Category: Important Questions

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 7 Body Fluids and Circulation which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 7 Body Fluids and Circulation

Answer the following.

Question 1.
In what way the circulatory system helps our body?
Answer:
The circulatory system helps to maintain the homeostasis of the body fluids and body temperature (heat exchange). The homeostatic regulation of the cardiovascular system maintains blood flow, or perfusion, to the heart and brain.

Question 2.
Why vasovagal syncope (fainting) occurs?
Answer:
Signals from the nervous system cause a sudden decrease in blood pressure, and the individual faints from lack of oxygen to the brain.

Question 3.
Name the three types of extracellular fluids.
Answer:
The three types of extracellular fluids are the interstitial fluid or tissue fluid (surrounds the cell), the plasma (fluid component of the blood) and lymph.

Question 4.
What are the four main types of plasma proteins?
Answer:
The four main types of plasma proteins synthesized in the liver are albumin, globulin, prothrombin and fibrinogen.

Question 5.
Write the functions of the plasma protein.
Answer:
Albumin maintains the osmotic pressure of the blood. Globulin facilitates the transport of ions, hormones, lipids and assists in immune function. Both Prothrombin and Fibrinogen are involved in blood clotting.

Question 6.
What happens to the constituents of blood immediately after a meal?
Answer:
The composition of plasma is not always constant. Immediately after a meal, the blood in the hepatic portal vein has a very high concentration of glucose as it is transporting glucose from the intestine to the liver where it is stored. The concentration of the glucose in the blood gradually falls after some time as most of the glucose is absorbed.

Question 7.
If too much protein consumed more than the normal need, what will happen to the excess of protein consumed?
Answer:
If too much protein is consumed, the body cannot store the excess amino acids formed from the digestion of proteins. The liver breaks down the excess amino acids and produces urea. Blood in the hepatic vein has a high concentration of urea than the blood in other vessels namely, the hepatic portal vein and hepatic artery.

Question 8.
What is the name of the pigment present in RBC? Write its important role.
Answer:
The red colour of the RBC is due to the presence of a respiratory pigment, haemoglobin dissolved in the cytoplasm. Haemoglobin plays an important role in the transport of respiratory gases and facilitates the exchange of gases with the fluid outside the cell.

Question 9.
What is haematocrit?
Answer:
Erythropoietin is a hormone secreted by the kidneys in response to low oxygen and helps in the differentiation of stem cells of the bone marrow to erythrocytes (erythropoiesis) in adults. The ratio of red blood cells to blood plasma is expressed as Haematocrit.

Question 10.
Write about the red cells of the blood.
Answer:

Red blood cells are abundant than other blood cells. There are about 5 million to 5.5 millions of RBC mm”3 of blood in a healthy man and 4.5 – 5.0 million RBC mnr3 in healthy women. The RBCs are very small with a diameter of about 7pm (micrometer). The structure of RBC is shown in Figure. The red colour of the RBC is due to the presence of a respiratory pigment, haemoglobin dissolved in the cytoplasm. Haemoglobin plays an important role in the transport of respiratory gases and facilitates the exchange of gases with the fluid outside the cell (tissue fluid). The biconcave shaped RBCs increases the surface area to volume ratio, hence oxygen diffuses quickly in and out of the cell. The RBCs are devoid of the nucleus, mitochondria, ribosomes and endoplasmic reticulum. The absence of these organelles accommodates more haemoglobin thereby maximising the oxygen-carrying capacity of the cell. The average life span of RBCs in a healthy individual is about 120 days after which they are destroyed in the spleen (graveyard/cemetery of RBCs) and the iron component returns to the bone marrow for reuse. Erythropoietin is a hormone secreted by the kidneys in response to low oxygen and helps in the differentiation of stem cells of the bone marrow to erythrocytes (erythropoiesis) in adults. The ratio of red blood cells to blood plasma is expressed as Haematocrit

Question 11.
Give an account of different types of white blood cell with suitable diagrams.
Answer:

White blood cells (leucocytes) are colourless, amoeboid, nucleated cells devoid of haemoglobin and other pigments. Approximately 6000 to 8000 per cubic mm of WBCs are seen in the blood of an average healthy individual. The different types of WBCs are shown in Figure. Depending on the presence or absence of granules, WBCs are divided into two types, granulocytes and agranulocytes. Granulocytes are characterised by the presence of granules in the cytoplasm and are differentiated in the bone marrow. The granulocytes include neutrophils, eosinophils and basophils.
Neutrophils are also called heterophils or polymorphonuclear (cells with 3-4 lobes of nucleus connected with delicate threads) cells which constitute about 60% – 65% of the total WBCs. They are phagocytic in nature and appear in large numbers in and around the infected tissues.
Eosinophils have a distinctly bilobed nucleus and the lobes are joined by thin strands. They are non – phagocytic and constitute about 2 – 3% of the total WBCs. Eosinophils increase during certain types of parasitic infections and allergic reactions.
Basophils are less numerous than any other type of WBCs constituting 0.5% – 1.0% of the total number of leucocytes. The cytoplasmic granules are large-sized but fewer than eosinophils. The nucleus is large-sized and constricted into several lobes but not joined by delicate threads. Basophils secrete substances such as heparin, serotonin and histamines. They are also involved in inflammatory reactions.
Agranulocytes are characterised by the absence of granules in the cytoplasm and are differentiated in the lymph glands and spleen. These are of two types, lymphocytes and monocytes. Lymphocytes constitute 28% of WBCs. These have a large round nucleus and a small amount of cytoplasm. The two types of lymphocytes are B and T cells. Both B and T cells are responsible for the immune responses of the body. B cells produce antibodies to neutralize the harmful effects of foreign substances and T cells are involved in cell-mediated immunity.
Monocytes (Macrophages) are phagocytic cells that are similar to mast cells and have kidney shaped nucleus. They constitute 1 – 3% of the total WBCs. The macrophages of the central nervous system are the ‘microglia’, in the sinusoids of the liver they are called ‘Kupffer cells’ and in the pulmonary region they are the ‘alveolar macrophages.

Question 12.
Which type of WBC increase in its amount during allergic reactions? Write a few words about it.
Answer:

1. Eosinophils increase during certain types of parasitic infections and allergic reactions.
2. Eosinophils have a distinctly bilobed nucleus and the lobes are joined by thin strands. They are non – phagocytic and constitute about 2 – 3% of the total WBCs.

Question 13.
Write a few lines about the two types of lymphocytes.
Answer:
The two types of lymphocytes are B and T cells. Both B and T cells are responsible for the immune responses of the body. B cells produce antibodies to neutralize the harmful effects of foreign substances and T cells are involved in cell-mediated immunity.

Question 14.
Write short note on platelets.
Answer:
Platelets are also called thrombocytes that are produced from megakaryocytes (special cells in bone marrow) and lack nuclei. Blood normally contains 1, 50,000 – 3, 50,000 platelets mm-3 of blood. They secrete substances involved in the coagulation or clotting of blood. The reduction in platelet number can lead to clotting disorders that result in excessive loss of blood from the body.

Question 15.
Tabulate the different types of blood group in human with the antigen and antibodies.
Answer:
Distribution of antigens and antibodies in different blood groups

Blood group	Agglutinogens (antigens) on the RBC	Agglutinin (antibodies) in the plasma
A	A	Anti B
B	B	Anti A
AB	AB	No antibodies
O	No antigens	Anti A and Anti B

Question 16.
How can we identify the ‘+ve’ or’-ve’ factor in the blood groups?
Answer:
Rh factor is a protein (D antigen) present on the surface of the red blood cells in the majority (80%) of humans. This protein is similar to the protein present in Rhesus monkey, hence the term Rh, Individuals who carry the antigen D on the surface of the red blood cells are Rh” (Rh-positive) and the individuals who do not carry antigen D, are Rh (Rh-negative).

Question 17.
What is erythroblastosis foetalis? How is it caused? Can it be treated?
Answer:
Rh factor compatibility is also checked before blood transfusion. When a pregnant woman is Rh^– and the foetus is Rh⁺ incompatibility (mismatch) is observed. During the first pregnancy, the Rh^– antigens of the foetus does not get exposed to the mother’s blood as both their blood are separated by the placenta. However, a small amount of the foetal antigen becomes exposed to the mother’s blood during the birth of the first child. The mother’s blood starts to synthesize D antibodies. But during subsequent pregnancies, the Rh antibodies from the mother (Rh^–) enters the foetal circulation and destroys the foetal RBCs. This becomes fatal to the foetus because the child suffers from anaemia and jaundice. This condition is called erythroblastosis foetalis. This condition can be avoided by the administration of anti D antibodies (Rhocum) to the mother immediately after the first childbirth.

Question 18.
Give a brief account of the general structure of blood vessels.
Answer:
The vessels carrying the blood are of three types; they are arteries, veins and capillaries. These vessels are hollow structures and have complex walls surrounding the lumen. The blood vessels in humans are composed of three layers, tunica intima, tunica media and tunica externa. The inner layer, tunica intima or tunica interna supports the vascular endothelium, the middle layer, tunica media is composed of smooth muscles and an extracellular matrix that contains a protein, elastin. The contraction and relaxation of the smooth muscles result in vasoconstriction and vasodilation. The outer layer, tunica externa or tunica adventitia is composed of collagen fibres.

Question 19.
Describe the process of clotting blood.
Answer:

The clotting process begins when the endothelium of the blood vessel is damaged and -the connective tissue in its wall is exposed to the blood. Platelets adhere to collagen fibres in the connective tissue and release substances that form the platelet plug which provides emergency protection against blood loss. Clotting factors released from the clumped platelets or damaged cells mix with clotting factors in the plasma. The protein called prothrombin is converted to its active form called thrombin in the presence of calcium and vitamin K. Thrombin helps in the conversion of fibrinogen to fibrin threads. The threads of fibrins become interlinked into a patch that traps blood cell and seals the injured vessel until the wound is healed. After some time fibrin fibrils contract, squeezing out a straw-coloured fluid through a meshwork called serum (Plasma without fibrinogen is called serum).

Question 20.
How blood in the blood vessels are liquid in nature without clotting?
Answer:
Heparin is an anticoagulant produced in small quantities by mast cells of connective tissue which prevents coagulation in small blood vessels.

Question 21.
What are anastomoses?
Answer:
These are connections of one blood vessel (arteries) with another blood vessel. They provide an alternate route of blood flow if the original blood vessel is blocked. For example, Arteries in the joints contain numerous anastomoses. This allows blood to flow freely even if one of the arteries closes during bending of the joints.

Question 22.
What are coronary blood vessels? Why did they call so?
Answer:
Blood vessels that supply blood to the cardiac muscles with all nutrients and remove wastes are the coronary arteries and veins. Heart muscle is supplied by two arteries namely the right and left coronary arteries. These arteries are the first branch of the aorta. Arteries usually surround, the heart in the manner of a crown, hence called the coronary artery.

Question 23.
What is the Law of Laplace? Where the law can be used in the biological system?
Answer:
The Law of Laplace is used to understand the structure and function of blood vessels and the heart. Laplace law states that the tension in the walls of the blood vessel is proportional to the blood pressure and vessel radius. Blood vessels such as the aorta that is subjected to high pressures have thicker walls than the arterioles that are subjected to low pressures.

Question 24.
Why vigorous exercise after a meal can cause indigestion or abdominal cramps?
Answer:
When exercising vigorously, blood is rerouted from the digestive organ (food or no food) to the capillary beds of the skeletal muscles where it is needed immediately. This rerouting explains why vigorous exercise after a meal can cause indigestion or abdominal cramps.

Question 25.
What is incomplete double circulation?
Answer:
Amphibians have two auricles and one ventricle and no interventricular septum whereas reptiles except crocodiles have two auricles and one ventricle and an incomplete interventricular septum. Thus mixing of oxygenated and deoxygenated blood takes place in the ventricles. This type of circulation is called incomplete double circulation.

Question 26.
Write an account on the L.S. of heart with a simple sketch.
Answer:
The heart is divided into four chambers, the upper two small auricles or atrium and the lower two large ventricles. The two auricles are separated by inter auricular septum and the two ventricles are separated by the interventricular septum The separation of chambers avoids mixing of oxygenated and deoxygenated blood. The auricle communicates with the ventricle through an opening called auriculo ventricular aperture which is guarded by the auriculo ventricular valves. The opening between the right atrium and the right ventricle is guarded by the tricuspid valve (three flaps or cusps), whereas* a bicuspid (two flaps or cusps) or mitral valve guards the opening between the left atrium and left ventricle. The valves of the heart allow the blood to flow only in one direction, i.e., from the atria to the ventricles and from the ventricles to the pulmonary artery or the aorta. These valves prevent the backward flow of blood. The opening of the right and left ventricles into the pulmonary artery and aorta are guarded by aortic and pulmonary valves and are called semilunar valves. The opening and closing of the semilunar valves are achieved by the chordae tendinae. The chordae tendinae are attached to the lower end of the heart by papillary muscles.

Question 27.
What does the following term define?

1. Heartbeat,
2. Systole,
3. Diastole,
4. ‘lub’ and ‘dub’ sound of the heart,
5. Cardiac cycle.

Answer:

1. Rhythmic contraction and expansion of the heart are called heartbeat.
2. The contraction of the heart is called systole.
3. The relaxation of the heart is called diastole.
4. The first heart sound (lub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves.
5. The events that occur at the beginning of the heartbeat and last until the beginning of the next beat is called the cardiac cycle.

Question 28.
Write about the series of event takes place in the cardiac cycle.
Answer:
The series of events takes place in a cardiac cycle.
PHASE 1: Ventricular diastole – The pressure in the auricles increases than that of the ventricular pressure. AV valves are open while the semilunar valves are closed. Blood flows from the auricles into the ventricles passively.
PHASE 2: Atrial systole – The atria contracts while the ventricles are still relaxed. The contraction of the auricles pushes the maximum volume of blood to the ventricles until they reach the end-diastolic volume (EDV). EDV is related to the length of the cardiac muscle fibre. The more the muscle is stretched, greater the EDV and the stroke volume.
PHASE 3: Ventricular systole (isovolumetric contraction) – The ventricular contraction forces the AV valves to close and increases the pressure inside the ventricles. The blood is then pumped from the ventricles into the aorta without a change in the size of the muscle fibre length and ventricular chamber volume (isovolumetric contraction).
PHASE 4: Ventricular systole (ventricular ejection) – Increased ventricular pressure forces the semilunar valves to open and blood is ejected out of the ventricles without backflow of blood. This point is the end of the systolic volume (ESV).
PHASE 5: (Ventricular diastole) – The ventricles begin to relax, pressure in the arteries exceeds ventricular pressure, resulting in the closure of the semilunar valves. The heart returns to phase 1 of the cardiac cycle.

Question 29.
Define cardiac output. How can it be calculated?
Answer:
The amount of blood pumped out by each ventricle per minute is called cardiac output(CO).
It is a product of heart rate (HR) and stroke volume (SV). Heart rate or pulse is the number of beats per minute.
Pulse pressure = systolic pressure – diastolic pressure.
Stroke volume (SV) is the volume of blood pumped out by one ventricle with each beat. SV depends on ventricular contraction.
CO = HR x SV
SV represents the difference between EDV (amount of blood that collects in a ventricle during diastole) and ESV (volume of blood remaining in the ventricle after contraction).
SV = EDV – ESV

Question 30.
Explain how the Frank-Starling effect protects the heart from an abnormal increase in blood volume.
Answer:
According to the Frank-Starling law of the heart, the critical factor controlling SV is the degree to which the cardiac muscle cells are stretched just before they contract. The most important factor stretching the cardiac muscle is the amount of blood returning to the heart and distending its ventricles, venous return. During vigorous exercise, SV may double as a result of venous return. The heart’s pumping action normally maintains a balance between cardiac output and venous return. Because the heart is a double pump, each side can fail independently of the other. If the left side of the heart fails, it results in pulmonary congestion and if the right side fails, it results in peripheral congestion. Frank-Starling effect protects the heart from an abnormal increase in blood volume.

Question 31.
What is blood pressure? What are its types? How is it caused?
Answer:

1. Blood pressure is the pressure exerted on the surface of blood vessels by the blood.
2. There are two types of pressure, systolic pressure and diastolic pressure.
3. Systolic pressure is the pressure in the arteries as the chambers of the heart contracts. Diastolic pressure is the pressure in the arteries when the heart chambers relax.

Question 32.
What is orthostatic hypotension?
Answer:
The primary reflex pathway for homeostatic control of mean arterial pressure is the baroreceptor reflex. The baroreceptor reflex functions every morning when you get out of bed. When you are lying flat the gravitational force is evenly distributed. When you stand up, gravity causes blood to pool in the lower extremities. The decrease in blood pressure upon standing is known as orthostatic hypotension.

Question 33.
What is ECG? How ECG graph helps the activity of the heart during one cardiac cycle?
Answer:
Electrocardiogram (ECG): An electrocardiogram (ECG) records the electrical activity of the heart over a period of time using electrodes placed on the skin, arms, legs and chest.
It records the changes in electrical potential across the heart during one cardiac cycle.
A normal ECG shows 3 waves designated as P wave, QRS complex and T wave.

P Wave (atrial depolarisation): It is a small upward wave and indicates the depolarization of the atria. This is the time taken for the excitation to spread through the atria from the SA node. Contraction of both atria lasts for around 0.8 – 1.0 sec.
PQ Interval (AV node delay): It is the onset of the P wave to the onset of the QRS complex. This is from the start of depolarization of the atria to the beginning of ventricular depolarization. It is the time taken for the impulse to travel from the atria to the ventricles (0.12 – 0.21 sec). It is the measure of AV conduction time.
QRS Complex (ventricular depolarisation): No separate wave for atrial depolarization in the ECG is visible. Atrial depolarization occurs simultaneously with ventricular depolarization. The normal QRS complex lasts for 0.06 – 0.09 sec. The QRS complex is shorter than the P wave because depolarization spreads through the Purkinje fibres. Prolonged QRS wave indicates delayed conduction through the ventricle, often caused due to ventricular hypertrophy or due to a block in the branches of the bundle of His.
ST-Segment: It lies between the QRS complex and the T wave. It is the time during which all regions of the ventricles are completely depolarised and reflects the long plateau phase before repolarisation. In the heart muscle, prolonged depolarization is due to retardation of K+ efflux and is responsible for the plateau. The ST segment lasts for 0.09 sec.
T wave (ventricular depolarisation): It represents ventricular depolarization. The duration of the T wave is longer than the QRS complex because repolarisation takes place simultaneously throughout the ventricular depolarisation.

Question 34.
Show the double circulation by way of the diagrammatic sketch.
Answer:

Question 35.
In what way systemic circulation is different from pulmonary circulation incomplete double blood circulation?
Answer:
In the systemic circulation, the oxygenated blood entering the aorta from the left ventricle is carried by a network of arteries, arterioles and capillaries to the tissues. The deoxygenated blood from the tissue is collected by venules, veins and vena cava and emptied into the right atrium. In pulmonary circulation, the blood from the heart (right ventricle) is taken to the lungs by the pulmonary artery and the oxygenated blood from the lungs is emptied into the left auricle by the pulmonary vein.

Question 36.
How cardiac activity is regulated?
Answer:
The type of heart in human is myogenic because the heartbeat originates from the muscles of the heart. The nervous and endocrine systems work together with paracrine signals (metabolic activity) to influence the diameter of the arterioles and alter the blood flow. Neuronal control is achieved through the autonomic nervous system (sympathetic and parasympathetic). Sympathetic neurons release norepinephrine and the adrenal medulla releases epinephrine. The two hormones bind to p – adrenergic receptors and increase the heart rate. The parasympathetic neurons secrete acetylcholine that binds to muscarinic receptors and decreases the heartbeat.

Question 37.
Write about the following disorders of the circulatory system.
(a) Hypertension
(b) Coronary heart disease.
(c) Angina pectoris.
(d) Stroke
(e) Myocardial infarction.
(f) Rheumatoid heart disease.
Answer:
Hypertension is the most common circulatory disease. The normal blood pressure in man is 120/80 mm Hg. In cases when the diastolic pressure exceeds 90 mm Hg and the systolic pressure exceeds 150 mm Hg persistently, the condition is called hypertension. Uncontrolled hypertension may damage the heart, brain and kidneys.
Coronary heart disease occurs when the arteries are lined by atheroma. The build-up of atheroma contains cholesterol, fibres, dead muscle and platelets and is termed Atherosclerosis. The cholesterol-rich atheroma forms plaques in the inner lining of the arteries making them less elastic and reduces the blood flow. Plaque grows within the artery and tends to form blood clots, forming coronary thrombus. A thrombus in a coronary artery results in a heart attack.
Angina pectoris (ischemic pain in the heart muscles) is experienced during the early stages of coronary heart disease. Atheroma may partially block the coronary artery and reduce the blood supply to the heart. As a result, there is tightness or choking with difficulty in breathing. This leads to angina or chest pain. Usually, it lasts for a short duration of time.
Stroke is a condition when the blood vessels in the brain bursts, (Brain haemorrhage) or when there is a block in the artery that supplies the brain, (atherosclerosis) or thrombus. The part of the brain tissue that is supplied by this damaged artery dies due to a lack of oxygen (cerebral infarction).
Myocardial infarction (Heart failure): The prime defect in heart failure is a decrease in cardiac muscle contractility. The Frank-Starling curve shifts downwards and towards the right such that for a given EDV, a failing heart pumps out a smaller stroke volume than a normal healthy heart.
When the blood supply to the heart muscle or myocardium is remarkably reduced it leads to the death of the muscle fibres. This condition is called a heart attack or myocardial infarction. The blood clot or thrombosis blocks the blood supply to the heart and weakens the muscle fibres. It is also called Ischemic heart disease due to a lack of oxygen supply to the heart muscles. If this persists it leads to chest pain or angina. Prolonged angina leads to death of the heart muscle resulting in heart failure.
Rheumatoid Heart Disease: Rheumatic fever is an autoimmune disease that occurs 2-4 weeks after throat infection usually a streptococcal infection. The antibodies developed to combat the infection cause damage to the heart. Effects include fibrous nodules on the mitral valve, fibrosis of the connective tissue and accumulation of fluid in the pericardial cavity.

Question 38.
What is CPR? Which case is more useful?
Answer:
CPR is a life-saving procedure that is done at the time of emergency conditions such as when a person’s breath or heartbeat has stopped abruptly in case of drowning, electric shock or heart attack. CPR includes rescue of breath, which is achieved by mouth to mouth breathing, to deliver oxygen to the victim’s lungs by external chest compressions which helps to circulate blood to the vital organs. CPR must be performed within 4 to 6 minutes after cessation of breath to prevent brain damage or death.

Question 39.
How varicose veins are formed? Where are they usual located?
Answer:
Varicose veins The veins are so dilated that the valves prevent the backflow of blood. The veins lose their elasticity and become congested. Common sites are legs, rectoanal regions (haemorrhoids), the oesophagus and the spermatic cord.

Question 40.
What is embolism?
Answer:
Embolism is the obstruction of the blood vessel by an abnormal mass of materials such as a fragment of the blood clot, bone fragment or an air bubble. An embolus may lodge in the lungs, coronary artery or liver and leads to death.

Question 41.
What is an aneurysm? How much it is dangerous?
Answer:
Aneurysm the weakened regions of the wall of the artery or veins bulges to form a balloon-like sac. An unruptured aneurysm may exert pressure on the adjacent tissues or may burst to cause a massive haemorrhage.

Question 42.
Mention the three layers of blood vessels.
Answer:
The blood vessels in humans are composed of three layers, tunica intima, tunica media and tunica externa.

Choose the correct answer.

1. The average blood volume in an adult is:
(a) 5000 ml
(b) 50000 ml
(c) 50 ml
(d) 5000 l
Answer:
(a) 5000 ml

2. Liver receives oxygenated blood from the heart through:
(a) hepatic portal vein
(b) hepatic vein
(c) hepatic artery
(d) Renal artery
Answer:
(c) hepatic artery

3. ………… plasma protein is responsible for maintaining the osmotic pressure of blood.
(a) Globulin
(b) Prothrombin
(c) Fibrinogen
(d) Albumin
Answer:
(d) Albumin

4. ………… plasma proteins are involved in blood clotting.
(a) Albumin and prothrombin
(b) Globulin and albumin
(c) Prothrombin and fibrinogen
(d) Globulin and fibrinogen.
Answer:
(c) Prothrombin and fibrinogen

5. Healthy man has ………. millions of RBC in mm-3 of blood.
(a) 5.0-5.5
(b) 5.0-6
(c) 4.5-5.0
(d) 5.5-5.6
Answer:
(a) 5.0-5.5

6. Healthy women has ……….. millions of RBC in mm-3 of blood.
(a) 5.0 – 5.5
(b) 5.5 – 6.0
(c) 4.5 – 5.5
(d) 4.5 – 5.0
Answer:
(d) 4.5 – 5.0

7. Human RBC has ………… number of nucleus.
(a) 3
(b) 2
(c) 0
(d) 1
Answer:
(c) 0

8. The average life span of RBCs in a healthy individual is:
(a) 102 days
(b) 201 days
(c) 120 days
(d) 21 days.
Answer:
(c) 120 days

9. After the life span of RBCs are over they get destroyed in the:
(a) kidneys
(b) bone marrow
(c) thymus
(d) spleen
Answer:
(d) spleen

10. Differentiation of stem cells of the bone marrow to erythrocytes in adults is called:
(a) erythropoiesis
(b) apoptosis
(c) coagulation
(d) embolus
Answer:
(a) erythropoiesis

11. ………….. are the types of WBCs which are involved in inflammatory reactions.
(a) Neutrophils
(b) Basophils
(c) Eosinophils
(d) Monocytes
Answer:
(b) Basophils

12. The macrophages of the liver are called:
(a) microglia
(b) alveolar macrophages
(c) kupffer cells
(d) lymphocytes
Answer:
(c) kupffer cells

13. Blood normally contains ……….. platelets mm-3 of blood.
(a) 1,00,000 – 2,00,000
(b) 1,50,000 – 3,50,000
(c) 3,50,000 – 4,00,000
(d) 2,50,000 – 3,50,000
Answer:
(b) 1,50,000 – 3,50,000

14. The Rh factor protein is otherwise called ………… present on the surface of the red blood cells, in the majority of human.
(a) B – antigen
(b) D – antigen
(c) C – antigen
(d) A – antigen
Answer:
(b) D – antigen

15. The capillaries doesn’t posses the ………… layer of the blood vessel.
(a) tunica media
(b) tunica intima
(c) tunica Externa
(d) epicardium
Answer:
(a) tunica media

16. Complete double circulation is seen in ………….
(a) amphibians
(b) crocodiles
(c) fishes
(d) annelids
Answer:
(b) crocodiles

17. ………… valve of the heart is otherwise called mitral valve.
(a) Semilunar valves
(b) Tricuspid valves
(c) Bicuspid valves
(d) Atrial valve
Answer:
(c) Bicuspid valves

18. The heart sounds can be heard through …………
(a) electrocardiogram
(b) sphygmomanometer
(c) haemocytometer
(d) stethoscope
Answer:
(d) stethoscope

19. The amount of blood pumped out by each ventricle per minute is called:
(a) pulse pressure
(b) cardiac output
(c) stroke volume
(d) heart rate.
Answer:
(b) cardiac output

20. The volume of blood pumped out by one ventricle with each beat is called:
(a) Cardiac output
(b) Heartbeat
(c) Stroke volume
(d) systolic pressure
Answer:
(c) Stroke volume

21. Stroke volume depends on:
(a) Ventricular contraction
(b) Ventricular relaxation
(c) Atrial contraction
(d) Atrial relaxation
Answer:
(a) Ventricular contraction

22. Blood pressure of man is measured by using:
(a) Electrocardiogram
(b) Sphygmomanometer
(c) Cytometer
(d) Computed tomography
Answer:
(b) Sphygmomanometer

23. Normal blood pressure in man is:
(a) 120/80 mm Hg
(b) 130/70 mm Hg
(c) 120/60 mm Hg
(d) 150/60 mm Hg
Answer:
(a) 120/80 mm Hg

24. The QRS complex wave of ECG lasts for ………… seconds.
(a) 0.05 -0.1
(b) 0.06 – 0.09
(c) 0.8
(d) 0.12-0.21
Answer:
(b) 0.06 – 0.09

25. The blood clot inside the blood vessel is called:
(a) thrombus
(b) embolus
(c) atheroma
(d) plaques
Answer:
(b) embolus

26. Cells found in the lymphatics are called:
(a) basophils
(b) neutrophils
(c) lymphocytes
(d) monocytes
Answer:
(c) lymphocytes

27. Plasma without fibrinogen is called:
(a) Plasma protein
(b) Serum
(c) Fibrin
(d) atheroma
Answer:
(b) Serum

28. The damaged tissues and dead leucocytes are removed from the body in the forms of:
(a) pus
(b) blood
(c) serum
(d) plasma
Answer:
(a) pus

29. Match:

Column – I	Column II
(i) Kupffer cells	(a) Heterophils
(ii) Neutrophils	(b) Ratio of RBC to blood plasma
(iii) Platelets	(c) Liver
(iv) White blood cells	(d) Thrombocytes
(v) Haematocrit	(e) Leucocytes

(a) (i)-(c); (ii)-(a); (iii)-(d); (iv)-(e); (v)-(b)
(b) (i)-(d); (ii)-(a); (iii)-(b); (iv)-(e); (v)-(c)
(c) (i)-(c); (ii)-(b); (iii)-(d); (iv)-(a); (v)-(e)
(d) (i)-(a); (ii)-(c); (iii)-(e); (iv)-(b); (v)-(d)
Answer:
(a) (i)-(c); (ii)-(a); (iii)-(d); (iv)-(e); (v)-(b)

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 6 Respiration which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 6 Respiration

Answer the following.

Question 1.
In what way the process of respiration is related with the process of release of energy from food?
Answer:
The process of breathing is connected to the process of release of energy from food. Oxygen is utilized by the organisms to breakdown the biomolecules like glucose and to derive energy. During this breakdown carbondioxide, which is a harmful gas is also released. It is very obvious that oxygen has to be provided to cells continuously and the CO₂ to be released immediately by the cells. So the need of a respiratory system is essential for life.

Question 2.
Define respiration.
Answer:
The term respiration refers to the exchange of oxygen and carbondioxide between environment and cells of our body where organic nutrients are broken down enzymatically to release energy.

Question 3.
Discuss the five primary functions of the respiratory system.
Answer:
The five primary functions of the respiratory system are:

1. To exchange O₂ and CO₂ between the atmosphere and the blood.
2. To maintain homeostatic regulation of body pH.
3. To protect us from inhaled pathogens and pollutants.
4. To maintain the vocal cords for normal communication (vocalization).
5. To remove the heat produced during cellular respiration through breathing.

Question 4.
The rate of breathing in aquatic organisms is much faster than land animals. Why?
Answer:
Different animals have different organs for exchange of gases, depending upon their habitats and levels of organization. The amount of dissolved oxygen is very low in water compared to the amount of oxygen in the air. So the rate of breathing in aquatic organisms is much faster than land animals.

Question 5.
Write the respiratory organs of the following.
Answer:
Coelenterates, earthworm, insects, molluscs, fishes, reptiles, frog.

1. Coelenterates the body surface by simple diffusion.
2. Earthworms use their moist skin.
3. Insects have tracheal tubes.
4. Gills are used as respiratory organs in most of the aquatic Arthropods and Molluscs.
5. Fishes use gills.
6. Reptiles and mammals have well vascularised lungs.
7. Frogs use their moist skin for respiration along with the lungs.

Question 6.
Write the respiratory passage of human.
Answer:
The respiratory system includes the external nostrils, nasal cavity, the pharynx, the larynx, the trachea, the bronchi and bronchioles and the lungs which contain the alveoli.

Question 7.
What is the use of epithelial cells lining the trachea.
Answer:

1. The ciliated epithelial cells lining the trachea, bronchi and bronchioles secrete mucus.
2. The mucus membrane lining the airway contains goblet cells which secrete mucus, a slimy material rich in glycoprotein.
3. Microorganisms and dust particles attach in the mucus films and are carried upwards to pass down the gullet during normal swallowing

Question 8.
Name the three layers of alveoli.
Answer:
The diffusion membrane of the alveolus is made up of three layers – the thin squamous epithelial cells of the. alveoli, the endothelium of the alveolar capillaries and the basement substance found in between them.

Question 9.
Name the cells and their nature that are present in the alveoli.
Answer:
The thin squamous epithelial cells of the alveoli are composed of Type I and Type II cells. Type I cells are very thin so that gases can diffuse rapidly through them. Type II cells are thicker, synthesize and secrete a substance called Surfactant.

Question 10.
Write about the location of the lungs and its structure briefly.
Answer:

1. The lungs are light spongy tissues enclosed in the thoracic cavity surrounded by an airtight space.
2. The thoracic cavity is bound dorsally by the vertebral column and ventrally by the sternum, laterally by the ribs and on the lower side by the dome-shaped diaphragm.
3. The lungs are covered by double-walled pleural membrane containing several layers of elastic connective tissues and capillaries, which encloses the pleural fluid. Pleural fluid reduces friction when the lungs expand and contract.

Question 11.
What are the characteristic features of the respiratory surface?
Answer:
Characteristic features of respiratory surface:

1. The surface area must be very large and richly supplied with blood vessels.
2. Should be extremely thin and kept moist.
3. Should be in direct contact with the environment.
4. Should be permeable to respiratory gases.

Question 12.
Write the steps involved in respiration.
Answer:
The steps involved in respiration are

1. The exchange of air between the atmosphere and the lungs.
2. The exchange of O₂ and CO₂ between the lungs and the blood.
3. Transport of O₂ and CO₂ by the blood.
4. Exchange of gases between the blood and the cells.
5. Uptake of O₂ by the cells for various activities and the release of CO₂.

Question 13.
What is meant by ‘SURFACTANT’?
Answer:
Surfactants are thin non-cellular films made of protein and phospholipids covering the alveolar membrane. The surfactant lowers the surface tension in the alveoli and prevents the lungs from collapsing. It also prevents pulmonary oedema.

Question 14.
What is ‘NRDS’?
Answer:
Premature Babies have low levels of surfactant in the alveoli may develop the new bom respiratory distress syndrome (NRDS) because the synthesis of surfactants begins only after the 25th week of gestation.

Question 15.
What is breathing? What are the steps involved?
Answer:
The movement of air between the atmosphere and the lungs is known as ventilation or breathing. Inspiration and expiration, are the two phases of breathing. Inspiration is the movement of atmospheric air into the lungs and expiration is the movement of alveolar air that diffuse out of the lungs.

Question 16.
What are the muscles involved in the action of the lungs?
Answer:
Lungs do not contain muscle fibres but expands and contracts by the movement of the ribs and diaphragm. The diaphragm is a sheet of tissue that separates the thorax from the abdomen.
In a relaxed state, the diaphragm is dome-shaped. Ribs are moved by the intercostal muscles. External and internal intercostal muscles found between the ribs and the diaphragm helps in creating pressure gradients.

Question 17.
How the difference of pressure gradient helps in inspiration?
Answer:
Inspiration occurs if the pressure inside the lungs (intrapulmonary pressure) is less than the atmospheric pressure likewise expiration takes place when the pressure within the lungs is higher than the atmospheric pressure.

Question 18.
What is inspiration?
Answer:
The increase in pulmonary volume and decrease in the intrapulmonary pressure forces the fresh air from outside to enter the air passages into the lungs to equalize the pressure. This process is called inspiration.

Question 19.
What is expiration?
Answer:
Relaxation of the diaphragm allows the diaphragm and sternum to return to its dome shape and the internal intercostal muscles contract, pulling the ribs downward reducing the thoracic volume and pulmonary volume. This results in an increase in the intrapulmonary pressure slightly above the atmospheric pressure causing the expulsion of air from the lungs. This process is called expiration.

Question 20.
Explain about the different respiratory volumes and capacities of lungs.
Answer:
Respiratory Volumes – Tidal Volume (TV): Tidal volume is the amount of air inspired or expired with each normal breath. It is approximately 500 mL., i.e. a normal human adult can inspire or expire approximately 6000 to 8000mL of air per minute. During vigorous exercise, the tidal volume is about 4-10 times higher.
Inspiratory Reserve volume (IRV): Additional volume of air a person can inspire by forceful inspiration is called Inspiratory Reserve Volume. The normal value is 2500 to 3000 mL.
Expiratory Reserve volume (ERV): Additional volume of air a person can forcefully exhale by forceful expiration is called Expiratory Reserve Volume. The normal value is 1000 -1100 mL.
Residual Volume (RV): The volume of air remaining in the lungs after a forceful expiration. It . is approximately 1100 – 1200 mL.
Respiratory capacities – Vital capacity (VC): The maximum volume of air that can be moved out during a single breath following a maximal inspiration. A person first inspires maximally then expires maximally.
VC = ERV + TV + IRV
Inspiratory capacity (IC): The total volume of air a person can inhale after normal expiration.
It includes tidal volume and inspiratory reserve volume. IC = TV + IRV.
Expiratory capacity (EC): The total volume of air a person can exhale after normal inspiration.
It includes tidal volume and expiratory reserve volume. EC = TV + ERV.
Total Lung Capacity (TLC): The total volume of air which the lungs can accommodate after forced inspiration is called Total Lung Capacity. This includes the vital capacity and the esidual volume. It is approximately 6000mL. TLC = VC + RV.
Minute Respiratory Volume: The amount of air that moves into the respiratory passage per minute is called minute respiratory volume.
Normal TV = 500mL; Normal respiratory rate = 12 times/minute. Therefore, minute respiratory volume = 6 Litres/minute (for a normal healthy man).

Question 21.
Define ‘Dead Space’.
Answer:
The inspired air never reaches the gas exchange areas but fills the respiratory passages where the exchange of gases does not occur. This air is called dead space. Dead space is not involved in gaseous exchange. It amounts to approximately 150mL.

Question 22.
Write in a tabular form giving information about the partial pressure of respiratory gases in different regions of the respiratory system and in tissues.
Answer:
Partial pressure of oxygen and carbon di oxide (in mmHg) in comparison to those gases in the atmosphere.

Question 23.
How the blood transports the oxygen to the tissues.
Answer:
Molecular oxygen is carried in blood in two ways: bound to haemoglobin within the red blood cells and dissolved in plasma. Oxygen is poorly soluble in water, so only 3% of the oxygen is transported in the dissolved form. 97% of oxygen binds with haemoglobin in a reversible manner to form oxyhaemoglobin (HbO₂). The rate at which haemoglobin binds with O₂ is regulated by the partial pressure of O₂. Each haemoglobin carries a maximum of four molecules of oxygen. In the alveoli high pO₂, low pCO₂, low temperature and less H⁺ concentration, favours the formation of oxyhaemoglobin, whereas in the tissues low pO₂, high pCO₂, high H⁺ and high temperature favours the dissociation of oxygen from oxyhaemoglobin.

Question 24.
What is a sigmoid curve and how is it obtained.
Answer:
A sigmoid curve (S-shaped) is obtained when the percentage saturation of haemoglobin with oxygen is plotted against pO₂. This curve is called the oxygen haemoglobin dissociation curve. This S-shaped curve has a steep slope for pO₂ values between 10 and 50mm Hg and then flattens between 70 and 100 mm Hg.

Question 25.
How blood transports CO₂ from the tissue cells.
Answer:
Blood transports CO₂ from the tissue cells to the lungs in three ways

1. Dissolved in plasma About 7 -10% of CO₂ is transported in a dissolved form in the plasma.
2. Bound to haemoglobin About 20 – 25% of dissolved CO₂ is bound and carried in the RBCs as carbaminohaemoglobin (Hb CO₂) CO₂ + Hb Hb CO₂
3. As bicarbonate ions in plasma about 70% of CO₂ is transported as bicarbonate ions.

Question 26.
Describe the process of transport of CO₂ carried out by haemoglobin.
Answer:

1. About 20 – 25% of dissolved CO₂ is bound — and carried in the RBCs as carbaminohaemoglobin (Hb CO₂).
2. This is influenced by pCO₂ and the degree of haemoglobin oxygenation. RBCs contain a high concentration of the enzyme, carbonic anhydrase, whereas small amounts of carbc’ric anhydrase is present in the plasma.
3. At the tissues, the pCO₂ is high due to catabolism and diffuses into the blood to form HCO₃^– and H⁺ ions.
4. When CO₂ diffuses into the RBCs, it combines with water forming carbonic acid (H₂CO₃) catalyzed by carbonic anhydrase.
5. Carbonic acid is unstable and dissociates into hydrogen and bicarbonate ions.
6. Carbonic anhydrase facilitates the reaction in both directions.
   
7. The HCO₃^– moves quickly from the RBCs into the plasma, where it is carried to the lungs. At the alveolar site where pCO₂ is low, the reaction is reversed leading to the formation of CO₂ and water. Thus CO₂ trapped as HCO₃^– at the tissue level it is transported to the alveoli and released out as CO₂.

Question 27.
Write the events that make differences between inspiration and expiration.
Answer:

Question 28.
Write short notes on nitrogen narcosis.
Answer:
When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume. This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation. This increase in blood nitrogen content can lead to a condition called nitrogen narcosis.

Question 29.
What is ‘bends’ and how it is risky in scuba divers.
Answer:
When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while still in the blood-forming bubbles. Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings. Decompression sickness is associated with pain in joints and muscles and neurological problems including a stroke. The risk of nitrogen narcosis and bends is common in scuba divers.

Question 30.
Define the following terms and write their symptoms also.

1. Asthma,
2. emphysema,
3. bronchitis,
4. pneumonia,
5. tuberculosis,
6. Occupational respiratory disorders.

Answer:

1. Asthma: It is characterized by narrowing and inflammation of bronchi and bronchioles and difficulty in breathing. Common allergens for asthma are dust, drugs, pollen grains, certain food items like fish, prawn and certain fruits etc.
2. Emphysema: Emphysema is chronic breathlessness caused by the gradual breakdown of the thin walls of the alveoli decreasing the total surface area of gaseous exchange, i.e., widening of the alveoli is called emphysema. The major cause of this disease is cigarette smoking, which reduces the respiratory surface of the alveolar walls.
3. Bronchitis: The bronchi when it gets inflated due to pollution smoke and cigarette smoking, causes bronchitis. The symptoms are cough, shortness of breath and sputum in the lungs.
4. Pneumonia: Inflammation of the lungs due to infection caused by bacteria or virus is called pneumonia. The common symptoms are sputum production, nasal congestion, shortness of breath, sore throat, etc.
5. Tuberculosis: Tuberculosis is caused by Mycobacterium tuberculae. This infection mainly occurs in the lungs and bones. The collection of fluid between the lungs and the chest wall is the main complication of this disease.
6. Occupational respiratory disorders: The disorders due to one’s occupation of working in industries like grinding or stone breaking, construction sites, cotton industries, etc. Dust produced affects the respiratory tracts. Long exposure can give rise to inflammation leading to fibrosis. Silicosis and asbestosis are occupational respiratory diseases resulting from inhalation of a particle of silica from sand grinding and asbestos into the respiratory tract. Workers, working in such industries must wear protective masks.

Question 31.
How smoking causes a series of effects in our organ system.
Answer:

1. Smoking is inhaling the smoke from burning tobacco.
2. There are thousands of known chemicals which include nicotine, tar, carbon monoxide, ammonia, sulphur dioxide and even small quantities of arsenic.
3. Carbon monoxide and nicotine damage the cardiovascular system and tar damages the gaseous exchange system.
4. Nicotine is the chemical that causes addiction and is a stimulant that makes the heart beat faster and the narrowing of blood vessels results in raised blood pressure and coronary heart diseases.
5. The presence of carbon monoxide reduces oxygen supply.
6. Lung cancer, cancer of the mouth and larynx is more common in smokers than non-smokers.
7. Smoking also causes cancer of the stomach, pancreas and bladder and lowers sperm count in men.
8. Smoking can cause lung diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis.
9. These two diseases along with asthma are often referred as Chronic Obstructive Pulmonary Disease (COPD).

Question 32.
What will happen when a person travels quickly from sea level to an elevation above 8000ft?
Answer:
When a person travels quickly from sea level to elevations above 8000ft, where the atmospheric pressure and partial pressure of oxygen are lowered, the individual responds with symptoms of acute mountain sickness (AMS) headache, shortness of breath, nausea and dizziness due to poor binding of O₂ with haemoglobin.

Answer the following.

1. The tracheal tube of a human is supported by:
(a) Bones
(b) Cartilaginous rings
(c) Tracheal filaments
(d) Epiglottis.
Answer:
(b) Cartilaginous rings

2 is used to measure the volume of air involved in breathing movement for clinical assessment.
(a) Sphygmomanometer
(b) Clinical thermometer
(c) Spirometer
(d) Haemocytometer.
Answer:
(c) Spirometer

3 . An average, healthy human breathes times/minute.
(a) 12-16
(b) 13-17
(c) 12-18
(d) 10-16
Answer:
(a) 12-16

4. During vigorous exercise, the tidal volume of each respiration is about this much times higher:
(a) 6 times
(b) 10-12 times
(c) 4-10 times
(d) more than 5 times
Answer:
(c) 4-10 times

5. Minute respiratory volume for a normal healthy man is:
(a) 12 liters/minute
(b) 150 ml/minute
(c) 5 liters/minute
(d) 6 liters/minute
Answer:
(d) 6 liters/minute

6. The partial pressure of O₂ in the oxygenated blood is:
(a) 104 mmHg
(b) 95 mmHg
(c) 159 mmHg
(d) 45 mmHg
Answer:
(b) 95 mmHg

7. Breakdown of thin wall of the alveoli will leads to:
(a) Emphysema
(b) Asthma
(c) Pneumonia
(d) Bronchitis
Answer:
(a) Emphysema

8. Tuberculosis is caused by:
(a) Vibrio cholerae
(b) Helicobacter pylori
(c) Mycobacterium tuberculae
(d) Mycoplasma
Answer:
(c) Mycobacterium tuberculae

9. The world TB day is:
(a) March 24th
(b) April 23rd
(c) October 12th
(d) March 12th
Answer:
(a) March 24th

10. When our brain sense the shortage of O₂, it sends a message to CNS to imbalance to 02 demand and trigger us to:
(a) Snore
(b) Hiccups
(c) Sneeze
(d) Yawn
Answer:
(d) Yawn

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 5 Digestion and Absorption which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 5 Digestion and Absorption

Answer the following.

Question 1.
What are the steps involved in the digestive process?
Answer:
The process of digestion involves intake of the food (Ingestion), breakdown of the food into micromolecules (Digestion), absorption of these molecules into the bloodstream (Absorption), the absorbed substances becoming components of cells (Assimilation) and elimination of the undigested substances (Egestion). The digestive system includes the alimentary canal and associated digestive glands.

Question 2.
Define the following terms: (i) Thecodont, (ii) Diphyodont, (iii) Heterodont.
Answer:

1. Each tooth is embedded in a socket in the jaw bone; this type of attachment is called the thecodont.
2. Human beings and many mammals form two sets of teeth during their lifetime, a set of 20 temporary milk teeth (deciduous teeth) which gets replaced by a set of 32 permanent teeth (adult teeth). This type of dentition is called diphyodont.
3. The permanent teeth are of four different types (heterodont), namely, Incisors (I) chisel-like cutting teeth, Canines (C) dagger-shaped tearing teeth, Premolars (PM) for grinding, and Molars (M) for grinding and crushing.

Question 3.
What is plaque? If it persists for a long time what will happen? What are the symptoms of it?
Answer:
Mineral salts like calcium and magnesium are deposited on the teeth and form a hard layer of ‘tartar’ or calculus called plaque. If the plaque formed on teeth is not removed regularly, it would spread down the tooth into the narrow gap between the gums and enamel and causes inflammation, called gingivitis, which leads to redness and bleeding of the gums and to bad smell.

Question 4.
What is the position of the tongue inside the mouth?
Answer:
The tongue is a freely movable muscular organ attached at the posterior end by the frenulum to the floor of the buccal cavity and is free in the front. It acts as a universal toothbrush and helps in intake.

Question 5.
What are the muscles present in the upper and lower regions of the stomach? What are its uses?
Answer:
A cardiac sphincter (gastro oesphageal sphincter) regulates the opening of oesophagus into the stomach. The opening of the stomach into the duodenum is guarded by the pyloric sphincter. It periodically allows partially digested food to enter the duodenum and also prevents regurgitation of food.

Question 6.
What is GERD?
Answer:
If the cardiac sphincter does not contract properly during the churning action of the stomach the gastric juice with acid may flow back into the oesophagus and cause heart bum, resulting in GERD (Gastro Oesophagus Reflex Disorder).

Question 7.
How the stomach can accommodate a large meal?
Answer:
The inner wall of the stomach has many folds called gastric rugae which unfolds to accommodate a large meal.

Question 8.
How the structure of the small intestine is helping the process of absorption of food after the digestion of food?
Answer:
The ileal mucosa of the small intestine has numerous vascular projections called villi which are involved in the process of absorption and the cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance that increases the surface area enormously.

Question 9.
Give an account of the structure of the large intestine and its associated structures.
Answer:
The large intestine consists of the caecum, colon and rectum. The caecum is a small blind pouch-like structure that opens into the colon and it possesses a narrow finger-like tubular projection called the vermiform appendix. Both caecum and vermiform appendix are large in herbivorous animal and act as an important site for cellulose digestion with the help, of symbiotic bacteria. The colon is divided into four regions – an ascending, a transverse, a descending part and a sigmoid colon. The colon is lined by dilations called haustra (singular – haustrum). The “S” shaped sigmoid colon (pelvic colon) opens into the rectum. The rectum is concerned with the temporary storage of faeces. The rectum opens out through the anus. The anus is guarded by two anal sphincter muscles. The anal mucosa is folded into several vertical folds and contains arteries and veins called anal columns. The anal column may get enlarged and causes piles or haemorrhoids.

Question 10.
Short notes on the following: (i) Serosa, (ii) Muscularis, (iii) Sub-mucose, (iv) Muscosa layers of the alimentary canal.
Answer:
The wall of the alimentary canal from the oesophagus to the rectum consists of four layers namely serosa, muscularis, sub-mucosa and mucosa.

1. The serosa (visceral peritoneum) is the outermost layer and is made up of thin squamous epithelium with some connective tissues.
2. Muscularis is made of smooth circular and longitudinal muscle fibres with a network of nerve cells and parasympathetic nerve fibres which controls peristalsis.
3. The submucosal layer is formed of loose connective tissue containing nerves, blood, lymph vessels and the sympathetic nerve fibres that control the secretions of intestinal juice.
4. The innermost layer lining the lumen of the alimentary canal is the mucosa which secretes mucous.

Question 11.
Write about the brief account on salivary and gastric glands of the digestive system.
Answer:
The stomach wall has gastric glands that secrete gastric juice and the intestinal mucosa secretes intestinal juice.
Salivary glands: There are three pairs of salivary glands in the mouth. They are the largest parotids gland in the cheeks, the sub-maxillary/ sub-mandibular in the lower jaw and the sublingual beneath the tongue. These glands have ducts such as Stenson’s duct, Wharton’s duct and Bartholin’s duct or duct of Rivinis respectively. The salivary juice secreted by the salivary glands reaches the mouth through these ducts. The daily secretion of saliva from salivary glands ranges from 1000 to 1500 ml.
Gastric glands: The wall of the stomach is lined by gastric glands. Chief cells or peptic cells or zymogen cells in the gastric glands secrete gastric enzymes and Goblet cells secrete mucus. The Parietal or oxyntic cells secrete HCl and an intrinsic factor responsible for the absorption of Vitamin B12 called Castle’s intrinsic factor.

Question 12.
Describe the structure of the liver, which is considered the largest gland in our body with a neat sketch.
Answer:
Liver: The liver, the largest gland in our body is situated in the upper right side of the abdominal cavity, just below the diaphragm. The liver consists of two major left and right lobes; and two minor lobes. These lobes are connected with the diaphragm. Each lobe has many hepatic lobules (functional unit of the liver) and is covered by a thin connective tissue sheath called the Glisson’s capsule. Liver cells (hepatic cells) secrete bile which is stored and concentrated in a thin muscular sac called the gall bladder. The duct of the gall bladder (cystic duct) along with the hepatic duct from the liver forms the common bile duct. The bile duct passes downwards and joins with the main pancreatic duct to form a common duct called the hepato-pancreatic duct. The opening of the hepato-pancreatic duct into the duodenum is guarded by a sphincter called the power of regeneration and liver cells are replaced by new ones every 3-4 weeks.

Question 13.
Apart from bile secretion, what are the other functions carried out by liver?
Answer:

1. Destroys ageing and defective blood cells.
2. Stores glucose in the form of glycogen or disperses glucose into the bloodstream with the help of pancreatic hormones.
3. Stores fat-soluble vitamins and iron.
4. Detoxifies toxic substances.
5. Involves in the synthesis of non-essential amino acids and urea.

Question 14.
What is the dual role performed by the Pancreas?
Answer:
The second-largest gland in the digestive system is the Pancreas, which is a yellow7 coloured, compound elongated organ consisting of exocrine and endocrine cells. It is situated between the limbs of the ‘U’ shaped duodenum. The exocrine portion secretes pancreatic juice containing enzymes such as pancreatic amylase, trypsin and pancreatic lipase and the endocrine part called Islets of Langerhans secretes hormones such as insulin and glucagon. The pancreatic duct directly opens into the duodenum.

Question 15.
Define the term digestion.
Answer:
The process of digestion converts solid food into absorbable and assimilable forms. This is accomplished by mechanical and chemical processes.

Question 16.
Name the types of Salivary glands.
Answer:
There are three pairs of salivary glands in the mouth. They are the largest parotids gland in the cheeks, the sub-maxillary/sub-mandibular in the lower jaw and the sublingual beneath the tongue.

Question 17.
State the functions of Salivary amylase on food.
Answer:
The mucus in saliva prepares the food for swallowing by moistening, softening, lubricating and adhering the masticated food into a bolus. About 30 per cent of polysaccharide, starch is hydrolyzed by the salivary amylase enzyme into disaccharides (maltose).

Question 18.
Define the term ‘Chyme’.
Answer:
Food remains in the stomach for 4 to 5 hours, the rhythmic peristaltic movement chums and mixes the food with gastric juice and make it into a creamy liquid called chyme.

Question 19.
Mention the names of the enzymes in the stomach.
Answer:
Inactive pepsinogen, Rennin

Question 20.
What is the role of HCl in the stomach?
Answer:
The HCl provides an acidic medium (pH 1.8) which is optimum for pepsin, kills bacteria and other harmful organisms and avoids putrefaction.

Question 21.
What is the function of mucus present in gastric juice?
Answer:
The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from the eroding nature of the highly acidic HCl.

Question 22.
Mention the names of Pancreatic enzymes.
Answer:
Bicarbonate neutralizes stomach acid. Trypsin and chymotrypsin digest proteins. Carboxypeptidase digests peptides. Amylase digests starch and glycogen. Lipase digests lipids. Nuclease digests nucleic acids.

Question 23.
Describe the role of intestinal juice in the digestive process.
Answer:
The enzymes in the intestinal juice such as maltase, lactase, sucrase (invertase), dipeptidases, lipases, nucleosidases act on the breakdown products of bile and pancreatic digestion.

Question 24.
Write a short note on Absorption and Assimilation.
Answer:
Absorption is a process by which the end product of digestion passes through the intestinal mucosa into the blood and lymph. The villi in the lumen of the ileum are the absorbing units, consisting of a lacteal duct in the middle surrounded by fine network of blood capillaries. The process of absorption involves active, passive and facilitated transport. Small amounts of glucose, amino acids and electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon concentration gradients. However, some of the substances like fructose are absorbed with the help of the carrier ions like Na+. This mechanism is called facilitated transport.
Nutrients like amino acids, glucose and electrolytes like Na+ are absorbed into the blood against the concentration gradient by active transport. The insoluble substances like fatty acids, glycerol and fat-soluble vitamins are first incorporated into small, spherical water soluble droplets called micelles and are absorbed into the intestinal mucosa where they are re-synthesized into protein-coated fat globules called chylomicrons which are then transported into the lacteals within the intestinal villi and eventually empty into the lymphatic duct. The lymphatic ducts ultimately release the absorbed substances into the bloodstream. While the fatty acids are absorbed by the lymph duct, other materials are absorbed either actively or passively by the capillaries of the villi. Water-soluble vitamins are absorbed by simple diffusion or active transport. The transport of water depends upon the osmotic gradient.

Question 25.
Write a note on Egestion.
Answer:
The digestive waste and unabsorbed substances in the ileum enter into the large intestine and it mostly contains a fibre called roughage. The roughage is utilized by symbiotic bacteria in the large intestine for the production of substances like vitamin K and other metabolites. All these substances are absorbed in the colon along with water. The waste is then solidified into faecal matter in the rectum. The faecal matter initiates a neural reflex causing an urge or desire for its removal. The egestion of faeces through the anal opening is called defaecation. It is a voluntary process and is carried out by a peristaltic movement.

Question 26.
What is the energy value of carbohydrate and their sources?
Answer:
Carbohydrates are sugar and starch. These are the major source of cellular fuel which provides energy. The caloric value of carbohydrate is 4.1 calories per gram and its physiological fuel value is 4 Kcal per gram.

Question 27.
Name the deficiency diseases of protein.
Answer:
Marasmus and Kwashiorkor.

Question 28.
Write a short note on the following:

1. Indigestion,
2. Constipation,
3. Vomiting,
4. Jaundice,
5. Liver cirrhosis,
6. Gall stones,
7. Appendicitis,
8. Hiatus hernia.

Answer:

1. Indigestion: It is a digestive disorder in which the food is not properly digested leading to a feeling of fullness of the stomach. It may be due to inadequate enzyme secretion, anxiety, food poisoning, overeating and spicy food.
2. Constipation: In this condition, the faeces are retained within the rectum because of irregular bowel movement due to poor intake of fibre in the diet and lack of physical activities.
3. Vomiting: It is reverse peristalsis. Harmful substances and contaminated food from the stomach are ejected through the mouth. This action is controlled by the vomiting centre located in the medulla oblongata. A feeling of nausea precedes vomiting.
4. Jaundice: It is the condition in which the liver is affected and the defective liver fails to break down haemoglobin and to remove bile pigments from the blood. Deposition of these pigments changes the colour of the eye and skin yellow. Sometimes, jaundice is caused due to hepatitis viral infections.
5. Liver cirrhosis: Chronic disease of the liver results in degeneration and destruction of liver cells resulting in an abnormal blood vessel and bile duct leading to the formation of fibrosis.
   It is also called deserted liver or scarred liver. It is caused due to infection, consumption of poison, malnutrition and alcoholism.
6. Gall Stones: Any alteration in the composition of the bile can cause the formation of stones in the gall bladder. The stones are mostly formed of crystallized cholesterol in the bile. The gall stone causes obstruction in the cystic duct, hepatic duct and also hepato-pancreatic duct causing pain, jaundice and pancreatitis.
7. Appendicitis: It is the inflammation of the vermiform appendix, leading to severe abdominal pain. The treatment involves the removal of the appendix by surgery. If treatment is delayed the appendix may rupture and results in infection of the abdomen, called peritonitis.
8. Hiatus hernia (Diaphragmatic hernia): It is a structural abnormality in which the superior part of the stomach protrudes slightly above the diaphragm. In some people, injury or other damage may weaken muscle tissue, by applying too much pressure (repeatedly) on the muscles around the stomach while coughing, vomiting, and straining during bowel movement and lifting a heavy object. Heart bum is also common in those with a hiatus hernia. In this condition, stomach contents travel back into the oesophagus or even into oral cavity and cause pain in the centre of the chest due to the eroding nature of acidity.

Question 29.
What is diarrhoea? Why diarrhoea is caused? What are its symptoms? How it can be treated.
Answer:
Diarrhoea is the most common gastrointestinal disorder worldwide. It is sometimes caused by bacteria or viral infections through food or water. When the colon is infected, the lining of the intestine is damaged by the pathogens, thereby the colon is unable to absorb fluid. The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. Unless the condition is treated, dehydration can occur. Treatment is known as oral hydration therapy. This involves drinking plenty of fluids – sipping small amounts of water at a time to rehydrate the body.

Question 30.
What is called Obesity? How can it be assessed?
Answer:
Obesity is caused due to the storage of excess body fat in adipose tissue. It may induce hypertension, atherosclerotic heart disease and diabetes. Obesity may be genetic or due to excess intake of food, endocrine and metabolic disorders.

Choose the correct answer.

1. Each tooth is embedded in a socket in the jaw bone called:
(a) thecodont
(b) diphyolont
(c) heterodont
(d) both (b) & (c)
Answer:
(a) thecodont

2. Tartar is the deposited minerals of on the teeth.
(a) Sodium and Calcium
(b) Calcium and Magnesium
(c) Sodium and Potassium
(d) Magnesium and Sodium
Answer:
(b) Calcium and Magnesium

3. A cartilaginous flap that prevents the entry of food in to the glottis is called:
(a) Buccal cavity
(b) Gullet
(c) Epiglottis
(d) Calculus
Answer:
(b) Gullet

4. The inner wall of the stomach has many folds called ………. Which unfold to accommodate a large meal.
(a) Villi
(b) Sigmoid colono
(c) Haustra
(d) Gastric rugae
Answer:
(d) Gastric rugae

5. The mucal layer of ileum has lymphoid tissue known as
(a) Peyer’s patches
(b) Crypts of Leiberkuhn
(c) Zymogen cells
(d) Haemmorhoids
Answer:
(a) Peyer’s patches

6. The duct of the parotid gland is called:
(a) Wharton’s duct
(b) duct of Rivinis
(c) Stenson’s duct
(d) Bartholin’s duct
Answer:
(c) Stenson’s duct

7. The opening of the bile duct into the duodenum is guarded by a sphincter called:
(a) cardiac sphincter
(b) pyloric sphincter
(c) anal sphincter
(d) sphincter of oddi
Answer:
(d) sphincter of oddi

8. Masticated food particles passed into pharynx and then into the oesophagus by the process called:
(a) deglutition
(b) dentition
(c) mastication
(d) putrifaction
Answer:
(a) deglutition

9. Who is known as the “Father of gastric physiology.
(a) Aristotle
(b) William Beaumont
(c) Alexis
(d) A.C Guyton and Hall
Answer:
(b) William Beaumont

10. The functional unit of liver is:
(a) Goblet cells
(b) Hepatic lobules
(c) micelles
(d) gastric glands
Answer:
(b) Hepatic lobules

11. The caloric value of carbohydrate is:
(a) 9 k cal
(b) 5.65 k cal
(c) 4.1k cal
(d) 3.2 k cal
Answer:
(c) 4.1k cal

12. Peptic ulcer is caused by the bacterium called:
(a) Streptococcus
(b) Vibrio cholerae
(c) Lactobacillus
(d) Helicobacter pylori
Answer:
(d) Helicobacter pylori

13. A normal BMI range for adult is:
(a) 19-25
(b) 18-26
(c) 19-28
(d) 17-25
Answer:
(a) 19-25

14. Match the following:

(i) ICMR	(a) Gastro Oesophagus Reflex Disorder
(ii) BMI	(b) World Health Organization
(iii) GERD	(c) Indian Council of Medical Research
(iv) WHO	(d) Body Mass Index

(a) (i)-(c); (ii)-(d); (iii)-(a); (iv)-(b)
(b) (i)-(a); (ii)-(b); (iii)-(d); (iv)-(c)
(c) (i)-(b); (ii)-(a); (iii)-(c); (iv)-(d)
(d) (i)-(d); (ii)-(c); (iii)-(a); (iv)-(b)
Answer:
(a) (i)-(c); (ii)-(d); (iii)-(a); (iv)-(b)

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 4 Organ and Organ Systems in Animals which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 4 Organ and Organ Systems in Animals

Answer the following.

Question 1.
Define the following: (i) Morphology, (ii) Anatomy.
Answer:

1. Morphology: Morphology refers to the study of form or externally visible features.
2. Anatomy: The word anatomy is used for the study of internal organs in animals.

Question 2.
Name some common Indian earthworms.
Answer:
The common Indian earthworms are Lampito mauritii (Syn. Megascolex mauritii), Perioynx excavatus and Metaphire posthuma.

Question 3.
Write down the classification of earthworms based on their ecological strategies.
Answer:
Earthworms are also conveniently classified based on their ecological strategies as epigeics, anecics and endogeics. Epigeics are surface dwellers, eg. Perionyx excavatus and Eudrilus eugeniae. Anecics are found in upper layers of the soil, eg. Lampito mauritii, Lumbricus Terrestris. Endogeics are found in deeper layers of the soil.
eg. Octochaetona thurstoni.

Question 4.
Describe the morphology of earthworms with a neat diagram.
Answer:

Lampito mauritii is commonly found in Tamil Nadu. It has a long and cylindrical narrow body that is bilaterally symmetrical. L. mauritii is 80 to 210 mm in length with a diameter of 3.5-5 mm, and is light brown in colour, with purplish tinge at the anterior end. This colour of the earthworm is mainly due to the presence of porphyrin pigment. The body of the earthworm is encircled by a large number of grooves which divides it into a number of compartments called segments or metameres. L. mauritii consists of about 165 – 190 segments. The dorsal surface of the body is marked by a dark mid-dorsal line (dorsal blood vessel) along the longitudinal axis of the body. The ventral surface is distinguished by the presence of genital openings. The mouth is found in the centre of the first segment of the body, called the peristomium. Overhanging the mouth is a small flap called the upper lip or prostomium. The last segment has the anus called the pygidium. In mature worms, segments 14 to 17 may be found swollen with a glandular thickening of the skin called the clitellum. This helps in the formation of the cocoon. Due to the presence of clitellum, the body of an earthworm is divided into pre clitellar region (1st – 13th segments), clitellar region (14th – 17th segments) and the post-clitellar region (after the 17th
segment). In all the segments of the body except the first, last and clitellum, there is a ring of chitinous body setae. This body setae arises from a setigerous sac of the skin and it is curved as S-shaped. Setae can be protruded or retracted and their principal role is in locomotion.

Question 5.
What is metameric segmentation in earthworm?
Answer:
The body of the earthworm is encircled by a large number of grooves which divides it into a number of compartments called segments or metameres. This is called metameric segmentation

Question 6.
Write the external apertures found in the body of the earthworm with a neat sketch.
Answer:
The external apertures are the mouth, anus, dorsal pores, spermathecal openings, genital openings and nephridiopores. The dorsal pores are present from the 10thsegment onwards. The coelomic fluid communicates to the exterior through these pores and keeps the body surface moist and free from harmful microorganisms. Spermathecal openings are three pairs of small ventrolateral apertures lying intersegmentally between the grooves of the segments 6/7, 7/8 and 8/9. The female genital aperture lies on the ventral side in the 14thsegment and a pair of male genital apertures are situated latero-ventrally in the 18thsegment.

Nephridiopores are numerous and found throughout the body of the earthworm except for a few anterior segments, through which the metabolic wastes are eliminated.

Question 7.
Differentiate pre-clitellar region and post-cliteliar region.
Answer:
Due to the presence of clitellum in the 14-17 segment, the body of an earthworm is divided into pre clitellar region (1st – 13th segments), clitellar region (14th – 17th segments) and the post-clitellar region (after the 17th segment). In all the segments of the body except the first, last and clitellum, there is a ring of chitinous body setae.

Question 8.
Write down any ten points to make morphological and anatomical differences between Lampito mauritti and Metaphire posthuma.
Answer:

Question 9.
Describe the anatomy of the earthworm.
Answer:
The body wall of the earthworm is very moist, thin, soft, skinny, elastic and consists of the cuticle, epidermis, muscles and coelomic epithelium. The epidermis consists of supporting cells, gland cells, basal cells and sensory cells. A spacious body cavity called the coelom is seen between the alimentary canal and the body wall. The coelom contains the coelomic fluid and serves as a hydrostatic skeleton, in which the coelomocytes are known to play a major role in regeneration, immunity and wound healing. The coelomic fluid of the earthworm is milky, which consists of granulocytes or eleocytes, amoebocytes mucocytes and alkaline, leucocytes.

Question 10.
Write about the digestive system of earthworm with neat diagram.
Answer:
The digestive system of the earthworm consists of the alimentary canal and the digestive glands. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus. The mouth opens into the buccal cavity which occupies the 1st and 2nd segments. The buccal cavity leads into a thick muscular pharynx, which occupies the 3rd and 4th segments and is surrounded by the pharyngeal glands.

A small narrow tube, the oesophagus lies in the 5th segment and continues into a muscular gizzard in the 6th segment. The gizzard helps in the grinding of soil particles and decaying leaves. The intestine starts from the 7th segment and continues till the last segment. The dorsal wall of the intestine is folded into the cavity as the typhlosole. This fold contains blood vessels and increases the absorptive area of the intestine. The inner epithelium consists of columnar cells and glandular cells. The alimentary canal opens to the exterior through the anus.

Question 11.
How is cutaneous respiration made possible in earthworm?
Answer:
The earthworm has no special respiratory organs like lungs or gills. Respiration takes place through the body wall. The outer surface of the skin is richly supplied with blood capillaries which aid in the diffusion of gases. Oxygen diffuses through the skin into the blood while carbon dioxide from the blood diffuses out. The skin is kept moist by mucous and coelomic fluid and facilitates the exchange of gases.

Question 12.
Write an account on the circulatory system of Lampito mauritii.
Answer:
Lampito mauritii exhibits a closed type of blood vascular system consisting of blood vessels, capillaries and lateral hearts. Two median longitudinal vessels run above and below the alimentary canal as dorsal and ventral vessels of the earthworm. There are paired valves in the dorsal vessels which prevent the backward flow of the blood. The ventral vessel has no valves and is non-contractile, allowing the backward flow of blood. In the anterior part of the body, the dorsal vessel is connected with the ventral vessel by eight pairs of commissural vessels or the lateral hearts lying in the 6thto 13thsegments. These vessels run on either side of the alimentary canal and pump blood from the dorsal vessel to the ventral vessel. The dorsal vessel receives blood from various organs in the body. The ventral vessel supplies blood to the various organs. Blood glands are present in the anterior segments of the earthworm. They produce blood cells and haemoglobin which is dissolved in the plasma and gives a red colour to the blood.

Question 13.
Give a short note on the nervous system of the earthworm.
Answer:
The bilobed mass of nervous tissue called supra-pharyngeal ganglia, which lies on the dorsal wall of the pharynx in the 3rd segment, is referred to as the “brain”. The ganglion found below the pharynx in the 4thsegment is called the sub-pharyngeal ganglion. The brain and the sub -pharyngeal ganglia are connected by a pair of circum-pharyngeal connectives. They run one on each side of the pharynx. Thus a nerve ring is formed around the anterior region of the alimentary canal. The double ventral nerve cord runs backwards from the sub – pharyngeal ganglion. The brain along with other nerves in the ring integrates sensory inputs and command muscular responses of the body.

Question 14.
Write about the type of receptors found in earthworm.
Answer:
The earthworm’s receptors are stimulated by a group of slender columnar cells connected with nerves. The Photoreceptors (sense of light) are found on the dorsal surface of the body. Gustatory (sense of taste) and olfactory receptors (sense of smell) are found in the buccal cavity. Tactile receptors (sense of touch), chemoreceptors (detect chemical changes) and thermoreceptors (changes in, temperature) are present in the prostomium and the body wall.

Question 15.
Describe the nephridial system of the common earthworm, and write how it is helpful in removing the waste materials.
Answer:
Excretion is the process of elimination of metabolic waste products from the body. In earthworm, excretion is affected by segmentally arranged, minute coiled, paired tubules called nephridia. There are three types of nephridia:

1. Pharyngeal or tufted nephridia – present as paired tufts in the 5th 9th segments.
2. Micronephridia or Integumentary nephridia – attached to the lining of the body wall from the 14th segment to the last which open on the body surface.
3. Meganephridia or septal nephridia – present as pair on both sides of intersegmental septa of the 19th segment to the last and open into the intestine. The meganephridium has an internal funnel-like opening called the nephrostome, which is fully ciliated.

The nephrostome consists of three distinct divisions, the ciliated, the glandular and the muscular region. The waste material collected through the ciliated funnel is pushed into the muscular part of nephridium by the ciliated region. The glandular part extracts the waste from the blood and finally the wastes exit out through the nephridiopore.
Besides nephridia, special cells on the coelomic wall of the intestine, called chloragogen cells are present. They extract the nitrogenous waste from the blood of the intestinal wall, into the body cavity to be sent out through the nephridia.

Question 16.
Write an account of the reproductive system of earthworm.
Answer:
Earthworms are hermaphrodites or monoecious i.e., male and female reproductive organs are found in the same individual.
In the male reproductive system, two pairs of testes are present in the 10th and 11th segments. The testes give rise to the germ cells or spermatogonia, which develops into spermatozoa in the two pairs of seminal vesicles. Two pairs of seminal funnels called ciliary rosettes are situated in the same segments as the testes. The ciliated funnels of the same side are connected to a long tube called vas deferens. The vasa deferentia run-up to the 18th segment where they open to the exterior through the male genital aperture. The male genital aperture contains two pairs of penial setae for copulation. A pair of prostate glands lie in the 18th – 19th segments. The secretion of the prostate gland serves to cement the spermatozoa into bundles known as spermatophores.

Question 17.
What is protandrous?
Answer:
Earthworms are hermaphrodites or monoecious i.e. male and female reproductive organs are found in the same individual. Self-fertilization is avoided because two sex organs mature at different times, which means the sperm develops earlier than the production of ova to promote cross-fertilization. This phenomenon is called Protandrous.

Question 18.
How cocoon is formed in earthworms and what is the use it.
Answer:
Mature egg cells in the nutritive fluid are deposited in the cocoons produced by the gland cells of the clitellum which also collects the partner’s sperms from the spermthecae. Fertilization and development occur within the cocoons, which are deposited in the soil. After about 2-3 weeks, each cocoon produces baby earthworms.

Question 19.
Why earthworms are called “Friends of Farmer?”
Answer:
Earthworms are known as “friends of the farmer” because they make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots.

Question 20.
What is “VERMITEC”.
Answer:
Vermiculture, vermicomposting, vermiwash and wormery are interlinked and interdependent processes, collectively referred to as Vermitech.

Question 21.
Define “vermiculture.”
Answer:
Lampito mauritti helps in recycling dead and decayed plant material by feeding on them. Artificial rearing or cultivation of earthworms involves new technology for the betterment of human beings. This process is known as Vermiculture.

Question 22.
What is vermicomposting?
Answer:
Lampito mauritti helps in recycling dead and decayed plant material by feeding on them. Artificial rearing or cultivation of earthworms involves new technology for the betterment of human beings. This process is known as Vermiculture. The process of producing compost ‘ using earthworms is called Vermicomposting.

Question 23.
What is vermiwash.
Answer:
Vermiwash is liquid manure or plant tonic obtained from an earthworm. It is used as a foliar spray and helps to induce plant growth. It is a collection of excretory products and mucus secretion of earthworms along with micronutrients from the soil organic molecules.

Question 24.
What is “wormery” or “worm bin”.
Answer:
Earthworms can be used for recycling waste food, leaf, litter and biomass to prepare a good fertilizer in a container known as wormery or worm bin.

Question 25.
Describe the economic importance of earthworm.
Answer:
Earthworms are known as “friends of the farmer” because they make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots. Vermiculture, vermicomposting, vermiwash and wormery are interlinked and interdependent processes, collectively referred to as Vermitech. Lampito mauritti helps in recycling dead and decayed plant material by feeding on them. Artificial rearing or cultivation of earthworms involves new technology for the betterment of human beings. This process is known as Vermiculture. The process of producing compost using earthworms is called Vermicomposting. Vermiwash is liquid manure or plant tonic obtained from an earthworm. It is used as a foliar spray and helps to induce plant growth. It is a collection of excretory products and mucus secretion of earthworms along with micronutrients from the soil organic molecules. Earthworms can be used for recycling of waste food, leaf, litter and biomass to prepare a good fertilizer in a container known as wormery or worm bin. It makes superior compost than conventional composting methods: Earthworms are also used as bait in fishing.

Question 26.
Why cockroaches are called vectors.
Answer:
Cockroach carries with them harmful germs of various bacterial diseases like cholera, diarrhoea, tuberculosis, and typhoid and hence are known as “Vectors”.

Question 27.
Describe the morphology of Periploneta Americana.
Answer:
The body of the cockroach is compressed dorso-ventrally, bilaterally symmetrical, segmented and is divisible into three distinct regions – head, thorax and abdomen. The entire body is covered by a hard, brown coloured, chitinous exoskeleton. In each segment, the exoskeleton has hardened plates called sclerites. The sclerites of the dorsal side are called tergites, those on the ventral side are called stemites and those of lateral sides are called pleurites.

The head of the cockroach is small, triangular lies at a right angle to the longitudinal body axis, the mouthparts are directed downwards so it is hypognathous. It is formed by the fusion of six segments and shows great mobility in all directions due to a flexible neck. The head capsule bears a pair of large, sessile, and reniform compound eyes, a pair of antennae and appendages around the mouth. Antennae have sensory receptors that help in monitoring the environment. The appendages from the mouthparts are of biting and chewing type (Mandibulate or Orthopterus type). The mouthparts consist of a labrum (upper lip), a pair of mandibles, a pair of maxillae, a labium (lower lip) and a hypopharynx (tongue) or lingua. The thorax consists of three segments – Prothorax, Mesothorax and Metathorax. The prothoracic segment is the largest. The head is connected with the thorax by a short extension of the prothorax called the neck or cervicum. Each thoracic segment bears a pair of walking legs.
The cockroach has two pairs of wings, the first pair arises from the mesothorax and protects the hind wings when at rest, and is called elytra or tegmina. The second pair of wings arise from the metathorax and are used in flight, The abdomen in both male and female consists of 10 segments. Each segment is covered by the dorsal tergum, the ventral sternum and between ‘ them a narrow membranous pleuron on each side.
In males, the genital pouch lies at the hind end of the abdomen. It contains the dorsal anus and ventral male genital pore. In both sexes, genital apertures are surrounded by sclerites called gonapophysis. Male bears a pair of short and slender anal styles in the 9th sternum which is absent in the female. In both sexes, the 10th segment bears a pair of jointed filamentous structures called anal cerci and bears a sense organ that is receptive to vibrations in air and land.

Question 28.
Name the parts of the mouth in cockroaches.
Answer:
The mouthparts consist of a labrum (upper lip), a pair of mandibles, a pair of maxillae, a labium (lower lip) and an Irypopharynx (tongue) or lingua.

Question 29.
Write down the difference between male and female cockroaches.
Answer:

Character	Male cockroach	Female cockroach
Abdomen	Long and narrow.	Short and broad.
Segments	In the abdomen, nine segments are visible.	In the abdomen, seven segments are visible.
Anal styles	Present.	Absent.
Terga	7th tergum covers 8th tergum.	The 7th tergum covers the 8th and 9th terga.
Brood pouch	Absent.	Present.
Antenna	Longer in length.	Shorter in length.
Wings	Extends beyond the tip of the abdomen.	Extends up to the end of the abdomen.

Question 30.
Write about the description of the digestive system of cockroaches with a neat sketch.
Answer:

The digestive system of cockroach consists of the alimentary canal and digestive glands. The alimentary canal is present in the body cavity and is divided into three regions: foregut, midgut and hindgut. The foregut includes the preoral cavity, mouth, pharynx and oesophagus. This in turn opens into a sac-like structure called the crop which is used for storing food. The crop is followed by the gizzard or proventriculus which has an outer layer of thick circular muscles and thick inner cuticle forming six highly chitinous plates called “teeth”. Gizzard helps in the grinding of the food particles. The midgut is a short and narrow tube behind the gizzard and is glandular in nature. At the junctional region of the gizzard are eight fingers like tubular blind 100 – 150 yellow coloured thin filamentous malphigian tubules which are helpful in the removal of the excretory products from the haemolymph. The hindgut is broader than the midgut and is differentiated into the ileum, colon, and rectum. The rectum opens out through the anus.
The digestive glands of cockroach consist of the salivary glands, the glandular cells and hepatic caecae. A pair of salivary glands are found on either side of the crop in the thorax. The glandular cells of the midgut and hepatic or gastric caecae produce digestive juices.

Question 31.
What are the digestive glands of cockroach?
Answer:
The digestive glands of cockroach consist of the salivary glands, the glandular cells and hepatic caecae. A pair of salivary glands is found on either side of the crop in the thorax. The glandular cells of the midgut and hepatic or gastric caecae produce digestive juices.

Question 32.
Describe the circulatory system of Periplaneta americana.
Answer:
Periplaneta has an open type of circulatory system. Blood vessels are poorly developed and opens into the haemoeoel in which the blood or haemolymph flows freely. Visceral organs located in the haemoeoel are bathed in blood. The haemolyph is colourless and consists of plasma and haemocytes which are ‘phagocytic’ in nature. The heart is an elongated tube with a muscular wall lying mid dorsally beneath the thorax. The heart consists of 13 chambers with Ostia on either side. The blood from the sinuses enters the heart through the ostia and is pumped anteriorly to the sinuses again. The triangular muscles that are responsible for blood circulation in the cockroach are called alary muscles (13 pairs). One pair of these muscles is found in each segment on either side of the heart. In a cockroach, there is an accessory pulsatile vesicle at the base of each antenna which also pumps blood.

Question 33.
Write an account on the nervous system of cockroach.
Answer:

The nervous system of cockroach consists of a nerve ring and a ganglionated double ventral nerve cord, sub-oesophageal ganglion, circum-oesophageal connectives and double ventral nerve cord. The nerve ring is present around the oesophagus in the head capsule and is formed by the supra-oesophagial ganglion called the ‘brain’, The brain is mainly a sensory and an endocrine centre and lies above the oesophagus. A sub-oesophageal ganglion is the motor centre that controls the movements of the mouthparts, legs and wings. It lies below the oesophagus and formed by the fusion of the paired gangalia of mandibular, maxillary and labial segments of the head. A pair of circum-oesophageal connectives is present around the oesophagus, connecting the supra-oesophageal ganglia with the sub-oesophageal ganglion. The double ventral nerve cord is solid, ganglionated and arises from the sub-oesophageal ganglion and extends up to the 7th abdominal segment. Three thoracic ganglia are present, one in each thoracic segment and six abdominal ganglia in the abdomen.

Question 34.
What are the sense organs present in Periplaneta americana and write the position and function of each?
Answer:
In cockroach, the sense organs are antennae, compound eyes, labrum, maxillary palps, labial palps and anal cerci. The receptor for touch (thigmo receptors) is located in the antenna, maxillary palps and cerci. The receptor for smell (olfactory receptors) is found on the antennae. The receptor for taste (gustatory receptors) is found on the palps of the maxilla and labium. Thermoreceptors are found on the first four tarsal segments on the legs. The receptor chordotonal is found on the anal cerci which respond to air or earth borne vibrations. The photoreceptors of the cockroach consist of a pair of compound eyes at the dorsal surface of the head. Each eye is formed of about 2000 simple eyes called the ommatidia (singular: ommatidium), through which the cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity but less resolution.

Question 35.
How do the malpighian tubules of cockroaches do the function of excretion?
Answer:
The malpighian tubules are thin, long, filamentous, yellow coloured structures attached at the junction of the midgut and hindgut. These are about 100-150 in number and are present in 6-9 bundles. Each tubule is lined by glandular and ciliated cells and the waste is excreted out through the hindgut. The glandular cells of the malpighian tubules absorb water, salts, and nitrogenous wastes from the haemolymph and transfer them into the lumen of the tubules. The cells of the tubules reabsorb water and certain inorganic salts. By the contraction of the tubules, nitrogenous waste is pushed into the ileum, where more water is reabsorbed. It moves into the rectum and almost solid uric acid is excreted along with the faecal matter.

Question 36.
What are the organs that constitute the male reproductive system in Periplaneta Americana?
Answer:
The male reproductive system consists of a pair of testes, vasa deferentia, and ejaculatory duct, utricular gland, phallic gland and external genitalia.

Question 37.
What, are the female reproductive organs of cockroach.
Answer:
The female reproductive system of cockroach consists of a pair of ovaries, vagina, genital pouch, collaterial glands, spermathecae and external genitalia.

Question 38.
Write short note on ootheca of cockroach.
Answer:
Ootheca is a dark reddish to blackish brown capsule about 12mm long which contains nearly 16 fertilized eggs and dropped or glued to a suitable surface, usually in a crack or crevice of high relative humidity near a food source. On average, each female cockroach produces nearly 15-40 oothecae in its life span of about one to two years. The embryonic development occurs in the ootheca, which takes nearly 5-13 weeks. The development of cockroach is gradual through the nymphal stages (paurometabolous). The nymph resembles the adult and undergoes moulting. The nymph grows by moulting or ecdysis about 13 times to reach the adult form.

Question 39.
Write about the features showing the morphology of the frog.
Answer:
The body of a frog is streamlined to help in swimming. It is dorsoventrally flattened and is divisible into the head and trunk. The body is covered by smooth, slimy skin loosely attached to the body wall. The skin is dark green on the dorsal side and pale ventrally. The head is almost triangular in shape and has an apex that forms the snout. The mouth is at the anterior end and can open widely. External nostrils are present on the dorsal surface of the snout, one on each side of the median line. Eyes are large and project above the general surface of the body. They lie behind the external nostrils and are protected by a thin movable lower eyelid, thick immovable upper eyelid and a third transparent eyelid called the nictitating membrane. This membrane protects the eye when the frog is underwater. A pair of tympanic membranes form the eardrum behind the eyes on either side. Frogs have no external ears, neck and tail are absent. Trunk bears a pair of forelimbs and a pair of hind limbs. At the posterior end of the dorsal side, between the hind limbs is the cloacal aperture. This is the common opening for the digestive, excretory and reproductive systems.

Forelimbs are short, stumpy, and helps to bear the weight of the body. They are also helpful for the landing of the frog after leaping. Each forelimb consists of an upper arm, forearm and hand. Hand bears four digits. Hind limbs are large, long and consist of thigh, shank and foot. Foot bears five long webbed toes and one small spot called the sixth toe. These are adaptations for leaping and swimming. When the animal is at rest, the hind limbs are kept folded in the form of the letter Z Sexual dimorphism is exhibited clearly during the breeding season. The male frog has a pair of vocal sacs and a copulatory or nuptial pad on the ventral side of the first digit of each forelimb (Figure 4.16). Vocal sacs assist in amplifying the croaking sound of the frog. Vocal sacs and nuptial pads are absent in the female frogs.

Question 40.
Differentiate frogs from toads.
Answer:

Characters	Frog	Toad
Family	Ranidae	Bufonidae
Body shape	Slender	Bulkier
Legs	Longer	Shorter
Webbed feet	Present	Absent
Skin	Smooth and moist skin	Dry skin covered with wart-like glands.
Teeth	Maxillary and vomerine teeth.	Teeth absent.
Egg formation	Lays eggs in clusters.	Lays eggs in strings.

Question 41.
Write the anatomy of the digestive system in Rono hexadactyla.
Answer:
The Digestive System: The alimentary canal consists of the buccal cavity, pharynx, oesophagus, duodenum, ileum and the rectum which leads to the cloaca and opens outside by the cloacal aperture. The wide mouth opens into the buccal cavity. On the floor of the buccal cavity lies a large muscular sticky tongue. The tongue is attached in front and free behind. The free edge is forked. When the frog sights an insect it flicks out its tongue and the insect gets glued to the sticky tongue. The tongue is immediately withdrawn and the mouth closes. A row of small and pointed maxillary teeth is found on the inner region of the upper jaw. In addition vomerine. teeth are also present as two groups, one on each side of the internal nostrils. The lower jaw is devoid of teeth. The mouth opens into the buccal cavity that leads to the oesophagus through the pharynx. The oesophagus is a short tube that opens into the stomach and continues as the intestine, rectum and finally opens outside by the cloaca. The liver secretes bile which is stored in the gall bladder. The pancreas, a digestive gland produces pancreatic juice containing digestive enzymes.

Question 42.
How digestion process takes place in common Indian frog.
Answer:
Food is captured by the bilobed tongue. Digestion of food takes place by the action of Hydrochloric acid and gastric juices secreted from the walls of the stomach. Partially digested food called chyme is passed from the stomach to the first part of the intestine, the duodenum. The duodenum receives bile from the gall bladder and pancreatic juices from the pancreas through a common bile duct. Bile emulsifies fat and pancreatic juices digest carbohydrates, proteins and lipids. Final digestion takes place in the intestine. Digested food is absorbed by the numerous finger-like folds in the inner wall of the intestine called villi and microvilli. The undigested solid waste moves into the rectum and passes out through the cloaca.

Question 43.
Describe the different blood vessels associated with the heart and its supply of blood to a different region in Rana hexadactyla.
Answer:
The Blood-Vascular System- The blood vascular system consists of a heart with three chambers, blood vessels and blood. The heart is covered by a double-walled membrane called the pericardium. There are two thin-walled anterior chambers called auricles (Atria) and a single thick-walled posterior chamber called the ventricle. Sinus venosus is a large, thin-walled, triangular chamber, which is present on the dorsal side of the heart. Truncus arteriosus is a thick-walled and cylindrical structure that is obliquely placed on the ventral surface of the heart. It arises from the ventricle and divides into the right and left aortic trunk, which is further divided into three aortic arches namely carotid, systemic and pulmo-cutaneous. The Carotid trunk supplies blood to the anterior region of the body. The Systemic trunk of each side is joined posteriorly to form the dorsal aorta. They supply blood to the posterior part of the body. Pulmo-cutaneous trunk supplies blood to the lungs and skin. Sinus venosus receives the deoxygenated blood from the body parts by two anterior precaval veins and one post caval vein. It delivers the blood to the right auricle; at the same time, the left auricle receives oxygenated blood through the pulmonary vein. Renal portal and hepatic portal systems are seen in the frog.

Question 44.
What are the three types of nervous system seen in frog?
Answer:
The Nervous system of the frog is divided into the Central Nervous System [CNS], the Peripheral Nervous System [PNS], and the Autonomous Nervous System [ANS].

Question 45.
Give a short note on the structure of the brain in Rana hexadactyla.
Answer:
The brain is situated in the cranial cavity and covered by two meninges called piamater and duramater. The brain is divided into forebrain, midbrain and hindbrain. Forebrain (Prosencephalon) is the hindbrain (Rhombencephalon) consists of the cerebellum and medulla oblongata. The cerebellum is a narrow, thin transverse band followed by medulla oblongata. The medulla oblongata passes out through the foramen magnum and continues as the spinal cord, which is enclosed in the vertebral column.

Question 46.
In frogs how the elimination of nitrogenous waste takes place with the help of excretory organs.
Answer:
Elimination of nitrogenous waste and salt and water balance is performed by a well developed excretory system. It consists of a pair of kidneys, ureters, urinary bladder and cloaca. Kidneys are dark red, long, flat organs situated on either side of the vertebral column in the body cavity. Kidneys are Mesonephric. Several nephrons are found in each kidney. They separate nitrogenous waste from the blood and excrete urea, so frogs are called ureotelic organisms.
A pair of ureters emerges from the kidneys and opens into the cloaca. A thin-walled unpaired urinary bladder is present ventral to the rectum and opens into the cloaca.

Question 47.
Write shortly about the metamorphosis process in the frog.
Answer:
In frogs within few days of fertilization, the eggs hatch into tadpoles. A newly hatched tadpole lives off the yolk stored in its body. It gradually grows larger and develops three pairs of gills.
The tadpole grows and metamorphosis into an air-breathing carnivorous adult frog. Legs grow from the body, and the tail and gills disappear. The mouth broadens, developing teeth and jaws, and the lungs become functional.

Question 48.
In what way the frogs are economically important.
Answer:
Economic importance of Frog:

1. The frog is an important animal in the food chain; it helps to maintain our ecosystem. So ‘frogs should be protected’.
2. Frog is beneficial to man since they feed on insects and helps in reducing the insect pest population.
3. Frogs are used in traditional medicine for controlling blood pressure and for their anti-ageing properties.
4. In the USA, Japan, China and the North East of India, frogs are consumed as delicious food as they have high nutritive value.

Question 49.
Which is the longest species of earthworm in South India and in Africa?
Answer:
In South India – Drawida nilamburansis.
In Africa – Microchaetus rappi.

Question 50.
How do the earthworms crawl?
Answer:
The earthworms normally crawl with the help of their body muscles, setae, and buccal chamber.
The outer circular and inner longitudinal muscle layers lies below the epidermis of the body wall. The contraction of circular muscles makes the body long and narrow, while that of the longitudinal muscle makes the body short and broad. The locomotion of the earthworm is brought about by the contraction and relaxation of the muscular body wall and is aided by the turgence of the coelomic fluid hence called the Hydrostatic skeleton. The alternate waves of extensions and contractions are aided by the leverage afforded by the buccal chamber and the setae.

Question 51.
Are the cockroaches are oviparous or viviparous? Give example.
Answer:
Cockroaches are generally oviparous, but the Diploptera punctata is a viviparous cockroach found in Myanmar, China, Fiji, Hawaii, and India.

Choose the correct answer.

1. The earthworm in gardens, can be traced by their faecal deposits known as:
(a) vermiwash
(b) vermicompost
(c) worm castings
(d) wormery
Answer:
(c) worm castings

2. The common Indian earthworms are:
(a) Perioynx excavatus
(b) Lampito mauritii
(c) Pheretima posthuma
(d) Eudrilus eugeniae
Answer:
(b) Lampito mauritii

3. The locomotory organ of earthworms is:
(a) appendages
(b) setae
(c) tubefeet
(d) legs
Answer:
(b) setae

4. The colour of the earthworm is mainly due to the presence of the pigment called:
(a) Haemoglobin
(b) Haemocyanin
(c) Porphyrin
(d) Chiorocruorin
Answer:
(c) Porphyrin

5. A small flap found hanging on the mouth of earthworm is called:
(a) Prostomium
(b) Peristomium
(c) Clitellum
(d) Pygidium
Answer:
(a) Prostomium

6. The muscle fold found in the dorsal wall in the intestine of the earthworm is:
(a) Diaphragm
(b) Typhiosole
(c) Myotome
(d) Ommatidium
Answer:
(b) Typhiosole

7. The chioracogen cells on the wall of the intestine of the earthworm meant for:
(a) Digestion
(b) Circulation
(c) Excretion
(d) Reproduction
Answer:
(c) Excretion

8. Match the following and choose the correct answer:
(a) Photoreceptors (i) detect chemical changes
(b) Gustatory (ii) sense of touch
(c) Chemoreceptors (iii) sense of smell
(di Olfactory receptors (iv) sense of taste
(e) Tactile receptors (v) sense of light
(a) (a)-(v), (b)-(iv), (c)-(i), (d)-(iii), (e)-(ii)
(b) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i), (e)-(v)
(e) (a)-(i), (b)-(ii), (c)-(iii), (d)-(v), (e)-(iv)
(d) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(v), (e)-(i)
Answer:
(a) (a)-(v), (b)-(iv), (c)-(i), (d)-(iii), (e)-(ii)

9. Development of sperm earlier than the production of ova is called:
(a) Protandrous
(b) Heterogametic
(c) Progaxnetic
(d) spermatophore
Answer:
(a) Protandrous

10. The zoological name of the cockroach is:
(a) Periplaneta americana
(b) Lampiro mauritti
(c) Rana hexadactyla
(d) Metaphireposthuma
Answer:
(a) Periplaneta americana

11. The respiratory organ of cockroach is:
(a) Lungs
(b) Gills
(c) Skin
(d) Trachea
Answer:
(d) Trachea

12. The triangular muscles of cockroaches responsible for blood circulation is:
(a) Sphincter muscles
(b) Pulsatile vesicle
(c) Alary muscles
(d) Smooth muscles
Answer:
(c) Alary muscles

13. Cockroaches excretes uric acid, so it is:
(a) Ureotelic
(b) Uricotelic
(c) Ammoniotelics
(d) Elimination
Answer:
(b) Uricotelic

14. The egg case of cockroach is called:
(a) Collateral glands
(b) spermathecae
(c) vagina
(d) Ootheca
Answer:
(d) Ootheca

15. The zoological name of common Indian green frog is:
(a) Rana hexadactyla
(b) Rana tigiris
(c) Toad
(d) Caecilian
Answer:
(a) Rana hexadactyla

16. During the aestivation and hibernation period of frog, gaseous exchange takes place through:
(a) gills
(b) nosthis
(c) skin
(d) lungs
Answer:
(c) skin

17. Frogs are:
(a) Ureotelic
(b) Uricotelic
(c) Ammoniotelics
(d) Dwellers
Answer:
(a) Ureotelic

18. The tadpoles of frog respire through:
(a) skin
(b) gills
(c) nostrils
(d) fins
Answer:
(b) gills

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 3 Tissue Level of Organisation which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 3 Tissue Level of Organisation

Answer the following.

Question 1.
Define tissues and write their types.
Answer:
Groups of cells that are similar in structure and perform common or related functions are called ‘tissues’.
Classification of Animal Tissues: Epithelial, Connective, Muscular, Neural.

Question 2.
Why tissues are called living fabrics.
Answer:
Tissues are organized in specific proportions and patterns to form organs like lungs, heart, stomach, kidneys, ovaries, testes etc., hence the tissues are called the ‘living fabrics’.

Question 3.
Classify animal tissues.
Answer:
Animal tissues are classified according to the size, shape and function of the cells. There are four primaries (basic) tissue types that interweave to form the ‘fabric’ of the body. They are the epithelial tissue (covering), the connective tissue (support), the muscle tissue (movement) and the nervous tissue (control).

Question 4.
Write about the special features of simple epithelium, its location, and its different types. Draw suitable diagrams also.
Answer:

1. Simple epithelium is composed of a single layer of cells. They are found in the organs of absorption, secretion and filtration. Simple epithelial tissue is further classified into squamous epithelium, cuboidal epithelium, columnar epithelium, ciliated epithelium and pseudostratified epithelium. The squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries. They are found in the kidney glomeruli, air sacs of lungs, lining of heart, blood vessels and lymphatic vessels and are involved in functions like forming a diffusion boundary and filtration in sites where protection is not important.
2. The cuboidal epithelium is made of a single layer of cube-like cells. This tissue is commonly found in the kidney tubules, ducts and secretory portions of small glands and the surface of the ovary. Its main functions are secretion and absorption.
3. The columnar epithelium is composed of a single layer of tall cells with round to oval nuclei at the base. It lines the digestive tract from the stomach to the rectum.
4. The functions of this epithelium include absorption, secretion of mucus, enzymes and other substances.
5. This ciliated type propels mucus by ciliary actions and it lines the small bronchioles, fallopian tubes and uterus. Non-ciliated type lines most of the digestive tract, gall bladder and secretory ducts of glands.
6. Pseudo-stratified epithelial cells are columnar but unequal in size. Although the epithelium is single-layered yet it appears to be multi-layered because the nuclei lie at different levels in different cells. Hence, it is also called pseudostratified epithelium and its functions are protection, secretion and absorption. Ciliated forms line the trachea and the upper respiratory tract. The non-ciliated forms line the epididymis, large ducts of glands and tracts of the male urethra.
   

Question 5.
Write about the modified columnar epithelium.
Answer:
Some of the cuboidal or columnar cells get specialized for secretion and are called the glandular epithelium. They are mainly of two types: unicellular, consisting of isolated glandular cells (goblet cells of the alimentary canal), and multicellular, consisting of a cluster of cells (salivary gland). On the basis of the mode of pouring of their secretions, glands are divided into two categories namely exocrine and endocrine glands. Exocrine glands secrete mucus, saliva, earwax, oil, milk, digestive enzymes and other cell products. These products are released through ducts or tubes. In contrast, endocrine glands do not have ducts. Their secretions called hormones are secreted directly into the fluid bathing the gland. The exocrine glands are classified as unicellular and multicellular glands. The multicellular glands are further classified based on the structure as simple and compound glands, based on their secretory units as tubular, alveolar, (Acinus) and tubuloalveolar.

Question 6.
What are the types of exocrine glands? Differentiate the both.
Answer:

1. The exocrine glands are classified as unicellular and multicellular glands.
2. Unicellular, consisting of isolated glandular cells (goblet cells of the alimentary canal), and multicellular, consisting of a cluster of cells (salivary gland).

Question 7.
Name the exocrine glands based on their mode of secretions.
Answer:
Based on the mode of secretion exocrine glands are classified as merocrine, holocrine and apocrine.

Question 8.
What is the main function of compound epithelium and what are its types.
Answer:

1. The compound epithelium is made of more than one layer (multi-layered) of cells and thus has a limited role in secretion and absorption.
2. Their main function is to provide protection against chemical and mechanical stresses.
   There are four types of compound epithelium namely, stratified squamous epithelium, cuboidal epithelium, columnar epithelium and transitional epithelium.

Question 9.
Write the location of compound epithelium.
Answer:
The compound epithelium covers the dry surface of the skin, the moist surface of the buccal cavity, pharynx, the inner lining of ducts of salivary glands and of pancreatic ducts.

Question 10.
Write about the different types of compound epithelium.
Answer:
There are four types of compound epithelium namely, stratified squamous epithelium, cuboidal epithelium, columnar epithelium and transitional epithelium.
Stratified squamous epithelium is of two types called keratinized type which forms the dry epidermis of the skin and the non keratinized type forms the moist lining of the oesophagus, mouth, conjunctiva of the eyes and vagina. Stratified cuboidal epithelium mostly found in the ducts of sweat glands and mammary glands. The stratified columnar epithelium has limited distribution in the body, found around the lumen of the pharynx, male urethra and lining of some glandular ducts.
Transitional Epithelium is found lining the ureters, urinary bladder and part of the urethra. This epithelium allows stretching and is protective in function.

Question 11.
What is cell junction? What are its different types?
Answer:
All cells of the epithelium are held together with little intercellular material. In most animal tissues, specialized junctions provide both structural and functional links between their individual cells. Three types of cell junctions are found in the epithelium and other tissues. These are called tight, adhering and gap junctions.

Question 12.
Name the different types of cell junction and their functions.
Answer:
Tight junctions help to stop substances from leaking across a tissue. Adhering junctions perform cementing to keep neighbouring cells together. Gap junctions facilitate the cells to communicate with each other by connecting the cytoplasm of adjoining cells, for rapid transfer of ions, small molecules and sometimes big molecules.

Question 13.
Name the three types of fibres found in the connective tissues matrix.
Answer:
The ‘Fibres’ of connective tissue provide support. Three types of fibres are found in the connective tissue matrix. They are collagen, elastic and reticular fibres.

Question 14.
Name the two broad types of connective tissues.
Answer:
Connective tissue is of two types namely, Loose connective tissues (Areolar, Adipose and Reticular) and Dense connective tissues (dense regular, dense irregular and elastic).

Question 15.
Write an account on different varieties of loose connective tissue.
Answer:

1. In this tissue, the cells and fibres are loosely arranged in semi-fluid ground substances.
2. It contains fibroblasts, macrophages, and mast cells.
3. The Areolar connective tissue beneath the skin acts as a support framework for epithelium and acts as a reservoir of water and salts for the surrounding body tissues, hence aptly called tissue fluid.
4. Adipose tissue is similar to areolar tissue in structure and function and located beneath the skin. Adipocytes commonly called adipose or fat cells predominate and account for 90% of this tissue mass.
5. While fasting, these cells maintain our life by producing and supplying energy as fuel. Adipose tissues are also found in subcutaneous tissue, surrounding the kidneys, eyeball, heart, etc. Adipose tissue is called ‘white fat’ or white adipose tissue. The adipose tissue which contains abundant mitochondria is called ‘Brown fat’ or Brown adipose tissue. White fat stores nutrients whereas brown fat is used to heat the bloodstream to warm the body. Brown fat produces heat by non-shivering thermogenesis in neonates.

Question 16.
Write the differences between dense regular connective tissues and dense irregular connective tissues.
Answer:
Dense regular connective tissues primarily contain collagen fibres in rows between many parallel bundles of tissues and a few elastic fibres. The major cell type is a fibroblast. It attaches muscles and bones and withstands great tensile stress when pulling force is applied in one direction. This connective tissue is present in tendons, that attach skeletal muscles to bones and ligaments attach one bone to another.
Dense irregular connective tissues have bundles of thick collagen fibres and fibroblasts which are arranged irregularly. The major cell type is the fibroblast. It is able to withstand tension exerted in many directions and provides structural strength. Some elastic fibres are also present. It is found in the skin as the leathery dermis and forms fibrous capsules of organs such as kidneys, bones cartilages, muscles, nerves and joints. Elastic connective tissue contains a high proportion of elastic fibres. It allows recoil of tissues following stretching. It maintains the pulsatile flow of blood through the arteries and the passive recoil of the lungs following inspiration. It is found in the walls of large arteries; ligaments associated with the vertebral column and within the walls of the bronchial tubes.

Question 17.
Name the divisions of specialised connective tissue.
Answer:
Specialised connective tissues are classified as cartilage, bones and blood,

Question 18.
Give an account of different varieties of specialised connective tissue.
Answer:

Specialised connective tissues are classified as cartilage, bones and blood. Cartilage is solid and pliable and resists compression. Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of the nose, outer ear joints, ear pinna, between adjacent bones of the vertebral column, limbs and hands in adults.
Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which gives strength to the bones. It is the main tissue that provides a structural frame to the body. Bones support and protect softer tissues and organs. The bone cells (osteocytes) are present in the spaces called lacunae. Limb bones, such as the long bones of the legs, serve weight-bearing functions. They also interact with skeletal muscles attached to them to bring about movements. The bone marrow in some bones is the site of the production of blood cells.
Blood is the fluid connective tissue containing plasma, red blood cells (RBC), white blood cells (WBC) and platelets. It functions as the transport medium for the cardiovascular system, carrying nutrients, wastes, respiratory gases throughout the body.

Question 19.
Mention the places where cartilage bones are located in the human body.
Answer:
Cartilage is present in the tip of the nose, outer ear joints, ear pinna, between adjacent bones of the vertebral column, limbs and hands in adults.

Question 20.
What is muscle tissue? What is its main function?
Answer:
Each muscle is made of many long, cylindrical fibres arranged in parallel arrays. These fibres are composed of numerous fine fibrils, called myofibrils. Muscle fibres contract (shorten) in response to stimulation, then relax (lengthen) and return to their uncontracted state in a coordinated fashion. In general, muscles play an active role in all the movements of the body.

Question 21.
Write an account on different types of muscles with a neat diagram, (or)
Write a short note of the following:- (i) Skeletal muscle, (ii) Smooth muscle, (iii) Cardiac muscle.
Answer:
Muscles are of three types, skeletal, smooth and cardiac. They are –

1. Skeletal muscle,
2. Smooth muscle,
3. Cardiac muscle.

(i) Skeletal muscle:
Skeletal muscle tissue is closely attached to skeletal bones. In a typical muscle such as the biceps, the striated (striped) skeletal muscle fibres are bundled together in a parallel fashion. A sheath of tough connective tissue encloses several bundles of muscle fibres.

(ii) Smooth muscle:
The smooth muscle fibres taper at both ends (fusiform) and do not show striations. Cell junctions hold them together and they are bundled together in a connective tissue sheath. The walls of internal organs such as the blood vessels, stomach and intestine contain this type of muscle tissue. Smooth muscles are “involuntary’ as their functions cannot be directly controlled. Unlike smooth muscles, skeletal muscles can not be controlled by merely thinking.

(iii) Cardiac muscle:
Cardiac muscle tissue is contractile tissue present only in the heart. Cell junctions fuse the plasma membranes of cardiac muscle cells and make them stick together. Communication junctions (intercalated discs) at some fusion points allow the cells to contract as a unit, i.e., when one cell receives a signal to contract, its neighbours are also stimulated to contract.

Question 22.
Differentiate simple epithelium from squamous epithelium.
Answer:

Simple Epithelium	Squamous epithelium
Simple epithelium is composed of a single layer of cells.	The squamous epithelium is made of a single thin layer of flattened cells with irregular boundaries.
They are found in the organs of absorption, secretion and filtration. Simple epithelial tissue is further classified into squamous epithelium, cuboidal epithelium, columnar epithelium, ciliated epithelium and pseudostratified epithelium.	They are found in the kidney glomeruli, air sacs of lungs, lining of heart, blood vessels and lymphatic vessels and are involved in functions like forming a diffusion boundary and filtration in sites where protection is not important.

Question 23.
Why Areolar connective tissue is called tissue fluids.
Answer:
The Areolar connective tissue beneath the skin acts as a support framework for epithelium and acts as a reservoir of water and salts for the surrounding body tissues, hence aptly called tissue fluid.

Question 24.
What is a biopsy?
Answer:
A biopsy is an examination of tissue or liquid removed from a living body to discover the presence, cause or extent of a disease.

Question 25.
What is an autopsy?
Answer:
An autopsy is a post-mortem (dissection of a dead body) examination to discover the cause of death or the extent of disease.

Question 26.
Write about the heritable connective tissue disorders?
Answer:
Ehler’s -Danlos syndrome: Defect in the synthesis of collagen in the joints, heart valves, organ walls and arterial walls.
Stickler syndrome: Affects collagen and results in facial abnormalities.
Rhabdomyosarcoma: Life-threatening soft tissue tumour of head, neck and urinogenital tract.

Question 27.
What are autoimmune connective tissue disorders? Write its symptoms.
Answer:
Rheumatoid arthritis: The immune cells attack and inflame the membranes around the joints.
It can also affect the heart, lungs and eyes.
Sjogren’s syndrome: Progressive inability to secrete saliva and tears.

Question 28.
Name the diseases of the nervous system and mention their symptoms.
Answer:
Parkinson’s disease: A degenerative disorder of the nervous system that affects movement, often including tremors.
Alzheimer’s disease: It is a chronic neurodegenerative disease that includes the symptoms of difficulty in remembering recent events, problems with language, disorientation and mood swings.

Question 29.
Mention the names of disorders of epithelial tissue.
Answer:
Eczema, Psoriasis, Epithelial Carcinoma and severe asthma.

Choose the correct answer.

1. Group of cells that are similar in structure and fûnction are called:
(a) Organ
(b) Organ system
(c) Tissues
(d) Germ layers
Answer:
(c) Tissues

2. ………… is a contractile tissue present only in the heart.
(a) Kidney
(b) digestive tract
(c) Heart
(d) Brain
Answer:
(c) Heart

3. ………. is closely attached to skeletal bones.
(a) Smooth muscle
(b) cardiac muscle
(c) Muscle fibers
(d) Striped muscle
Answer:
(d) Striped muscle

4. …………. is a fluid cãnnective tissue.
(a) Lymph
(b) Blood
(c) Mucus
(d) Saliva
Answer:
(b) Blood

5. The study of tissues is known as:
(a) Histology
(b) Anatomy
(c) Physiology
(d) Genetics
Answer:
(a) Histology

6. …………… is the type of modified columnar epithelium which secretes the protective lubricating mucus.
(a) Sebaceous gland
(b) Oil glands
(c) Mast cells
(d) Goblet cells
Answer:
(d) Goblet cells

7. The secretions of endocrine glands ari called:
(a) Enzymes
(b) Hormones
(c) Granules
(d) Cell fragments
Answer:
(b) Hormones

8. Dry epidermis of the skin is lined with:
(a) Keratinized type of stratified squamous epithelium
(b) Stratified cuboidal epithelium
(c) Non keratinized type of stratified squamous epithelium
(d) Transitional epithelium
Answer:
(a) Keratinized type of stratified squamous epithelium

9. ……….. helps to stop substances from leaking across a tissue.
(a) Gap junction
(b) Tissue junction
(c) Tight junction
(d) Adhering junction
Answer:
(c) Tight junction

10. The adipose tišsue which contains abundant mitochondria is called:
(a) white fat
(b) fat cells
(c) brown fat
(d) white adipose tissue.
Answer:
(c) brown fat

11. The skeletal muscles and bones are attached by:
(a) fibroblast
(b) stroma
(c) tendons
(d) cartilages
Answer:
(c) tendons

12. The cartilage cells are called:
(a) Chondrocytes
(b) Osteocytes
(e) Melanocytes
(d) Lymphocytes
Answer:
(a) Chondrocytes

13. The …………. of bones is the site of production of blood cells.
(a) Osteocytes
(b) Bone marrow
(c) Collagen fibres
(d) Chondrocytes
Answer:
(b) Bone marrow

14. These are the fundamental units of neural tissues:
(a) Cartilages
(b) Nephrons
(c) Tendons
(d) Neurons
Answer:
(d) Neurons

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 2 Kingdom Animalia which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 2 Kingdom Animalia

Answer the following.

Question 1.
Define levels of Organization.
Answer:
All members of Kingdom Animalia are metazoans (multicellular animals) and exhibit different patterns of cellular organization. The cells of the metazoans are not capable of independent existence and exhibit division of labour. Among the metazoans, cells may be functionally isolated, or similar kinds of cells may be grouped together to form tissues, organ, and organ systems.

Question 2.
Name the layers of cells found in sponges.
Answer:
In sponges, the outer layer is formed of pinacocytes (plate-like cells that maintain the size and structure of the sponge) and the inner layer is formed of choanocytes. These are flagellated collar cells that create and maintain water flow through the sponge thus facilitating respiratory and digestive functions.

Question 3.
What is the tissue level of Organization?
Answer:
In some animals, cells that perform similar functions are aggregated to form tissues. The cells of a tissue integrate in a highly coordinated fashion to perform a common function, due to the presence of nerve cells and sensory cells. This tissue level of organization is exhibited in diploblastic animals like cnidarians. The formation of tissues is the first step towards the evolution of body plan in animals (Hydra – Coelenterata).

Question 4.
What is the organ level of the Organization from which level of animals show this type of Organization?
Answer:
Different kinds of tissues aggregate to form an organ to perform a specific function. Organ, level of organization is a further advancement over the tissue level of organization and appears for the first time in the Phylum Platyhelminthes and seen in other higher phyla.

Question 5.
Make out the difference between the incomplete and complete digestive systems.
Answer:
The digestive system of Platyhelminthes has only a single opening to the exterior which serves as both mouth and anus and hence called an incomplete digestive system. From Aschelminthes to Chordates, all animals have a complete digestive system with two openings, the mouth, and the anus.

Question 6.
Name the embryonic layers of animals and on the basis of the origin and development.
Answer:
During embryonic development, the tissues and organs of animals originate from two or three embryonic germ layers. On the basis of origin and development, animals are classified into two categories: Diploblastic and Triploblastic.

Question 7.
What are diploblastic animals?
Answer:
Animals in which the cells are arranged in two embryonic layers the external ectoderm and internal endoderm are called diploblastic animals. In these animals, the ectoderm gives rise to the epidermis (the outer layer of the body wall) and the endoderm gives rise to the gastrodermis (tissue lining the gut cavity). An undifferentiated layer present between the ectoderm and endoderm is the mesoglea (Corals, Jellyfish, Sea anemone).

Question 8.
Define triploblastic animals for example.
Answer:
Animals in which the developing embryo has three germinal layers are called triploblastic, animals and consists of outer ectoderm (skin, hair, neuron, nail, teeth, etc), inner endoderm (gut, lung, liver), and middle mesoderm (muscle, bone, heart). Most of the triploblastic animals show organ system level of organization (Flatworms to Chordates).

Question 9.
What is meant by symmetry?
Answer:
Symmetry is the body arrangement in which parts that lie on the opposite side of an axis are identical. An animal’s body plan results from the animal’s pattern of development.

Question 10.
Define asymmetry for example.
Answer:
The simplest body plan is seen in sponges. They do not display symmetry and are asymmetrical. Such animals lack a definite body plan or are irregular shaped and any plane passing through the center of the body does not divide them into two equal halves (Sponges).

Question 11.
Write about the radial symmetry with a suitable diagram.
Answer:
Symmetrical animals have paired body parts that are arranged on either side of a plane passing through the central axis. When any plane passing through the central axis of the body divides an organism into two identical parts, it is called radial symmetry. Such radially symmetrical animals have a top and bottom side but no dorsal (back) and ventral (abdomen) side, no right and left side. They have a body plan in which the body parts are organized in a circle around an axis. It is the principal symmetry in diploblastic animals. Cnidarians such as sea anemone and corals are radially symmetrical. However, triploblastic animals like echinoderms eg. Starfish have five planes of symmetry and show Pentamerous radial symmetry:

Question 12.
What is bilateral and biradial symmetry? Give examples.
Answer:
Animals that possess two pairs of symmetrical sides are said to be biradially symmetrical. Biradial symmetry is a combination of radial and bilateral symmetry as seen in ctenophores. There are only two planes of symmetry, one through the longitudinal and sagittal axis and the other through the longitudinal and transverse axis. eg. Comb jellyfish – Pleurobrachia.
Animals that have two similar halves on either side of the central plane show bilateral symmetry. It is an advantageous type of symmetry in triploblastic animals, which helps in seeking food, locating mates and escaping from predators more efficiently. Animals that . have dorsal and ventral sides, anterior and posterior ends, right and left sides are bilaterally symmetrical.

Question 13.
Which animals are called acoelomates? Give example.
Answer:
Animals that do not possess a body cavity are called acoelomates. Since there is nobody cavity in these animals their body is solid without a perivisceral cavity, this restricts the free movement of internal organs (eg, Flatworms).

Question 14.
What is called coelomates?
Answer:
Eucoelom or true coelom is a fluid-filled cavity that develops within the mesoderm and is lined by mesodermal epithelium called the peritoneum. Such animals with a true body cavity are called coelomates or coelomates.

Question 15.
What is notochord? How the animals are classified based on. this presence or absence of notochord?
Answer:
The notochord is a mesodermally derived rod-like structure formed on the dorsal side during embryonic development in some animals. Based on the presence or absence of notochord, animals are classified as chordates (Cephalochordates, Urochordates, Pisces to Mammalia) and non chordates (Porifera to Hemichordata).

Question 16.
How the kingdom Animalia is classified broadly into sub-kingdoms?
Answer:
The animal kingdom is divided into two sub-kingdoms, the Parazoa and Eumetazoa based on their organization.

1. Parazoa: These include the multicellular sponges and their cells are loosely aggregated and do not form tissues or organs.
2. Eumetazoa: These include multicellular animals with well-defined tissues, which are organized as organs and organ systems. Eumetazoans include two taxonomic levels called grades. They include Radiata and Bilateria.

Question 17.
Write about the ‘Division level’ Classification of Bilateria.
Answer:
The eumetazoans other than Radiata, show organ level of organization and are bilaterally symmetrical and triploblastic. The grade Bilateria includes two taxonomic levels called Division.
Division: 1. Protostomia (Proto: first; stomium: mouth): Protostomia includes the eumetazoans in which the embryonic blastopore develops into the mouth. This division includes three subdivisions namely acoelomata, pseudocoelomata and schizocoelomata.
Division: 2. Deuterostomia (deuteron: secondary; stomium: mouth):, Eumetazoans in which anus is formed from or near the blastopore and the mouth is formed away from the blastopore.
It includes only one subdivision Enterocoelomata. They have a true coelom called enterocoel, formed from the archenteron.

Question 18.
Write about the special features of the phylum Porifera.
Answer:

1. These pore-bearing animals are commonly called sponges. They are aquatic, mostly marine, asymmetrical and a few species live in freshwaters.
2. They are primitive, multicellular, sessile animals with a cellular level of organization in which the cells are loosely arranged. They are either radially symmetrical or asymmetrical animals.
3. They possess a water transport system or canal system where water enters through minute pores called Ostia lining the body wall through which the water enters into a central cavity (spongocoel) and goes out through the osculum.
4. This water transport system is helpful in food gathering, circulation, respiration, and removal of waste.
5. Choanocytes or collar cells are special flagellated cells lining the spongocoel and the canals.
6. Nutrition is holozoic and intracellular.
7. All sponges are hermaphrodites.
8. They also reproduce asexually by fragmentation or gemmule formation and sexually by the formation of gametes.
9. Development is indirect with different types of larval stages such as parenchyma and amphiblastula.
10. eg. Sycon (Scypha), Spongilla (freshwater sponge).

Question 19.
Write about the canal system found in Porifera.
Answer:
They possess a water transport system or canal system where water enters through minute pores called Ostia lining the body wall through which the water enters into a central cavity (spongocoel) and goes out through the osculum.
This water transport system is helpful in food gathering, circulation, respiration, and removal of waste.

Question 20.
What is the function of cnidoblasts in phylum Cnidaria?
Answer:
The name Cnidaria is derived from cnidocytes or cnidoblasts with stinging cells or nematocyst on tentacles. Cnidoblasts are used for anchorage, defense, and capturing prey.

Question 21.
Differentiate polyp and medusa of coelenterates or cnidarians.
Answer:
Cnidarians exhibit two basic body forms, polyp, and medusa. The polyp forms are sessile and cylindrical (eg. Hydra, Adamsia), whereas the medusa is umbrella-shaped and free swimming.
The polyp represents the asexual generation and medusa represents the sexual generation. Polyps produce medusa asexually and medusa forms polyps sexually.

Question 22.
Write short notes about Ctenophora.
Answer:

1. Ctenophora are exclusively marine, radially symmetrical, diploblastic animals with tissue level of organization.
2. Though they are diploblastic, their mesoglea is different from that of cnidaria. It contains amoebocytes and smooth muscle cells.
3. They have eight external rows of ciliated comb plates (comb jellies) which help in locomotion, hence commonly called comb jellies or sea walnuts.
4. Bioluminescence (the ability of a living organism to emit light) is well marked in ctenophores.
5. They possess special cells called lasso cells or colloblasts which help in food capture.
6. Digestion is both extracellular and intracellular.
7. Sexes are not separate (monoecious). They reproduce only by sexual means.
8. (Fertilization is external and development is indirect and includes a larval stage called cydippid larva, eg. Pleurobrachia and Ctenoplana.

Question 23.
What is the type of digestive system found among flatworms?
Answer:
Some of the parasitic flatworms absorb nutrients directly from the host through their body surface. However, flatworms like liver fluke have an incomplete digestive system which means it has an only a single opening to the exterior which serves as both mouth and anus.

Question 24.
List out the characters of the phylum Annelida.
Answer:

1. Annelids are the first segmented animals to evolve.
2. They are aquatic or terrestrial, free-living but some are parasitic.
3. They are triploblastic, bilaterally symmetrical, schizocoelomates, and exhibit organ system level of body organization.
4. The coelom with coelomic fluid creates a hydrostatic skeleton and aids in locomotion.
5. Their elongated body is metamerically segmented and the body surface is divided into segments or metameres.
6. Internally the segments are divided from one another by partitions called septa. This phenomenon is known as metamerism.
7. Aquatic annelids like Nereis have lateral appendages called parapodia, which help in swimming.
8. The circulatory system is of closed type and the respiratory pigments are hemoglobin and chlorocruorin.
9. The nervous system consists of paired ganglion connected by the lateral nerves to the double ventral nerve cord.
10. They reproduce sexually. Development is direct or indirect and includes a trochophore larva, eg. Lampito mauritii (Earthworm).

Question 25.
Write an account on the common characters of Arthropoda.
Answer:

1. This is the largest phylum of the Kingdom Animalia and includes the largest class called Insecta.
2. They are bilaterally symmetrical, segmented, triploblastic and schizocoelomate animals with organ system grade of body organization.
3. They have jointed appendages that are used for locomotion feeding and are sensory in function,
4. The body is covered by cjiitinous exoskeleton for protection and to prevent water loss, It is shed off periodically by a process called molting or ecdysis.
5. The body consists of a head, thorax, and abdomen with a body cavity called haemocoel.
6. Respiratory organs are gills, book gills, book lungs or trachea.
7. The circulatory system is of open type.
8. Sensory organs like antennae, eyes (compound and simple), statocysts (organs of balance/ equilibrium) are present.
9. Excretion takes place through malpighian tubules, green glands, coxal glands, etc.
10. They are mostly dioecious and oviparous; fertilization is usually internal.
11. Life history includes many larval stages followed by metamorphosis. eg. Limulus.

Question 26.
Define metamerism in Annelids.
Answer:
The body of the annelids are metamerically segmented and the body surface is divided into segment or metameres. Internally the segments are divided from one another by partitions called septa. This phenomenon is known as metamerism.

Question 27.
Define moulting or ecdysis.
Answer:
Body is covered by chitinous exoskeleton for protection and to prevent water loss, It is shed off periodically by a process called moulting or ecdysis.

Question 28.
Write few examples for the phylum, Arthopoda.
Answer:

1. Limulus (King crab, a living fossil)
2. Palamnaeus (Scorpion)
3. Eupagarus (Hermit crab)
4. Apis (Honey bee)
5. Musca (House fly)
6. Vectors – Anopheles, Culex, Aedes (mosquitoes)
7. Economically important insects – Apis- (Honey bee), Bombyx (Silk worm)
8. Laccifer (Lac insects)
9. Living fossils Limulus-(King crab)
10. Gregarious pest – Locusta (Locust)

Question 29.
Write the salient features of the phylum mollusca.
Answer:

1. This is the second-largest animal phylum.
2. Molluscs are terrestrial or aquatic (marine or fresh water) and exhibit organ system level of body organisation.
3. They are bilaterally symmetrical (except univalves), triploblastic and coelomate animals.
4. Body is covered by a calcareous shell and is unsegmented with a distinct head, muscular foot and a visceral hump or visceral mass.
5. A soft layer of skin forms a mantle over the visceral hump. The space between the visceral mass and mantle (pallium) is called the mantle cavity.
6. A number of feather like gills (ctenidia) are present, which are respiratory in function.
7. The digestive system is complete and mouth contains a rasping organ called radula with transverse rows of chitinous teeth for feeding (radula is absent in bivalves.
8. The sense organs are tentacles, eyes and ospharidium (to test the purity of water and present in bivalves and gastropods).
9. Excretory organs are nephridia.
10. Open type of circulatory system is seen except for cephalopods such as squids, cuttle fishes and octopuses. Blood contains haemocyanin, a copper containing respiratory pigment.
11. Development is indirect with a veliger larva.
12. eg. Pila (Apple snail).

Question 30.
Write the functions of the following: (i) Ctenidia, (ii) Ospharidiam.
Answer:

1. Ctenidia: Mandle cavity has a number of feather like gills (ctenidia) are present, which are respiratory in function.
2. Ospharidiam: The anterior head regim of molluscs has ospharidium to test the purity of water and presence in bivalves and gasropods.

Question 31.
Why ore certain marine animals termed as echinoderms?
Answer:
All Echinoderms are marine animals. These animals have a mesodermal endoskeleton of calcareous ossicles and hence the name Echinodermata.

Question 32.
What is the most distinctive feature of echinoderms and write its importance in them?
Answer:
The most distinctive feature of echinoderms is the presence of the water vascular system or ambulacral system with tube feet or podia, which helps in locomotion, capture and transport of food and respiration.

Question 33.
What are the fundamental distinct features of all chordates?
Answer:
All chordates possess three fundamental distinct features at some stage of their life cycle, they are:

1. Presence of elongated rod like notochord below the nerve cord and above the alimentary canal. It serves as a primitive internal skeleton. It may persist throughout life in lancelets and lampreys. In adult vertebrates, it may be partially or completely replaced by backbone or vertebral column.
2. A dorsal hollow or tubular fluid filled nerve cord lies above the notochord and below the dorsal body wall. It serve’s to integrate and co-ordinate the body functions. In higher chordates, the anterior end of the nerve cord gets enlarged to form the brain and the posterior part becomes the spinal cord, protected inside the vertebral column.
3. Presence of pharyngeal gill slits or clefts in all chordates at some stage of their life cycle.
   It is a series of gill slits or clefts that perforates the walls of pharynx and appears during the development of every chordate. In aquatic forms, pharyngeal gill slits are vascular, lamellar and form the gills for respiration. In terrestrial chordates, traces of non-functional gill clefts appear during embryonic developmental stages and disappear later. Besides the above said features, chordates are bilaterally symmetrical, triploblastic, coelomates with organ system level of organisation; they possess post anal tail, closed circulatory system with a ventral myogenic heart except in Amphioxus.

Question 34.
Give a brief account on the characters of Tunicates.
Answer:
They are exclusively marine and are commonly called sea squirts. Mostly sessile, some pelagic or free swimming, exist as solitary and colonial forms. Body is unsegmented and covered by a test or tunic. Adult forms%re sac like. Coelom is absent, but has an atrial cavity surrounding the pharynx. Notochord is present only in the tail region of the larval stage, hence named urochordata. Alimentary canal is complete and circulatory system is of open type. The heart is ventral and tubular.

Respiration is through gill slits and clefts. Dorsal tubular nerv e cord is present only in the larval stage and a single dorsal ganglion is present in the adults. Mostly hennaphrodites, development indirect and includes a free swimming tadpole larva with chordate characters. Retrogressive metamorphosis is seen. eg. Ascidia.

Question 35.
Give an account of the General features of subphylum caphalochorelata.
Answer:
Cephalochordates are marine forms, found in shallow waters, leading a burrowing mode of life. They are small fish like coelomate forms with chordate characters such us notochord, dorsal tubular nerve cord and pharyngeal gill slits throughout their life. Closed type of circulatory system is seen without heart. Excretion is by protonephridia. Sexes are separate, Fertilization is external, eg. Branchiostoma.

Question 36.
How are the vertebrates are further divided by divisions? Write the differences between them.
Answer:

1. Subphylum Vertebrata is divided into two divisions, Agnatha and Gnathostomata.
2. Agnatha includes jawless fish-like aquatic vertebrates without paired appendages. Notochord persists in the adult.
3. Agnatha includes one important class – Cyclostomata.
4. Gnathostomata includes jawed vertebrates with paired appendages. Notochord is replaced partly or wholly by the vertebral column.
5. Gnathostomata includes jawed fishes (Pisces) and Tetrapoda ( amphibia, reptilia, aves and mammals).The superclass Pisces includes all fishes which are essentially aquatic forms with paired fins for swimming and gills for respiration. Pisces includes cartilaginous fishes (Chondrichthyes) and bony fishes (Osteicthyes).

Question 37.
What is the special character seen in cyclostomes during spawning?
Answer:
Cyclostomes are marine but migrate to fresh waters for spawning (anadromous migration). After spawning within a few days they die. The larvae (ammocoete) after metamorphosis returns to the ocean.

Question 38.
Give some examples for bony fishes coming under Osteichthyes.
Answer:
Exocoetus, Labeo, Catla, Echeneis, Pterophyllam.

Question 39.
Write the general characters of amphibians with some examples.
Answer:

1. Amphibians are the first vertebrates and tetrapods to live both in aquatic as well as terrestrial habitats.
2. They are poikilothermic.
3. Their body is divisible into the head and trunk and most of them have two pairs of limbs; tail may or may not be present.
4. Their skin is smooth or rough, moist, pigmented and glandular.
5. Eyes have eyelids and the tympanum represents the ear.
6. Respiration is by gills, lungs and through the skin.
7. The heart is three-chambered.
8. Kidneys are mesonephric. Sexes are separate and fertilization is external.
9. They are oviparous and development is indirect. They show hibernation and aestivation.
10. eg. Bufo (Toad), Rana (Frog).

Question 40.
What are the general features of Reptilians?
Answer:

1. They are mostly terrestrial animals and their body is covered by dry, and comified skin with epidermal scales or scutes.
2. Reptiles have three chambered heart but four chambered in crocodiles.
3. All are cold blooded amniotes.
4. Most reptiles lay cleidoic eggs with extraembryonic membranes like amnion, allantois, chorion and yolk sac.
5. Excretion by metanephric kidneys and are uricotelic.
6. They are monoecious. Internal fertilization takes place and all are oviparous.
7. eg. Calotes (Garden Lizard), Draco (Flying Lizard), Crocodilus (Crocodile).

Question 41.
Describe the special characteristic features of Aves.
Answer:

1. Aves are commonly known as birds. The characteristic feature of Aves is the presence of feathers and the ability to fly except for flightless birds, eg. Ostrich, Kiwi.
2. The forelimbs are modified into wings, and the hind limbs are adapted for walking, running, swimming and perching.
3. The skin is dry and devoid of glands except the oil gland or preen gland at the base of the tail.
4. The exoskeleton consists of epidermal feathers, scales, claws on legs and the homy, covering on the beak.
5. The endoskeleton is fully ossified (bony) and the long bones are hollow with air cavities (pneumatic bones).
6. The pectoral muscles of flight (pectoralis major and pectoralis minor) are well developed.
7. Respiration is by compact, elastic, spongy lungs that are continuous with air sacs to supplement respiration.
8. The heart is four chambered.
9. The urinary bladder is absent.
10. Sexes are separate with well marked sexual dimorphism.
11. All birds are oviparous. Eggs are megalecithal and cleidoic. Fertilization is internal.
12. eg. Corvus (Crow), Columba (Pigeon), Pavo (Peacock).

Question 42.
Write the general characters of the class mammalia.
Answer:

1. Their body is covered by hair, a unique feature of mammals. Some of them are adapted to fly or live in water.
2. The presence of mammary glands is the most unique feature of mammals.
3. They have two pairs of limbs adapted for walking, running, climbing, burrowing, swimming and flying.
4. Their skin is glandular in nature, consisting of sweat glands, scent glands and sebaceous glands.
5. Exoskeleton includes homy epidermal horns, spines, scales, claws, nails, hooves and bony dermal plates.
6. Teeth are thecodont, heterodont and diphyodont.
7. External ears or pinnae are present.
8. The heart is four-chambered and possesses a left systematic arch. Mature RBCs are circular, biconcave and non nucleated.
9. Mammals have a large brain when compared to other animals.
10. They show greatest intelligence among all animals. Their kidneys are metanephric and are ureotelic.
11. All are homeothermic, sexes are separate and fertilization is internal.
12. eg. Platypus, Kangaroo, Monkey, Elephant.

Question 43.
Name two different types of larval stages in Porifera.
Answer:
Two different types of larval stages of porifera are parenchymula and amphiblastula.

Choose the correct answer.

1. Biradial symmetry is seen in:
(h) Star fish
(b) Comb jelly fish
(c) Sea anemone
(d) Sponge
Answer:
(b) Comb jelly fish

2. The special flagellated cells lining the spongocoel is:
(a) Choanocytes
(b) Cridocytes
(c) Nematocyst
(d) Lasso cells
Answer:
(a) Choanocytes

3. The minute pores lining the body wall of Porifera are called:
(a) Osculum
(b) Podia
(c) Ostia
(d) Gills
Answer:
(c) Ostia

4. The central body cavity of poriferans are:
(a) Gastrocoel
(b) Coelom
(c) Haemocoel
(d) Spongocoel
Answer:
(d) Spongocoel

5. The free-swimming ciliated larval form of criclaria is:
(a) Planula larva
(b) Parenchymula larva
(c) Amphiblastula larva
(d) Veliger larva
Answer:
(a) Planula larva

6. In case of flatworms the specialized excretory cells are named as:
(a) Nematocysts
(b) Flame cells
(c) Nephridia
(d) Malphigian tubules
Answer:
(b) Flame cells

7. Nereis have lateral appendages called:
(a) Parapodia
(b) Body setae
(c) Foot
(d) Tube feet
Answer:
(a) Parapodia

8. The special cells of ctenophora helps in food capture is:
(a) Cnidoblasts
(b) Choanocytes
(c) Flamecells
(d) Colloblasts
Answer:
(d) Colloblasts

9. ………… are the organs of balance in Arthropods.
(a) Nematocysts
(b) Statocysts
(c) Choanocytes
(d) Cochlea
Answer:
(b) Statocysts

10. ……….. is the largest phylum of the kingdom Animalia.
(a) Annelida
(b) Arthropoda
(c) Aschelminthes
(d) Echinodermata
Answer:
(b) Arthropoda

11. The second largest animal phylum is:
(a) Ctenophora
(b) Arthropoda
(c) Mollusca
(d) Coelenterata
Answer:
(c) Mollusca

12. The anterior head region of molluscs has got this organ which helps to test the purity of water:
(a) Ostia
(b) Ospharidiam
(c) Ossicles
(d) Gills
Answer:
(b) Ospharidiam

13. The larva of Nereis is:
(a) Planula
(b) Tomaria larva
(c) Trocophore larva
(d) Miracidium
Answer:
(c) Trocophore larva

14. Presence of water vascular system is the most distinctive feature of the Phylum:
(a) Mollusca
(b) Sponges
(c) Echinodermata
(d) Arthropoda
Answer:
(c) Echinodermata

15. The mantle cavity of the molluscs has got number of feather like gills, which are respiratory and excretory in function are:
(a) Book lungs
(b) Trachea
(c) Ambulacral system
(d) Ctenidia
Answer:
(d) Ctenidia

16. The phylum Hemichordata are mostly tubiculous and commonly called:
(a) Flat worms
(b) Round worms
(c) Tongue worms
(d) Parasitic worms
Answer:
(c) Tongue worms

17. The free swimming larva of Hemichordata are called:
(a) Tomaria larvae
(b) Planula larvae
(c) Trochophore larvae
(d) Cercaria larvae
Answer:
(a) Tomaria larvae

18. The tunicates are normally called:
(a) Sea squirts
(b) Sea anemone
(c) Sea-walnuts
(d) Sea urchin
Answer:
(a) Sea squirts

19. The chondrichthyes has got this type of gills helps for respiration:
(a) Filamentous gills
(b) Lamelliform gills
(c) Filiform gills
(d) Ambnlacral system
Answer:
(b) Lamelliform gills

20. The excretory organ of Chondrichthyes are:
(a) Pronephric kidneys
(b) Opisthonephric kidneys
(c) Mesonephric kidneys
(d) Metanephric kidneys
Answer:
(b) Opisthonephric kidneys.

21. The excretory organ of Osteichthyes are:
(a) Mesonephnc kidneys
(b) Opisthonephric kidneys
(c) Holonephric kidneys
(d) Metanephric kidneys
Answer:
(a) Mesonephnc kidneys

22. The eggs of Aves are of type.
(a) Microlecithal
(b) Mesolecithal
(c) Megalecithal
(d) Homolecithal
Answer:
(c) Megalecithal

23. The mature RBCs are non – nucleated in:
(a) Molluscs
(b) Mammals
(c) Fishes
(d) Birds
Answer:
(b) Mammals

24. Match:

(i) Scoliodon	(a) Electric rav
(it) Trygon	(b) Sting ray
(iii) Torpedo	(c) Lamprey
(iv) Pristis	(d) Dogfish
(v) Petromyzon	(e) Sawfish

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(e), (v)-(c)
(b) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a), (v)-(c)
(c) (i)-(a), (ii)-(c), (iii)-(e), (iv)-(b), (v)-(d)
(d) (i)-(c), (ii)-(e), (iii)-(d), (iv)-(b), (v)-(a)
Answer:
(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(e), (v)-(c)

25. Match:

(i) Sycon	(a) Bath sponges
(ii) Euspongia	(b) Glassrope sponge
(iii) Euplectella	(c) Scypha
(iv) Hyalonema	(d) deadman’s finger
(v) Chalina	(e) Venus flower baskets

(a) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(e), (v)-(a)
(b) (i)-(c), (ii)-(a), (iii)-(e), (iv)-(b), (v)-(d)
(c) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d), (v)-(e)
(d) (i)-(e), (ii)-(d), (iii)-(c), (iv)-(b), (v)-(a)
Answer:
(b) (i)-(c), (ii)-(a), (iii)-(e), (iv)-(b), (v)-(d)

Students get through the TN Board 11th Bio Zoology Important Questions Chapter 1 The Living World which is useful for their exam preparation.

TN State Board 11th Bio Zoology Important Questions Chapter 1 The Living World

Answer the following.

Question 1.
Define Ecosystem?
Answer:
The ecosystem is a community of living organisms (plants and animals), a non-living environment (including minerals, climate, soil, water; sunlight), and their interrelationships.

Question 2.
Define biological diversity?
Answer:
The presence of a large number of species in a particular ecosystem is called ‘biological diversity or in short ‘biodiversity’.

Question 3.
What is called classification?
Answer:
Classification is a process by which things are grouped in convenient categories, based on easily observable characters. The scientific term used for these categories is taxa.

Question 4.
What are the key characters of living organisms?
Answer:
The key characters of living organisms are cellular organization, nutrition, respiration, metabolism, growth, response to stimuli, movement, reproduction, excretion, adaptation, and homeostasis.

Question 5.
Define taxonomy?
Answer:
Based on their characteristics, all living organisms can be classified into different taxa. This science of classification is called taxonomy.

Question 6.
What are the basic principles of classification?
Answer:

1. To identify and differentiate closely related species:
2. To know the variation among the species.
3. To understand the evolution of the species.
4. To create a phylogenetic tree among the different groups
5. To conveniently study the living organisms

Question 7.
What are the main criteria for systematics?
Answer:
The main criteria of systematics are identifying, describing, naming, arranging, preserving, and documenting the organisms. Apart from the above-said features, the evolutionary history of the species and the environmental adaptations and interrelationship between species are also being investigated in systematics.

Question 8.
How did Aristotle classify the animals?
Answer:
Based on the presence or absence of red blood cells. Aristotle classified the animals into two as Enaima with blood and those without blood as Anaima.

Question 9.
What is cladistics?
Answer:
Biologists initiated studies on the evolutionary and genetic relationships among organisms, which led to the emerge of phylogenetic classification or cladistics. It is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the genetic differences between all species in the ‘phylogenetic tree’.

Question 10.
Define cladogram?
Answer:
Ernst Haeckal introduced the method of representing evolutionary relationships with the help of a tree diagram known as cladogram.

Question 11.
What are the points to be followed to draw a cladogram?
Answer:
Phylogenetic system of classification takes into account ancestral characters (traits of basic body design which would be in the entire group) and derived characters (traits whose structure and functions differ from those of ancestral characters). One or more derived characters which appeared during evolution resulted in the formation of new subspecies. In a cladogram, each evolutionary step produces a branching and all the members of the branch would possess the derived character which will not be seen in organisms below the particular branch point. Arranging organisms on the basis of their similar or derived characters which differ from the ancestral characters produced a phylogenetic tree or cladogram.

Question 12.
What is the Five Kingdom classification? Who proposed this?
Answer:
Five kingdom classification defined by R.H. Whittaker as Monera, Protista, Fungi, Plantae, and Animalia based on the cell structure, mode of nutrition, mode of reproduction, and phylogenetic relationships.

Question 13.
Which aspect paved the way for three domain classification?
Answer:
Classification has come a long way and now takes into account even molecular level DNA and RNA identification. The advancement in molecular techniques and biochemical assays has led to a new classification – The “Three Domain” classification.

Question 14.
Who proposed three-domain classification? What are the features of this classification?
Answer:
Three domain classification was proposed by Carl Woese. This system emphasizes the separation of prokaryotes into two domains. Bacteria and Archaea, and all the eukaryotes are placed into the domain Eukarya. Archaea appears to have more in common with die Eukarya than the Bacteria. Archaea differ from bacteria in cell wall composition and differs from bacteria and eukaryotes in membrane composition and rRNA types.

Question 15.
What are the three domains of living things? Write it with examples.
Answer:
Three domains are Archaea, Bacteria, Eukarya.

(Three Domains)
ARCHAEA (Extremophiles)	BACTERIA	EUKARYA (Eukaryotes)
Methanogens, Halophiles, Thermoacidophiles	Cyanobactiera & Eubacteria, beneficial & pathogenic	Protista, Fungi, Plants Animals

Question 16.
Why domain Archaea are called extremophiles?
Answer:
The domain Archaea includes single-celled organisms, the prokaryotes which have the ability to grow in extreme conditions like volcano vents, hot springs, and polar ice caps, hence are also called extremophiles. They are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulfide and other chemicals from volcanic vents.

Question 17.
Define the following terms – (i) Halophiles, (ii) Methanogens, (iii) Thermoacidophiles.
Answer:

1. Halophiles: Organisms that live in salty environments are called Halophiles.
2. Methanogens: Some of the extremophiles that produce methane are called methanogens.
3. Thermoacidophiles: Thermoacidophiles thrive in acidic environments and at high temperatures.

Question 18.
Name the two broad classifications of Bacteria?
Answer:
Bacteria are of two types. They are beneficial probiotic bacteria and harmful pathogenic bacteria.

Question 19.
What is the role of Cyanobacteria dulling eat ly geologic periods?
Answer:
Cyanobacteria are photosynthetic blue-green algae during geological periods which produce oxygen. These had played a key role in the changes of atmospheric oxygen levels from anaerobic to aerobic during the early geologic periods.

Question 20.
Classify the organisms as per the seven kingdom system of classification?
Answer:
In 1987, Cavalier-Smith revised the six kingdom system to the Seven Kingdom system. According to that, the classification is divided into two Super Kingdoms (Prokaryota and Eukaryota) and seven kingdoms, two Prokaryotic Kingdoms (Eubacteria and Archaebacteria) and five Eukaryotic Kingdoms (Protozoa, Chromista, Fungi, Plantae, and Animalia).

Question 21.
Define species?
Answer:
Species are the basic unit of classification in the taxonomic hierarchical system. It is a group of animals having similar morphological features (traits) and is reproductively isolated to produce fertile offspring.

Question 22.
Differentiate monotypic and polytypic genus?
Answer:
Genus: It is a group of closely related species which have evolved from a common ancestor. In some genus, there are only one species which is called a monotypic genus.
eg. Red panda is the only species in the genus Ailurus: Ailurus fulgens. If there are more than one species in the genus it is known as polytypic genus, eg. ‘Cats’ come under the genus Felis, which has a number of closely related species.

Question 23.
Define family?
Answer:
It is a taxonomic category that includes a group of related genera with less similarity as compared to genus and species. For example, the family Felidae includes the genus Felis (cats) and the genus, Panthera (lions, tigers, leopards).

Question 24.
Define class?
Answer:
This category includes one or more related orders with some common characters, eg. order Primata comprising monkeys, apes, and man is placed in the Class Mammalia, along with the order Carnivora which includes dogs and cats.

Question 25.
Write the resultant animal when there is a cross between (i) a Male donkey with a female horse, (ii) a male lion with a female tiger, (iii) a male tiger with a female lion.
Answer:

1. Male donkey with female horse results in Mule. (Sterile)
2. Male lion with female tiger results in Liger.
3. Male tiger with female lion results in Tigon.

Question 26.
What is Binomial nomenclature? What is its significance?
Answer:
Biologists follow universally accepted principles to provide scientific names to known organisms. Each name has two components, a generic name, and a specific epithet. This system of naming the organism is called Binomial Nomenclature.
Classification and grouping were done to facilitate a deeper understanding of the unique characteristics of each organism and its interrelationship among closely related species.
It plays a vital role in the arrangement of known species based on their similarities and dissimilarities.

Question 27.
Write down the binomial names for the following (i) National Bird of India, (ii) National Animal of India, (iii) Tamil Nadu State Bird.
Answer:

1. National Bird of India – Pavo cristatus (Indian Peafowl)
2. National Animal of India Panthera tigris (tiger)
3. Tamil Nadu State Bird – Chalcophaps indica (Emerald dove)

Question 28.
When does the system of trinomial nomenclature is followed?
Answer:
When members of any species have large variations then a trinomial system is used. On the basis of dissimilarities, this species gets classified into subspecies.

Question 29.
What are the basic rules followed for naming the animals scientifically? (or) Describe the rules of Nomenclature.
Answer:

1. The scientific name should be italicized in printed form and if handwritten, it should be underlined separately.
2. The generic name’s (Genus) first alphabet should be in uppercase.
3. The specific name (species) should be in lowercase.
4. The scientific names of any two organisms are not similar.
5. The name dr abbreviated name of the scientist who first publishes the scientific name may be written after the species name along with the year of publication, eg. Lion-Felis Leo Linn., 1758 or Felis Leo L., 1758.
6. If the species name is framed after any person’s name the name of the species shall end with ‘i’, ‘ii’ or ‘ae’. eg. A new species of a ground-dwelling lizard (Cyrtodactylus) has been discovered and named after scientist Varad Giri, Cyrtodactylus varadgirii.

Question 30.
What are the tools used for the study of taxonomy?
Answer:
Tools and taxonomical aids may be different for the study of plants and animals. Herbarium and Botanical garden may be used as tools for the study of plant taxonomy. In the case of animal studies, the classical tools are Museum, Taxonomical Keys, and Zoological and Marine parks.

Question 31.
What are the components of taxonomical tools?
Answer:
The important components of the taxonomical tools are field visits, survey, identification, classification, preservation, and documentation.

Question 32.
Differentiate museum from the zoological park?
Answer:
Museum: Biological museums have a collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.
Zoological parks: These are places where wild animals are kept in protected environments under human care which enables us to study their food habits and behavior.

Question 33.
What are the recent molecular tools used for taxonomical studies?
Answer:
DNA hybridization (measures the degree of genetic similarity between pools of DNA sequences), DNA fingerprinting (to identify an individual from a sample of DNA by looking at unique patterns in their DNA), Restriction Fragment Length Polymorphisms (RFLP) analysis (difference in homologous DNA sequences that can be detected by the presence of fragments of different lengths after digestion of the DNA samples), and Polymerase Chain Reaction (PCR) sequencing (to amplify a specific gene or portion of a gene,) are used as taxonomical tools.

Question 34.
Give the abbreviations for the following: ALIS, DAISY, ABIS, SPIDA, Draw wing.
Answer:

ALIS	Automated Leafhopper Identification System.
DAISY	Digital Automated Identification System.
ABIS	Automatic Bee Identification System.
SPIDA	Species Identified Automatically (spiders, wasp, and bee wing characters).
Draw Wing	Honey Bee wing identification.

Question 35.
Write about some taxonomical tools.
Answer:

1. Neo taxonomical tools: This is based on Electron Microscopy images to study the molecular structures of cell organelles.
2. Ethology of taxonomical tools: Based on the behavior of the organisms it can be classified. For example sound of birds, bioluminescence, etc.
3. e-Taxonomic resources: INOTAXA is an electronic resource for digital images and descriptions of the species which was developed by the Natural History Museum, London. INOTAXA means Integrated Open TAXonomic Access.

Question 36.
What is Tautonymy?
Answer:
The practice of naming the animals in which the generic name and species name are the same is called Tautonymy. eg. Naja naja (The Indian Cobra).

Question 37.
What kind of adaptations extremophiles have to live in extreme conditions.
Answer:

1. Extremophiles are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulfide and other chemicals from volcanic vents.
2. Methanogens produced methane.
3. Halophiles live in salty environments and thermoacidophiles thrive in acidic environments and high temperatures.

Question 38.
Why T. aquaticus is used in PCR?
Answer:
T. aquaticus: Thermus aquaticus.
PCR: Polymerase chain reaction.
The DNA polymerase found in Thermus aquaticus remains stable even at very high temperatures.
Because of this stability, it can be used in the process known as PCR.

Question 39.
If you discover a new species how will you name them nominally?
Answer:
Before naming a new species there are certain pre-requisites that need to be followed. Initially, a holotype must be designated. A holotype is nothing but a single specimen that serves as the identifier for the entire new Species. The holotype must possess the key features that differentiate the new species from the existing ones i.e., it must be unique. The details of the origin of the holotype, the environment in which it was collected and its paratypes must be provided.
Secondly, a description of the species must be provided. The definition must be in terms of behavioral, anatomical, and genetic marks of the new species.
Thirdly, the species must be allocated a new name. Species are always identified by both a generic name and a species name. The generic name of the species is the genus to which it belongs to and then the new species name is added. There are several rules and conventions involved in creating and managing the names of the new species.
Finally, the details of the newly found species must be published in an internationally accessible format and archived in multiple locations.

Question 40.
Why pig-nosed frog in India is called Bhupathy’s purple frog?
Answer:
Scientists have discovered a new and unusual species of frog in the Western Ghats mountain. range in India.
The frog has shiny, purple skin. The scientists have called the new species as Bhupathy’s purple frog, in honor of their colleague Dr. Subramaniam Bhupathy, a respected herpetologist who lost his life in the Western Ghats in 2014.

Question 41.
Where did they discover a freshwater jellyfish in Tamilnadu?
Answer:
Freshwater jellyfish are extremely rare. Ishaan Abraham Pichamuthu, the nine-year-old boy, discovered a new species of freshwater jellyfish thriving in the depths of Kodaikanal lake.

Question 42.
What are the newly discovered species in India?
Answer:

1. 55 vertebrates and 258 invertebrates are the new animal species discovered in India.
2. Insects 97 species. Fishes 27 species, Amphibians 12 species, Platyhelminthes 10 species, Crustacea 9 species, Reptiles 6 species have been discovered and described by the scientist.
3. Moths and butterflies 61 species, Beetles 38 species also identified.
4. Most of these new species were from biological hotspots of the country, – the Himalayas, the Western Ghats, and the Andaman and Nicobar Islands.

Question 43.
How to save endangered species.
Answer:

1. Preventing listed species from being killed or harmed.
2. Protecting habitat essential to the survival of endangered species.
3. Creating plans to restore a healthy population of endangered animals.
4. Never purchase products made from threatened or endangered species.
5. Don’t use toxic herbicides or pesticides.
6. Support zoos and other wildlife parks.
7. Defending and strengthening the endangered species act.
8. Advocating for increased funding for conservation programs that benefits endangered species.
9. Slow down while driving in forest areas and watch out for wild animals cross the road.
10. Preserving water bodies in habitats of endangered species to ensure then sustaining.

Question 44.
What are the different taxonomical tools? Explain in detail.
Answer:
Tools and taxonomical aids may be different for the study of plants and animals. Herbarium and Botanical garden may be used as tools for the study of plant taxonomy. In the case of animal studies, the classical tools are Museum, Taxonomical Keys, and Zoological and Marine parks.
The important components of the taxonomical tools are field visits, survey, identification, classification, preservation, and documentation. Many tools are being used for taxonomical studies, amongst them, some of the important tools are discussed below:
The classical taxonomical tools:
Taxonomical Keys: Keys are based on a comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.
Museum: Biological museums have a collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.
Zoological parks: These are places where wild animals are kept in protected environments under human care which enables us to study their food habits and behavior.
Marine parks: Marine organisms are maintained in a protected environment. Printed taxonomical tools consist of identification cards, descriptions, field guides, and manuals.
Molecular taxonomical tools:
Technological advancement has helped to evolve molecular taxonomical tools from classical tools to molecular tools. The accuracy and authenticity is more significant in the molecular tools. The following methods are being used for taxonomical Classification.
Molecular techniques and approaches such as DNA barcoding (short genetic marker in an organism’s DNA to identify it as belonging to a particular species), DNA hybridization (measures the degree of genetic similarity between pools of DNA sequences), DNA fingerprinting (to identify an individual from a sample of DNA by looking at unique patterns in their DNA), Restriction Fragment Length Polymorphisms (RFLP) analysis (difference in homologous DNA sequences that can be detected by the presence of fragments of different lengths after digestion, of the DNA samples), and Polymerase Chain Reaction (PCR) sequencing (to amplify a specific gene or portion of a gene,) are used as taxonomical tools.
Automated species identification tools:
It consists of Cyber tools, eg. DAISY, ALIS, ABIS, SPIDA, Draw wing, etc.

ALIS	Automated Leafhopper Identification System.
DAISY	Digital Automated Identification System.
ABIS	Automatic Bee Identification System.
SPIDA	Species Identified Automatically (spiders, wasp, and bee wing characters).
Draw Wing	Honey Bee wing identification.

Neo taxonomical tools: This is based on Electron Microscopy images to study the molecular structures of cell organelles.
Ethology of taxonomical tools: Based on the behavior of the organisms it can be classified. For example sound of birds, bioluminescence, etc.
e-Taxonomic resources: INOTAXA is an electronic resource for digital images and descriptions of the species which was developed by the Natural History Museum, London. INOTAXA means Integrated Open TAXonomic Access.

Choose the correct answer.

1. The term bio-diversity was first introduced by:
(a) A. G. Tansley
(b) Walter Rosen
(c) Aristotle
(d) AP de Candle
Answer:
(b) Walter Rosen

2 is called as the father of taxonomy:
(a) Carolus Linnaeus
(b) Aristotle
(c) Theophrastus
(d) John Ray
Answer:
(b) Aristotle

3. The word taxonomy was coined by:
(a) AP de Candolle
(b) Ernst Haeckel
(c) Carl Woese
(d) Smith
Answer:
(a) AP de Candolle

4. The father f modern taxonomy is:
(a) John Ray
(b) Huxley
(c) Darwin
(d) Carolus Linnaeus
Answer:
(d) Carolus Linnaeus

5. According to Aristotle, the animals with red blood cellš are called:
(a) Anaima
(b) chromista
(c) Enaima
(d) Protozoa
Answer:
(c) Enaima

6. Five kingdom classification was proposed by:
(a) Walter Rosen
(b) R.H. Whittaker
(c) AG Tansley
(d) Bauhin
Answer:
(b) R.H. Whittaker

7. Three domain classification was proposed by:
(a) Carl Woese
(b) Huxley
(c) Varad Gin
(d) Aristotle
Answer:
(a) Carl Woese

8. Prokaryotes that have the ability to grow in extreme conditions are called:
(a) Probiotic
(b) Pathogenic
(c) Extremophiles
(d) Archaea
Answer:
(c) Extremophiles

9. The cell wall of bacteria contains:
(a) Glycogen
(b) Peptidoglycans
(c) Polypeptides
(d) Histones
Answer:
(b) Peptidoglycans

10. Seven Kingdom system was proposed by:
(a) Cavalier-Smith
(b) Adams
(c) AG Tansley
(d) Walter Rosen
Answer:
(a) Cavalier-Smith

11. Mating between the male lion and female tiger results in:
(a) Mule
(b) Tigon
(c) Liger
(d) Feus
Answer:
(c) Liger

12. The scientffic name of human is:
(a) Homo erectus
(b) Homo sapiens
(c) Emberi
(d) Umano
Answer:
(b) Homo sapiens

13. The binomial name of Indian National Bird is:
(a) Pavo cristatus
(b) Homo sapiens
(c) Ailurusfulgens
(d) Fells maigarita
Answer:
(a) Pavo cristatus

14. The binomial name of National animal of India is:
(a) Panthera tigris
(b) Corvus splendens
(c) Felis domestica
(d) Liger
Answer:
(a) Panthera tigris

15. ‘Origin of species’ the book that explains the evolutionary connections of species by the process of natural selection, written by:
(a) Linnaeus
(b) Aristotle
(c) Darwin
(d) Lamark
Answer:
(c) Darwin

16. India has about …………. number of threatened species of animals as per the data given IUCN.
(a) 172
(b) 72
(c) 127
(d) 721
Answer:
(a) 172

Students get through the TN Board 11th Bio Botany Important Questions Chapter 15 Plant Growth and Development which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 15 Plant Growth and Development

Very short answer questions

Question 1.
Give the definition for Growth.
Answer:
Growth is defined as an irreversible permanent increase in size, shape, number, volume, and dry weight.

Question 2.
Explain the term ‘Monocarpic perennial’.
Answer:
Monocarpic perennials produce flowers only once during their lifetime but the plants survive for many years. Example: Bamboo.

Question 3.
Define the grand period of growth and mention its phases.
Answer:
The total period from the initial to the final stage of growth is called the grand period of growth. They are:

1. Lag phase,
2. Log phase,
3. Decelerating phase and
4. Maturation phase.

Question 4.
Plot a graph depicting the constant linear growth.
Answer:

Question 5.
Compare absolute growth rate with relative growth rate.
Answer:
Comparison between Absolute and Relative Growth Rates

Absolute Growth Rate	Relative Growth Rate
Increase in total growth of two organs measured and compared per unit time is called absolute growth rate.	The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
Define Differentiation.
Answer:
The process of maturation of meristematic cells to specific types of cells performing specific functions is called differentiation.

Question 7.
What is Re-differentiation?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation.

Question 8.
What are Plant Growth Regulators (PGR’s)?
Answer:
Plant Growth Regulators are defined as organic substances which are synthesized in minute quantities in one part of the plant body and transported to another part where they influence specific physiological processes.

Question 9.
Mention any four Phytohormones.
Answer:
Auxins, gibberellins, cytokinin, ethylene, abscisic acid.

Question 10.
Define the term Bioassay.
Answer:
Bioassay means-testing of substances for their activity in causing a growth response in a living plant or its part.

Question 11.
Explain Anti-auxins.
Answer:
Anti-auxin compounds when applied to the plant inhibit the effect of auxin. Example: 2, 4, 5-Tri Iodine Benzoic Acid (TIBA) and Napthylpthalamine.

Question 12.
Apical dominance – Explain.
Answer:
Suppression of growth in the lateral bud by apical bud due to auxin produced by apical bud is termed as apical dominance.

Question 13.
Mention the precursors of (a) Auxin, (b) Gibberellin, (c) Cytokinin and (d) Abscisic acid.
Answer:

S. No.	Plant Hormones	Precursors
(a)	Auxin	Tryptophan
(b)	Gibberellin	Acetate
(c)	Cytokinin	Adenine
(d)	Abscisic acid	Xanthophylls

Question 14.
How Gibberellins are transported from their site of production?
Answer:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through the phloem and also occur in the xylem due to lateral movement between vascular bundles.

Question 15.
What do you mean by the term – Basipetal transport and Acropetal transport?
Answer:
Basipetal means transport through the phloem from shoot to root and acropetal means transport through the xylem from root to shoot.

Question 16.
Define Bolting.
Answer:
Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of a stem followed by flowering is called bolting.

Question 17.
Which plants are affected by Bakanae’s disease? Name the causative organism.
Answer:
Bakanae’s disease affects the rice plants. It is caused by a fungus called Gibberella fujikuroi.

Question 18.
Biennials usually flower at the second year of growth. Is it possible to make them flower in the first year itself? How?
Answer:
Many biennials usually flower during the second year of their growth. For flowering to take place, these plants should be exposed to the cold season. Such plants could be made to flower without exposure to the cold season in the first year itself when they are treated with gibberellins.

Question 19.
Where cytokinin is synthesized in plants?
Answer:
Cytokinin is formed in root apex, shoot apex, buds, and young fruits.

Question 20.
Give an account on neem Coty Ledon assay.
Answer:
Bioassay (Neem Cotyledon Assay): Neem cotyledons are measured and placed in cytokinin solution as well as in ordinary water. Enlargement of cotyledons is an indication of cytokinin activity.

Question 21.
State Richmond Lang effect.
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as the Richmond Lang effect.

Question 22.
Write a note on Bioassay of Ethylene.
Answer:
Ethylene can be measured by gas chromatography. This technique helps in the detection of the exact amount of ethylene from different plant tissues like lemon and orange.

Question 23.
What are climacteric fruits? Give example.
Answer:
In most of the plants, there is a sharp rise in respiration rate near the end of the development of fruit, called climacteric rise. Such fruits are called climacteric fruits. Example: Tomato, Apples, Banana, Mango.

Question 24.
ABA is called the stress hormone – Justify.
Answer:
It inhibits the shoot growth and promotes the growth of the root system. This character protects the plants from water stress. Hence, ABA is called the stress hormone.

Question 25.
Define photoperiodism. Name the person who coined this term.
Answer:
The physiological change on flowering due to relative length of light and darkness (photoperiod) is called Photoperiodism. The term photoperiodism was coined by Gamer and Allard.

Question 26.
Define the term photo neutrals.
Answer:
There are a number of plants that can flower in all possible photoperiods. They are also called photo neutrals or indeterminate plants. Example: Potato, Rhododendron, Tomato, and Cotton.

Question 27.
Point out the importance of photoperiodism.
Answer:
Importance of photoperiodism

1. The knowledge of photoperiodism plays an important role in hybridization experiments.
2. Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 28.
Vernalisation – Define.
Answer:
Many species of biennials and perennials are induced to flower by low-temperature exposure ‘ (0°C to 5°C). This process is called Vernalization.

Question 29.
How vernalisation is carried out?
Answer:
The seeds are first soaked in water and allowed to germinate at 10°C to 12°C. Then seeds are transferred to low temperature (3°C to 5°C) from few days to 30 days. Germinated seeds after this treatment are allowed to dry and then sown. The plants will show quick flowering when compared to untreated control plants.

Question 30.
Define Seed germination and state its types.
Answer:
The activation and growth of an embryo from seed into seedling during favorable conditions is called seed germination. There are two methods of seed germination. Epigeal and hypogeal.

Question 31.
What are photoelastic seeds?
Answer:
There are many seeds that respond to light for germination and these seeds said to be photoelastic.

Question 32.
How can we overcome dormancy in photoelastic seeds?
Answer:
The dormancy of photoelastic seeds can be broken by exposing them to red light.

Question 33.
Senescence-Define.
Answer:
Old age is called senescence in plants. Senescence refers to all collective, progressive and deteriorative processes which ultimately lead to complete loss of organization and function.

Question 34.
What field do phytogerontology deal with?
Answer:
The branch of botany that deals with aging, abscission, and senescence is called Phytogerontology.

Question 35.
Mention the types of senescence.
Answer:
Leopold (1961) has recognized four types of senescence:

1. Overall senescence
2. Top senescence
3. Deciduous senescence
4. Progressive senescence

Question 36.
Where the abscission zone is formed?
Answer:
Leaf abscission takes place at the base of the petiole which is marked internally by a distinct zone of few layers of thin-walled cells arranged transversely. This zone is called the abscission zone or abscission layer. An abscission layer is greenish-grey in color and is formed by rows of cells of 2 to 15 cells thick.

Question 37.
Expand PCD and define it.
Answer:
Death of the plant or its parts consequent to senescence is called Programmed Cell Death (PCD).

Short answer questions

Question 1.
Draw a graph of the sigmoid curve.
Answer:

Question 2.
Correlate the terms light & etiolation in the plant.
Answer:
Light has its own contribution to the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. The absence of light may lead to yellowish in color. This is called etiolation.

Question 3.
List out the internal factors that affect growth.
Answer:
a. Genes are intracellular factors for growth.
b. Phytohormones are intracellular factors for growth. Example: auxin, gibberellin, cytokinin.
c. C/N ratio.

Question 4.
Give an account on Dedifferentiation.
Answer:
The living differentiated cells which had lost capacity to divide, regain the capacity to divide under certain conditions. Hence, dedifferentiation is the regaining of the ability of cell division by the differentiated cells. Example: Interfascicular cambium and Vascular cambium.

Question 5.
Explain synergistic & Antagonistic effects of plant hormones.
Answer:
Synergistic and Antagonistic effects

1. Synergistic effects: The effect of one or more substances in such a way that both promote each other’s activity. Example: Activity of auxin and gibberellins or cytokinins.
2. Antagonistic effects: The effect of two substances in such a way that they have opposite effects on the same process. One accelerates and the other inhibits. Example: ABA and gibberellins during seed or bud dormancy. ABA induces dormancy and gibberellins break it.

Question 6.
Name any three synthetic auxins.
Answer:
1. 2,4-Dichloro Phenoxy Acetic Acid (2,4-D)
2. 2,4,5-Trichloro Phenoxy Acetic Acid (2,4,5-T)
3. Naphthalene Acetic Acid (NAA).

Question 7.
Describe the procedure of the Avena curvature test.
Answer:
When the Avena seedlings have attained a height of 15 to 30 mm, about 1mm of the coleoptile tip is removed. This apical part is the source of natural auxin. The tip is now placed on agar blocks for few hours. During this period, the auxin diffuses out of these tips into the agar. The auxin containing agar block is now placed on one side of the decapitated stump of Avena coleoptile. The auxin from the agar blocks diffuses down through coleoptile along the side to which the auxin agar block is placed. An agar block without auxin is placed on another decapitated coleoptile. Within an hour, the coleoptiles with auxin agar block bend on the opposite side where the agar block is placed. This curvature can be measured.

Question 8.
How auxin is useful in tissue culture technique?
Answer:
Auxin is responsible for the initiation and promotion of cell division in the cambium, which is responsible for secondary growth and tumor. This property of induction of cell division has been exploited for tissue culture techniques and for the formation of callus.

Question 9.
Mention the role of ethylene in the agriculture field.
Answer:

1. Ethylene normally reduces flowering in plants except in Pineapple and Mango.
2. It increases the number of female flowers and decreases the number of male flowers.
3. Ethylene spray in the cucumber crops produces female flowers and increases the yield.

Question 10.
Give an answer for the following with regard to abscisic acid.
(a) Chemical structure (b) Precursor (c) Bioassay
Answer:
(a) Chemical structure – carotenoid structure
(b) Precursor – Mevalonic acid pathway or Xanthophyll
(c) Bioassay – Rice coleoptile.

Question 11.
How ABA involves stomatal closure.
Answer:
ABA helps in reducing the transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in the closure of stomata.

Question 12.
Enumerate the practical application of vernalization.
Answer:

1. Vernalization shortens the vegetative period and induces the plant to flower earlier.
2. It increases the cold resistance of the plants.
3. It increases the resistance of plants to fungal disease.
4. Plant breeding can be accelerated.

Question 13.
Write a note on Epigeal & hypogeal germination.
Answer:

1. Epigeal Germination: During epigeal germination, cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. Example: Castor and Bean.
2. Hypogeal Germination: During hypogeal germination cotyledons remain below the soil due to rapid elongation of epicotyls. Example: Maize.

Question 14.
Discuss seed viability.
Answer:
Viability: Usually seeds remain viable or living only for a particular period. Viability of seeds ranges from a few days (Example: Oxalis) to more than a hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

Question 15.
Comment on seed dormancy and its reason.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental conditions is called seed dormancy. There are two main reasons for the development of dormancy: Imposed dormancy and innate dormancy. Imposed dormancy is due to low moisture and low temperature. Innate dormancy is related to the properties of the seed itself.

Question 16.
Which is the final stage of senescence? Define it.
Answer:
The final stage of senescence is abscission. Abscission is a physiological process of shedding of organs like leaves, flowers, fruits, and seeds from the parent plant body.

Question 17.
Mention the role of hormones in Abscission.
Answer:
All naturally occurring hormones influence the process of abscission. Auxins and cytokinins retard abscission, while abscisic acid (ABA) and ethylene induce it.

Question 18.
Why Abscission has to take place?
Answer:
Significance of abscission

1. Abscission separates dead parts of the plant, like old leaves and ripe fruits.
2. It helps in the dispersal of fruits and continuing the life cycle of the plant.
3. Abscission of leaves in deciduous plants helps in water conservation during summer.
4. In lower plants, shedding of vegetative parts like gemmae or plantlets helps in vegetative reproduction.

Long answer questions

Question 1.
Enumerate the characteristics of growth.
Answer:

1. Growth increases in protoplasm at the cellular level.
2. Stem and roots are indeterminate in growth due to continuous cell division and are called an open form of growth.
3. The primary growth of the plant is due to the activity of apical meristem where new cells are added to the root and shoot apex causing linear growth of the plant body.
4. The secondary vascular cambium and cork cambium add new cells to cause an increase in girth.
5. Leaves, flowers, and fruits are limited in growth or of determinate or closed-form growth.

Question 2.
Explain the three phases of growth.
Answer:
Phases of growth
There are three phases of growth, 1. Formative phase, 2. Elongation phase and 3. Maturation phase

1. Formative phase: Growth in this phase occurs in meristematic cells of the shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus, and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain the capability of cell division while other cells enter the next phase of growth.
2. Elongation Phase: Newly formed daughter cells are pushed out of the meristematic zone
   and increase the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), the addition of protoplasm, and development of central vacuole take place.
3. Maturation Phase: During this stage cells attain mature form and size. Thickening and differentiation take place. After differentiation, the cells do not grow further.

Question 3.
Give a detailed account of the geometric growth rate.
Answer:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In-plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. Example: Round three in the given figure 15.5, produces 8 cells as 23 5 8 and after round 20 there are 220 5 1,048,576 cells. The large, plant, or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small.

An exponential growth curve can be expressed as,
W₁ = W₀e^rt
W₁ = Final size (weight, height and number)
W₀ = Initial size at the beginning of the period
r = Growth rate
t = Time of growth
e = Base of the natural logarithms
Here r is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W₁ depends on the initial size W₀.

Question 4.
List out the characteristics of phytohormones.
Answer:
Characteristics of phytohormones

1. Usually produced in tips of roots, stems, and leaves.
2. The transfer of hormones from one place to another takes part through conductive systems.
3. They are required in trace quantities.
4. All hormones are organic in nature.
5. There are no specialized cells or organs for their secretion.
6. They are capable of influencing physiological activities leading to promotion, inhibition, and modification of growth.

Question 5.
List out the physiological effects of Auxin.
Answer:
Physiological Effects

1. They promote cell elongation in stem and coleoptile.
2. At higher concentrations, auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root-only at extremely low concentrations.
3. Suppression of growth in the lateral bud by apical bud due to auxin produced by apical bud is termed as apical dominance.
4. Auxin prevents abscission.
5. It is responsible for the initiation and promotion of cell division in the cambium, which is responsible for secondary growth and tumor. This property of induction of cell division has been exploited for tissue culture techniques and for the formation of callus.
6. Auxin stimulates respiration.
7. Auxin induces vascular differentiation.

Question 6.
Mention the role of Auxin in Agri-field.
Answer:

1. It is used to eradicate weeds. Example: 2,4-D and 2,4,5-T.
2. Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
3. It is used to break the dormancy in seeds.
4. Induce flowering in Pineapple by NAA & 2,4-D.
5. Increase the number of female flowers and fruits in cucurbits.

Question 7.
Mention the role of Gibberellin in agriculture.
Answer:

1. The formation of seedless fruits without fertilization is induced by gibberellins. Example: Seedless tomato, apple, and cucumber.
2. It promotes the formation of male flowers in Cucurbitaceae. It helps in crop improvement.
3. Uniform bolting and increased uniform seed production.
4. Improves the number and size of fruits in grapes. It increases yield.
5. Promotes elongation of inter-node in sugarcane without decreasing sugar content.
6. Promotion of flowering in long-day plants even under short-day conditions.
7. It stimulates seed germination.

Question 8.
Explain the physiological effect of Ethylene.
Answer:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in stem and root and inhibits linear growth.
• It breaks the dormancy of buds, seeds, and storage organiser.
• It stimulates the formation of the abscission zone in leaves, flowers, and fruits. This makes the leaves shed prematurely.
• Inhibition of stem elongation (shortening the internode).
• In low concentrations, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

Question 9.
List out the physiological effects of Abscisic acid.
Answer:

• It helps in reducing the transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in the closure of stomata.
• It spoils chlorophylls, proteins, and nucleic acids of leaves making them yellow.
• Inhibition of cell division and cell elongation.
• ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
• It induces bud and seed dormancy.
• It promotes the abscission of leaves, flowers, and fruits by forming abscission layers.
• ABA plays an important role in plants during water stress and during drought conditions. It results in loss of turgor and closure of stomata.
• It has anti-auxin and anti-gibberellin properties.
• Abscisic acid promotes senescence in leaves by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis.

Question 10.
Classify and explain the plants based on photoperiodism.
Answer:

1. Long day plants: The plants that require long critical day length for flowering are called long-day plants or short night plants. Example: Pea, Barley, and Oats.
2. Short long-day plants: These are long-day plants but should be exposed to short-day lengths during the early period of growth for flowering. Example: Wheat and Rye.
3. Short-day plants: The plants that require a short critical day length for flowering are called short-day plants or long night plants. Example: Tobacco, Cocklebur, Soybean, Rice, and Chrysanthemum.
4. Long short day plants: These are actually short-day plants but they have to be exposed to long days during their early periods of growth for flowering. Example: Some species of Bryophyllum and Night jasmine.
5. Intermediate day plants: These require a photoperiod between long days and short days for flowering. Example: Sugarcane and Coleus.
6. Day-neutral plants: There are a number of plants that can flower in all possible photoperiods. They are also called photo neutrals or indeterminate plants. Example: Potato, Rhododendron, Tomato, and cotton.

Question 11.
Describe the concept of Phytochrome.
Answer:
Phytochrome is a bluish biliprotein pigment responsible for the perception of light in a photo physiological process. Butler et al, (1959) named this pigment and it exists in two interconvertible forms: (i) red light-absorbing pigment which is designated as P_r and (ii) far-red light-absorbing pigment which is designated as P_fr. The P_r form absorbs red light in 660 nm and changes to P_fr. The P_fr form absorbs far-red light in 730 nm and changes to P_r. The P form is biologically inactive and it is stable whereas P_fr form is biologically active and it is very unstable. In short-day plants, P_r promotes flowering and P_fr inhibits the flowering whereas in long-day plants flowering is promoted by P_fr and inhibited by P_r form. P_fr is always associated with a hydrophobic area of membrane systems while P_r is found in the diffused state in the cytoplasm. The interconversion of the two forms of phytochrome is mainly involved in flower induction and also additionally plays a role in seed germination and changes in membrane conformation.

Question 12.
Explain the theories involved in the mechanism of vernalization.
Answer:
Mechanism of Vernalization:
Two main theories to explain the mechanism of vernalization is:
i. Hypothesis of phasic development,
ii. The hypothesis of hormonal involvement.

1. The hypothesis of phasic development: According to Lysenko, the development of an annual seed plant consists of two phases. The first phase is a thermostat, which is a vegetative phase requiring low temperature and suitable moisture. The next phase is the photo stage which requires high temperature for the synthesis of florigen (flowering hormone).
2. The hypothesis of hormonal involvement: According to Purvis (1961), the formation of a substance A from its precursor, is converted into B after chilling. The substance B is unstable. At a suitable temperature, B is converted into stable compound D called Vemalin. Vemalin is converted to F (Florigen). Florigen induces flower formation. At high-temperature B is converted to C and devemalization occurs.

Question 13.
Discuss the external factors that affect seed germination.
Answer:
External Factors:
a. Water: It activates the enzymes which digest the complex reserve foods of the seed. If the water content of the seed goes below a critical level, seeds fail to germinate.
b. Temperature: Seeds fails to germinate at very low and high temperature. The optimum temperature is 25°C to 35°C for most tropic species.
c. Oxygen: It is necessary for germination. Since aerobic respiration is a physiological requirement for germination most will germinate well in air containing 20% oxygen.
d. Light: There are many seeds that respond to light for germination and these seeds said to be photoelastic.
e. Soil Conditions: Germination of seed in its natural habit is influenced by soil conditions such as holding capacity, mineral composition, and aeration of the soil.

Question 14.
Illustrate the various methods of breaking seed dormancy.
Answer:
Methods of breaking dormancy:
The dormancy of seeds can be broken by different methods. These are:

1. Scarification: Mechanical and chemical, treatments like cutting or chipping of hard tough
   seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification. •
2. Impaction: In some seeds, water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Impaction.
3. Stratification: Seeds of rosaceous plants (Apple, Plum, Peach, and Cherry) will not germinate until they have been exposed to well-aerated, moist conditions under low temperatures (0°C to 10°C) for weeks to months. Such treatment is called Stratification.
4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating
   daily temperatures. An alternation of low and high temperature improves the germination of seeds. .
5. Light: The dormancy of photoblastic seeds can be broken by exposing them to red light.

Question 15.
Give a detailed account of different types of senescence.
Answer:

1. Overall senescence: This kind of senescence occurs in annual plants when the entire plant gets affected and dies. Example: Wheat and Soybean. It also occurs in few perennials also. Example: Agave and Bamboo.
2. Top senescence: It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable. Example: Banana and Gladiolus.
3. Deciduous senescence: It is common in deciduous plants and occurs only in the leaves of plants, the bulk of the stem and root system remains alive. Example: Elm and Maple.
4. Progressive senescence: This kind of senescence is gradual. First, it occurs in old leaves followed by new leaves then stems, and finally root system. It is common in annuals.

Question 16.
Explain the physiology of senescence.
Answer:
Physiology of Senescence

• Cells undergo changes in structure.
• Vacuole of the cell acts as a lysosome and secretes hydrolytic enzymes.
• The starch content is decreased in the cells.
• Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
• There is a marked decrease in protein content in the senescing organ.
• RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
• DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

Question 17.
What are the physiological changes occurring in plants during Abscission?
Answer:
Morphological and Anatomical changes during abscission:
Leaf abscission takes place at the base of the petiole which is marked internally by a distinct zone of few layers of thin-walled cells arranged transversely. This zone is called the abscission zone or abscission layer. An abscission layer is greenish-grey in color and is formed by rows of cells of 2 to 15 cells thick. The cells of the abscission layer separate due to the dissolution of the middle lamella and primary wall of cells by the activity of enzymes pectinase and cellulase resulting in the loosening of cells. Tyloses are also formed blocking the conducting vessels. Degrading of chlorophyll occurs leading to the change in the color of leaves, leaf detachment from the plant, and leaf fall. After abscission, the outer layer of cells becomes suberized by the development of periderm.

Higher Order Thinking Skills (HOTs)

Question 1.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth hormone can be applied to achieve this.
Answer:
Auxin can be used to increase the number of female flowers.

Question 2.
In the figure of the sigmoid and curve given below, label the segments A, B, and C.

Answer:
A = lag phase, B = log phase and C = Steady state phase.

Question 3.
In most plants, the terminal bud suppresses the development of lateral buds. What is this phenomenon called? Name the phytohormone that promotes this phenomenon.
Answer:
Suppression in the development of lateral buds due to terminal bud is called apical dominance. Auxins are the phytohormones that can promote this phenomenon.

Question 4.
The physiological effects of ethylene are both positive and negative. Comment.
Answer:

Positive Aspect of Ethylene	The negative aspect of Ethylene
Ethylene plays a crucial role in the ripening of fruits. Hence called as fruit ripening hormone.	It inhibits the longitudinal growth of stem and root.
It breaks the dormancy in seeds and buds.	It stimulates abscission causing leaves, flowers and fruits to shed prematurely.

Question 5.
Light plays an important role in the life of all organisms. Name any two physiological processes in plants that are affected by light.
Answer:
In plants, light plays a vital role in photosynthesis and growth.

Question 6.
Classify the following plants into long-day plants, short-day plants, and day-neutral plants, Wheat, sunflower, maize, tobacco, oats, chrysanthemum.
Answer:
Long Day plants – Wheat, Oats.
Short Day plants – Tobacco, chrysanthemum.
Day Neutral plants – Sunflower, maize.

Choose the correct answer.

1. Identify the monocarpic perennial plant from the following list.
(a) Paddy
(b) Bean
(c) Bamboo
(d) Coconut
Ans.
(c) Bamboo

2. Growth rate is maximum in ……….. phase.
(a) Lag
(b) Log
(c) Decelerating
(d) Maturation
Ans.
(b) Log

3. Proper plant growth occurs at a temperature between …………
(a) 25°C to 28°C
(b) 28°C to 45°C
(c) 25°C to 35°C
(d) 28°C to 30°C
Ans. (d) 28°C to 30°C

4. The process by which differentiated cell regain the ability of cell division ………….
(a) Differentiation
(b) Dedifferentiation
(c) Redifferentiation
(d) Totipotency
Ans.
(b) Dedifferentiation

5. The term Auxin was first used by ………….
(a) Kogl Smith
(b) Darwin
(c) F.W. Went
(d) Kurosawa
Ans.
(c) F.W. Went

6. Auxin was first isolated from ……….
(a) Com grain oil
(b) Human blood
(c) Human urine
(d) Rice bran oil
Ans.
(c) Human urine

7. Which is NOT a natural auxin?
(a) IAA
(b) PAA
(c) IPA
(d) NAA
Ans.
(d) NAA

8. The amino which is a precursor of IAA is ……….
(a) Methionine
(b) Valine
(c) Isoleuine
(d) Tryptophan
Ans.
(d) Tryptophan

9. Identify the wrong statement regarding the physiological effects of Auxin.
(i) Auxin prevents abscission
(ii) Auxin inhibits respiration
(iii) Auxin promotes cell elongation
(iv) Auxin breaks seed dormancy
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iv)
Ans.
(b) (ii) only

10. How many number of giberellins were discovered so far?
(a) 50
(6) 70
(c) 100
(d) 80
Ans.
(c) 100

11. Foolish seedling disease affects ………….
(a) Maize
(b) Rice
(c) Sorghum
(d) Wheat
Ans.
(b) Rice

12. Bakanae’s disease was first noticed by …………
(a) F.W. Went
(b) Kurosawa
(c) Cocken
(d) Denny
Ans.
(b) Kurosawa

13. Apical dominance is due to the effect of ………….
(a) Auxin
(b) Gibberellin
(c) Ethylene
(d) Cytokinin
Ans.
(a) Auxin

14. Match the following:
(a) a- iii, b – iv, c – ii, d – i
(b) a – i, b- iii, c – iv, d – ii
(c) a – iii, b – ii, c – iv, d – i
(d) a – ii, b – iii, c – iv, d – i
Ans.
(a) a- iii, b – iv, c – ii, d – i

15. The most widely occuring cytokinin in plants is ………..
(a) Iso propyl adenine
(b) Iso pentenyl adenine
(c) Indole propionic acid
(d) Iso propionic adenine
Ans.
(b) Iso pentenyl adenine

16. Which of the following does not act as a precursor of ethylene?
(a) Fumaric acid
(b) Malic acid
(c) Linolenic acid
(d) Methionine
Ans.
(b) Malic acid

17. Identify the non-climacteric fruit.
(a) Tomato
(b) Grapes
(c) Apples
(d) Mango
Ans.
(b) Grapes

18. ……….. is a stress hormone.
(a) Ethylene
(b) Cytokinin
(c) Auxin
(d) Abscissic acid
Ans.
(d) Abscissic acid

19. Which of the following plant hormone functions against auxin?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Ans.
(d) Abscissic acid

20. Closure of stomato can be induced by ………..
(a) Abscissic acid
(b) Ethylene
(c) NAA
(d) Cytokinin
Ans.
(a) Abscissic acid

21. Maryland mammoth requires …………. hours of light.
(a) 8
(b) 10
(c) 12
(d) 15.05
Ans.
(c) 12

22. Identify the day neutral plants.
(a) Pea
(b) Wheat
(c) Tomato
(d) Soyabean
Ans.
(c) Tomato

23. The Pr form absorbs red light in ………… nm.
(a) 730
(b) 620
(c) 660
(d) 635
Ans.
(c) 660

24. The term vernalisation was first used by …………
(a) Chailakyan
(b) Gamer & Allard
(c) Lysenko
(d) F.W. Went
Ans.
(c) Lysenko

25. …………. seeds show a maximum viability of 1000 years.
(a) Orange
(b) Oxalis
(c) Lotus
(d) Coconut
Ans.
(c) Lotus

26. Shaking the seeds vigorously to remove the plug in the microphyle is the method of …………..
(a) Stratification
(b) Impaction
(c) Scarification
(d) Emaciation
Ans.
(b) Impaction

27. After abicission, outer layer of cells becomes ………….. by periderm.
(a) Lignified
(b) Pectinised
(c) Suberized
(d) Stratified
Ans.
(c) Suberized

Students get through the TN Board 11th Bio Botany Important Questions Chapter 14 Respiration which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 14 Respiration

Very short answer questions

Question 1.
Define the term Respiration.
Answer:
During the night, plants take up oxygen and release carbon dioxide and as a result, carbon dioxide will be abundant around the tree. This process of CO₂ evolution is called respiration.

Question 2.
What do you mean by Respiratory substrate? Give example.
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins, and fats take place and as a result of this, energy is produced where O₂ is taken in and CO₂ is liberated. The organic substances which are oxidised during respiration are called respiratory substrates.

Question 3.
Write the overall equation of respiration.
Answer:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy
(686 K cal or 2868 KJ)

Question 4.
How Blackman classified respiration? Name them.
Answer:
Depending upon the nature of respiratory substrate, Blackman divided respiration into,

1. Floating respiration and
2. Protoplasmic respiration.

Question 5.
Compensation point – Define.
Answer:
At dawn and dusk, the intensity of light is low. The point at which CO₂ released in respiration is exactly compensated by CO₂ fixed in photosynthesis that means no net gaseous exchange takes place, it is called the compensation point. At this moment, the amount of oxygen released from photosynthesis is equal to the amount of oxygen utilized in respiration. The two common factors associated with the compensation point are CO₂ and light (Figure 14.2). Based on this there are two types of compensation points. They are CO₂ compensation point and light compensation point. C₃ plants have compensation points ranging from 40-60 ppm (parts per million) CO₂ while those of C₄ plants range from 1-5 ppm CO₂.

Question 6.
Name the types of compensation points.
Answer:
(a) CO₂ compensation point, (b) Light compensation point.

Question 7.
ATP is the universal energy currency of the cell – Justify.
Answer:
Respiration is responsible for the generation of ATP. The discovery of ATP was made by Karl Lohman (1929). ATP is a nucleotide consisting of a base-adenine, a pentose sugar-ribose, and three phosphate groups. Out of three phosphate groups, the last two are attached by high-energy-rich bonds (Figure 14.3). On hydrolysis, it releases energy (7.3 Kcal or 30.6 KJ/ATP) and it is found in all living cells and hence it is called the universal energy currency of the cell. ATP is an instant source of energy within the cell. The energy contained in ATP is used in Lipman (1941).

Question 8.
Who discovered ATP? Name any two other high-energy compounds in the cell similar to ATP?
Answer:
ATP was discovered by Karl Lohman in 1929. GTP (Guanosine Triphosphate), UTP (Uridine Triphosphate) are some other energy-rich compounds in the cell.

Question 9.
Expand the term (a) ATP, (b) ADP.
Answer:
AMP – Adenosine Monophosphate, ADP – Adenosine Diphosphate.

Question 10.
Compare Aerobic respiration with anaerobic respiration.
Answer:

Aerobic Respiration	Anaerobic Respiration
Respiration occurs in the presence of oxygen.	Respiration occurs in the absence of oxygen.
It occurs to all higher plants and animals.	It occurs in microorganisms like yeast.

Question 11.
Mention the stages of aerobic respiration.
Answer:

1. Glycolysis-conversion of glucose into pyruvic acid in the cytoplasm of the cell.
2. Link reaction-conversion of pyruvic acid into acetyl coenzyme-A in the mitochondrial matrix.
3. Krebs cycle-conversion of acetyl coenzyme A into carbon dioxide and water in the mitochondrial matrix.

Question 12.
Where does glycolysis occur? Mention the scientists who described it?
Answer:
Glycolysis occurs in the cytoplasm of the cell. Glycolysis was described by Gustav Embden, Otto Meyerhoff, and J. Pamas.

Question 13.
Mention the two phases of Glycolysis.
Answer:
(a) Preparatory phase or hexose phase, (b) Pay off phase or triose phase.

Question 14.
Define Glycolysis.
Answer:
Glycolysis: (Gr: Glykos 5 Glucose, Lysis 5 Splitting) Glycolysis is a linear series of reactions in which 6-carbon glucose is split into two molecules of 3-carbon pyruvic acid. The enzymes which are required for glycolysis are present in the cytoplasm (Figure 14.6). The reactions of glycolysis were worked out in yeast cells by three scientists Gustav Embden (German), Otto Meyerhoff (German), and J Parnas (Polish) and so it is also called as EMP pathway.

Question 15.
What is Substrate/Trans-phosphorylation?
Answer:
The direct transfer of phosphate moiety from substrate molecule to ADP and is converted into ATP is called substrate phosphorylation or direct phosphorylation or trans-phosphorylation.

Question 16.
Define the term Enolation.
Answer:
In glycolysis, at step nine, 2-phosphoglycerate is dehydrated into phosphoenol pyruvate by the enzyme enolase. As a result, an enol group is formed within the molecule. This process is called enolation.

Question 17.
List the products of Glycolysis.
Answer:
(a) Pyruvic acid (2 molecules), (b) 2 ATP molecules and (c) 2 NADH, molecules.

Question 18.
Who discovered the TCA cycle? Where does it occur?
Answer:
The tricarboxylic Acid (TCA) cycle was discovered by Sir Hans Adolf Krebs in 1937. TCA cycle occurs in the mitochondrial matrix.

Question 19.
Draw the structure of oxysomes.
Answer:

Question 20.
Why TCA cycle is called so?
Answer:
TCA cycle starts with the condensation of acetyl CoA with oxaloacetate in the presence of water to yield citrate or citric acid. Therefore, it is also known as the citric acid Cycle (CAC) or Tri Carboxylic Acid (TCA) cycle.

Question 21.
Krebs cycle is an amphibolic pathway – Comment.
Answer:
The Krebs cycle is primarily a catabolic pathway, but it provides precursors for various biosynthetic pathways thereby an anabolic pathway too. Hence, it is called the amphibolic pathway. It serves as a pathway for the oxidation of carbohydrates, fats, and proteins.

Question 22.
Mention the four multi-protein complexes in ETC.
Answer:
(a) Complex – I – NADH dehydrogenase
(b) Complex – II – Succinic dehydrogenase
(c) Complex – III – Cytochrome be, complex
(d) Complex – IV – Cytochrome c oxidase

Question 23.
Write a brief note on Ubiquinone.
Answer:
Ubiquinone and cytochrome bc₁, the complex is structurally and functionally similar to plastoquinone and cytochrome b₆, f complex respectively in the photosynthetic electron transport chain.

Question 24.
Mitochondria are called powerhouses of the cell. Why?
Answer:
Complete oxidation of a glucose molecule in aerobic respiration results in the net gain of 36 ATP molecules in plants. Since a huge amount of energy is generated in mitochondria in the form of ATP molecules they are called the powerhouse of the cell. In the case of aerobic prokaryotes due to lack of mitochondria, each molecule of glucose produces 38 ATP molecules.

Question 25.
Name any two ETC inhibitors.
Answer:
2, 4 DNP – cyanide.

Question 26.
Define Respiratory Quotient.
Answer:
The ratio of the volume of carbon dioxide given out and the volume of oxygen taken in during respiration is called Respiratory Quotient or Respiratory ratio. RQ value depends upon respiratory substrates and their oxidation.

Question 27.
Calculate the respiratory quotient for the following equation.
C₆H₁₂O₆ → 2CO₂2 + 2C₂H₅OH + Energy
Answer:
Volume of O₂ used = O. Volume of CO₂ released = 2

Question 28.
What is the significance of the respiratory quotient?
Answer:
Significance of RQ

1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
2. It also helps to know which type of respiratory substrate is involved.

Question 29.
Define fermentation and mention its types.
Answer:
Some organisms can respire in the absence of oxygen. This process is called fermentation or anaerobic respiration. There are three types of fermentation:

1. Alcoholic fermentation
2. Lactic acid fermentation
3. Mixed acid fermentation

Question 30.
Give an account on lactic acid fermentation.
Answer:
Some bacteria (Bacillus), fungi and muscles of vertebrates produce lactic acid from pyruvic acid:

Question 31.
Observe the given diagram given below and answer the questions.

(a) Name the apparatus set up.
(b) Mention the purpose of the apparatus.
Answer:
(a) The apparatus set up is Kuhne’s fermentation tube.
(b) Kuhne’s fermentation tube is used to demonstrate alcoholic fermentation.

Question 32.
Name the scientists who discovered HMP shunt. Also, expand the term HMP shunt.
Answer:
HMP shunt refers to Hexose Monophosphate shunt which was discovered by Warburg, Dickens, and Lipmann.

Short answer questions

Question 1.
Compare floating respiration with protoplasmic respiration.
Answer:

S. No.	Floating Respiration	Protoplasmic Respiration
a.	Carbohydrates, fat, or organic acids act as respiratory substrates.	Protein acts as a respiratory substrate.
b.	It does not produce any toxic product.	It produces toxic ammonia and also depletes structural and functional proteins of protoplasm.
c.	It is a common mode of respiration.	It is a rare form of respiration.

Question 2.
Draw the structure of an ATP molecule.
Answer:

Question 3.
Describe the structure of ATP.
Answer:
Respiration is responsible for the generation of ATP. The discovery of ATP was made by Karl Lohman (1929). ATP is a nucleotide consisting of a base-adenine, a pentose sugar-ribose, and three phosphate groups.
Out of three phosphate groups, the last two are attached by high-energy-rich bonds (Figure 14.3). On hydrolysis, it releases energy (7.3 Kcal or 30.6 KJ/ATP) and it is found in all living cells and hence it is called the universal energy currency of the cell. ATP is an instant source of energy within the cell. The energy contained in ATP is used in the synthesis of carbohydrates, proteins, and lipids. The energy transformation concept was established by Lipman (1941).

Question 4.
How Redox reaction occurs in NADH₂ and FADH₂.
Answer:
NAD⁺ + 2e⁺ + 2H⁺ → NADH + H⁺
FAD + 2e^– + 2H⁺ → FADH₂
When NAD⁺ (Nicotinamide Adenine Dinucleotide-oxidised form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H⁺ and FADH₂ respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD⁺ and FAD gain (reduction) or lose (oxidation) electrons is called redox reaction (Oxidation-reduction reaction). These reactions are important in cellular respiration.

Question 5.
Explain simply the stages of respiration.
Answer:
Stages of Respiration:

1. Glycolysis-conversion of glucose into pyruvic acid in the cytoplasm of the cell.
2. Link reaction-conversion of pyruvic acid into acetyl coenzyme-A in the mitochondrial matrix.
3. Krebs cycle-conversion of acetyl coenzyme A into carbon dioxide and water in the mitochondrial matrix.
4. Electron transport chain and oxidative phosphorylation remove hydrogen atoms from the products of glycolysis, link reaction, and Krebs cycle release water molecule with energy in the form of ATP in the mitochondrial inner membrane.

Question 6.
Write the overall equation for glycolysis.
Answer:

Question 7.
Give an account of pyruvic oxidation.
Answer:
Two molecules of pyruvate formed by glycolysis in the cytosol enter into the mitochondrial matrix. In aerobic respiration, this pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. This reaction is irreversible and produces two molecules of NADH + H⁺ and 2CO₂. It is also called transition reaction or Link reaction. The reaction of pyruvate oxidation is

Question 8.
Draw and label the structure of mitochondria.
Answer:

Question 9.
How fats and protein enter the Krebs cycle?
Answer:
When fats are a respiratory substrate they are first broken down into glycerol and fatty acid. Glycerol is converted into DHAP and acetyl CoA. This acetyl CoA enters into the Krebs cycle. When proteins are the respiratory substrate they are degraded into amino acids by proteases. The amino acids after deamination enter into the Krebs cycle through pyruvic acid or acetyl CoA and it depends upon the structure.

Question 10.
Mention the types of electron transport chain inhibitors and their action.
Answer:
Electron transport chain inhibitors:

1. 2,4 DNP (Dinitrophenol) – It prevents the synthesis of ATP from ADP, as it directs electrons from Co Q to Q₂
2. Cyanide – It prevents the flow of electrons from Cytochrome a₃ to O₂
3. Rotenone – It prevents the flow of electrons from NADH + H⁺ / FADH₂ to Co Q
4. Oligomycin – It inhibits oxidative phosphorylation.

Question 11.
Give a brief account of Alcoholic fermentation.
Answer:
The cells of roots in waterlogged soil respire by alcoholic fermentation because of lack of oxygen by converting pyruvic acid into ethyl alcohol and CO₂. Many species of yeast (Saccharomyces) also respire anaerobically.
This process takes place in two steps:

Question 12.
List out the industrial uses of alcoholic fermentation.
Answer:

1. In bakeries, it is used for preparing bread, cakes, biscuits.
2. In beverage industries for preparing wine and alcoholic drinks.
3. In producing vinegar and in tanning, curing of leather.
4. Ethanol is used to make gasohol (a fuel that is used for cars in Brazil).

Question 13.
Enumerate the characteristics of anaerobic respiration.
Answer:

1. Anaerobic respiration is less efficient than aerobic respiration.
2. a Limited number of ATP molecules is generated per glucose molecule.
3. It is characterized by the production of CO₂ and is used for Carbon fixation in photosynthesis.

Question 14.
Give a comparative study on glycolysis and fermentation.
Answer:
Comparison between glycolysis and fermentation:

Glycolysis	Fermentation
Glucose is converted into pyruvic acid.	Starts from pyruvic acid and is converted into alcohol or lactic acid.
It takes place in the presence or absence of oxygen.	It takes place in the absence of oxygen.
Net gain is 2ATP.	No net gain of ATP molecules.
2NADH + H+ molecules are produced.	2NADH + H+ molecules are utilized.

Question 15.
Point out the importance of the pentose phosphate pathway.
Answer:

1. HMP shunt is associated with the generation of two important products, NADPH and pentose sugars, which play a vital role in anabolic reactions.
2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen free radicals
3. Ribose-5-phosphate and its derivatives are used in the synthesis of DNA, RNA, ATP, NAD⁺, FAD, and Coenzyme A.
4. Erythrose is used for the synthesis of anthocyanin, lignin, and other aromatic compounds.

Long answer questions

Question 1.
Explain the phases of glycolysis.
Answer:
1. Preparatory phase: Glucose enters the glycolysis from sucrose which is the end product of photosynthesis. Glucose is phosphorylated into glucose-6- phosphate by the enzyme hexokinase and subsequent reactions are carried out by different enzymes. At the end of this phase fructose-1, 6 – bisphosphate is cleaved into glyceraldehyde-3- phosphate and dihydroxyacetone phosphate by the enzyme aldolase. These two are isomers. Dihydroxyacetone phosphate is isomerized into glyceraldehyde-3- phosphate by the enzyme triosephosphate isomerase, now’ two molecules of glyceraldehyde 3 phosphate enter into pay off phase. During the preparatory phase, two ATP molecules are consumed in step-1 and step-3.
2. Pay-off phase: Two molecules of glyceraldehyde-3- phosphate oxidatively phosphorylated into two molecules of 1,3 – bisphosphate glycerate. During this reaction, 2NAD⁺ is reduced to 2NADH + H⁺ by glyceraldehyde- 3- phosphate dehydrogenase at step 6. Further reactions are carried out by different enzymes and at the end, two molecules of pyruvate are produced. In this phase, 2ATPs are produced at step 7 and 2 ATPs at step 10. The direct transfer of phosphate moiety from substrate molecule to ADP and is converted into ATP is called substrate phosphorylation or direct phosphorylation or transphosphorylation. During the reaction at step 9, 2phospho glycerate dehydrated into Phospho enol pyruvate a water molecule is removed by the enzyme enolase. As a result, the enol group is formed within the molecule. This process is called Enolation.

Question 2.
Draw a Flow chart depicting the steps of glycolysis.
Answer:

Question 3.
Point out the significance of Kreb’s Cycle.
Answer:
Significance of Krebs cycle:

1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
2. It provides carbon skeleton or raw material for various anabolic processes.
3. Many intermediates of the TCA cycle are further metabolized to produce amino acids, proteins, and nucleic acids.
4. Succinyl CoA is the raw material for the formation of chlorophylls, cytochrome, phytochrome, and other pyrrole substances.
5. α-ketoglutarate and oxaloacetate undergo reductive animation and produce amino acids.
6. It acts as a metabolic sink which plays a central role in intermediary metabolism.

Question 4.
Draw a Flow chart of Kreb’s cycle.
Answer:

Question 5.
Compare alcoholic fermentation with lactic acid fermentation.
Answer:
Comparison of alcoholic fermentation and lactic acid fermentation

Alcoholic fermentation	Lactic acid fermentation
It produces alcohol and releases CO2 from pyruvic acid.	It produces lactic acid and does not release CO2 from pyruvic acid.
It takes place in two steps.	It takes place in a single step.
It involves two enzymes, pyruvate decarboxylase with Mg+ + and alcohol dehydrogenase.	It uses one enzyme, lactate dehydrogenase with Zn+ +.
It forms acetaldehyde as an intermediate compound.	Does not form an intermediate compound.
It commonly occurs in yeast.	Occurs in bacteria, some fungi and vertebrate muscles.

Question 6.
Demonstrate alcoholic fermentation using Kuhne’s apparatus.
Answer:
Demonstration of alcoholic fermentation:
Take a Kuhne’s fermentation tube which consists of an upright glass tube with a side bulb. Pour 10% sugar solution mixed with baker’s yeast into the fermentation tube the side tube is filled plug the mouth with a lid. After some time, the glucose solution will be fermented. The solution will give out an alcoholic smell and the level of solution in the glass column will fall due to the accumulation of CO₂ gas. It is due to the presence of the zymase enzyme in yeast which converts the glucose solution into alcohol and CO₂. Now introduce a pellet of KOH into the tube, the KOH will absorb CO₂, and the level of solution will rise in the upright tube.

Question 7.
What are the factors that affect respiration?
Answer:
Factors Affecting Respiration Answer: Internal Factors

1. The amount of protoplasm and its state of activity influence the rate of respiration.
2. The concentration of respiratory substrate is proportional to the rate of respiration.

External Factors

1. Wounding of plant organs stimulates the rate of respiration in that region.
2. Some chemical substance acts as inhibitors. Example: Cyanides.
3. Rate of respiration decreases with decreasing amount of water. Proper hydration is essential for respiration.
4. Light is an indirect factor affecting the rate of respiration.
5. The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
6. When a sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called the Extinction point.
7. A high concentration of CO₂ reduces the rate of respiration.
8. A plant or tissue transferred from water to salt solution will increase the rate of respiration. It is called salt respiration.

Higher Order Thinking Skills (HOTs)

Question 1.
Glycolysis and Kreb’s cycle both are energy-yielding pathways in aerobic respiration. How these two pathways differ among themselves?
Answer:

Glycolysis	Krebs Cycle
It occurs in the cytoplasm.	It occurs in mitochondria.
CO2 is not released.	CO2 is released.
Linear pathway consisting of nine steps.	Cyclic pathway consisting of eight steps.
End products are two molecules of pyruvic acid.	End products are CO2 and H2O and energy molecules.

Question 2.
Anaerobic respiration is usually noticed in lower organisms. A human being is an advanced organism whether anaerobic respiration occurs in a man? If so when and where does it takes place.
Answer:
Yes, anaerobic respiration occurs man at certain cases. Anaerobic respiration (lactic acid fermentation) occurs in the muscle cells of man during heavy exercise, continuous workouts running, etc.

Question 3.
Complete the formula of the respiratory quotient by naming A and B. Also, mention what type of substrate have a respiratory question of 1 and > 1?
Respiratory Quotient = \(\frac{A}{B}\)
Answer:

Respiratory Quotient will be equal to unity if the substrate is a carbohydrate.
Respiratory Quotient will be more than unity if the substrate is an organic acid.

Question 4.
The Flow Chart given below depicts the preparatory phase of the glycolysis pathway. Complete it by filling the missing steps A, B, C and also indicate whether ATP is being used up or released at Step D.

Answer:
A = Glucose – 6 – Phosphate.
B = Fructose -1,6- Bisphosphate.
C = Dihydroxy Acetone Phosphate.
In step D, ATP is utilized to phosphorylate Fructose – 6 – phosphate to Fructose —1,6 — Bisphosphate.

Question 5.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration – Discuss.
Answer:
In aerobic respiration, oxidation of one molecule of glucose generates 36 ATP molecules. On the other hand, during anaerobic respiration, the glucose is incompletely oxidized producing only 2 ATP molecules. Thus it is clear that aerobic respiration is more ATP beneficial than anaerobic respectively.

Question 6.
Answer the following questions in concern with the TCA cycle.
(a) Where does it take place?
(b) When and by whom it was discovered?
(c) Name the first formed product of the cycle.
(d) How many ATP molecules are generated per cycle?
(e) TCA cycle is an amphibolic pathway. Say yes or no. Why?
Answer:
(a) TCA cycle takes place in mitochondrion.
(b) TCA cycle was discovered by Sir Hans Adolf Krebs in 1937.
(c) The first formed product of the TCA cycle is citric acid or citrate.
(d) Only one ATP molecule is generated per TCA cycle.
(e) Yes. TCA is an amphibolic pathway because it involves both anaerobic and catabolic reactions.

Question 7.
Mention the fate of pyruvic acid in a cell, under the given circumstances.

Answer:
A = Acetyl Coenzyme A
B = Ethyl alcohol.
C = Lactic acid.

Choose the correct answer.

1. Which of the following is NOT a respiratory substrate in floating respiration?
(a) Carbohydrate
(b) Fat
(c) Organic acids
(d) Protein
Answer:
(d) Protein

2. The term respiration was coined by
(a) Pepys
(b) Calvin-Benson
(c) Meyerhoff
(d) Karl Lohman
Answer:
(a) Pepys

3. Amount of energy released when an ATP molecule is hydrolysed is …………
(a) 2.7 kcal
(b) 7.3 kcal
(c) 2.8 kcal
(d) 6.5 kcal
Answer:
(b) 7.3 kcal

4. ………… is called as the universal energy currency of the cell.
(a) ATP
(b) GTP
(c) UTP
(d) AMP
Answer:
(a) ATP

5. Which step is irrelevant with respect to aerobic respiration?
(a) Glycolysis
(b) Pyruvate oxidate
(c) Fermentation
(d) TCA cycle
Answer:
(c) Fermentation

6. Anaerobic respiration occurs only in …………
(a) Mitochondria
(b) Golgi bodies
(c) Nucleus
(d) Cytoplasm
Answer:
(d) Cytoplasm

7. In glycolysis, Glucose is phosphorylated to glucose-phosphate by the enzyme ………….
(a) Aldolase
(b) Phosphofructo isomerase
(c) Hexokinase
(d) Exolase
Answer:
(c) Hexokinase

8. Glucose is a ………….. carbon compound.
(a) Six
(b) Five
(c) Three
(d) Four
Answer:
(a) Six

9. Which of the following step is common in both aerobic and anaerobic respiration?
(a) Pyruvate oxidation
(b) Glycolysis
(c) ETC
(d) TCA cycle
Answer:
(6) Glycolysis

10. Net gain of ATP’s at the end of glycolysis is ………….
(a) 2
(b) 4
(c) 6
(d) 0
Answer:
(a) 2

11. Which statement is NOT – correct in concern with glycolysis?
(i) Preparatory phase is also called the hexose phase.
(ii) Pay off phase is also called the hexose phase.
(iii) Two ATP’s are consumed in the preparatory phase.
(iv) Glycolysis is also called EMP pathway.
(a) Only (i)
(b) Only (ii)
(c) (iii) and (iv)
(d) Only (iii)
Answer:
(b) Only (ii)

12. Pyruvic oxidation occurs in ……….
(a) Cytoplasm
(b) Mitochondrial matrix
(c) Inner membrane of mitochondria
(d) Both cytoplasm and mitochondrial membrane
Answer:
(b) Mitochondrial matrix

13. Kreb’s cycle is ……… in nature.
(a) Anabolic
(b) Catabolic
(c) Amphoteric
(d) Amphibolic
Answer:
(d) Amphibolic

14. F1 particles are also referred as ……….
(a) Polysomes
(b) Oxysomes
(c) Mesosomes
(d) Liposomes
Answer:
(b) Oxysomes

15. On oxidation in mitochondrian one molecule of NADH₂ yield …………. ATP’s.
(a) 2
(b) 4
(c) 3
(d) 1
Answer:
(c) 3

16. Which cell organelles are referred as “powerhouses”?
(a) Golgi bodies
(b) Endoplasmic reticulum
(c) Nucleus
(d) Mitochondria
Answer:
(d) Mitochondria

17. Identify the electron transport chain inhibitors prevent e^– flow from cytochrome a₃ to O₂.
(a) 2, 4 DNP
(b) Cyanide
(c) Oligomycin
(d) Rotenone
Answer:
(b) Cyanide

18. Pick out the correct statements.

(a) Only (i)
(b) Only (b)
(c) Both (i) and (iii)
(d) Both (b) and (iv)
Answer:
(a) Only (i)

19. Respiratory quotient of glucose in presence of oxygen is …………
(a) Unity
(b) Infinity
(c) Less than unity
(d) Zero
Answer:
(a) Unity

20. Number of CO₂ molecules generated in Kreton cycle is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(b) 4

21. Calculate the respiratory quotient from the following equation
C₄H₆O₅ + 3O₂ → 4CO₂ + 3H₂O + Energy
(a) Unity
(b) More than unity
(c) Less than unity
(d) Zero
Answer:
(b) More than unity

22. Identify the wrong statement regarding fermentation.
(i) Fermentation can also be called anaerobic respiration.
(ii) In anaerobic respiration, O₂ is not evolved.
(iii) Sargassum undergoes anaerobic respiration.
(iv) In anaerobic respiration, CO₂ is evolved.
(a) Only (iv)
(b) Only (ii)
(c) Only (iii)
(d) Both (ii) & (iv)
Answer:
(c) Only (iii)

23. ………. is an alternate way for glucose break down.
(a) Glycolysis
(b) Fermentation
(c) Respiration
(d) HMP shunt
Answer:
(d) HMP shunt

24. …….. is used for the synthesis of anthocyanin.
(a) Ribulose
(b) Erythrose
(c) Sedoheptulose
(d) Xylulose
Answer:
(b) Erythrose

25. Mention the optimum temperature for respiration?
(a) 25°C
(b) 30°C
(c) 26°C
(d) 28°C
Answer:
(b) 30°C

Students get through the TN Board 11th Bio Botany Important Questions Chapter 13 Photosynthesis which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 13 Photosynthesis

Very short answer questions

Question 1.
Who was Stephen Hales?
Answer:
Stephen Hales is the father of plant physiology. He was the person to propound that plants obtain nourishment from air & light.

Question 2.
Give the overall equation of photosynthesis.
Answer:

Question 3.
What happens to water & carbon dioxide during photosynthesis?
Answer:
Water is oxidised to oxygen. CO₂ is reduced to carbohydrates.

Question 4.
Define anaerobic photosynthesis.
Answer:
In some bacteria, oxygen is not evolved and is called non-oxygenic and anaerobic photosynthesis. Examples: Green sulphur, Purple sulphur and green filamentous bacteria.

Question 5.
What is Bioluminescence?
Answer:
Bioluminescence is the production and emission of light by a living organism. Bioluminescence is rare in true plants.

Question 6.
What are Quantasomes?
Answer:
In chloroplast, Inner surface of lamellar membrane consists of a small spherical structure called as Quantasomes.

Question 7.
Define the term photosynthetic pigment.
Answer:
A photosynthetic pigment is a pigment that is present in chloroplasts or photosynthetic bacteria ‘ which captures the light energy necessary for photosynthesis.

Question 8.
Apart from chlorophyll, other pigments are called accessory pigments. Why?
Answer:
Chlorophyll ‘a’ is the primary pigment that acts as a reaction centre and all other pigments act as accessory pigments and trap solar energy and then transfer it to chlorophyll ‘a’.

Question 9.
What is a Phytol tail? Mention its role.
Answer:
a. Phytol tail is the lipophilic tail of chlorophyll molecule,
b. It helps in anchoring the chlorophyll to lamellae.

Question 10.
Mention the minerals used in the biosynthesis of Chlorophyll a.
Answer:
Mg, Fe, Cu, Zn, Mn, K and Nitrogen.

Question 11.
How chlorophyll b differs from chlorophyll a?
Answer:
Chlorophyll ‘b’ differs from Chlorophyll ‘a’ in having CHO (aldehyde) group instead of CH₃ (Methyl) group at the 3rd C atom in II Pyrrole ring.

Question 12.
Carotenoids are shield pigments – Comment.
Answer:
Carotenoids are yellow to orange pigments, mostly tetraterpens and these pigments absorb light strongly in the blue to the violet region of the visible spectrum. These pigments protect chlorophyll from photooxidative damage. Hence, they are called shield pigments.

Question 13.
Give an account of Xanthophylls.
Answer:
Yellow (C₄₀H₅₆O₂) pigments are like carotenes but contain oxygen. Lutein is responsible for the yellow colour change of leaves during the autumn season. Examples: Lutein, Violaxanthin and Fucoxanthin.

Question 14.
Name the two forms of phycobilins and also give an example.
Answer:
Phycobilins exist in two forms. They are:
a. Phycocyanin found in Cyanobacteria.
b. Phycoerythrin found in red algae.

Question 15.
Define Quantum.
Answer:
Light as a particle is called a photon. Each photon contains an amount of energy known as quantum.

Question 16.
How will you define Quantasomes?
Answer:
Quantasomes are the morphological expression of physiological photosynthetic units, located on the inner membrane of thylakoid lamellae.

Question 17.
Mention the events occurring in the photo-oxidation phase of light reaction.
Answer:
Photo-oxidation Phase:

• Absorption of light energy.
• Transfer of energy from accessory pigments to the reaction centre.
• Activation of Chlorophyll ‘a’ molecule.

Question 18.
Mention the events of the Photochemical phase of light reaction.
Answer:
Photo Chemical Phase:

• Photolysis of water and oxygen evolution
• Electron transport and synthesis of assimilatory power.

Question 19.
Define Photophosphorylation.
Answer:
Phosphorylation takes place with the help of light generated electron and hence it is known as
photophosphorylation.

Question 20.
Compare fluorescence with Phosphorescence.
Answer:

Fluorescence	Phosphorescence
Fluorescence is the immediate emission of absorbed radiation.	Phosphorescence is the delayed emission of absorbed radiation.

Question 21.
Draw the diagram representing the oxygen-evolving complex.
Answer:

Question 22.
Compare oxidative phosphorylation with substrate-level phosphorylation.
Answer:
Phosphorylation taking place during respiration is called oxidative phosphorylation and ATP produced by the breakdown of the substrate is known as substrate-level phosphorylation.

Question 23.
State Chemiosmotic theory.
Answer:
The chemiosmotic theory was proposed by P. Mitchell (1966). According to this theory, electrons are transported along the membrane through PS I and PS II and connected by the Cytochrome b6-f complex.

Question 24.
What are the assimilatory powers produced during the light reaction?
Answer:
ATP and NADPH + H⁺

Question 25.
Why PCR cycle is called as C₃ cycle?
Answer:
The first product of the PCR pathway is a 3- carbon compound (Phospho Glyceric Acid) and so it is also called as C₃ Cycle. It takes place in the stroma of the chloroplast.

Question 26.
Name the three stages of dark reaction.
Answer:

1. Carboxylation (fixation),
2. Reduction (Glycolytic Reversal) and
3. Regeneration.

Question 27.
Give an account on RUBISCO.
Answer:
RUBISCO – RUBP Carboxylase Oxygenase enzyme, is the most abundant protein found on earth. It constitutes 16 % of the chloroplast protein. It acts as carboxylase in the presence of CO₂ and oxygenase in the absence of CO₂.

Question 28.
C₄ plants are of ecological benefit – Comment.
Answer:
C₄ plants account for about 30% of terrestrial carbon fixation. Increasing the proportion of C₄ plants on earth could assist the biosequestration of CO₂ and represent an important climate change avoidance strategy.

Question 29.
Compare the features of dimorphic chloroplasts.
Answer:
The characteristic feature of C₄ plants is the presence of dimorphic chloroplast: Bundle sheath chloroplast: Larger chloroplast, thylakoids not arranged in granum and rich in starch. Mesophyll Chloroplast: Smaller chloroplast, thylakoids arranged in granum and less starch.

Question 30.
Define Photorespiration.
Answer:
Photorespiration is the excess respiration taking place in photosynthetic cells due to the absence of CO₂ and an increase of O₂.

Question 31.
What is CO₂ compensation plant?
Answer:
When the rate of photosynthesis equals the rate of respiration, there is no exchange of oxygen and carbon dioxide and this is called a carbon dioxide compensation point.

Question 32.
State Blackman’s law of limiting factor.
Answer:
“At any given point of time, the lowest factor among essentials will limit the rate of photosynthesis”.

Question 33.
List out the external and internal factors that affect photosynthesis.
Answer:

• External factors: Light, carbon dioxide, temperature, water, mineral and pollutants.
• Internal factors: Pigments, protoplasmic factor, accumulation of carbohydrates, anatomy of leaf and hormones.

Question 34.
Enumerate the anatomical features of the leaf that affects photosynthesis.
Answer:
The thickness of cuticle and epidermis, distribution of stomata, presence or absence of Kranz anatomy and relative proportion of photosynthetic cells affect photosynthesis.

Question 35.
Name the photosynthetic apparatus of bacteria.
Answer:
Bacteria have a special type of photosynthetic apparatus called chlorosomes and chromatophores.

Short answer questions

Question 1.
Give an account of the biosynthesis of chlorophyll.
Answer:
Chlorophyll is synthesized from intermediates of respiration and photosynthesis. Succinic acid an intermediate of the Krebs cycle is activated by the addition of coenzyme A and it reacts with a simple amino acid glycine and the reaction goes on to produce chlorophyll ‘a’.
Biosynthesis of chlorophyll ‘a’ requires Mg, Fe, Cu, Zn, Mn, K and nitrogen. The absence of any one of these minerals leads to chlorosis.

Question 2.
List out the types of radiations in the Electromagnetic spectrum.
Answer:
The electromagnetic spectrum consists of 8 types of radiations such as cosmic rays, gamma rays, X rays, U-V rays, Visible light spectrum, infrared rays, electric rays and radio rays.

Question 3.
Write a brief note on Emerson’s Red Drop.
Answer:
Emerson conducted an experiment in Chlorella using only one wavelength of light (monochromatic light) at a time and he measured quantum yield. He plotted a graph of the quantum yield in terms of O, evolution at various wavelengths of light. His focus was to determine at which wavelength the photochemical yield of oxygen was maximum. He found – that in the wavelength of 600 to 680 the yield was constant but suddenly dropped in the region above 680 nm (red region). The fall in the photosynthetic yield beyond the red region of the spectrum is referred to as Red drop or Emerson’s first effect.

Question 4.
Explain the term Emerson’s Enhancement Effect.
Answer:
Emerson found that the monochromatic light of longer wavelength (far red light), when supplemented with a shorter wavelength of light (red light), enhanced photosynthetic yield and recovered red drop. This enhancement of photosynthetic yield is referred to as Emerson’s Enhancement Effect.

Question 5.
Give the conclusions of Hills Reaction.
Answer:
Conclusions of Hill’s Reaction:

1. During photosynthesis, oxygen is evolved from water.
2. Electrons for the reduction of CO₂ are obtained from water.
3. A reduced substance produced, later helps to reduce CO₂.
   2H₂O + 2A → 2 AH₂ + O₂

Question 6.
Define Dark Reaction.
Answer:
Fixation and reduction of CO₂ into carbohydrates with the help of assimilatory power produced during light reaction. This reaction does not require light and is not directly light-driven. Hence, it is called as Dark reaction or Calvin-Benson cycle.

Question 7.
Illustrate S’ state mechanism.
Answer:
S’ State Mechanism consists of a series of 5 states called as S₀, S₁, S₂, S₃ and S₄. Each state acquires a positive charge by a photon (hv) and after the S₄ state, it acquires 4 positive charges, four electrons and the evolution of oxygen. Two molecules of water go back to the S₀. At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.

Question 8.
Kranz Anatomy – Define.
Answer:
Kranz Anatomy is a German term meaning a halo or wreath. In C₄ plants vascular bundles are surrounded by a layer of bundle sheath. The bundle sheath is surrounded by a ring of mesophyll cells.

Question 9.
Point out the significances of C₄ cycle.
Answer:
Significance of C₄ cycle

1. Plants having C₄ cycle are mainly of tropical and sub-tropical regions and are able to survive in an environment with low CO₂ concentration.
2. C₄ plants are partially adapted to drought conditions.
3. Oxygen has no inhibitory effect on C₄ cycle since PEP carboxylase is insensitive to O₂.
4. Due to the absence of photorespiration, CO₂ Compensation Point for C₄ is lower than that of C₃ plants.

Question 10.
Point out the significances of photorespiration.
Answer:
Significance of photorespiration

1. Glycine and Serine synthesised during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
2. It consumes excess NADH + H⁺ generated.
3. Glycolate protects cells from Photooxidation.

Question 11.
Mention the role of light intensity in photosynthesis.
Answer:
The intensity of light plays a direct role in the rate of photosynthesis. Under low intensity the photosynthetic rate is low and at higher intensity photosynthetic rate is higher. It also depends on the nature of plants. Heliophytes (Bean Plant) require higher intensity than Sciophytes (Oxalis).

Question 12.
Classify photosynthetic bacteria and give example.
Answer:
Photosynthetic bacteria are classified into three groups:

1. Green sulphur bacteria. Example: Chlorobacterium and Chlorobium.
2. Purple sulphur bacteria. Example: Thiospirillum and Chromatium.
3. Purple non-sulphur bacteria. Example: Rhodopseudomonas and Rhodospirillum.

Long answer questions

Question 1.
Mention the significance of photosynthesis.
Answer:
Significance of Photosynthesis:

1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
3. Photosynthesis balances the oxygen and carbon cycle in nature.
4. Fuels such as coal, petroleum and other fossil fuels are from preserved photosynthetic plants.
5. Photosynthetic organisms are the primary producers on which all consumers depend for energy.
6. Plants provide fodder, fibre, firewood, timber, useful medicinal products and these sources come by the act of photosynthesis.

Question 2.
Describe the structure of chloroplasts.
Answer:
Chloroplasts are the main site of photosynthesis and both the energy-yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in the chloroplast. It is a double-wall membrane-bounded organelle, discoid or lens-shaped, 4 -10 pm in diameter and 1-33 pm in thickness. The membrane is a unit membrane and the space between them is 100 to 200 A. A colloidal and proteinaceous matrix called stroma is present inside.
A sac-like membranous system called thylakoid or lamellae is present in the stroma and they are arranged one above the other forming a stack of a coin-like structure called granum (plural grana). Each chloroplast contains 40 to 80 grana and each granum consists of 5 to 30 thylakoids. Thylakoids found in granum are called grana lamellae and in the stroma are called stroma lamellae. Thylakoid disc size is 0.25 to 0.8 micron in diameter. A thinner lamella called the Fret membrane connects grana. Pigment system I is located on the outer thylakoid membrane facing stroma and Pigment system II is located on the inner membrane facing the lumen of the thylakoid. Grana lamellae have both PS I and PS II whereas stroma lamellae have only PS I. Chloroplast contains 30-35% Proteins, 20-30% phospholipids, 5-10% chlorophyll, 4-5% Carotenoids, 70S ribosomes, circular DNA and starch grains. The inner surface of the lamellar membrane consists of a small spherical structure called as Quantasomes. The presence of 70S ribosome and DNA gives them the status of semi-autonomy and proves the endosymbiotic hypothesis which says chloroplast evolved from bacteria. Thylakoid contains pigment systems that produce ATP and NADPH + H⁺ using solar energy. The stroma contains an enzyme which reduces carbon dioxide into carbohydrates.

Question 3.
Tabulate the different types of photosynthetic pigments.
Answer:

Question 4.
Give an account of Chlorophyll.
Answer:
Chlorophyll ‘ a’ is the primary pigment that acts as a reaction centre and all other pigments act as accessory pigments and trap solar energy and then transfer it to chlorophyll ‘a’. Chlorophyll molecules have a tadpole-like structure. It consists of the Mg-Porphyrin head (Hydrophilic Head) and (Lipophilic tail) Phytol tail. The Porphyrin head consists of four pyrrole rings linked together by C-H bridges. Each pyrrole ring comprises four carbons and one nitrogen atom. Porphyrin ring has several side groups which alter the properties of the pigment. Different side groups are indicative of various types of chlorophyll. The Phytol tail made up of 20 carbon alcohol is attached to carbon 7 of the Pyrrole ring IV. It has a long propionic acid ester bond. A long lipophilic tail helps in anchoring chlorophyll to the lamellae.

Question 5.
Explain the steps involved in paper chromatography.
Answer:
Separation of Chloroplast pigments by paper Chromatography method:
Step 1. Extract chlorophyll pigment from the leaves using 80% Acetone.
Step 2. Allow concentrating by evaporation.
Step 3. Apply few drops on one end above 2 cm from the edge of a chromatographic paper.
Step 4. A solvent with a mixture of Petroleum ether and acetone in the ratio of 9:1 is prepared and poured into the development chamber.
Step 5. Place the strip above the solvent by placing one end of the strip touching the solvent. Observation: After one hour of observing the chromatographic paper. You can find the pigments being separated into four distinct spots.

Question 6.
Enumerate the properties of light.
Answer:
Properties of Light

1. Light is a transverse electromagnetic wave.
2. It consists of oscillating electric and magnetic fields that are perpendicular to each other and perpendicular to the direction of propagation of the light.
3. Light moves at a speed of 3 x 108 ms^-1.
4. Wavelength is the distance between successive crests of the wave.
5. Light as a particle is called a photon. Each photon contains an amount of energy known as quantum.
6. The energy of a photon depends on the frequency of the light.

Question 7.
Differentiate between Photosystem I and Photosystem II.
Answer:

Photosystem I	Photosystem II
The reaction centre is P700.	The reaction centre is P680.
PS I is involved in both cyclic and non-cyclic.	PS II participates in the Non-cyclic pathway.
Not involved in the photolysis of water and evolution of oxygen.	Photolysis of water and the evolution of oxygen takes place.
It receives electrons from PS II during non-cyclic photophosphorylation.	It receives electrons by the photolysis of water.
Located in unstacked region granum facing chloroplast stroma.	Located in the stacked region of the thylakoid membrane facing the lumen of the thylakoid.
The chlorophyll and Carotenoid ratio is 20 to 30:1.	The chlorophyll and Carotenoid ratio is 3 to 7:1.

Question 8.
Explain the various complexes in the Electron transport chain.
Answer:
The electron transport chain in each photosystem involves four complexes:

• Core Complex (CC): CC I in PS I the reaction centre is P700, CC II in PS II the reaction centre is P680
• Light-Harvesting Complex or Antenna complex (LHC):
• Two types: LHC I in PS I and LHC II in PS II.
• Cytochrome b₆ f complex: It is the non-pigmented protein complex connecting PS I and PS II. Plastoquinone (PQ) and Plastocyanin (PC) are intermediate complexes acting as mobile or shuttle electron carriers of the Electron Transport Chain. PQ acts as a shuttle between PS II and Cytochrome b₆– f complex and PC connects
• Cytochrome b₆-f and PS I complex.
• AT Pave complex or Coupling factor: It is found on the surface of the thylakoid membrane. This complex is made up of CF₁ and CF₀ factors. This complex utilizes energy from ETC and converts ADP and inorganic phosphate (Pi) into ATP.

Question 9.
Explain cyclic photophosphorylation.
Answer:
Cyclic photophosphorylation refers to the electrons ejected from the pigment system I (Photosystem I) and again cycled back to the PS I. When the photons activate P700 reaction centre photosystem II is activated. Electrons are raised to a high energy level. The primary electron acceptor is Ferredoxin Reducing Substance (FRS) which transfers electrons to Ferredoxin (Fd), Plastoquinone (PQ), cytochrome b₆-f complex, Plastocyanin (PC) and finally back to chlorophyll P700 (PS I). During this movement of electrons, Adenosine Di Phosphate (ADP) is phosphorylated, by the addition of inorganic phosphate and generates Adenosine Tri Phosphate (ATP). Cyclic electron transport produces only ATP and there is no NADPFI + H⁺ formation. At each step of electron, transport, the electron loses potential energy and is used by the transport chain to pump H⁺ ions across the thylakoid membrane. The proton gradient triggers ATP formation in the ATP synthase enzyme situated on the thylakoid membrane. Photosystem I need the light of a longer wavelength (> P700 nm). It operates under low light intensity, less CO₂ and under anaerobic conditions which makes it considered as earlier in evolution.

Question 10.
Explain the stages occurring during non-cyclic electron transport.
Answer:
In oxygenic species, non-cyclic electron transport takes place in three stages.

1. Electron transport from water to P680: Splitting of water molecule produce electrons, protons and oxygen. Electrons lost by the PS II (P680) are replaced by electrons from the splitting of a water molecule.
2. Electron transport from P680 to P700: Electron flow starts from P680 through a series of electron carrier molecules like pheophytin, plastoquinone (PQ), cytochrome b6 – f complex, plastocyanin (PC) and finally reaches P700 (PS I).
3. Electron transport from P700 to NADP⁺: PS I(P700) is excited now and the electrons pass to a high energy level. When electron travels downhill through ferredoxin, NADP+ is reduced to NADPH + H⁺.

Question 11.
Point out the Bio-energetics of light reaction.
Answer:
Bio energetics of light reaction:

• To release one electron from the pigment system requires two quanta of light.
• One quantum is used for the transport of electron from water to PS I.
• The second quantum is used for the transport of electron from PS I to NADP
• Two electrons are required to generate one NADPH + H⁺.
• During Non-Cyclic electron transport two NADPH + H⁺ are produced and it requires 4 electrons.
• Transportation of 4 electrons requires 8 quanta of light.

Question 12.
Differentiate between Cyclic & Non-cyclic photophosphorylation.
Answer:
Differences between Cyclic Photophosphorylation and Non-Cyclic Photophosphorylation

Cyclic Photophosphorylation	Non-Cyclic Photophosphorylation
PS I only involved.	PS I and PS II involved.
The reaction centre is P700.	The reaction centre is P680.
Electrons released are cycled back.	Electron released are not cycled back.
Photolysis of water does not take place.	Photolysis of water takes place.
Only ATP synthesized.	ATP and NADPH + H+ are synthesized.
Phosphorylation takes place at two places.	Phosphorylation takes place at only one place.
It does not require an external electron donor.	Requires external electron donor like H20 or H2S.
It is not sensitive to dichloro dimethyl urea (DCMI).	It is sensitive to DCMI and inhibits electron flow.

Question 13.
Draw a Flow chart depicting Calvins cycle.
Answer:

Question 14.
Explain the three phases of the Dark reaction.
Answer:
Phase 1- Carboxylation (Fixation)
The acceptor molecule Ribulose 1,5 Bisphosphate (RUBP) a 5 carbon compound with the help of RUBP carboxylase oxygenase (RUBISCO) enzyme accepts one molecule of carbon dioxide to form an unstable 6 carbon compound. This 6C compound is broken down into two molecules of 3-carbon compound phospho glyceric acid (PGA).

Phospho glyceric acid is phosphorylated by ATP and produces 1,3 bis phospho glyceric acid by PGA kinase. 1,3 bis phospho glyceric acid is reduced to glyceraldehyde 3 Phosphate (G-3-P) by using the reducing power NADPH + H⁺. Glyceraldehyde 3 phosphate is converted into its isomeric form dihydroxy acetone phosphate (DHAP).

Phase 3 – Regeneration: Regeneration of RUBP involves the formation of several intermediate compounds of 6-carbon, 5-carbon, 4-carbon and 7- carbon skeleton. Fixation of one carbon dioxide requires 3 ATPs and 2 NADPH + H⁺, and the fixation of 6 CO₂ requires 18 ATPs and 12 NADPH + H⁺ during C₃ cycle. One 6 carbon compound is the net gain to form hexose sugar.

Question 15.
Explain the phases of C₄ pathway.
Answer:
C₄ pathway is completed in two phases, first phase takes place in the stroma of mesophyll cells, where the CO₂ acceptor molecule is a 3-Carbon compound, phosphoenol pyruvate (PEP) to form 4-carbon Oxalo acetic acid (OAA). The first product is a 4-carbon and so it is named as C₄ cycle. Oxalo acetic acid is a dicarboxylic acid and hence this cycle is also known as a dicarboxylic acid pathway. Carbon dioxide fixation takes place in two places one in mesophyll and another in bundle sheath cell (dicarboxylation pathway). It is the adaptation of tropical and subtropical plants growing in warm and dry conditions. Fixation of C02 with minimal loss is due to the absence of photorespiration. C₄ plants require 5 ATP and 2 NADPH + H⁺ to fix one molecule of CO₂.

Oxaloacetic acid (OAA) is converted into malic acid or aspartic acid and is transported to the bundle sheath cells through plasmodesmata.
Stage: II Bundle Sheath Cells: Malic acid undergoes decarboxylation and produces a 3 carbon compound Pyruvic acid and CO₂. The released CO₂ combines with RUBP and follows the Calvin cycle and finally, sugar is released to the phloem. Pyruvic acid is transported to the mesophyll cells.

Question 16.
Differentiate between C₃ & C₄ plants.
Answer:
Differences between C₃ and C₄ plants

C3 Plants	C4 Plants
CO2, fixation takes place in mesophyll cells only.	CO2 fixation takes place mesophyll and bundle sheath.
CO2 acceptor is RUBP only.	PEP in mesophyll and RUBP in bundle sheath cells.
The first product is 3C- PGA.	The first product is 4C- OAA.
Kranz anatomy is not present.	Kranz anatomy is present.
Granum is present in mesophyll cells.	Granum is present in mesophyll cells and absent in bundle sheath.
Normal Chloroplast.	Dimorphic chloroplast.
Optimum temperature 20° to 25°C.	Optimum temperature 30° to 45°C.
Fixation of CO2 at 50 ppm.	Fixation of CO2 even less than 10 ppm.
Less efficient due to higher photorespiration.	More efficient due to less photorespiration.
RUBP carboxylase enzyme used for fixation.	PEP carboxylase and RUBP carboxylase used.
18 ATPs used to synthesize one glucose.	Consumes 30 ATPs to produce one glucose.
Efficient at low CO2	Efficient at higher CO2
Example: Paddy, Wheat, Potato and so on.	Example: Sugar cane, Maize, Sorghum, Amaranthus and so on.

Question 17.
Explain in detail about Crassulacean Acid Metabolism.
Answer:
Crassulacean Acid Metabolism or CAM cycle:
It is one of the carbon pathways identified in succulent plants growing in semi-arid or xerophytic condition. This was first observed in Crassulaceae family plants like Bryophyllum, Sedum, Kalanchoe and is the reason behind the name of this cycle. It is also noticed in plants from other families. Examples: Agave, Opuntia, Pineapple and Orchids. The stomata are closed during the day and are open during the night (Scotoactive). This reverse stomatal rhythm helps to conserve water loss through transpiration and will stop the fixation of CO₂ during the daytime. At night time CAM plants fix CO₂ with the help of Phospho Enol Pyruvic acid (PEP) and produce oxalo acetic acid (OAA). Subsequently, OAA is converted into malic acid like C₄ cycle and gets accumulated in vacuole increasing the acidity. During the daytime stomata are closed and malic acid is decarboxylated into pyruvic acid resulting in the decrease of acidity. CO₂ thus formed enters into Calvin Cycle and produces carbohydrates.
Significance of CAM Cycle

1. It is advantageous for succulent plants to obtain CO₂ from malic acid when stomata are, closed.
2. During the daytime, stomata are closed and CO₂ is not taken but continue their photosynthesis.
3. Stomata are closed during the daytime and help the plants to avoid transpiration and water loss.

Question 18.
State the differences between photorespiration and dark respiration.
Answer:
Differences between Photorespiration and Dark Respiration

Photorespiration	Dark respiration
It takes place in photosynthetic green cells.	It takes place in all living cells.
It takes place only in the presence of light.	It takes place all the time.
It involves chloroplast, peroxisome and mitochondria.	It involves only mitochondria.
It does not involve Glycolysis, Kreb’s Cycle, and ETS.	It involves glycolysis, Kreb’s Cycle and ETS.
The substrate is glycolic acid.	The substrate is carbohydrates, protein or fats.
It is not essential for survival.	Essential for survival.
No phosphorylation and yield of ATP.	Phosphorylation produces ATP energy.
NADH2 is oxidised to NAD+.	NAD+ is reduced to NADH.
Hydrogen peroxide is produced.	Hydrogen peroxide is not produced.
End products are CO2 and PGA.	End products are CO2 and water.

Question 19.
Explain the Internal factors that affect photosynthesis.
Answer:
Internal Factors

1. Photosynthetic Pigments: It is an essential factor and even a small quantity is enough to carry out photosynthesis.
2. Protoplasmic factor: Hydrated protoplasm is essential for photosynthesis. It also includes enzymes responsible for Photosynthesis.
3. Accumulation of Carbohydrates: Photosynthetic end products like carbohydrates are accumulated in cells and if translocation of carbohydrates is slow then this will affect the rate, of photosynthesis.
4. Anatomy of leaf: Thickness of cuticle and epidermis, distribution of stomata, presence or absence of Kranz anatomy and relative proportion of photosynthetic cells affect photosynthesis.
5. Hormones: Hormones like gibberellins and cytokinin increase the rate of photosynthesis.

Question 20.
Compare photosynthesis in plants & bacteria.
Answer:
Differences between Photosynthesis in Plants and Photosynthesis in Bacteria

Photosynthesis in Plants	Photosynthesis in Bacteria
Cyclic and non-cyclic phosphorylation takes place.	Only cyclic phosphorylation takes place.
Photosystem I and II involved.	Photosystem I only involved.
The electron donor is water.	The electron donor is H2S.
Oxygen is evolved.	Oxygen is not evolved.
Reaction centres are P700 and P680.	The reaction centre is P870.
The reducing agent is NADPH + H+.	The reducing agent is NADPH + H+.
PAR is 400 to 700 nm.	PAR is above 700 nm.
Chlorophyll, carotenoid and xanthophyll.	Bacterio chlorophyll and bacterio viridin.
Photosynthetic apparatus – chloroplast.	It is chlorosomes and chromatophores.

Higher Order Thinking Skills (HOTs)

Question 1.
The picture given below is an organelle of a plant cell. Identify the picture and answer the questions.

(a) Name the organelle.
(b) Mention the role of the organelle in the cell.
(c) How do you call the stack of coin-like structures present on it?
(d) Whether it shows semi-autonomy? If yes how? If no, why?
Answer:
(a) Chloroplast.
(b) Chloroplast is the site of photosynthesis.
(c) Grana made of thylakoids.
(d) Yes, the presence of DNA, 70S ribosomes gives them the status of semi-autonomy.

Question 2.
Succulents are known to keep their stomata closed during the day to check transpiration.
How do they meet their photosynthetic CO₂ requirements?
Answer:
Succulents undergo a special carbon pathway called Crassulacean Acid Metabolism (CAM pathway). At night time, CAM plants for CCE with the help of Phosphoenol Pyruvic Acid (PEP) and produce Oxalo Acetic Acid (OAA). Subsequently, OAA is converted into malic acid like C₄ cycle and produces CO₂. The CO₂ thus formed enters the Calvins cycle and produces carbohydrates.

Question 3.
An increase in temperature decreases photosynthetic rale – Justify.
Answer:
In general, the optimum temperature for photosynthesis is 25°C to 35°C. An increase in temperature will lead to the closure of stomata thereby inhibiting the gaseous exchange and also inactivate the enzymes responsible for photosynthesis.

Question 4.
“Photosynthesis is a redox reaction”. Comment.
Answer:
Photosynthesis is an oxidation and reduction process where, water is oxidised to release oxygen and CO₂ is reduced to form carbohydrates.

Question 5.
Carrots, Capsicum, Tomatoes etc are organe/red coloured fruits.
(a) Name the pigment responsible for the colour.
(b) State whether it is a photosynthetic pigment.
(c) Mention its role in plants.
Answer:
(a) Carotenoids are the yellow to orange pigments responsible for fruit colours.
(b) Yes, it is a photosynthetic pigment.
(c) They absorb the light energy and transfer it to the chlorophyll. They also protect chloro¬phyll from photo-oxidative damage. Hence also called shield pigments.

Question 6.
Paddy is a C₃ plant. C₃ plants utilise ATP and NADPH₂ molecules to generate oxygen molecules. How many ATP and NADPH₂ molecules would a C₃ plant consume to generate 12 molecules of oxygen.
Answer:
C₃ plants utilise 3 ATP’s and 2 NADPH₂ molecules to evolve one oxygen molecule. To generate 12 molecules of oxygen, 36 ATP’s and 24 NADPH₂ molecules are utilised.

Question 7.
Give the meaning for the following terminologies:
(a) Decarboxylation (b) Phosphorylation (c) Photolysis
Answer:
(a) Decarboxylation : Removal of C02 from a molecule.
(b) Phosphorylation: Synthesis of ATP by the addition of inorganic phosphate to ADP.
(c) Photolysis: Splitting of water molecules using light energy.

Question 8.

The above equation represents the first step of a cyclic pathway occurring in plant cells.
(а) Name the pathway.
(б) Where does this pathway occur inside the cell?
(c) How many carbon molecules are seen in RUBP & PGA.
(d) State the role of RUBISCO in this step.
(e) Under which condition does this pathway proceed.
Answer:
(a) Photorespiration or C₂ cycle.
(b) C₂ cycle takes place in chloroplast, peroxisome and mitochondrion.
(c) RUBP is a 5C compound and PGA is a 3C compound
(d) RUBISCO plays the role of oxygenase enzyme.
(e) C₂ cycle occurs in photosynthetic cells due to the absence of CO₂ and increased O₂.

Choose the correct answer.

1. Photosynthetic organisms used only percent of solar light.
(a) 0.2
(b) 0.6
(c) 0.1
(d) 0.8
Answer:
(a) 0.2

2. ………… is famously called as father of plant physiology.
(a) Joseph Priestley
(b) Lavoisier
(c) Stephen Hales
(d) Van Helmont
Answer:
(c) Stephen Hales

3. In Green sulphur bacteria, is the hydrogen donor.
(a) H₂O
(b) H₂S
(c) H₂O₂
(d) HCl
Answer:
(b) H₂S

4. Ruben & Kamen used radioactive oxygen to prove the evolution of oxygen from water.
(a) 180
(b) 160
(c) 140
(d) 120
Answer:
(a) 180

5. In photosynthesis ………….. is reduced into carbohydrates.
(a) H₂O
(b) Chlorophyll
(c) CO₂
(d) O₂
Answer:
(c) CO₂

6. Which of the following performs anaerobic photosynthesis?
(a) Green sulphur bacteria
(b) Cyanobacteria
(c) Purple sulphur bacteria
(d) Green filamentous bacteria
Answer:
(b) Cyanobacteria

7. The colloidal proteinaceous matrix of chloroplast is
(a) Thylakoid
(b) Stroma
(c) Grana
(d) Lamellae
Answer:
(b) Stroma

8. Each granum has thylakoids.
(a) 8-30
(b) 40 – 80
(c) 5-30
(d) 40 – 70
Answer:
(c) 5-30

9. Which of the following is a primary pigment?
(a) Chlorophyll a
(b) Carotene
(c) Xanthophyll
(d) Phycoerythrin
Answer:
(a) Chlorophyll a

10. The chlorophyll pigment found in xanthophycean algae is
(a) Chlorophyll b
(b) Chlorophyll c
(c) Chlorophyll d
(d) Chlorophyll e
Answer:
(d) Chlorophyll e

11. The nature of phytol tail in chlorophyll is
(a) Hydrophilic
(b) Lipophilic
(c) Hydrophobic
(d) Lipophobic
Answer:
(b) Lipophilic

12. The porphyrin head of chlorophyll has ………. pyrrole rings.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

13. Which of the following mineral is NOT required for the biosynthesis of chlorophyll a?
(a) Mn
(b) Mg
(c) Mo
(d) Cu
Answer:
(c) Mo

14. In chlorophyll C, which of the following component is absent?
(a) Porphyrin head
(b) Phytol tail
(c) Pyrrole ring
(d) Methyl group
Answer:
(b) Phytol tail

15. …………. pigments are also called as shield pigments.
(a) Carotenoids
(b) Chlorophyll b
(c) Phycobilins
(d) Phycoerythrin
Answer:
(a) Carotenoids

16. ………….. is responsible for the yellow colour change of leaves during autumn.
(a) Cutin
(b) Lutein
(c) Phycobilin
(d) Carotein
Answer:
(b) Lutein

17. The colour of the light is determined by its …………..
(a) Intensity
(b) Refractive power
(c) Wavelength
(d) Reflection
Answer:
(c) Wavelength

18. Visible spectrum ranges between ………….
(a) 390 – 763 nm
(b) 370 – 700 nm
(c) 450 – 700 nm
(d) 357 – 736 nm
Answer:
(a) 390 – 763 nm

19. Electromagnetic spectrum consists of …………. types of radiation.
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

20. Light as a particle is called …………
(a) Neutron
(b) Quantum
(c) Photon
(d) Quantasome
Answer:
(c) Photon

21. Usually …………… chlorophyll molecules are considered as physiological units of photosynthesis.
(a) 300 – 700
(b) 200-300
(c) 240-750
(d) 200-280
Answer:
(b) 200-300

22. Who coined the term Quantasomes?
(a) Steinmann
(b) Park & Biggins
(c) Emerson & Arnold
(d) Von Mayer
Answer:
(b) Park & Biggins

23. In Emerson’s first effect, the photosynthetic yield was dropped in the region above ………..
(a) 720 nm
(b) 620 nm
(c) 680 nm
(d) 600 nm
Answer:
(c) 680 nm

24. In photosynthetic reactions ………… is considered as assimilatory power.
(a) NADPH
(b) FADPH
(c) ATP
(d) GTP
Answer:
(c) ATP

25. Antenna molecules refers to …………
(a) Light-harvesting complex
(b) Central core complex
(c) PS II
(d) Oxygen evolving complex
Answer:
(a) Light-harvesting complex

26: Chlorophyll and carotenoid ratio in PS II is …………
(a) 3 to 7:1
(b) 20 to 30:1
(c) 7 to 30:1
(d) 10 to 30:1
Answer:
(a) 3 to 7:1

27. Phosphorylation taking place during respiration is called …………
(a) Substrate level phosphorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) Photophosphorylation
Answer:
(b) Oxidative phosphorylation

28. Which of the following statement is NOT true regarding cyclic photophosphorylation?
(a) The primary electron acceptor is FRS
(b) It produces only ATP molecules
(c) It produces only NADPH + H⁺ molecules.
(d) Electrons ejected from PSI again cycled back to PSI
Answer:
(c) It produces only NADPH + H⁺ molecules

29. Chemiosmotic theory was proposed by …………
(a) Mitchell
(b) Hatch & Slack
(c) Calvin
(d) Priestley
Answer:
(a) Mitchell

30. How many ATP molecules are utilized by C₃ plants to evolve one oxygen molecule.
(a) 3
(b) 4
(c) 8
(d) 12
Answer:
(a) 3

31. The first formed product of C₃ cycle is …………
(a) Succinic acid
(b) Phosphoglyceric acid
(c) Oxalo Acetic acid
(d) Malic acid
Answer:
(6) Phosphoglyceric acid

32. RUBP is a ……….. carbon compound.
(a) Three
(b) Four
(c) Five
(d) Seven
Answer:
(c) Five

33. Number of dicot species performing C₄ pathway is …………
(a) 200
(b) 300
(c) 800
(d) 1000
Answer:
(b) 300

34. The CO₂ acceptor molecule in C₃ plants is …………
(a) PEP
(b) PGA
(c) OAA
(d) RUBP
Answer:
(d) RUBP

35. C₂ cycle refers to
(a) CAM cycle
(b) PCO cycle
(c) PCR cycle
(d) DCA cycle
Answer:
(b) PCO cycle

36. Identify the mismatched pair:
(i) Phosphorylation reaction – phosphorous
(ii) Photolysis of water – Manganese & Chlorine
(iii) Plastocyanin formation – Copper and Zinc
(iv) Chlorophyll formation – Magnesium, Iron, Nitrogen
(a) (i) only
(b) (iii) only
(c) Both (i) & (iii)
(d) All the above
Answer:
(b) (iii) only

37. Photosynthetically Active Radiation (PAR) is between
(a) 700 – 760 nm
(b) 400-700 nm
(c) 500-600 nm
(d) 350 – 760 nm
Answer:
(b) 400 – 700 nm

38. In atmosphere the percentage of CO₂ is …………
(a) 0.3%
(b) 0.7%
(c) 0.1%
(d) 0.6%
Answer:
(a) 0.3%

39. The inhibitory effect of oxygen in photosynthesis was first discovered by ……………..
(a) Warburg
(b) Van Helmont
(c) Dutrochet
(d) Desaussure
Answer:
(a) Warburg

40. In general, the optimum temperature for photosynthesis is …………
(a) 26°C to 39°C
(b) 25°C to 40°C
(c) 55°C
(d) 25°C to 35°C
Answer:
(d) 25°C to 35°C