Prasanna – Maharashtra Board Solutions

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

January 8, 2025January 7, 2025 by Prasanna

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

Respiration and Circulation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 8 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 8 Exercise Solutions

1. Multiple choice questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2. Match the columns

Question 1.
Respiratory surface Organism

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

Answer:

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

3. Very Short Answer Questions

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

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

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

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

4. Short Answer Questions

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

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

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

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

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

Question 4.
Blood plasma.
Answer:

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

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

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

Question 6.
Describe pericardium.
Answer:

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

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

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

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

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

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

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

5. Give Scientific Reason

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

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

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

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

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

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

Question 4.
Arteries are thicker than veins.
Answer:

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

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

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

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

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

Question 2.
Arteries and veins.
Answer:

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

Question 3.
Blood and Lymph.
Answer:

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

Question 4.
Blood capillary and Lymph capillary.
Answer:

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

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

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

7. Long Answer Questions

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

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

The breathless persons can be treated by following method:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 8.
Describe histological structure of artery, vein and capillary.

Answer:
Histological structure of artery and vein.

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

Histology of Capillaries:

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

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

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

2. Measurement of blood pressure:

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

3. Factors affecting blood pressure:

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

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

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

(5) Red blood corpuscles or Erythrocytes:

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

Functions of RBCs:

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

(6) White blood corpuscles / Leucocytes:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

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

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

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

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

(I) Neutrophils:

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

(II) Eosinophils or acidophils:

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

(III) Basophils:

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

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

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

(II) Monocytes:

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

(7) Thrombocytes/Platelets:

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

(8) Functions of blood:

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

12th Std Biology Questions And Answers:

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

January 8, 2025January 7, 2025 by Prasanna

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

Control and Co-ordination Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 9 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 9 Exercise Solutions

1. Multiple choice questions

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

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

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

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

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

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

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

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

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

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

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

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

2. Very very short answer questions.

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

4. Very short answer questions.

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

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

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

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

5. Short answer questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Gonadotropins-

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

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

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

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

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

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

(2) Goitre is of two types.

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

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

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

6. Answer the following.

Complete the table.

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

Answer:

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

7. Long answer questions.

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

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

Question 2.
Draw the neat labelled diagrams.
a. Human ear.
Answer:

b. Sectional view of human eye.
Answer:

c. Draw the neat labelled diagram of sagittal section or L.S. of human brain
Answer:

d. Draw the neat labelled diagram of Multipolar Neuron.
Answer:

Question 3.
Answer the questions after observing the diagram given below.

a. What do the synaptic vesicles contain?
Answer:
Synaptic vesicles contain a neurotransmitter – acetyl choline.

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

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

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

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

Question 4.
Explain the Reflex Pathway with the help of a neat labelled diagram.
OR
With the help of a neat and labelled diagram, describe reflex arc.
Answer:

I. Reflex action : Reflex action Is defined as a quick, automatic involuntary and often unconscious action brought about when the receptors are stimulated by external or internal stimuli.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Following are the releasing and inhibiting factors produced by hypothalamus:

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

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

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

It secretes following hormones:

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

Gonadotropins-

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

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

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

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

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

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

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

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

II. Protein hormones:

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

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

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

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

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

12th Std Biology Questions And Answers:

12th Biology Chapter 2 Exercise Reproduction in Lower and Higher Animals Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

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

Reproduction in Lower and Higher Animals Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 2 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 2 Exercise Solutions

1. Multiple choice questions

Question 1.
The number of nuclei present in a zygote is ……………….
(a) two
(b) one
(c) four
(d) eight
Answer:
(b) one

Question 2.
Which of these is the male reproductive organ in human?
(a) Sperm
(b) Seminal fluid
(c) Testes
(d) Ovary
Answer:
(c) Testes

Question 3.
Attachment of embryo to the wall of the uterus is known as ……………….
(a) fertilization
(b) gestation
(c) cleavage
(d) implantation
Answer:
(d) implantation

Question 4.
Rupturing of follicles and discharge of ova is known as ……………….
(a) capacitation
(b) gestation
(c) ovulation
(d) copulation
Answer:
(c) ovulation

Question 5.
In human females, the fertilized egg gets implanted in uterus ……………….
(a) after about 7 days of fertilization
(b) after about 30 days of fertilization
(c) after about two months of fertilization
(d) after about 3 weeks of fertilization
Answer:
(a) after about 7 days of fertilization

Question 6.
Test tube baby technique is called ……………….
(a) In vivo fertilization
(b) In situ fertilization
(c) In Vitro Fertilization
(d) Artificial Insemination
Answer:
(c) In Vitro Fertilization

Question 7.

The given figure shows a human sperm. Various parts of it are labelled as A, B, C, and D. Which labelled part represents acrosome?
(a) B.
(b) C
(c) D
(d) A
Answer:
(d) A

Question 8.
Presence of beard in boys is a ……………….
(a) primary sex organ
(b) secondary sexual character
(c) secondary sex organ
(d) primary sexual character
Answer:
(b) secondary sexual character

2. Very short answer questions

Question 1.
What is the difference between a foetus and an embryo?
Answer:
Embryo is a growing egg after fertilization until the main parts of the body and the internal organs have started to take shape while foetus is a stage which has the appearance of a fully developed offspring.

Question 2.
Outline the path of sperm up to the urethra.
Answer:
The path of sperm up to the urethra in male is as follows :
Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory ducts Urethra.

Question 3.
Which glands contribute fluids to the semen?
Answer:
The glands which contribute fluids to the semen are seminal vesicles, prostate gland.

Question 4.
Name the endocrine glands involved in maintaining the sexual characteristics of males.
Answer:
Interstitial cells of Leydig which lie in between the seminiferous tubules are involved in maintaining the sexual characteristics of male by secreting the male hormone androgen or testosterone. Adenohypophysis also regulates this secretion from the testis.

Question 5.
Where does fertilization and implantation occur?
Answer:
Fertilization of ovum takes place in the ampulla region of fallopian tube whereas implantation occur in the endometrium of uterus.

Question 6.
Enlist the external genital organs in female.
Answer:
The external genital organs in female include the following parts such as vestibule, labia minora, clitoris, labia majora and mons Veneris.

Question 8.
What is the difference between embryo and zygote?
Answer:
Zygote is the unicellular diploid structure formed as a result of fusion of sperm and ovum whereas embryo is a multicellular structure formed from zygote in the uterus 3 weeks after fertilization.

3. Fill in the blanks

Question 1.
The primary sex organ in human male is ……………….
Answer:
testis

Question 2.
The ……………… is also called the womb.
Answer:
uterus

Question 3.
Sperm fertilizes ovum in the ……………….. of fallopian tube.
Answer:
ampulla

Question 4.
The disc like structure which helps in the transfer of substances to and from the foetus’s body is called ………………..
Answer:
placenta

Question 5.
Gonorrhoea is caused by ……………….. bacteria.
Answer:
Neisseria gonorrhoeae

Question 6.
The hormone produced by the testis is ……………………
Answer:
testosterone / androgen.

4. Short Answer Questions

Question 1.
Budding in Hydra.
Answer:

Budding is a type of asexual reproduction method seen in Hydra.
Budding takes place during favourable period.
Towards the basal end of the body, small outgrowth is produced which is called a bud.
It grows and forms tentacles and gradually forms a new individual.
The young Hydra after complete development detaches from the parent and becomes an independent new organism.

Question 2.
Explain the different methods of reproduction occurring in sponges.
Answer:

Sponges reproduce both asexually and sexually and they also possess the power of regeneration. Their sexual reproduction is similar to higher animals even though their body organization is primitive type.
Asexual reproduction in sponges takes place by regeneration, budding and gemmule formation.
In sponges, during unfavourable period, gemmule is produced. It is an internal bud.
Archaeocytes which are dormant cells are seen in the aggregation in gemmule. These cells are capable of developing into a new organism.
Amoebocytes are other cells which secrete thick resistant layer of secretion which is coated around archaeocytes.
When favourable conditions of water and temperature return back, the gemmules can develop into new individuals by hatching, e.g. Spongilla.

Question 3.
IVF.
Answer:

In laboratory under sterile conditions, oocyte and sperms are placed in a test tube or glass plate to form a zygote. This process is called In Vitro Fertilization.
The zygote with 8 blastomeres is then transferred into the fallopian tube for further development.
IVF technique is used when childless couple wants to have a baby, but there are issues of sterility.
IVF is also called test tube baby technique.

Question 4.
Comment on any two mechanical contraceptive methods.
Answer:
Two mechanical contraceptive methods are as follows:
A. Condom or Nirodh:

It is a protective barrier in the form of thin rubber sheath which is used by male partner during the sexual coitus. It covers the penis and does not allow semen to flow during copulation.
Thus the entry of ejaculated semen into the female reproductive tract is obstructed. This can prevent conception. It is a simple and effective method and has no side effects.
“Nirodh” is a condom, most widely used in India as a contraceptive by males.
Condom also protects both the partners against sexually transmitted diseases such as AIDS and others.

B. Diaphragm, cervical caps and vaults:

Diaphragm and cervical caps are to be used by females as mechanical contraceptive measures.
They are made up of rubber. They are fitted on the cervix in vagina so that they prevent the entry of sperms into the uterus.
They are kept at least six hours after sexual intercourse in order to inhibit sperms from entering female genital tract.

Question 5.
Tubectomy.
Answer:

The permanent birth control method in women, is called tubectomy.
It is a surgical method, also called sterilization.
In tubectomy, a small part of the fallopian tube is tied and cut.
Tubectomy blocks transport of oocytes and also blocks sperms, thus preventing fertilization from reaching the oocyte.

Question 6.
Give the name of causal organism of Syphilis and write on its symptoms.
Answer:
1. Syphilis is a sexually transmitted veneral disease caused by a Spirochaete bacterium Treponema pallidum.

2. The site of infection is the mucous membrane in genital, rectal and oral region.

3. Symptoms of syphilis:

Primary lesion known as chancre at the site of infection.
They are seen on the external genitalia in males and inside the vagina in females.
Skin rashes accompanied by fever, inflammed joints and loss of hair.
Paralysis
Degenerative changes in the heart and brain.

Question 7.
What is colostrum?
Answer:

The fluid secreted by the mammary glands soon after childbirth is called colostrum.
Colostrum is the sticky and yellow fluid. It contains proteins, lactose and mother’s antibodies, e.g. IgA.
The fat content in colostrum is low.
The antibodies present in colostrum helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

5. Answer the Following Questions

Question 1.
Describe the phases of menstrual cycle and their hormonal control.
Answer:
Menstrual cycle (Ovarian cycle):
i. Menstrual cycle involves a series of cyclic, changes in the ovary and uterus. The cyclic events are regulated by gonadotropins from pituitary and the hormones from ovary.
ii. The cyclic events in woman are repeated within approximately 28 days.
iii. Menstrual cycle is divided into following phases, viz.

Menstrual phase (Day 1-5)
Follicular phase in ovary that coincides with proliferative phase in uterus. Post menstrual phase (Day 5-14)
Ovulatory phase (Day 14-15)
Luteal phase in ovary which coincides with secretory phase in uterus (Day 16 to 28).

1. Menstrual Phase:

Menstrual phase occurs in the absence of fertilization.
During menstruation, uterine endometrium is sloughed off. Level of progesterone and estrogen decrease during this phase resulting into release of prostaglandins which cause this rupture.
Blood about 45-100 ml, tissue fluid, mucus, endometrial lining and unfertilized oocyte and other cellular debris is discharged through vagina as a menstrual flow. The endometrial lining becomes about 1 mm thin.
Fibrinolysin does not allow blood to clot during this period.
Pituitary starts secreting FSH, which further makes many primordial follicles to develop into primary and few of them into secondary follicles.

2. Proliferative phase/Follicular phase/Post menstrual phase:

During this phase in the ovary the follicles develop while in uterus the endometrium starts proliferating. 6 to 12 secondary follicles start developing but usually only one of them becomes Graafian follicle due to action of FSH.
Developing secondary follicles secrete the hormone estrogen.
Estrogen brings about regeneration of endometrium. Further proliferation of endometrium causes formation of endothelial cells, endometrial or uterine glands and network of blood vessels. Endometrium’s thickness becomes 3-5 mm.

(3) Ovulatory phase:

Ovulation occurs in this phase. Mature Graafian follicle ruptures and secondary oocyte is released into the pelvic region of abdomen.
Ovulation occurs due to surging quantity of LH from pituitary.

(4) Luteal phase/Secretary phase :
(i) Since the empty Graafian follicle converts itself into corpus luteum under the influence of LH, this phase is called luteal phase in ovary. At the same time, the uterine endometrium thickens and becomes more secretory and hence it is called secretory phase in uterus.

(ii) Corpus luteum secretes progesterone, some amount of estrogens and inhibin. These hormones stimulate the growth of endometrial glands which later start uterine secretions.

(iii) Endometrium becomes more vascularized becomes 8-10 mm. in thickness. These changes are the preparation for the implantation of the ovum if fertilization occurs.

(iv) In absence of fertilization, corpus luteum can survive for only two weeks and then degenerate into a non-secretory white scar called corpus albicans.

(v) If ovum is fertilized, woman becomes pregnant and hormone hCG (human Chorionic Gonadotropin) is secreted by chorionic membrane of embryo which keeps corpus luteum active till the formation of placenta.

Question 2.
Explain the steps of parturition.
Answer:
Parturition involves the following three steps:
1. Dilation stage:

Dilation stage means dilating the birth canal or passage though which baby is pushed out. In the beginning uterine contractions start from top and baby is moved to cervix. Due to compression of blood vessels and movements of flexible joints in pelvic girdle, mother experiences labour pains.
Oxytocin is secreted later in more amount causing severe uterine contractions. This pushes baby in a head down position and closer to cervix.
Cervix and vagina both are dilated.
This stage lasts for about 12 hours.
At the end, amniotic sac ruptures and amniotic fluid is passed out.

2. Expulsion stage:

During second stage of about 20 to 60 minutes, the uterine and abdominal contractions become stronger.
Foetus moves out with head down position through cervix and vagina.
The umbilical cord which connects the baby to placenta is tied and cut off close to the baby’s navel.

3. After birth or placental stage : In the last stage of 10 to 45 minutes, once the baby is out then the placenta is also separated from uterine wall and is expelled out as “after birth”. This is accompanied by severe contractions of the uterus.

Question 3.
Explain the histological structure of testis.
Answer:
Histological structure of testis:

The external covering of testis is a fibrous connective tissue called tunica albuginea.
Then there is an incomplete peritoneal covering called tunica vaginalis.
Interior to this there is a covering called tunica vascularis formed by capillaries.
The testis is composed of many seminiferous tubules that are lined by cuboidal germinal epithelial cells.
In the seminiferous tubules various stages of developing sperms are seen as spermatogenesis takes place here. These stages are spermatogonia, primary and secondary spermatocytes, spermatids and sperms.
Large, pyramidal sub tentacular cells, nurse cells or Sertoli cells are present between germinal epithelium. Sperm bundles remain attached to Sertoli cells with their heads.
Seminiferous tubules form sperms whereas Sertoli cells provide nourishment to the sperms till maturation.
In between the seminiferous tubules there are interstitial cells of Leydig which are endocrine in nature. They secrete testosterone.

Question 4.
Describe the structure of blastocyst or blastulation
Answer:

The outer layer of cells of the morula is called trophoblast or trophoectoderm. This layer absorbs the nutritive fluid secreted by uterine endometrial membrane.
As more and more fluid is absorbed by trophoblast cells, the cells become flat and a cavity called blastocyst cavity or blastocoel or segmentation cavity is formed.
This causes trophoblast cells to get separated from inner cell mass except at one side.
The trophoblast cells in contact with embryonal knob are known as cells of Rauber. As the quantity of fluid increases, the morula enlarges rapidly and assumes the shape of a cyst. This stage is called blastocyst.
The side of the blastocyst to which embryonal knob is attached is known as the embryonic or animal pole and the opposite side as abembryonic pole.
The trophoblast produces extra embryonic membranes and does not participate in the formation of embryo proper.
Zona pellucida disappears allowing the blastula to increase in size and volume. The blastocyst stage is reached in about five days after fertilization.
Blastocyst depends on mother for nutrition which it obtained through placenta.

Question 5.
Explain the histological structure of ovary in human.
Answer:
Histological structure of ovary:
(1) Each ovary is a compact structure differentiated into a central part called medulla and the outer part called cortex.

(2) The cortex is covered externally by a layer of germinal epithelium while the medulla contains the stroma or loose connective tissue with blood vessels, lymph vessels and nerve fibres.

(3) Different stages of developing ovarian follicles are seen in the cortex. Each primordial follicle has at its centre a large primary oocyte (2n) surrounded by a single layer of flat follicular cells, then gradually it matures.

(4) In the ovary during each menstrual cycle there is a maturation of primordial follicles into multilayered primary, secondary and Graafian follicles.

(5) Every Graafian follicle has three layers, viz. theca externa, theca interna and membrana granulosa which are from outer to inner side. A space called antrum filled with liquor folliculi is present inside the follicle. There is a small hillock of cells called cumulus oophours or discus proligerus over which the ovum is lodged. The ovum in turn is covered by vitelline membrane, zona pellucida and corona radiata from inner side to outer surface.

(6) Ovarian cortex also shows corpus luteum, or yellow body formed from empty Graafian follicle after ovulation. Corpus luteum is converted into corpus albicans or white body in case of absence of conception.

Question 6.
Describe the various methods of birth control to avoid pregnancy.
Answer:
Birth control/Contraceptive methods are of two main types, viz. temporary and permanent.
A. Temporary methods:
(1) Natural method/Safe period/Rhythm method : A week before and a week after menstrual bleeding is considered the safe period for sexual intercourse. It is based on the fact that ovulation occurs on the 14th day of menstrual cycle.

(2) Coitus Interruptus or withdrawal : In this method, the male partner withdraws his penis from the vagina before ejaculation, so as to avoid insemination. This method also has some drawbacks, as the pre-ejaculation fluid may contain sperms and this can cause fertilization.

(3) Lactational amenorrhoea (absence of menstruation) : This method is based on the fact that ovulation does not occur during the period of intense lactation following parturition so chances of conception are almost negligible. However, this method also has high chances of failure.

(4) Chemical means (spermicides) : In this method chemicals like foam, tablets, jellies and creams are introduced into the vagina before sexual intercourse, they adhere to the mucous membrane, immobilize and kill the sperms.

(5) Mechanical means/Barrier methods:
(i) Condom : It is a thin rubber sheath that is used to cover the penis of the male. Condom should be used before starting coital activity. It also prevents STDs and AIDS.

(ii) Diaphragm, cervical caps and vaults : These devices made of rubber are inserted into the female reproductive tract to cover the cervix diming copulation. They prevent conception by blocking the entry of sperms through the cervix.

(iii) Intra-uterine devices (IUDs) : These are plastic or metal objects placed in the uterus by a doctor. These include Lippes loop, copper releasing IUDs (Cu-T, Cu 7, multiload 375) and hormone releasing IUDs (LNG-20, progestasert). They prevent fertilization of the egg or implantation of the embryo.

(6) Physiological (Oral) Devices : Birth control pills (oral contraceptive pills) check ovulation as they inhibit the secretion of follicle stimulating hormone (FSH) and luteinizing hormone (LH) that are necessary for ovulation. The pill ‘Saheli’ is taken weekly.

(7) Other contraceptives : The birth control implant is similar to that of pills in their mode of action. It is implanted under the skin of the upper arm of the female.

B. Permanent methods surgical operations : In men surgical operation is called vasectomy and in women it is called tubectomy. This method blocks gamete transport and prevent pregnancy.

Question 7.
What are the goals of RCH programmes?
Answer:
Goals of RCH programmes are as follows:

Various aspects related to reproduction are made aware to general public.
Facilities are provided to people to understand and build up reproductive health.
Support is given for building up a reproductively healthy society.
Three critical health indicators, i.e. reducing total infertility rate, infant mortality rate and maternal mortality rate are well looked after.

Question 8.
What is parturition? Which hormones are involved in parturition?
Answer:

Parturition is the act of expelling out the mature foetus from the uterus of mother via the vagina.
When the foetus is fully mature, it starts secreting ACTH (Adreno Cortico Trophic Hormone) from its pituitary.
ACTH stimulates adrenal glands of foetus to produce corticosteroids.
These corticosteroids diffuse from foetal blood to mother’s blood across the placenta. Corticosteroids accumulate in mother’s blood that results in decreased amount of progesterone. Corticosteroids also increase secretion of prostaglandins.
Simultaneously estrogen levels rise bringing about initation of contractions of uterine muscular wall.
Reduced progesterone level and increased estrogen level cause secretion of oxytocin from mother’s pituitary. This causes greater stimulation of myometrium of uterus.
Prostaglandins cause increased forceful contraction of uterus which expels the foetus out of the uterus.
Hormone relaxin secreted by the placenta makes the pubic ligaments and sacroiliac joints of the mother loosen. This causes widening of birth canal which facilitates the normal birth of the baby.

Question 9.
What are the functions of male accessory glands?
OR
Write a brief account of accessory sex glands associated with human male reproductive system.
Answer:
Seminal Vesicles, prostate gland and Cowper’s glands are associated with human male reproductive system.
(i) Seminal Vesicles:

Seminal vesicles occur in pair present on the posterior side of urinary bladder. Its secretion consists about 60% of the total volume of the semen. The secretion is an alkaline seminal fluid containing fructose, fibrinogen and prostaglandins.
Fructose helps in the movement of sperms by providing energy to them.
Semen is coagulated in bolus by fibrinogen. This helps in faster movements of sperms in vagina after insemination.
Reverse peristalsis in vagina and uterus for faster movement of sperms towards the egg in the female body is elided by prostaglandins.

(ii) Prostate gland:

It is a single gland located under the urinary bladder. It has about 20 to 30 separate lobes which open separately into the urethra.
Prostatic fluid secreted by this gland is milky white and slightly acidic. It forms 30 % of the semen and is secreted in urethra.
Its contents are citric acid, acid phosphatase and various other enzymes.
The sperms are protected from the acidic environment of vagina by acid phosphatase.

(iii) Cowper’s glands (Bulbo-urethral glands):

Cowper’s glands occur in pair on either side of urethra. They are small and pea shaped.
Cowper’s glands secrete an alkaline, viscous, mucous-like fluid. It helps as lubricant during copulation.

Question 10.
What is capacitation? Give its importance.
Answer:

Capacitation is the process by which the sperms are made capable to swim up to the fallopian tubes. This process takes place in 5-6 hours.
50% of ejaculated sperms die due to unfavourable vaginal and uterine conditions.
The remaining sperms are capacitated with the help of prostaglandin and vestibular secretions of female tract. It involves the changes in the membrane covering the acrosome.
Due to capacitation, acrosome membrane becomes thin, Calcium ions enters the sperm and their tail begin to show rapid whiplash movements.
Sperms become extra active and then they ascend upwards to reach fallopian tubes.
After capacitation the sperms swim through the vagina and uterus and reach ampulla of fallopian tube within 5 minutes.

Long answer questions

Question 1.
Explain the following parts of male reproductive system along with labelled diagram showing these parts – Testis, vasa deferentia, epididymis, seminal vesicle, prostate gland and penis.
OR
With the help of a neat, labelled diagram, describe the human male reproductive system.
Answer:

(i) Testis, the male gonad, the accessory ducts and glands along with external genitalia form the male reproductive system.

(ii) Testes:

Testes are male gonads with dimensions of about 4.5. cm length, 2.5 cm width and 3 cm thickness.
There are about 200 to 300 lobules in each testis in which there are seminiferous tubules that form rete testis.
Testes produce sperms and secrete male sex hormone, androgen or testosterone.

(iii) Accessory ducts : Rete testis, vasa efferentia, epididymis, vas deferens, ejaculatory duct and urethra together form the accessory ducts of male reproductive system.
1. Vasa efferentia : Vasa efferentia are 12-20 fine tubules. They arise from rete testis and end into the epididymis. The sperms from the testis are carried by these ducts to the epididymis.

2. Epididymis : Epididymis are long and coiled tubes having three parts, viz. caput, corpus and cauda epididymis. They are located on the posterior border of each testis. The sperms undergo maturation in epididymis.

3. Vasa deferentia:

Vasa deferentia are a pair of 40 cm long tubular structures that arise from cauda epididymis.
Each vas deferens enters the abdominal cavity through the inguinal canal and then ascends in the form of spermatic cord.
Vas deferens of each side is joined by the duct from seminal vesicle to form ejaculatory duct.

4. Ejaculatory duct : About 2 cm long pair of ducts formed by joining of vas deferens and a duct of seminal vesicle are the ejaculatory ducts. Both ejaculatory ducts open into urethra near the prostate gland. Seminal fluid containing spermatozoa are carried by ejaculatory duct to the urethra.

5. Urethra : The male urethra provides a common passage for the urine and semen hence is also called urinogenital duct.

(iv) Accessory glands : Associated with male reproductive system are : (a) Seminal vesicles (b) Prostate gland and (c) Cowper’s or Bulbourethral glands. Every accessory gland has secretion which helps in functions of reproductive system.

(v) External genitalia : External genitalia consists of penis and scrotum.
1. Penis:

Penis is the copulatory organ used for insemination or deposition of sperms in female genital tract.
It is cylindrical, erectile and pendulous organ through which passes the urethra.
It contains three columns of erectile tissues which has abundant blood sinuses.
The tip is called glans penis while the retractible fold of skin on penis is called prepuce.

2. Scrotum : The scrotum is a pouch of pigmented skin arising from lower abdominal wall. It protects testes within it. Scrotum acts as thermoregulator. Testis are suspended in scrotum by spermatic cord.

Question 2.
Describe female reproductive system of human
Answer:

The female reproductive system consists of internal organs and external genitalia.

Internal organs are pair of ovaries and pair of fallopian ducts or oviducts, single median uterus and vagina. External genitalia is called vulva. There are a pair of vestibular glands in external genitalia. Mammary glands or breasts are also associated with reproductive system of female.

(1) Ovaries:

Ovaries are situated in the abdomen in upper lateral part of the pelvis near the kidneys. Their dimensions are about 3 cm in length, 1.5 cm in breadth and 1.0 cm thick. They are solid, oval or almond shaped organs.
Ovaries produce ova and they are also endocrine in nature as they produce estrogen, progesterone, relaxin, activin and inhibin.
Ovarian hormones bring about secondary sexual characters. They also control menstrual cycle, pregnancy and parturition.

(2) Fallopian tubes/oviducts:
(i) Fallopian tubes lie horizontally over peritoneal cavity. These are about 10 to 12 cm long, narrow, muscular structure lined by ciliated epithelium.
(ii) They transport the ovum after ovulation from the ovary to the uterus.
(iii) Fallopian tube can be subdivided into the following three parts:

The infundibulum which bears a number of finger-like processes called fimbriae at its free border.
Infundibulum is funnel-shaped having ostium which receives ova released from the ovary.
The second part is the ampulla where the fertilization takes place.
The last part is short cornua or isthmus which opens into the uterus.

(3) Uterus/Womb:
(i) Uterus is a pear-shaped, highly muscular, thick walled, hollow organ measuring about 8 cm in length, 5 cm in width and 2 cm in thickness.

(ii) Uterus has the following three parts : Fundus, Body or corpus and Cervix.

(iii) The cervix communicates above with the body of the uterus by an aperture, the internal os and with vagina below by an opening the external os.

(iv) Uterus has three-layered wall. These layers are:

Perimetrium : An outer serous layer.
Myometrium : The middle thick muscular layer of smooth muscles.
Endometrium : The inner highly vascular mucosa that has many uterine glands.

(v) Uterus receives the ovum from fallopian tube. It develops placenta during pregnancy for the nourishment of foetus. At the time of parturition, it expels the young one at birth.

(4) Vagina:

Vagina is a highly distensible fibro-muscular tube that lies between the cervix and the vestibule.
It is about 7 to 9 cm in length and is internally lined by stratified and non- keratinised epithelium. The vaginal wall has inner mucous lining.
Vagina acts as a birth canal as well as copulatory passage. It also allows passage of menstrual flow.
Vagina opens into vestibule by vaginal orifice which may be covered with hymen which is also a mucous membrane.

(5) External genitalia or vulva or pudendum : The external genitalia consists of five parts; viz. labia majora, labia minora, mons veneris, clitoris and vestibule.

(6) A pair of vestibular glands / Bartholin’s glands : These glands open into the vestibule and release a lubricating fluid.

(7) A pair of mammary glands/breasts : These are the accessory organs of female reproductive system for production and release of milk after parturition.

Question 3.
Describe the process of fertilization.
Answer:
(1) Fertilization is the process of fusion of the haploid male and female gametes which results in the formation of a diploid zygote (2n).

(2) In human beings fertilization is internal. Sperms deposited in vagina, swim across the uterus and fertilize the ovum in ampulla of the fallopian tube.

(3) Fertilization involves the following events:
(i) Insemination : Discharge of semen into the vagina at the time of copulation is called insemination.

(ii) Movement of sperm towards egg : Sperms reaching the vagina undergo capacitation process for 5-6 hours. During capacitation acrosomal membrane of sperm becomes thin and Ca⁺⁺ enters the sperm making it extra active. Sperms reach up to the ampulla by swimming aided with contraction of uterus and fallopian tubes. These contractions are stimulated by oxytocin of female. By capacitation sperm can reach ampulla within 5 minutes, they remain 5 viable for 24 to 48 hours, whereas ovum remains viable for 24 hours.

(iii) Entry of sperm into the egg : Though many sperms reach the ampulla, only a single sperm fertilizes the ovum. The acrosome of sperm after coming in contact with the ovum, releases lysins; hyaluronidase and corona penetrating enzymes. Due to these enzymes cells of corona radiata are separated and dissolved. The sperm head then passes through zona pellucida of egg. The zona pellucida has glycoprotein fertilizin receptor proteins. These bind to specific acid protein-antifertilizin of sperm. This makes sperm and ovum to come together. Fertilizin-Antifertilizin interaction is species- specific.

(iv) Acrosome reaction : When the sperm head comes in contact with the zona pellucida, its acrosome covering ruptures to release lytic enzymes, acrosin or zona lysin. These enzymes dissolve plasma membrane of egg so that the sperm nucleus and the centrioles enter the egg, while other parts remain outside. Now the vitelline membrane of egg changes into fertilization membrane which prevents any further entry of other sperms into the egg, thus polyspermy is prevented.

(v) Activation of ovum : After the entry of sperm head into ovum, it gets activated to resume and complete its meiosis-II. With this it gives out the second polar body. The germinal vesicle organises into female pronucleus. At this stage, it is true ovum.

(vi) Fusion of egg and sperm : The coverings of male and female pronuclei degenerate results in the formation of a synkaryon by a process called syngamy or karyogamy. The zygote is thus formed.

Question 4.
Explain the process by which zygote divides and redivides to form the morula.
Answer:
(1) Cleavage is a rapid mitotic division to form a blastula. These divisions takes place immediately after fertilization. The cells formed by cleavage are called blastomeres.

(2) The type of cleavage in human is holoblastic, i.e. the whole zygote gets divided, radial and indeterminate, i.e. fate of each blastomere is not predetermined.

(3) Cleavage show faster synthesis of DNA and high consumption of oxygen.

(4) Since there is no growth phase between the cleavages, the size of blastomeres will be reduced with every successive cleavage.

(5) The cleavages occur as follows:

(6) Successive divisions produce a solid ball of cells called morula of 16 cells. It consists of an outer layer of smaller clearer cells and an inner mass of larger cells.

(7) Morula reaches the uterus about 4-6 days after fertilization.

12th Std Biology Questions And Answers:

12th Biology Chapter 14 Exercise Ecosystems and Energy Flow Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 14 Ecosystems and Energy Flow Textbook Exercise Questions and Answers.

Ecosystems and Energy Flow Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 14 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 14 Exercise Solutions

1. Multiple choice questions

Question 1.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers

Question 2.
The second trophic level in a lake is ……………………
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton

Question 3.
Secondary consumers are …………………….
(a) Herbivores
(b) Producers
(c) Carnivores
(d) Autotrophs
Answer:
(c) Carnivores

Question 4.
What is the % of photosynthetically active radiation in the incident solar radiation?
(a) 100%
(b) 50%
(c) 1-5%
(d) 2-10%
Answer:
(b) 50%

Question 5.
Give the term used to express a community in its final stage of succession?
(a) End community
(b) Final community
(c) Climax community
(d) Dark community
Answer:
(c) Climax community

Question 6.
After landslide which of the following type of succession occurs?
(a) Primary
(b) Secondary
(c) Tertiary
(d) Climax
Answer:
(a) Primary

Question 7.
Which of the following is most often a limiting factor of the primary productivity in any ecosystem?
(a) Carbon
(b) Nitrogen
(c) Phosphorus
(d) Sulphur
Answer:
(c) Phosphorus

2. Very short answer question.

Question 1.
Give an example of ecosystem which shows inverted pyramid of numbers.
Answer:
Number of insects dependent on a single tree, is an example of ecosystem having inverted pyramid of numbers.

Question 2.
Give an example of ecosystem which shows inverted pyramid of biomass.
Answer:
Oceanic ecosystem has inverted pyramid of biomass.

Question 3.
Which mineral acts as limiting factor for productivity in an aquatic ecosystem?
Answer:
Phosphorus acts as limiting factor for productivity in an aquatic ecosystem.

Question 4.
Name the reservoir and sink of carbon in carbon cycle.
Answer:
Atmosphere is the reservoir of carbon cycle, while fossil fuels embedded in ocean and oceanic waters are the sink of carbon in carbon cycle.

3. Short answer questions.

Question 1.
Upright and inverted pyramid of biomass.
Answer:

Upright pyramid	Inverted pyramid
1. In upright pyramid, the number and biomass of the organisms which are at first trophic level of producers is high.	1. In inverted pyramid, the number and biomass of organisms at first trophic levels of producers is lowest.
2. The biomass goes on decreasing at each trophic level.	2. The biomass foes on increasing at each trophic level.
3. The base of the pyramid is always in large number of producers.	3. The base of pyramid is always in small numbers of producers.
4. Pyramid is always upright.	4. Pyramid is always inverted.

Question 2.
Food chain and Food web.
Answer:

Food chain	Food web
1. Food chain is the linear sequence of organisms for feeding purpose.	1. Food web is interconnections between many small food chains.
2. In food chain the flow of energy is through a single straight pathway from the lower trophic level to the higher trophic level.	2. In food web, the energy flow is interconnected through numerous food chains in the ecosystem.
3. In a food chain, members present at higher trophic level feeds on only single type of organisms.	3. In a food web, one organism can feed on multiple types of organisms.
4. Energy flow can be easily calculated in food chain.	4. Energy flow is difficult to calculate in a food web.
5. In food chain there is increased instability due to increasing number of separate and confined food chains.	5. In food web there is increased stability due to the presence of the complex food chains.
6. The whole food chain gets affected even if one group of an organism is disturbed.	6. The food web does not get disturbed by the removal of one group of organisms.
7. Member of higher trophic level depends or feed upon the single type of organisms of the lower trophic level.	7. The members of higher trophic level depend or feed upon many different types of the organism of the lower trophic level.
8. Food chain consists of only 4-6 trophic levels of different species.	8. Food web contains numerous trophic levels and also of different populations of species.
9. Competition is seen in members of same trophic level.	9. Competition is seen in members of same as well as different trophic levels.
10. Food chains are of two types: 1. Grazing food chain 2. Detritus food chain.	10. In food web there are no types.

4. Long answer questions

Question 1.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
1. Ecological Pyramids : Ecological Pyramids are the representation of relationships between different components of ecosystem at successive trophic levels.

2. Pyramid of numbers:

Pyramid of numbers is the diagrammatic representation which shows the relationship between producers, herbivores and carnivores at successive trophic levels in terms of their numbers.
As we go up the trophic levels, the interdependent organisms keep on reducing in their numbers.
For example, the number of grasses are more than the number of herbivores which eat them. The number of herbivores such as rabbits would be lesser than grass but greater than the carnivores that are dependent upon the population of rabbits.
Thus, the producers would be more than primary consumers and primary consumers would be more than secondary consumers. The top level consumers would be least in their numbers. This pyramid shows upright nature.

3. Pyramid of biomass:
(1) Pyramid of biomass are constructed by taking into consideration the different biomass in every successive trophic level.
(2) Pyramid of biomass in seas in inverted as the biomass of fishes is more than the biomass of phytoplankton.

Question 2.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
(1) Primary Productivity : The rate of generation of biomass in an ecosystem which is expressed in units of mass per unit surface (or volume) per unit time, for instance grams per square metre per day (g/m²/day) is called primary productivity.

(2) Primary productivity is described as gross primary productivity (GPP) and net primary productivity (NPP).

(3) The rate of production of organic matter during photosynthesis is called gross primary productivity of an ecosystem. Of this the amount of energy lost through respiration of plants is called respiratory losses.

(4) Gross primary productivity minus respiratory losses gives the net primary productivity (NPP).

(5) Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores, carnivores and decomposers).

(6) Factors affecting primary productivity: Gross primary productivity (GPP) depends on the following factors:

Plant species inhabiting a particular area.
Variety of environmental factors such as temperature, sunlight, salinity, oxygen and carbon dioxide content, etc.
Availability of nutrients and
Photosynthetic capacity of plants.

Question 3.
Define decomposition and describe the processes and products of decomposition.
Answer:

Decomposition is the process carried out by the decomposer organisms.
Most of the bacteria, actinomycetes and fungi are decomposers. They convert the dead and decaying organic matter into simpler compounds. These simpler inorganic substances return back to the environment.
Decomposition takes place through detritus food chain. It starts from the dead organic matter. Detritus eating organisms called detritivores like earthworm, etc. breakdown the detritus into smaller fragments. Therefore, this first step of decomposition is called fragmentation.
Water soluble inorganic nutrients seep into the soil after fragmentation. These nutrients get precipitated as salts. Therefore, this second step of decomposition is called leaching.
The third step of decomposition is called catabolism. In this step, fungal and bacterial enzymes degrade the detritus into simple inorganic substances.
The partially decomposed organic matter is called humus which is formed by the process of humification. Humus is a dark coloured amorphous substance which is the reservoir of nutrients.
Humus too undergoes decomposition by bacterial action at a very slow rate and ultimately releases inorganic matter. This process is therefore called mineralization.
Decomposition requires oxygen in greater amount. The rate of decomposition is dependent upon the temperature and the humidity of the environment.

Question 4.
Write important features of a sedimentary cycle in an ecosystem.
Answer:

Reservoir of sedimentary cycles is earth’s crust.
The nutrients such as phosphorus which show sedimentary cycle, moves through hydrosphere, lithosphere and biosphere.
There is no respiratory release of nutrients into the atmosphere which show sedimentary cycle.
Natural reservoir of such nutrients are usually in the form of rocks. The rocks upon weathering release such nutrients into circulation.
Sedimentary cycles are very slow in their reactions.

Question 5.
Describe carbon cycle and add a note on the impact of human activities on carbon cycle.
Answer:
I. Carbon cycle:
(1) The entire carbon cycle has following basic processes viz. Photosynthesis, Respiration, Decomposition, Sedimentation and Combustion.

(2) Carbon is an important element as it forms 49% of the dry weight of all organisms. 71% of global carbon is present in the oceans. Therefore, ocean is the major reservoir of carbon. Carbon is also present in all fossil fuels. This is long term storage places or sinks for carbon which is in the form of coal, natural gas, etc.

(3) Respiration and photosynthesis are the two events that keep the carbon in cyclic circulation. During respiration, oxygen is used for combustion of carbohydrates as a result of which carbon dioxide and water are formed with the release of energy. The process of photosynthesis utilizes carbon dioxide and water vapour liberating oxygen and producing carbohydrates at the same time.

(4) Solar energy is stored in the carbon-carbon bonds of carbohydrates during photosynthesis whereas respiration releases the same stored energy.

(5) The main reservoirs for carbon dioxide are in the oceans and in rocks. Carbon dioxide is highly soluble in water and forms mild carbonic acid upon dissolving. This dissolved carbon dioxide precipitate as a solid rock or limestone which is calcium carbonate. This reaction in the seas is aided by corals and algae which in turn builds the coral reefs made up of limestone.

(6) Carbon moves through food chains. Autotrophic green plants on land and in water take up carbon dioxide and manufacture carbohydrates by the process of photosynthesis. The carbon stored in plants has three different fates, viz. liberation into atmosphere, consumption by animals upon feeding, storage in the plant till the plant dies.

(7) Animals get their carbon requirement through their food. When autotrophs are consumed, the heterotrophs obtain carbon. Carbon in animals also has three fates, viz. release back into the atmosphere in the process of respiration, release of stored carbon from the body by the action of decomposers or conversion into fossil fuels if buried intact.

(8) Fossil fuels such as coal, oil, natural gas, etc. can be mined and burned for energy purposes. This burning releases carbon dioxide back into the atmosphere.

(9) Carbon from limestone can also be released if pushed to the surfaces and slowly weathered away. Subducting and volcanic eruptions can also release the stored carbon from sediments.

II. Impact of human activities on carbon cycle:
(1) Excessive burning of fossils fuels for power plants, industrial processes and vehicular traffic, adds excessive carbon dioxide into atmosphere. When fossil fuels burn to run factories, power plants, motor vehicles, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

(2) Each year, 5.5 billion tonnes of carbon is released through combustion of fossil fuels. Of this massive amount, 3.3 billion tonnes stays in the atmosphere.

(3) Rapid deforestation also increases carbon dioxide. Since plants absorb carbon dioxide for their photosynthesis, they always reduce the concentration of CO₂. But deforestation upsets this balance.

(4) Massive burning of fossil fuel for energy and transport, have significantly increased the rate of release of carbon dioxide into the atmosphere which is causing global warming and resultant climate change.

12th Std Biology Questions And Answers:

12th Biology Chapter 10 Exercise Human Health and Diseases Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 10 Human Health and Diseases Textbook Exercise Questions and Answers.

Human Health and Diseases Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 10 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Which of the following is NOT caused by unsterilized needles?
(a) Elephantiasis
(b) AIDS
(e) Malaria
(d) Hepatitis
Answer:
(a) Elephantiasis

Question 2.
Opium derivative is …………………
(a) Codeine
(b) Caffeine
(c) Heroin
(d) Psilocybin
Answer:
(c) Heroin

Question 3.
The stimulant present in tea is …………………
(a) tannin
(b) cocaine
(C) caffeine
(d) crack
Answer:
(c) caffeine

Question 4.
WhIch of the following Is caused by smoking?
(a) Liver cirrhosis
(b) Pulmonary tuberculosis
(c) Emphysema
(d) Malaria
Answer:
(c) Emphysema

Question 5.
An antibody is …………………
(a) molecuic that binds specifically an antigen
(b) WBC which invades bacteria
(c) secretion of mammalian RBC
(d) cellular component of blood
Answer:
(a) molecule that binds specifically an antigen

Question 6.
The antiviral proteins released by a virus-infected cell are called …………………
(a) histamines
(b) interferons
(c) pyrogens
(d) allergens
Answer:
(b) interferons

Question 7.
Both B-cells and T-cells are derived from …………………
(a) lymph nodes
(b) thymus glands
(c) liver
(d) stem cells in bone marrow
Answer:
(b) thymus glands

Question 8.
Which of the following diseases can be contracted by droplet infection?
(a) Malaria
(b) Chicken pox
(c) Pneumonia
(d) Rabies
Answer:
(c) Pneumonia

Question 9.
Confirmatory test used for detecting HIV infection is …………………
(a) ELISA
(b) Western blot
(c) Widal test
(d) Eastern blot
Answer:
(b) Western blot

Question 10.
Elephantiasis is caused by …………………
(a) W. barterofti
(b) P. vivax
(c) Bedbug
(d) Elephant
Answer:
(a) W. bancrofti

Question 11.
Innate immunity is provided by …………………
(a) phagocytes
(b) antibody
(c) T-lymphocytes
(d) B-lymphocytes
Answer:
(c) T-lymphocytes

2. Short Answer Questions

Question 1.
What is the source of cocaine?
Answer:
Source of cocaine is coca plant – Erythroxylum coca.

Question 2.
Name one disease caused by smoking.
Answer:
Emphysema. (Damaged and enlarged lungs causing breathlessness)

Question 3.
Which cells stimulate B-cells to form antibodies ?
Answer:
Helper T-cells stimulate B-cells to form antibodies.

Question 4.
What does the abbreviation AIDS stand for?
Answer:
AIDS stands for Acquired Immuno Deficiency Syndrome.

Question 5.
Name the causative agent of typhoid fever.
Answer:
Salmonella typhi

Question 6.
What is Rh factor?
Answer:
Antigen ‘D’ present on the surface of RBCs is known as Rh factor.

Question 7
What is schizont?
Answer:
Schizont is a ring-like form produced from merozoites inside the erythrocytes of human beings, infected by Plasmodium, which again forms new merozoites.

Question 8.
Name the addicting component found in tobacco.
Answer:
Nicotine

Question 9.
Name the pathogen causing Malaria.
Answer:
Plasmodium vivax

Question 10.
Name the vector of Filariasis.
Answer:
Female Culex mosquito

Question 11.
Name of the causative agent of ringworm.
Answer:
Trichophyton

Question 12.
Health
Answer:
Health is defined as the state of complete physical, mental and social well¬being and not merely the absence of disease or infirmity.

3. Short Answer Questions

Question 1.
What are acquired diseases?
Answer:
Diseases which are developed after the birth of an individual are called acquired diseases. These are of two types, viz. (a) Communicable or infectious diseases and (b) Non- communicable or Non-infectious diseases. Communicable or infectious diseases are transmitted from infected person to another healthy person either directly or indirectly. They are caused due to pathogens like viruses, bacteria, fungi, helminth worms, etc. Non-communicable or Non-infectious diseases cannot be transmitted from infected person to another healthy one either directly or indirectly.

Question 2.
Antigen and antibody.
Answer:

Antigen	Antibody
1. Antigens are foreign proteins which are capable of producing infection.	1. Antibodies are immunoglobulins produced by the body to act against the antigens.
2. The structure of antigens is variable dependent upon the type of pathogen.	2. The structure of antibody is Y-shaped.
3. The antigen is the ‘non-self’ molecule.	3. The antibody is ‘self’ molecule.
4. The antigens have epitope sites which bind with the antibody molecule.	4. The antibodies have paratope sites which bind with the antigen molecule.

Question 3.
Name the infective stage of Plasmodium. Give Symptoms of malaria
Answer:
Sporozoite
I. Symptoms of malaria:

Fever accompanied by shivering.
Joint pain or arthralgia.
Vomiting.
Anaemia caused due to rupture of RBCs or haemolysis.
Haemoglobinuria.
Retinal damage.
Convulsions.
Cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting for four to six hours. This is called a classic symptom of malaria.
Splenomegaly or enlarged spleen, severe headache, cerebral ischemia, hepatomegaly, i. e. enlarged liver, hypoglycaemia and haemoglobinuria with renal failure may occur in severe infections.

II. Spread / Transmission of malaria:

Malaria parasite is transmitted through the female Anopheles mosquito and hence it is known as mosquito-borne disease. Mosquito acts as a vector.
There are four species of Plasmodium, viz., P. vivax, P. falciparum, P. ovale and P. malariae which transmit malaria.

Question 4.
Explain the mode of infection and cause of elephantiasis.
Answer:
Mode of infection, i.e. transmission:

The parasite Wuchereria bancrofti is transmitted from a patient to other normal human being by female Culex mosquito.
The filarial larvae leave mosquito body and arrive on the human skin where they penetrate the skin and enter inside.
They undergo two moultings to become adults. Later they settle in the lymphatic system. They incubate for about 8-16 months.
When they settle in lymphatic system, this infection is called lymphatic filariasis.
The worms start infecting lymphatic circulation resulting into enlargement of lymph vessels and lymph nodes. The extremities like legs or limbs become swollen which resembles elephant legs. Therefore it is called elephantiasis.
This condition is lymphoedema, i.e. accumulation of lymph fluid in tissue causing swelling.

Question 5.
Why is smoking a bad habit?
Answer:

Smoking involves inhaling the cigarette smoke which contains nicotine and other toxic substances like N-nitrosodimethlene. There is some amount of carbon monoxide.
All these substances affect the normal respiratory health.
Smoking invites problems like asthma, hypertension, heart disease, stroke, lung damage.
The worst impact is that these substances are carcinogenic and hence can cause cancer of larynx, trachea, lung, etc.
Smoking not only affects the smokers but also has bad effect on others due to passive smokers.
In women, smoking is still hazardous as their ovaries can undergo mutations due to mutagenic chemicals found in smoke.
Therefore, smoking is a very bad habit.

Question 6.
What do the abbreviations AMIS and CMIS denote?
Answer:
AMIS is Antibody-mediated immune system or humoral immunity and CMIS is cell- mediated immune system.

Question 7.
What is a carcinogen? Name one chemical carcinogen with its target tissue.
Answer:

Carcinogen is the substance or agent that causes cancer.
Urinary bladder cancer caused by 2-naphthylamine and 4-aminobiphenyl.

Question 8.
Active immunity and passive immunity.
Answer:

Active immunity	Passive immunity
1. Active immunity is produced in response to entry of pathogens and their antigenic stimuli.	1. Passive immunity is produced due to antibodies that are transferred to the body.
2. Active immunity is the long lasting immunity.	2. Passive immunity is short-lived immunity.
3. In active immunity, the body produces its own antibodies.	3. In passive immunity, antibodies are given to the body from outside.
4. Natural acquired active immunity is obtained due to infections by pathogens.	4. Natural acquired passive immunity is obtained through antibodies of mother transmitted- to baby by placenta or colostrum.
5. Artificial acquired active immunity is obtained through vaccinations. These vaccines contain dead or live but attenuated pathogens.	5. Artificial acquired passive immunity is also obtained through vaccinations, but here the vaccines contain the readymade antibodies which are prepared with the help of other animals such as horses.

4. Short Answer Questions

Question 1.
B-cells and T-cells.
Answer:

B-cells	T-cells.
1. B-cells are type of lymphocyte whose origin is in bone marrow but maturation is in blood.	1. T-cells are type of lymphocytes which originate in bone marrow but maturation occurs in thymus.
2. B-cells Eire type of lymphocytes which are involved in humoral mediated immunity.	2. T-cells are type of lymphocytes which are involved in cell-mediated immunity.
3. 20% of lymphocytes present in the blood are B-cells.	3. 80% of lymphocytes present in the blood are T-cells.
4. Two types of B-cells are Memory cells and Plasma cells.	4. T-cells are of following subtypes : Cytotoxic T-cells, helper T-cells, suppressor T-cells.
5. They are involved in antibody mediated immunity. (AMI)	5. They are involved in cell-mediated immunity (CMI).
6. B-cells produced antibodies with which they fight against pathogens.	6. T-cells do not produce antibodies.
7. B-cells have membrane bound immunoglobulins located on the surface.	7. There is a presence of T cell receptors on the T-cell surface.

Question 2.
What are the symptoms of malaria? How does malaria spread?
Answer:
Symptoms of malaria:

Fever accompanied by shivering.
Joint pain or arthralgia.
Vomiting.
Anaemia caused due to rupture of RBCs or haemolysis.
Haemoglobinuria.

Question 3.
AIDS.
Answer:
(1) AIDS or the acquired immuno deficiency syndrome, is fatal viral disease caused by a retrovirus (ss RNA) known as the human immuno deficiency virus (HIV) which weakens the body’s immune system. It is called a modern pandemic.

(2) The HIV attacks the immune system which in turn causes many opportunistic infections, neurological disorders and unusual malignancies ultimately leading to death.

(3) AIDS was first noticed in USA in 1981 whereas in India, first confirmed case of AIDS was in April 1986 from Tamil Nadu.

(4) HIV is transmitted through body fluids such as saliva, tears, nervous system tissue, spinal fluid, blood, semen, vaginal fluid and breast milk. However, only blood, semen, vaginal secretions and breast milk generally transmit infection to others.

(5) The transmission of HIV occurs by sexual contact, through blood and blood products and by contaminated syringes, needles, etc. There is also transplacental transmission or through breast milk at the time of nursing.

(6) Accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs are some of the rare occasions of transmission of HIV.

(7) HIV infection is not spread by casual contact such as hugging, bite of mosquitoes or using other objects touched by a patient.

(8) Acute HIV infection progresses over time to asymptomatic HIV infection and then to early symptomatic HIV infection. Later, it progresses to full blown AIDS when patient shows advanced HIV infection with CD4 T-cell count below 200 cells/mm.

Question 4.
Give the symptoms of cancer.
Answer:
Symptoms of cancer:

Presence of lump or tumour.
White patches in the mouth.
Change in a wart or mole on the skin.
Swollen or enlarged lymph nodes.
Vertigo, headaches or seizures if cancer affect the brain.
Coughing and shortness of breath if lungs are affected due to cancer.

Question 5.
Antigens on blood cells.
Answer:

There are about 30 known antigens on the surface of human red blood cells. They decide the type of blood group such as ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay.
The different blood groups are determined genetically due to presence of a particular antigen.
Landsteiner found two antigens or agglutinogens on the surface of human red blood cells which are named as antigen A and antigen B.
There is another antigen called Antigen D which decides the Rh status of the blood. If Antigen D is present, the person is said to be RH positive and when it is lacking, the person is Rh negative.
These antigens are responsible for types of blood group and the specific transfusions.
Antigens present on the RBCs and antibodies present in the serum can cause agglutination reactions if they are non-compatible. Therefore, at the time of transfusion blood groups are checked properly.

Question 6.
Antigen-antibody complex:
Img 1
Answer:

Between antigen and antibody there is specificity.
Each antibody is specific for a particular antigen.
On the antigens there are combining sites which are called antigenic determinants or epitopes.
Epitopes react with the corresponding antigen binding sites of antibodies which are called paratopes.
The antigen binding sites are located on the variable regions of the antibody. Variable regions have small variations which make each antibody highly specific for a particular antigen.
Owing to variable region the antibody can recognize the specific antigen.
Antibody thus binds to specific antigen in a lock and key manner, forming an antigen- antibody complex.

Question 7.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.
Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.
One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.
Proper diet and exercise should be followed to improve one’s own immunity.
Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.
Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.
One should have responsible sexual behaviour to avoid sexually transmitted diseases.
Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.
Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Question 8.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis:
Answer:
Amoebiasis is usually transmitted by the following ways:

The faecal-oral route.
Through contact with dirty hands or objects.
By anal-oral contact.
Through contaminated food and water.

(b) Malaria:
Answer:
Symptoms of malaria:

Fever accompanied by shivering.
Joint pain or arthralgia.
Vomiting.
Anaemia caused due to rupture of RBCs or haemolysis.
Haemoglobinuria.

Unsafe and unhygienic food and drinks contaminated with the eggs of Ascaris are the main mode of transmission.
Eggs hatch inside the intestine of the new host.
The larvae pass through various organs and settle as adults in the digestive system.

(d) Pneumonia:
Answer:

Pneumonia usually spreads by direct person to person contact.
It is also spread via droplet infection, i.e. droplets released by infected person.
Using clothes and utensils of the patient.

Question 9.
What measures would you take to prevent water-borne diseases?
Answer:

To prevent water-borne diseases, use of safe, clean and potable water is a must. Water should be filtered, then boiled and stored in covered container. If possible water purifier systems should be installed at home.
One should preferably use bottled water or carry our own water container while travelling.
Cleaning of water containers and maintaining personal hygiene near water storage is a must.
Megacities offer chlorinated and purified water for citizens. But villages and smaller rural set ups use river water which may be highly contaminated with pathogens. Such water should be purified before consumption to prevent water-borne diseases.

Question 10.
Typhoid.
Answer:
Typhoid is an infective disease caused by Gram-ve bacterium, Salmonella typhi.
(1) It is food and water-borne infection. In the intestinal lumen of infected person this bacteria is found.

(2) The bacterium has “O” – antigen, which is a lipopolysaccharide (LPS), present on surface coat and its flagella has “H” – antigen. Thus it becomes pathogenic.

(3) Signs and Symptoms of typhoid are as follows:
Prolonged and high fever with nausea, fatigue, headache.
Abdominal pain, constipation or diarrhoea. In severe cases rose-coloured rash is seen on skin. Tongue shows white coating and there is cough. Anorexia or loss of appetite is seen. In chronic cases there is breathlessness, irregular heartbeats and haemorrhage.

(4) Poor hygiene habits and poor sanitation and insects like houseflies and cockroaches spread typhoid.

(5) Typhoid is diagnosed by Widal test.

(6) Antibiotics like Chloromycetin can cure typhoid. Preventive vaccines such as oral Ty21a vaccine and injectable typhim vi and typherix against typhoid are also available. Chronic cases need surgical removal of gall bladder.

5. Match the following.

Column I	Column II
(a) AIDS	(i) Antibody production
(b) Lysozyme	(ii) Activation of B-cells
(c) B-cells	(iii) Immunoglobulin
(d) T-helper cells	(iv) Tears
(e) Antibody	(v) Immuno deficiency

Answer:

Column I	Column II
(a) AIDS	(v) Immuno deficiency
(b) Lysozyme	(iv) Tears
(c) B-cells	(i) Antibody production
(d) T-helper cells	(ii) Activation of B-cells
(e) Antibody	(iii) Immunoglobulin

6. Long Answer Questions

Question 1.
Describe the structure of antibody.
Answer:
Img 2

Antibodies are highly specific to specific antigens. They are glycoprotein called immunoglobulins (Igs.).
They are produced by plasma cells. Plasma cells are in turn formed by B-lymphocytes.
About 2000 molecules of antibodies are formed per second by the plasma cells.
Antibody is a ‘Y’-shaped molecule. It has four polypeptide chains, two heavy or H-chains and two light or L-chains.
Disulfide bonds (-s-s-) hold the polypeptide chains together to form a ‘Y’-shaped structure.
The region holding arms and stem of antibody is termed as hinge. Each chain of the antibody has two distinct regions, the variable region and the constant region.
Variable regions have a paratope which is an antigen-binding site. This part of antibody recognizes and binds to the specific antigen forming an antigen-antibody complex.
Antibodies are called bivalent as they carry two antigen binding sites.

Question 2.
Vaccination.
Answer:

Vaccines are prepared from inactivated pathogen, in the form of protein or sugar from pathogen or dead form of pathogen or toxoid from pathogens or attenuated pathogen.
These when they are administered to a person to protect against a particular pathogen, it is called vaccination.
Vaccination ’teaches’ the immune system to recognize and eliminate pathogenic organism. Because, already in the body the vaccine is injected and body has made antibodies in response to it. Thus, body is prepared before the attack, if at all it is exposed to pathogen.
Thus, it is an important form of primary prevention, which reduces the chances of illness by protecting people. It works by exposing the pathogen in a safe form.
Vaccinations control spread of diseases like measles, polio, tetanus and whooping cough that once threatened many lives.
Vaccination controls the epidemic outbreak of diseases, if all the people Eire pre-vaccinated.
Some hazardous diseases like small, pox and polio have been completely eradicated by the vaccination.

Question 3.
What is cancer? Differentiate between benign tumour and malignant tumour. The main five types of cancer
Answer:
I. Cancer : Cancer is a disease caused by uncontrolled cell division due to disturbed cell cycle.

II. Difference between benign tumour and malignant tumour:

Benign tumour	malignant tumour
1. Benign tumour is localized and it does not spread to neighbouring areas.	1. Malignant tumour starts as local but spreads rapidly to neighbouring areas.
2. Benign tumour is enclosed in connective tissue sheath.	2. Malignant tumour is not enclosed in connective tissue sheath.
3. Benign tumour compresses the surrounding normal tissue.	3. Malignant tumour invades and destroys the surrounding tissue.
4. Benign tumours can be removed surgically.	4. Malignant tumours need further treatment after removal.
5. Except for brain tumour, benign tumours are usually not fatal.	5. Malignant tumours are fatal.
6. Benign tumours do not show metastasis.	6. Malignant tumours show metastasis.
7. Benign tumours are well differentiated.	7. Malignant tumours are poorly differentiated.
8. Benign tumours show slow and progressive growth.	8. Malignant tumours show rapid and erratic growth.

III. The main five types of cancer:

Types of Cancer : According to the tissue affected, the cancers are classified into five main types. These are as follows:

Carcinoma : Cancer of epithelial tissue covering or lining the body organs is known as carcinoma. E.g. breast cancer, lung cancer, cancer of stomach, skin cancer, etc.
Sarcoma : Cancer of connective tissue is called sarcoma. Following are the types of sarcoma osteosarcoma (bone cancer), myosarcoma (muscle cancer),
chondrosarcoma (cancer of cartilage) and liposarcoma (cancer of adipose tissue).
Lymphoma : Cancer of lymphatic tissue is called lymphoma. Lymphatic nodes, spleen and tissues of immune system are affected due to lymphoma.
Leukaemia : Leukaemia is blood cancer. In this condition, excessive formation of leucocytes take place in the bone marrow. There are millions of abnormal immature leucocytes which cannot fight infections. Monocytic leukaemia, lymphoblastic leukaemia, etc. are the types of leukaemia.
Adenocarcinoma : Cancer of glandular tissues such as thyroid, pituitary, adrenal, etc. is called adenocarcinoma.

Question 4.
Describe the different type of immunity.
Answer: There are two basic types of immunity, viz. innate immunity and acquired immunity.
(A) Innate immunity:

Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.
Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.
Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity:

The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.
Acquired immunity takes long time for its activation.
This type of immunity is seen only in vertebrates.
Due to acquired immunity, the body is able to defend against any invading foreign agent.

Question 5.
Describe the ill-effects of alcoholism on health.
Answer:

Alcohol in any form is toxic for the body. Hence as soon as alcohol is consumed, the liver tries to detoxify it.
In low doses it acts as a stimulant but in high dose, it acts on central nervous system, especially the cerebrum and cerebellum. Still higher dose can induce a comatose condition.
Alcohol affect the gastrointestinal tract by causing inflammation and damage to gastric 4 mucosa. Ulceration and painful condition arises in alcoholics.
Excessive doses of alcohol induce vomiting.
The worst effect of alcohol is on liver causing diseases like cirrhosis.
Alcohol induces hypertension and cardiac problems.
Apart from physical effect, it causes deterioration of mental health and emotional well-being.
Alcoholic person cannot think due to numbness in his/her cerebrum.
The social health is greatly affected as the alcoholic can cause problems to his family, friends and society in general.

Question 6.
In your view, what motivates the youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
I. Taking drugs or alcohol:

Youngsters are at the vulnerable age, where they lack the planning about their future.
If they fall into bad company or are facing parental neglect, they get hooked on to alcohol or drugs.
Some common causes for addiction among youngsters are insufficient parental supervision and monitoring or excessive pressure and expectations from them. Lack of communication between an adolescent and parents.
Poorly defined rules for the family. Continuous family conflicts.
Favourable parental attitudes towards alcohol and drug uses. Many a times, at home children are exposed to such habits.
Inability to cope up with present and hence switching to the addictions. Risk taking behaviour which is common among youngsters.

II. Methods/measures to avoid drug abuse:

There should be complete acceptance for the child, because the adolescent phase is the most crucial phase when the children should be treated with love, care and respect.
Many physical, hormonal and psychological transformations are taking place in this phase. Therefore child suffers from stressful situation.
Wrong company and bad influence of peer group can trap the child in bad addictive habits. Thus, family should be supportive and communicative to help such children.
The sexual thoughts should be sublimed by channelizing energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.
Ill-effects of drugs or alcohol should be told to youngsters.
Education and counselling can control the children from getting hooked on to the addictions.

Question 7.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Friends can influence one to take alcohol and drugs, if a boy or girl is timid and non-communicative with his or her parents and teachers. It also depends on the personality of the indtvidual. In the adolescent age, many fall in trap due to such peer pressure. The confusion in the mind and role of hormones playing on the psyche and thought process makes one unable to understand the hazards of such habits. Also there is curiosity to do these experimentations due to bad influence of media.

If there is complete trust and friendship with sensible parents, then such influence does not work. One should protect himself or herself by a strong denial. Communicating such incidents to an elder in whom a boy or girl can confide, is very important. One should tell his or her friends about the ill-effects of alcohol and drugs. He should be made aware of these aspects that he or she has learnt in this lesson.

12th Std Biology Questions And Answers:

12th Biology Chapter 12 Exercise Biotechnology Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 12 Biotechnology Textbook Exercise Questions and Answers.

Biotechnology Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 12 Exercise Solutions

1. Multiple choice questions

Question 1.
MU The bacterium which causes a plant disease called crown gall is ………………..
(a) Helicobacter pylori
(b) Agrobacterium tumifaciens
(c) Thermophilus aquaticus
(d) Bacillus thuringienesis
Answer:
(b) Agrobacterium tumtfaciens

Question 2.
The enzyme nuclease hydrolyses ……………….. of polynucleotide chain of DNA.
(a) hydrogen bonds
(b) phosphodiester bonds
(c) glycosidic bonds
(d) peptide bonds
Answer:
(b) phosphodiester bonds

Question 3.
In vitro amplification of DNA or RNA segment is known as ………………..
(a) chromatography
(b) southern blotting
(c) polymerase chain reaction
(d) gel electrophoresis
Answer:
(c) polymerase chain reaction

Question 4.
Which of the following is the correct recognition sequence of restriction enzyme hind III.
(a) 5′ —A-A-G-C-T-T— 3′
3′ —T-T-C-G-A-A—5′
(b) 5′ — G-A-A-T-T-C—3′
3′ — C-T-T-A-A-G—5′
(c) 5′ — C-G-A-T-T-C—3′
3′ — G-C-T-A-A-G—5′
(d) 5′ — G-G-C-C—3′
3′ — C-C-G-G—5′
Answer:
(a) 5’ —A-A-G-C-T-T—3’
3’ —T-T-C-G-A-A—5’

Question 5.
Recombinant protein ……………….. is used to dissolve blood clots present in the body.
(a) insulin
(b) tissue plasminogen activator
(c) relaxin
(d) erythropoietin
Answer:
(b) tissue plasminogen activator

Question 6.
Recognition sequence of restriction enzymes are generally ……………….. nucleotide long.
(a) 2 to 4
(b) 4 to 8
(c) 8 to 10
(d) 14 to 18
Answer:
(b) 4 to 8

2. Very short answer questions

Question 1.
Name the vector which is used in production of human insulin through recombinant DNA technology.
Answer:
PBR 322

Question 2.
Which cells from Langerhans of pancreas do produce a peptide hormone insulin?
Answer:
cells of islets of Langerhans of a peptide hormone insulin.

Question 3.
Give the role of Ca⁺⁺ ions in the transfer of recombinant vector into bacterial host cell.
Answer:
Ca⁺⁺ ions promotes binding of plasmid DNA to lipo polysaccharides on bacterial cell surface. Then plasmid can enter the cell on heat shock.

Question 4.
Expand the following acronyms which are used in the held of biotechnology:

YAC
RE
dNTP
PCR
GMO
MAC
CCMB.

Answer:

YAC : Yeast Artificial chromosome
RE : Restriction Endonuclease
dNTP : Deoxyribonucleoside triphosphates
PCR : Polymerase Chain Reaction
GMO : Genetically Modified Organisms
MAC : Mammalian Artificial Chromosome
CCMB : Centre for Cellular and Molecular Biology

Question 5.
Fill in the blanks and complete the chart.

GMO	Purpose
(i) Bt cotton	———–
(ii) ———-	Delay the softening of tomato during ripening
(iii) Golden rice	———–
(iv) Holstein cow	———–

Answer:

GMO	Purpose
(i) Bt cotton	Insect resistance
(ii) Flavr savr Tomato	Delay the softening of tomato during ripening
(iii) Golden rice	Rich in vitamin A
(iv) Holstein cow	High milk productivity

3. Short answer type questions.

Question 1.
Explain the properties of a good or ideal cloning vector for r-DNA technology.
Answer:
Desired characteristics of ideal cloning vector are as follows:

Vector should be able to replicate independenly (through ori gene), so that as vector replicates, multiple copies of the DNA insert are also produced.
It should be able to easily transferred into host cells.
It should have suitable control elements like promoter, operator, ribosomal binding sites, etc.
It should have marker genes for antibiotic resistance and restriction enzyme recognition sites within them.

Question 2.
A PCR machine can rise temperature up to 100 °C but after that it is not able to lower the temperature below 70 °C automatically. Which step of PCR will be hampered first in this faulty machine? Explain why?
Answer:

If the faulty machine is not able to lower the temperature below 70 °C, then the primer annealing step will be hampered first.
Each primer has a specific annealing temperature, depending upon its A, T, G, C content.
For most of the primers annealing temperature is about 40-60 °C.
Hence, if temperature is more than primers annealing temperature, it will be able to pair with its complementary sequence in ssDNA.

Question 3.
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA then which problem will arise in the formation of r-DNA or chimeric DNA? Explain.
Answer:
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA, then it will result in fragments with different sticky ends which will not be complementary to each other.

Question 4.

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(a) Anemia
(2) a-antitrypsin	(b) Cystic fibrosis
(3) Relaxin	(c) Haemophilia A
(4) Eryhthropoietin	(d) Diabetes
(5) Factor VIII	(e) Emphysema
(6) DNA ase	(f) Parturition
	(g) Atherosclerosis

Answer:

Recombinent protein	Its use in or for
(1) Platelet derived growth factor	(g) Atherosclerosis
(2) a-antitrypsin	(e) Emphysema
(3) Relaxin	(f) Parturition
(4) Eryhthropoietin	(a) Anemia
(5) Factor VIII	(c) Haemophilia A
(6) DNA ase	(b) Cystic fibrosis

4. Long answer type questions.

Question 1.
(i) Define and explain the terms Bioethics.
Answer:

Bioethics is the study of moral vision, decision and policies of human behaviour in relation to biological phenomena or events.
Bioethics deals with wide range of reactions on new developments like cloning, transgenic, gene therapy, eugenics, r-DNA technology, in vitro fertilization, sperm bank, gene therapy, euthanasia, death, maintaining those who are in comatose state, prenatal genetic selection, etc.
Bioethics also includes the discussion on subjects like what should and should not be done in using recombinant DNA techniques.

Ethical aspects pertaining to the use of biotechnology are:

Use of animals cause great sufferings to them.
Violation of integration of species caused due to transgenosis.
Transfer of human genes into animals and vice versa.
Indiscriminate use of biotechnology pose risk to the environment, health and biodiversity.
The effects of GMO on non-target organisms, insect resistance crops, gene flow, the loss of diversity.
Modification process disrupting the natural process of biological entities.

(ii) Define and explain the term Biopiracy.
Answer:

Biopiracy is defined as ‘theft of various natural products and then selling them by getting patent without giving any benefits or compensation back to the host country’.
It is unauthorized misappropriation of any biological resource and traditional knowledge.
It is bio-patenting of bio-resource or traditional knowledge of another nation without proper permission of the concerned nation or unlawful exploitation and use of bioresources without giving compensation.

Following are the examples of biopiracy:
(a) Patenting of Neem (Azadirachta indica):

Pirating India’s traditional knowledge about the properties and uses of neem, the USDA and an American MNC W.R. Grace sought a patent from the European Patent Office (EPO) on the “method for controlling on plants by the aid of hydrophobic extracted neem oil,” in the early 90s.
The patenting of the fungicidal properties of Neem, was an example of biopiracy.

(b) Patenting of Basmati:

Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.
This is a case of biopiracy as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent.
Very broad claims about “Inventing” the said rice was the basis of patent application.
The UPSTO has rejected all the claims due to people movement against Rice Tec in March 2001.

A patent claim about the healing properties of Haldi was made by two American researchers of Indian origin of the University of Mississippi Medical Center, to the US Patent and Trademark Office.
They were granted a patent in March 1995.
This is an example of biopiracy because healing properties of Haldi is not a new discovery, but it is a traditional knowledge in ayurvedas for centuries.
The Council of Scientific and Industrial Research (CSIR) applied to the US Patent Office for a reexamination and they realized the mistake and cancelled the patent.

(iii) Define and explain the term Biopatent.
Answer:

Biopatent is a biological patent awarded for strains of microorganisms, cell lines, genetically modified strains, DNA sequences, biotechnological processes, product processes, product and product applications.
It allows the patent holder to exclude others from making, using, selling or importing protected invention for a limited period of time.
Duration of biopatentis five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.
Awarding biopatents provides encouragement to innovations and promote development of scientific culture in society. It also emphasizes the role of biology in shaping human society.
First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.
Patent jointly issued by Delta and Pineland company and US department of agriculture having title ‘control of plant gene expression’, is based on a gene that produces a protein toxic to plant and thus prevents seed germination.

This patent was not granted by Indian government. Such a patent is considered morally unacceptable and fundamentally unequitable. Such patents would pose a threat to global food security as financially powerful corporations would acquire monopoly over biotechnological process.

Question 2.
Explain the steps in process of r-DNA technology with suitable diagrams.
Answer:

The steps involved in gene cloning are as follows:
(1) Isolation of DNA (gene) from the donor organism:

To obtain the desired gene to be cloned, the cells of the donor organism are sheared with the blender and treated with suitable detergent. Genetic material is then isolated and purified.
Isolated purified DNA is then cleaved using restriction Endonucleases.
Restriction fragment containing desired gene is isolated and selected for cloning. This is now called foreign DNA or passanger DNA.
A desired gene can also be obtained directly from genomic library or c-DNA library.

(2) Insertion of desired foreign gene into a cloning vector (vehicle DNA):

The foreign DNA or passanger DNA is inserted into a cloning vector (vehicle DNA) like bacterial plasmids and the bacteriophages like lamda phage and M13. The most commonly used plasmid is pBR 322.
Plasmids are isolated from the bacteria and are cleaved by using same RE which is used in the isolation of the desired gene from the donor.
Enzyme DNA ligase is used to join foreign DNA and the plasmid DNA.
Plasmid DNA containing foreign DNA is called recombinant DNA (r-DNA) or chimeric DNA.

(3) Transfer of r-DNA into suitable competent host or cloning organism:

The r-DNA is introduced into a competent host cell, which is mostly a bacterium.
Host cell takes up naked r-DNA by process of ‘transformation’ and incorporates it into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.
The transfer of r-DNA into a bacterial cell is assisted by divalent Ca⁺⁺.
The cloning organisms are E.coli and Agrobacterium tumifaciens.
The competent host cells which have taken up r-DNA are called transformed cells.
By using techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.
In plant biotechnology the transformation is through Ti plasmids of A. tumifaciens.

(4) Selection of the transformed host cell:

For isolation of recombinant cell from non-recombinant cell, marker gene of plasmid vector is employed.
For example, pBR322 plasmid vector contains different marker genes like ampicillin resistant gene and tetracycline resistant gene. When pstl RE is used, it knocks out ampicillin resistant gene from the plasmid, so that the recombinant cells become sensitive to ampicillin.

(5) Multiplication of transformed host cell:

The transformed host cells are introduced into fresh culture media where they divide.
The recombinant DNA carried by them also multiplies.

(6) Expression of gene to obtain desired product. Then desired products like enzymes, antibiotiocs etc. separated and purified through down stream processing using bioreactors.

Question 3.
Explain the gene therapy. Give two types of it.
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.
Types of gene therapy:
(a) Germ line gene therapy:

In this germ cells are modified genetically to correct a genetic defect.
Normal gene is introduced into germ cells like sperms, eggs, early embryos.
It allows transmission of the modified genetic information to the next generation.
Although it is highly effective in treatment of the genetic disorders, its use is not preferred in human beings because of various technical and ethical reasons.

(b) Somatic cell gene therapy:

In this somatic cells are modified genetically to correct a genetic defect.
Healthy genes are introduced in somatic cells like bone marrow cells, hepatic cells, fibroblasts endothelium and pulmonary epithelial cells, central nervous system, endocrine cells and smooth muscle cells of blood vessel walls.
Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.
Somatic cell gene therapy is the only feasible option and the clinical trials have already employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 4.
How are the transgenic mice used in cancer research?
Answer:

Transgenic mice are used in various research areas of cancer research.
Transgenic mice containing a particular oncogene (cancer causing gene) develop specific cancer.
They are used to study the relationship between oncogenes and cancer development, cancer treatment and prevention of malignancy.
The transgenic mouse model for the investigation of the breast cancer was developed in the laboratory of Philip Leder in Harvard (USA).
Transgenic mice containing oncogenes myc and ras were analyzed to find out role of these genes in the development of breast cancer.

Question 5.
Give the steps in PCR or polymerase chain reaction with suitable diagrams.
Answer:

(1) The DNA segment and excess of two primer molecules, four types of dNTPs, the thermostable DNA polymerase are mixed together in ‘eppendorf tube’.

(2) One PCR cycle is of 3-4 minutes duration and it involves following steps:

Denaturation : The reaction mixture is heated at 90-98°C. Due to this hydrogen bonds in the DNA break and two strands of DNA separate. This is called denaturation.
Annealing of primer : When the reaction mixture is cooled to 40-60°C, the primer pairs with its complementary sequences in ssDNA. This is called annealing.
Extension of primer : In this step, the temperature is increased to 70-75°C. At this temperature thermostable Taq DNA polymerase adds nucleotides to 3’end of primer using single-stranded DNA as template. This is called primer extension. Duration of this step is about two minutes.

(3) In an automatic thermal cycler, the above three steps are automatically repeated 20-30 times.
(4) Thus, at the end of ‘n’ cycles 2n copies of DNA segments, get synthesized.

Question 6.
What is a vaccine? Give advantages of oral vaccines or edible vaccines.
Answer:

A vaccine is a biological preparation that provides active acquired immunity against a certain disease.
Vaccine is often made from a weakened or killed form of the microorganism, its toxins or one of its surface protein antigens.
Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
Immunogenic protein of certain pathogens are active when’administered orally.
When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.
Oral or edible vaccines have low cost, they are easy to administer and store.

Question 7.
Enlist different types of restriction enzymes commonly used in r-DNA technology? Write on their role.
Answer:

Different restriction enzymes commonly used in r-DNA technology are Alu I, Bam HI, Eco RI, Hind II, Hind III, Pst I, Sal I, Taq I, Mbo II, Hpa I, Bgl I, Not I, Kpn I, etc.
They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.
The sites recognized by them are called recognition sequences or recognition sites.
Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.
Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).
Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.
Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.
Type I restriction endonucleases fuction simultaneously as endonuclease and methylase e.g. EcoK.
Type II restriction endonucleases have separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the palindrome.
Type III restriction endonucleases cut DNA at specific non-palindromic sequences e.g. Hpal, MboII.
In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Question 8.
Enlist and write in brief about the different biological tools required in r-DNA technology.
Answer:
The biological tools used in r-DNA technology are various enzymes, cloning vectors and competent hosts.
(1) Enzymes:

Enzymes like lysozymes, nucleases (exonucleases and endonucleases), DNA ligase, reverse transcriptase, DNA polymerase, alkaline phosphatases, etc. are used in r-DNA technology.
The restriction endonucleases are used as biological or molecular scissors. They are able to cut a DNA molecule at a specific recognition site.

(2) Vectors:

Vectors are DNA molecules which carry foreign DNA segment and replicate inside the host cell.
Vectors may be plasmids, bacteriophages (M13, lambda virus), cosmid, phagemids, BAC (bacterial artificial chromosome), YAC (yeast artificial chromosome), transposons, baculoviruses and mammalian artificial chromosomes (MACs).
Most commonly used vectors are plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lamda phage, M13 phage).

(3) Competent host cells:

They are bacteria like Bacillus haemophilus, Helicobacter pyroliand E. coli.
Mostly E. coli is used for the transformation with recombinant DNA.

12th Std Biology Questions And Answers:

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

January 8, 2025January 7, 2025 by Prasanna

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

Molecular Basis of Inheritance Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 4 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 4 Exercise Solutions

1. Multiple Choice Questions

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

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

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

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

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

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

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

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

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

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

2. Very Short Answer Questions

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

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

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

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

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

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

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

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

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

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

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

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

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

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

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

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

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

4. Short Answer Questions

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

2. The main aims:

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

3. Significance:

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

Question 2.
Describe the structure of operon.
Answer:

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

Question 3.
In the figure below A, B and C are three types of

Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.

Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

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

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

Answer:

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

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:

DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

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

Point of Origin or Initiation point:

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

Unwinding of DNA molecule:

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

Replicating fork:

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

Synthesis of new strands:

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

Leading and Lagging strand:

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

Formation of two daughter DNA molecules:

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

Question 2.
Describe the process of transcription in protein synthesis.
Answer:

Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

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

(2) Elongation:

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

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

Question 3.
Describe the process of translation in protein synthesis.
Answer:

Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

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

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

ii. Peptide bond formation.

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

iii. Translocation.

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

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

Question 4.
Describe Lac ‘Operon’.
Answer:

Lac operon consists of the following components:
(1) Regulator gene:

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

(2) Promoter gene:

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

(3) Operator gene:

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

(4) Structural gene:

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

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

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

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

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

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

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

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

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

Answer:

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

12th Std Biology Questions And Answers:

12th Biology Chapter 15 Exercise Biodiversity, Conservation and Environmental Issues Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues Textbook Exercise Questions and Answers.

Biodiversity, Conservation and Environmental Issues Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 15 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 15 Exercise Solutions

1. Multiple choice questions

Question 1.
Observe the graph and select correct option.

(a) Line A represents, S = CA²
(b) Line B represents, log C = log A + Z log S
(c) Line A represents, S = CA^Z
(d) Line B represents, log S = log Z + C log A
Answer:
(c) Line A represents, S = CA^Z

Question 2.
Select odd one out on the basis of Ex situ conservation.
(a) Zoological park
(b) Tissue culture
(c) Sacred groves
(d) Cryopreservation
Answer:
(a) Zoological park

Question 3.
Which of the following factors will favour species diversity?
(a) Invasive species
(b) Glaciation
(c) Forest canopy
(d) Co-extinction
Answer:
(a) Invasive species

Question 4.
The term “terror of Bengal’ is used for
(a) algal bloom
(b) water hyacinth
(c) increased BOD
(d) eutrophication
Answer:
(b) water hyacinth

Question 5.
CFC are air polluting agents which are produced by
(a) Diesel trucks
(b) Jet planes
(c) Rice fields
(d) Industries
Answer:
(b) Jet planes

2. Very short answer type questions.

Question 1.
Give two examples of biodegradable materials released from sugar industry.
Answer:

Molasses
Bagasse.

Question 2.
Name any two modern techniques of protection of endangered species.
OR
Two modern methods of ex-situ conservation of species
Answer:

Tissue culture
In vitro fertilization of eggs
Cryopreservation.

Question 3.
Where was ozone hole discovered?
Answer:
Ozone hole was discovered in Antarctica.

Question 4.
Give one example of natural pollutant.
Answer:
Volcanic ash is a natural pollutant.

Question 5.
What do you understand by EW category of living being?
Answer:
A species which becomes extinct in the wild (EW) is called EW category, their members are seen only in captivity or as a naturalized population outside its historic range due to massive habitat loss.

3. Short answer type questions.

Question 1.
Dandiya raas is not allowed after 10.00 pm. Why?
Answer:
Dandiya rass involves blaring loudspeakers which cause noise pollution. It is undesired loud sound which could be hazardous for ears and general health. In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant. As per law noise after 10 pm is not allowed as many people may be resting. Therefore, Dandiya Raas is not allowed after 10 pm.

Question 2.
Tropical regions exhibit species richness as compared to polar regions. Justify.
Answer:

Tropical regions are bestowed by thicker vegetation and ample food due to available sunlight and humidity.
Polar regions are covered over with snow, with almost no vegetation.
Only handful species of animals can survive here due to their adaptations.
Species richness always shows latitudinal gradient for many plants and animal species. It is high at lower latitudes and there is a steady decline towards the poles. Therefore, tropical regions show more species richness.

Question 3.
How does genetic diversity affect sustenance of a species?
Answer:

Genetic diversity develops the capability of the species to adapt to the varying changes in the environment.
The large variation of the different gene sets allows an individual or the whole population to have the capacity to endure environmental stress in any form.
Some individuals have, a better capacity to endure the increasing pollution in the environment whereas some do not have it.
Those that do not have show infertility or even death from the same conditions.
Those who are able to endure and adapt to this change survive and live in a better way.
This is called natural selection which leads to a loss of genetic diversity in particular habitats.
Thus, due to genetic diversity can affect sustenance of some species.

Question 4.
Greenhouse effect is boon or bane? Give your opinion.
Answer:
(1) The natural greenhouse effect is good, it is a boon but human enhanced greenhouse effect is a bane.

(2) In the absence of an atmosphere, Earth’s surface temperature would be about -18 °C, or 0 °F, which is too cold for sustaining life.

(3) Earth is habitable because of the natural greenhouse effect. Heating of Earth’s atmosphere due to the presence of greenhouse gases such as water vapour, carbon dioxide (CO₂), methane (CH₄) and oxides of nitrogen (NO₂).

(4) Greenhouse gases have just the right molecular structure to absorb infrared radiation that the Earth emits. It re-emits most of that infrared energy in all directions, warming the atmosphere to its comfortable average temperature of 15 °C (60 °F). So, the greenhouse effect was a boon in olden days before industrialization and invention of automobiles.

(5) However, due to human impact, the proportion of greenhouse gases has increased tremendously causing global warming. Thus, now greenhouse effect has become a bane.

Question 5.
State the effects of CO in human body.
OR
How does CO cause giddiness and exhaustion?
Answer:
Effects of Carbon monoxide:

Carbon monoxide is tasteless, colourless and odourless gas, therefore its presence goes unnoticed.
It can inhibit the blood’s ability to carry oxygen to body tissues.
Supply of oxygen to vital organs such as.the heart and brain is affected due to presence of CO.
When CO is inhaled, it combines with the oxygen carrying haemoglobin of the blood to form carboxyhaemoglobin. Once combined with the haemoglobin, that haemoglobin is no longer available for transporting oxygen.
The symptoms of CO poisoning are headache, nausea, giddiness, etc.

Question 6.
Name two types of particulate pollutants found in air. Add a note on ill effects of the same on human health.
OR
Describe any 2 particulate and gaseous pollutants.
Answer:
I. Types of gaseous pollutants include CO₂, CO, SO₂, NO, NO₂, etc.
(1) Carbon dioxide : It is a greenhouse gas. It is produced in excess due to human activities such as burning of fossil fuels. It is also rising due to increasing deforestation. The natural cycle of Carbon dioxide is disturbed due to human interference. Otherwise, the process of photosynthesis can balance CO₂ : O₂ ratio of the air. Aeroplane traffic such as a jet plane also emits lots of CO₂.

(2) Carbon monoxide (CO) : CO is produced due to incomplete combustion of fuels. It is a toxic gas. Vehicular exhausts produce lot of CO.

II. Types of particulate pollutants are mist, dust, fume and smoke particles, smog, pesticides, heavy metals and radioactive elements, etc.
(1) Dust are fine particles which enter the respiratory passage and can cause damage to delicate tissues in the lungs. Various processes such as construction work, demolition of buildings and traffic can cause dust pollution. There are natural causes of release of dust too, through wind or volcanic eruption.

(2) Smoke and smog are worst type of particulate air pollutants which can cause many respiratory problems like emphysema or asthma.

4. Long answer type questions.

Question 1.
Montreal Protocol is an essential step. Why is it so?
Answer:

Montreal Protocol was an international treaty signed at Montreal in Canada in 1987.
Later many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries.
All these efforts were for reducing emission of CFCs and other ozone depleting chemicals.
All nations realized that ozone depletion can cause penetration of harmful UV radiations to the earth’s surface. This is very hazardous, for flora, fauna and for mainly human beings. Therefore, urgent action was needed to combat this effect.
Montreal Protocol was a very positive move because after 1987, there have been much better condition of ozone layer.

Question 2.
Name any 2 personalities who have contributed to control deforestation in our country. Elaborate on importance of their work.
Answer:
Two personalities who have contributed to control deforestation in our country are:
Saalumara Thimmakka from Karnataka and Moirangthem Loiya from Manipur.
1. Saalumara Thimmakka :

Saalumara Thimmakka is the best example of peoples’ participation in reforestation.
She is an Indian environmentalist from Karnataka. She has taken up work of planting and tending to 385 banyan trees along a 4 km stretch of highway between Hulikal and Kudur. Other 800 trees are also planted by her.
She is honoured with the National Citizens Award of India and Padma Shri in 2019.

2. Moirangthem Loiya :

Moirangthem Loiya is from Manipur who has restored Punshilok forest. For last 17 years he is planting trees after leaving his job.
He brought the lost glory back for the 300 acres forest land. He planted a variety of trees like, bamboo, oak Ficus, teak, jackfruit and Magnolia.
This forest now has over 250 varieties of plants including 25 varieties of bamboo along with many animals making the forest rich in biodiversity.

Question 3.
How BS emission standards changed over time? Why is it essential?
Answer:

BS emission standards changed over the time due to changing city life and more vehicular traffic on the road, especially in the megacities.
Since capital city of Delhi was declared as worst polluted city as far as its air quality is concerned, various measures were taken by the Government of India. There was new fuel policy declared, in which Bharat stage emission standards (BS) were set.
These norms were set to reduce sulphur and aromatic content of petrol and diesel. Also the vehicular engines were upgraded.
Bharat stage emission standards (BS) are standards which are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.
Since population of Delhi was to be saved, in 2001, Bharat stage II emission norms were set for CNG and LPG vehicles.
This helped in reduced emission of sulphur which was controlled at 50 ppm in diesel and 150 ppm in petrol. Also aromatic hydrocarbons were reduced at 42% in concerned fuel according to norms.
Because, in spite of all the efforts, Delhi was declared as worst air-polluted city in the world in 2016, therefore, Government of India directly adapted BS VI in the year 2018, skipping BS V These efforts decreased the levels of CO₂ and SO₂ in Delhi.

Question 4.
During large public gatherings like Pandharpur vari, mobile toilets are deployed by the government. Explain how this organic waste is disposed.
Answer:

The toilets deployed at Pandharpur at the time of vari are of the Ecosan type.
Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
When the pit of an Ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
In this time the faeces get completely composted to organic manure. In this way the organic waste can be disposed.
It is a practical, efficient and cost-effective solution for human waste disposal.
Also, open-air defecation is prohibited which can cause health problems. Therefore, during large public gatherings like Pandharpur vari mobile toilets like Ecosan are deployed by the government.

Question 5.
How Indian culture and traditions helped in bio-diversity conservation? Give importance of conservation in terms of utilitarian reasons.
Answer:
In Indian culture and traditions in different religions, biodiversity is protected and conserved. Few examples of worship of animals and plants can be given here.

Nagpanchami festival is towards the respect of snakes. They are worshipped on that day and the local people are aware of their role in ecosystem of control of rat population.
Vatapournima festival is worshipping a banyan tree.
Various other festivals teach the value of plants and animals surrounding us. Even the cattle are worshipped on a particular day as a tradition.
Jain religion strongly advocates protection of all animals through vegetarianism.

Conservation in terms of utilitarian reasons:
The conservation of biodiversity can be done in utilitarian way or for ethical reasons. Utilitarian reasons are further classified into narrowly utilitarian and broadly utilitarian reasons:

I. Narrowly utilitarian reasons:

Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.
Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.
For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.
Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.
Around 25000 species are used for traditional medicines by tribal population worldwide.
Bioprospecting which is a systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganisms, and other valuable products from nature which is of economically important species is also due to biodiversity.

II. Broadly utilitarian reasons:

Production of oxygen done by all green plants helps human beings to thrive. Amazon forest alone gives 25% of the oxygen to the entire world.
Insects carry out pollination and seed dispersal.
If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.
Biodiversity also is useful in recreation of human beings.

III. Taking all these aspects in consideration, conservation of biodiversity becomes essential. Therefore, to protect and conserve our rich biodiversity on the planet, we have to remember all the utilitarian reasons.

12th Std Biology Questions And Answers:

12th Biology Chapter 3 Exercise Inheritance and Variation Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 3 Inheritance and Variation Textbook Exercise Questions and Answers.

Inheritance and Variation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 3 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 3 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Phenotypic ratio of incomplete dominance in Mirabilis jalapa.
(a) 2 : 1 : 1
(b) 1 : 2 : 1
(c) 3 : 1
(d) 2 : 2
Answer:
(b) 1 : 2 : 1

Question 2.
In dihybrid cross, F₂ generation offspring show four different phenotypes while the genotypes are ……………….
(a) six
(b) nine
(c) eight
(d) sixteen
Answer:
(b) nine

Question 3.
A cross between an individual with unknown genotype for a trait with recessive plant for that trait is ……………….
(a) back cross
(b) reciprocal cross
(C) monohybrid cross
(d) test cross
Answer:
(d) test cross

Question 4.
When phenotypic and genotypic ratios are the same, then it is an example of ……………….
(a) incomplete dominance
(b) complete dominance
(c) multiple alleles
(d) cytoplasmic inheritance
Answer:
(a) incomplete dominance

Question 5.
If the centromere is situated near the end of the chromosome, the chromosome is called ……………….
(a) Metacentric
(b) Acrocentric
(c) Sub-Metacentric
(d) Telocentric
Answer:
(d) Telocentric

Question 6.
Chromosomal theory of inheritance was proposed by ……………….
(a) Sutton and Boveri
(b) Watson and Crick
(c) Miller and Urey
(d) Oparin and Halden
Answer:
(a) Sutton and Boveri

Question 7.
If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have least probability of being inherited together ?
(a) p and q
(b) r and s
(c) s and t
(d) p and s
Answer:
(d) p and s

Question 8.
Find the mismatched pair:
(a) Down’s syndrome = 44 + XY
(b) Turner’s syndrome = 44 + XO
(c) Klinefelter’s syndrome = 44 + XXY
(d) Super female = 44 + XXX
Answer:
(a) Down’s syndrome = 44 + XY

Question 9.
A colourblind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is ……………….
(a) 0%
(b) 25%
(c) 50%
(d) 100%
Answer:
(a) 0%

2. Very Short Answer Questions

Question 1.
Explain the statements
a. Test cross is back cross but back cross is not necessarily a test cross.
b. Law of dominance is not universal.
Answer:
a. (1) Test cross is the cross between F₁ hybrid and its homozygous recessive parent.
(2) Back cross is the cross of offspring with any one of the parents, either dominant or recessive.
(3) Therefore, test cross can be a back cross – but back cross cannot be a test cross.

b. (1) There are many traits in many organisms which show dominance. For example, widow’s peak in human beings is a dominant trait. Yellow seed colour and round seed shape are dominant traits in pea plant.
(2) However, there are characters which are either co-dominant, such as genes for human blood group A and B or incompletely dominant as in flower colour of Mirabilis jalapa.
(3) Therefore the law of dominance is not universally applicable.

Question 2.
Define the following terms:
a. Dihybrid cross
b. Homozygous
c. Heterozygous
d. Test cross
Answer:
a. A cross between parents differing in two heritable traits is called dihybrid cross.
b. An individual possessing identical alleles for a particular trait is called homozygous or pure for that trait. E.g. TT for tallness and tt for dwarfness.
c. An individual possessing contrasting allele for a particular trait is called heterozygous. E.g. Tt showing tallness.
d. The cross of F₁ progeny with homozygous recessive parent is called a test cross.

Question 3.
What are allosomes?
Answer:
Allosomes are the chromosomes which decide the sex of an organism.

Question 4.
What is crossing over?
Answer:
Crossing over is the process of forming new recombinations by interchanging and exchanging non-sister chromatid arms of the homologous chromosomes.

Question 5.
Give one example of autosomal recessive disorder.
Answer:
Thalassemia is an example of autosomal recessive disorder.

Question 6.
What are X-linked genes?
Answer:
Genes located on the non-homologous region of X chromosome are called X-linked genes.

Question 7.
What are holandric traits?
Answer:
Genes located on the non-homologous region of Y chromosome are called Y-linked genes. The traits due to such genes are called holandric traits which are seen only in male sex.

Question 8.
Give an example of chromosomal disorder caused due to non-disjunction of autosomes.
Answer:
Down’s syndrome is an example of chromosomal disorder caused due to non-disjunction of autosomes.

Question 9.
Give one example of complete sex linkage.
Answer:
Sex linkage can be complete X linkage and complete Y linkage. X linkage is haemophilia and Y linkage is hypertrichosis.

3. Short Answer Questions

Question 1.
Enlist seven traits of pea plant selected / studied by Mendel.
Answer:
Seven traits in pea selected by Mendel:

Tall habit versus dwarf habit (Height of the plant).
Purple flowers versus white flowers. (Colour of flowers)
Yellow seeds versus green seeds. (Colour of seeds)
Round seeds versus wrinkled seeds. (Shape of seeds)
Green pods versus yellow pods. (Colour of pods)
Inflated pods versus constricted pods. (Shape of pods)
Axial flower versus terminal flower. (Position of a flower)

Question 2.
Why law of segregation is also called the law of purity of gametes?
Answer:
(1) Mendel’s law of segregation is also called Law of purity of gametes because, during formation of gametes, the alleles separate/ segregate from each other and only one allele enters a gamete.

(2) The separation of one allele does not affect other. Since single allele enters a gamete means gametes will be pure for a trait.
E.g. The contrasting characters such as tall (T) and dwarf (t) present in F₁ hybrid (Tt) segregate during the formation of gametes.

(3) Owing to this, two types of gametes i.e. T and t are formed which are pure for the characters which they carry.
(4) Thus for example:

Question 3.
Pleiotropy.
Answer:

When a single gene controls two or more different traits, it is called a pleiotropic gene and the phenomenon is known as pleiotropy or pleiotropism.
The pleiotropic ratio is always 1 : 2 instead . of normal 3 : 1.
Sickle-cell anaemia is caused by the gene HbS. The healthy or normal gene which is dominant is HbA. The heterozygotes or carriers i.e., HbA/Hbs show anaemia as there is deficiency of haemoglobin due to sickling of RBCs. Abnormally low concentration of oxygen can cause sickling of RBCs.
The homozygotes possessing the recessive gene HbS die because of fatal anaemia because the gene for sickle-cell anaemia is lethal in homozygous condition and causes sickle-cell trait in heterozygous carrier.
Thus a single gene produces two different expressions.
When two carriers are married they will produce normal carriers and Sickle-cell anaemic children in the ratio of 1 : 2 : 1. Out of these three children sickle-cell anaemic child will die leaving the ratio 1 : 2 instead of 3 : 1.

Question 4.
What are the reasons of Mendel’s success?
Answer:
Reasons for Mendel’s success:

Mendel planned his experiments carefully and these experiments consisted of large sample.
He always recorded the results of number of plants of each type and their ratios.
The contrasting characters that he chose were easily recognizable.
The seven pairs of contrasting characters that he selected were under control of a single factor each. They were present on separate chromosomes and were transmitted from one generation to the next.
Mendel studied and introduced concept of dominance and recessiveness.

Question 5.
“Father is responsible for determination of sex of child and not the mother”. Justify.
Answer:

Human made is heterogame tic, i.e. he produces two different types of sperms. One is bearing X chromosome along with 22 autosomes and the other is Y bearing sperm with 22 autosomes.
Mother, on the other hand, is homogametic, producing all similar types of ova, i.e 22 + X chromosomal combination.
If 22+X bearing sperm fertilise an egg, female child is formed while if Y bearing sperm fertilizes an egg, male child is formed.
Thus the sex of the child is dependent upon type of sperm that father gives, therefore, it is said that father is responsible for determination of sex of a child and not the mother.

Question 6.
What is linkage? How many linkage groups do occur in human being and maize?
Answer:

Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome. Linkage group is group of genes situated on a chromosome.
Humans have 23 linkage groups because they have 23 pairs of chromosomes.
Maize plant has 10 linkage groups because they have 10 pairs of chromosomes.

Question 7.
PKU.
Answer:

PKU means phenylketonuria which is an autosomal recessive inborn error.
In this disorder the metabolism of phenylalanine does not occur due to deficiency of phenylalanine hydroxylase (PAH) enzyme.
This enzyme is necessary to metabolize the amino acid phenylalanine to the amino acid tyrosine.
When PAH activity is reduced, phenylalanine accumulates in blood and cerebrospinal fluid and is converted into phenylpyruvate or phenyl-ketone which is a toxic compound. This may cause mental retardation. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.
PKU is caused by mutations in the PAH gene on chromosome no. 12.
Untreated PKU causes abnormal phenotype which includes growth failure, poor skin pigmentation, microcephaly, seizures, global developmental delay and severe intellectual impairment. However, at birth if an infant is checked for PKU, the further abnormalities can be avoided.

Question 8.
Compare X-chromosome and Y-chromosome.
Answer:

X-chromosome	Y-chromosome
1. X-chromosome is straight, rod like and longer 1. than Y chromosome. It is metacentric.	1. Y-chromosome is shorter chromosome which is acrocentric.
2. X-chromosome has large amount of euchromatin and small amount of heterochromatin.	2. Y-chromosome has small amount of euchromatin and large amount of heterochromatin.
3. X-chromosome has large amount of DNA, hence it is genetically active due to more genes.	3. Y-chromosome has less amount of DNA, hence it is genetically less active or inert due to lesser genes.
4. Non-homologous region of X-chromosome is longer and contains more genes.	4. Non-homologous region of Y-chromosome is shorter and contains lesser genes.
5. Contains X-linked genes on non-homologous region.	5. Contains Y-linked genes on non-homologous region.
6. X-chromosome is present in men as well as women.	6. Y-chromosome is present only in men.

Question 9.
Explain the chromosomal theory of inheritance.
Answer:
Chromosomal theory of inheritance was put forth by Sutton and Boveri after studying paraillel behaviour of genes and chromosomes during meiotic division. This theory states following points:

Chromosomal theory identifies chromosomes as the carrier of genetic material.
All the hereditary characters are transmitted by gametes. Nucleus of gametes, i.e. sperms and ova of the parents contain chromosomes which transmit the heredity to offspring.
Chromosomes are found in pairs in somatic or diploid cells.
During gamete formation, homologous chromosomes pair and segregate independently at meiosis. The diploid condition is converted into haploid condition. Thus each gamete contains only one chromosome of a pair.
During fertilization, the union of sperm and egg restores the diploid number of chromosomes.

Question 10.
Observe the given pedigree chart and answer the following questions

(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a trait in human beings which shows such a pattern of inheritance.
Answer:
Pedigree given above shows:

First Generation : Carrier woman marrying a sufferer man. Their three children are in following birth order.
Second generation : First son is normal, second daughter is carrier and third daughter is sufferer.
Third generation : The sufferer daughter marries a normal man. Her children are normal daughter and sufferer son.

(a) The above pedigree show sex-linked (X-linked) trait. Since criss-cross inheritance is seen in the trait, it must be sex-linked inheritance.
(b) Such trait and its inheritance can be seen in colour blindness.

4. Match the Columns

rewrite the matching pairs.

Column I	Column II
(1) 21 trisomy	(a) Turner’s syndrome
(2) X-monosomy	(b) Klinefelter’s syndrome
(3) Holandric traits	(c) Down’s syndrome
(4) Feminized male	(d) Hypertrichosis

Answer:

Column I	Column II
(1) 21 trisomy	(c) Down’s syndrome
(2) X-monosomy	(a) Turner’s syndrome
(3) Holandric traits	(d) Hypertrichosis
(4) Feminized male	(b) Klinefelter’s syndrome

5. Long Answer Questions

Question 1.
What is dihybrid cross? Explain with suitable example and checker board method.
Answer:
1. A cross which involves two pairs of alleles is called a dihybrid cross. A phenotypic ratio of 9 : 3 : 3 : 1 obtained in the F₂ generation of a dihybrid cross is called a dihybrid ratio.

(2) Thus for example, when we cross a true breeding pea plant bearing round and yellow seeds with a true breeding pea plant bearing wrinkled and green seeds we get pea plants bearing round and yellow seeds in the F₁ generation.

(3) When F₁ plants are selfed, we get a ratio of 9 : 3 : 3 : 1 in the F₂ generation, where 9 plants bear yellow round seeds, 3 plants bear yellow wrinkled seeds, 3 plants bear green round seeds and 1 plant bears green wrinkled seeds.

(4) Parents (P₁) : RRYY × rryy
Gametes of P₁ RY and ry
F₁ generation : RrYy(Yellow round)
On selfing F₁ : RrYy × RrYy
Gametes of F₁ : RY, Ry, rY, ry

P₂ generation:

Round Yellow : 9 Round green : 3 Wrinkled yellow : 3 Wrinkled green : 1
Phenotypic ratio : 9 : 3 : 3 : 1
Genotypic ratio : 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1

Question 2.
Explain with suitable example an independent assortment.
Answer:
(1) The law of independent assortment states that when hybrid possessing two or more pairs of contrasting characters bearing alleles form gametes, the alleles in each pair segregate independently of the other pair. Therefore, the inheritance of one pair of characters is independent of that of the other pair of characters.
(2) For example, when we cross a pea plant which is tall and having purple flowers with dwarf plant having white flowers we obtain all tall plants with purple flowers in F₁ generation. When F₁ generation are selfed, 9 : 3 : 3 : 1 ratio was obtained in F₂ generation with 9 tall and purple flower, 3 tall with white flowers, 3 dwarf with purple flowers and 1 which was dwarf and white. Tallness and purple colour are dominant traits while dwarfness and white colour are recessive traits.

(i) Homozygous tall purple – TTPP
(ii) Homozygous dwarf white – ttpp

Tall purple = 9. Tall white = 3
Dwarf purple = 3, Dwarf white = 1,
Phenotypic ratio = 9 : 3 : 3 : 1
Results : The offspring of F₁ generation will be in the proportion of 9 : 3 : 3 : 1, where 9 are tall purple, 3 are tall white, 3 are dwarf purple and 1 is dwarf white.

Question 3.
Define test cross and explain its significance.
Answer:
1. Definition of test cross : A cross between F₁ offspring and its homozygous recessive parent is called a test cross.
2. Significance of test cross:

Test cross can be used to find out the genotype of any plant which shows dominant characters.
Whether the plant is homozygous or heterozygous can be understood by performing test cross.
Test cross is used to introduce useful recessive traits in the hybrids of self- pollinated plants.
Test cross is quicker method to improve the variety of crop plants and thus it is useful for breeders and geneticists.
Test cross can be used for verifying the laws of inheritance.

Question 4.
What is parthenogenesis? Explain the haplodiploid method of sex determination in honey bee.
Answer:
I. Parthenogenesis is a natural form of asexual reproduction in which growth and development of embryos occur without fertilization by sperm. In some insects like honey bees, parthenogenesis means development of an embryo from an unfertilized egg cell.

II. In honey bee:

Sex determination is by haplodiploid system.
Sex is determined by the number of sets of chromosomes received by an individual.
The egg which is fertilized by sperm, becomes diploid and develops into female.
The egg which is not fertilized develops by parthenogenesis and develops into a male.
The queen and worker bee therefore contain 32 chromosomes. The drone, i.e. male bears 16 chromosomes.
The sperms are produced by mitosis while eggs are produced by meiosis.

Question 5.
In the answer for inheritance of X-linked. genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.
Answer:
Males can never be carriers. They have single X and other Y chromosome. In X linked inheritance, the genes are present on the non-homologous region of X chromosome. Males do not have other X and hence if the genes are present on his X chromosome, they will not be suppressed in them. The Y chromosome does not have dominant gene to hide this expression as there is no homolorous region too. But in case of females, there are double X chromosomes and hence if X-linked gene is recessive, the other X can hide the expression of such X-linked gene.

Thus she becomes a carrier without showing any physical characters. She is physically normal and does not suffer from such X-linked recessive disorder. Thus, Madhav will get his answer wrong due to incorrect concept.

Question 6.
With the help of neat labelled diagram, describe the structure of chromosome.
Answer:

(1) A chromosome is best visible during metaphase, when it is highly condensed.

(2) Chromosome shows two identical halves, called sister chromatids. Chromatids are held together at centromere which is also called primary constriction.

(3) Primary constriction has disc shaped plate called kinetochore. This plate is useful for attachment of spindle fibres at the time of cell division.

(4) Additional narrow areas called secondary constrictions are seen in some chromosomes which are known as nucleolar organizers. They help in the formation of nucleolus. At secondary constriction (i) there is nucleolar organising region. Secondary constriction (ii) shows attachment of satellite body or SAT body.

(5) Each chromatid is made up of sub¬chromatids called chromonemata. Each chromonema consists of a long, unbranched, slender, highly coiled DNA thread. This double stranded DNA molecule extends throughout the length of the chromosome.

(6) The ends of the chromatid arms are called telomeres.

Question 7.
What is criss-cross inheritance? Explain with suitable example.
Answer:
Criss-cross inheritance is the type of inheritance in which the genes are passed on from father to daughter and then to her son, i.e. from male to female and from female to male (grandson). In other words, it is also said that the transmission is from the grandfather to his grandson through his daughter.

I. Inheritance of Colour blindness show criss-cross pattern.
(1) Colour blindness is a sex-linked disorder in which the person concerned cannot distinguish between red and green colours.

(2) It is recessively X-linked disorder, which is expressed in males. It is rarely seen in females.

(3) The genes for normal vision are dominant whereas those for colour blindness are recessive.

(4)

Gene for normal vision : X^C
Gene for colour blindness : X^c
Normal female : X^CX^C
Normal male : X^CY
Colour blind female : X^cX^c
Carrier female : X^CX^c
Colour blind male : X^c Y

II. Crosses showing the inheritance of colour blindness:
(i) A cross between normal female and colour-blind male.

(ii) A cross of carrier female with normal male.

(1) Normal female with Colour blind male. Such cross produces 50% carrier daughters and 50% normal sons.

(2) Carrier female with normal male. Such a cross produces 25% normal daughters, 25% normal sons, 25% carrier daughters and 25% colour blind sons.

(3) Colour blind father transmits the disorder to his grandson through his carrier daughter. The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance.

Question 8.
Describe the different types of chromosomes.
Answer:
I. Chromosomes are classified into the following four types according to the position of the centromere in them:
(1) Metacentric : In metacentric chromosome, the centromere is situated in the middle of the chromosome. The two arms of the chromosome are nearly equal. It appears ‘V’-shaped during anaphase.

(2) Sub-metacentric : In sub-metacentric chromosome, the centromere is situated some distance away from the middle. Due to this, one arm of the chromosome is shorter than the other. It appears T-shaped during anaphase.

(3) Acrocentric : In acrocentric chromosome, the centromere is situated near the end of the chromosome. One arm of the acrocentric chromosome is very short while the other is long making it appear like ‘J’-shaped during anaphase.

(4) Telocentric : In telocentric chromosome, the centromere is situated at the tip of the chromosome. Telocentric chromosome has only one arm thus it appears rod-shaped.

II. Based on the functions, chromosomes are divided into autosomes and allosomes. Autosomes are somatic chromosomes which decide the body characters. Allosomes are sex chromosomes which decide the sex of the individual.

12th Std Biology Questions And Answers:

12th Biology Chapter 11 Exercise Enhancement of Food Production Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 11 Enhancement of Food Production Textbook Exercise Questions and Answers.

Enhancement of Food Production Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 11 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 11 Exercise Solutions

1. Multiple choice questions

Question 1.
Antibiotic Chloromycetin is obtained from ………………….
(a) Streptomyces erythreus
(b) Penicillium chrysogenum
(c) Streptomyces venezuelae
(d) Streptomyces griseus
Answer:
(c) Streptomyces venezuelae

Question 2.
Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called ………………….
(a) primary treatment
(b) secondary treatment
(c) final treatment
(d) amplification
Answer:
(a) primary treatment

Question 3.
Which one of the following is free living bacterial biofertilizer?
(a) Azotobacter
(b) Rhizobium
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(a) Azotobacter

Question 4.
Most commonly used substrate for industrial production of beer is ………………….
(a) barley
(b) wheat
(c) corn
(d) sugar cane molasses
Answer:
(a) barley

Question 5.
Ethanol is commercially produced through a particular species of ………………….
(a) Aspergillus
(b) Saccharomyces
(c) Clostridium
(d) Trichoderma
Answer:
(b) Saccharomyces

Question 6.
One of the free-living anaerobic nitrogen- fixers is ………………….
(a) Azotobacter
(b) Beijerinckia
(c) Rhodospirillum
(d) Rhizobium
Answer:
(c) Rhodospirillum

Question 7.
Microorganisms also help in production of food like ………………….
(a) bread
(b) alcoholic beverages
(c) vegetables
(d) pulses
Answer:
(a) bread

Question 8.
MOET technique is used for ………………….
(a) production of hybrids
(b) inbreeding
(c) outbreeding
(d) outcrossing
Answer:
(a) production of hybrids

Question 9.
Mule is the outcome of ………………….
(a) inbreeding
(b) artificial insemination
(c) interspecific hybridization
(d) outbreeding
Answer:
(c) interspecific hybridization

2. Very Short Answer Questions

Question 1.
What makes idlis puffy?
Answer:
During preparation of idlis, rice and black gram flour is fermented by air borne Leuconostoc and Streptococcus bacteria. CO₂ produced during fermentation makes them puffy.

Question 2.
Bacterial biofertilizers.
Answer:
Rhizobium, Frankia, Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Azotobacter, Costridium, Beijerinkia, Klebsiella.

Question 3.
What is the microbial source of vitamin B₁₂?
Answer:
The microbial source of vitamin B₁₂ is Pseudomonas denitrificans.

Question 4.
What is the microbial source of enzyme invertase?
Answer:
The microbial source of enzyme invertase is Saccharomyces cerevisiae.

Question 5.
Milk starts to coagulate when Lactic Acid Bacteria (LAB) are added to warm milk as a starter. Mention any two other benefits of LAB.
Answer:
Lactic Acid Bacteria (LAB) check the growth of disease causing microbes and produce vitamin B.

Question 6.
Name the enzyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer:
The enzyme produced by Streptococcus spp. is streptokinase. It is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 7.
What is breed?
Answer:
Breed is a group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc.

Question 8.
Estuary
Answer:
Estuary is a place where river meets the sea.

Question 9.
What is shellac?
Answer:
Shellac is the pure form of lac obtained by washing and filtering.

3. Short Answer Questions.

Question 1.
Many microbes are used at home during preparation of food items. Comment on such useful ones with examples.
Answer:

Many food preparations made at home involves the use of microorganisms.
The microbes Lactobacilli are used in the preparation of dhokla from gram flour and buttermilk by the process of fermentation.
Dosa and idlis are prepared by using batter of rice and black gram which is fermented by air borne Leuconostoc and Streptococcus bacteria.
Large, fleshy fruiting bodies of some mushrooms and truffles are directly used as food. It is sugar free, fat free food rich in proteins, vitamins, minerals and amino acids. It is food with low calories.
Curd is prepared by inoculating milk with Lactobacillus acidophilus. Lactic acid produced during fermentation causes coagulation and partial digestion of milk protein casein and milk turns into curd.
Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.

Question 2.
What is biogas? Write in brief about the production process.
Answer:
Biogas is a mixture of methane CH₄ (50-60%), CO₂ (30-40%), H₂S (0-3%) and other gases (CO, N₂, H₂) in traces.

Biogas production process:
a. A typical biogas plant consists of digester (made up of concrete bricks and cement or steel and is partly buried in the soil) and gas holder (a cylindrical gas tank to collect gases).
b. Raw materials like cow dung is mixed with water in equal proportion to make slurry which is fed into the digester’ through a side opening (charge pit).

Anaerobic digestion involves following processes:
i. Hydrolysis or solubilization:
Anaerobic hydrolyzing bacteria like Clostridium and Pseudomonas hydrolyse carbohydrates into simple sugars, proteins into amino acids and lipids into fatty acids.

ii. Acidogenesis:
Facultative and obligate anaerobic, acidogenic bacteria convert simple organic substances into acids like formic acid, acetic acid, H₂ and CO₂

iii. Methanogenesis:
Anaerobic methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂ and CO₂ into Methane, CO₂ and H₂O and other products.
12 mol CH₃COOH → 12CH₄ + 12CO₂ 4mol H.COOH → CH₄ + 3CO₂ + 2H₂O CO₂ + 4H₂ → CH₄ + 2H₂O

Question 3.
Biocontrol agents.
Answer:
(1) Biocontrol agents are the organisms like (bacteria, fungi, viruses and protozoans) act which are employed for controlling pathogens, pests and weeds.

(2) They cause the disease to the pest or compete or kill them.

(3) The use of biocontrol measures greatly reduces use of toxic chemicals and pesticides that are harmful to human beings and also pollute our environment.

(4) Biocontrol agents and their hosts.

Bacteria (Bacillus thuringiensis, B. papilliae and B. lentimorbus Hosts : Caterpillars, cabbage worms, adult beetles
Fungi (Beauveria bassiana, Entomophothora, pallidaroseum, Zoophthora radicans) Host : Aphid crocci, A. unguicilata, mealy bugs, mites, white flies, etc.
Protozoans (Nosema locustae) Host: Grasshoppers, caterpillars, crickets
Viruses (Nucleopolyhedro virus-NPV, Granulovirus-GV) Host : Caterpillars, Gypsy moth, ants and beetles.

(5) Some examples:

Bacillus thuringiensis (Bt) is a microbiai pesticide used to get rid of butterfly, caterpillars.
Trichoderma fungus is an effective biocontrol agent against soil borne fungal plant pathogens which infect roots and rhizomes.
Phytophthora palmiuora is a mycoherbicide that controls milk weed in orchards.
Pseudomonas spp. is a bacterial herbicide that attacks several weeds.
Tyrea moth controls the weed Senecio jacobeac.

Question 4.
Name any two enzymes and antibiotics with their microbial source.
Answer:

Microbial source of Chloromycetin. – Streptomyces venezuelae
Microbial source of Erythromycin. – Streptomyces erythreus
Microbial source of Penicillin. – Penicillium chrysogenum
Microbial source of Streptomycin. – Streptomyces griseus
Microbial source of Griseofulvin. – Penicillium griseojulvum
Microbial source of Bacitracin. – Bacillus licheniformis
Microbial source of Oxytetracyclin / Terramycin. – Streptomyces aurifaciens
The enzyme produced by Streptococcus bacterium. – Streptokinase
Microbial source of Invertase. – Saccharomyces cerevisiae
Microbial source of Pectinase. – Sclerotinia libertine, Aspergillus niger
Microbial source of Lipase. – Candida lipolytica
Microbial source of Cellulase. – Trichoderma konigii

Question 5.
Write the principles of farm management.
Answer:
The principles of farm management are as follows:

Selection of high-yielding breeds.
Understanding the feed requirements of farm animals.
Supply of adequate nutritional sources for the animals.
Maintaining the cleanliness of environment.
Maintenance of health with the help of veterinary supervision.
Undertaking vaccination programmes.
Development of high-yielding cross-bred varieties.
Making various products and their preservation.
Distribution and marketing of the farm produce.

Question 6.
Give the economic importance of fisheries.
Answer:
Economic importance of fisheries is as follows:

Fish is a nutritious food and thus is a source of many vitamins, minerals and nutrients.
Commercial products such as fish oil, fish meal and fertilizers, fish guano, fish glue, isinglass are prepared from fish.
These by-products are used in paints, soaps, oils and medicines.
Some organisms like prawns and lobsters have high export value and market price.
Fish farming and other fishery trades provide job opportunity and self-employment
Productivity and national economy is improved through fishery practices.

Question 7.
Enlist the species of honey bee mentioning their specific uses.
Answer:
(1) The four species of honey bees commonly found in India : Apis dorsata (rock bee, or wild bee), Apis jlorea (little bee), Apis mellifera (European bee) and Apis indica (Indian bee).

(2) Uses:

Rock bee : They produce 36 kg of honey per comb per year. They produce bee wax.
Little bee : They produce half kg of honey per hive per year.
European bee : The average production per colony per year is 25 to 40 kg.
Indian bee : The average production per colony per year is 6 to 8 kg.

Question 8.
What are A, B, C, D in the table given below.

Types of microbe	Name	Commercial Product
Fungus	A	Penicillin
Bacterium	Acetobacter aceti	B
C	Aspergillus niger	Citric acid
Yeast	D	Ethanol

Answer:
A : Penicillium chrysogenum
B : Vinegar (Acetic acid)
C : Fungus
D : Sachharomyces cerevisiae var. ellipsoidis

4. Long Answer Questions.

Question 1.
Explain the process of sewage water treatment before it can be discharged into natural bodies. Why is this treatment essential?
Answer:
Sewage treatment includes following steps:
(1) Preliminary Treatment:

Screening: The larger suspended or floating objects are filtered and removed in screening chambers by passing the sewage through screens or net in the chambers.
Grit Chamber : Filtered sewage is passed into series of grit chambers which contain large stones (pebbles) and brick-ballast. Coarse particles which settle down by gravity are removed.

(2) Primary treatment (physical treatment):

The sewage water is pumped into the primary sedimentation tank where 50-70% of the suspended solid or organic matter get sedimented and about 30-40% (in number) of coliform organisms are removed.
The organic matter which is settled down is called primary sludge.
Primary sludge is removed by mechanically operated devices.
Dissolved organic matter and micro-organisms in the supernatant (effluent) are then removed by the secondary treatment.

(3) Secondary treatment (biological treatment):

The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
The mesh like masses of aerobic bacteria, slime and fungal hyphae, known as floes are formed.
Aerobic microbes consume most of the organic matter and this reduces BOD (Biochemical Oxygen Demand) of the effluent.

(4) Tertiary treatment:

Once the BOD is sufficiently reduced, waste water is passed into a settling tank where the floes are allowed to sediment.
The sediment is called activated sludge.
Small part of activated sludge is transferred to aeration tank and the major part is pumped in to large anaerobic sludge digesters.
In these tanks, anaerobic bacteria grow and digest the bacteria and fungi in the sludge and gases like methane, hydrogen sulphide, CO₂, etc. are released.
Effluents from these digesters are released in natural water bodies like rivers and streams after chlorination which kills pathogenic bacteria.
Digested sludge is then disposed.

Question 2.
Lac culture.
Answer:

Lac is a pink coloured resin secreted by dermal glands of female lac insect (Trachardia lacca) that hardens on coming in contact with air forming lac.
Lac is a complex substance having resin, sugar, water, minerals and alkaline substances.
Lac insect is colonial in habit and it feeds on succulent twigs like ber, peepal, palas, kusum, babool,
These plants are artificially inoculated in order to get better and regular supply of good quality and quantity of lac.
Natural lac is always contaminated and pure form of lac obtained by washing and filtering is called as shellac.
Lac is used to make bangles, toys, woodwork, inks, mirrors, etc.
India’s share is 85% of total lac produced in the world.

Question 3.
Describe various methods of fish preservation.
Answer:

Fish is a highly perishable commodity.
After catching the fish it immediately starts spoilage process.
In order to prevent this process, the fish preservation is done.

The different methods of fish preservation are as follows:

Chilling : This involves covering the fish with layers of ice. Ice is effective for short term preservation. It inhibits the activity of autolytic enzymes.
Freezing : It is a long duration preservation method. Fish are freezed at 0°C to -20°C. This also inhibit autolytic enzyme activities and slows down bacterial growth.
Freeze drying : The deep frozen -fish at -20°C are dried by direct sublimation of ice to water vapour with any melting into liquid water. This is achieved by exposing the frozen fish to 140°C in a vacuum chamber. The fish is then packed or canned in dried condition.
Sun drying : This inhibits the growth of microorganisms that spoil the fish.
Smoke drying : Smoke is prepared by burning woods with less resinous matter. Bacteria are destroyed by the acid content of the smoke. Smoking also give the characteristic colour, taste and odour to fish.
Salting : Salt removes the moisture from the fish tissues by osmosis. High salt concentration destroys autolytic enzymes and halts bacterial activity.
Canning : Canning involves sealing the food in a container, heat ‘sterilising’ the sealed unit and cooling it to ambient temperature for subsequent storage.

Question 4.
Give an account of poultry diseases.
Answer:
Various poultry diseases are as follows:

Viral diseases : Ranikhet, Bronchitis, Avian influenza (bird flu), etc. Bird flu had serious impact on poultry farming and also caused human infection.
Bacterial diseases : Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis, etc.
Fungal diseases : Aspergillosis, Favus and Thrush.
Parasitic diseases : Lice infection, round worm, caecal worm infections, etc.
Protozoan diseases : Coccidiosis.

Question 5.
Give an account of mutation breeding with examples.
Answer:

Mutations are sudden heritable changes in the genotype.
Natural mutations occur at a very slow rate.
Natural physical mutagens include exposure to high temperature, high concentration of C02, X-rays, UV rays.
Mutations can be induced by using various mutagens.
Mutagens cause gene mutations and chromosomal aberrations.
Chemical mutagens include nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.
Seedlings or seeds are irradiated by using CO60 or UV bulbs or X-ray machines.
The mutated seedlings are then screened for resistance to diseases/pests, high yield, etc.
Examples of mutant varieties in different crops are Jagannath (rice), NP 836 (rust resistant wheat variety), Indore-2 (cotton variety resistant to bollworm), Regina-II (cabbage variety resistant to bacterial rot).

Question 6.
Describe briefly various steps of plant breeding methods.
Answer:
The main steps of the plant breeding program (Hybridization) are as follows:

(1) Collection of variability:

Germplasm collection is the entire collection of all the diverse alleles for all genes in a given crop.
Wild species and relatives of the cultivated species having desired traits are collected and preserved.
Forests and natural reserves are the means of in situ conservation of germplasm.
Botanical gardens, seed banks, etc. are means of ex situ conservation of germplasm.

(2) Evaluation and selection of parents:

The collected germplasm is evaluated to identify healthy and vigorous plants with desirable and complementary characters.
Selected parents are selfed for three to four generations to increase homozygosity.
Only pure lines are selected, multiplied and used in the hybridization.

(3) Hybridization:

The variety showing maximum desirable features is selected as female (recurrent) parent and the other variety which lacks good characters found in recurrent parent is selected as male parent (donor).
The pollen grains from anthers of male parent are artificially dusted over stigmas of emasculated flowers of female parent.
Hybrid seeds are collected and sown to grow F₁ geneartion.

(4) Selection and Testing of Superior Recombinants:

The F₁ hybrid plants which are superior to both the parents and having high hybrid vigour, are selected and selfed for few generations to make them homozygous for the said desirable characters.
This ensures that there is no further segregation of the characters.

(5) Testing, release and commercialization of new cultivars:

The newly selected lines are evaluated for the productivity and desirable features like disease resistance, pest resistance, quality, etc.
They are initially grown under controlled conditions of water, fertilizers, etc. and their performance is recorded.
The selected lines are then grown for at least three generations in natural field, in different agroclimatic zones.
Finally variety is released as new variety for use by the farmers.

12th Std Biology Questions And Answers: