K × [H₂O]² = [H₃O⁺] × [OH^–]
Since K and active mass of pure water [H₂O] are constant we can write,
K × [H₂O] = K_w,
∴ K_w= [H₃O⁺] × [OH^–]
where K_w is called ionic product of water. At 25 °C,
K_w= 1 × 10^-14.

Question 23.
Define ionic product of water.
Answer:
Ionic product of water : It is defined as the product of molar concentrations of hydronium ions (or hydrogen ions) and hydroxyl ions at equilibrium in pure water at constant temperature.
It is represented as,
K_w = [H₃O⁺] × [OH^–]
At 25 °C, K_w= 1 × 10^-14.

Question 24.
Define the following :
(i) pH (2) pOH. (2 marks)
Ans.
(1) pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H⁺ is known as the pH of a solution.
PH = -log₁₀ [H⁺]
(2) pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH^– is known as the pOH of a solution.
pOH = -log₁₀ [OH^–]

Question 25.
What are approximate concentrations of H₃O⁺ and OH^– in, (a) pure water or neutral solution, (b) acidic solution and (c) basic solution ? Also mention pH values.
Answer:

Question 26.
Write a note on pH scale.
Answer:
Most of the chemical reactions and industrial processes are carried out in aqueous solutions, hence there is a need to know concentration of H⁺ and OH^– ions in the solution.
Sorensen developed a convenient scale to represent the acidic, basic or neutral nature of the solution.
The pH scale is used to express the concentration of H⁺ and OH^– along with pH and pOH of the solution.
According to Sorensen,
pH = -log₁₀ [H+], pOH = -log₁₀ [OH^–]
pH + pOH = 14.

Acids, basic and neutral solutions.

Question 27.
Solve the following :

(1) At 50 °C the value of ionic product of water is 5.5 × 10^-14. What are the concentrations of [H₃O⁺] and OH^– in a neutral solution at 50°C temperature ?
Solution :
Given : at 50 °C
Ionic product of water = K_w = 5.5 × 10^-14
[H₃O⁺] = ? OH^– = ?
Water at any temperature will be neutral.
Hence, [H₃O⁺] = [OH^–] = x mol dm^-3
[H₃O⁺] × [OH^–] = K_w
x × x = 5.5 × 10^-14
∴ x = 2.345 × 10^-7
∴ [H⁺] = [OH^–] = 2.345 × 10^-7 M
Ans. Concentrations : [H₃O⁺] = [OH^–]
= 2.345 × 10^-7 M

(2) The concentration of H+ ion in lemon juice is 2.5 × 10^-3 M. Calculate the OH^– ion concentration and classify the solution as acidic, basic or neutral.
Solution :
Given : [H₃O⁺] = 2.5 × 10^-3 M, K_w = 1 × 10^-4
[OH^–] = ?
By ionic product of water,
[H₃O⁺] × [OH^–] = K_w
∴ [OH^–] = \(\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\)
= \(\frac{1 \times 10^{-14}}{2.5 \times 10^{-3}}\)
Ans. Concentration of OH^–
= [OH^–] = 4 × 10^-12 M
Hence the solution of lemon juice is acidic.

(3) Calculate pH and pOH of 0.02 M HCl solution.
Solution :
Given : C = 6.02 M HCl; pH = ? pOH = ?
\(\begin{aligned}
\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow & \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \\
& 0.02 \mathrm{M}
\end{aligned}\)
[H⁺] = [H₃O⁺] = 0.02 M
PH= -log₁₀ [H₃O⁺]
= -log₁₀ 0.02
= -(\(\overline{2} .3010\))
= 2 – 0.3010 = 1.699
pH + pOH = 14
∴ pOH = 14 – pH
= 14 – 1.699
= 12.3010
Ans. pH = 1.6990; pOH = 12.3010.

(4) The pH of a solution is 6.06. Calculate its [H₃O⁺] ion concentration.
Solution :
Given : pH = 6.06, [H₃O⁺] = ?
PH = -log₁₀ [H₃O⁺]
∴ log₁₀ [H₃O⁺] = -pH
∴ [H₃O⁺] = Antilog – pH
= Antilog – 6.06
= Antilog \(\overline{7} .94\)
= 8.714 × 10^-7 M
Ans. [H₃O⁺] = 8.714 × 10^-7 M.

(5) Calculate number H⁺ ions present in 1 mL of 0.01 M H₂SO₄ solution.
Solution :
Given : C = 0.01 M H₂SO₄; Y = 1 mL
Number of H⁺ ions = ?
\(\begin{aligned}
\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \longrightarrow & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \\
& 0.01 \times 2 \mathrm{M}
\end{aligned}\)
∵ 1000 mL solution contains 0.02 mol H⁺
∴ 1 mL solution contains \(\frac{0.02}{1000}\) mol
= 2 × 10^-5 mol H⁺
∴ Number of H⁺ ions = 2 × 10^-5 × 6.022 × 10²³
= 1.204 × 10¹⁹
Ans. Number of H⁺ ions = 1.204 × 10¹⁹

(6) The pH of a 0.1 M monoacidic base is 11.11. What is the percent dissociation of base ?
Solution :
Given : pH = 11.11; per cent Dissociation of base = ?
c = 0.1 M
\(\begin{gathered}
\mathrm{BOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{B}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \\
c(l-\alpha) \quad c \alpha \quad c \alpha
\end{gathered}\)
pH + pOH = 14
∴ pOH = 14 – pH= 14 – 11.11 = 2.89
POH = -log₁₀ [OH^–]
∴ [OH^–] = Antilog – pOH
= Antilog – 2.89
= Antilog \(\overline{3} .11\)
= 1.29 × 10^-3 M
∵ [OH^–] = cα
∴ α = \(\frac{\left[\mathrm{OH}^{-}\right]}{c}=\frac{1.29 \times 10^{-3}}{0.1}\) = 1.29 × 10^-2
∴ Per cent dissociation = α × 100
= 1.29 × 10^-2 × 100 = 1.29
Ans. Per cent dissociation = 1.29.

(7) Calculate the pH of dedmolar solution of sulphuric add aqueous solution. Assuming complete Ionization of sulphuric add.
Solution:
Given : Concentration of H₂SO₄ = decimolar
= 0.1 M
H₂SO₄ → 2H⁺ + \(\mathrm{SO}_{4}^{-2}\)
∴ [H⁺] = 2 × 0.1
= 0.2 M
PH = -log₁₀ [H⁺]
= -log₁₀ 0.2 M
= \(-[\overline{1} .3010]\)
= 1 – 0.3010 = 0.6990
Ans. pH of H₂SO₄ solution = 0.6990.

(8) The pH of a 0.2 M solution of ammonia is 10.78. Calculate (i) OH^– ions concentration (ii) the degree of dissociation (iii) the dissociation constant.
Solution :
Given: pH = 10.78, C = 0.02 M, [OH^–] = ? K_b = ?
As NH₃ is a base,
pOH = 14 – pH
pH = 10.78
∴ pOH = 14 – 10.78 = 3.22
pOH = -log₁₀ [OH^–]
∴ 3.22 = -log₁₀ [OH^–]
∴ -3.22 = log₁₀ [OH^–]
[OH^–] = antilog (-3.22)
∴ [OH^–] = antilog \((\overline{4} .78)\)
= 6.026 × 10^-4 M
As NH₄OH is a monoacidic base, [OH^–]

(9) NH₄OH is 4.3% ionised at 298 K in 0.01 M solution. Calculate the ionization constant and pH of NH₄OH.
Solution :
Given : Per cent dissociation = 4.3,
C = 0.01 M, K_b = ?, pH = ?
The degree of dissociation and dissociation constant of NH₄OH are related to each other by the formula :
K_b = α²C
K_b = Dissociation constant of NH₄OH = ?
α = Degree of dissociation of NH₄OH = 4.3%
= 4.3 × 10^-2
C = Molar concentration of NH₄OH = 0.01 M
∴ K_b = (4.3 × 10^-2)² × 0.01
= 18.49 × 10^-4 × 10^-2
∴ K_b = 1.849 × 10^-5
Since NH₄OH is a monoacidic base,
[OH^–] = αC
= 4.3 × 10^-2 × 0.01
= 4.3 × 10^-4 mol dm^-3
pOH = -log₁₀ [OH-]
= -log₁₀ 4.3 × 10^-4
= -[0.6335 – 4] = 3.3665
pH + pOH = 14
pH = 14 – 3.3665 = 10.6335
Ans. K_b = 1.849 × 10^-5, pH = 10.6335

Question 28.
Define hydrolysis.
Answer:
Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.

Question 29.
What are the types of the salts? Give examples.
Answer:
A salt is formed by the reaction between equivalent amounts of an acid and a base. According to the nature of an acid and a base, there are four types of the salts as follows :
(1) Salt of a strong acid and a strong base :

(2 ) Salt of a weak acid and a strong base :

(3) Salt of a strong acid and a weak base :

(4) Salt of a weak acid and a weak base : CH₃COONH₄ :

Question 30.
A salt of strong acid and strong base does not undergo hydrolysis. Explain.
OR
An aqueous solution of sodium chloride is neutral. Explain.
Answer:
(1) Sodium chloride is a salt of strong acid HCl and strong base NaOH.
(2) In water, it reacts forming HCl and NaOH.
(3) As both are strong, they dissociate almost completely to liberate H⁺ and OH^– ions, respectively.
(4) H⁺ and OH^– ions combine together to form weakly dissociating H₂O. As there are no free H⁺ ions and OH^– ions, the solution is neutral and the salt does not undergo hydrolysis.
NaCl + H₂O ⇌ NaOH + HCl
Ionic equation :
Na⁺ + Cl^– +H₂O ⇌ Na⁺ + OH^– + H⁺ + Cl^–
H₂O ⇌ H⁺ + OH^–
Since the solution contains equal number of H⁺ and OH^– ions, it is neutral.
Hence the salt of strong acid and strong base does not undergo hydrolysis.

Question 31.
Explain the hydrolysis of the salt of strong acid and weak base.
OR
A solution of CuSO₄ reacts acidic. Explain.
Answer:
(1) Consider a salt of strong acid and weak base, like CuSO₄ obtained from strong acid H₂SO₄ and weak base Cu(OH)₂.
(2) When it is dissolved in water it undergoes hydrolysis as follows :

(3) Since [H₃O⁺] > [OH^–], the solution is acidic in nature.

Question 32.
Explain the hydrolysis of a salt of weak acid and strong base.
OR
A solution of sodium acetate, CH₃COONa reacts basic explain.
Answer:
(1) Consider a salt of weak acid and strong base like CH₃COONa. In aqueous solution it undergoes hydrolysis as follows :

(2) Since the base NaOH is strong, it dissociates completely while acid CH₃COOH being weak dissociates partially.
(3) Hence [OH^–] > [H₃O⁺] in the solution and the solution reacts basic.

Question 33.
Explain the hydrolysis of the salt of weak acid and weak base.
Answer:
(1) Consider a salt BA of weak acid (HA) and weak base (BOH).
(2) In aqueous solution it undergoes hydrolysis as follows :

(3) The nature of the solution will depend upon relative strength of weak acid and weak base, hence will depend upon their dissociation constants K_a and K_b.

(i) A salt of weak acid and weak base for which K_a > K_b :
Consider hydrolysis of NH₄F.

Since K_a (7.2 × 10^-4) for HF is greater than K_b (1.8 × 10^-5) for NH₄OH, the acid dissociates partially more than the base, hence, [H₃O⁺] > [OH^–] and the solution reacts acidic after hydrolysis.

(ii) A salt of weak acid and weak base for which K_a < K_b :
Consider hydrolysis of NH₄CN.

Since K_a (4 × 10^-10) for HF is less than K_b (1.8 × 10^-5) for NH₄OH, the base dissociates more than acid and hence [H₃O⁺] < [OH^–] and the solution reacts basic after hydrolysis.

(iii) A salt of weak acid and weak base for which K_a = K_b:
Consider hydrolysis of CH₃COONH₄.

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^–] and the solution reacts neutral after hydrolysis.

Question 34.
Define buffer solution.
OR
What is buffer solution?
Answer:
Buffer solution : It is defined as a solution which resists the change in pH even after the addition of a small amount of a strong acid or a strong base or on dilution or on addition of water.

Question 35.
What are the types of buffer solutions ?
Answer:
These are two types of buffer solutions :
(A) Acidic buffer solution : it is a solution containing a weak acid e.g. (CH₃COOH) and its salt of a strong base. e.g. (CH₃COONa).
pH of an acidic buffer is given by following Henderson Hasselbalch equation,
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}\)
where pK_a = -log₁₀ K_a
and K_a is the dissociation constant of weak acid.

(B) Basic buffer solutions : It is a solution containing a weak base (e.g. NH₄OH) and its salt of strong acid, (e.g. NH₄Cl).
pOH of a basic buffer is given by Henderson Hassebalch equation,
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}\)
where pK_b = -log10 K_b
and K_b is the dissociation constant of a weak base.

Question 36.
Explain a buffer action of an acidic buffer.
Answer:
Mechanism of action of an acidic buffer :
(1) An acidic buffer is a mixture of a weak acid and its salt with a strong base. The weak acid dissociates feebly, but the salt dissociates almost completely. Moreover, due to the common ions, largely supplied by the salt, dissociation of the weak acid is further suppressed.
(CH₃COOH + CH₃COONa) :
CH₃COONa_(aq) → CH₃COO_(aq) + \(\mathrm{Na}_{(\mathrm{aq})}^{+}\) (Complete)

(2) When a small quantity of strong acid (H⁺) is added to this mixture, hydrogen ions combine with acetate ions to form undissociated acetic acid. Thus, addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COOH}_{(\mathrm{aq})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base (OH^–) is added, the hydroxide ions react with the acid producing the corresponding anions and water. Thus, the concentrations of H⁺ and OH^– in the solution do not change and the pH remains constant.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})}+\mathrm{OH}_{(\mathrm{aq})}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added OH^– is called reserved acidity.

Question 37.
Explain a buffer action of a basic buffer.
Answer:
Mechanism of action of a basic buffer :
(1) A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. Moreover, due to the presence of the common ion, largely supplied by the salt, the dissociation of the base is further suppressed. (NH₄OH + NH₄Cl) :
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \rightarrow \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\) (Complete)

(2) When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water. Thus, the addition of an acid does not change the pH of the buffer.
\(\mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)
This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base is added, the hydroxide ions combine with \(\mathrm{NH}_{4}^{+}\) ions to form undissociated NH₄OH. As a result, the hydrogen or hydroxyl ion concentration does not change. Thus, the pH of the solution does not change.
\(\mathrm{OH}_{(\mathrm{aq})}^{-}+\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}\)
This removal of added OH^– is called reserved acidity.

Question 38.
What are properties of a buffer solution ?
OR
What are the advantages of a buffer solution ?
Answer:
Properties (or advantages) of a buffer solution :

1. The pH of a buffer solution is maintained appreciably constant.
2. By addition of a small amount of an acid or a base pH does not change.
3. On dilution with water, pH of the solution doesn’t change.

Question 39.
What are the applications of a buffer solution ?
Answer:
Buffer solutions have many applications as follows :
(1) In a biochemical system : Blood in our body has pH 7.36 – 7.42 due to (\(\mathrm{HCO}_{3}^{-}+\mathrm{H}_{2} \mathrm{CO}_{3}\)) and little change of 0.2 pH unit may be fatal. For example, saline solution used in intravenous injection contains a buffer solution maintaining pH of the blood in the required range.

(2) Agriculture : The properties of soil depend upon its pH. The salts present in soil such as phosphates, carbonates, bicarbonates and organic acids impart definite pH to the soil. Depending on pH the fertilizers are selected.

(3) Industry : In many industries, buffer solutions are used to carry out chemical processes very effectively, such as the industries of paper, dye, paints, drugs, ink, etc.

(4) Medicines : Many medicines particularly in the liquid state have a good stability and optimum activity at a definite pH, for which buffer solutions are used. For example penciline preparations are carried out in the presence of a buffer of sodium citrate. A buffer solution of magnesium citrate is prepared by adding citric acid to Mg(OH)₂.

(5) Analytical chemistry : In a qualitative analysis, the precipitation of groups, the chemical tests for detection of ions, etc. are carried out at a definite pH. For example, precipitation of cations of IIIA are carried in the presence of a basic buffer of pH 8 – 10 obtained by using NH₄OH and NH₄Cl.

Solved Examples 3.8

Question 40.
Solve the following :

(1) Calculate the pH of a buffer solution containing 0.1 M CH₃COOH and 0.05 M CH₃COONa. Dissociation constant of CH₃COOH is 1.8 × 10^-5 at 25 °C.
Solution :
Given : [CH₃COOH] = 0.1 M,
[CH₃COONa] = 0.05 M; K_a = 1.8 × 10^-5; pH = ?
pK_a = -log₁₀ K_a
= -log₁₀ 1.8 × 10^-5
= -(\(\overline{5} \cdot 2553\))
= 5 – 0.2553
= 4.7447
pH = \(\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
= \(4.7447+\log _{10} \frac{0.05}{0.1}\)
= 4.7447 + \((\overline{1} .6990)\)
= 4.7447 + (-1 + 0.6990)
= 4.7447 – 0.3010
= 4.4437
Ans. pH = 4.4437

(2) A buffer solution contains 0.3 M NH₄OH and 0.4 M NH₄Cl. If K_b for NH₄OH is 1.8 × 10^-5, calculate pH of the solution.
Solution :
Given : [NH₄OH] = 0.3 M; [NH₄Cl] = 0.4 M
K_b =1.8 × 10^-5; pH = ?
pK_b = -log₁₀ K_b
= -log₁₀ 1.8 × 10^-5
= \(-(\overline{5} .2553)\)
= 4.7447
pOH = \(\mathrm{p} K_{\mathrm{b}}+\log _{10} \frac{[\text { Salt }]}{[\mathrm{Base}]}\)
= \(4.7447+\log _{10} \frac{0.4}{0.3}\)
= 4.7447 + log₁₀ 1.333
= 4.7447 + 0.1248
= 4.8695
∵ pH + pOH = 14
∴ pH = 14 – pOH
= 14 – 4.8695
= 9.1305
Ans. pH = 9.1305.

(3) 0.2 dm³ acidic buffer solution contains 1.18 g acetic acid and 2.46 g sodium acetate. If K_a for acetic acid is 1.8 × 10^-5 at 25 °C, find pH of the solution.
Solution :

(4) A basic buffer solution contains 0.3 M NH₄OH and 0.2 M (NH₄)₂SO₄. If K_b for NH₄OH at a certain temperature is 2 × 10^-5, what is the pH of the solution ?
Solution :
Given : [NH₄OH] = 0.3 M
[NH₄⁺ ] = 2 × 0.2 = 0.4 M

pH + pOH = 14
∴ pH = 14 – pOH = 14 – 4.8239 = 9.1761
Ans. pH = 9.1761

Question 41.
Define solubility. How is it expressed ?
Answer:
Solubility : It is defined as the maximum amount of a substance in moles, that can be dissolved at constant temperature to give one litre of its saturated solution.
It is expressed in moles per litre or moles per decimeter cube of a saturated solution at given temperature.

Question 42.
Derive a relationship between solubility and solubility product.
Answer:
Consider a saturated solution of a spraingly soluble electrolyte (or salt) AxBy at a given constant temperature. Let S mol dm^-3 be the solubility of AxBy.
A following heterogeneous ionic equilibrium exists.

By a law of mass action, the equilibrium constant K will be represented as,

Since the active mass (concentration) of pure solid, A_xB_y(s) is treated as constant, [A_xB_y(s)] = K’
K × [A_xB_y(s)] = K × K’ = K_(sp)
Therefore, K_sp = [A^y+]^x × [B^x-]^y
where K_sp is called solubility product of A_xB_y.
At equilibrium the concentrations are,
[A^y+] = xS mol dm^-3
[B^x-]^y = yS mol dm^-3
∴ K_sp = [A^y+]^x × [B^x-]^y
= (xS)^x × (yS)^y
∴ K_sp = x^x.y^y.(S)^x+y ……….(1)
Hence solubility S is given by,
S = \(\left(\frac{K_{(\mathrm{sp})}}{x^{x} \cdot y^{y}}\right)^{\frac{1}{x+y}} \mathrm{~mol} \mathrm{dm}^{-3}\) …………(2)
The above equations, (1) and (2) give the relationship between solubility and solubility product.
Here x and y represent number of cations and anions respectively from the electrolyte.

Question 43.
Write expression for solubility and solubility product of following sparingly soluble salts : (1) AgBr (2) PbI₂ (3) Al(OH)₃
Answer:
In general, for a sparingly soluble salt A_xB_y,
K_sp = x^x.y^y.(S)^x+y

Question 44.
What is ionic product?
Answer:
Ionic product (IP) : It is defined as the product of concentrations in mol dm^-3 of ions of an electrolyte in the solution and denoted by IP.
In a saturated solution,
IP = K_sp where K_sp is the solubility product of the electrolyte.

Solved Examples 3.9

Question 45.
Solve the following :

(1) The solubility of AgBr in water is 1.28 × 10^-5 mol/dm³ at 298 K. Calculate the solubility product of AgBr at the same temperature.
Solution :
Given : S = 1.28 × 10^-5 mol dm^-3; K_sp = ?
AgBr dissociates as,
\(\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Br}_{(\mathrm{aq})}^{-}\)
K_sp = [Ag⁺] [Br^–]
As the solubility of AgBr in water is 1.28 × 10^-5 moles/dm³,
[Ag⁺] = [Br^–] = 1.28 × 10^-5 mol dm³
∴ K_sp = [1.28 × 10^-5] [1.28 × 10^-5]
= 1.638 × 10^-10
Ans. Solubility product of AgBr = 1.638 × 10^-10.

(2) The solubility of lead sulphate is 3.03 × 10^-5 kg/dm³. Calculate its solubility product. [Molecular mass of PbSO₄ = 303]
Solution :
Given : S = 3.03 × 10^-5 kg dm^-3, K_sp = ?
Lead sulphate dissociates as
\(\begin{aligned}
&\mathrm{PbSO}_{4} \rightleftharpoons \mathrm{Pb}^{2+}+\mathrm{SO}_{4}^{2-} \\
&\text { (solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
Molecular weight of PbSO₄ = 303
= 303 × 10^-3 kg
The solubility of PbSO₄ is 3.03 × 10^-5 kg/dm³.
Solubility in mol dm^-3
Weight of PbSO₄ per dm³ Molecular weight 3.03 × 10^-5

(3) The solubility product of AgBr is 3.3 × 10^-12 at 298 K. What concentration of Br^– ion is needed to precipitate AgBr from solution of 0.01 M Ag⁺?
Solution :
Given : K_sp = 3.3 × 10; [Ag⁺] = 1 × 10^-2 M;
[Br^–] = ?
AgBr dissociates as
\(\begin{aligned}
&\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-} \\
&\text {(solid) } \quad \text { (aq) } \quad \text { (aq) }
\end{aligned}\)
K_sp = [Ag⁺] [Br^–]
K_sp = Solubility product = 3.3 × 10^-12
[Ag⁺] = Concentration of Ag⁺
= 0.01 = 1.0 × 10^-2 M
[Br^–] = Concentration of Br^– = ?
∴ 3.3 × 10^-12 = 1.0 × 10^-2 × [Br^–]
∴ [Br^–] = 3.3 × 10^-10 mol/dm³
Ans. The concentration of Br^– required for precipitation of AgBr should be greater than 3.3 × 10^-10 mol/dm³.

(4) The solubility product of magnesium hydroxide is 1.4 × 10^-11. Calculate the solubility of magnesium hydroxide.
Solution :
Given : K_sp =1.4 × 10^-11; S = ?
Magnesium hydroxide dissociates as shown below :
Mg(OH)₂ ⇌ Mg²⁺ + 2(OH)^–
K_sp = [Mg²⁺] [OH^–]²
Let the solubility of Mg(OH)₂ be S mol dm^-3.
∴ [Mg²⁺] = Concentration of Mg²⁺ ions
= S mol dm^-3
∴ [OH^–] = Concentration of OH^– ions
= 2S mol dm^-3
∴ K_sp = S × (25)² = 4S³
K_sp = 1.4 × 10^-11
∴ 1.4 × 10^-11 =4S³
1.4 × 10^-11 = 4S³.

S = 1.518 × 10^-4 mol dm^-3
Ans. Solubility of Mg(OH)₂
= 1.518 × 10^-4 mol dm^-3

(5) The solubility of silver chloride is 1.562 × 10^-10 mol dm^-3 at 298 K. Find its solubility in g dm^-3 at the same temperature.
Solution :
Given :
Solubility of AgCl = S = 1.562 × 10^-10 mol dm^-3
Solubility of AgCl in g dm^-3 = ?
Molar mass of AgCl = M = 143.5 g mol^-1
Solubility in gram per dm³
= solubility in mol dm^-3 × molar mass
= 1.562 × 10^-10 × 143.5
= 2.241 × 10^-5 g dm-3
Ans. Solubility of AgCl = 2.241 × 10^-8 g dm^-3

(6) The solubility of PbSO₄ in water is 0.038 g dm^-3 at room temperature. Calculate its solubility and solubility product at the same temperature. (Atomic weights : Pb = 207.3, S = 32, O = 16)
Solution :
Given : Solubility of PbSO₄ = 0.038 g/dm^-3
Molar mass of PbSO₄ = 303.3 g mol^-1
Solubility in mol dm^-3 = ?
K_sp =?
Solubility in mol dm^-3 = \(\frac{0.038}{303.3}\)
= 1.253 × 10^-4 mol dm^-3

Ans. S = 1.253 × 10^-4 mol dm^-3;
K_sp = 1.57 × 10^-8

(7) The solubility product of PbS at 298 K is 4.2 × 10^-28, The concentration of Pb⁺⁺ ion is 0.001 M. Calculate S^2- ion concentration at which PbS just gets precipitated.
Solution :
Given :
Solubility product of PbS = K_sp = 4.2 × 10^-28
Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M
Concentration of S^{– –} = [S^{– –}] = ?
For PbS,
\(\mathrm{PbS}_{(\mathrm{s})} \rightleftharpoons \mathrm{Pb}^{++}+\mathrm{S}^{–}\)
∴ K_sp = [Pb⁺⁺] × [S^{– –}]
∴ [S–] = \(\frac{K_{\mathrm{sp}}}{\left[\mathrm{Pb}^{++}\right]}\)
= \(\frac{4.2 \times 10^{-28}}{0.001}\)
= 4.2 × 10^-25 M
To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28.
Hence, [S^{– –}] > 4.2 × 10^-25 M.
Ans. Concentration of S^{– –} required > 4.2 × 10^-25 M

(8) At 298 K, the solubility of silver sulphate is 1.85 × 10^-2 mol dm^-3. Calculate the solubility product of silver sulphate.
Solution :
Given : Silver sulphate dissociates as follows :

Solubility of Ag₂SO₄ = 1.85 × 10^-2 mol dm^-3
K_sp = Solubility product of Ag₂SO₄ = ?
[Ag⁺] = Concentration of Ag⁺ ion
= 2 × 1.85 × 10^-2 = 3.70 × 10^-2 mol dm^-3
[latex]\mathrm{SO}_{4}^{2-}[/latex] = Concentration of \(\mathrm{SO}_{4}^{2-}\)
= 1.85 × 10^-2 mol dm^-3
∴ K_sp = (3.70 × 10^-2)² × (1.85 × 10^-2)
= 13.69 × 10^-4 × 1.85 × 10^-2
= 25.33 × 10^-6
= 2.533 × 10^-5
Ans. Solubility product of Ag₂SO₄
= 2.533 × 10^-5

Question 46.
What is common ion?
Answer:
Common ion : An ion common to two electrolytes is called common ion. This is generally applicable to a mixture of a strong and a weak electrolyte. For example, a solution containing weak electrolyte CH₃COOH and strong electrolyte salt CH₃COONa.
CH₃COONa → CH₃COO^– + Na⁺;
CH₃COOH ⇌ CH₃COO^– + H⁺
Hence CH₃COOH and CH₃COONa have a common ion CH₃COO^–.

Question 47.
Define the term common ion effect.
Answer:
Common ion effect : The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte having an ion in common with the weak electrolyte is called common ion effect. For example, CH₃COOH and CH₃COONa have common ion CH₃COO^–.

Question 48.
Explain common ion effect with suitable example.
Answer:
A weak electrolyte dissociates partially in aqueous solution to produce cations and anions. Equilibrium exists between ions thus formed and the undissociated molecules.
BA ⇌ B⁺ +A^–
For such an equilibrium, the dissociation constant K is defined as
K = \(\frac{\left[\mathrm{B}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{BA}]}\)
K is constant for the weak electrolyte at a given temperature.
Now, if another electrolyte BC or DA is added to the solution BA, having a common ion either B⁺ or A^–, then the concentration of either B⁺ or A^– is increased. However, as K is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect. For example, the dissociation of a weak acid CH₃COOH is suppressed by adding CH₃COONa having common ion CH₃COO^–.
CH₃COOH ⇌ CH₃COO^– + H⁺
CH₃COONa → CH₃COO^– + Na⁺

Question 49.
Explain the common ion effect on dissociation of a weak acid.
Answer:
(1) Consider the dissociation or ionisation of a weak acid, CH₃COOH in its solution.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+}\)
The dissociation constant Ka for CH₃COOH will be,
K_a = \(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\)
K_a is constant for CH₃COOH at constant temperature.

(2) If a strong electrolyte like salt CH₃COONa is added to the solution of CH₃COOH, then on dissociation it gives a common ion CH₃COO^–.
CH₃COONa → CH₃COO^– + Na⁺

(3) Due to common ion CH₃COO^–, overall concentration of CH₃COO^– in the solution is increased, which increases the ratio,
\(\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \times\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}\). In order to keep this ratio constant, the concentration of H⁺ is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier’s principle.

(4) Thus the ionisation of a weak acid is suppressed by a common ion.

Question 50.
Explain the effect of common ion on the dissociation of weak base.
Answer:
(1) Consider the dissociation or ionisation of a weak base, NH₄OH in its dilute solution. \(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
The dissociation constant K_b for NH₄OH will be,
\(K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}\)

(2) If a strong electrolyte like salt NH₄Cl is added to the solution of NH₄OH, then it gives common ion \(\mathrm{NH}_{4}^{+}\).
NH₄Cl → \(\mathrm{NH}_{4}^{+}\) + Cl^–

(3) Due to common ion \(\mathrm{NH}_{4}^{+}\), overall concentration of \(\mathrm{NH}_{4}^{+}\) is increased, which increases the ratio \(\left[\mathrm{NH}_{4}^{+}\right]\) × [OH^–]/[NH₄OH],
In order to keep this ratio constant, the equilibrium is shifted to the left hand side which satisfies Le Chatelier’s principle.

(4) Thus the ionisation of a weak base is suppressed by a common ion.

Multiple Choice Questions

Question 51.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. According to Lowry-Bronsted concept, base is a substance which acts as –
(a) a proton donor
(b) an electron donor
(c) a proton acceptor
(d) an electron acceptor
Answer:
(c) a proton acceptor

2. BF₃ is a
(a) Lewis acid
(b) Lewis base
(c) amphoteric compound
(d) Electrolyte only
Answer:
(a) Lewis acid

3. Which of the following is a conjugate acid-base pair ?
(a) HCl, NaOH
(b) KCN, HCN
(c) NH₄Cl, NH₄OH
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)
Answer:
(d) H₂SO₄, \(\mathrm{HSO}_{4}^{-}\)

4. The conjugate acid of \(\mathrm{NH}_{2}^{-}\) is
(a) NH₃
(b) NH₂OH
(c) \(\mathrm{NH}_{4}^{+}\)
(d) N₂H₄
Answer:
(a) NH₃

5. Which of the following molecules is not a Lewis base?
(a) H₂O
(b) BF₃
(c) NH₃
(d) CO
Answer:
(b) BF₃

6. In the following reaction
\(\mathrm{HC}_{2} \mathrm{O}_{4(\mathrm{aq})}^{-}+\mathrm{PO}_{4}^{3-} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{C}_{2} \mathrm{CO}_{4}^{2-}\) Which of two are Lowry-Bronsted bases ?
(a) \(\mathrm{HC}_{2} \mathrm{C}_{4}^{-} \text {and } \mathrm{PO}_{4}^{3-}\)
(b) \(\mathrm{HPO}_{4}^{2-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
(c) \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} \text {and } \mathrm{HPO}_{4}^{2-}\)
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)
Answer:
(d) \(\mathrm{PO}_{4}^{3-} \text { and } \mathrm{C}_{2} \mathrm{O}_{4}^{2-}\)

7. According to the Arrhenius theory,
(a) an acid is a proton donor
(b) an acid is an electron pair acceptor
(c) a hydrogen ion exists freely in an aqueous solution
(d) a hydrogen ion is always hydrated to form a hydrogen ion
Answer:
(c) a hydrogen ion exists freely in an aqueous solution

8. The species which will behave both as a conjugate acid and base is
(a) NH₄OH
(b) \(\mathrm{CO}_{3}^{–}\)
(c) \(\mathrm{HSO}_{4}^{-}\)
(d) H₂SO₄
Answer:
(c) \(\mathrm{HSO}_{4}^{-}\)

9. According to the Lewis theory, an acid is
(a) nucleophile
(b) an electrophile
(c) a proton acceptor
(d) an electron donor
Answer:
(b) an electrophile

10. If a 0.1 M solution of HCN is 0.01% dissociated, the dissociation constant for HCN is,
(a) 10^-3
(b) 10⁺³
(c) 10^-7
(d) 10^-9
Answer:
(d) 10^-9

11. The pH of decimolar solution KOH is
(a) 1
(b) 4
(c) 10
(d) 13
Answer:
(d) 13

12. Ostwald’s dilution law is applicable in case of dilute solution of
(a) HCl
(b) H₂SO₄
(c) NaOH
(d) CH₃COOH
Answer:
(d) CH₃COOH

13. The degree of dissociation of a 0.1 M monobasic acid is 0.4%. Its dissociation constant is
(a) 0.4 × 10^-4
(b) 4.0 × 10^-4
(c) 1.6 × 10^-6
(d) 0.8 × 10^-5
Answer:
(c) 1.6 × 10^-6

14. The ionic product of water will increase, if
(a) Pressure is decreased
(b) H⁺ ions are added
(c) OH^– ions are added
(d) Temperature is increased
Answer:
(d) Temperature is increased

15. The [OH^–] for a weak base of dissociation constant K_b and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{b}}}{C}}\)
(b) KbC
(c) \(\sqrt{K_{\mathrm{b}} C}\)
(d) \(\frac{C}{K_{\mathrm{b}}}\)
Answer:
(c) \(\sqrt{K_{\mathrm{b}} C}\)

16. The [H⁺] for a weak acid of dissociation constant K_a and concentration C is nearly equal to
(a) \(\sqrt{\frac{K_{\mathrm{a}}}{C}}\)
(b) \(\sqrt{K_{\mathrm{a}} C}\)
(c) \(\frac{K_{\mathrm{a}}}{\sqrt{c}}\)
(d) \(\frac{c}{K_{\mathrm{a}}}\)
Answer:
(b) \(\sqrt{K_{\mathrm{a}} C}\)

17. Which of the following solution with same concentration will have highest pH
(a) Al(OH)₃
(b) K₂CO₃
(c) NH₄OH
(d) NaOH
Answer:
(d) NaOH

18. 10 ml of 0.1 M H₂SO₄ is mixed with 20 ml of 0.1 M KOH, the pH of resulting solution will be
(a) 0
(b) 7
(c) 2
(d) 9
Answer:
(b) 7

19. The gastric juice in our stomach contains enough hydrochloric acid to make the hydrogen ion concentration 0.01 mol/dm³. The pH of gastric juice is-
(a) 0.01
(b) 1
(c) 2
(d) 14
Answer:
(c) 2

20. If the hydrogen ion concentration of an acid is decreased ten times, its pH will be
(a) increased by one
(b) decreased by one
(c) remains unchanged
(d) increase by 10
Answer:
(a) increased by one

21. Which of the following metal sulphide is precipitated in an acidic medium ?
(a) NiS
(b) CoS
(c) CuS
(d) MnS
Answer:
(c) CuS

22. The relationship between the solubility and solubility product for silver carbonate is
(a) K_sp = s²
(b) \(\sqrt{K_{\mathrm{sp}}}\) = 4s²
(c) K_sp = 27S⁴
(d) K_sp = 4s³
Answer:
(d) K_sp = 4s³

23. If ‘S’ is solubility in mol dm^-3 and K_sp is solubility product of BA₂ type of salt, then relation between them is
(a) S = \(\sqrt{K_{\mathrm{sp}}}\)
(b) K_sp = 4S³
(c) K_sp = S³
(d) S = K_sp
Answer:
(b) K_sp = 4S³

24. The addition of solid sodium carbonate to pure water results in
(a) an increase in H⁺ ion concentration
(b) an increase in pH
(c) no change in pH
(d) a decrease in OH^– concentration
Answer:
(b) an increase in pH

25. Which of the following salt, when dissolved in water will hydrolyse ?
(a) NaCl
(b) NH₄Cl
(c) KCl
(d) Na₂SO₄
Answer:
(b) NH₄Cl

26. A solution of blue vitriol is acidic in nature because
(a) CuSO₄ reacts with water
(b) Cu²⁺ ions reacts with water
(c) SO₄^2- ions reacts with water
(d) CuSO₄ removes OH^– ions from water
Answer:
(a) CuSO₄ reacts with water

27. What is the nature of the solution of salt FeCl₃ ?
(a) Acidic
(b) Basic
(c) Neutral
(d) Amphoteric
Answer:
(a) Acidic

28. Which of the following salts does not hydrolyse in water ?
(a) Sodium acetate
(b) Sodium carbonate
(c) Sodium nitrate
(d) Sodium cyanide
Answer:
(c) Sodium nitrate

29. An aqueous solution of magnesium chloride changes blue litmus red due to
(a) the formation of Cl^– ions
(b) the formation Mg²⁺ ions
(c) reaction of Cl^– ions with water
(d) hydrolysis of the salt
Answer:
(d) hydrolysis of the salt

30. An aqueous solution of which of the following salts is basic ?
(a) CH₃COONa
(b) NH₄Cl
(c) KNO₃
(d) CuSO₄
Answer:
(a) CH₃COONa

31. The number of moles of hydroxide ions (OH^–) produced from 2 moles of Na₂CO₃ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

32. The POH value for solution is 4, its hydrogen ion concentration will be
(a) 10^-4
(b) 10^-10
(c) 10¹⁰
(d) 10⁴
Answer:
(b) 10^-10

33. If an acid is diluted
(a) pH increases
(b) pH decreases
(c) no change occurs
(d) can vary depending on an acid
Answer:
(a) pH increases

34. pH of a solution is 13. H⁺ ions present in 1 cm³ of the solution is
(a) 6.023 × 10¹⁰
(b) 6.023 × 10⁷
(c) 6.023 × 10^-10
(d) 6.023 × 10^-7
Answer:
(b) 6.023 × 10⁷

35. pH of blood is maintained constant by mechanism of
(a) common ion effect
(b) buffer
(c) solubility
(d) all of these
Answer:
(b) buffer

36. The pH of 0.05 M solution of dibasic acid is
(a) +1
(b) -1
(c) +2
(d) -2
Answer:
(a) +1

37. The pH of a 0.63% nitric acid solution is (Equivalent weight of nitric acid is 63)
(a) 6
(b) 7
(c) 1
(d) 9
Answer:
(c) 1

38. 100 ml of 0.01 M solution of NaOH is diluted to 1 dm³. What is the pH of the dilute solution?
(a) 12
(b) 11
(c) 2
(d) 3
Answer:
(b) 11

39. If the H⁺ ion concentration in a solution is 0.01 M, the pOH of the solution is
(a) 12
(b) 10^-10
(c) 2
(d) 14
Answer:
(a) 12

40. If the pH value of a solution is zero, the solution is
(a) a strong acid
(b) a very weak acid
(c) neutral
(d) a base
Answer:
(a) a strong acid

41. The pH of a solution is 5, when the hydroxyl ion concentration is
(a) 10^-5 mol/dm³
(b) 10^-7 mol/dm³
(c) 10^-9 mol/dm³
(d) 10^-14 mol/dm³
Answer:
(c) 10^-9 mol/dm³

42. The pH of human blood in a normal person is approximately
(a) 4.7
(b) 6.04
(c) 7.40
(d) 8.74
Answer:
(c) 7.40

43. If molarity of NaOH is 3.162 × 10^-3 M, its pH is
(a) 8.5
(b) 9.5
(c) 10.5
(d) 11.5
Answer:
(d) 11.5

44. The common ion effect is based on
(a) Sorensen’s principle
(b) Le Chatelier’s principle
(c) Heisenberg’s principle
(d) Freundlich’s principle
Answer:
(b) Le Chatelier’s principle

45. The ion that cannot be precipitated by both HCl and H2S is
(a) Pb²⁺
(b) Cu²⁺
(c) Ag⁺
(d) Ca²⁺
Answer:
(d) Ca²⁺

46. The correct representation for solubility product of SnS2 is
(a) [Sn⁴⁺] [S^2-]²
(b) [Sn⁴⁺] [S^2-]
(c) [Sn⁴⁺] [2S^2-]
(d) [Sn⁴⁺] [2S^2-]
Answer:
(a) [Sn⁴⁺] [S^2-]²

47. The solubility product of a salt BA at room temperature is 1.21 × 10^-6. Its molar solubility is
(a) 1.21 × 10^-3 M
(b) 1.1 × 10^-4 M
(c) 1.1 × 10^-3 M
(d) 1.21 × 10^-2 M
Answer:
(c) 1.1 × 10^-3 M

48. Among the following hydroxides, the one which has the lowest value of solubility product at temperature 298 K is,
(a) Mg(OH)₂
(b) Ca(OH)₂
(c) Ba(OH)₂
(d) Be(OH)₂
Answer:
(d) Be(OH)₂

49. A solution becomes unsaturated when
(a) ionic product = solubility product
(b) ionic product < solubility product
(c) ionic product > solubility product
(d) ionic product ≥ solubility product
Answer:
(b) ionic product < solubility product

50. The solubility product of Fe(OH)₃ is
(a) [latex]\mathrm{F}_{\mathrm{e}}^{2+}[/latex] [OH^–]³
(b) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]²
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³
(d) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex]³ [OH^–]³
Answer:
(c) [latex]\mathrm{F}_{\mathrm{e}}^{3+}[/latex] [OH^–]³

51. The solubility product of PbS in 4.2 × 10^-28 at 300 K. The sulphide ions concentration required to precipitate PbS from a solution containing 0.001 M of lead ion is
(a) ≥ 2.1 × 10^-14 mol/dm³
(b) ≥ 4.2 × 10^-14 mol/dm³
(c) ≥ 4.2 × 10^-25 mol/dm³
(d) ≤ 4.2 × 10^-28 mol/dm³
Answer:
(c) ≥ 4.2 × 10^-25 mol/dm³

52. 0.025 M CH₃COOH is dissociated 9.5%. Hence the pH of the solution is
(a) 2.6244
(b) 3.128
(c) 2.988
(d) 2.267
Answer:
(a) 2.6244

53. 0.1 M HCN is dissociated 0.01%. The dissociation constant of HCN is
(a) 1.1 × 10^-6
(b) 1 × 10^-8
(c) 1 × 10^-9
(d) 1 × 10^-7
Answer:
(c) 1 × 10^-9

54. The solubility of product of a sparingly soluble salt AB₂ is 3.2 × 10^-11. If solubility in mol dm^-3 is
(a) 4 × 10^-4
(b) 3.2 × 10^-4
(c) 1 × 10^-5
(d) 2 × 10^-4
Answer:
(d) 2 × 10^-4

Maharashtra State Board HSC 12th Commerce Book Keeping & Accountancy Important Questions and Answers

• Chapter 1 Introduction to Partnership and Partnership Final Accounts Important Questions
• Chapter 2 Accounts of ‘Not for Profit’ Concerns Important Questions
• Chapter 3 Reconstitution of Partnership (Admission of Partner) Important Questions
• Chapter 4 Reconstitution of Partnership (Retirement of Partner) Important Questions
• Chapter 5 Reconstitution of Partnership (Death of Partner) Important Questions
• Chapter 6 Dissolution of Partnership Firm Important Questions
• Chapter 7 Bills of Exchange Important Questions
• Chapter 8 Company Accounts – Issue of Shares Important Questions
• Chapter 9 Analysis of Financial Statements Important Questions
• Chapter 10 Computer in Accounting Important Questions

Maharashtra Board 12th HSC Important Questions

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 12 Correspondence with Statutory Authorities

1A. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Headquarters of Ministry of Corporate Affairs.
Answer:
New Delhi

Question 2.
The number of Regional Directors.
Answer:
Seven

Question 3.
Total number of Registrar of Companies.
Answer:
Twenty-two

Question 4.
The number of ROC’s – cum Official liquidator.
Answer:
Nine

Question 5.
Officer appointed by High Court to look into the matter of winding up of a company.
Answer:
Official Liquidator

Question 6.
Principal Bench of NCLT.
Answer:
New Delhi

1B. State whether the following statements are True or False.

Question 1.
MCA operates with the help of 10 Regional Directors.
Answer:
False

Question 2.
NCLT works through 25 benches.
Answer:
False

Question 3.
MCA supervises the working of professional bodies like ICAI, and ICSI.
Answer:
True

1C. Complete the sentences.

Question 1.
To issue an order for seizure of books and papers, ROC has to seek _____________
Answer:
Special Court

Question 2.
MCA is concerned with administration of _____________
Answer:
The Companies Act, 2013

Question 3.
MCA supervises professional body like _____________
Answer:
Institute of Secretaries of India (ICSI)

Question 4.
The headquarter of MCA is at _____________
Answer:
New Delhi

Question 5.
An organization who act as a middleman between capital provider and capital seekers is called as _____________
Answer:
Market Intermediaries

Question 6.
Informing about difficulty in uploading e-form means _____________
Answer:
Raising a ticket

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Public Deposits	……………………….
(2) ………………………	The agent between capital provider and seeker

(Market Intermediaries, 36 Months)
Answer:

Group ‘A’	Group ‘B’
(1) Public Deposits	36 Months
(2) Market Intermediaries	Agents between capital provider and seller

1E. Answer in one sentence.

Question 1.
Name the authority who supervises the working of ROC and the Official Liquidator.
Answer:
The authority who supervises the working of ROC’s and the Official Liquidator is Regional Director (RD).

Question 2.
Who can extend the period of helding the Annual General Meeting?
Answer:
Registrar of Company can extend the period of helding the Annual General Meeting.

Question 3.
Where can an appeal be made against the order of NCLAT?
Answer:
An appeal against the order of NCLAT can be made to Supreme Court within 60 days of receipt of the order of NCLAT.

Question 4.
What do you mean by Market Intermediaries?
Answer:
People or organizations who act as a middleman between the capital provider and capital seeker are called as Market Intermediaries. E.g. Stock Brokers, bankers, underwriters, etc.

1F. Correct the underlined word and rewrite the following sentences.

Question 1.
An appeal can be made against the order issued by NCLAT within 90 days.
Answer:
An appeal can be made against the order issued by NCLAT within 60 days.

Question 2.
An appeal against the order issued by NCLAT can be made to High Court.
Answer:
An appeal against the order issued by NCLAT can be made to Supreme Court.

Question 3.
MCA conducts its operations through 11 Regional Directors.
Answer:
MCA conducts its operations through 7 Regional Directors.

2. Answer in brief.

Question 1.
Give a flow chart to explain the organizational setup to administer the Companies Act, 2013.
Answer:
Organizational set up to administer the Companies Act, 2013.

3. Justify the following statements.

Question 1.
Explain the reasons for not holding Annual General Meeting by a company on time.
Answer:
Reasons for not holding Annual General Meeting on time:

• Directors traveling abroad, so cannot attend Annual General Meeting.
• Employees Strike in a company.
• A raid by Income Tax Department.
• Annual Financial statement not approved or not audited or incomplete audit or loss of financial data due to natural calamity.

Question 2.
Under what circumstances, Secretary has to correspond with Statutory authorities.
Answer:
Under the following circumstances, Secretary has to correspond with a few of the Statutory authorities:

• To correspond with ROC for seeking extension of time for holding Annual General Meeting.
• To correspond with MCA for ‘Ticket Raising’ or other service-related technical complaints.
• To correspond with SEBI in reply to the complaint by the investor.
• To correspond with NCLT seeking extension of time to repay Public deposits.

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 15 Introduction to Polymer Chemistry

Question 1.
What are polymers?
Answer:
Polymers are high molecular mass macromolecules (10³ – 10⁷ u) and consist of repeating units of monomers.

Question 2.
Write the classification of polymers based on source. Give examples.
OR
What are natural and synthetic polymers? Give two examples of each type.
Answer:
Based on the source polymers are classified as natural, semisynthetic and synthetic polymers.

1. Natural polymers : These polymers are obtained either from plants or animals, e.g., cellulose, jute, linene, rubber, silk.
2. Semisynthetic polymers : The fibres obtained by giving special chemical treatment to natural fibres (cellulose) and further regenerated are called semisynthetic polymers e.g., acetate rayon, viscose rayon, cuprammonium silk.
3. Synthetic polymers : The man made fibres prepared by polymerization of one monomer or copolymerization of
   two or more monomers are called synthetic polymers, e.g., nylon, terylene, polythene, etc.

Question 3.
Write the classification of polymers based on structure. Give examples
OR
Write the reactions involved in the preparation of (1) Polyvinyl chloride (PVC) (2) Polypropylene.
Answer:
Based on structure polymers are classified as linear chain polymers, branched chain polymers and network or cross linked polymers.

(1) Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(2) Branched-chain polymers : These polymers consist of long and straight chain with smaller side chains give rise to branched-chain polymers. They have low density. They have lower melting points and tensile strength. Polypropylene having methyl groups as branches.

(3) Network or cross-linked polymers : These polymers consist of cross-linking of chains by strong covalent bonds leading to a network-like structure. Cross-linking results from polyfunctional monomers, e.g., melamine, bakelite, vulcanization of rubber. These polymers are hard rigid and brittle.

Question 3.
How are polymers classified on the basis of mode of polymerization?
OR
Write the classification of polymers based on mode of polymerization.
Answer:
There are three modes of polymerization depending upon the types of reactions taking place between the monomers.

1. Addition polymerization (or chain growth polymerization)
2. Condensation polymerization (or step growth polymerization)
3. Ring opening polymerization
4. Addition polymerization or chain growth polymerization : It is a process of polymers by the repeated addition of a large number of monomers is called addition polymerization (like alkenes) without loss of any small molecules.
   Example : Formation of polyethylene from ethylene is well known example of addition polymerization. It is a chain growth polymerization.
5. Condensation polymerization or step growth polymerization : The process of formation of polymers from polyfunctional monomers with the elimination of some small molecules such as water, hydrochloric acid, methanol, ammonia is called condensation polymerization.
   
   Example : The formation of terylene, a polyester polymer, from ethylene glycol and terephthalic acid. Condensation polymerization is a step growth polymerization.
6. Ring opening polymerization : The process of formation of polymers from cyclic compounds (like lactams, cyclic ethers, etc.) by ring opening is called ring opening polymerization.
   Example : Polymerization of e-caprolactam.
   Ring opening polymerization proceeds by addition of a single monomer unit to the growing chain molecules. It is a step growth polymerization.

Question 4.
Classify the polymers given below as addition, condensation and ring opening polymers:
PVC, polythene, cyclic ethers, polyester, polyacrylonftrile. polystyrene.
Answer:

• Addition polymers: PVC, polythene. polyacrylonitrile. polystyrene.
• Condensation polymers: Polyester.
• Ring opening polymers : Cyclic ethers

Question 5.
Write the classification of polymers based on intermolecular forces. Give examples.
OR
In which dasses, are the polymers classified on the basis of Inter molecular forces?
Answer:
Molecular forces bind the polymer chains either by hydrogen bonds or Vander Waal’s forces. These forces are called intermolecular forces. On the basis of magnitude of intcrmolccular forces, polymers are further classified as ebstomers, fibres, thermoplastic polymers. thermosetting polymers.

(1) Elastomers: Weak van der Waals type of intermolecular forces of attraction between the polymer chains are observed in cbstomcrs. When polymer is stretched, the polymer chain stretches and when the strain is relieved the chain returns to its odginal position, Thus, polymer shows elasticity and is called elastomers. Elastomers, the elastic polymers, have weak van der Waals type of intermolecular forces which permit the polymer to be stretched. Lilastorners are soft and stretchy and used in making rubber bands. E.g.. neoprene, vulcanized rubber, buna.S, buna-N.

(2) Fibres : It consists of strong intermolecular forces of’ attraction due to hydrogen bonding and strong dipole-dipole forces. These polymers possess high tensile strength. Due to these strong intermolecular forces the fibres are crystalline in nature. They are used in textile industries, strung tyres. etc.. e.g., nylon, terylene.

(3) Thermoplastic polymers: These polymer possess moderately strong intermolecular forces of attraction between those of elastomers and fibres. These polymers arc called thermoplastic because they become soft on heating and hard on cooling. They are either linear or branched chain polymers. They can be remoulded and recycled. E.g. polyethenc, PVC, polystyrene.

(4) Thermosetting polymers: These polymers are cross linked or branched molecules and are rigid polymers. During their formation they have property of being shaped on heating. but they get hardened while hot. Once hardened these become infusible, cannot be softened by heating and therefore, cannot be remoulded and recycled.
This shows extensive cross linking by covalent bonds formed in the moulds during hardening/setting process while hot. E.g. Bakelite, urea formaldehyde resin.

Question 6.
What is meant by the term homopolymer?
Answer:
A polymer made from only one type of repeating units is called homopolymer. in some cases the repeating unit is formed by condensation of two distinct monomers. Examples. Polythene, PVC. Nylon-6.

Question 7.
What is meant by the term copolymer?
Answer:
A polymer made from two or more types of repeating units is called a copolymer. The different monomer units are randomly sequenced in the copolymer, e.g., Terylene, Nylon-6, 6, Buna-S, Buna-N.

Question 8.
Write the classification of polymers on the basis of biodegradability?
OR
(1) What are biodegradable polymers?
(2) What are nonbiodegradable polymers?
Answer:
Based un biodegradability, polymers are classified as biodegradable polymer and nonbiodegradable polymers.

(1) Biodegradable polymers: Polymers that are affected by microbes or disintegrate by themselves afler a certain period of time due to environmental degradation are called biodegradable polymers.

Examples: PHBV i.e., Polyhydroxy butyrate-CO-β-hydroxy valerate Dextron. Nylon-2-nylon-6.

(2) Non biodegradable polymers: Synthetic polymers do not disintegrate by themselves after a certain period or not affected by microbes, are called nonbiodegradhle polymers.

Examples: Bakelite, Nylon, Terylene.

Question 9.
Explain the following terms :
Answer:

1. Branched chain polymer : The polymer consists of long and straight chain with smaller side chains give rise to branched chain polymers, e.g. Polypropylene
2. Addition polymer : The polymer formed by the repeated addition of a large number of monomers (like alkenes) without loss of any small molecules are called addition polymers, e.g. polythene -[-CH₂ – CH₂-]_n. It is a chain growth polymerization.
3. Condensation polymer : The polymers formed by the repeated condensation reaction between polyfunctional monomers with the elimination of some molecules such as water, hydrochloric acid, methanol, ammonia are called condensation polymers, e.g. Nylon-6, 6.
4. Elastomers : Polymers in which the intermolecular forces of attraction between the polymer chains are the weakest. When polymer is stretched, it has ability to stretch and when the strain is relieved it returns to its original position. Thus, polymer shows elasticity and is called elastomers, e.g. natural rubber, neoprene, vulcanized rubber.
5. Homopolymer : A polymer made from only one type of repeating unit of one monomer is called homopolymer e.g. Polythene, PVC, etc.
6. Biodegradable polymer : Polymers which are affected by microbes or disintegrate by themselves after a certain period of time due to environmental degradation are called biodegradable polymers.
   Example : PHSV i.e. Polyhydroxy butyrate -CO-β-hydroxy valerate Dextron.

Question 10.
What is natural rubber?
Answer:
Natural rubber is a high molecular mass linear polymer of isoprene (2-methyl-1, 3-butadiene).

Question 11.
Write a note on natural rubber.
Answer:
Natural rubber is manufactured from rubber latex obtained from the rubber tree in the form of colloidal suspension. Reaction involved in the formation of natural rubber by the process of addition polymerization is as follows :

Natural rubber is -1, 4- polyisoprene. It exhibits elastic properties.

Question 12.
State properties of natural rubber.
Answer:

1. Polyisoprene molecule has cis configuration of the C = C double bond. It consists of various chains held together by weak van der Waals forces and has coiled structure.
2. It can be stretched like a spring and exhibits elastic property.
3. Its molecular mass varies from 130,000 u to 340,000 u.

Question 13.
Write a note on vulcanization of rubber. OR Discuss the main purpose of vulcanization of rubber.
Answer:
The tensile strength, toughness and elasticity of natural rubber can be increased by adding 3 to 5% sulphur and heating at 100-150°C, resulting in cross linking of cis-1, 4-polypropene chains through disulphide bonds, (-S-S-). This process is known as vulcanization of rubber. The physical properties of rubber can be changed by controlling the amount of sulphur in the vulcanization process. Rubber made with 20-30% sulphur is hard, 3 to 10% sulphur is little harder and is used in making tyres and 1 to 3% sulphur is used in making rubber bands.

Question 14.
Write the name and structure of one of the initiators used in free radical polymerisation.
Answer:
The initiator used in free radical polymerisation is acetyl peroxide.

Question 15.
What is LDP? How is it prepared? Give its properties and uses.
Answer:
LDP means low-density polyethene. LDP is obtained by polymerization of ethylene under high pressure (1000 – 2000 atm) and temperature (350 – 570 K) in presence of traces of O₂ or peroxide as initiator.

The mechanism of this reaction involves free radical addition and H-atom abstraction. The latter results in branching. As a result the chains are loosely held and the polymer has low density.

Properties of LDP :

• LDP films are extremely flexible, but tough chemically inert and moisture resistant.
• It is poor conductor of electricity with melting point 110 °C.

Uses of LDP :

• LDP is mainly used in preparation of pipes for agriculture, irrigation, domestic water line connections as well as insulation to electric cables.
• It is also used in submarine cable insulation.
• It is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc.

Question 16.
Whatis HDP ? How is it prepared ? Give its properties and uses ?
Answer:
HDP means high density polyethylene. It is a linear polymer with high density due to close packing.

HDP is obtained by polymerization of ethene in presence of Zieglar-Natta catalyst which is a combination of triethyl aluminium with titanium tetrachloride at a temperature of 333K to 343K and a pressure of 6-7 atm.

Properties of HDP :

• HDP is crystalline, melting point in the range of 144 – 150 °C.
• It is much stiffer than LDP and has high tensile strength and hardness.
• It is more resistant to chemicals than LDP.

Uses of HDP :

• HDP is used in manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, etc.
• It is used to prepare laboratory wares and other objects where high tensile strength and stiffness is required.

Question 17.
How is polyacrylonitrile (PAN) prepared? Give its uses.
Answer:
Acrylonitrile (monomer) on polymerization (addition polymerization) in the presence of peroxide initiator gives polyacrylonitrile.

Uses : Polyacrylonitrile resembles wool and is used as wool substitute and for making orlon or acrilan.

Question 18.
How is nylon-6 prepared?
Write the reaction for the preparation of nylon 6.
Answer:
When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

Question 19.
Draw the structures of polymers formed using the following monomers.

Answer:

Question 20.
How is Novolac prepared?
OR
Write the reaction to prepare Novolac polymer.
Answer:
The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

Question 21.
How is bakelite prepared?
Answer:
In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 22.
How is a melamine-formaldehyde polymer (melamine) prepared?
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

Question 23.
Write the preparation of the following synthetic rubbers and give their uses :
(1) Buna-S or styrene-butadiene rubber (SBR) (2) Neoprene rubber
Answer:
(1) Buna-S rubber : Its trade name is SBR (Styrene-butadiene rubber) Buna-S is a copolymer of styrene and 1, 3-butadiene. When 75 parts of butadiene and 25 parts of styrene subjected to addition polymerization by the action of sodium.

Uses : Buna-S is superior to natural rubber with regard to mechanical strength and has abrasion resistance. Hence, it is used in tyre industry.

(2) Neoprene : Neoprene, a synthetic rubber, is a condensation polymer of chloroprene (2-chloro-l, 3-butadiene), Chloroprene polymerizes rapidly in presence of oxygen.

Vulcanization of neoprene takes place in presence of magnesium oxide.

Uses : Neoprene is resistant to petroleum, vegetable oils, light as well as heat. It is used in making hose pipes for transport of gasoline and making gaskets. It is used for manufacturing insulator cable, jackets, belts for power transmission and conveying.

Question 24.
Write structure of natural rubber and neoprene rubber along with the name and structure of their monomers.
Answer:

Question 25.
Write the preparation of viscose rayon.
Answer:
Viscose rayon is a semisynthetic fibre. It is a regenerated cellulose. The molecular formula of cellulose is (C₆H₁₀O₅)_n. A modified representation of the molecular formula of cellulose Cell-OH is used in the reactions involved in viscose formation.

Cellulose (wood pulp) is treated with the concentrated NaOH to form alkali cellulose. It is then converted to xanthate by treating with CS₂. Further xanthate is mixed with dilute NaOH to form viscose solution which is extruted through spinnerates of spinning machine into acid bath, when regenerated cellulose fibres precipitate, i.e. viscose rayon.

Question 26.
How is PHBV polymer prepared?
Answer:
It is a copolymer. The monomers β-hydroxy butyric acid (3-hydroxy butanoic acid) and β-hydroxy valeric acid (3-hydroxy pentanoic acid) undergo polymerization to form PHBV polymer. It has an ester linkage. Hydroxyl group of one monomer forms ester link by reacting with carboxyl group of the other. Thus PHBV is an aliphatic polyester i.e. poly β-hydroxybutyrate-co-β-hydroxy valerate (PHBV). It is a biodegradable polymer.

Question 27.
Write the name/s of monomer/s, polymer structure and one use of each of the following polymers (trade name) :
Answer:

Question 28.
Write the names of monomers used in preparing following polymers :
(1) Dacron.
Answer:
Monomers : Ethylene glycol and Dmiethyl terephthalate (DMT)

(2) Bakelite.
Answer:
Monomers : o-hydroxy methyl phenol and formaldehyde.

(3) Nylon-6, 8.
Answer:
Monomers : Hexamethylene diamine and Hexamethylene dicarboxylic acid.
H₂N-(CH₂)₆-NH₂ HOOC-(CH₂)₆-COOH

(4) Melamine.
Answer:
Monomers : Melamine and Formaldehyde

(5) Buna-S.
Answer:

(6) Buna-N.
Answer:

(7) Butyl rubber.
Answer:

(8) Teflon.
Answer:
Monomers : F₂C = CF₂ Tetrafluoroethene

(9) Natural rubber.
Answer:
Monomers : 1,3-Butadiene
CH₂ = CH – CH = CH₂

(10) Neoprene.
Answer:
Monomers : Chloroprene or 2-Chloro-l,3-butadiene

Question 29.
Write the structures of monomers used in the preparation of following polymers.

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomers used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer are :

Answer:
The monomer used in the preparation of above polymer is

Question 30.
Following monomers are used to prepare polymers. Predict the structures of polymers:

(1) Ethylene glycol.
Answer:
Ethylene glycol is used in the preparation of polyester (terylene or dacron).

Terylene is polyester fibre formed by the polymerization of terephthalic acid and ethylene glycol.

Terylene is obtained by condensation polymerization of ethylene glycol and terephthalic acid in presence of catalyst zinc acetate and antimony trioxide at high temperature.

Properties :

• Terylene has relatively high melting point (265 °C)
• It is resistant to chemicals and water.

Uses :

• It is used for making wrinkle-free fabrics by blending with cotton (terycot) and wool (terywool), and also as glass reinforcing materials in safety helmets.
• PET is the most common thermoplastic which is another trade name of the polyester polyethylenetereph- thalate.
• It is used for making many articles like bottles, jams, packaging containers.

(2) ε-Caprolactam.
Answer:
ε-caprolactam is used in the preparation of nylon-6.

When epsilon (ε)-caprolactam is heated with water at high temperature it undergoes ring opening polymerization to form the polyamide polymer called nylon-6.

The name nylon-6 is given on the basis of six carbon atoms present in the monomer unit. Nylon-6 has high tensile strength and luster, nylon-6 fibres are used for manufacture of tyre cords, fabrics and ropes.

(3) Ethene.
Answer:
Ethene is used in the preparation of polythene
Polymer:

Linear chain polymers : When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight-chain or long-chain polymer, e.g., polythene, PVC.

They have high melting points; high densities and high tensile strength.

(4) Formaldehyde.
Answer:
Formaldehye is used in the preparation of bakelite.

The monomers phenol and formaldehyde undergo polymerisation in the presence of alkali or acid as catalyst.

Phenol reacts with formaldehyde to form ortho or p-hydroxy methyl phenols, which further reacts with phenol to form a linear polymer called Novolac. It is used in paints.

In the third stage, various articles are shaped from novolac by putting it in appropriate moulds and heating at high temperature (138 °C to 176 °C) and at high pressure forms rigid polymeric material called bakelite. Bakelite is insoluble and infusible and has high tensile strength.

Bakelite is used in making articles like telephone instrument, kitchenware, electric insulators frying pans, etc.

Question 31.
Classify the following polymers as step growth or chain growth polymers :
(1) Nylon-6,6
(2) Terylene
(3) Polyethene (4) PVC.
Answer:
Step growth polymers : Nylon-6,6, terylene
Chain growth polymers : Polythene, PVC.

Question 32.
Classify the following as linear, branched or cross linked polymers :
(1) Bakelite
(2) Starch
(3) Polythene
(4) Nylon.
Answer:
Linear polymers : Polythene, nylon.
Cross-linked polymers : Bakelite, starch.

Question 33.
Classify the following as addition and condensation polymers :
(1) Bakelite
(2) polyvinyl chloride
(3) polythene
(4) terylene.
Answer:
Addition polymers : Polyvinyl chloride, polythene
Condensation polymers : Bakelite, terylene.

Question 34.
Arrange the polymers in increasing order of their intermolecular forces :
Nylon-6,6, Polythene, Buna-S.
Answer:
The increasing order of their intermolecular forces of attraction follows the order :
Buna-S, Polythene, Nylon-6,6.

Question 35.
Classify the following as natural, synthetic and semisynthetic polymers :
Terylene, cuprammonium silk, jute, melamine
Answer:
Natural polymers : Jute
Synthetic polymers : Terylene, melamine
Semisynthetic polymers : Cuprammonium silk

Question 36.
Complete the following statements :
(1) Chemically teflon is …………………………. .
(2) …………………………. is the catalyst used to obtain HDP by polymerisation of ethene.
(3) Viscose rayon is a …………………………. .
Answer:
(1) polytetrafluoroethylene
(2) Zieglar-Natta
(3) semisynthetic fibre (regenerated fibre).

Question 37.
Answer the following in one sentence each.

(1) Name a polymer used for making LCD screen?
Answer:
The polymer used for making LCD screen is Perspex.

(2) Which of the two is a condensation polymer? Bakelite or Polythene?
Answer:
The condensation polymer is bakelite.

(3) Which of the two is a linear polymer? Nylon or Starch.
Answer:
The linear polymer is nylon.

(4) Which of the two is a step growth polymer? Nylon-6,6 or PVC.
Answer:
The step growth polymer is Nylon-6,6.

(5) Write the use of polyacrylamide gel.
Answer:
Polyacrylamide gel is used in electrophoresis.

(6) Write the use of urea formaldehyde resin.
Answer:
Urea formaldehyde resin is used in making unbreakable dinnerware and decorative laminates.

(7) Give an example of semisynthetic polymer.
Answer:
Semisynthetic polymer : Viscose rayon, cuprammonium silk.

(8) Write the monomer unit of teflon.
Answer:
Monomer unit of teflon : Tetrafluoroethene (F₂C = CF₂).

(9) Write the equation for the preparation of polypropylene.
Answer:

(10) Name a synthetic polymer which contains amide linkage.
Answer:
Polymer that contains amide linkage : Nylon-6; Nylon 6,6.

(11) Name a synthetic polymer which contains ester linkage.
Answer:
Polymer that contains ester linkage : Terylene or Dacron.

(12) Name one thermosetting and one thermoplastic polymer.
Answer:
Thermosetting polymer : Bakelite.
Thermoplastic polymer : Polythene, polystyrene.

(13) State the uses of biodegradable polymers.
Answer:
Biodegradable polymers are used as orthopaedic devices, implants, sutures and drug release matrices.

(14) Name a copolymer which is used for making nonbreakable crockeries.
Answer:
The polymer used in making nonbreakable crockeries : Melamine formaldehyde polymer.

(15) Write the structure of monomer used in the preparation of
Answer:
The structure of monomer :

(16) Write the structure of melamine.
Answer:
The monomers formaldehyde and melamine undergo condensation polymerisation to form cross-linked melamine formaldehyde. It is used in making crockeries, decorative table tops like formica and plastic dinner-ware.

(17) What does SBR stand for?
Answer:
SBR stand for styrene(S)butadiene (B) rubber (R).

(18) Draw the structure of repeating unit in nylon-6,10.
Answer:
The structure of repeating unit in nylon-6,10 is

(19) What are the monomers used to prepare nylon given below?

Answer:
Monomers used in the preparation of nylon are

(20) Write the monomers used to prepare nylon given below :

Answer:
Monomers used in the preparation of nylon are

(21) Name the catalyst which is formed from titanium chloride and triethyl aluminium.
Answer:
The catalyst Zieglar Natta is formed from titanium chloride and triethyl aluminium.

(22) Define molecular mass of polymer.
Answer:
Molecular mass of a polymer is an average of the molecular masses of the constituent molecules.

(23) Which factor determines the molecular mass of polymer.
Answer:
Molecular mass of polymer depends upon the degree of polymerization (DP). DP is the number of monomer units in a polymer molecule.

Multiple Choice Questions

Question 38.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. Which one is the natural polyamide polymer?
(a) Cuprammonium silk
(b) Wool
(c) Perlon-L
(d) Jute
Answer:
(b) Wool

2. The synthetic fibres are prepared from
(a) cellulose
(b) starch
(c) chemical compounds
(d) polymers
Answer:
(c) chemical compounds

3. Which of the following is NOT a vegetable fibre?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

4. Which of the following fibres are made up of polyamides?
(a) Dacron
(b) Rayon
(c) Nylon
(d) Terylene
Answer:
(c) Nylon

5. An example of natural cellulose fibre is
(a) cotton
(b) wool
(c) silk
(d) rayon
Answer:
(a) cotton

6. Cellulose is the main constituent of
(a) nylon-6
(b) cotton
(c) terylene
(d) wool
Answer:
(b) cotton

7. Of the following, which group contains two cellulosic fibres and one protein fibre?
(a) Cotton, keratin, wool
(b) Linen, keratin, wool
(c) Cotton, linen, rayon
(d) Cotton, keratin, linen
Answer:
(d) Cotton, keratin, linen

8. Which of the following is not a vegetable fibres?
(a) Hemp
(b) Jute
(c) Cotton
(d) Keratin
Answer:
(d) Keratin

9. Which of the following is polyamide?
(a) Teflon
(b) Nylon 6, 6
(c) Terylene
(d) Bakelite
Answer:
(b) Nylon 6, 6

10. The monomer of e-caprolactam is
(a) styrene
(b) amino acid
(c) aminocaproic acid
(d) adipic acid O
Answer:
(c) aminocaproic acid

11. is the formula of _ Jn
(a) Nylon-66 salt
(b) Nylon-66
(c) DMT
(d) Nylon-6
Answer:
(d) Nylon-6

12. Nylon-6 is a
(a) polyester fibre
(b) protein fibre
(c) poly caprolactum fibre
(d) poly amine fibre
Answer:
(c) poly caprolactum fibre

13. The condensation product of e-caprolactum is
(a) teflon
(b) nylon-6
(c) nylon-66
(d) bakelite
Answer:
(b) nylon-6

14. \(\left[\overline{\mathrm{O}} \mathrm{OC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{COO}-\mathrm{N} \mathrm{H}_{3}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{N} \mathrm{H}_{3}\right]\) is the formula of
(a) nylon-6
(b) nylon-6 salt
(c) nylon-66 salt
(d) nylon-66
Answer:
(b) nylon-6 salt

15. The starting material required for the preparation of Nylon-66 is
(a) glycol
(b) α-amino acid
(c) adipic acid and hexamethylene diamine
(d) dimethyl terephthalate and ethylene glycol
Answer:
(c) adipic acid and hexamethylene diamine

16. Terylene is also known as
(a) styrene
(b) butadiene
(c) dacron
(d) teflon
Answer:
(c) dacron

17. Terylene is
(a) vegetable fibre
(b) protein fibre
(c) polyester fibre
(d) polyamide fibre
Answer:
(c) polyester fibre

18. Terylene polymer is formed in
(a) the presence of nitrogen
(b) the presence of hydrogen
(c) the presence of oxygen
(d) the vacuum
Answer:
(d) the vacuum

19. The by-product formed during the preparation of terylene fibre is
(a) glycerol
(b) propylene glycol
(c) ethylene glycol
(d) ethyl alcohol
Answer:
(c) ethylene glycol

20. Nylon polymer cannot be used for making
(a) tyre cords
(b) films
(c) dress materials
(d) fishing nets
Answer:
(b) films

21. Glycol is an important constituent of
(a) nylon-6
(b) nylon-66
(c) terylene
(d) hexamethylene diammonium adipate
Answer:
(c) terylene

22. Terylene is prepared by the process of
(a) halogenation
(b) condensation
(c) esterification
(d) hydrogenation
Answer:
(b) condensation

23. What are the steps during polymerisation to form terylene?
(a) Terephthalic acid is condensed with ethylene glycol at 420 K-460 K.
(b) Ethylene glycol displaces methanol to form dihydroxy diethyl terephthalic acid
(c) Zinc acetate – antimony trioxide is used as catalyst
(d) All of these
Answer:
(d) All of these

24. During polymerisation of nylon salt to nylon-66, the conditions are
(a) room temperature and pressure
(b) temperature 503 K
(c) temperature 553 K in presence of Nitrogen
(d) heating in an autoclave at 373 K
Answer:
(c) temperature 553 K in presence of Nitrogen

25. Which one of the following is a condensation polymer?
(a) Nylon
(b) Polythene
(c) PVC
(d) Teflon
Answer:
(a) Nylon

26. Which one of the following is an addition polymer?
(a) Bakelite
(b) Nylon-6,6
(c) Polystyrene
(d) Terylene
Answer:
(c) Polystyrene

27. A polymer of butadiene and Acrylonitrile is called
(a) Buna-S
(b) Buna-N
(c) Buna-B
(d) Buna-A
Answer:
(b) Buna-N

28. Natural rubber is a polymer of
(a) Styrene
(b) Butadiene
(c) Vinyl chloride
(d) Isoprene
Answer:
(d) Isoprene

29. In which of the following pairs both are copolymers?
(a) PHBV, bakelite
(b) Polythene, terylene
(c) Polyacrylonitrile, nylon-6,6
(d) Polystyrene, melamine
Answer:
(a) PHBV, bakelite

30. The polymer used in paints is
(a) Nylon
(b) Glyptal
(c) Neoprene
(d) Terylene
Answer:
(b) Glyptal

31. Which of the following contains biodegradable polymers only?
(a) Cellulose, dextron, PHBV
(b) Starch, PHBV, PVC
(c) Bakelite, nylon-2-nylon-6, nylon-6,6
(d) Cellulose, starch, terylene
Answer:
(a) Cellulose, dextron, PHBV

32. Thermosetting polymer is
(a) Nylon-6
(b) Nylon-6,6
(c) Bakelite
(d) SBR
Answer:
(c) Bakelite

33. Nylon thread contains the polymer
(a) Polyamide
(b) Polyvinyl
(c) Polyester
(d) Polyethylene
Answer:
(a) Polyamide

34. Polythene, PVC, teflon and neoprene are all
(a) Monomers
(b) Homopolymers
(c) Copolymers
(d) Condensation polymers
Answer:
(b) Homopolymers

35. Which one of the following is NOT a biodegrad¬able polymer?
(a) Starch
(b) Cellulose
(c) Dextron
(d) Decron
Answer:
(d) Decron

36. The polymer used in making blankets (artificial wool) is
(a) Polyester
(b) Polyacrylonitrile
(c) Polythene
(d) Polystyrene
Answer:
(b) Polyacrylonitrile

37. Which one of the following is a linear polymer?
(a) Bakelite
(b) LDP
(c) Nylon
(d) Formaldehyde melamine polymer
Answer:
(c) Nylon

38. Which one of the following is a branched polymer?
(a) PVC
(b) Polyester
(c) Nylon
(d) Polypropylene
Answer:
(d) Polypropylene

39. The polymer used to make non-stick cookware is
(a) Polyethene
(b) Polystyrene
(c) Polytetrafluoroethylene
(d) Polyvinyl chloride
Answer:
(c) Polytetrafluoroethylene

40. The monomer used to prepare orlon is
(a) CH₂ = CH-CN
(b) CH₂ = CHCl
(c) CH₂ = CH-F
(d) CH₂ = CF₂
Answer:
(a) CH₂ = CH-CN

41. Buna-N rubber is a copolymer of

Answer:
(b)

42. Bakelite is a polymer of
(a) Formaldehyde and phenol
(b) Benzaldehyde and phenol
(c) Formaldehyde and benzyl alcohol
(d) Acetaldehyde and phenol
Answer:
(a) Formaldehyde and phenol

43. The process involving heating of natural rubber with sulphur is known as
(a) Sulphonation
(b) Vulcanisation
(c) Galvanisation
(d) Calcination
Answer:
(b) Vulcanisation

44. A polymer of bisphenol and phosgene is called
(a) Polyamide
(b) Glyptal
(c) Polycarbonate
(d) Polystyrene
Answer:
(c) Polycarbonate

45. Thermocol is a homopolymer of
(a) terephthalic acid
(b) acrylonitrile
(c) methyl a-cyanoacrylate
(d) styrene
Answer:
(d) styrene

46. The polymer is used to prepare shatter resistant glass is called
(a) Perspex/acrylic glass
(b) Soda glass
(c) Buna N
(d) Polyacrylamide
Answer:
(a) Perspex/acrylic glass

47. A polymer used in making shoe soles is
(a) Glyptal
(b) Buna-N
(c) Buna-S
(d) Poly carbonate
Answer:
(b) Buna-N

48. The Zieglar-Natta catalyst is used in the preparation of
(a) LDPE
(b) PHBV
(c) PAN
(d) HDPE
Answer:
(d) HDPE

49. Which of the following is natural rubber?
(a) cis-1, 4-polyisoprene
(b) neoprene
(c) Trans-1. 4-polyisoprene
(d) Butyl rubber
Answer:
(a) cis-1, 4-polyisoprene

50. Which one from the following is the Terylene polymer?

Answer:
(b)

51. Equivalent amount of Hexamethylene diamine and adipic acid on complete neutralization produces :

Answer:
(a)

52. Polyhydroxy butyrate-CO-β-hydroxy valerate represents
(a) Dextron
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) PHBV
Answer:
(d) PHBV

53. Among the following, the biodegradable polymer is
(a) PVC
(b) Nylon-6, 6
(c) Nylon-2-nylon-6
(d) Neoprene
Answer:
(c) Nylon-2-nylon-6

54. The monomers of Nylon-2-nylon-6 are
(a) glycine and ω-amino caproic acid
(b) lactic acid and glycolic acid
(c) glycolic acid and co-amino caproic acid
(d) isobutylene and isoprene
Answer:
(a) glycine and ω-amino caproic acid

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 11 Correspondence with Banks Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 11 Correspondence with Banks

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Negotiable Instrument which can be discounted with the bank _____________
(a) Cheque
(b) Bills of Exchange
(c) Bank Draft
Answer:
(b) Bills of Exchange

Question 2.
Bank overdraft is a _____________ term facility provided by bank.
(a) medium
(b) short
(c) long
Answer:
(b) short

Question 3.
A loan for a period of more than 5 years is called as _____________ loans.
(a) short term
(b) long term
(c) medium-term
Answer:
(b) long term

Question 4.
Purchase and sale of securities is _____________ function of Bank.
(a) primary
(b) utility
(c) agency
Answer:
(c) agency

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Current Account	(1) Time Deposits
(b) Financial Arrangement	(2) ATM, Credit Card and Debit Card
(c) Agency Function	(3) Demand Deposits
(d) Utility Function	(4) Collection of Dividend and Interest
(e) Cash Credit	(5) Bank Overdraft
(f) Bill of Exchange	(6) Advance against the stock of raw material
(g) Letter of Credit	(7) Non-negotiable instrument
	(8) Bank Draft
	(9) Negotiable Instrument
	(10) International trade transaction

Answer:

Group ‘A’	Group ‘B’
(a) Current Account	(3) Demand Deposits
(b) Financial Arrangement	(5) Bank Overdraft
(c) Agency Function	(4) Collection of Dividend and Interest
(d) Utility Function	(2) ATM, Credit Card, and Debit Card
(e) Cash Credit	(6) Advance against the stock of raw material
(f) Bill of Exchange	(9) Negotiable Instrument
(g) Letter of Credit	(10) International trade transaction

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Deposits generating saving habits among the people.
Answer:
Saving Deposits

Question 2.
Deposits are not repayable on demand.
Answer:
Time Deposits

Question 3.
Deposits repayable on demand.
Answer:
Demand Deposits

Question 4.
The loan was provided for a period of less than one year.
Answer:
Short Term Loan

Question 5.
The loan is provided for a period between 1 to 5 years.
Answer:
Medium-Term Loan

Question 6.
The loan was provided for a period of more than 5 years.
Answer:
Long Term Loan

1D. State whether the following statements are True or False

Question 1.
Letter of credit is issued by banks for domestic trade transactions.
Answer:
False

Question 2.
Bill of Exchange is a negotiable instrument.
Answer:
True

Question 3.
The bank acts as a trustee, attorney, and executor of the will.
Answer:
True

Question 4.
RTGS is an agency function of the Bank.
Answer:
False

Question 5.
The bank plays the role of Depository Participant (DP).
Answer:
True

1E. Find the odd one.

Question 1.
Loan, Deposit, Cash credit, Overdraft facility.
Answer:
Deposit

Question 2.
Cheque, Withdrawal slip, Pay in slip.
Answer:
Pay in slip

Question 3.
Travellers Cheque, Safe Deposit Vault, NEFT, Transfer of Money.
Answer:
Transfer of Money

Question 4.
Collection of Dividend, Collection of cheque, Buying, and Selling of Securities, Payment of Electricity Bill.
Answer:
Buying and Selling of Securities

1F. Complete the sentences.

Question 1.
The appointment of bankers of a company is made by the _____________
Answer:
Board of Directors

Question 2.
Banker is a dealer in _____________
Answer:
Money

Question 3.
Bank account in which money is deposited at regular interval is a _____________
Answer:
Recurring Deposit Account

Question 4.
The instrument which can be discounted with the bank is a _____________
Answer:
Bills of Exchange

Question 5.
Businessman, Companies etc, usually open a _____________ account with a bank.
Answer:
Current

Question 6.
There is no restriction on the withdrawals from _____________ account.
Answer:
Current

Question 7.
Interest is not paid on the _____________ account.
Answer:
Current

Question 8.
_____________ account is suitable for salaried people.
Answer:
Saving

Question 9.
Stop payment letter is sent, when the cheque is _____________
Answer:
lost/misplaced

Question 10.
Withdrawals are not permitted from the _____________ accounts.
Answer:
Fixed Deposits

Question 11.
The facility given by the bank to draw more money than the actual balance in the credit is called _____________
Answer:
Overdraft facility

Question 12.
A deposit which is kept for a fixed period in a bank is a _____________
Answer:
Fixed Deposit

Question 13.
A type of bank account which is generally opened by a businessman for his business transactions is a _____________
Answer:
Current account

Question 14.
A type of account in which the interest is paid at higher rate is a _____________
Answer:
Fixed Deposit Account

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) Loans not more than one year	……………………….
(2) ……………………..	Cash Credit
(3) Collection of cheques	………………………..
(4) ……………………..	Letter of Credit

(Interest charged on, actual amount withdrawn, Agency function, Utility function, Short Term loans)
Answer:

Group ‘A’	Group ‘B’
(1) Loans not more than one year	Short Term loans
(2) Interest charged on actual amount withdrawn	Cash Credit
(3) Collection of cheques	Agency function
(4) Utility function	Letter of Credit

1H. Answer in one sentence.

Question 1.
What is a demand deposit?
Answer:
Deposits that are repayable on demand are called demand deposits. There are 2 types of demand deposits i.e Saving Deposits and Current Deposits.

Question 2.
What is a time deposit?
Answer:
Deposits that are repayable after a specific period of time are called time deposits. There are 2 types of time deposits i.e. Fixed Deposits and Recurring Deposits.

Question 3.
What do you mean by loan?
Answer:
Any amount granted or lent for a specific period of time against personal security, gold or silver, or other movable or immovable assets is called a loan.

Question 4.
What is the Bill of Exchange?
Answer:
It is a written unconditional order by the seller to the buyer, to pay a certain sum of money on a future fixed date for payment of goods or services received.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Short-term loans are of more than 5 years.
Answer:
Long-term loans are of more than 5 years.

Question 2.
Bills of Exchange are the non-negotiable instrument.
Answer:
Bills of exchange is a negotiable instrument.

Question 3.
The collection of cheques and bills is a utility function of the bank.
Answer:
The collection of cheques and bills is the agency function of the bank.

Question 4.
The rate of interest is high for saving deposits.
Answer:
The rate of interest is high for fixed deposits.

Question 5.
Deposits that are repayable after the specific period is Demand Deposit.
Answer:
Deposits that are repayable after the specific period are Time Deposits.

2. Answer in brief.

Question 1.
Explain the types of term loans.
Answer:
A loan granted for a specific time period against personal security, gold or silver, and movable and immovable assets is called a term loan.
Types of term loans:

• Short-term loan: A loan repayable within 1 year is called a short-term loan. It is generally taken by businessmen to meet their working capital requirements.
• Medium-term loan: A loan repayable within 2 years to 5 years period is called a medium-term loan.
• Long-term loan: A loan repayable after 5 years is called a long-term loan. It is taken by businessmen for the growth and development of the business.

Question 2.
Explain the types of Advances.
Answer:
Advance is a credit facility provided by the bank to its customers. Advances are for a shorter period than loans.
Type of Advances:

• Overdraft: It is an arrangement in which a customer i.e. Current A/c holder, is allowed to overdrew money in excess of the credit balance in his account. It is allowed against collateral securities like – shares, F.D.R., Government securities, etc.
• Cash credit: In this mode of advance, a separate bank account called ‘Cash Credit Account is opened in the name of the borrower. He can withdraw from this account as and when required. It is allowed against security like – stock of raw materials, finished goods, etc.
• Discounting of Bills: Discounting of bills means encashing the Bills of Exchange before its due date with the banker. The bank charges a certain amount of interest for this service which is called a discounting rate.

3. Justify the following statements.

Question 1.
In cash credit, the customer’s account is credited by a bank with the sanctioned amount.
Answer:
Cash credit is another kind of credit facility given by the bank to its customers including businessmen, companies, etc.

• A separate bank account known as a “Cash Credit Account” is required to be opened in the name of the borrower.
• The bank credits the account as per the sanctioned cash limit from which the customer can utilize funds whenever required for cash credit.
• Generally, the security of tangible assets like goods, finished stock is required to be kept.
• The interest is charged only on the actual amount utilized by the customer.
• Cash credit arrangement is for a longer period as compared to overdraft.
• This system of lending is prevalent in India only.
• Thus, in cash credit, the customer’s account is credited by a bank with the sanctioned amount.

Question 2.
Bank correspondence should be brief and to the point.
Answer:

• The bank is a financial institution.
• A company secretary has to conduct bank correspondence as per the instruction of the Board.
• Bank correspondence needs careful and cautious drafting.
• The company secretary has to use his knowledge, skill, and experience while conducting bank correspondence,
• The company secretary has to conduct bank correspondence promptly and accurately.
• Mistakes and delays in bank correspondence may bring financial loss to the company.
• Bank correspondence should be always brief, compact, and precise.
• Unnecessary or irrelevant information should be avoided in bank correspondence.
• Thus, bank correspondence should be brief and to the point.

Question 3.
No interest is paid by the bank on the current account.
Answer:

• The Current account is normally opened by businessmen, firms, or companies.
• A current account is a running account and in practice it never becomes time-barred.
• This account is opened with a minimum deposit.
• There is no limit on the amount or number of withdrawals.
• Interest is not payable on this account.
• Overdraft facility is given only to current depositors after following the prescribed bank procedure.
• Hence, interest is paid only in the case of recurring, fixed, and saving accounts and not in the case of the current account.
• Thus, no interest is paid by the bank on the current account.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Multiple-choice questions

Question 1.
Gemmule formation takes place in ……………….
(a) Hydra
(b) Spongilla
(c) Planaria
(d) Human being
Answer:
(b) Spongilla

Question 2.
Which part of ovary in mammals acts as an endocrine gland after ovulation?
(a) stroma
(b) germinal epithelium
(c) vitelline membrane
(d) graafian follicle
Answer:
(d) graafian follicle

Question 3.
Cessation of menstrual cycle in female is called ……………….
(a) lactation
(b) ovulation
(c) menarche
(d) menopause
Answer:
(d) menopause

Question 4.
Capacitation of sperms occurs in ……………….
(a) vas deferens
(b) vas efferens
(c) vagina
(d) ejaculatory duct
Answer:
(c) vagina

Question 5.
How many sperms are formed from a secondary spermatocyte ?
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(c) 2

Question 6.
The middle piece of the sperm contains ……………….
(a) proximal centriole
(b) nucleus
(c) mitochondria
(d) distal centriole
Answer:
(c) mitochondria

Question 7.
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH surge) normally occurs ?
(a) 14th day
(b) 20th day
(c) 5th day
(d) 11th day
Answer:
(a) 14th day

Question 8.
Morphogenetic movements occur during ……………….
(a) blastulation
(b) gastrulation
(c) fertilization
(d) cleavage
Answer:
(b) gastrulation

Question 9.
The technique used to block the passage of sperm in male is ……………….
(a) tubectomy
(b) vasectomy
(c) coitus interruptus
(d) rhythm method
Answer:
(b) vasectomy

Question 10.
Planaria reproduces asexually through ……………….
(a) budding
(b) gemmule formation
(c) regeneration
(d) binary fission
Answer:
(c) regeneration

Question 11.
The role of Leydig cells is ……………….
(a) nourishment of sperms
(b) to give motility to sperms
(c) synthesis of testosterone
(d) to undergo spermatogenesis
Answer:
(c) synthesis of testosterone

Question 12.
Chancre are the primary lesions caused by ……………….
(a) Neisseria gonorrhoeae
(b) Treponema pallidum
(c) Plasmodium vivax
(d) Salmonella typhi
Answer:
(b) Treponema pallidum

Question 13.
Smooth muscles lining the wall of scrotum are called ……………….
(a) detrusor muscles
(b) dartos muscles
(c) gluteal muscles
(d) latissimus dorsi muscles
Answer:
(b) dartos muscles

Question 14.
The trophoblast cells in contact with embryonal knob are called ……………….
(a) inner mass cells
(b) blastomere
(c) amniogenic cells
(d) cells of Rauber
Answer:
(d) cells of Rauber

Question 15.
The external layer of collagenous connective tissue of human testis is ……………….
(a) tunica vasculosa
(b) tunica vaginalis
(c) tunica granulosa
(d) tunica albuginea
Answer:
(d) tunica albuginea

Question 16.
Which of the following is mesodermal in origin ?
(a) Retina
(b) Enamel of teeth
(c) Heart
(d) Liver
Answer:
(c) Heart

Question 17.
Pregnancy in second trimester is maintained by ……………….
(a) LH (luteinizing hormone)
(b) progesterone
(c) estrogen
(d) hCG (human Chorionic Gonadotropin)
Answer:
(b) progesterone

Question 18.
In human foetus, the heart begins to beat at developmental age of ……………….
(a) 4th week
(b) 3rd week
(c) 6th week
(d) 8th week
Answer:
(c) 6th week

Question 19.
……………… contribute about 60% of the total volume of the semen.
(a) Prostate gland
(b) Cowper’s glands
(c) Seminal vesicles
(d) Bartholin’s glands
Answer:
(c) Seminal vesicles

Question 20.
Which of the following is hormone releasing IUD?
(a) Lippes loop
(b) Cu 7
(c) LNG 20
(d) Multiload 375
Answer:
(c) LNG 20

Question 21.
Which of the following is incorrect regarding vasectomy?
(a) Vasa deferentia is cut and tied
(b) Irreversible sterility
(c) No sperm occurs in seminal fluid
(d) No sperm occurs in epididymis
Answer:
(d) No sperm occurs in epididymis

Question 22.
The test-tube baby programme employs which one of the following techniques?
(a) Gamete intra fallopian transfer (GIFT)
(b) Zygote intra fallopian transfer (ZIFT)
(c) Intra cytoplasmic sperm injection (ICSI)
(d) Intra uterine insemination (IUI)
Answer:
(b) Zygote intra fallopian transfer (ZIFT)

Question 23.
Medical Termination of Pregnancy (MTP) is considered safe up to how many weeks of pregnancy?
(a) 8 weeks
(b) 12 weeks
(c) 24 weeks
(d) 6 weeks
Answer:
(b) 12 weeks

Question 24.
‘Saheli? an oral contraceptive pill is to be taken ……………….
(a) daily
(b) weekly
(c) quarterly
(d) monthly
Answer:
(b) weekly

Question 25.
The role of copper releasing IUDs is to ……………….
(a) inhibit ovulation
(b) prevent fertilization
(c) inhibit implantation of blastocyst
(d) inhibit gametogenesis
Answer:
(b) prevent fertilization

Question 26.
The phenomenon of nuclear fusion of sperm and egg is known as ……………….
(a) karyogamy
(b) parthenogenesis
(c) vitellogenesis
(d) oogenesis
Answer:
(a) karyogamy

Question 27.
Acrosome of spermatozoa is formed from ……………….
(a) lysosomes
(b) Golgi bodies
(c) ribosomes
(d) mitochondria
Answer:
(b) Golgi bodies

Question 28.
Which of the following undergoes spermiogenesis ?
(a) Spermatids
(b) Spermatogonia
(c) Primary spermatocytes
(d) Secondary spermatocytes
Answer:
(a) Spermatids

Question 29.
In mammals, the estrogens are secreted by the graafian follicle from its ……………….
(a) theca externa
(b) theca interna
(c) membrane granulosa
(d) corona radiata
Answer:
(b) theca interna

Question 30.
Which hormone is essential for maintenance of the endometrium of uterus?
(a) FSH
(b) LH
(c) Progesterone
(d) Estrogen
Answer:
(c) Progesterone

Question 31.
Which of the following cells during gametogenesis is normally diploid?
(a) Spermatid
(b) Spermatogonia
(c) Second polar body
(d) Secondary oocyte
Answer:
(b) Spermatogonia

Question 32.
Fertilization takes place at ……………….
(a) cervix
(b) ampulla
(c) isthmus
(d) vagina
Answer:
(b) ampulla

Question 33.
In mammals, failure of testes to descend into scrotum is known as ……………….
(a) paedogenesis
(b) castration
(c) cryptorchidism
(d) impotency
Answer:
(c) cryptorchidism

Question 34.
Polar body is produced during the formation of ……………….
(a) sperm
(b) secondary oocyte
(c) oogonium
(d) spermatocytes
Answer:
(b) secondary oocyte

Question 35.
Menstrual flow occurs due to lack of ……………….
(a) vasopressin
(b) progesterone
(c) FSH
(d) oxytocin
Answer:
(b) progesterone

Question 36.
Approximately how many eggs are produced by a normal healthy human female up to the age of 25 years if the age of menarche is 12 years?
(a) 169
(b) 416
(c) 240
(d) 100
Answer:
(a) 169

Question 37.
In humans, at the end of the first meiotic division, the male germ cells differentiate into the ……………….
(a) spermatids
(b) spermatozoa
(c) primary spermatocytes
(d) secondary spermatocytes
Answer:
(d) Secondary spermotocytes

Question 38.
The part that carries sperms from testis to epididymis is ……………….
(a) rete testis
(b) vasa efferentia
(c) vasa differentia
(d) ejaculatory ducts
Answer:
(c) vasa differentia

Question 39.
Which period of menstrual cycle is called risky period of conception?
(a) 3rd to 7th day
(b) 7th to 13th day
(c) 10th to 17th day
(d) 15th to 25th day
Answer:
(c) 10th to 17th day

Question 40.
Which hormone confirms pregnancy?
(a) Progesterone
(b) Estrogen
(c) hCG
(d) LH
Ans
(c) hCG

Match the columns

Question 1.

Column I [Organs]	Column II [Functions]
(1) Epididymis	(a) Transport of sperms
(2) Sertoli cells	(b) Copulatory organ
(3) Vas deferens	(c) Nourishment to developing sperms
(4) Penis	(d) Maturation of sperms

Answer:

Column I [Organs]	Column II [Functions]
(1) Epididymis	(d) Maturation of sperms
(2) Sertoli cells	(c) Nourishment to developing sperms
(3) Vas deferens	(a) Transport of sperms
(4) Penis	(b) Copulatory organ

Question 2.

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(a) hCG
(2) Interstitial cells / Leydig’s cells	(b) Estrogen
(3) Placenta	(c) Progesterone
(4) Graafian follicle	(d) Testosterone

Answer:

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(c) Progesterone
(2) Interstitial cells / Leydig’s cells	(d) Testosterone
(3) Placenta	(a) hCG
(4) Graafian follicle	(b) Estrogen

Question 3.

Column I	Column II
(1) Graafian follicle	(a) Site of implantation
(2) Uterus	(b) Birth canal
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(d) Release of secondary oocyte

Answer:

Column I	Column II
(1) Graafian follicle	(d) Release of secondary oocyte
(2) Uterus	(a) Site of implantation
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(b) Birth canal

Question 4.

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(a) Rapid secretion of LH
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(c) Increased level of progesterone
(4) Secretory phase	(d) Decrease in progesterone and estrogen

Answer:

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(d) Decrease in progesterone and estrogen
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(a) Rapid secretion of LH
(4) Secretory phase	(c) Increased level of progesterone

Question 5.

Column I	Column II
(1) Acrosome	(a) Completion of IInd meiotic division of secondary oocyte
(2) Penetration of sperm into ovum	(b) Dissolution of zona pellucida
(3) Formation of fertilization membrane	(c) Secretion of Hyaluronidase
(4) Acrosin / Zona lysine	(d) Prevention of polyspermy

Answer:

Column I	Column II
(1) Acrosome	(c) Secretion of Hyaluronidase
(2) Penetration of sperm into ovum	(a) Completion of IInd meiotic division of secondary oocyte
(3) Formation of fertilization membrane	(d) Prevention of polyspermy
(4) Acrosin / Zona lysine	(b) Dissolution of zona pellucida

Question 6.

Column I	Column II
(1) Parturition	(a) Attachment of embryo to endometrium
(2) Gestation	(b) Release of egg from Graafian follicle
(3) Ovulation	(c) Delivery of baby from uterus
(4) Implantation	(d) Duration between pregnancy and birth

Answer:

Column I	Column II
(1) Parturition	(c) Delivery of baby from uterus
(2) Gestation	(d) Duration between pregnancy and birth
(3) Ovulation	(b) Release of egg from Graafian follicle
(4) Implantation	(a) Attachment of embryo to endometrium

Question 7.

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(a) Prevents sperms reaching cervix
(2) Condom	(b) Prevents implantation
(3) Vasectomy	(c) Prevents ovulation
(4) Copper T	(d) Semen contains no sperms

Answer:

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(c) Prevents ovulation
(2) Condom	(a) Prevents sperms reaching cervix
(3) Vasectomy	(d) Semen contains no sperms
(4) Copper T	(b) Prevents implantation

Question 8.

Column I	Column II
(1) Mechanical means	(a) Saheli
(2) Physiological device	(b) Jellies
(3) Chemical device	(c) Vasectomy
(4) Permanent method	(d) Diaphragm

Answer:

Column I	Column II
(1) Mechanical means	(d) Diaphragm
(2) Physiological device	(a) Saheli
(3) Chemical device	(b) Jellies
(4) Permanent method	(c) Vasectomy

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

Question 1.
Classify the following contraceptives given below as per Column ‘A’ and complete Column ‘B’. Select from the given options:
(i) Foams
(ii) Lippe’s loop
(iii) Cervical caps
(iv) Multiload 375
(v) Diaphragms
(vi) Jellie

Column A	Column B
(1) Mechanical means	————–, ————
(2) Chemical means	————-, ————-
(3) Intra-uterine device	————-, ————

Answer:

Column A	Column B
(1) Mechanical means	Cervical caps, Diaphragms
(2) Chemical means	Foams, Jellies
(3) Intra-uterine device	Lippe’s loop, Multiload 375

Question 2.
Classify the following components of semen given below as per Column ‘A’ and complete the Column ‘B’. Select from the given options
(i) Acid phosphatase
(ii) Mucous like fluid
(iii) Prostaglandins
(iv) Citric acid
(v) Fructose
(vi) Fibrinogen

Column A	Column B
(1) Seminal fluid	————–, ————
(2) Prostatic fluid	————-, ————-
(3) Fluid from Cowper’s gland	————-, ————

Answer:

Column A	Column B
(1) Seminal fluid	Prostaglandins, Fructose, Fibrinogen
(2) Prostatic fluid	Acid phosphatase, Citric acid
(3) Fluid from Cowper’s gland	Mucous like fluid

Very short answer questions

Question 1.
How many sperms are present in single ejaculation?
Answer:
A single ejaculation contains about 400 millions of sperms.

Question 2.
What is gemmule? How is gemmule formed ?
Answer:
Gemmule is an internal bud formed by aggregation of archeocytes in sponges to overcome unfavourable season.

Question 3.
What is cryptorchidism?
Answer:
Failure of testis to descend into scrotum leading to sterility is called cryptorchidism.

Question 4.
What is the beginning of the menstrual cycle and cessation of menstrual cycle respectively called?
Answer:
The beginning of the menstrual cycle is called menarche while cessation of menstrual cycle is called menopause.

Question 5.
Which men have an increased risk of prostate cancer?
Answer:
Men who are over 50 years of age and have a daily high consumption of fat have an increased risk of prostate cancer.

Question 6.
What is capacitation with reference to sperm?
Answer:
Changes in a mammalian sperm which prepare it for fertilization of ovum is called capacitation.

Question 7.
Give any two examples each of seasonal breeders and continuous breeders among sexually reproducing animals.
Answer:
Example of seasonal breeders : Goat, Sheep and Donkey.
Example of continuous breeders : Humans, apes.

Question 8.
What does IUCD indicate?
Answer:
IUCD means Intra Uterine Contraceptive Device.

Question 9.
What is full form of IVF?
Answer:
IVF means In Vitro Fertilization.

Question 10.
From which germinal layers the nervous system is derived?
Answer:
The nervous system is derived from ectoderm.

Question 11.
A mother of a one-year-old child wanted to space her second child. Her doctor suggested ‘Copper-T’. Explain its contraceptive action.
Answer:
Copper ions released from ‘Copper-T’ suppress sperm motility and the fertilizing capacity of sperms.

Question 12.
Which options are available for infertile couples to have child?
Answer:
Infertile couples have many options to have a child such as fertility drugs, modern techniques such as IVE ZIFT, GIFT, ICSI, artificial insemination, IUI, using surrogate mother or taking the sperm from sperm bank.

Question 13.
How many primary spermatocytes and oocytes are required for the formation of 100 spermatozoa and ova?
Answer:
25 Primary spermatocytes and 100 primary oocytes will be required for the formation of 100 spermatozoa and ova respectively.

Question 14.
The entrance of fallopian tube of a lady is blocked. She wants motherhood. Which method will help her?
Answer:
The method of GIFT or Gamete Intra Fallopian Transfer is the method that will help the lady to have a child.

Question 15.
What is the role of birth control pills?
Answer:
Birth control pills are contraceptive pills that check the ovulation by inhibiting the secretion of FSH and LH.

Question 16.
In T.S. of ovary, can all stages of follicles be seen simultaneously?
Answer:
In T.S. of ovary, all the stages of follicles cannot be seen simultaneously. The stage of follicles develop alternately in the ovary as per timing of menstrual cycle under the influence of hormones of pituitary and ovaries.

Question 17.
What will be marriageable age for boy and girl as per the Indian law?
Answer:
As per the Hindu Marriage Act, minimum age for boy must be 21 years and for a girl 18 years, at the time of marriage.

Question 18.
What is MTP Act?
Answer:
MTP Act is for reducing the incidences of illegal abortions and maternal mortalities.

Question 19.
Which is the time period legally allowed by MTP ACT for terminating pregnancy?
Answer:
According to MTP Act, pregnancy may be terminated within first 12 weeks, more than 12 weeks but lesser than 20 weeks.

Give definitions of the following

Question 1.
Amphimixis
Answer:
It is the process which involves the production of offspring by the formation and fusion of gametes.

Question 2.
Gametogenesis
Answer:
The gametogenesis is the process of formation of gametes in sexually reproducing animals.

Question 3.
Spermiogenesis
Answer:
The process of transformation of non-motile and non¬functional spermatid into a functional and motile spermatozoa is called spermiogenesis.

Question 4.
Insemination
Answer:
The process of deposition of semen into the vagina of the female at the time of coitus or sexual intercourse is called insemination.

Question 5.
Cleavage
Answer:
The process of early mitotic division of the zygote to form a multicellular morula stage is called cleavage.

Question 6.
Implantation
Answer:
The process by which the blastocyst after its formation, gets implanted or embedded into the endometrium of the uterus is called implantation.

Question 7.
Gestation
Answer:
The condition of carrying one or more embryos in the uterus is called gestation.

Question 8.
Placenta
Answer:
A flattened, discoidal organ present in the uterus of pregnant mother and which acts as endocrine source and nutrition provider for growing foetus is called placenta.

Question 9.
Lactation
Answer:
The process of secretion of milk in the mammary glands and expelling it through nipples out to provide nourishment to the growing baby is called lactation.

Question 10.
Parturition
Answer:
Parturition is the process of giving birth to a baby.

Question 11.
Amniocentesis
Answer:
Amniocentesis is a process in which amniotic fluid containing foetal cells is collected using a hollow needle inserted into the uterus under ultrasound guidance.

Question 12.
Infertility
Answer:
Infertility is defined as the inability to conceive naturally after (one year of) regular unprotected intercourse.

Question 13.
IVF (In Vitro Fertilization)
Answer:
It is a process of fertilization where an egg is combined with sperm outside the body in a test tube or glass plate to form a zygote under simulated conditions in the laboratory.

Question 14.
Artificial Insemination (AI)
Answer:
It is the technique during which the sperms are collected from the male and artificially introduced into the cervix of female, for the purpose of achieving a pregnancy through in vivo fertilization (inside the body).

Question 15.
Adoption
Answer:
Adoption is a legal process by which a couple or a single parent gets legal rights, privileges and responsibilities that are associated to a biological child for the upbringing of the adopted child.

Give functions of the following

Question 1.
Corpus luteum.
Answer:
Corpus luteum is a secondary endocrine source that produces progesterone for maintaining pregnancy.

Question 2.
Scrotum.
Answer:
Scrotum protects the testis and also acts as thermoregulator.

Question 3.
Acrosome of sperm.
Answer:
Acrosome of the sperm releases hyaluronidase which digests the zona pellucida surrounding the ovum due to which sperm can fertilize the ovum.

Question 4.
Sertoli cells.
Answer:
Sertoli cells provide nourishment and surface to the sperm bundles during their development.

Question 5.
Interstitial cells / Leydig’s cells.
Answer:
Interstitial cells / Leydig’s cells secrete testosterone or androgen which is a male sex hormone.

Question 6.
Prostate gland.
Answer:
Prostate gland secretes prostatic fluid which forms 30% of semen, Citric acid and acid phosphatase present in this fluid protects the sperms from acidic environment of vagina.

Question 7.
Bulbourethral glands.
Answer:
Bulbourethral glands secrete alkaline, viscous mucus like fluid which provides lubrication during copulation.

Question 8.
Bartholin’s glands.
Answer:
Bartholin glands secrete lubricating mucus like fluid which is released in vestibule.

Question 9.
Uterus.
Answer:

1. Uterus receives ovum from fallopian tubes, develops placenta and provides site for implantation of embryo.
2. It provides protection and nourishment to the developing embryo.
3. It also provides path for sperms to ascend.
4. Due to contractions of uterus, baby is expelled out at the time of parturition.

Question 10.
Vagina
Answer:

1. Vagina acts as a copulatory passage.
2. It acts as a birth canal during parturition in normal delivery
3. It provides the passage for menstrual flow.

Name the following

Question 1.
The canal through which the testes descend into scrotum just before birth in human male child.
Answer:
Inguinal canal

Question 2.
The structure where sperms are matured.
Answer:
Epididymis

Question 3.
The part where the sperms are produced in the testes.
Answer:
Germinal epithelium of seminiferous tubules.

Question 4.
The gland in females homologous to Cowper’s gland.
Answer:
Bartholin’s glands or Vestibular glands.

Question 5.
Type of cleavage in human zygote
Answer:
Holoblastic, radial and indeterminate

Question 6.
The developmental stage of human being which gets implanted in the endometrium of uterus.
Answer:
Blastocyst

Question 7.
Name the primates who show presence of menstrual cycle.
Answer:
Human being and Apes like gorilla, chimpanzee, orangutan, etc.

Question 8.
Structures which help in transport of secondary oocyte through uterine tube.
Answer:
Ciliated epithelium

Question 9.
Hormones produced in women only during pregnancy.
Answer:
hCG, HPL (Human Placental Lactogen) and relaxin.

Question 10.
The oral contraceptive pill which is now. a part of the National Family Programme in India.
Answer:
Saheli

Question 11.
The scientific term for the animals giving birth to live young ones.
Answer:
Viviparous

Question 12.
The site of fertilization in woman.
Answer:
Ampulla of fallopian tube

Question 13.
The trophoblast cells lying over the embryonal knob.
Answer:
Cells of Rauber

Question 14.
The muscles which form the wall of scrotum.
Answer:

1. Dartos muscles
2. Cremaster muscles

Question 15.
Names of erectile tissues in penis.
Answer:

1. Corpora cavernosa
2. Corpus spongiosum

Question 16.
Any two copper releasing IUD.
Answer:

1. Copper-T, Cu 7
2. Multiload 375

Question 17.
Any two hormone-releasing IUDs.
Answer:

1. LNG-20
2. Progestaert

Question 18.
Two methods of birth control which have high chances of failure.
Answer:

1. Safe period
2. Lactational amenorrhea

Question 19.
Uterine walls.
Answer:

1. Perimetrium
2. Myometrium
3. Endometrium

Question 20.
Regions of the uterus.
Answer:

1. Fundus
2. Body
3. Cervix

Question 21.
Parts of fallopian tubes.
Answer:

1. Infundibulum
2. Ampulla
3. Isthmus

Question 22.
Layers of Graafian follicle which enclose antrum.
Answer:

1. Theca externa
2. Theca interna
3. Membrana granulosa

Question 23.
Stages of cells in spermatogenesis.
Answer:

1. Spermatogonia
2. Primary spermatocytes
3. Secondary spermatocytes
4. Spermatids
5. Sperms

Question 24.
Stages of cells in oogenesis.
Answer:

1. Oogonia
2. Primary oocytes
3. Secondary oocytes
4. Ootid
5. Ovum

Give significance of the following

Question 1.
Fertilization.
Answer:
Significance of fertilization:

1. Fertilization forms the zygote which eventually produces new offspring.
2. Fertilization restores diploid number of chromosomes in the zygote as two haploid gametes come together in a zygote.
3. During fertilization, centrioles are passed on to the ovum, due to this secondary oocyte can complete meiosis-II. The fertilization thus concludes the process of oogenesis.
4. By fertilization the genetic characters of two parents are mixed. This leads to variation and has significance in evolution.
5. Due to fertilization the sex of young one is determined.

Question 2.
Implantation.
Answer:
Gestation becomes possible due to implantation. Implantation protects the embryo and helps it to derive nourishment from the mother’s body through placenta.

Question 3.
Corpus luteum.
Answer:

1. Corpus luteum is the temporary source of female hormone, progesterone.
2. Corpus luteum is formed from empty Graafian follicle after the process of ovulation.
3. Due to progesterone secreted from corpus luteum, the endometrial wall of uterus undergoes repair and increase in thickness.
4. Progesterone is a gestational hormone and thus pregnancy is maintained if corpus luteum is well functional.

Question 4.
Fertilization membrane.
Answer:
Fertilization membrane prevents any further entry of other sperms into the egg, i.e. polyspermy is avoided.

Question 5.
Gastrulation.
Answer:

1. Due to the process of gastrulation, three germinal layers, viz. ectoderm, mesoderm and endoderm are formed.
2. Cells of embryonal knob become embryonic disc which develop into embryo due to gastrulation.
3. Gastrulation is necessary for the formation of amniotic cavity which is filled with amniotic fluid.

Question 6.
Trophoblast of blastocyst.
Answer:

1. Trophoblast cells help in absorbing nutrition for the developing embryo.
2. Trophoblast cells at the embryonal knob (cells of Rauber) help in implantation of blastocyst.
3. Synctiotrophoblast helps in implantation of fertilized ovum in the uterine endometrium.

Question 7.
hCG [human chorionic gonadotropin].
Answer:
hCG [human chorionic gonadotropin] is secreted in the pregnant female to extend the life of corpus luteum and stimulates its secretory activity. Presence of hCG in maternal blood and urine is an indication of pregnancy.

Question 8.
Colostrum.
Answer:

1. Colostrum is the first milk which is sticky and yellowish secreted by the mammary glands soon after the parturition.
2. Being high protein in its content, it nourishes the newly born child.
3. The antibodies present in it helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

Distinguish between the following

Question 1.
Asexual reproduction and Sexual reproduction.
Answer:

Sexual reproduction	Asexual reproduction
1. Asexual reproduction requires single parent.	1. Sexual reproduction needs two different parents.
2. Meiosis does not take place in asexual reproduction. Only mitosis takes place.	2. Sexual reproduction involves meiosis and mitosis.
3. Gamete formation, fertilization and zygote formation does not take place.	3. Gamete formation, fertilization and zygote formation are important processes in sexual reproduction.
4. Progeny and parent Eire identical genetically.	4. Progeny and parents are genetically dissimilar.
5. Large number of progeny is developed by asexual reproduction. E.g. Spore formation, gemmule formation, budding, regeneration are the types of a sexual reproduction.	5. Limited number of progeny is developed by sexual reproduction. E.g. Sexual reproduction is only by a single method.

Question 2.
Primary sex organs and Secondary sex organs.
Answer:

Primary sex organs	Secondary / Accessory sex organs
1. Primary sex organs produce gametes.	1. Secondary sex organs do not produce gametes.
2. Primary sex organs secrete sex hormones.	2. Secondary sex organs do not secrete sex hormones.
3. Development of these organs is under the control of Gonadotropins released from Pituitary. E.g. Testes in male and Ovaries in females.	3. Development of these organs is under the control of estrogen and progesterone in females and testosterone in males. Eg. Prostate, seminal vesicles, vas deferens in males. Fallopian tubes, uterus and vagina in females.

Question 3.
Vasa efferentia and Vasa deferentia.
Answer:

Vasa efferentia	Vasa deferentia
1. Vasa efferentia arise from the rete testes and enter the epididymis.	1. Vasa deferentia arise from the epididymis and form ejaculatory duct after the union with seminal duct.
2. They are present in 15-20 number and are fine convoluted ductules.	2. They are thick and coiled ductules present in a single pair.
3. The spermatozoa are carried from rete testis to epididymis by vasa efferentia.	3. The spermatozoa are carried from epididymis to ejaculatory ducts by vasa deferentia.

Question 4.
Graafian follicle and Corpus luteum.
Answer:

Graafian follicle	Corpus luteum
1. Graafian follicle is produced by the maturation of the primary follicle.	1. Corpus luteum is produced by the cells of ruptured Graafian follicle.
2. It is formed in the ovary before ovulation.	2. It is formed in the ovary after ovulation.
3. It produces the hormone estrogen.	3. It produces the hormone progesterone.
4. It has secondary oocyte surrounded by follicle cells.	4. It has only follicle cells.

Question 5.
Menarche and Menopause.
Answer:

Menarche	Menopause
1. Menarche is the beginning of menstrual cycle.	1. Menopause is the stoppage of menstrual cycle.
2. Menarche is at the age of 10 to 14.	2. Menopause is at the age of 45 to 50.
3. Menarche begins with secretion of FSH and LH.	3. Menopause is caused due to decline of FSH and LH secretion.
4. Menarche is the beginning of the reproductive period.	4. Menopause is the end of the reproductive period.

Question 6.
Proliferative Phase and Secretory Phase.
Answer:

Proliferative Phase	Secretory Phase
1. Proliferative phase begins with the repair of endometrium.	1. Secretory phase begins with ovulation.
2. Time required for proliferative phase is 5th to 13th day of menstrual cycle.	2. Time required for secretory phase is 15th to 28th day of menstrual cycle.
3. Proliferative phase always ends with ovulation.	3. Secretory phase ends with menstruation if egg is not fertilized. It continues further if egg is fertilized.
4. Proliferative phase is in uterus which coincides . with follicular phase in ovary during which there is formation of Graafian follicle.	4. Secretory phase is in uterus which coincides with luteal phase in ovary during which there is formation of corpus luteum.
5. Proliferative phase is controlled by FSH from anterior pituitary.	5. Secretory phase is controlled by LH from anterior pituitary.
6. Hormone estrogen is secreted during this phase.	6. Hormone progesterone is secreted during this phase.
7. It causes the development of blood vessels and thickening of endometrium of uterus.	7. It causes further thickening and secretory activity of the glands of endometrium of uterus.

Question 7.
Spermatogenesis and Oogenesis.
Answer:

Spermatogenesis	Oogenesis
1. Spermatogenesis takes place in testis in mature and fertile males.	1. Oogenesis takes place in ovaries in mature and fertile females.
2. From one spermatogonium four haploid sperms are formed during spermatogenesis.	2. From one oogonium one haploid ovum and a polar body is formed during oogenesis.
3. Spermatid developed undergoes metamorphosis in the process of spermiogenesis.	3. There is no such process of metamorphosis in oogenesis.
4. Spermatid development takes place which later becomes a functional sperm.	4. Ootid development does not take place during oogenesis. It develops only after fertilization.
5. Spermatogonia, primary and secondary spermatocytes and spermatid are the stages of sperms formed during spermatogenesis.	5. Oogonia, primary and secondary oocytes are the stages formed during oogenesis. Ootid formation occurs only after fertilization.

Question 8.
Zona pellucida and Corona radiata.
Answer:

Zona pellucida	Corona radiata
1. Zona pellucida is inner, thin and transparent layer surrounding the secondary oocyte.	1. Corona radiata is the outer thick layer surrounding the secondary oocyte.
2. Zona pellucida is a non-cellular layer.	2. Corona radiata is a cellular layer.
3. Zona pellucida is secreted by the ovum itself.	3. Corona radiata is formed by follicular cells which are glued together by hyaluronic acid.
4. Zona pellucida is retained for more time after fertilization till the ovum gets implanted in the uterus.	4. Corona radiata is retained till the ovum gets fertilized.
5. Zona pellucida is digested by zona lysine or acrosin at the time of fertilization.	5. Corona radiata is digested by hyaluronidase enzyme at the time of fertilization.

Question 9.
Morula and Blastula.
Answer:

Morula	Blastula
1. Morula is the embryonic stage formed after the completion of cleavage.	1. Blastula is the embryonic stage formed after the completion of blastulation.
2. Morula is formed 4 to 6 days after the fertilization.	2. Blastula is formed 6 to 7 days after the fertilization.
3. Morula consists of 16 cells.	3. Blastula consists of more than 64 cells.
4. Morula is solid ball of cells.	4. Blastula is a hollow ball of cells.
5. Morula stage is passed in fallopian tube, once it reaches uterus, it starts developing into the next stage.	5. Blastula after reaching the uterus is implanted on the wall of uterus.
6. Morula does not have any distinction of its inner cell structure.	6. Blastula has a blastocoel, trophoblast and inner cell mass.

Question 10.
Blastula and Gastrula
OR
Give two differences between blastula and gastrula.
Answer:

Blastula	Gastrula
1. Blastula is formed from morula on 7th day after fertilization.	1. Gastrula is formed from blastula 15 days after fertilization.
2. Blastula has a blastocoel.	2. Gastrula has a gastrocoel or archenteron.
3. Blastula is produced by the process of blastulation.	3. Gastrula is produced by the process of gastrulation.
4. Blastula undergoes implantation followed by gastrulation.	4. Gastrula undergoes morphogenesis and then forms germs layers.
5. During blastula formation there is no movement of cells.	5. Gastrula formation results from the morphogenetic movement of cells.

Give reasons

Question 1.
Testes are located outside the body cavity in scrotal sacs.
Answer:

1. During early foetal life, the testes develop in the lumbar region of the abdominal cavity just below the kidney but during seventh month of development, they descend permanently into the respective scrotal sacs through a passage called inguinal canal.
2. For the development of the sperm, lesser temperature than the body temperature is required.
3. If the testes remain in the abdominal cavity, then the sperm production does not take place.
4. This may result in impotency. Therefore, testes are located outside the body cavity.

Question 2.
Urethra is also called urinogenital duct in males.
Answer:

1. Urinogenital duct means common duct for urine and the genital products.
2. In males, the penis lodges urethra throughout its entire length, through which urine as well as semen are given out of the body during urination or copulation.
3. Since the urethra carries both urine and semen, it is called urinogenital duct.

Question 3.
Proliferative phase is also called follicular phase.
Answer:

1. Proliferative phase means there is proliferation of endometrial cells in the uterus. Follicular means there is growth of ovarian follicles in the ovaries. Both these phases are simultaneous.
2. The follicular phase of ovaries is due to effect of FSH from adenohypophysis.
3. The ovaries follicles grow due to FSH and start secreting estrogen.
4. This estrogen from ovaries bring proliferative effect on the uterus.

Question 4.
Missing of menses is the first indication of pregnancy.
Answer:

1. Menstruation occurs if there is no fertilization of ovum.
2. The endometrium of uterus along with unfertilized egg is given out in the form of menstrual flow.
3. The sloughing off uterine endometrium takes place due to degeneration of corpus luteum.
4. In the absence of functional corpus luteum progesterone levels fall down. However, if the ovum is fertilized, the corpus luteum is maintained and it secretes progesterone which maintains the uterine endometrium. In such case, further growth of ovarian follicles and ovulation remains suspended and woman is said to be pregnant.
5. Endometrial wall of uterus now thickens and helps in the growth of placenta. Thus during pregnancy, menses will not take place.

Question 5.
Progesterone is called pregnancy hormone.
Answer:

1. Progesterone is secreted from corpus luteum which is formed from empty ovarian follicle after the ovulation.
2. Progesterone has the capacity to maintain pregnancy.
3. It acts on uterine endometrium and causes it to proliferate and develop in thickness.
4. Corpus luteum keeps on secreting progesterone till the placenta takes up the function of secreting the same.

Question 6.
Human female has restricted reproductive life.
Answer:

1. In human female, the reproductive period is about 30 – 33 years.
2. There is menarche at the age of about 13 and menopause at the age of 45-50.
3. During this span of 30 years, ovaries secrete sex hormones like estrogen and progesterone. After menopause this secretion is suspended.
4. Due to changes in hormonal level, human females cannot produce eggs later. Moreover, eggs in her ovaries are utilized by the age of 45.
5. Human female, therefore, has restricted reproductive period.

Question 7.
Zona pellucid is retained for sometime after fertilization.
Answer:

1. Fertilization of the ovum takes place in fallopian tube where it starts cleavages immediately.
2. Zona pellucida which remains on the surface of the ovum prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Zona pellucida keeps the sticky and phagocytic trophoblast cells unexposed till the ovum reaches the uterine lumen.
4. Zona pellucida also protects the ovum. Therefore zona pellucida is retained for some time after fertilization.

Question 8.
The acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.
Answer:

1. The enzyme hyaluronidase secreted by acrosome of sperm dissolves the membranous covering of the ovum to facilitate the entry of sperm into the ovum.
2. It is a lytic enzyme causing lysis of egg membrane.
3. Owing to this, the acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.

Question 9.
The middle part of the human sperm is characterized by the presence of a number of mitochondria.
Answer:

1. Mitochondria provide energy required by sperms for their agile movement.
2. The agile movement of sperms helps them to reach the vicinity of the ovum at the time of fertilization.
3. Owing to this, the middle part of the human sperm is characterised by the presence of a number of mitochondria.

Question 10.
The size of morula remains almost same as that of ovum.
Answer:

1. The layer zone pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula.
2. This layer is important because it prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Though the number of blastomeres increase, the size of morula remains almost same as that of ovum till it reaches the uterus by the end of the day 4.

Question 11.
Placenta serves as the nutritive, respiratory and excretory organ of the embryo.
Answer:

1. Between the foetus and mother there is exchange of several materials. Food in the form of glucose, amino acids, simple proteins, lipids, mineral, salts, vitamins and hormones, antibodies, etc. is sent to foetus by maternal circulation.
2. Oxygen from mother’s blood is also given to the foetus.
3. The foetal metabolic wastes such as carbon dioxide, urea and water pass from foetus into the maternal blood.
4. This exchange takes place through the placenta.
5. In the placenta, foetal blood comes very close to maternal blood to permit these exchanges. Therefore placenta is said to serve as the nutritive, respiratory and excretory organ of the embryo.

Write short notes

Question 1.
Graafian follicle.
Answer:

1. Graafian follicle is a mature ovarian follicle.
2. There are following protective layers on the Graffian follicle : The outermost protective and fibrous covering, theca externa. Theca interna is the next layer which can secrete hormone estrogen.
3. Next to theca interna, there is membrana granulosa which forms discus proligerous and the corona radiata layer.
4. Graafian follicle contains an eccentric secondary oocyte. The oocyte is surrounded by a vitelline membrane which produces zona pellucida layer.
5. In the centre there is antrum which is filled with liquor folliculi fluid.

Question 2.
Mammary glands.
Answer:

1. Mammary glands are accessory organs of female reproductive system. These glands are essential for lactation after parturition.
2. They are modified sweat glands present in the subcutaneous tissue of the anterior thorax. They are in the pectoral region in the location between 2nd to 6th rib.
3. Each mammary gland consists of fatty connective tissue and many lactiferous ducts.
4. Each breast has glandular tissue which is divided into 15-20 irregularly shaped mammary lobes. Each lobe has an alveolar glands and lactiferous duct.
5. Milk is secreted by alveolar glands and it is stored in the lumen of alveoli. The alveoli open into mammary tubules and these in turn forms a mammary duct.
6. All the lactiferous ducts converge towards the nipple.
7. Nipple is surrounded by a dark brown coloured and circular area of the skin called areola.

Question 3.
Structure of sperm.
Answer:

1. Sperm is microscopic, elongated haploid motile male gamete produced by spermatogenesis.
2. It measures to about 0.055 mm or 60y in length.
3. The sperm consists of head, neck, middle piece and tail.

Head:

1. Head is the main part which is flat and oval and has a large nucleus and an acrosome.
2. Acrosome is formed from Golgi complex. It secretes enzyme hyaluronidase which helps in penetration of the egg during fertilization.
3. The acrosome and anterior half of nucleus is covered by a fibrillar sheath.

Neck : Neck is short region having two centrioles.

1. The proximal centriole plays a role in first cleavage of zygote.
2. The axial filament of the sperm is formed by the distal centriole.

Middle piece:

1. Middle piece acts as a power house for sperm.
2. It bears many spirally coiled mitochondria or Nebenkern around the axial filament.
3. The mitochondria supply energy for the sperm to swim in the female genital tract with a speed of about 1.5 to 3 mm per minute.
4. Posterior half of nucleus, neck and middle piece of sperm are covered by a sheath.

Tail:

1. The tail is formed of cytoplasm and is long, slender and tapering structure.
2. The axial filament is a fine thread-like structure that arises from the distal centriole and traverses the middle piece and tail.
3. Nine accessory fibres are present surrounding the two central longitudinal axial filaments.
4. Tail lashes and helps the spermatozoa to swim.

Question 4.
Structure of secondary oocyte.
Answer:

1. The unfertilized egg released through ovary at the time of ovulation is a secondary oocyte.
2. It is rounded, non-motile and haploid, non- cleidoic and microlecithal female gamete.
3. The size is approximately 0.1 mm (100 microns).
4. It has abundant cytoplasm called ooplasm which contains a large eccentric and prominent nucleus called germinal vesicle.
5. Centrioles are absent in secondary oocyte.
6. Various coverings seen around the oocyte are (i) vitelline membrane (ii) zona pellucida (iii) Corona radiata.
7. The cells are glued together with hyaluronic acid. Between vitelline membrane and zona pellucida, there is perivitelline space which lodges first polar body. This end is called 5 animal pole and the opposite is called vegetal pole.

Question 5.
Implantation
Answer:

1. Implantation is the process by which the blastocyst is embedded into the endometrium of uterus in the fundus region.
2. The implantation starts 7 days after fertilization and completed by the end of 10th day.
3. The trophoblast cells of blastocyst at the embryonal knob can stick to the uterine endometrium. The trophoblast layer then divides into inner cytotrophoblast and outer syncytiotrophoblast due to contact with endometrial cells.
4. Cytotrophoblast is the inner layer whose cells retain their cell boundaries.
5. Syncytiotrophoblast is the outer layer of cells without plasma membrane. The cells of syncytiotrophoblast appear multinucleate. This layer projects invasively into the endometrium and destroys endometrial cells by releasing lytic enzymes. Due to this blastocyst is buried deeply in the endometrium.

Question 6.
Fate of three germinal layers.
Answer:
Fate of germinal layers : The embryo after gastrulation develops the three germ layers, viz., ectoderm, mesoderm and endoderm. Later a process of histogenesis starts which leads to the development of different tissues and organs.
(i) Fate of ectoderm : Following tissues, structures and organs develop from the ectoderm : Epidermis of the skin, epidermal derivatives such as

1. hair and nails
2. sweat glands
3. conjunctiva
4. cornea
5. lens
6. retina
7. internal and external ear
8. enamel of teeth
9. nasal cavity
10. adrenal medulla
11. stomodaeum and proctodaeum
12. neurohypophysis and
13. entire nervous system.

(ii) Fate of mesoderm : The mesoderm forms the following derivatives:

1. All types of muscles
2. connective tissue
3. dermis of skin
4. adrenal cortex
5. kidney
6. circulatory system
7. heart
8. blood vessels
9. blood
10. lymphatic vessels
11. middle ear and
12. dentine of teeth.

(iii) Fate of endoderm : The following organs develop from the endoderm:

1. Epithelium of gut from pharynx to colon
2. glands of stomach and intestine
3. tongue and tonsils
4. lungs, trachea, bronchi, larynx, etc.
5. urinary bladder, vestibule and vagina
6. liver and pancreas
7. adenohypophysis
8. thymus, thyroid and parathyroid
9. eustachian tube
10. epithelium of urethra and associated glands.

Question 7.
Placenta.
Answer:

1. Placenta is a temporary organ derived from the tissues of the foetus as well as mother.
2. Human placenta is called chorionic placenta as it is made up of chorion which is an extra-embryonic membrane.
3. Blood vessels from the allantois vascularize the placenta. Branching villi emerge from the chorion and penetrate in the corresponding pits which are located in the uterine wall.
4. There are two parts of placenta, viz. foetal placenta and maternal placenta.
5. Foetal placenta is formed of chorionic villi.
6. Maternal placenta is formed of uterine wall which is in intimate contact with the chorionic villi.
7. Chorionic villi receive the blood from the embryo by umbilical artery. Umbilical vein returns the blood back to the embryo.
8. Human placenta is said to be haemochorial because a part of placenta is from foetus which has chorionic villi. The other highly vascularized part is from uterine wall of mother. Thus foetal and maternal placenta together is called haemocorial placenta.

Question 8.
Intratuterine devices (IUDs).
Answer:

1. IUDs are plastic or metal objects which act as contraceptive devices. They are placed into the uterus by a doctor or trained nurse.
2. E.g. Lippe’s loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375) and hormone releasing IUDs (LNG-20, Progestaert).
3. Plastic double ‘S’ loop is called Lippe’s loop which stimulates accumulation of macrophages in the uterine cavity by attracting them. As phagocytosis increases the sperms are destroyed. Thus it acts as a contraceptive.
4. Copper releasing IUDs suppress sperm motility and the fertilizing capacity of sperms.
5. The hormone releasing IUDs make the : uterus unsuitable for implantation and ; cervix hostile to the sperms.
6. Their presence in the uterus acts as a minor irritant and thus makes the ovum to move quickly out of the body.
7. However, IUD can cause infection and occasional haemorrhage. It can cause discomfort for woman and may get spontaneously expelled out.

Question 9.
Physiological (Oral) Contraceptive « Devices.
Answer:

1. Physiological devices are in the form of oral contraceptive pills or birth control pills. 5 They are hormonal preparations and check ovulation by inhibiting the secretion of follicle stimulating hormone and luteinizing hormone.
2. Woman who is using pills does not release ovum at the time of ovulation and therefore conception does not occur.
3. Birth control pills have side effects such as nausea, breast tenderness, weight gain and ‘break through’ bleeding, i.e. slight bleeding between the menstrual periods. These health hazards are due to synthetic hormones.
4. These pills also alter the quality of cervical J mucus to prevent the entry of sperms.
5. The birth control pills contain progesterone and estrogen. Mala-D to be taken daily and Saheli to be taken weekly are two common birth control pills in India. These pills are non-steroidal.

Question 10.
Fate of trophoblast cells of blastocyst.
Answer:

1. Trophoblast cells do not form any part of the embryo proper.
2. They form ectoderm of the extra-embryonic membrane called chorion.
3. Chorion helps in supply of oxygen and nutrients to foetus from mother’s body. CO₂ and nitrogenous wastes are collected from foetus and passed in mother’s blood.
4. Thus, these cells have an important role in formation of placenta.

Question 11.
Medical Termination of Pregnancy (MTP).
Answer:

1. MTP or Medical Termination of Pregnancy is voluntary termination of pregnancy under medical supervision. It is an induced abortion.
2. Only during first trimester, MTP is safe for mother’s health.
3. Upon amniocentesis examination, if abnormality is detected, usually MTP is performed.
4. Government of India has legalized MTP There was MTP Act in 1971, which was later amended in 2017, to prevent its misuse, especially female foeticide should never be done through MTP
5. As per MTP Act, the procedure can be done only in first 12 weeks and never after 20 weeks of pregnancy.

Question 12.
Amniocentesis
Answer:

1. In amniocentesis, amniotic fluid containing foetal cells is collected using a hollow needle. This needle is inserted into the uterus of pregnant mother, under ultrasound guidance.
2. The chromosomes from the foetal cells are sujected to karyotyping. This helps to detect abnormalities in the developing foetus.
3. Amniocentesis is misused to determine the sex of the unborn child. This is illegal in India because it results into female foeticide.
4. Another risks involved in amniocentesis are miscarriage, needle injury to foetus, leaking amniotic fluid, infection, etc.
5. As per MTP Act (1971) the misuse of amniocentesis is curtailed.

Question 13.
ZIFT [Zygote Intra Fallopian Transfer].
Answer:

1. If there is a blockage in the fallopian tubes due to which fertilization is prevented, then ZIFT treatment is used.
2. The oocyte is removed form woman’s ovary. This oocyte is fertilized outside the body under sterile conditions with the known sperms. This forms zygote. This is In Vitro Fertilization (IVF).
3. Later the zygote is transferred in fallopian tube to achieve pregnancy.

Question 14.
GIFT [Gamete Intra Fallopian Transfer].
Answer:

1. When the oocyte is collected from donor and transferred into the fallopian tube of another female, the technique is called GIFT. This female provides suitable environment for further development.
2. When the entrance or upper segments of the fallopian tubes is blocked, this technique is used.
3. Ooocytes and sperms are directly injected into regions of the fallopian tubes. Here fertilization takes place forming a blastocyst. It later enters the uterus for implantation.
4. GIFT is successful in only 30 per cent cases.

Question 15.
Sterilization operations.
Answer:

1. Sterilization operations are the permanent means for the birth control. These can be performed on both the sexes. Usually these are performed after the couple does not desire another child.
2. These surgical interventions block the gamete transport and thus prevents the pregnancy.
3. Sterilization operation in males is called vasectomy while in females it is called tubectomy.
4. In vasectomy the vas deferens are tied and cut. In tubectomy fallopian tubes are ligated or cut.

Question 16.
Gonorrhoea.
Answer:
(1) Gonorrhoea is a sexually transmitted veneral disease caused by Diplococcus bacterium, Neisseria gonorrhoea.

(2) The incubation period is 2 to 14 days in males and 7 to 21 days in females.

(3) Infection sites are mucous membrane of urino-genital tract, rectum, throat and eye.

(4) Males show following symptoms : Partial blockage of urethra and reproductive ducts, pus from penis, pain and burning sensation : during urination, arthritis, etc.

(5) Symptoms in female include, pelvic inflammation of urinary tract, sterility, arthritis. The children born to affected mother suffer from gonococcal ophthalmia, In girl-child, there is occurrence of gonococcal vulvovaginitis before puberty.

(6) Preventive measures for gonorrhoea are as follows:

• Sexual hygiene
• Use of condom during coitus.
• Sex with unknown partner or multiple partners should be avoided.

(7) Gonorrhoea can be treated with Cefixime which is antibiotic.

Short-answer Questions

Question 1.
What are sexual dimorphic characters? Enlist these characters in human male and female.
Answer:
Sexual dimorphism is the phenomena in which the sexes of the individual can be identified externally. In human beings, even in infancy there is sexual dimorphism, by which one can identify the sex of the infant.

But when the male or female reaches puberty, then secondary sexual characters are developed due to sex hormones. These characters are called sexual dimorphic characters.
(i) Secondary sexual characters in males:

1. Presence of beard, Moustache.
2. Hair on the Chest, Axillary and Pubic Region.
3. Muscular body.
4. Enlarged larynx (Adam’s apple).

(ii) Secondary sexual characters in females:

1. Breast development.
2. Broadening of pelvis.
3. High pitched voice.

Question 2.
Describe the duct system that transports the sperms from seminiferous tubules to the exterior.
Answer:
(1) All the seminiferous tubules present in the testis show posterior network of tubules called rete testis. Vasa efferentia are the fine tubules which are 12-20 in number, are seen arising from rete testis. From testis to epididymis, the sperm transport is done by vasa efferentia.

(2) Epididymis has three parts, caput, corpus and cauda epididymis. In this long and highly coiled tube sperms undergo physiological maturation.

(3) Then from here sperms enter into vas deferens, which is a tube that arise from epididymis enters the abdominal cavity. On its course, later it joins the duct of seminal vesicle. Both together form the ejaculatory duct.

(4) Ejaculatory duct passes through the prostate gland and then opens into the urethra. Urethra is a common passage for urine and semen and hence it is also called urinogenital duct.

(5) Urethra passes through penis and opens to the outside by an opening called the urethral meatus or urethral orifice.

(6) Thus sperms are transported through vas deferens into urethra via ejaculatory duct and then to the outside through urethral orifice.

Question 3.
What is semen? Describe the composition of semen.
Answer:
(1) Semen is the viscous, alkaline and milky fluid having pH 7.2 to 7.7 ejaculated during sexual intercourse by male.

(2) A single ejaculation of semen i.e. 2.5 to 4 ml semen contains about 400 millions of sperms.

(3) Semen consists of sperms suspended in secretions of the epididymis and the accessory glands (seminal vesicles, prostate gland and Cowper’s gland). The semen nourishes the sperms by fructose, neutralizes acidity by Ca⁺⁺, ions and bicarbonates and also activates them for movement due to prostaglandins.

Question 4.
Describe in detail the external genitalia of human female reproductive system.
Answer:
The external genital organs of female are located external to the vagina. They have collective name, ‘vulva’ or pudendum. Following are the parts of vulva.
(1) Labia majora : Labia majora are homologous to scrotum of males. They are two large folds which form the boundary of the vulva. They are composed of skin, fibrous tissue and fat. These Eire prominent and longitudinal folds on right and left sides of the vestibule.

(2) Labia minora : Smaller and thinner lip-like folds located just medially are labia minora. Posteriorly the labia minora are fused together to form the fourchette.

(3) Mons veneris : Mons veneris is fleshy elevation above the labia majora.

(4) Clitoris : It is present at the anterior end of the labia minora. It shows the presence of erectile tissues.

(5) Vestibule : Vestibule is a median vertical depression of vulva enclosing vagina and urethral opening.

(6) Hymen : Hymen is a thin layer of mucous membrane which partially occludes the opening of the vagina.

(7) Vestibular glands:

1. Vestibular glands or Bartholin’s glands are homologous to the Cowper’s glands of the male.
2. These are paired glands situated on either side of the vaginal opening, secreting lubricating fluid.

Question 5.
How is puberty attained in females? Will a female normally remain reproductively capable even after age 50? If not then what makes her incapable?
Answer:
(1) Puberty is achieved due to gonadotropins such as FSH and LH secreted by the anterior pituitary. These hormones stimulate the ovaries. The ovaries in turn produce estrogen and progesterone, which brings about secondary sexual characters in female. Thus she attains the puberty. The beginning of menstrual cycle or menarche takes place due to these hormonal changes at about 10 to 14 years.

(2) But the women do not remain reproductively active after the age of 50 due to hormonal imbalance. This is called menopause or cessation of reproductive cycles. Absence of enough gonadotropins and unresponsive ovarian cells cause menopause at 45 to 50 years of age.

Question 6.
Why is menstruation painful in some women?
Answer:

1. The menstruation is painful in some women as the muscles in the uterus contract or tighten.
2. Women who experience painful periods can have higher levels of prostaglandins, a natural body chemical that causes contractions of the uterus and blood vessels.
3. Some women have a build-up of prostaglandins which means they experience stronger contractions and therefore due to spasmodic pain in some women menstruation is more painful.
4. Endometrial sloughing that takes at the time of menstruation also causes painful discomfort.

Question 7.
Why is it said that consumption of mother’s milk is safety for the newborn?
Answer:
Consumption of mother’s milk is safety for the newborn because of the following reasons:

1. Mother’s milk is the perfect food for babies in the first months of their lives. With the exception of vitamin D, it contains all the nutrients an infant needs.
2. Mother’s milk supplies antibodies [IgA] that protect the baby’s body organs from infections. Mother’s milk provides immunity and also help in maturation of the infant’s immune system which are lacking in ordinary milk. Natural acquired passive immunity is obtained only through mother’s milk.
3. Feeding of mother’s milk reduces the risk of overweight and obesity during childhood.
4. It also creates the bond between mother and child.

Question 8.
Which hygienic practices should be followed by the female during menstruation ?
Answer:
The following personal hygienic practices should be followed by the female during menstruation:

1. Keeping the pubic area clean.
2. Changing the sanitary napkin every 4-5 hours.
3. Reducing risk of infections by maintaining hygiene.
4. Proper disposal of soiled sanitary napkin.
5. Not to use damp and dirty clothes which can cause infections and bad odour. A sanitary napkin which is not changed in time can act as a perfect environment for rapid growth of infectious bacteria.

Question 9.
How can the goals of RCH be achieved?
Answer:
The goads of RCH can be achieved by the following ways:

1. Sex education in schools is introduced. Proper and scientific information about sexual organs and safe sexual acts should be given to students. They should be made aware of sexually transmitted diseases (STD, AIDS), and problems related to adolescence.
2. Audio-visual and the print media should be used by government and non-government organisations for creating awareness about reproductive health.
3. Younger generation should be educated about family planning measures, pre-natal and post-natal care of women and care of infant with knowledge about importance of breastfeeding.
4. Awareness should be spread about problems arising due to uncontrolled population growth, sex abuse and sex related crimes. Necessary steps to prevent these to be taken.
5. Statutory ban on amniocentesis for sex determination is practised. This should be known by all.
6. Details of child immunization programmes should be understood.
7. New parents should get the training for new born care so that infant and maternal mortality rate can be reduced.

Question 10.
How do addictions like smoking, alcoholism and drug abuse contribute in causing infertility in men?
Answer:

1. Tobacco, marijuana and other drugs, smoking may cause infertility in both men and women.
2. Nicotine blocks the production of sperm and decreases the size of testicles.
3. Alcoholism by men interferes with the synthesis of testosterone and has an impact on sperm count.
4. Use of cocaine or marijuana may temporarily reduce the number and quality of sperm.

Question 11.
Jayesh, a young married man of 26 years is suffering from T.B. for the last 2 years. He and his wife are desirous of a child but unable to have one, what could be the possible reason? Explain.
Answer:
Jayesh, though young, is suffering from TB for last 2 years. His wife is unable to conceive the child may be due to following reasons:

1. Tuberculosis disrupts sexual and reproductive function in patients.
2. Moreover T.B. patients have to take not less than 4 anti-TB drugs simultaneously for a long time.
3. These are basically a very high dose antibiotics which may hamper formation of sperms. In this way the anti-tuberculosis drugs may negatively influence on sexual function.
4. Pulmonary TB patient shows, deterioration of all parameters of copulatory act, from sexual desire to orgasm and thus the couple is unable to conceive.
5. Infertility is one of the most common symptoms of genital tuberculosis.

Question 12.
Neeta is 45 years old and the doctor advised her not to go for such a late pregnancy. She however wants to be the biological mother of a child without herself getting pregnant. Is this possible and how?
Answer:
(1) Neeta being 45 years old, she is approaching menopause. Therefore, she will be advised by the doctor to take the help of the modern remedial technique called surrogacy.

(2) In this technique the embryo is formed using intended father’s sperm and intended mother’s egg by In Vitro Fertilization (IVF) technique and then that embryo is implanted in a surrogate mother, sometimes called a gestational carrier.

(3) In surrogacy there is legal arrangement where the surrogate mother agrees to bear child for a couple. Remains pregnant with all the care and nourishment. Later she delivers a baby and hands it over to biological mother.

Chart based /Table based questions

Question 1.
Complete the following chart and rewrite

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	————–
2. ————-	Bulbourethral glands/ Cowper’s gland
3. Clitoris	—————

Answer:

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	Scrotum
2. Bartholin’s gland/ Vestibular gland	Bulbourethral glands/ Cowper’s gland
3. Clitoris	Penis

Question 2.

Hormones	Functions
1. Testosterone	————–
2. ————-	Stimulates contractions uterine during parturition
3. Progesterone	—————

Answer:

Hormones	Functions
1. Testosterone	Stimulates spermatogenesis
2. Oxytocin	Stimulates contractions uterine during parturition
3. Progesterone	Maintain endometrium of uterus during secretory phase and gestation.

Question 3.
Complete the following chart and rewrite

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	————–
2. Secretion of endometrial glands and increased secretion of progesterone	————–
3. Breakdown of endometrium in absence of fertilization	————-

Answer:

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	Proliferative phase / Follicular phase
2. Secretion of endometrial glands and increased secretion of progesterone	Secretory phase / Luteal phase
3. Breakdown of endometrium in absence of fertilization	Menstrual phase

Diagram based questions

Question 1.
Sketch and label Human male reproductive system.
Answer:

Question 2.
Label the given male reproductive system you have studied.

Answer:

1. Seminal vesicle
2. Ejaculatory duct
3. Cowper’s glands
4. Urethra
5. Epididymis
6. Testis
7. Urinary bladder
8. Prostate gland
9. Vas deferens
10. Penis

Question 3.
Sketch and label human female reproductive system.
Answer:

Question 4.
Give labels to given diagram of female reproductive system.

Answer:

1. Fallopian tube
2. Fundus of Uterus
3. Ampulla of fallopian tube
4. Ovarian ligament
5. Uterus
6. Ovary
7. Infundibulum with fimbriae
8. Endometrium of uterus
9. Cervix
10. Vagina

Question 5.
Sketch and label Seminiferous tubules as seen in T.S. of testis.
Answer:

Question 6.
Identify ‘A’ and ‘B’ in the diagram below and mention their functions.

Answer:
(1) A : Seminiferous tubule
Function : Seminiferous tubules produce sperms by spermatogenesis.

(2) B : Vas deferens
Function : Vas deferens carry sperm from epididymis to ejaculatory duct.

Question 7.
Sketch and label – T.S. of ovary.
Answer:

Question 8.
Sketch and label sectional view of mammary gland.
Answer:

Question 9.
Sketch and label – Graafian follicle.
Answer:

Question 10.
Sketch and label – Process spermatogenesis.
Answer:

Question 11.
Sketch and label process of oogenesis.
Answer:

Question 12.
Give the name and functions of ‘A’ and ‘B’ from the diagram given below

Answer:
(1) A is acrosome.
Function of acrosome : Acrosome produces lytic enzyme, hyalourinidase and thus helps in the penetration of the egg during fertilization.

(2) B is tail of the human sperm.
Function of tail : Tail lashes continuously and helps the movement of the sperm in the female genital tract.

Question 13.
The diagram represents a surgical sterilization method in males. Study the same and answer the questions that follow

1. Give the name of the surgical method represented in the diagram.
2. Which part is ligated or cut.
3. Name the corresponding surgical method conducted in females.
4. Name the part which is ligated in females and why?

Answer:

1. Vasectomy
2. Vas deferens
3. Tubectomy
4. Fallopian tubes are ligated so that the egg may not meet with the sperms.

Question 14.
Given below is the figure of an important structure developed during pregnancy.

1. Name the structure and its type.
2. Identify ‘A’. In which technique it is used.
3. Identify ‘B’ What is its function?

Answer:
(1) The given figure is placenta. The type of placenta in humans is haemochorial placenta.

(2) A is amnio tic fluid. Amnio tic fluid is withdrawn in amniocentesis technique. From this fluid foetal cells can be obtained, which are examined for any chromosomal abnormality by karyotyping.

(3) B is umbilical cord. This is the connection between placenta of mother and growing foetus. Through the umbilical cord, foetus gets nutrition and oxygen. Nitrogenous wastes and carbon dioxide is collected from foetus and brought into maternal circulation.

Question 15.
The diagram given below is that of a intra-uterine contraceptive device. Study the same and then answer the questions that follows:

1. Give the name of intra-uterine contraceptive device shown in the diagram.
2. What is its mode of action?

Answer:

1. The Intra-uterine contraceptive device shown in the diagram is Lippes loop.
2. It is a plastic double ‘S’ loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and acts as a contraceptive.

Question 16.
Identify A in the given diagram. Write the function of the same.

Answer:
A in the above diagram is intrauterine device or IUD. It is a contraceptive device inserted in the uterus of woman. This is a hormone releasing IUD. It acts as a mechanical means of contraception and avoids pregnancy.

Long answer questions

Question 1.
With the help of diagrammatic representation, explain the process of spermatogenesis.
Answer:

(1) The process of spermatogenesis takes place in the male gonads or testis. The cells of germinal epithelium that line the seminiferous tubules undergo spermatogenesis.

(2) Primordial germ cells or germinal cells pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation.

• Multiplication phase : Primordial germ cells undergo mitotic divisions to produce many diploid (2n) spermatogonia.
• Growth phase : Spermatogonium accumulates nutrients and grows in size, giving rise to primary spermatocyte (2n).
• Maturation phase : The primary spermatocyte undergoes first meiotic division or maturation division. Exchange of genetic material occurs between homologous chromosomes in each spermatocyte.

(3) The meiotic division gives rise to secondary spermatocyte which is haploid (n). At the end of first meiotic division two secondary spermatocytes are formed while at the end of second meiotic division four haploid spermatids are formed.

(4) Spermatids are non-motile. They undergo spermiogenesis and form motile spermatozoan (sperm).

(5) The changes taking place during spermiogeneis are as follows:

• Increase in length.
• Formation of proximal and distal centriole.
• Distal centriole forms the axial filament.
• Mitochondria become spirally coiled.
• Acrosome is formed from Golgi complex.

Question 2.
What is oogenesis? Describe it briefly.
Answer:

1. Oogenesis is the process of formation of the haploid female gamete, i.e. ovum.

2. The process of oogenesis takes place in the follicular cells inside the ovaries. The germinal epithelium cells undergo oogenesis.

3. They pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation at the time of oogenesis.

• Multiplication phase : Germinal cells undergo mitotic divisions and produce large number of diploid (2n) oogonia. Oogonia are present in the ovaries of female even before she is born.
• Growth phase : During puberty changes, the FSH from pituitary makes one oogonium to develop at a time. The growth takes place as the follicle matures and larger primary oocyte (2n) is produced inside the Graafian follicle.
• Maturation phase : The primary oocyte undergoes first meiotic division. There are equal nuclear divisions during meiosis but the cytoplasm is unequally divided.

4. By the end of first meiotic division, larger haploid secondary oocyte and smaller haploid polar body are produced. Since the embryo develops from the egg, there is provision for more food in the secondary oocyte.

(5) The second meiotic division takes place in the secondary oocyte and polar body. But this division is arrested during metaphase.

(6) The secondary oocyte is released from the ovary in the process of ovulation. Remaining division takes place if and only if ovum is fertilized.

(7) The division is unequal and form functional female gamete or ovum at the time of fertilization.

Question 3.
What is gastrulation? What are the changes that are brought about by gastrulation?
Answer:
(1) Gastrulation : The process of formation of three germ layers by morphogenetic movements and rearrangements of the cells in blastula leading to the formation of gastrula is known as gastrulation.

(2) Cells on the free end of inner cell mass called hypoblasts (primitive endoderm) become flat, divide and grow towards the blastocoel to form endoderm.

(3) Endodermal cells grow within the blastocoel to form a Yolk sac.

(4) The remaining cell of the inner cell mass, in contact with cells of Rauber are called epiblasts (primary ectoderm) which further differentiate to form ectoderm.

(5) Cells of ectoderm divide and re-divide and move in such a way that they enclose the amniotic cavity. The floor of this cavity has the embryonal disc while roof is lined by amniogenic cells. Amnion is an extra embryonic membrane that surrounds and protects the embryo.

(6) Actual gastrulation occurs about days after fertilization.

(7) Trilaminar embryonic disc begins with the formation of primitive streak and a shallow groove on the surface called primitive groove. From the site of primitive streak, a third layer of cells called mesoderm extends between ectoderm and endoderm. Anterior end of the primitive groove communicates with yolk sac by an aperture called blastopore (future anus).

(8) The embryonal knob thus finally differentiate into three layers – ectoderm, mesoderm and endoderm.

Question 4.
Explain the major changes taking place during the three trimesters of pregnancy in woman.
Answer:
The pregnancy period of approximately nine months (280 days) is divided into three trimesters of three months each.
1. First Trimester : (From fertilization to 12th week)

• During first trimester there are radical changes in the body of mother as well as in the embryo.
• The embryo receives nutrients in the first 2-4 weeks directly from the endometrium.
• It is the main period of organogenesis and the development of body organs.
• By the end of eight weeks, the major structures found in the adult are formed in the embryo in a rudimentary form. It is now called foetus and is about 3 cm long.
• Arms, hands, fingers, feet, toes, CNS, excretory and circulatory system including heart are formed and begins to work.
• Progesterone level becomes high and menstrual cycle is suspended till the end of pregnancy.
• At the end of first trimester foetus is about 7-10 cm long.
• The maternal part of placenta grows, the uterus becomes larger. In this period, the mother experiences morning sickness, (nausea, vomiting, mood swings, etc.)

2. Second Trimester: (From 13th to 26th week)

• The foetus is very active and grows to about 30 cm.
• The uterus grows enough for the pregnancy to become obvious.
• Hormone levels stabilize as hCG declines, the corpus luteum deteriorates and the placenta completely takes over the production of progesterone which maintains the pregnancy.
• Head has hair, eyebrows and eyelashes appear, pinnae are distinct. Baby’s movement can be easily felt by the mother.
• The baby reaches half the size of a new born.

3. Third Trimester: (From 27th week till the parturition)

• Foetus grows to about 50 cm in length and about 3-4 kg in weight.
• As the foetus grows, the uterus expands around it, the mother’s abdominal organs become compressed and displaced, leading to frequent urination, digestive blockages and strain in the back muscles.
• At the end of third trimester the foetus becomes fully developed and ready for parturition.

Balbharti Maharashtra State Board Class 11 Secretarial Practice Important Questions Chapter 10 Correspondence with Directors Important Questions and Answers.

Maharashtra State Board 11th Secretarial Practice Important Questions Chapter 10 Correspondence with Directors

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Secretary is ____________ to Directors.
(a) owner
(b) servant
(c) member
Answer:
(b) servant

Question 2.
Directors are the ____________
(a) owners
(b) representative of shareholders
(c) creditors of the company
Answer:
(b) representative of shareholders

Question 3.
The provision regarding qualification shares of a director is contained in the ____________
(a) Articles of Association
(b) Memorandum of Association
(c) Prospectus
Answer:
(a) Articles of Association

1B. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
A person is elected by the shareholders of a company.
Answer:
Director

Question 2.
Maximum information in minimum words.
Answer:
Brevity

1C. Complete the sentences.

Question 1.
The Directors are ____________ of shareholders.
Answer:
representative

Question 2.
The Directors are responsible for making ____________
Answer:
decision, plans and policies

Question 3.
The report prepared by the directors at the end of every Financial Year is called ____________
Answer:
Directors report

Question 4.
The gap between two consecutive Board meetings should not be more than ____________
Answer:
120 days

1D. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) ………………………	The link between members and the directors
(2) Brevity	……………………………..

(Secretary, Concise and compact information)
Answer:

Group ‘A’	Group ‘B’
(1) Secretary	The link between members and the directors
(2) Brevity	Concise and compact information

1E. Correct the underlined word and rewrite the following sentences.

Question 1.
Brevity means not using harsh words while writing a letter.
Answer:
Politeness means not using harsh words while writing a letter.

Question 2.
The First Board meeting should be held with 60 days from the date of its incorporation.
Answer:
The First Board meeting should be held with 30 days from the date of its incorporation.

Question 3.
Accuracy means avoiding unnecessary details and irrelevant information in a letter.
Answer:
Brevity means avoiding unnecessary details and irrelevant information in a letter.

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 15 Excretion and Osmoregulation Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 15 Excretion and Osmoregulation

Question 1.
What are metabolic waste products?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-products are called metabolic waste products.

Question 2.
Define excretion.
Answer:
The process of eliminating waste products from the body is called excretion.

Question 3.
Where are metabolic wastes produced?
Answer:
Metabolic wastes are produced inside body cells.

Question 4.
Enlist the various excretory products produced in the human body.
Answer:
The various excretory products produced by the human body are as follows:

1. Fluids such as water; gaseous wastes like CO₂; nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium, calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
2. Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
3. The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
4. Volatile substances present in spices (eliminated through lungs).

Question 5.
Write a note on deamination.
Answer:

1. Deamination is the process of breakdown of excess amino acids.
2. It is an essential process, since the body of an organism is unable to store excess amino acids.
3. In this process, amino group is separated from the amino acid and ammonia is formed.
4. Toxic ammonia is either excreted or converted to less toxic forms like urea or uric acid before excretion.

Question 6.
Availability of water plays a key role in deciding the mode of excretion of an organism. Justify.
Answer:

1. Ammonia is the basic product of deamination process.
2. Ammonia is highly toxic and needs to be diluted immediately.
3. If there is no or limited access to water, need for conversion of ammonia becomes necessary.
   Hence, the availability of water plays a key role in deciding the mode of excretion of an organism.

Question 7.
What are the three main modes of excretion in animals?
Answer:
The three main modes of excretion in animals are as follows:
i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

1. Ammonotelism:
   • Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
   • Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
   • Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
   • Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
   • This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
   • Animals that follow this mode of excretion are known as ammonotelic animals.
   • 1 gm ammonia needs about 300 – 500 ml of water for elimination.
   • Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
     e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.
2. Ureotelism:
   • Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
   • Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
   • The body requires less water for elimination.
   • Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
   • It takes about 50 ml H2O for removal of 1 gm NH2 in form of urea.
   • Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
     e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.
3. Uricotelism:
   • Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
   • Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
   • It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
   • Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
   • Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

Question 8.
Fill in the blanks:
i. ________ is the basic product of deamination process.
ii. Aquatic amphibians excrete nitrogenous waste in the form of _______.
iii. Uricotelic organisms, convert ammonia to urea in the _______ by operating _____ cycle.
iv. Ammonia is converted into uric acid by ______ pathway in birds.
Answer:
i. Ammonia
ii. Ammonia
iii. liver, omithine/urea
iv. inosinic acid

Question 9.
Explain the following sentences,
i. Humans are ureotelic.
Answer:

• Urea is comparatively less toxic and less water – soluble than ammonia. Hence, it can be concentrated to some extent in the body.
• The body requires less water for elimination of urea.
  c. Due to these properties, ureotelism is suitable for animals which need to conserve water to some extent.
  Thus, humans are ureotelic.

ii. Sharks retain more urea in their blood.
Answer:

• Sharks retain more urea in their body fluid (blood) to make their blood isotonic to surrounding marine water (in order to maintain osmotic balance).
• This helps them to prevent possible loss of water by exosmosis.

Question 10.
Distinguish between Ureotelism and Uricotelism.
Answer:

No.	Ureotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of urea.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of urea requires less (moderate ) amount of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of urea requires 50 ml of water.	Removal of 1 gm of uric acid requires 5 – 10 ml of water, j
iv.	Urea is less toxic.	Uric acid is least toxic.
e.g.	It is generally seen in terrestrial animals. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most adult amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

Question 11.
Distinguish between Ammonotelism and Uricotelism.
Answer:

No.	Ammonotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of ammonia.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of ammonia requires plenty of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of ammonia requires 300 – 500 ml of water.	Removal of 1 gm of uric acid requires 10ml of water.
iv.	Ammonia is very toxic.	Uric acid is less toxic.
e.g.	It is found in aquatic invertebrates, bony fishes and aquatic/ larval amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

Question 12.
Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic. Why?
Answer:

1. Ammonia is highly toxic to animals.
2. An animal requires large amount of water to dissolve and eliminate ammonia.
3. Terrestrial animals cannot lose such a large amount of water.
4. Ureotelic and uricotelic animals require less amount of water for removal of nitrogenous waste. Hence, to conserve water, ureotelism and uricotelism is adapted by terrestrial animals.

Question 13.
What is plasma creatinine? Why is it used as an index of kidney function?
Answer:

1. Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
2. It provides a ready source of high energy phosphate.
3. Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
4. Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.
   [Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate.]

Question 14.
How can excretion play a role in homeostasis?
Answer:

1. Homeostasis is the maintenance of constant internal environment of the body.
2. Homeostasis is however dependent on osmoregulation, which is the process of controlling solute concentrations and water balance.
3. The composition of blood and therefore the internal environment is highly dependent on what the excretory organs retain in the body.
   Hence, excretion plays an important role in homeostasis.

Question 15.
How do different organisms carry out excretion?
Answer:
Different organisms carry out excretion in the following manner:

1. Unicellular organisms have contractile vacuoles which collect and discharge waste products outside the cell.
2. Excretion in sponges takes place by diffusion of waste material in water. This waste is discharged through the osculum.
3. True organs of excretion are found in those animals that show bilateral symmetry.
4. The most common type of excretory organ is a simple or branching tube that opens to the exterior, through pores called nephridiopores. This system is generally found in some annelids, Amphioxus, earthworms, etc.
5. In most of the insects, excretion takes place by a set of blind ended tubules called malpighian tubules.
6. Crustaceans have green glands as excretory organs.
7. Members of phylum Echinodermata do not have any specialised excretory organs. Waste materials directly diffuse into water or are excreted through tube feet.
8. The mammalian kidneys are a collection of functional units called nephrons, which are well designed to excrete metabolic waste.

Question 16.
What are nephridia? Explain the major types of nephridia in detail.
Answer:
Nephridia are simple or branching tubules used for excretion which open to the exterior through pores called nephridiopores.
Two major types of nephridia are as follows:
i. Protonephridia:
These are network of dead end tubes called flame cells. They are mostly found in animals that lack a true body cavity, e.g. Platyhelminthes, rotifers, some annelids and Amphioxus.

ii. Metanephridia:
These are unbranched coiled tubes that are connected to the body cavity through funnel like structures called nephrostomes. Body fluid enters the nephridium through nephrostome and gets discharged ‘ through nephridiopore. e.g. Earthworms.

Question 17.
Distinguish between the Ureter and Urethra.
Answer:

No.	Ureter	Urethra
i.	Ureters are two duct-like structures arising from the hilum of the kidney.	Urethra is a single tube-like structure arising from the urinary bladder.
ii.	Ureter carries urine from the kidney to the urinary bladder.	Urethra carries urine from the urinary bladder to the exterior of the body.
iii.	Ureters are paired structures.	Urethra is unpaired structure.

Question 18.
Write a short note on micturition.
Answer:

1. The process of release of urine from the urinary bladder is called micturition.
2. The average capacity of urinary bladder is 700 ml.
3. When urinary bladder is almost half filled, stretch receptors in urinary bladder transmit impulses to spinal cord, initiating a conscious desire to expel urine.
4. Micturition reflex center of spinal cord transmit impulses to the wall of urinary bladder and internal urethral sphincter.
5. Bladder muscles contract and muscles of internal urethral sphincter relax.
6. The external sphincter receives impulses from conscious centre of brain and relaxes.
7. This leads to elimination of urine from the bladder.

Question 19.
Explain the L.S of kidney with a neat and labelled diagram.
Answer:

1. Each kidney is covered by three layers of tissue, namely the outermost renal fascia, middle adipose capsule and innermost renal capsule.
   • The outermost layer, renal fascia is made up of a thin layer of fibrous connective tissue. It anchors the kidney to the abdominal wall as well as surrounding tissue.
   • The middle layer is a mass of fatty tissue called adipose capsule. It protects the kidneys by shock absorption.
   • The innermost layer, renal capsule is a smooth transparent fibrous membrane that is continuous with outer layer of ureters. It acts as a barrier against spread of infections in kidney.
2. The L.S. of kidney shows two distinct regions within the capsule. Histologically, kidney is divisible into two regions as renal cortex and renal medulla.
   • Renal cortex is the outer / peripheral, red coloured and granular region. It contains Malpighian bodies, convoluted tubules and blood vessels.
   • Medulla is inner region of kidney with pale red colour and striated appearance. Medulla mainly consists of Loops of Henle and collecting ducts. All these are arranged in conical manner to form renal pyramids.
   • Cortex extends in medulla as columns of Bertini / renal columns between pyramids. Narrow tip of pyramid is called as renal papilla. There are several pyramids.
   • Renal papilla open into the minor calyx. Minor calyces merge together to form major calyces and major calyces unite together to form renal pelvis.
   • Renal pelvis (renal sinus) is funnel-shaped area in the region of medulla of kidney. Renal pelvis
     

Question 20.
What is nephrology?
Answer:
Nephrology is branch of biology that deals with the structure, function and disorders of male and female urinary system.

Question 21.
Write a note on nephron.
Answer:

1. Nephrons are structural and functional units of kidney.
2. Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
3. The wall of the renal tubule is made up of a single layer of epithelial cells.
4. Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
5. The nephron is divisible into Bowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
6. The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question 22.
With the help of a well labelled diagram, describe the structure of nephron.
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:
i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.
a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).
All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:
a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question 23.
What is the difference between Cortical nephrons and Juxtamedullary nephrons.
Answer:

	Cortical nephrons	Juxtamedullary nephrons
i.	They have a shorter loop of Henle.	They have a longer loop of Henle.
ii.	Loop of Henle of these nephrons extends very little into the medulla.	Loop of Henle of these nephrons run deep into the medulla.
iii.	Most nephrons are cortical nephrons.	Few nephrons are juxtamedullary nephrons.
iv.	Efferent arteriole forms peritubular capillary network around DCT, PCT and Henle’s loop of cortical nephrons	Efferent arteriole forms loop-shaped vasa recta around  Henle’s loop of juxtamedullary nephrons.

Question 24.
Sketch and label Malpighian body.
Answer:

Question 25.
What are podocytes? In which part of the nephron are they present?
Answer:
Podocytes are a special type of squamous cells that have a foot-like pedicel. They are present in the visceral wall of the Bowman’s capsule and are in close contact with the walls of capillaries of glomerulus

Question 26.
Write a short note on Juxta Glomerular Apparatus.
Answer:

1. Some smooth muscle cells of the wall of afferent arteriole are modified in such a way that their sarcoplasm is granular. These cells are called juxtaglomerular (JG) cells.
2. In each nephron, initial part of DCT makes contact with the afferent arteriole of same nephron.
3. Cells in the wall of DCT in this region are packed more densely than those in other region of DCT. This is called macula densa.
4. Macula densa and the JG cells together form Juxta Glomerular Apparatus (JGA).
5. The JGA plays an important role in blood pressure regulation within the kidney.

Question 27.
Explain the mechanism of urine formation in detail.
Answer:
Process of urine formation is completed in three steps, namely;
i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 +15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).

Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.

Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.

This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.
Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.

Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 28.
Marine bony fishes and desert amphibians rely on which process of excretion?
Answer:
Tubular secretion

Question 29.
Where does selective reabsorption of glomerular filtrate take place?
Answer:
Selective reabsorption of glomerular filtrate takes place in the proximal convoluted tubule (PCT).

Question 30.
What is glomerular filtration pressure or net effective filtration pressure?
Answer:
Glomerular filtration pressure (GFP)/ Effective filtration pressure (EFP) is the difference between the hydrostatic pressure (GHP) and the sum of osmotic pressure of blood and capsular pressure (CHP).
It can be represented as:
EFP = Glomerular hydrostatic pressure – (Osmotic pressure of blood + Filtrate Flydrostatic pressure)
= 55 – (30 + 15)
= 10 mmHg
[Note: Net filtration pressure = Glomerular blood hydrostatic pressure – (Capsular hydrostatic pressure + Blood colloid osmotic pressure)
Source: Tort or a, G., Derrickson, B. Principles of Anatomy and Physiology. 11^th Edition.]

Question 31.
Distinguish between Selective reabsorption and Tubular secretion.
Answer:

No.	Selective rcabsorption	Tubular secretion
i.	Selective reabsorption is concerned with the selective absorption of useful substances from the glomerular filtrate.	Tubular secretion is transfer of materials from peritubular capillaries to the renal tubular lumen.
ii.	Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca++, K+, Na+, Cl– are absorbed actively, against concentration gradient	In this process, substances like urea. amino acids, glucose, pigments, and inorganic ions arc removed from the blood and discharged along with the urine.
iii.	Selective reabsorption occurs in Proximal convoluted tubule, Henle’s loop, Distal convoluted tubule and collecting duct.	Tubular secretion occurs in Distal convoluted tubules only.

Question 32.
Explain the process of concentration of urine in deai1.
OR
Explain counter current mechanism in detail.
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L, than the blood (300 mOsm/L). hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullarv nephrons and vasa recta is as follows:
i. It involves the passage of fluid from descending to ascending limb of Henle’s loop.

ii. This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.

iii. In case of the vasa recta, blood flows from ascending to descending parts of itself.

iv. Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
v. The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.

vi. Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.

vii. Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.

viii. Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.

ix. Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.

x. Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.

xi. This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.

xii. However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.

xiii. Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.

xiv. Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.

xv. This osmotic gradient is maintained by vasa recta by operating counter current exchange system.

xvi. Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.

xvii. As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.

xviii. Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.

xix. Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.

Question 33.
Camel excretes concentrated urine. Give reason.
Answer:
In order to reabsorb water to maximum capacity, loop of Henle is longer in desert mammals like camel. Hence, camel excretes concentrated urine.

Question 34.
Why is urine yellow in colour?
Answer:
Normal urine is pale yellow coloured transparent liquid, due to the pigment urochrome.

Question 35.
How is the composition of urine regulated?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

1. Regulating water reabsorption through ADH
2. Electrolyte reabsorption though RAAS
3. Atrial Natriuretic Peptide

i) Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.

In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii) Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii) Atrial natriuretic peptide (ANP):
A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na+ and Cl reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na+ in urine) and diuresis.

Question 36.
What is renin? Give its function.
Answer:
Renin is an enzyme secreted by juxtaglomerular cells of afferent arteriole.
Function: It activates Angiotensinogen to Angiotensin-I.

Question 37.
What is the function of Angiotensin II?
Answer:
Functions of Angiotensin II:

1. It constricts arterioles in kidney thereby reducing blood flow and increasing blood pressure.
2. It stimulates PCT cells to enhance reabsorption of Na⁺, Cl^– and water.
3. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

Question 38.
Which hormones and factors are involved in regulation of kidney function?
Answer:
Hormones like ADH, Renin, Angiotensin and Atrial Natriuretic Peptide (ANP) are involved in regulation of kidney function.

Question 39.
Can improper kidney function lead to brittle bones? Justify your answer.
Answer:
Yes, improper kidney function lead to brittle bones.

1. Kidneys participate in synthesis of calcitriol, the active form of Vitamin D which is needed for absorption of dietary calcium.
2. Deficiency of calcitriol can lead to brittle bones.

Question 40.
Do organs other than kidney participate in excretion? Explain.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:
i. Skin:
Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.
a. Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions. These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
b. Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum. It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:
Lungs are the accessory excretory organs. They help in excretion of volatile substances like C02 and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Question 41.
Enlist the human excretory organs and their excretory products.
Answer:
Excretory OrgAnswer:

1. Lungs: Remove CO₂ and also water vapour to a considerable extent. Volatile substances present in spices and other food stuff are excreted through lungs
2. Kidneys: Remove nitrogenous waste products like ammonia, urea and uric acid, creatinine. They also remove excessive amount of water, salts and certain minerals.
3. Skin: Remove water, NaCl, lactic acid and urea by through of sweat.

Question 42.
What is albuminaria? What are its causes?
Answer:

• Albuminaria is the presence of excess albumin in the urine.
• Causes: Injury to the endothelial capsular membrane as a result of increased blood pressure, injury or irritation of kidney cells by substances such as toxins or heavy metals.

Question 43.
A patient report indicates presence of excessive quantities of ketone bodies in the urine. What does this indicate? How is it caused?
Answer:
Presence of excessive quantities of ketone bodies in the urine indicates that the patient is suffering from diabetes mellitus, starvation or too little carbohydrates in the diet.

Question 44.
Sheela is suffering from kidney infection. Presence of which type of cells in the urine can be indicative of this?
Answer:
Presence of leucocytes in the urine can be indicative of infection of kidney or other urinary organs.

Question 45.
Enlist the various disorders and diseases related to the excretory system.
Answer:
Some disorders and diseases related to the excretory system are as follows:

1. Kidney stones
2. Uremia
3. Nephritis
4. Renal Failure
5. Ketonuria
6. Albuminaria

Question 46.
Write a note on kidney stones with reference to types, symptoms and diagnosis.
Answer:
Kidney stones are also called renal calculi. They may be formed in any portion of urinary tract, from kidney tubules to external opening.
Types:
Depending on their composition, kidney stones are classified into the following types.

1. Calcium stones : These are usually calcium oxalate or calcium phosphate stones.
2. Struvite stones : These are formed in response to bacterial infection caused by urea – splitting bacteria. They grow in size quickly and become quite large.
3. Uric acid stones : These stones usually affect people drinking less water or consuming high protein diet.
4. Cystine stones : It is a genetic disorder that causes the kidney to excrete too much of certain amino acid.

Symptoms:
Intermittent pain below rib cage in back and side ways. Hazy, brownish/reddish/ pinkish urine. Frequent urge to pass urine. Pain during micturition.

Diagnosis:
Uric acid content of blood, colour of urine, kidney X-ray, sonography of kidney are different diagnostic tests prescribed depending on symptoms.

Question 47.
What is uremia?
Answer:
If the level of urea in blood rises above 0.05%, the condition is known as uremia. It may lead to kidney failure.

Question 48.
What is the normal content of urea in blood?
Answer:
The normal content of urea in blood is 0.01 to 0.03 %.

Question 49.
Write a note on nephritis.
Answer:

1. Nephritis is the inflammation of kidneys characterised by proteinuria.
2. It is caused due to increased permeability of glomerular capsular membrane, permitting large amounts of proteins to escape from blood to urine.
3. This leads to change in blood colloidal osmotic pressure, leading to movement of fluid from blood to interstitial spaces.
4. It is reflected as edema.

Question 50.
What is renal failure? Describe its types.
Answer:
Renal failure is the decrease or cessation of glomerular filtration and ¡s classified into two types.
i) Acute Renal failure (ARF):
ARF is sudden worsening of renal function that most commonly happens after severe bleeding. There is a decrease in urine output (oligouriaf scanty urine i.e., less than 400 mi/day or less than 0.5 ml/kg/li in children). Other causes of ARF may include acute obstruction of both ureters or nephrotoxic drugs. ARF can be detected biochemically by elevated serum creatinine levels.

ii) Chronic kidney disease (CKD) :
It is the progressive and generally irreversible decline in glomerular filtration rate (GFR). It may be caused due to chronic glomerulonephritis. It can be detected by reduced kidney size and possibility of anaemia.

Question 51.
When does a patient need to undergo haemodialysis? Explain the process in detail.
Answer:

1. When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition, the person has to go for artificial means of filtration of blood i.e. haemodialysis.
2. In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
3. In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
4. The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
5. Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
6. Filtered blood is returned to vein.
7. In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
8. Also, the blood has to move slowly through the tube and hence the process is slow.

Question 52.
Write a note on peritoneal dialysis.
Answer:

1. In this method, the dialyzing fluid is introduced in abdominal cavity or peritoneal cavity.
2. The peritoneal membrane acts as semipermeable dialyzing membrane.
3. Toxic wastes and extra solutes pass into the fluid.
4. This fluid is drained out after a prescribed period of time.
5. Peritoneal dialysis can be repeated as per the need of the patient.
6. It can be carried out at home at work or while travelling. But it is not as efficient as haemodialysis.

Question 53.
What are the drawbacks of haemodialysis?
Answer:

• Kidneys are associated with secretion of erythropoietin, renin and calcitriol which is not possible using dialysis machine.
• During dialysis, the blood has to move slowly through the tube and hence the process is slow.

Question 54.
What is kidney transplant?
Answer:

1. It is the organ transplant of a healthy kidney into a patient with end – stage renal disease.
2. Kidney transplantation is classified as cadaveric (deceased donor) or living donor kidney transplant.
3. Living donor kidney transplant are further classified as genetically related (living-related) or non-related (living non-related) transplants.

Question 55.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it.
i)

Answer:

Skin:
Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.
a. Sweat glands are distributed all over the skin.
They are abundant in the palm and facial regions. These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCI, lactic acid and urea.

b. Sebaceous glands are present at the neck of hair follicles.
They secrete oily substance called sebum. It forms a Lubricating layer on skin making it softer. It protects skin from infection and injury.

ii)

Answer:

1. In this method, the dialyzing fluid is introduced in abdominal cavity or peritoneal cavity.
2. The peritoneal membrane acts as semipermeable dialyzing membrane.
3. Toxic wastes and extra solutes pass into the fluid.
4. This tluid is drained out after a prescribed period of time.
5. Peritoneal dialysis can be repeated as per the need of the patient.
6. It can be carried out at home at work or while travelling. But it is not as efficient as haemodialysis.

Question 55.
Invertebrates, bony fishes, tadpoles, etc. are ammonotelic. Whereas birds, reptiles, land snails, etc. are uricotelic. What would be the probable reason for the difference in mode of excretion?
i. Ammonotelic organisms conserve water as conversion of ammonia to uric acid requires large amount of water.
ii. Elimination of ammonia requires large quantity of water, thus ammonotelism is seen in aquatic animals.
iii. Uricotelic organisms require moderate water for eliminating urea and formation of ammonia requires expenditure of energy. Thus, to conserve water and energy, these animals have uricotelism mode of excretion.
iv. Aquatic animals can retain ammonia and store it in the body for long time.
(A) i and iii
(B) only ii
(C) only iv
(D) ii and iii
Answer:
(B)

Question 56.
A lab technician was evaluating blood reports of some patients. She observed the following values of the report:

Sr. No.	Patient	Comments on urine/ blood report
i.	A	High levels of glucose in urine
ii.	B	Presence of excess quantities of ketone in urine
iii.	C	Presence of leucocytes in urine
iv.	D	0.06% urea in blood
V.	E	High level of proteins in blood

Read the given comments and discuss what disorder/disease the patient must be suffering from.

Answer:

Sr. No.	Patient	Comments on urine/ blood report	Disorder/ Disease
i.	A	High levels of glucose in urine	Glucosuria
ii.	B	Presence of excess quantities of ketone in urine	Ketonuria (Indicative of diabetes mellitus)
iii.	C	Presence of leucocytes in urine	Infection of kidney or urinary organs
iv.	D	0.06% urea in blood	Uremia
V.	E	High level of proteins in blood	Proteinuria/ Nephritis

Multiple Choice Questions

Question 1.
Uric acid is the main nitrogenous waste in
(A) birds
(B) cartilaginous fish
(C) mammals
(D) larval amphibians
Answer:
(A) birds

Question 2.
Protonephridia is the excretory organ of
(A) platyhelminthes
(B) coelenterates
(C) arthropods
(D) aschelminthes
Answer:
(A) platyhelminthes

Question 3.
The glomerulus receives blood through the
(A) vasa recta
(B) renal artery
(C) afferent arteriole
(D) efferent arteriole
Answer:
(C) afferent arteriole

Question 4.
More ADH results in
(A) reduced permeability of DCT
(B) dilute urine
(C) reduced blood pressure
(D) concentrated urine
Answer:
(D) concentrated urine

Competitive Corner

Question 1.
Match the items in Column-I with those in Column-II [NEET Odisha 2019]

	Column-I		Column-II
i.	Podocytes	a.	Crystallised oxalates
ii.	Protonephridia	b.	Annelids
iii.	Nephridia	c.	Amphioxus
iv.	Renal calculi	d.	Filtration slits

Select the correct option from the following:
(A) i-d, ii-b, iii-c, iv-a
(B) i-c, ii-d, iiii-b, iv-a
(C) i-c, ii-b, iii-d, iv-a
(D) i-d, ii-c, iii-b, iv-a
Answer:
(D) i-d, ii-c, iii-b, iv-a

Question 2.
Match the following parts of a nephron with their function: [NEET Odisha 2019]

i.	Descending limb of Henle’s loop	a.	Reabsorption of salts only
ii.	Proximal convoluted tubule	b.	Reabsorption of water only
iii.	Ascending limb of Henle’s loop	c.	Conditional reabsorption of sodium ions and water
iv.	Distal convoluted tubule	d.	Reabsorption of ions, water and organic nutrients

Select the correct option from the following:
(A) i-d, ii-a, iii-c, iv-b
(B) i-a, ii-c, iii-b, iv-d
(C) i-b, ii-d, iii-a, iv-c
(D) i-a, ii-d, iii-b, iv-c
Answer:
(C) i-b, ii-d, iii-a, iv-c

Question 3.
Which of the following factors is responsible for the formation of concentrated urine? [NEET(UG) 2019]
(A) Secretion of erythropoietin by Juxtaglomerular complex.
(B) Hydrostatic pressure during glomerular filtration.
(C) Low levels of antidiuretic hormone.
(D) Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
Answer:
(D) Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.

Question 4.
Portions of renal cortex, which are projected into renal medulla, among the renal pyramids are called as _______. [MHT CET 2019]
(A) Columns of Bertini
(B) Columnae Camae
(C) Ampullae
(D) Ducts of Bellini
Answer:
(A) Columns of Bertini

Question 5.
Renal failure is typically detected by biochemical analysis of blood which shows _______. [MHT CET 2019]
(A) increased level of amino acids
(B) lower level of uric acid
(C) lower level of serum creatinine
(D) higher level of serum creatinine
Answer:
(D) higher level of serum creatinine

Question 6.
Which of the following statements is CORRECT with reference to nephron? [MHT CET 2019]
(A) ADH hormone increases permeability of PCT cells to reabsorb water.
(B) Efferent arteriole forms peritubular network all around tubule.
(C) Podocytes occur in ascending limb of loop of Henle.
(D) Descending limb of Henle’s loop is impenneable to water.
Answer:
(B) Efferent arteriole forms peritubular network all around tubule.

Question 7.
Match the items given in Column I with those in Column li and select the correct option given below. [NFET (UG) 2018]

	Column I (Function)		Column II (Part of excretory system)
i.	Ultrafiltration	a.	Henle’s loop
ii.	Concentration of urine	b.	Ureter
iii.	Transport of urine	c.	Urinary bladder
iv.	Storage of urine	d.	Malpighian corpuscle
v.		e.	Proximal convoluted tubule

(A) i-e, ii-d, iii-a, iv-b
(B) i-d, ii-a, iii-b, iv-c
(C) i-d, ii-e, iii-b, iv-c
(D) i-e, ii-d, iii-a, iv-c
Answer:
(B) i-d, ii-a, iii-b, iv-c

Question 8.
Match the items given in Column I with those in Column II and select the correct option given below. [NEET (UG) 2018]

	Column I		Column II
i.	Glycosuria	a.	Accumulation of uric acid in joints
ii.	Gout	b.	Mass of crystalised salts within the kidney
iii.	Renal calculi	c.	Inflammation of glomeruli
iv.	Glomerular nephritis	d.	Presence of glucose in urine

(A) i-b, ii-c, iii-a, iv-d
(B) i-a, ii-b, iii-c, iv-d
(C) i-c, ii-b, iii-d, iv-a
(D) i-d, ii-a, iii-b, iv-c
Answer:
(D) i-d, ii-a, iii-b, iv-c

Question 9.
In the given diagram of Malpighian body, blood is filtered from part labelled ________

(A) L
(B) M
(C) N
(D) O
Hint: L – Afferent arteriole
M – Efferent arteriole
N – Glomerulus
O – Proximal convoluted tubule
Blood filtration occurs in glomerulus. Afferent arteriole is the blood vessel leading to glomerulus. Efferent arteriole carries blood away from the glomerulus. Proximal convoluted tubule is involved in reabsorption of useful substances from the filtrate.
Answer:
(C) N

Question 10.
Majority of kidney stones consist crystals of [MHT CET 2018]
(A) calcium oxalate, sodium bicarbonate
(B) calcium oxalate, calcium phosphate
(C) calcium phosphate, sodium chloride
(D) calcium carbonate, copper sulphate
Answer:
(B) calcium oxalate, calcium phosphate

Question 11.
Which of the following group of animals is guanotelic? [MHT CEE 2018]
(A) Labeo, turtle, camel
(B) Lizard, snake, scorpion
(C) Penguin, spider, scorpion
(D) Spider, scorpion, snake
Answer:
(C) Penguin, spider, scorpion

Question 12.
Which of the following statements is CORRECT? [NEET (UG) 2017]
(A) The ascending limb of loop of Henle is impermeable to water.
(B) The descending limb of loop of Henle is impermeable to water.
(C) The ascending limb of loop of Henle is permeable to water.
(D) The descending limb of loop of Henle is permeable to electrolytes.
Hint: Descending limb of loop of Henle is permeable to water but impermeable to electrolytes. While ascending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Answer:
(A) The ascending limb of loop of Henle is impermeable to water.

Question 13.
A decrease in blood pressure/volume will not cause the release of [NEET (UG) 2017]
(A) Renin
(B) Atrial Natriuretic Factor
(C) Aldosterone
(D) ADH
Hint: Decrease in blood pressure or volume stimulates the release of renin, aldosterone and ADH. ANF is released when blood pressure increases or blood volume increases. Release of ANF causes vasodilation and also inhibit renin angiotensin mechanism that decreases blood pressure and blood volume.
Answer:
(B) Atrial Natriuretic Factor

Question 14.
Formation of urea takes place in the _________. [MHT CET 2017]
(A) Heart
(B) Kidney
(C) Liver
(D) Lung
Answer:
(C) Liver

Question 15.
The yellow colour of normal urine is due to [MHT CET 2017]
(A) Bilirubin
(B) Biliverdin
(C) Urochrome
(D) Uric acid
Answer:

Question 16.
Uremia is indicated when the blood urea level rises above _______ [MHT CET 2017]
(A) 0.05%
(B) 0.04%
(C) 0.03%
(D) 0.02 %
Answer:
(A) 0.05%

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 3 Kingdom Plantae Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 3 Kingdom Plantae

Question 1.
What is the basis of classification of kingdom Plantae?
Answer:
Kingdom plantae is classified on the basis of characteristics like absence or presence of seeds, vascular tissues, differentiation of plant body, etc.

Question 2.
What are Phanerogams and Cryptogams?
Answer:
1. Phanerogams are seed producing plants. These plants produce special reproductive structures that are visible.
2. Cryptogams are spore producing plants. These plants do not produce seed and flowers. They reproduce sexually by gametes, however their sex organs are concealed.

Question 3.
Write a short note on Chlorophyceae.
Answer:

1. Chlorophyceae includes green algae.
2. These are mostly fresh water (few brackish water and marine).
3. Plant body is unicellular, colonial or filamentous.
4. Cell wall contains cellulose.
5. Chloroplasts are of various shapes like discoid, plate-like, reticulate, cup-shaped, ribbon-shaped or spiral with chlorophyll a and b.
6. Reserved food is in the form of starch.
7. Pyrenoids are located in the chloroplast.
8. Green algae like Chlorella are rich in protein, hence used as food even by space travelers, e.g. Chlamydomonas, Spirogyra, Chara, Volvox, Ulothrix, etc.

Question 4.
Observe the given figure of Chara and identify the parts labelledas.

Answer:
X: Oogonium (contains egg)
Y: Antheridium (contains sperms)

Question 5.
Internet my friend (Textbook page no. 20)
Make a list of green algae with their characteristic shape of chloroplast.
Answer:
Green algae with their characteristic shapes of chloroplast:

1. Chlamydomonas – Cup-shaped
2. Spirogyra – Spiral or ribbon-shaped
3. Oedogonium – Reticulate
4. Zygnema – Stellate or Star-shaped

[Students are expected to search for more information regarding green algae with their characteristic shape of chloroplast from internet.]

Question 6.
Write the characteristics of Phaeophyceae.
Answer:
Characteristics of Phaeophyceae (Brown algae):

1. These algae are mostly marine, rarely fresh water.
2. Plant body is simple branched, filamentous (e.g. Ectocarpus) or profusely branched (e.g. Petalonia).
3. Cell wall has cellulose, fucans and algin.
4. Photosynthetic pigments like chlorophyll-a, chlorophyll-c and fucoxanthin are present.
5. Mannitol, laminarin are stored food materials. Body is usually differentiated into holdfast, stalk called stipe and leaf-like photosynthetic organ called frond.
6. Many species of marine algae are used as food. e.g. Laminaria, Sargassum.
7. Some species are used for the production of hydrocolloids (water holding substances), e.g. Ectocarpus, Fucus, etc.

Question 7.
Identify the given figure of a algae and explain the characteristics of its class with the help of following points:
Habitat, Plant body, photosynthetic pigments, cell wall, stored food

Answer:
The given figure is of Gracillaria. It belongs to class Rhodophyceae (Red algae).
Characteristics of Rhodophyceae:

1. Habitat: These are found in marine as well as fresh water on the surface, deep sea and brackish water.
2. Plant body: Plant body is thalloid.
3. Photosynthetic pigments: Cells contain chlorophyll-a, chlorophyll-d and phycoerythrin.
4. Cell wall: Cell wall is made up of cellulose and pectin glued with other carbohydrates.
5. Stored food: Stored food is in the form of Floridean starch.

Question 8.
What is the commercial use of red algae?
Answer:
Red algae like Gelidium and Gracilaria are used to obtain agar-agar which is used as solidifying agent in tissue culture medium.

Question 9.
Differentiate between red algae and brown algae.
Answer:
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-d and phycoerythrin. Photosynthetic pigments are chlorophyll – a, chlorophyll-c and fucoxanthin.
2. Reserve food is Floridean starch. Reserve food is mannitol and laminarin.
e.g. Porphyra, Gracilaria, Gelidium, Polysiphonia, etc. Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Question 10.
How rhizoids in liverworts differ from that of mosses?
Answer:
Rhizoids are unicellular in liverworts while they are multicellular in mosses.

Question 11.
Explain the thallus structure in lower members of Bryophyta. Give its two examples.
Answer:
1. Liverworts (Hepaticeae) are known as lower members of Bryophyta.
2. Gametophyte possesses flat plant body called thallus.
The thallus is green, dorsiventral, prostrate with unicellular rhizoids.
Examples: Riccia, Marchantia.

Question 12.
What are Hornworts? Give one example.
Answer:
Hornworts (Anthocerotae) are bryophytes which have flattened thallus that produces hornlike structures called as sporophytes. e.g. Anthoceros. In liverworts, asexual reproduction occurs by fragmentation of thalli or with the help of specialized structures called as gemmae. These are green, multicellular, asexual buds which grow in receptacles called gemma cup located on thalli. These gemmae detach from the thallus and germinate to form new individual.

Question 13.
Explain alternation of generation in life cycle of Bryophyta.
Answer:

1. Life cycle of Bryophytes shows sporophytic and gametophytic stages.
2. They alternate with each other to complete their life cycle.
3. Gametophyte is haploid, thalloid or leafy and dominant, (photosynthetic, independent thalloid or erect phase)
4. Sporophyte is short lived, multicellular and depends totally or partially on gametophyte for nutrition and anchorage.

Question 14.
Explain in detail the two stages of gametophytic phase in life cycle of Mosses.
Answer:

1. Gametophytic phase of the life cycle of Mosses (Musci) includes two stages namely; protonema stage and leafy stage.
2. The protonema is prostrate green, branched and filamentous (it is also called juvenile gametophyte). It bears many buds.
3. Leafy stage is produced from each bud.
4. Vegetative reproduction takes place by fragmentation and budding in secondary protonema.
5. The leafy stage has erected, slender stem like (Cauloid) main axis bearing spiral leaf like structures (Phylloid).
6. It is fixed in soil by multicellular branched rhizoids.
7. Leafy stage bears sex organs.

Question 15.
1. Name the two groups of Bryophytes.
2. Give the role of rhizoids in Bryophytes.
Answer:
1. Liverworts and mosses
2. Rhizoids absorb water and minerals and also help in fixation of thallus to the substratum.

Question 16.
Write economic importance of Bryophytes.
Answer:
Economic importance of Bryophytes:

1. Some mosses provide food for herbivorous mammals, birds, etc.
2. Species of Sphagnum, a moss; provides peat used as fuel.
3. Mosses are also used as packing material for transport of living materials because they have significant water holding capacity.
4. Mosses along with lichens are the first living beings to grow on rocks. They decompose rocks to form soil and make them suitable for growth of higher plants.
5. Dense layers of mosses help in prevention of soil erosion, thus act as soil binders.

Question 17.
Which group of plant is known as first vascular and true land plants? Write their characteristics in detail.
Answer:

1. Pteridophytes are known as first vascular and true land plants.
2. Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).
3. Plant body: It is differentiated into root, stem and leaves.
4. Primary root: The primary root is short lived and is soon replaced by adventitious roots.
5. Stem: The stem may be aerial or underground.
6. Leaves: This group contains plants with pinnate (feather – like) leaves. Leaves may be scaly (e.g. Equisetum), simple and sessile (e.g. Lycopodium), small (microphylls e.g. Selaginella) or large (macrophylls) and pinnately compound (e.g. Nephrolepis l Ferns).
7. Vascular tissues: In these members xylem consists of only tracheids and phloem consists of only sieve cells.
8. Secondary growth: Secondary growth is not seen in pteridophytes due to absence of cambium.
9. Alternation of generations: Pteriodphytes show heteromorphic alternation of generations in which the sporophyte is diploid, dominant, autotrophic and independent. Gametophyte is haploid multicellular, generally autotrophic and short lived.

Question 18.
Match the columns.

Column I	Column II
1. Psilopsida	(a) Selaginella
2. Lycopsida	(b) Equisetum
3. Sphenopsida	(c) Adiantum
	(d) Psilotum

Answer:

Column I	Column II
1. Psilopsida	(d) Psilotum
2. Lycopsida	(a) Selaginella
3.Sphenopsida	(b) Equisetum

Question 19.
Write economic importance of Pteridophytes.
Answer:
1. Pteridophytes are used for medicinal purpose and as soil binders.
2. Many varieties are grown as ornamental plants.

Question 20.
Compare the gametophyte and sporophyte of Bryophytes with that of Pteridophytes.
Answer:

	Bryophytes	Pteridophytes
Gametophyte	It is haploid, dominant, photosynthetic, independent, thalloid or erect.	It is haploid, multicellular, generally autotrophic and short lived.
Sporophyte	It is short lived, multicellular and depends totally or partially on gametophyte for nutrition and anchorage.	It is dominant, independent and | vascular plant body.    i

Question 21.
Explain the given figure.

Answer:
1. The given figure represents megasporophyll of Cycas.
2. Megasporophyll of Cycas:
Megasporophylls are usually arranged in compact structures called female cones or female strobili. Megasporophyll contains megasporangia (ovule) which produce megaspores.
[Students are expected to collect more information about coralloid roots, scale leaf and megasporophyll of Cycas.]

Question 22.
Give the economic importance of Cycas and Pinus.
Answer:
1. Cycas is grown as an ornamental plant.
2. Pinus is used as source of pine wood, turpentine oil and pine resin.

Question 23.
Name the following:

Question 1.
Smallest gymnosperm
Answer:
Zamiapygmaea

Question 2.
The plant known as the ‘Coast red wood of California’.
Answer:
Sequoia sempervirens

Question 24.
Ginlcgo biloba is called as living fossil. Why?
Answer:
Ginkgo biloba is called as living fossil, because this plant is found in living as well as fossil form and the number of fossil forms is much more than the living forms.

Question 25.
Which of the following nuts will not be enclosed in fruits?
Betel nut/ Areca nut, pine nut, walnut, almond, cashew nut, nutmeg.
Answer:
1. Pine nuts are edible seeds of pines which are not enclosed in a fruit. It belongs to class gymnospermae thus, seeds are not enclosed within the fruit.
2. Nuts like betel nut/ areca nut, walnut, almond, cashew nut, nutmeg will be enclosed in fruits. It is because these plants belong to class angiospermae in which seeds are enclosed within the fruit.

Question 26.
Name various groups of vascular plants. Give one characteristic feature of each group.
Answer:
There are 3 groups of vascular plants:
1. Pteridophytes
2. Gymnosperms
3. Angiosperms
Characteristics of Pteridophytes: Pteridophytes are the only cryptogams with vascular tissue. Characteristics of Gymnosperms: Gymnosperms are the plants which possess naked seeds and also known as phanerogams without ovary.
Characteristics of Angiosperms: Angiosperms are the flowering plants in which the seeds remain enclosed within the fruits. Double fertilization is the unique feature of angiosperms. [Any one feature]

Question 27.
Classify the given plants into their respective groups and complete the given table.
Equisetum, Chara, Marchantia, Ginkgo biloba, Riccia, Spirogyra, Adiantum, Sorghum
Answer:

Chlorophyceae	Liverworts	Pteridophyta	Gymnosperms	Monocotyledonae
Chara, Spirogyra	Riccia, Marchantia	Equisetum, Adiantum	Ginkgo biloba	Sorghum

Question 28.
Match the columns.

Column I	Column II
1. Bryophyta	(a) 70 genera and 1000 living species
2. Pteridophyta	(b) 32 genera and 80 species
3. Gymnospermae	(c) 960 genera and 25000 species
	(d) 400 genera and 11000 species

Answer:

Column I	Column II
1. Bryophyta	(c) 960 genera and 25000 species
2. Pteridophyta	(d) 400 genera and 11000 species
3. Gymnospermae	(a) 70 genera and 1000 living species

[Source: Textbook of Biology, standard XI, First Edition; 2019, page no. 21,22,23.]

Question 29.
Identify the plants in the given figure and match the columns.


Answer:
1. c – 1
2. d – 2
3. a – 4
4. b – 3

Question 30.
Write a short note on Haplontic life cycle.
Answer:
1. In haplontic life cycle mitosis occurs in haploid cells.
2. It results in the formation of a single celled haploid or a multicellular haploid organism.
3. These forms produce the gametes through mitosis.
4. Zygote is formed after fertilization. This cell is the only diploid cell in the entire life cycle of the organism.
5. Thus, the same zygotic cell later undergoes meiosis.
6. This type of life cycle observed in some algae and fungi.
[Note: Haplontic life cycle is observed in many algae]

Question 31.
Observe the given figure and explain in detail.

Answer:

1. The given figure indicates diplontic life cycle.
2. Here, mitotic division occurs only in diploid cells.
3. Gametes formed through meiosis are haploid in nature.
4. The diploid zygote formed after fertilization divides mitotically.
5. In this process, production of multicellular diploid organism or the production of many diploid single cells takes place.
6. Animals show diplontic life cycle.

[Note: Diplontic type of life cycle is commonly observed in animals and all seed-bearing plants i.e. gymnosperms and angiosperms.]

Question 32.
Explain the term: Haplo-diplontic life cycle.
Answer:
1. In haplo-diplontic life cycle, mitosis occur in both diploid and haploid cells.
2. These organisms undergo through a phase in which they are multicellular and haploid (the gametophyte), and a phase in which they are multicellular and diploid (the sporophyte).
3. It is observed in land plants and in many algae.
[Note: It is commonly observed in bryophytes and pteridophytes.]

Question 33.
Fill in the blanks.
1. In haplo-diplontic life cycle, mitosis occurs in cells.
2. In diplontic life cycle, mitosis occurs in cells.
3. In haplontic life cycle, mitosis occurs in cells.
Answer:
1. diploid and haploid cells
2. diploid cells
3. haploid cells

Question 34.
Practical/Project:

Question 1.
Visit any nursery or botanical garden. Observe some older leaves of fern plant. You can observe some brown spots on back side of the leaflets as shown in the picture given below. Collect more information about it.
Answer:
1. The brown spots on the back side of older leaves of fern are sori.
2. They reproduce asexually by spores produced within sporangia, which are present in sori. These sori are located along the posterior surface of leaflets.

Question 35.
Read the given points.
1. A plant shows thalloid body.
2. A plant shows presence of rhizoids instead of true roots.
3. A plant needs external water for fertilization.
4. Vascular tissues are absent.
Identify the division of the plant described above.
Answer:
The plant belongs to division Bryophyta.

Question 36.
If a person wants to obtain agar for tissue culture, which plant group he should search?
Answer:
A person should search Rhodophyceae. It is because, ‘agar’ which is used as solidifying agent in tissue culture is obtained from red algae-Gelidium and Gracilaria.

Question 37.
Vinaya while playing in garden observed a pond with a green coloured covering which was floating on the surface of water? Next day she asked her teacher about the same. What her teacher must have told her?
Answer:
Vinaya’s teacher must have told her that the green coloured covering floating on the surface of pond water can be green algae like Spirogyra, Chlorella, Chlamydomonas, etc.

Question 38.
Identify the following:

1. These plants belong to thallophyta and grow upto 100 meters in height.
2. Plants used to obtain a product which is used a solidifying agent in preparation of ice-creams and jellies.
3. Gymnosperm which has girth of about 125 feet.
4. Xerophytic fern which belongs to sphenopsida.
5. Unicellular motile alga which belongs to Chlorophyceae and shows cup-shaped chloroplast.

Answer:

1. Kelps
2. Gelidium, Gracilaria
3. Taxodium mucronatum
4. Equisetum
5. Chlamydomonas

Question 39.
Quick Review:

Question 40.
Exercise:

Question 1.
Name the group of spores producing plants in
which sex organs are concealed.
Answer:
Cryptogams are spore producing plants. These plants do not produce seed and flowers. They reproduce sexually by gametes, however their sex organs are concealed.

Question 2.
Name the two divisions of phanerogams.
Answer:
v – phanerogamae

Question 3.
Complete the given flow chart.
Answer:
Quick Review

Question 4.
Define phanerogams.
Answer:
Phanerogams are seed producing plants. These plants produce special reproductive structures that are visible.

Question 5.
Write any two examples of phaeophyceae.
Answer:
Examples of phaeophyceae

Question 6.
Enlist the accessory pigments of algae.
Answer:
Various types of photosynthetic pigments are found in algae.
1. The accessory pigments are chlorophyll-b, chlorophyll-c, chlorophyll-d, carotenes, xanthophylls and phycobilins. Phycobilins are of two types, i.e. phycocyanin and phycoerythrin.
[Students are expected to collect more information about pigments found in algae from internet.]

Question 7.
Bryophytes are the amphibians of the plant kingdom. Justify
Answer:
Members of Bryophyta are mostly terrestrial plants which depend on water for fertilization and completion of their life cycle. Hence, they are called ‘amphibians of Plant Kingdom’.

Question 8.
Distinguish between Rhodophyceae and phaeophyceae with respect to photosynthetic pigments and reserve food.
Answer:
1. Photosynthetic pigments are chlorophyll-a, chlorophyll-d and phycoerythrin. Photosynthetic pigments are chlorophyll-a, chlorophyll-c and fucoxanthin.
2. Reserve food is Floridean starch. Reserve food is mannitol and laminarin.
e.g. Porphyra, Gracilaria, Gelidium, Polysiphonia, etc. Ectocarpus, Sargassum, Fucus, Laminaria, etc.

Question 9.
Write the characteristics of division that includes members like Chlamydomonas, Fucus, Gelidium, etc.
Answer:
Algae belongs to division Thallophyta.
Salient features of algae:

1. Habitat: Algae are mostly aquatic, few grow on other plants as epiphytes and some grow symbiotically. Some algae are epizoic i.e. growing or living non-parasitically on the exterior of living organisms.
   Aquatic algae grow in marine or fresh water. Most of them are free-living while some are symbiotic.
2. Structure: Plant body is thalloid i.e. undifferentiated into root, stem and leaves. They may be small, unicellular, microscopic like Cblorella (non-motile), Chlamydomonas (motile). They can be multicellular, unbranched, filamentous like Spirogyra or branched and filamentous like Chara. Sargassum is a huge macroscopic sea weed which measures more than 60 meters in length.
3. Cell wall: The algal cell wall contains either polysaccharides like cellulose / glucose or a variety of proteins or both. Reserve food material: Reserve food is in the form of starch and its other forms.
4. Photosynthetic pigments: Photosynthetic pigments like chlorophyll – a, chlorophyll – b, chlorophyll – c, chlorophyll – d, carotenes, xanthophylls, phycobilins are found in algae.
5. Reproduction: Reproduction takes place by vegetative, asexual and sexual method.
6. Life cycle: The life cycle shows phenomenon of alternation of generation, dominant haploid and reduced diploid phases.

Question 10.
Name the two algae from which agar is obtained.
Answer:
Red algae like Gelidium and Gracilaria are used to obtain agar-agar which is used as solidifying agent in tissue culture medium.

Question 11.
Identify the incorrectly labelled part in the figure of Funaria.
Answer:

Question 12.
Which are the first terrestrial plants to possess xylem and phloem?
Answer:
Pteridophytes are known as first vascular and true land plants.

Question 13.
Explain in detail three classes of algae.
Answer:

1. Chlorophyceae includes green algae.
2. These are mostly fresh water (few brackish water and marine).
3. Plant body is unicellular, colonial or filamentous.
4. Cell wall contains cellulose.
5. Chloroplasts are of various shapes like discoid, plate-like, reticulate, cup-shaped, ribbon-shaped or spiral with chlorophyll a and b.
6. Reserved food is in the form of starch.
7. Pyrenoids are located in the chloroplast.
8. Green algae like Chlorella are rich in protein, hence used as food even by space travelers, e.g. Chlamydomonas, Spirogyra, Chara, Volvox, Ulothrix, etc.

Characteristics of Phaeophyceae (Brown algae):

1. These algae are mostly marine, rarely fresh water.
2. Plant body is simple branched, filamentous (e.g. Ectocarpus) or profusely branched (e.g. Petalonia).
3. Cell wall has cellulose, fucans and algin.
4. Photosynthetic pigments like chlorophyll-a, chlorophyll-c and fucoxanthin are present.
5. Mannitol, laminarin are stored food materials. Body is usually differentiated into holdfast, stalk called stipe and leaf-like photosynthetic organ called frond.
6. Many species of marine algae are used as food. e.g. Laminaria, Sargassum.
7. Some species are used for the production of hydrocolloids (water holding substances), e.g. Ectocarpus, Fucus, etc.

Question 14.
Write ecological importance of Bryophytes.
Answer:
Economic importance of Bryophytes:
1. Some mosses provide food for herbivorous mammals, birds, etc.
2. Mosses along with lichens are the first living beings to grow on rocks. They decompose rocks to form soil and make them suitable for growth of higher plants.
3. Dense layers of mosses help in prevention of soil erosion, thus act as soil binders.

Question 15.
Mention one example each of aquatic and xerophytic pteridophytes.
Answer:
Habitat: Pteridophytes grow in moist and shady places, e.g. Ferns, Horsetail. Some are aquatic (Azolla, Marsilea), xerophytic (Equisetum) and epiphytic (Lycopodium).

Question 16.
State the uses of algae.
Answer:
(a) Many species of algae are used as food. For e.g. Chlorella (rich in cell proteins hence used as food supplement, even by space travelers), Sargassum, Laminaria, Porphyra, etc.
(b) Alginic acid is produced commercially from Kelps.
(c) Hydrocolloids like algin and carrageen are obtained from brown algae and red algae respectively.
(d) ‘Agar’ which is used as solidifying agent in tissue culture is obtained from red algae like Gelidium and Gracilaria.
(e) Brown algae like sea weeds are used a fodder for sheep, goat, etc.
[Students are expected to collect more information about the economic importance of algae.]
(f) Role of algae in environment.
Answer:
(a) Being photosynthetic, algae help in increasing the level of dissolved oxygen in their immediate environment.
(b) Algae are primary producers of energy rich compounds which forms the basis of food cycles in aquatic animals.
[Students are expected to find out more information about the role of algae in environment on internet.]

Question 17.
Mosses are used as packing material during transport of living material. Give reason.
Answer:
Mosses are also used as packing material for transport of living materials because they have significant water holding capacity.

Question 18.
Write the important characteristics of gymnosperms with respect to following points:
1. Vascular tissues
2. Roots
3. Spores
4. Leaves
Answer:
(b) Vascular tissues: They are vascular plants having xylem with tracheids and phloem with sieve cells.
(e) Roots: The root system is tap root type. In some gymnosperms, the roots form symbiotic association with other life forms. Coralloid roots of Cycas show association with blue green algae and roots of Pinus show association with endophytic fungi called mycorrhizae.
(g) Leaves: The leaves are dimorphic. The foliage leaves are green, simple needle like or pinnately compound, whereas scale leaves are small, membranous and brown.
(h) Spores: Spores are produced by microsporophyll (Male) and megasporophyll (Female).

Question 19.
What are the essential and accessory whorls in flower?
Answer:
Flower: Besides the essential whorls of microsporophylls (androecium) and megasporophylls (gynoecium), there are accessory whorls namely, calyx (sepals) and corolla (petals) arranged together to form flowers.

Question 20.
Write the characteristics of the class which includes Helianthus annuus.
Answer:
Habitat: Angiosperms is a group of highly evolved plants, primarily adapted to terrestrial habitat.

Question 21.
Secondary growth is absent in monocotyledonous plants. Justify.
Answer:
(a) In dicots, vascular bundles are conjoint, collateral and open type. Cambium is present between xylem and phloem for secondary growth.
(b) Whereas in monocots, vascular bundles are conjoint, collateral and closed type. Thus, due to absence of cambium, secondary growth does not occur in majority of monocots.

Question 22.
State characteristic of class monocotyledonae.
Answer:
b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Question 23.
Draw a neat labelled diagram of:
1. Helianthus annuus (sunflower) plant.
2. Maize Plant.
Answer:
Two classes of Angiosperms are Dicotyledonae and Monocotyledonae.
а. Dicotyledonae:

1. These plants have two cotyledons in their embryo.
2. They have a tap root system and the stem is branched.
3. Leaves show reticulate venation.
4. Flowers show tetramerous or pentamerous symmetry.
5. Vascular bundles are conjoint, collateral and open type.
6. Cambium is present between xylem and phloem for secondary growth.
7. In dicots, secondary growth is commonly found.
   e. g. Helianthus annuus (Sunflower)

b. Monocotyledonae:

1. These plants have single cotyledon in their embryo.
2. They have adventitious root system and stem is rarely branched.
3. Leaves generally have sheathing leaf base and parallel venation.
4. Flowers show trimerous symmetry.
5. The vascular bundles are conjoint, collateral and closed type.
6. Cambium is absent between xylem and phloem.
7. In Monocots, except few plants secondary growth is absent, e.g. Zea mays (Maize)

Question 24.
Which is the diploid phase in life cycle of a plant?
Answer:
The life cycle of a plant includes two generations, sporophytic (diploid = 2n) and gametophytic (haploid = n)

Question 25.
Multiple Choice Questions:

Question 1.
Which of the following is not included in sub-kingdom Cryptogamae?
(A) Thallophyta
(B) Dicotyledonae
(C) Pteridophyta
(D) Bryophyta
Answer:
(B) Dicotyledonae

Question 2.
Unicellular, non-motile alga is
(A) Chara
(B) Chlorella
(C) Funaria
(D) Chlamydomonas
Answer:
(B) Chlorella

Question 3.
Which of the following is a brown algae?
(A) Laminaria
(B) Pteris
(C) Ulothrix
(D) Gelidium
Answer:
(A) Laminaria

Question 4.
Agar is obtained from group of algae.
(A) Rhodophyceae
(B) Chlorophyceae
(C) Phaeophyceae
(D) Both (A) and (C)
Answer:
(A) Rhodophyceae

Question 5.
In Chlamydomonas, pyrenoid is located in
(A) nucleus
(B) mitochondria
(C) chloroplast
(D) flagella
Answer:
(C) chloroplast

Question 6.
In bryophytes, represents sporophytic
generation.
(A) rhizoids
(B) thalloid
(C) capsule
(D) leafy plant body
Answer:
(C) capsule

Question 7.
Which of the following is an example of liverwort?
(A) Funaria
(B) Marchantia
(C) Polytrichum
(D) Sphagnum
Answer:
(B) Marchantia

Question 8.
The late Paleozoic era is regarded as the age of ______ .
(A) Thallophytes
(B) Gymnosperms
(C) Pteridophytes
(D) Angiosperms
Answer:
(C) Pteridophytes

Question 9.
Which of the following is an epiphytic pteridophyte?
(A) Azolla
(B) Equisetum
(C) Marsilea
(D) Lycopodium
Answer:
(D) Lycopodium

Question 10.
Complete the given analogy:
Lycopsida: _______:: Pteropsida: Pteris
(A) Adiantum
(B) Selaginella
(C) Equisetum
(D) Psilotum
Answer:
(B) Selaginella

Question 11.
Bryophytes differ from Pteridophytes in being
(A) vascular
(B) seeded
(C) non-vascular
(D) sporophytic
Answer:
(C) non-vascular

Question 12.
Endophytic fungi or mycorrhizae are found in the roots of
(A) Cycas
(B) Pinus
(C) Equisetum
(D) Hibiscus
Answer:
(B) Pinus

Question 13.
Gymnosperms are characterized by the absence of
(A) tracheids in xylem
(B) sieve cells in phloem
(C) heterosporous condition
(D) fruit formation
Answer:
(D) fruit formation

Question 14.
Complete the given analogy:
Tallest angiosperm : Eucalyptus :: Smallest angiosperm : _________ .
(A) Zanta pygmaea
(B) Sequoia sempervirens
(C) Taxodium mucronatum
(D) Wolffia
Answer:
(D) Wolffia

Question 15.
Select the INCORRECT statement with respect to angiosperms.
(A) Seeds are enclosed within a fruit.
(B) These plants show heteromorphic alternation of generation.
(C) Megaspores are borne on highly specialized microsporophyll.
(D) They are most advanced group of flowering plants.
Answer:
(C) Megaspores are borne on highly specialized microsporophyll.

Question 16.
Parallel venation is a characteristic feature of
(A) Monocotyledons
(B) Dicotyledons
(C) Pteridophytes
(D) Bryophytes
Answer:
(A) Monocotyledons

Question 17.
In gymnosperms and angiosperms _______ is much reduced.
(A) gametophyte
(B) root
(C) sporophyte
(D) vascular bundle
Answer:
(A) gametophyte

Question 18.
Presence of rhizoids in place of true roots is a characteristic of
(A) Gymnosperms
(B) Bryophyta
(C) Pteridophyta
(D) Angiosperms
Answer:
(B) Bryophyta

Question 19.
Competitive Corner:

Question 1.
Which one of the following statements is wrong?
(A) Laminaria and Sargassum are used as food.
(B) Algae increase the level of dissolved oxygen in the immediate environment.
(C) Algin is obtained from red algae, and carrageen from brown algae.
(D) Agar-agar is obtained from Gelidium and Gracilaria.
Hint: Algin is obtained from brown algae and carrageenan from red algae.
Answer:
(C) Algin is obtained from red algae, and carrageen from brown algae.

Question 2.
Select the CORRECT statement.
(A) Sequoia is one of the tallest trees.
(B) The leaves of gymnosperms are not well adapted to extremes of climate.
(C) Gymnosperms are both homosporous and heterosporous.
(D) Salvinia, Ginkgo and Pinus all are gymnosperms.
Hint: The leaves of gymnosperms are well adapted to withstand extremes of climate. Gymnosperms are heterosporous. Salvinia is a Pteridophyte.
Answer:
(A) Sequoia is one of the tallest trees.

Question 3.
In bryophytes and pteridophytes, transport of male gametes requires
(A) Birds
(B) Water
(C) Wind
(D) Insects
Answer:
(B) Water