Category: Class 11 Biology

NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination.

NCERT Exemplar Class 11 Biology Chapter 19 Excretory Products and Their Elimination

Multiple Choice Questions

Q1. The following substances are the excretory products in animals. Choose the least toxic from among them.
(a) Urea (b) Uric acid
(c) Ammonia (d) Carbon dioxide
Ans: (b) Ammonia is the most toxic form and requires large amount of water for its elimination, whereas uric acid, being the least toxic, can be removed with a minimum loss of water.

Q2. Filtration of the blood takes place at
(a) PCT (b) DCT
(c) Collecting ducts (d) Malpighian body
Ans: (d) Filtration of the blood takes place at malpighian body.

Q3. Which of the following statements is incorrect?
a. ADH—prevents conversion of angiotensinogen in blood to angiotensin
b. Aldosterone—facilitates water reabsorption
c. ANF—enhances sodium reabsorption
d. Renin—causes vasodilation
Ans: (a) ADH is a hormone released from the posterior pituitary gland that causes an increase in blood pressure through reabsorption of water.

Q4. A large quantity of one of the following is removed from our body by lungs
(a) C02 only (b) H20 only
(c) C02 and H20 (d) Ammonia
Ans: (a) A large quantity of C02 is removed from our body by lungs.

Q5. The pH of human urine is approximately
(a) 6.5 (b) 7 (c) 6 (d) 7.5
Ans: (c) The pH of human urine is approximately 6.

Q6. Different types of excretory structures and animals are given below. Match them appropriately and mark the correct answer from among those given below:

Excretory structure/ organ		Animals
A.	Protonephridia	(i)	Prawn
B.	Nephridia	(ii)	Cockroach
C.	Malpighian tubules	(iii)	Earthworm
D.	Green gland or Antennal gland	(iv)	Flatworms

Options:
(a) D—(i), (C)—(ii), B—(iii), A—(iv)
(b) B— (i), (C)—(ii), A—(iii), B—(iv)
(c) D—(i), (C)—(ii), A—(iii), B-(iv)
(d) B—(i), (C)—(ii), B—(iii), D—(iv)

Ans. (a)

Excretory structure/ organ		Animals
A.	Protonephridia	(iv)	Flatworms
B.	Nephridia	(iii)	Earthworm
C.	Malpighian tubules	(ii)	Cockroach                      •
D.	Green gland or Antennal gland	(i)	Prawn

 

Q7. Which one of the following statements is incorrect?
(a) Birds and land snails are uricotelic animals.
(b) Mammals and frogs are ureotelic animals.
(c) Aquatic amphibians and aquatic insects are ammonotelic animals.
(d) Birds and reptiles are ureotelic.
Ans: (d) Reptiles (snakes and lizards), birds, land snails and insects excrete nitrogenous wastes as uric acid in the form of pellet or paste with a minimum loss of water and are called uricotelic animals.

Q8. Which of the following pairs is wrong?
(a) Uricotelic — Birds (b) Ureotelic — Insects
(c) Ammonotelic — Tadpole (d) Ureotelic — Elephant
Ans: (b) Insects — Uricotelic

Q9. Which one of the following statements is incorrect?
(a) The medullary zone of kidney is divided into a few conical masses called medullary pyramids projecting into the calyces.
(b) Inside the kidney the cortical region extends in between the medullary pyramids as renal pelvis
(c) Glomerulus along with Bowman’s capsule is called the renal corpuscle
(d) Renal corpuscle, proximal convoluted tubule (PCT) and distal convoluted tubule (DCT) of the nephron are situated in the cortical region of kidney
Ans: (b) The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.

Q10. The condition of accumulation of urea in the blood is termed as
(a) Renal Calculi (b) Glomerulonephritis
(c) Uremia (d) Ketonuria
Ans: (c) The condition of accumulation of urea in the blood is termed as uremia.

Q11. Which one of the following is also known as antidiuretic hormone?
(a) Oxytocin (b) Vasopressin (c) Adrenaline (d) Calcitonin
Ans: (b) Vasopressin is also known as antidiuretic hormone (ADH).

Q12. Match the terms given in Column I with their physiological processes given in Column II and choose the correct answer

Column I		Column II
A.	Proximal convoluted tubule	(i)	Formation of concentrated urine
B.	Distal convoluted tubule	(ii)	Filtration of blood
C.	Henle’s loop	(iii)	Reabsorption of 70-80% of electrolytes
D.	Counter-current mechanism	(iv)	Ionic balance
E.	Renal corpuscle	(v)	Maintenance of concentration gradient in medulla

Options:

(a) A—(iii), B—(v), C—(iv), D—(ii), E—(i)
(b) A—(iii), B—(iv), C—(i), D—(v), E—(ii)
(c) A—(i), B—(iii), C—(ii), D—(v), E—(iv)
(d) A—(iii), B—(i), C—(iv), D—(v), E—(ii)

Ans: (b)

A.	Proximal convoluted tubule	(iii)	Reabsorption of 70-80% of electrolytes
B.	Distal convoluted tubule	(iv)	Ionic balance
C.	Henle’s loop	(i)	Formation of concentrated urine
D.	Counter-current mechanism	(v)	Maintenance of concentration gradient in medulla
E.	Renal corpuscle	(ii)	Filtration of blood

Q13. Match the abnormal conditions given in Column A with their explanations given in Column B and choose the correct option.

Column A		Column B
A.	Glycosuria	(i)	Accumulation of uric acid in joints
B.	Renal calculi	(ii)	Inflammation in glomeruli
C.	Glomerulonephritis	(iii)	Mass of crystallised salts within the kidney
D.	Gout	(iv)	Presence of glucose in urine

Options:
(a) A—(i), B—(iii), C—(ii), D—(iv)
(b) A—(iii), B—(ii), C—(iv), D—(i)
(c) A—(iv), B—(iii), C—(ii), D—(i)
(d) A—(iv), B—(ii), C—(iii), D—(i)

Ans. (c)

A.	Glycosuria	(iv)	Presence of glucose in urine
B.	Renal calculi	(iii)	Mass of crystallised salts within the kidney
C.	Glomerulonephritis	(ii)	Inflammation in glomeruli
D.	Gout	(i)	Accumulation of uric acid in joints

 

Q14. We can produce concentrated/dilute urine. This is facilitated by a special mechanism. Identify the mechanism.
(a) Reabsorption from PCT
(b) Reabsorption from collecting duct
(c) Reabsorption/secretion in DCT
(d) Counter current mechanism in Henle’s loop/Vasa recta
Ans: (d) We can produce concentrated/dilute urine. This is facilitated by a special mechanism called counter current mechanism in Henle’s loop/Vasa recta.

Q15. Dialysing unit (artificial kidney) contains a fluid which is almost same as plasma except that it has
(a) High glucose
(b) High urea
(c) No urea
(d) High uric acid
Ans: (c) Dialysing fluid = Plasma – nitrogenous wastes (urea)

Very Short Answer Type Questions
Q1. Where does the selective reabsorption of Glomerular filtrate take place?
Ans: DCT

Q2. What is the excretory product from kidneys of reptiles?
Ans: Uric acid

Q3. What is the composition of sweat produced by sweat glands?
Ans: Water, minerals, lactic acid and urea.

Q4. Identify the glands that perform the excretory function in prawns.
Ans: Antennal glands or green glands

Q5. What is the excretory structure in amoeba?
Ans: Contractile vacuole

Q6. The following abbreviations are used in the context of excretory functions, what do they stand for?
a.ANF
b. ADH
c. GFR
d. DCT
Ans: a. ANF—Atrial Natriuretic factor
b. ADH—Antidiuretic hormone
c. GFR—Glomerular Filtration Rate
d. DCT—Distal Convoluted Tubule
Q7. Differentiate Glycosuria from Ketonuria.
Ans: Glycosuria—Presence of glucose in urine.
Ketonuria—Presence of ketone bodies in urine.

Q8. What is the role of sebaceous glands?
Ans: Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Q9. Name two actively transported substances in Glomerular filtrate.
Ans: Glucose and aminoacids

Q10. Mention any two metabolic disorders, which can be diagnosed by analysis of urine.
Ans: Glycosuria and ketonuria

Q11. What are the main processes of urine formation?
Ans: The main processes are filtration, reabsorption, secretion and concentration/ dilution

Q12. Sort the following into actively or passively transported substances during reabsorption of GFR:
glucose, aminoacids, nitrogenous wastes, Na⁺, water
Ans: Actively transported—Glucose, aminoacids and Na⁺
Passively transported—Nitrogenous wastes and water

Q13. Complete the following:
a. Urinary excretion = Tubular reabsorption + tubular secretion
b. Dialysis fluid = Plasma
Ans: a. Urinary excretion = Glomerular filtration – tubular reabsorption + tubular secretion
b. Dialysis fluid = Plasma – nitrogenous wastes

Q14. Mention the substances that exit from the tubules in order to maintain a concentration gradient in the medullary interstitium.
Ans: NaCl and Urea.

Q15. Fill in the blanks appropriately                  .

       Organ                              Excretory wastes

1. Kidneys _____________________
2. Lungs _____________________
3. Liver _____________________
4. Skin _____________________

Ans: a. Kidneys — Urea
b. Lungs — C0₂ and H_zO
c. Liver — Bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs
d. Skin — Sweat (NaCl, urea, lactic acid) and sebum (sterols, hydrocarbons and waxes).

Short Answer Type Questions
Q1. Show the structure of a renal corpuscle with the help of a diagram.
Ans:

Q2. What is the role played by Renin-Angiotensin in the regulation of kidney function?
Ans: Renin is released from JGA on activation due to fall in the glomerular blood pressure/flow. Renin converts angiotensinogen in blood to angiotensin-I and further to angiotensin-II. Angiotensin-II being a powerful vasoconstrictor, increase the glomerular blood pressure and thereby GFR. Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na⁺ and water from the distal parts of the tubule. This also leads to.an increase in blood pressure and thereby GFR. This is generally known as the Renin-Angiotensin mechanism.

Q3. Aquatic animals generally are ammonotelic in nature whereas terrestrial forms are not. Comment.
Ans: Ammonia is the most toxic form and requires large amount of water for its elimination, terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes like urea and uric acid for conservation of water. Mammals, many terrestrial amphibians and marine fishes mainly excrete urea and are called ureotelic animals. Ammonia produced by metabolism is converted into urea in the liver of these animals and released into the blood which is filtered and excreted out by the kidneys.

Q4. The composition of glomerular filtrate and urine is not same. Comment.
Ans: A comparison of the volume of the filtrate formed per day (180 litres per
day) with that of the urine released (1.5 litres), suggest that nearly 99% of the filtrate has to be reabsorbed by the renal tubules. This process is called reabsorption. For example, substances like glucose, amino acids, Na+, etc.,
in the filtrate are reabsorbed actively so, these substances are not present in urine.

Q5. What is the procedure advised for the correction of extreme renal failure? Give a brief account of it.
Ans: Kidney transplantation is the ultimate method in the correction of acute renal failures (kidney failure). A functioning kidney is used in transplantation from a donor, preferably a close relative, to minimise its chances of rejection by the immune system of the host. Modem clinical procedures have increased the success rate of such a complicated technique.

Q6. How have the terrestrial organisms adapted themselves for conservation of water?
Ans: Terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes like urea and uric acid for conservation of water. Mammals, many terrestrial amphibians and marine fishes mainly excrete urea and are called ureotelic animals. Ammonia produced by metabolism is converted into urea in the liver of these animals and released into the blood which is filtered and excreted out by the kidneys. Some amount of urea may be retained in the kidney matrix of some of these animals to maintain a desired osmolarity. Reptiles, birds, land snails and insects excrete nitrogenous wastes as uric acid ‘ in the form of pellet or paste with a minimum loss of water and are called uricotelic animals.

Q7. Label the parts in the following diagram

Q8. Explain, why a haemodialysing uhit called’artificial kidney?
Ans: Malfunctioning of kidneys can lead to accumulation of urea in blood, a condition called uremia, which is highly harmful and may lead to kidney failure. In such patients, urea can be removed by a process called hemodialysis. Blood drained from a convenient artery is pumped into a dialysing unit (also called artificial kidney) after adding an anticoagulant like heparin. The unit contains a coiled cellophane tube surrounded by a fluid (dialysing fluid) having the same composition as that of plasma except the nitrogenous wastes. The porous cellophane membrane of the tube allows the passage of molecules based on concentration gradient. As nitrogenous wastes are absent in the dialysing fluid, these substances freely move out, thereby clearing the blood. The cleared blood is pumped back to the body through a vein after adding anti-heparin to it.

Q9. Comment upon the hormonal regulation of selective reabsorption.
Ans: Osmoreceptors in the body are activated by changes in blood volume, body fluid volume and ionic concentration. An excessive loss of fluid from the body can activate these receptors which stimulate the hypothalamus to release antidiuretic hormone (ADH) or vasopressin from the neurohypophysis. ADH facilitates water reabsorption from latter parts of the tubule, thereby preventing diuresis.

Long Answer Type Questions

Q1. Explain the mechanism of formation of concentrated urine in mammals.
Ans: Mechanism of Concentration of the Filtrate:
Mammals have the ability to produce a concentrated urine. The Henle’s loop and vasa recta play a significant role in this. The flow of filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter current in them help in maintaining an increasing osmolarity towards the inner medullary interstitium, i.e., from 300 mOsmo1L in the cortex to about 1200 mOsmolLin the inner medulla. This gradient is mainly caused by NaC1 and urea. NaC1 is transported by the ascending limb of Henle’s loop which is exchanged with the descending limb of vasa recta. NaCl is returned to the interstitium by the ascending portion of vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s ioop which is transported back to the interstitium by the collecting tubule. The above described transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the counter current mechanism. This mechanism helps to maintain a concentration gradient in the medullary interstitium. Presence of such interstitial gradient helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine). Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Q2. Draw a labelled diagram shewing reabsorption and secretion of major substances at different parts of the nephron.
Ans:

Q3. Explain briefly, micturition and disorders of the excretory system.
Ans:
• Micturition: Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This signal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine. The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex.
• Disorders of the Excretory System: Malfunctioning of kidneys can lead to accumulation of urea in blood, a condition called uremia, which is highly harmful and may lead to kidney failure. In such patients, urea can be removed by a process called hemodialysis. Kidney transplantation is the ultimate method in the correction of acute renal failures (kidney failure).
• Renal calculi: Stone or insoluble mass of crystallised salts (oxalates, etc.) formed within the kidney.
• Glomerulonephritis: Inflammation of glomeruli of kidney.

Q4. How does tubular secretion help in maintaining ionic and acid-base balance in body fluids?
Ans: During urine formation, the tubular cells secrete substances like H⁺, K⁺ and ammonia into the filtrate. Tubular secretion is also an important step in urine formation as it helps in the maintenance of ionic and acid base balance of body fluids.              .

• PCT helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate
• DCT is also capable of selective secretion of hydrogen and potassium ions and NH₃ to maintain the pH and sodium-potassium balance in blood.
• Collecting duct also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺

Q5. The” glomerular filtrate in the loop of Henle gets concentrated in the descending and then gets diluted in the ascending limbs. Explain.
Ans: A hairpin shaped Henle’s loop has a descending and an ascending limb. Reabsorption is minimum in its ascending limb. However, this region plays a significant role in the maintenance of high osmolarity of medullary interstitial fluid. The descending limb of loop of Henle is permeable to water but almost impermeable to electrolytes. This concentrates the filtrate as it moves down. The ascending limb is impermeable to water but allows transport of electrolytes actively or passively. Therefore, as the concentrated filtrate pass upward, it gets diluted due to the passage of electrolytes to the medullary fluid.

Q6. Describe the structure of a human kidney with the help of a labelled diagram.
Ans: In humans, the excretory system consists of a pair of kidneys, one pair of
ureters, a urinary bladder and a urethra. Kidneys are reddish brown, bean shaped structures situated between the levels of last thoracic and third lumbar vertebra close to the dorsal inner wall of the abdominal cavity. Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g. Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter. Inner to the hilum is a broad funnel shaped space called the renal pelvis with projections called calyces. The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and an inner meditlla. The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces (sing.: calyx). The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini. Each kidney has nearly one million complex tubular structures called nephrons, which are the functional units.

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NCERT Exemplar Class 11 Biology Chapter 2 Biological Classification are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 2 Biological Classification.

NCERT Exemplar Class 11 Biology Chapter 2 Biological Classification

Multiple Choice Questions
Single Correct Answer Type

Q1. All eukaryotic unicellular organisms belong to
(a) Monera (b) Protista (c) Fungi (d) Bacteria
Ans: (b) Monera—Kingdom of prokaryotes
• All eukaryotic unicellular organisms belong to protista.
Q2. The five kingdom classification was proposed by
(a) R.H. Whittaker . (b) C. Linnaeus
(c) A. Roxberg (d) Virchow
Ans: (a) This phylogenetic classification was proposed by R.H. Whittaker (1969). He created new kingdom ‘Fungi’. The five kingdom classification are as follows: 1. Plantae, 2. Animalia, 3. Protista. 4. Monera and 5. Fungi.
Whittaker has used 5 criteria for the 5 kingdom classification and are as follows:
1. Reproduction,
2. Cell structure,
3. Phylogenetic relationships,
4. Mode of nutrition,
5. Thallus organisation

Q3. Organisms living in salty areas are called as
(a) Methanogens (b) Halophiles
(c) Heliophytes (d) Thermoacidophiles
Ans: (b)
• Halophiles: Bacteria living in extremely salty areas.
• Thermoacidophiles: Bacteria living in hot springs/deep sea water.
E.g.: Thermococcus
• Methanogens: Bacteria living in marshy areas and produce methane gas.
• Heliophytes: Sun loving plants

Q4. Naked cytoplasm, multinucleated and saprophytic are the characteristics of
(a) Monerans (b) Protists (c) Fungi (d) Slime moulds
Ans: (d) Slime moulds are saprophytic protists, without cell walls. The spores
of slime moulds possess true walls. Thalloid multinucleate body of a slime mould is called plasmodium. Spores are dispersed by air currents.
E.g.: Acellular slime mould—Physarum, Cellular slime mould—Dictyostelium.

Q5. An association between roots of higher plants and fungi is called .
(a) Lichen (b) Fern (c) Mycorrhiza (d) BGA
Ans: (c) Lichens are symbiotic associations, i.e. mutually useful associations, between algae and fungi. This relationship is best known as helotism.

Q6. A dikaryon is formed when
(a) Meiosis is arrested
(b) The two haploid cells do not fuse immediately
(c) Cytoplasm does not fuse
(d) None of the above
Ans: (b) In some fungi the fusion of two haploid cells immediately results in diploid cells (2n). However, in other fungi (ascomycetes and basidiomycetes), an intervening dikaryotic stage (n + n, i.e. two nuclei per cell) occurs. Such a condition is called a dikaryon and the phase is called dikaryophase of fungus. A dikaryotic cell has two dissimilar haploid nuclei.

Q7. Contagium vivum fluidum was proposed by
(a) D. J. Ivanowsky (b) M. W. Beijerinck
(c) Stanley (d) Robert Hooke
Ans: (b) D.J. Ivanowsky (1892) discovered the virus and has recognised certain microbes as causal organism of the mosaic disease of tobacco. M.W. Beijerinck (1898) demonstrated that the extract of the infected plants of tobacco could cause infection in healthy plants and called the fluid as Contagium vivum fluidum (infectious living fluid). W. M. Stanley (1935) first time showed that viruses could be crystallised and crystals consist largely of proteins.

Q8. Association between mycobiont and phycobiont are found in
(a) Mycorrhiza (b) Root (c) Lichens (d) BGA
Ans. (c) Association between mycobiont and phycobiont are found in lichens.

Q9. Difference between virus and viroid is
(a) Absence of protein coat in viroid but present in virus
(b) Presence of low molecular weight RNA in virus but absent in viroid
(c) Both (a) and (b)
(d) None of the above
Ans: (a) Viroids are smaller than viruses and the cause of potato spindle tuber disease, chrysanthemum stunt disease. It was found to be a free RNA and lacked the protein coat that is found in viruses, hence the name viroid.

Q10. With respect to the fungal sexual cycle, choose the correct sequence of events.
(a) Karyogamy, plasmogamy and meiosis
(b) Meiosis, plasmogamy and karyogamy
(c) Plasmogamy, karyogamy and meiosis
(d) Meiosis, karyogamy and plasmogamy
Ans: (c) The sexual cycle involves the following three steps:
1. Fusion of protoplasms between two motile or non-motile gametes called
plasmogamy.
• Plasmogamy is fusion of two haploid cells without nuclear fusion.
2. Fusion of two nuclei is called karyogamy.
3. Meiosis in zygote resulting in haploid spores.

Q11. Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to?
(a) Monera (b) Protista (c) Fungi (d) None of these
Ans: (d) Viruses did not find a place in classification since they are not truly ‘living’ if we understand living as those organisms that have a cell structure. Viruses are neither prokaryotes nor eukaryotes. They are inert outside their specific host cell and cannot multiply of their own because they lack cellular machinery to use its genetic material. Viruses can only multiply in host or living cell.

Q12. Members of Phycomycetes are found in
(1) Aquatic habitats
(ii) On decaying wood
(iii) Moist and damp places
(iv) As obligate parasites on plants Choose from the following options.
(a) None of the above (b) (i) and (iv)
(c) (ii) and (iii) (d) All of the above
Ans: (d) Members of Phycomycetes are found in aquatic habitats, on decaying wood, moist and damp places and as obligate parasites on plants.

Very Short Answer Type Questions
Q1. What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
Ans: Cyanobacteria (BGA) are autotrophic microbes. Cyanobacteria are widely distributed in aquatic and terrestrial environments. Nostoc, Ariabaena and Oscillatoria are BGA that can fix atmospheric nitrogen. These organisms can fix atmospheric nitrogen in specialised cells called heterocysts, e.g., Nostoc and Anabaena. In paddy fields cyanobacteria serve as an important biofertiliser. BGA also add organic matter to the soil and increase its fertility.

Q2. Suppose you accidentally find an old preserved permanent slide without a label. In your effort to identify it, you place the slide under microscope and observe the following features:
a. Unicellular ‘
b. Well defined nucleus
c. Biflagellate—one flagellum lying longitudinally and the other transversely.
What would you identify it as? Can you iiame the kingdom it belongs to?
Ans: Dinoflagellates are unicellular eukaryotes. Most of them have two flagella; one lies longitudinally and the other transversely in a furrow between the wall plates. Dinoflagellates belongs to kingdom protista.

Q3. How is the five-kingdom classification advantageous over the two kingdom classification?
Ans: Two Kingdom system of classification with Plantae and Animalia kingdoms.
1. Two kingdom classification did not distinguish between the prokaryotes and eukaryotes. E.g.: It brought together the prokaryotic bacteria and blue green algae with other groups (like plant, fungi and animals) which were eukaryotic.
2. Two kingdom classification did not distinguish between the unicellular and multicellular organisms. E.g:: Chlamydomonas (unicellular) and Spirogyra (multicellular) were placed together under algae.
3. This system did not distinguish between autotrophic/photosynthetic (green algae and plants) and the heterotrophic/non-photosynthetic organisms (fungi).
Five Kingdom Classification:
1. Fungi were placed in a separate kingdom—kingdom fungi.
2. It has put together organisms which were placed in different kingdoms in earlier classifications.
Kingdom Protista brought together Chlamydomonas, Chlorella (earlier placed in algae within plants and both having cell walls) with Amoeba and Paramoecium (earlier plaeed in the animal kingdom and both lacking cell walls).
3. Animal and plant kingdoms become more homogenous than they are in the two kingdom classification. So, it is the advantageous over the two kingdom classification.

Q4. Polluted water bodies have usually very high abundance of plants like Nostoc and Oscillitoria. Give reasons.
Ans: Polluted water bodies (Ponds, ditches and rivers etc.) have usually nutrient contents (such as nitrate, phosphates) domestic sewage primarily contains biodegradable organic matter. Presence of large amount of nutrients in waters also causes excessive growth of Planktonic (free floating algae) called an algal bloom, which imparts a distinct colour to water bodies.
Algal bloom causes deterioration of the water quality and fish mortality. Some bloom forming algae mainly Nostoc and Oscillitoria are extremely toxic to human beings and animals.

Q5. Are chemosynthetic bacteria—autotrophic or heterotrophic?
Ans: Chemosynthetic bacteria oxidise various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for their ATP production. So chemosynthetic bacteria are autotrophic in nature.

Q6. The common name of pea is simpler than its botanical (scientific) name Pisum sativum. Why then is the simpler common name not used instead of the complex scientific/botanical name in biology?
Ans: As we know that pea (vernacular name or local name) is simpler than its botanical (scientific) name Pisum sativum. These local names would vary from place to place, even within a country. Probably one would recognise the confusion that would be created if we did not find ways and means to talk to each other, to refer to organisms we are talking about.
Hence, there is a need to standardise the naming of living organisms such that a particular organism is known by the same name all over the world.

Q7. A virus is considered as a living organism and an obligate parasite when inside a host cell. But virus is not classified along with bacteria or fungi. What are the characters of virus that are similar to non-living objects?
Ans: Virus are living organisms inside a host cell. But virus is not classified along with bacteria or fungi because they having an inert crystalline structure outside the living cell. They are inert outside their specific host cell and cannot multiply of their own because they lack cellular machinery to use its genetic material.

Q8. In the five kingdom system of Whittaker, how many kingdoms are eukaryotes?
Ans: In the five kingdom system of Whittaker, four kingdoms (Protista, fungi,
plantae and animalia) belong to eukaryotes.

Short Answer Type Questions

Q1. Diatoms are also called as ‘pearls of ocean’, why? What is diatomaceous earth?
Ans: The diatoms are the unique organisms, because of their distinctive cell walls. The walls are embedded with silica and thus the walls are indestructible. It show sculpturing and ornamentation that why Diatoms are also called as ‘Pearls of Ocean’.
Diatoms have left behind large amount of cell wall deposits in their habitat; this accumulation over billions of years is referred to as ‘diatomaceous earth’. Being gritty this soil is used in polishing, filtration of oils and syrups. Diatoms are the chief ‘producers’ in the oceans.

Q2. There is a myth that immediately after heavy rains in forest, mushrooms appear in large number and make a very large ring or circle, which may be several metres in diameter. These are called as ‘Fairy rings’. Can you explain this myth of fairy rings in biological terms?
Ans: After heavy rains in forest, moisture and nutrients pass down in soil and activates the growth of mushroom mycelium. The basidiocarps of Agaricus (mushroom) arise from the mycelium present in the soil. They appear in a circle like a ring. As these basidiocarps resemble buttons and grow in rings, they are known as fairy rings.

Q3. Neurospora—an ascomycetes fungus has been used as a biological tool to understand the mechanism of plant genetics much in the same way as Drosophila has been used to study animal genetics. What makes Neurospora so important as a genetic tool?
Ans: Neurospora is used as a genetic tool because it is easy to grow and has a haploid life cycle that makes genetic analysis simple since recessive traits will show up in the offspring. Beadle and Tatum exposed Neurospora crassa to X-rays, causing mutations. This led them to propose the “one gene, one en2yme” hypothesis that specific genes code for specific proteins.

Q4. Cyanobacteria and heterotrophic bacteria have been clubbed together in Eubacteria of kingdom Monera as per the “Five Kingdom Classification” even though the two are vastly different from each other. Is this grouping of the two types of taxa in the same kingdom justified? If so, why?
Ans: Cyanobacteria and heterotrophic bacteria have been clubbed together in Eubacteria of Kingdom Monera as per the “Five Kingdom Classification” because they do not have nuclear envelope and membrane bound organelles. Their genetic material is naked. They have 70S type of ribosomes. So, cyanobacteria and heterotrophic bacteria are prokaryotes and belong, to Kingdom Monera.

Q5. At a stage of their cycle, ascomycetes fungi produce the fruiting bodies like apothecium, perithecium or cleistothecium. How are these three types of fruiting bodies different from each other?
Ans: An apothecium is a wide, open, saucer-shaped or cup-shaped fruit body. It is sessile and fleshy. A cleistothecium is a globose, completely closed fruit body with no special opening to the outside. Perithecium are flask shaped structures opening by a pore or ostiole (short papilla opening by a circular pore).

Q6. What observable features in Trypanosoma would make you classify it under Kingdom Protista?
Ans: Trypanosoma is classified under the Kingdom Protista because it is unicellular eukaryotes. It has well defined nucleus with nuclear envelope, membrane bound organelles, 80S ribosomes and flagella with 9 + 2 organisation.

Q7. Fungi are cosmopolitan, write the role of fungi in your daily life.
Ans: Dough which is used for making bread, is fermented by fungi Saccharomyces cerevisiae (Baker’s yeast).
Roquefort cheese are ripened by growing a specific fungi on them, which gives them a particular flavour.
Microbes mainly yeasts used for the production of beverages like wine, beer, whisky, brandy or rum. For this purpose the yeast (Saccharomyces cerevisiae) used for fermenting malted cereals and juices to produce ethanol and commonly called Brewer’s yeast

Antibiotics produced by Fungi:

1.	Penicillin	Penicillium notatum and Penicillium chrysogenum
2.	Griseofulvin	Penicillium griseofulvum
3.	Cephalosporin	Cephalosporium acremonium
4.	Gentomycin	Micromonospora purpurea

• Cyclosporin A is produced by Trichoderma polysporum (Fungus). Cyclosporin A is used as an immunosuppressive agent in organ transplant patients.
• Statins produced by Monascus purpureus(Yeast). Statins used as blood- cholesterol lowering agent.
• Mushrooms, morels (Morchella) and truffles are edible fungi.
• Fungi causes several diseases in plants and animals including human beings.

Long Answer Type Questions
Q1. Algae are known to reproduce asexually by variety of spores under different environmental conditions. Name these spores and the conditions under which they are produced.
Ans: Asexual reproduction is by the production of different types of spores, the most common being the zoospores. They are flagellated (motile) and on germination gives rise to new plants.
Types of asexual reproduction:
1. Zoospores: Motile and formed in favourable condition.
2. Aplanospores: Thin walled, non-motile and formed in unfavourable condition.
3. Hypnospore: Thick walled, non-motile and formed in unfavourable condition.
4. Akinete: Under unfavourable condition, entire cell becomes thick.
5. Palmella stage: In condition of drought, protoplast is surrounded by gelatinous covering.

Q2. Apart from chlorophyll, algae have several other pigments in their chloroplast. What pigments are found in blue-green, red and brown algae that are responsible for their characteristic colours?
Ans: Apart from chlorophyll, algae have several other pigments in their chloroplast like carotenoids, xanthophylis (fucoxanthin) and r-phycoerythrin. In blue- green algae phycocyanin and r-phycoerythrin pigments are present beside chlorophyll a.
• Brown algae possess chlorophyll a, c, carotenoids and xanthophylls. They vary in colour from olive green to various shades of brown depending upon the amount of the xanthophyll pigment, fucoxanthin present in them.
• Red algae possess chlorophyll a, d and phycoerythrin in their body. The members of rhodophyceae are commonly called red algae because of the predominance of the red pigment, r-phycoerythrin in their body.

Q3. Make a list of algae and fungi that have commercial value as source of food,
chemicals, medicines and fodder.
Ans:
A. Economic importance of Algae:
1. Many species of Porphyra, Laminaria and Sargassum are among the 70 species of marine, algae used as food. Chlorella and Spirullina are unicellular algae, rich in proteins and are used as food supplements even by space travellers.
2. Certain marine brown and red algae produce large amounts of hydrocolloids (water holding substances) or phycocolloids, e.g.: algin (brown algae) and carrageen (red algae) are used commercially. Agar, one of the commercial products obtained from Gelidium and Gracilaria are used to grow microbes and in preparations of ice-creams and jellies.
3. Bromine is obtained from red algae Polysiphonia. Macrocystis is the source of Potash. Laminaria and Fucus are the source of Iodine.
B. Economic importance of Fungi:
1. Mushrooms, morels (Morchella) and truffles are edible fungi.
2. Microbes mainly yeasts used for the production of beverages like wine, beer, whisky, brandy or rum. For this purpose the yeast {Saccharomyces cerevisiae) used for fermenting malted cereals and juices to produce ethanol and commonly called Brewer’s yeast.
• Cyclosporin A is produced by Trichoderma polysporum (Fungus). Cyclosporin A is used as an immunosuppressive agent in organ transplant patients.
• Statins produced by Monascus purpureus (Yeast). Statins used as the blood-cholesterol lowering agent.

Penicillin	Penicillium notatum and Penicillium chrysogenum
Griseofulvin	Penicillium griseofulvum
Cephalosporin	Cephalosporium acremonium
Gentomycin	Micromonospora purpurea

Q4. ‘Peat’ is an important source of domestic fuel in several countries. How is ‘peat’ formed in nature?
Ans: Species of Sphagnum, a moss, provide peat that have long been used as fuel and because of their capacity to hold water as packing material for trans-shipment of living material. Peat forms when plant material decaying fully by acidic and anaerobic conditions. Peat is soft and easily compressed. Under pressure, water in the peat is forced out. Upon drying, peat can be used as fuel.

Q5. Biological classification is a dynamic and ever evolving phenomenon which keeps changing with our understanding of life forms. Justify the statement taking any two examples.
Ans: Kingdom Protista brought together Chlamydomonas, Chlorella (earlier placed in algae within plants and both having cell walls) with Amoeba and Paramoecium (earlier placed in the animal kingdom and both lacking cell walls). Five kingdom classification has put together organism (like Chlamydomonas and Amoeba) which were placed in different kingdoms in earlier classifications. This change happened because the criteria for classification changed. This kind of changes will take place in future too depending on the improvement in our understanding of characteristics and evolutionary relationships. So, biological classification is a dynamic and ever evolving phenomenon which keeps changing with our understanding of life forms.

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NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants.

NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants

Multiple Choice Questions

Q1. Which metal ion is a constituent of chlorophyll?
(a) Iron (b) Copper (c) Magnesium (d) Zinc
Ans: (c) Magnesium ion is a constituent of chlorophyll.

Q2. Which pigment acts directly to convert light energy to chemical energy?
(a) Chlorophyll a (b) Chlorophyll b
(c) Xanthophyll (d) Carotenoid
Ans: (a) Chlorophyll a pigment acts directly to convert light energy to chemical energy.

Q3. Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)?
(a) 100-390 (b) 390-430 (c) 400-700 (d) 760-10000
Ans: (c) 400-700 range of wavelength (in nm) is called photosynthetically active radiation (PAR).

Q4. Which light range is least effective in photosynthesis?
(a) Blue (b) Green (c) Red (d) Violet
Ans: (b) Green light range is least effective in photosynthesis.

Q5. Chemosynthetic bacteria obtain energy from
(a) Sun (b) Infrared rays
(c) Organic substances (d) Inorganic chemicals
Ans: (d) Chemosynthetic bacteria obtain energy from inorganic chemicals.

Q6. Energy required for ATP synthesis in PSII comes from
(a) Proton gradient (b) Electron gradient
(c) Reduction of glucose (d) Oxidation of glucose
Ans: (a) Energy required for ATP synthesis in PSII comes from proton gradient.

Q7. During light reaction in photosynthesis, the following are formed
(a) ATP and sugar                                 
(b) Hydrogen, 0₂ and sugar
(c) ATP, hydrogen donor and 0₂         
(d) ATP, hydrogen and 0₂ donor
Ans: (c) During light reaction in photosynthesis the following are formed ATP, hydrogen donor/(NADPH) and 0₂.

Q8. Dark reaction in photosynthesis is called so because
(a) It can occur in dark also
(b) It does not directly depend on light energy
(c) It cannot occur during day light
(d) It occurs more rapidly at night
Ans: (b) Dark reaction in photosynthesis is called so because it does not directly depend on light energy.

Q9. PEP is primary C0₂ acceptor in
(a) C₄ plants                                            
(b) C₃ plants
(c) C₂-plants                
 (d) Both C₃ and C₄ plants
Ans: (a) PEP is primary C0₂ acceptor in C₄ plants.

Q10. Splitting of water is associated with
(a) Photosystem I
(b) Lumen of thylakoid
(c) Both Photosystem I and II
(d) Inner surface of thylakoid membrane
Ans: (d) Splitting of water is associated with inner surface of thylakoid membrane.

Q11. The correct sequence of flow of electrons in the light reaction is
(a) PSII, plastoquinone, cytochromes, PSI, ferredoxin
(b) PSI, plastoquinone, cytochromes, PSII, ferredoxin
(c) PSI, ferredoxin, PSII
(d) PSI, plastoquinone, cytochromes, PSII, ferredoxin
Ans: (a) The correct sequence of flow of electrons in the light reaction is PSII, plastoquinone, cytochromes, PSI and ferredoxin.

Q12. The enzyme that is not found in a C₃ plant is
(a)    RuBP Carboxylase                      
(b)     PEP Carboxylase
(c)     NADP reductase                           
(d)     ATP synthase
Ans: (b) The enzyme that is not found in a C₃ plant is PEP Carboxylase.

Q13. The reaction that is responsible for the primary fixation C0₂ is catalysed by
(a) RuBP carboxylase
(b) PEP carboxylase
(c) RuBP carboxylase and PEP carboxylase
(d) PGA synthase
Ans: (c) The reaction that is responsible for the primary fixation of C0₂ is catalysed by RuBP carboxylase and PEP carboxylase.

Q14. When C0₂ is added to PEP, the first stable product synthesised is
(a)    Pyruvate                                        
(b)     Glyceraldehyde-3-phosphate
(c)     Phosphoglycerate                        
 (d)     Oxaloacetate
Ans: (d) When C0₂ is added to PEP, the first stable product synthesised is oxaloacetate.

Very Short Answer Type Questions

Q1. Examine the figure.

a. Is this structure present in animal cell or plant cell?
b. Can these be passed on to the progeny? How?
c. Name the metabolic processes taking place in the places marked (1) and (2).
Ans: a. Plant cell.
b. Yes, through female gametes.
c. In part (1)— Photophosphorylation. In part (2)—Calvin cycle.

Based on the above equation, answer the following questions:
a. Where does this reaction take place in plants?
b. What is the significance of this reaction?
Ans: a. Lumen of the thylakoids.
b. 0₂ is evolved during this reaction; moreover, electrons are made available to PS-II continuously.

Q3. Cyanobacteria and some other photosynthetic bacteria do not have chloroplasts. How do they conduct photosynthesis?
Ans: Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll.

Q4. a. NADP reductase enzyme is located on
b. Breakdown of proton gradient leads to release of
Ans: a. Grana-lamellae.
b. Energy.

Q5. Can girdling experiments be done in monocots? If yes, how? If no, why not?
Ans: No, because vascular bundles are scattered in monocot.

Analyse the above reaction and’answer the following questions:
a. How many molecules of ATP and NADPH are required to fix one molecule of C02?
b. Where in the chloroplast does this process occur?
Ans: a. Three molecules of ATP and two molecules of NADPH are required to fix one molecule of C02
b. Stroma of chloroplast

Q7. Does moonlight support photosynthesis? Find out.
Ans: As the intensity of moonlight is much less than the sunlight, so it does not support photosynthesis.

Q8. Some of these terms/chemicals are associated with the C₄ Explain.
a. Hatch and Slack pathway
b. Calvin cycle
c. PEP carboxylase
d. Bundle sheath cells
Ans: Though C₄ plants have C₄ oxaloacetic acid as is the first C0₂ fixation product they use the C₃ pathway or Calvin cycle as the main biosynthetic pathway. C₄ pathway is also called Hatch and Slack Pathway

• 1° C0₂ acceptor in C₄ plants is a 3-C molecule PEP (phosphoenol pyruvate) and is present in the mesophyll cells. The enzyme responsible for the fixation is PEPcase (PEP carboxylase) is found only in mesophyll cells. Bundle sheath cells lack PEPcase enzyme.
• C₄ acid (OAA) is formed by carboxylation is mesophyll cells; therefore, initial carboxylation reaction occurs in mesophyll cells (also in C₃ pathway). OAA forms other 4-C compounds like malic acid or aspartic acid in the mesophyll cells itself, which are transported to the bundle sheath cells.
• C0₂ released in the bundle sheath cells enters the C₃ or the Calvin pathway, a pathway common to all plants. The bundle sheath cells are rich in RuBisCO enzyme (necessary for the C₃ or the Calvin cycle), but lack PEPcase.
• Calvin pathway in C₄ plants takes place only in bundle sheath cells (because RuBisCO is present) but does not take place in the mesophyll cells because lack of RuBisCO enzyme in mesophyll cells of C₄ plants like maize, sorghum, sugarcane, Jowar, Euphorbia, Atriplex,

Q9. Where is NADP reductase enzyme located in the chloroplast? What is the role of this enzyme in proton gradient development?
Ans: The NADP reductase enzyme is located on the stroma side of the membrane. Along with electron that come from the acceptor of electrons of PSI protons are necessary for the reduction of NADP⁺ to NADPH + H⁺. These protons are also removed from the stroma.

Q10. ATPase enzyme consists of two parts. What are those parts? How are they arranged in the thylakoid membrane? Conformational change occurs in which part of the enzyme?
Ans: ATPase enzyme consists of two parts:
i.One portion called F₀ is imbedded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane.
ii.The other portion is called ‘F_l and protrudes on the outer surface-of the thylakoid membrane on the side that faces stroma.

The breakdown of the gradient provides enough energy to cause a conformational change in the F, particle of the ATPase, which makes the enzyme synthesise several molecules of energy-packed ATP.

Q11. Which products formed during the light reaction of photosynthesis are used to drive the dark reaction?
Ans: ATP and NADPH

Q12.What is the basis for designating C₃ and C₄ pathways of photosynthesis?
Ans:The number of carbon atoms in first stable product of carbondioxide fixation is the basis for designating C₃ and C₄ pathways of photosynthesis.

Short Answer Type Questions
Q1.Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic C0₂ requirements?
Ans: Succulent (water storing) plants such as cacti, euphorbias fix C0₂ into organic compound using PEP carboxylase at night, when the stomata are open.

The. organic compound (malic acid) accumulates throughout the night and is decarboxylated during the day to produce C0₂.

Q2.Chlorophyll a is the primary pigment for light reaction. What are accessory pigments? What is their role in photosynthesis?
Ans: Accessory pigments are those pigments, which assist in photosynthesis by capturing energy from light of different wavelengths, e.g., chlorophyll b, Xanthophylls and carotenoids.

Role in Photosynthesis:

1. They absorb wavelength of light not absorbed by chlorophyll a and transfer the energy to chlorophyll.
2. They also protect chlorophyll a from photo-oxidation.

Q3. Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Ans :ATP and NADPH are used to drive the processes leading to the synthesis of food, more accurately, sugars. This is the biosynthetic phase or dark reaction of photosynthesis. This process does not directly depend on the presence of light but is dependent on the products of the light reaction, i.e., ATP and NADPH, besides C0₂ and H₂0.

Q4. How are photosynthesis and respiration related to each other?
Ans: Photosynthesis and respiration are related to each other as

1. Both processes take place in double membrane bound organelles.
2. In both processes ATP synthesis takes place.
3. In both processes electron transport system requires.

Q5.   If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as a supplement to maintain its growth or survival?
Ans: No, this plant cannot photosynthesise in the absence of light. Only sunlight can be given as supplement to maintain its growth or survival.

Q6.Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions?\
Ans: Photosynthetic organisms occur at different depths in the ocean. They do not receive qualitatively and quantitatively the same light. The spectral quality of solar radiation is also important for life. The UV component of the spectrum is harmful to many organisms while not all the colour components of the visible spectrum are available for marine plants living at different depths of the ocean. Plants at great depth contains some accessory pigments that can easily capture the light.

Q7. In tropical rain forests, the canopy is thick and shorter plants growing below it, receive filtered light. How are they able to carry out photosynthesis?
Ans: In tropical rain forests, the canopy is thick and shorter plants growing below it called sciophytes (shade loving plants). They can photosynthesise in very low light conditions. They have larger photosynthetic units and hence they are able to carry out photosynthesis in filtered light.

Q8. What conditions enable RuBisCO to function as an oxygenase? Explain the ensuing process.
Ans: In the first step of the Calvin pathway RuBP combines with C0₂ to form 2 molecules of 3PGA, that is catalysed by RuBisCO.

Q9.Why does the rate of photosynthesis decrease at higher temperatures?
Ans: The rate of photosynthesis decreases at higher temperatures because at high temperatures the enzymes become denatured (destroy).

Q10. Explain how during light reaction of photosynthesis, ATP synthesis is a chemiosmotic phenomenon.
Ans: In the light reaction within the chloroplast, protons in the stroma decrease in number, while in the lumen there is accumulation of protons. This creates a proton gradient across the thylakoid membrane as well as a measurable decrease in pH in the lumen. This gradient is important because it is the breakdown of this gradient that leads to release of energy. The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the F₀ of the ATPase. The ATPase enzyme consists of two parts: one part called the F₀ is embedded in the membrane and forms a transmembrane channel that carries out facilitated diffusion of protons across the membrane. The other portion is called F1 and protrudes on the outer surface of the thylakoid membrane on the side that faces the stroma. The breakdown of the gradient provides enough energy to cause a conformational change in the F_i particle of the ATPase, which makes the enzyme synthesise several molecules of energy-packed ATP.

Q11. Find out how Melvin Calvin worked out the complete biosynthetic pathway for synthesis of sugar.
Ans: Just after World War II, among the several efforts to put radioisotopes to beneficial use, the work of Melvin Calvin is exemplary. The use of radioactive C¹⁴ by him in algal photosynthesis studies led to the discovery that the first C0₂ fixation product was a 3-carbon organic acid. He also contributed to working out the complete biosynthetic pathway; hence, it was called Calvin cycle after him. The first product identified was 3-phosphoglyceric acid or in short PGA.

Q12. Six turns of Calvin cycle are required to generate one mole of glucose. Explain.
Ans: The fixation of 6 molecules of C0₂ and 6 turns of the cycle are required for the removal of one molecule of glucose from the pathway. Hence for every C0₂ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH are required. To make one molecule of glucose 6 turns of the cycle are required.

In	Out
Six C02	One glucose
18 ATP	18ADP
12 NADPH	12 NADP

 

Q13. Complete the flow chart for cyclic photophosphorylation of the photosystem-I

Q14. In what kind of plants do you come across ‘Kranz’ anatomy? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Ans: On studying vertical sections of leaves, one of a C₃ plant and the other of a C₄ plant. The particularly large cells around the vascular bundles of the C₄ pathway plants are called bundle sheath cells, and the leaves which have such anatomy are said to have ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces. Leaves of C₄ plants – maize or sorghum lack photorespiration. In addition these plants show tolerance to higher temperatures. Plants that are adapted to dry tropical regions have the C₄ pathway.

Q15. A process is occurring throughout the day, in ‘X’ organism. Cells are participating in this process. During this process ATP, C0₂ and water are evolved. It is not a light-dependent process.
a.Name the process.
b. Is it a catabolic or an anabolic process?
c .What could be the raw material of this process?
Ans: ‘ a. Respiration
b.Catabolic process (actually amphibolic pathway)
c.Glucose

Q16. Tomatoes, carrots and chillies are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Ans: Tomatoes, carrots and chillies are red in colour due to the presence of carotene pigment. It is an accessory photosynthetic pigment.

Q17. Why do we believe chloroplast and mitochondria to be semi-autonomous organelle?
Ans: Mitochondria and Chloroplast are semi-autonomous organelles or endosymbionts of cells because they
i. Possess their own nucleic acid (DNA molecule).
ii. Can form some of the required protein but for most of the proteins these are dependent on nuclear DNA and cytoplasmic ribosome.
iii. Do not arise de novo.
iv. Have membrane similar to those of bacteria.

Q18. Observe the diagram and answer the following.

a. Which group of plants exhibits these two types of cells?
b. What is the first product of C4 cycle?
c. Which enzyme is there in bundle sheath cells and mesophyll cells?
Ans: a. C4 plants
b. OAA (Oxaloacetic acid)
c. Phosphoenol pyruvate (PEP) is present in the mesophyll cells. Enzyme Ribulose bisphosphate carboxylase-oxygenase (RuBisCO) is present in bundle sheath cells.

Q19. A cyclic process is occurring in C3 plant, which is light dependent, and needs O2 This process does not produce energy rather it consumes energy.
a. Can you name the given process?
b. Is it essential for survival?
c. What are the end products of this process?
d. Where does it occur?
Ans: a. Photorespiration
b. No
c. C02 and NH3
d. Photorespiration involves a complex network of enzyme reactions that exchange metabolites between chloroplasts, leaf peroxisomes and mitochondria.

Q20. Suppose Euphorbia and maize are grown in the tropical area.
a. Which one of them do you think will be able to survive under such conditions?
b. Which one of them is more efficient in terms of photosynthetic activity?
c. What difference do you think are there in their leaf anatomy?
Ans: a. Euphorbia is a CAM plant while maize is a C4 plant. Both of them will be able to survive in the tropical areas.
b. Maize (as it is a C4 plant)
c. Leaves of maize plant show Kranz anatomy which is absent in Euphorbia leaves.

Long Answer Type Questions

Q1. Is it correct to say that photosynthesis occurs only in leaves of a plant? Besides leaves, what are the other parts that may be capable of carrying out photosynthesis? Justify.
Ans: Photosynthesis does take place in the green leaves of plants but it does so also in other green parts of the plants. The mesophyll cells in the leaves, have a large number of chloroplasts. Usually the chloroplasts align themselves along the walls of the mesophyll cells, such that they get the optimum quantity of the incident light.
• Photosynthetic or Assimilatory roots: They are green roots which are capable of PHS, e.g., Trapa bispinosa (water chestnut = Singhara), Tmospora (Gillow or Gurcha), Podostemum.
• Some plants of arid regions modify their stems into flattened (Opuntia), or fleshy cylindrical (Euphorbia) structures. These modified stems of indefinite growth are called phylloclades. They contain chlorophyll and carry out photosynthesis.
• One intemode long phylloclade or stem which is leaf like is called cladode. Cladode is capable of photosynthesis. Cladode is found in certain xerophytes, e.g., Ruscus and Asparagus.

Q2.The entire process of photosynthesis consists of a number of reactions. Where in the cell do each of these take place?
a. Synthesis of ATP and NADPH _________
b. Photolysis of water _________
c. Fixation of C02 _________
d. Synthesis of sugar molecule _________
e. Synthesis of starch _________
Ans: a. Synthesis of ATP and NADPH: Membrane system (Grana)
b. Photolysis of water: Inner side of the membrane of thylakoid
c. Fixation of C02: Stroma of chloroplast
d. Synthesis of sugar molecule: Stroma of chloroplast
e. Synthesis of starch: Stroma of chloroplast

Q3. Which property of the pigment is responsible for its ability to initiate the process of photosynthesis? Why is the rate of photosynthesis higher in the red and blue regions of the spectrum of light?
Ans: Pigments are substances that have an ability to absorb light, at specific wavelengths. This property of the pigment is responsible for its ability to initiate the process of photosynthesis.
The wavelengths at which there is maximum absorption by chlorophyll a, i. e. in the blue and the red regions, also shows higher rate of photosynthesis.

Q4. What can we conclude from the statement that the action and absorption spectrum of photosynthesis overlap? At which wavelength do they show peaks?
Ans: The wavelengths at which therfe is maximum absorption by chlorophyll a, i.e. in the blue and the red regions, also shows higher rate of photosynthesis. Hence, we can conclude that chlorophyll a is the chief pigment associated with photosynthesis.
The action spectrum of photosynthesis superimposed on absorption spectrum of chlorophyll a.

Q5. Under what conditions are C4 plants superior to C3?
Ans: C4 plants are special:
i. They have a special type of leaf anatomy. ‘
ii: They tolerate higher temperatures. _
iii. They show a response to high light intensities.
iv. They lack a process called photorespiration.
v. They have greater productivity of biomass.

Q6. In the figure given below, the light line indicates action spectrum for photosynthesis and the black line indicates the absorption spectrum of chlorophyll a, answer the following:

a. What does the action spectrum indicate? How can we plot an action spectrum? Explain with an example.
b. How can we derive an absorption spectrum for any substance?
c. If chlorophyll a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Ans: a. Action spectrum of photosynthesis superimposed on absorption spectrum of chlorophyll a. The wavelengths at which there is maximum . absorption by chlorophyll a, i.e. in the blue and the red regions, also shows higher rate of photosynthesis. Hence, one can conclude that chlorophyll a is the chief pigment associated with photosynthesis.
b. Absorption spectrum for any substance can be derived by plotting the different wavelengths of light.
c. Though chlorophyll a is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to chlorophyll a. Indeed, they not only enable a wider range of wavelength of incoming light to be utilised for photosynthesis but also protect chlorophyll a from photo-oxidation. Hence, the action spectrum and absorption spectrum not overlap.

Q7. What are the important events and end products of the light reaction?
Ans. Light reactions or the ‘Photochemical’ phase include light absorption, water splitting, oxygen release, and the formation of high-energy chemical intermediates, ATP and NADPH. Several complexes are involved in the process. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction. The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments (except one molecule of chlorophyll a) forming a light harvesting system also called antennae. These pigments help to make photosynthesis more efficient by absorbing different wavelengths of light. The single chlorophyll a molecule forms the reaction centre. The reaction centre is different in both the photosystems. In PS I the reaction centre chlorophyll a has an absorption peak at 700 nm, hence is called P700,while in PS II it has absorption maxima at 680 nm, and is called P680.

Q8. In the diagram shown below label A, B, C. What type of phosphorylation is possible in this?


Q9. Why is the RuBisCo enzyme more appropriately called RUBP Carboxylase- Oxygenase and what important role does it play in photosynthesis?
Ans: For ease of understanding, the Calvin cycle can be described under three stages: carboxylation, reduction and regeneration. Carboxylation is the fixation of C02 into a stable organic intermediate. Carboxylation is the most crucial step of the Calvin cycle where C02 is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase which results in the formation of two molecules of 3-PGA. Since this enzyme also has an oxygenation activity it would be more correct to call it RuBP carboxylase-oxygenase or RuBisCO.

Q10. What special anatomical features are displayed by leaves of C4 plants? How do they provide advantage over the structure of C3 plants?
Ans: Study vertical sections of leaves, one of a C3 plant and the other of a C4 plant. The particularly large cells around the vascular bundles of the C4 pathway plants are called bundle sheath cells, and the leaves which have such anatomy are said to have ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces. C4 plants lack photorespiration. In addition these plants show tolerance to higher temperatures. Plants that are adapted to dry tropical regions have the C4 pathway.
C4 plants are special:
i. They have a special type of leaf anatomy.
ii. They tolerate higher temperatures.
iii. They show a response to high light intensities.
iv. They lack a process called photorespiration.
v. They have greater productivity of biomass.

Q11. Name the two important enzynies of C3 and C4 pathway, respectively? What important role do they play in fixing C02?
Ans: The important enzyme of C3 pathway is RuBisCO and that of C4 pathway is PEPcase.
Carboxylation in the C3 pathway is the fixation of C02 into a stable organic intermediate. Carboxylation is the most crucial step of the Calvin cycle, where C02 is utilised for the carboxylation of RuBP. This reaction is catalysed by the enzyme RuBP carboxylase which results in the formation of two molecules of 3-PGA.
The primary C02 acceptor in the C4 pathway is a 3-carbon molecule phosphoenol pyruvate (PEP) and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase or PEPcase.

Q12. Why is RuBisCo enzyme the most abundant enzyme in the world?
Ans: RuBisCo enzyme is the most abundant enzyme in the world because this enzyme is responsible for photosynthesis and present in all green parts of the plants including leaves.

Q13. Why does not photorespiration take place in C4 plants?
Ans: In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C02 at the enzyme site. This takes place when the C4 acid from the mesophyll is broken down in the bundle sheath cells to release C02 – this results in increasing the intracellular concentration of C02. In turn, this ensures that the RuBisCO functions as a carboxylase minimising the oxygenase activity.

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We hope the NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 13 Photosynthesis in Higher Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 8 Cell The Unit of Life are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 8 Cell The Unit of Life.

NCERT Exemplar Class 11 Biology Chapter 8 Cell The Unit of Life

Multiple Choice Questions

Q1. A common characteristic feature of plant sieve tube cells and most of mammalian erythrocytes is
(a) Absence of mitochondria (b) Presence of cell wall
(c) Presence of haemoglobin (d) Absence of nucleus
Ans: (d) A common characteristic feature of plant sieve tube cells and most of mammalian erythrocytes is absence of nucleus.

Q2. Select one which is not true for ribosomes.
(a) Made of two subunits (b) Form polysome
(c) May attach to mRNA (d) Have no role in protein synthesis
Ans: (d) Ribosomes is made of two subunits, form polysome and may attach to mRNA. Ribosomes are the site of protein synthesis.

Q3. Which one of these is not a eukaryote? .
(a) Euglena (b) Anabaena (c) Spirogyra (d) Agaricus
Ans: (b) Anabaena is a cynobacterium (prokaryote).

Q4. Which of the following dyes is not used for staining chromosomes?
(a) Basic Fuchsin (b) Salfanin
(c) Methylene green (d) Carmine
Ans: (b) Saffanin stain is not used for staining chromosomes while Basic Fuchsin, Methylene green and Carmine are used for staining chromosomes.

Q5. Different cells have different sizes. Arrange the following cells in an ascending order of their size. Choose the correct option among the following:
(i) Mycoplasma
(ii) Ostrich eggs
(iii) Human RBCs
(iv) Bacteria
(a) (i), (iv), (iii), (ii) (b) (i), (iii), (iv), (ii)
(c) (ii), (i), (iii), (iv) (d) (iii), (ii), (i), (iv)
Ans: (a) Ascending order of size:
Mycoplasma < Bacteria < Human RBCs < Ostrich eggs.

Q6. Which of the following features is common to prokaryotes and many eukaryotes?
(a) Chromatin material present
(b) Cell wall present
(c) Nuclear membrane present
(d) Membrane-bound subcellular organelles present
Ans: (b) Cell wall is present in all prokaryotes (except mycoplasma) and many eukaryotes (like plants and fungi).

Q7. Who proposed the fluid mosaic model of plasma membrane?
(a) Camillo Golgi (b) Schleiden and Schwann
(c) Singer and Nicolson (d) Robert Brown
Ans: (c) An improved model of the structure of cell membrane was proposed by
S.J. Singer and G.L. Nicolson (1972) widely accepted as fluid mosaic model.

Q8. Which of the following statements is true for a secretory cell?
(a) Golgi apparatus is absent.
(b) Rough Endoplasmic reticulum (RER) is easily observed in the cell.
(c) Only Smooth endoplasmic reticulum (SER) is present.
(d) Secretory granules are formed in nucleus.
Ans: (b) RER is frequently observed in the cells actively involved in protein synthesis and secretion. RER is well developed in cells engaged in synthesis of secretory products.

Q9. What is a tonoplast?
(a) Outer membrane of mitochondria
(b) Inner membrane of chloroplast
(c) Membrane boundary of the vacuole of plant cells
(d) Cell membrane of a plant cell.
Ans: (c) The vacuole is the membrane-bound space found in the cytoplasm. The vacuole is bound by a single membrane called tonoplast.

Q10. Which of the following is not true for an eukaryotic cell?
(a) Cell wall is made up of pepticjoglycans
(b) It has 80S type of ribosome present in the cytoplasm
(c) Mitochondria contain circular DNA
(d) Membrane bound organelles are present
Ans: (a) In bacteria (prokaryotes) cell wall is made up of peptidoglycan.

Q11. Which of the following statements is not true for plasma membrane?
(a) It is present in both plant and animal cell.
(b) Lipid is present as bilayer in it. .
(c) Proteins are present integrated as well as loosely associated with the lipid bilayer.
(d) Carbohydrates are never found in it.
Ans: (d) Chemical studies showed that the cell membrane is composed of lipids that are arranged in a bilayer. Later, biochemical investigation clearly revealed that the cell membranes also possess protein and carbohydrate.

Q12. Plastids differ from mitochondria on the basis of following features? Mark the right answer.
(a) Presence of two layers of membrane
(b) Presence of ribosome
(c) Presence of thylakoids
(d) Presence of DNA
Ans: (c) Thylakoids are present in plastids but not in mitochondria. Both plastids and mitochondria are similar in presence of two layers of membrane, presence of ribosome and presence of DNA.

Q13. Which of the following is not a function of cytoskeleton in a cell?
(a) Intracellular transport
(b) Maintenance of cell shape and structure
(c) Support of the organelle
(d) Cell motility
Ans: (a) The cytoskeleton in a cell are involved in many functions such as mechanical support, motility, maintenance of the shape of the cell.

Q14. The stain used to visualise mitochondria is
(a) Fast green (b) Saffanin (c) Aceto carmine(d) Janus green
Ans: (d) Janus green stain is used to visualise mitochondria.

Very Short Answer Type Questions
Q1. What is the significance of vacuole in a plant cell?
Ans: Vacuole in plant cells help in the storage, waste disposal and cell elongation and protection.

Q2. What does ‘S’ refer in a 70S and an 80S ribosome?
Ans: Svedberg’s Unit or sedimentation coefficient.

Q3. Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Ans: Lysosome

Q4. What are gas vacuoles? State their functions. ,
Ans: Gas vacuoles are aggregates of hollow cylindrical structures called gas vesicles. They are located inside some bacteria. The inflation and deflation of the vesicles provides buoyancy, allowing the bacterium to float at a desired depth in the water.

Q5. What is the function of a polysome? .
Ans: Several ribosomes may attach to a single mRNA and form a chain called polyribosome or polysome. The ribosomes of a polysome translate the mRNA into proteins.

Q6. What is the feature of a metacentric chromosome?
Ans: The metacentric chromosome has middle (medial) centromere forming two equal arms of the chromosome. Shape of metacentric chromosome is V-shaped.

Q7. What is referred to as satellite chromosome?
Ans: Sometimes a few chromosomes have non-staining secondary constrictions at a constant location. This gives the appearance of a small fragment called the satellite or trabant. These chromosomes are called sat (satellite) chromosome. Nucleolus is formed by sat chromosome.

Short Answer Type Questions
Q1. Discuss briefly the role of nucleolus in the cells actively involved in protein synthesis.
Ans: Nucleolus is a site for active ribosomal RNA synthesis. Larger and more numerous nucleoli are present in cells actively carrying out protein synthesis.

Q2. Explain the association of carbohydrate to the plasma membrane and its significance.
Ans: Carbohydrates forms glycoproteins and glycolipids by glycosylation. Glycoproteins and glycolipids are biochemicals that involved in cell recognition and adhesion.

Q3. Comment on the cartwheel structure of centriole.
Ans: Centrosome is an organelle usually containing two cylindrical structures called centrioles. Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like the cartwheel. They are made up of nine evenly spaced peripheral fibrils of tubulin protein. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The central part of the proximal region of the centriole is also proteinaceous and called the hub.

Q4. Briefly describe the cell theory.
Ans: Schleiden and Schwann together formulated the cell theory (1838-39). This theory, however, did not explain as to how new cells were formed. Rudolf Virchow (1855) first explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e cellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape. Cell theory as understood today is
(i) All living organisms are composed of cells and products of cells.
(ii) All cells arise from pre-existing cells.

Q5. Differentiate between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER).
Ans: The ER often shows ribosomes attached to their outer surface. The endoplasmic reticulum bearing ribosomes on their surface is called rough endoplasmic reticulum (RER). In the absence of ribosomes they appear smooth and are called smooth endoplasmic reticulum (SER). RER is frequently observed in the cells actively involved in protein synthesis and secretion. They are extensive and continuous with the outer membrane of the nucleus. The smooth endoplasmic reticulum is the major site for synthesis of lipid. In animal cells lipid-like steroidal hormones are synthesised in SER.

Q6. Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Ans: The detailed structure of the membrane was studied only after the advent of the electron microscope in the 1950s. Meanwhile, chemical studies on the cell membrane, especially in human red blood cells (RBCs), enabled the scientists to deduce the possible structure of plasma membrane. These studies showed that the cell membrane is composed of lipids that are arranged in a bilayer. Also, the lipids are arranged within the membrane with the polar head towards the outer sides and the hydrophobic tails towards the inner part. This ensures that the nonpolar tail of saturated hydrocarbons is protected from the aqueous environment. The lipid component of the membrane mainly consists of phosphoglycerides. Later, biochemical investigation clearly revealed that the cell membranes also possess protein and carbohydrate. The ratio of protein and lipid varies considerably in different cell types. In human beings, the membrane of the erythrocyte has approximately 52 per cent protein and 40 per cent lipids.

Q7. What are plasmids? Describe their role in bacteria?
Ans: In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA are called plasmids. The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. This plasmid DNA is used to monitor bacterial transformation with foreign DNA.

Q8. What are histones? What are their functions?
Ans: In eukaryotes there is a set of positively charged, basic proteins called histones. Histones are rich in the basic amino acid residues lysines and arginines. Both the amino acid residues carry positive charges in their side chains. Histones are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix.

Long Answer Type Questions
Q1. What structural and functional attributes must a cell have to be called a living cell?
Ans: All cells have an outer membrane called the cell membrane. Inside each cell is a dense membrane bound structure called nucleus. This nucleus contains the chromosomes which in turn contain the genetic material, DNA. Cells that have membrane bound nuclei are called eukaryotic whereas cells that lack a membrane bound nucleus are prokaryotic. In both prokaryotic and eukaryotic cells, a semi-fluid matrix called cytoplasm occupies the volume of the cell. The cytoplasm is the main arena of cellular activities in both the plant and animal cells. Various chemical reactions occur in it to keep the cell in the ‘living state’.
Besides the nucleus, the eukaryotic cells have other membrane bound distinct structures called organelles like the endoplasmic reticulum (ER), the golgi complex, lysosomes, mitochondria, raicrobodies and vacuoles. The prokaryotic cells lack such membrane bound organelles. Ribosomes are non-membrane bound organelles found in all cells—both eukaryotic as well as prokaryotic.

Q2. Briefly give the contributions of the following scientists in formulating the cell theory
a. Rudolf Virchow
b. Schielden and Schwann
Ans: In 1838, Matthias Schleiden, a German botanist, examined a large number of plants and observed that all plants are composed of different kinds of cells which form the tissues of the plant. At about the same time, Theodore Schwann (1839), a British Zoologist, studied different types of animal cells and reported that cells had a thin outer layer which is today known as the ‘plasma membrane’. He also concluded, based on his studies on plant tissues, that the presence of cell wall is a unique character of the plant cells. On the basis of this, Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells. Schleiden and Schwann together formulated the cell theory. This theory however, did not explain as to how new cells were formed. Rudolf Virchow (1855) first explained that cells divided and new cells are formed from pre-existing cells (Omnis cellula-e c’ellula). He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape. Cell theory as understood today is :
(i) all living organisms are composed of cells and products of cells.
(ii) all cells arise from pre-existing cells.

Q3. Is extra genomic DN A present in prokaryotes and eukaryotes? If yes, indicate their location in both the types of organisms.
Ans: Yes, extra genomic DNA present in prokaryotes and eukaryotes. In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria (prokaryotes) have small circular DNA outside the genomic DNA. These smaller extra genomic DNA are called plasmids. The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. This plasmid DNA is used to monitor bacterial transformation with foreign DNA. In eukaryotes, extra genomic DNA is present in two organelles- mitochondria and plastids. ’

Q4. Structure and function are correlatable in living organisms. Can you justify this by taking plasma membrane as an example?
Ans: The shape of the cell may vary with the function they perform. For example, RBCs are round and biconcave to pass through capillaries and carry more Oz. WBCs are amoeboid to do phagocytosis and diapedesis.
The quasi-fluid nature of lipid enables lateral movement of proteins within the overall bilayer. This ability to move within the membrane is measured as its fluidity. The fluid nature of the membrane is also important from the point of view of functions like cell growth, formation of intercellular junctions, secretion, endocytosis, cell division etc.

Q5. Eukaryotic cells have organelles which may
a. not be bound by a membrane
b. bound by a single membrane
c. bound by a double membrane
Group the various sub-cellular organelles into these three categories.
Ans: a. Non-membrane bound cell organelles—Ribosome, Centrosome (Centriole), Nucleolus, Cytoskeletal structures.
b. Single membrane bound cell organelles—ER, GB, Lysosome, Vacuoles, Microbodies (Glyoxysomes and Peroxisomes), Thylakoid.
c. Double membrane bound cell organelles—Plastid, Mitochondria and Nucleus.

Q6. The genomic content of the nucleus is constant for a given species whereas the extra chromosomal DNA is found to be variable among the members of a population. Explain.
Ans: The genomic content of the nucleus is constant for a given species whereas the extra chromosomal DNA is found to be variable among the members of a population. For humans (Homo sapiens) the genomic content of the nucleus is constant, i.e. 46 chromosomes. But extra chromosomal DNA is found to be variable among the members of the population like different humans have different amount of extra chromosomal DNA in their mitochondria.

Q7. Justify the statement, “Mitochondria are power houses of the cell”.
Ans: Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e. the outer compartment and the inner compartment. The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organelle. The inner membrane forms a number of infoldings called the cristae (sing.: crista) towards the matrix. The cristae increase the surface area. The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration. They produce cellular energy in the form of ATP, hence they are called ‘power houses’ of the cell.

Q8. Is there a species specific or region specific type of plastids? How does one distinguish one from the other?
Ans: Yes, plastids are species specific or region specific. Plastids are found in all plant cells and in euglenoids. These are easily observed under the microscope as they are large. They bear some specific pigments, thus imparting specific colours to the plants. Based on the type of pigments plastids can be classified into chloroplasts, chromoplasts and leucoplasts. The chloroplasts contain chlorophyll and carotenoid pigments which are responsible for trapping light energy essential for photosynthesis. In the chromoplasts fat soluble carotenoid pigments like carotene, xanthophylls and others are present. This
gives the part of the plant a yellow, orange or red colour. The leucoplasts are the colourless plastids of varied shapes and sizes with stored nutrients: Amyloplasts store carbohydrates (starch), e.g., potato; elaioplasts store oils and fats whereas the aleuroplasts store proteins.

Q9. Write the functions of the following:
a. Centromere
b. Cell wall
c. Smooth ER
d. Golgi Apparatus
e. Centrioles
Ans: a. Centromere: Every chromosome essentially has a primary constriction or the centromere. Two sister chromatids are joined together at the centromere.
b. Cell wall: Cell wall not only gives shape to the cell and protects the cell from mechanical damage and infection, it also helps in cell-to-cell interaction and provides barrier to undesirable macromolecules.
c. Smooth ER: The smooth endoplasmic reticulum is the major site for synthesis of lipid. In animal cells lipid-like steroidal hormones are synthesised in SER.
d. Golgi Apparatus: Golgi apparatus is the important site of formation of glycoproteins and glycolipids.
e. Centrioles: The centrioles form the basal body of cilia or flagella, and spindle fibres that give rise to spindle apparatus during cell division in animal cells.

Q10. Are the different types of plastids interchangeable? If yes, give examples where they are getting converted from one type to another.
Ans: Yes, different types of plastids are interchangeable.
Conversion of green tomatoes (or chilli) into red form is due to formation of chromoplasts from chloroplasts. Chromoplasts also formed from leucoplasts by development of some pigments (like carotenes in carrot).

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We hope the NCERT Exemplar Class 11 Biology Chapter 8 Cell The Unit of Life help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 8 Cell The Unit of Life, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 5 Morphology of Flowering Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 5 Morphology of Flowering Plants.

NCERT Exemplar Class 11 Biology Chapter 5 Morphology of Flowering Plants

Multiple Choice Questions

Q1. Rearrange the following zones choose the correct option as seen in the root in vertical section and choose the correct option
(A) Root hair zone
(B) Zone of meristems
(C) Root cap zone
(D) Zone of Maturation
(E) Zone of elongation

(a) C, B, E, A, D
(b) A,B,C,D,E
(c) D , E , A , C , B
(d) E , D ,C ,B ,A

Ans: (a) (C) Root cap zone, (B) Zone of meristems, (E) Zone of elongation, (A) Root hair zone, (D) Zone of maturation

Q2. In an inflorescence where flowers are borne laterally in an acropetal succession, the position of the youngest floral bud shall be
(a) Proximal (b) Distal (d) Intercalary (d) Anywhere
Ans: (b) In racemose type of inflorescences the main axis continues to grow, the flowers are borne laterally in an acropetal succession, i.e. youngest flower is present at apex and oldest flower is present at the base. In racemose, inflorescence the growth of floral axis is unlimited or indefinite.
In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipetal. order, i.e. youngest flower is present at the base and oldest flower is present at the apex. In cymose inflorescence oldest flower remains in center and youngest towards the periphery. This type of arrangement is called centrifugal.

Q3. The mature seeds of plants such as gram and peas, possess no endosperm, because
(a) These plants are not angiosperms
(b) There is no double fertilization in them
(c) Endosperm is not formed in them
(d) Endosperm gets used up by the developing embryo during seed development.
Ans: (d) The mature seeds of plants such as gram and peas, possess no endosperm, because endosperm gets used up by the developing embryo during seed development.

Q4. Roots developed from parts of the plant other than radicle are called
(a) Tap roots (b) Fibrous roots
(c) Adventitious roots (d) Nodular roots
Ans: (c) Roots developed from parts of the plant other than radicle are called adventitious roots.

Q5. Venation is a term used to describe the pattern of arrangement of
(a) Floral organs (b) Flowers in inflorescence
(c) Veins and veinlets in a lamina (d) All of them
Ans: (c) Venation is a term used to describe the pattern of arrangement of veins and veinlets in a lamina.
Q6. Endosperm, a product of double fertilization in angiosperms is absent in the seeds of
(a) Coconut (b) Orchids (c) Maize (d) Castor
Ans: (b) Endosperm, a product of double fertilization in angiosperms is absent in the seeds of orchids. .

Q7. Many pulses of daily use belong to one of the families below (tick the correct
answer). –
(a) Solanaceae (b) Fabaceae (c) Liliaceae (d) Poaceae
Ans: (b) Many pulses of daily use belong to one of the family fabaceae. Solanaceae (potato family)
Liliaceae (lily family)
Poaceae (cereal or grass family). „

Q8. The placenta is attached to the developing seed near the
(a) Testa (b) Hilum (c) Micropyle (d) Chalaza
Ans: (b) The placenta is attached to the developing seed near the hilum.

Q9. Which of the following plants is used to extract the blue dye?
(a) Trifolium (b) Indigofera (c) Lupin (d) Cassia
Ans: (b) Blue dye is obtained from Indigofera tinctoria which belongs to family fabaceae.

Q10. Match the followings and choose the correct option.

Group A		Group B
A.	Aleurone layer	(i)	Without fertilization
B.	Parthenocarpic fruit	(ii)	Nutrition
C.	Ovule	(iii)	Double fertilization
D.	Endosperm	(iv)	Seed

Options:

• A—(i), B—(ii), C—(iii), D—(iv)
• A—(ii), B—(i), C—(iv), D—(iii)
• A—(iv), B—(ii), C—(i), D—(iii)
• A—(ii), B—(iv), C—(i), D—(iii)

Ans. (b)

A.	Aleurone layer	(ii)	Nutrition
B.	Parthenocarpic fruit	(i)	Without fertilization
C.	Ovule	(iv)	Seed
D.	Endosperm	(iii)	Double fertilization

Very Short Answer Type Questions

Q1. Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of 02, root growth is restricted or completely stopped. How do the plants growing in marshlands or swamps obtain their 02 required for root respiration?
Ans: In some plants such as Rhizophora and Sonneratia (mangrove plant) growing in swampy areas near river mouths (saline marshy soil or halophytes), many roots come out of the ground and grow vertically upwards (negatively geotropic: against gravitational force). Such roots, called pneumatophores or breathing roots or respiratory roots, help to get oxygen for respiration.

Q2. Write floral formula for a flower which, is bisexual; actinoiflorphic; sepals five, twisted aestivation, petals five; valvate aestivation; stamens six; ovary tricarpellary, syncarpous, superior, trilocular with axile placentation.

Q3. In Opuntia the stem is modified into a flattened green structure to perform the function of leaves (i.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosynthesis.
Ans: Some plants of arid regions modify their stems into flattened {Opuntia), or fleshy cylindrical {Euphorbia) structures. These modified stems of indefinite growth are called phylloclades. They contain chlorophyll and carry out photosynthesis.

Q4. In swampy areas like the Sunderbans in West Bengal, plants bear special kind
of roots called __________ .
Ans: Pneumatophores

Q5. In aquatic plants like Pistia and Eichhomia, leaves and roots are found near
Ans: Node

Q6. Reticulate and parallel venation are characteristic of ________ and ______ respectively.
Ans: Dicotyledons and monocotyledons

Q7. Which parts in ginger and onion are edible?
Ans: Ginger: rhizome and onion: fleshy leaves

Q8. In epigynous flower, ovary is situated below the _________.
Ans: Calyx, corolla and androecium.

Q9. Add the missing floral organs of the given floral formula of Fabaceae.

Short Answer Type Questions

Q1. Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle.
Ans: The root that arise from parts of plant other than radicle are called adventitious roots.
Pneumatophores—for respiration Stilt roots—for support Prop roots—for support

Q2. The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants? How are roots of aquatic plants and terrestrial plants different?
Ans: The roots of aquatic plants help in balancing and anchorage. In terrestrial plants, functions of roots are anchorage and absorption of water and minerals.

Q3. Draw diagrams of a typical monocot and dicot leaves.to show their venation pattern.

Q4. A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.
Ans: A typical flower has four different kinds of whorls arranged successively on the swollen end of the stalk or pedicel, called thalamus or receptacle. These are calyx, corolla, androecium and gynoecium. The calyx is the outermost whorl of the flower and the members are called sepals. Corolla is the second outermost whorl composed of petals. Androecium is the second innermost whorl composed of stamens. Gynoecium is the innermost whorl which is female reproductive part of the flower and is made up of one or more carpels.

Q5. Given below are a few floral formulae of some well known plants. Draw floral diagrams from these formulae.

Q6. Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalization.
Ans: Smilax and Dioscorea are monocots having reticulate venation. Calophyllum and Eryngium are dicots having parallel venation.

Q7. You have heard about several insectivorous plants that feed on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in marsh lands. What part of the plant is modified into a ‘pitcher’? How does this modification help the plant for food even though it can photosynthesize like any other green plant?
Ans: In Nepenthes, picher is modified lamina. Insectivorous plants capable of growing in nitrogen or nitrate deficient soil. Pitcher help in providing the nitrogen to the plant. Leaves also photosynthesize their food.

Q8. Mango and coconut are ‘drupe’ type of fruits. In mango fleshy mesocarp is edible. What is the edible part of coconut? What does milk of tender coconut represent?
Ans: Edible part of coconut is endosperm or seed. Milk of tender coconut represent free nuclear liquid endosperm.

Q9. How can you differentiate between free central and axile placentation?
Ans: When the ovules are borne on central axis and septa are absent, as in Dianthus
and Primrose the placentation is called free central.
When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, as in China rose, tomato and lemon.

Q10. Tendrils are found in the following plants. Identify whether they are stem tendrils or leaf tendrils,
a.Cucumber
b. Peas
c. Pumpkins 
. Grapevine
e. Watermelons
Ans. a. Cucumber—stem tendrils
b. Peas Leaf Tendrils
c. Pumpkins—stem tendrils
d. Grapevine – stem tendrils
e. Watermelons—stem tendrils

Q11. Why is maize grain usually called as a fruit and not a seed?
Ans: Maize is an example of caryopsis fruit which is a simple, one seeded dry, indehiscent fruit in which pericarp and testa are fused. Pericarp is also called fruit wall which is present in maize grain, hence maize grain is actually a fruit and not a seed.

Q12. Tendrils of grapevines are homologous to the tendril of pumpkins but are analogous to that of pea. Justify the above statement.
Ans: Tendrils of grapevines are homologous to the tendril of pumpkins because both are the modification of stem (i.e., stem tendrils). Homologous structure have same origin.
Analogous structure have different origin but similar function. Tendrils of grapevines are analogous to that of pea because tendrils of grapevines are modification of stem while tendrils of pea are modification of leaves. Both have different origin but performing same function, i.e., support.

Q13. Rhizome of ginger is like the roots of other plants that grows underground. Despite this fact ginger is a stem and not a root. Justify. .
Ans:Rhizome is a underground stem growing parallel to soil surface. Ginger is a stem which can be differentiated from root as it has nodes and intemodes.

Q14. Differerifiate between.
a. Bract and Bracteole
b. Pulvinus and petiole
c. Pedicel and peduncle
d. Spike and spadix
e. Stamen and staminoide
f. Pollen and pollenium
Ans: a. Reduced leaf found at the base of the pedicel is called bract while leaf like structure present between bract and flower is called bracteole.
b. In some leguminous plants the leafbase may become swollen is called pulvinus. The petiole (mesopodium) connect the leafbase with lamina and help hold the blade to light.
c. Pedicie is the stalk of flower while peduncle is the stalk of inflorescence.
d. In spike sessile flowers are attached on elongated peduncle. E.g.: Achyranthes. Spadix is a special type of spike having a fleshy peduncle and a large brightly coloured bract called spathe. E.g.: Palm, Colocasia.
e. Androecium is composed of stamens. Each stamen represents the male reproductive organ. A sterile stamen is called staminode.
f. Male gametophyte of angiosperms is called pollen or microspore. A group of pollens forms the pollinium (present in Calotropis).

Long Answer Type Questions
1. Distinguish between families Fabaceae, Solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures), Also write economic importance of any one of the above family.
Ans: a. Gynoecium
i. Fabaceae— Monocarpellary, ovary unilocular, marginal placentation

ii. Solgnaceae—Bicarpellary,* syncarpous, carpels placed obliquely, bilocular, axile placentation

iii. Liliaceae—Tricarpellary, syncarpous, ovary superior, axile placentation

b. Economic importance of fabacae:
i. Source of pulses (gram, arhar)
ii. Edible oil (soyabean, groundnut)
iii. Dye (Indigofera)
iv. Fibres (sunhemp)
v. Fodder (Sesbania, Trifolium)
vi. Ornamental {Lupin)
vii. Medicine (mulaithi)

Q2. Describe various stem modifications associated with food storage, climbing and protection.
Ans: The stem may not always be typically like what they are expected to be. They are modified to perform different functions. Underground stems of potato, ginger, turmeric, zaminkand, Colocasia are modified to store food in them. They also act as organs of perennation to tide over conditions unfavourable for growth. Stem tendrils which develop from axillary buds, are slender and spirally coiled and help plants to climb such as in gourds (cucumber, pumpkins, watermelon) and grapevines. Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thoms are found in many plants such as Citrus, Bougainvillea. They protect plants from browsing animals. Some plants of arid regions modify their stems into flattened {Opuntia), or fleshy cylindrical {Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.

Q3. Stolon, offset and rhizome are different forms of stem modifications. How can these modified forms of stem be distinguished from each other?
Ans: Underground stems called stolon of some plants spread to new niches and when older parts die new plants are formed.
OfFest: A lateral branch with short intemodes and each node bearing a rosette of leaves and a tuft of roots is found in aquatic plants like Pistia and Eichhornia.
Rhizome is underground stem growing parallel to soil surface. In ginger, turmeric, Gloriosa and lotus stem is called rhizome which is differentiated from roots in having scale leaves and axillary buds and helps in vegetative propagation.

Q4. The mode of arrangements of sepals or petals in a floral bud is known as aestivation. Draw the various types of aestivation possible for a typical pentamerous flower.
Ans: The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation. The main types of aestivation are valvate, twisted, imbricate and vexillary. When sepals or petals in a whorl just touch one another at the margin, without overlapping, as in Calotropis, it is said to be valvate. If one margin of the appendage overlaps that of the next one and so on as in china rose, lady’s finger and cotton, it is called twisted. If the margins of sepals or petals overlap one another but not in any particular direction as in Cassia and gulmohur, the aestivation is called imbricate. –
In pea, bean and Crotolaria flowers, there are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel); this type of aestivation is known as vexillary or papilionaceous (or descending imbricate).

Q5. The arrangements of ovules within the ovary is known as placentation. What does the term placenta refer to? Name and draw various types of placentations in the flower as seen in T.S. or V.S.
Ans: The ovules are female reproductive structures and borne in the ovary of the flower. The number, structure, their position in the ovary varies in different plants. They also differ in mode of attachment with the ovary wall. At the point of attachment there is a cellular ridge or cushion of cells called placenta. The mode of attachment of ovule to the placenta is known as placentation which is of the following types: (a) Parietal (b) Marginal (c) Axile (d) Free central (e) Basal.

 

Q6. Sunflower is not a flower. Explain.
Ans: Sunflower is actually an inflorescence (capitulum), it is not a flower.
Capitulum (Head): Here the peduncle is flattened which is called receptacle. Inflorescence is surrounded by whorl of bracts collectively called involucre. On receptacle two kinds of florets are present: .
1. Ray florets: Zygomorphic, unisexual female, corolla ligulate or strap shaped.
ii. Disc florets: Actinomorphc, bisexual, corolla tubular, present in centre. Ray florets present towards periphery and disc florets present in the centre, e.g.: Compositae (Asteraceae) family—Sunflower (Helianthus), Tagetus (Marigold) and Dahlia.

Q7. How do you distinguish between hypogeal germination and epigeal germination? What is the role of cotyledon (s) and the endosperm in the germination of seeds?
Ans: 1. Hypogeal Germination: When the epicotyl grows first, only the plumule is pushed out of the soil. While cotyledons and all other parts remain under the soil, the germination is called hypogeal. E.g.: Most of the monocots (Maize^’ Rice and Coconut) and some of the dicots (Pea, Gram, Broad bean = Vicia faba, Mango).
2. Epigeal Germination: When hypocotyl grows first, it pushes the cotyledons and other parts of the seed out of the soil. This germination is called epigeal. E.g.: Helianthus (sunflower), mustard, cucurbits, castor, Onion, Tamarindus, French bean, Alisma.

Q8. Seeds of some plants germinate immediately after shedding from the plants while in other plants they require a period of rest before germination. The later phenomena is called as dormancy. Give the reasons for seed dormancy and some methods to break it.
Ans: There are certain seeds which fall to germinate even when external conditions are favourable. Such seeds are understood to be undergoing a period of dormancy which is controlled not by external environment but are under endogenous control or conditions within the seeds itself.
Reasons for seed dormancy:
• Impermeable and hard seed coat; presence of chemical inhibitors such as abscisic acid (ABA), phenolic acid; para-ascrobic acid; and immature embryos are some of the reasons which causes seed dormancy.
Methods of breaking seed dormancy:
• Seed dormancy allows the plants to overcome unfavourable climatic conditions. This dormancy however can be overcome through natural means and various other man-made measures. For example, the seed coat barrier in some seeds cane be broken by mechanical abrasions are caused by microbial action, and passage through digestive tract of animals. Weakening of hard seed coat with sharp edge or knives is called scarification. Prechilling treatment to break seed dormancy is called stratification. Effect of inhibitory substances can be removed by subjecting the seeds to chilling conditions or by application of certain chemical like gibberellic acid and nitrates. Seed domancy is also breaked by auxin and cytokinin. Changing the environment conditions, such as light and temperature and other methods to overcome seed dormancy.

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We hope the NCERT Exemplar Class 11 Biology Chapter 5 Morphology of Flowering Plants help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 5 Morphology of Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division.

NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division

Multiple Choice Questions

Q1. Meiosis in diploid organisms results in
(a) Production of gametes
(b) Reduction in the number of chromosomes
(c) Introduction of variation
(d) All of the above
Ans: (d)‘ Meiosis in diploid organisms results in production of gametes, reduction in the number of chromosomes and introduction of variation.

Q2. At which stage of meiosis does the genetic constitution of gametes is finally
decided? –
(a) Metaphase-I (b) Anaphase-II (c) Metaphase-II (d) Anaphase-I
Ans: (d) At anaphase-I, stage of meiosis the genetic constitution of gametes is finally decided.

Q3. Meiosis occurs in organisms during
(a) Sexual reproduction
(b) Vegetative reproduction
(c) Both sexual and vegetative reproduction
(d) None of these
Ans: (a) Meiosis occurs in organisms during sexual reproduction. The production of offspring by sexual reproduction includes the fusion of two gametes, each with a complete haploid set of chromosomes. Gametes are produced through meiosis.

Q4. During anaphase-I of meiosis
(a) Homologous chromosomes separate
(b) Non-homologous chromosomes separate
(c) Sister chromatids chromosomes separate
(d) Non Sister chromatids chromosomes separate
Ans: (a) The homologous chromosomes separate, while sister chromatids remain associated at their centromeres. Separation of homologous chromosomes at anaphase is called disjunction.

Q5. Mitosis is characterised by
(a) Reduction division
(b) Equal division
(c) Both reduction and equal division
(d) Pairing of homologous chromosomes
Ans: (b) Mitosis is the most dramatic period of the cell cycle, involving a major reorganisation of virtually all components of the cell. Since the number of chromosomes in the parent and progeny cells is the same, it is also called as equational division.

Q6. A bivalent of meiosis-I consists of
(a) Two chromatids and one centromere
(b) Two chromatids and two centromeres
(c) Four chromatids and two centromeres
(d) Four chromatids and four centromeres.
Ans: (c) A bivalent of meiosis-I consists of four chromatids and two centromeres.

Q7. Cells which are not dividing are likely to be at
(a) G, (b) G2 (C) G0 (d) S phase
Ans: (c) These cells that do not divide further exit G{ phase to enter an inactive stage is called quiescent stage (G0) of the cell cycle. G0 stage of cell denotes exit “of cell from cell cycle. During G0 stage of cell cycle, cell decides to undergo differentiation. Cells in G0 stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Q8. Which of the events listed below is not observed during mitosis?
(a) Chromatin condensation
(b) Movement of centrioles to opposite poles
(c) Appearance of chromosomes with two chromatids joined together at the centromere
(d) Crossing over
Ans: (d) Crossing over occurs in pachytene (it is a phase of meiosis-I). Crossing over is the exchange of genetic material (genes) between two homologous chromosomes. Crossing over is also an enzyme-mediated process and the enzyme involved is called recombinase. Crossing over leads to recombination of genetic material on the two chromosomes. Exchange of paternal and maternal chromosome material during pachytene is called crossing over.

Q9. Identify the wrong statement about meiosis.
(a) Pairing of homologous chromosomes
(b) Four haploid cells are formed
(c) At the end of meiosis number of chromosomes are reduced to half
(d) Two cycles of DNA replication occur.
Ans: (d) Meiosis involves two sequential cycles of nuclear and cell division called meiosis-I and meiosis-II but only a single cycle of DNA replication.

Q10. Select the correct statement about G1 phase.
(a) Cell is metabolically inactive
(b) DNA in the cell does not replicate
(c) It is not a phase of synthesis of macromolecules
(d) Cell stops growing.
Ans: (b) During Gj phase the ceil is metabolically active and continuously grows but does not replicate its DNA but proteins and RNA are synthesized.

Very Short Answer Type Questions
Q1. Between a prokaryote and an eukaryote, which cell has a shorter cell division time?
Ans: Prokaryotic cells has shorter cell division time than eukaryotic cells. A typical eukaryotic cell cycle is illustrated by human cells in culture. These cells divide once in approximately every 24 hours. In bacteria (E.coli) cell cycle is of 20 minutes.

Q2. Which of the phases of cell cycle is of longest duration?
Ans: Interphase

Q3. Name a stain commonly used to colour chromosomes.
Ans: Basic fuchsin, acetocarmine etc.

Q4. Which tissue of animals and plants exhibits meiosis?
Ans: Gohads (testes and ovary) in animals and sporangium in plants.

Q5. Given that the average duplication time of E.coli is 20 minutes, how much time will two E.coli cells take to become 32 cells?

For formation of 32 cells, two E.coli cells takes 4 cycles. So total time will be 4 x 20 = 80 minutes

Q6. Which part of the human body should one use to demonstrate stages in mitosis?
Ans: Nail base or any somatic cell (diploid cell).

Q7. What attributes does a chromatid require to be classified as a chromosome?
Ans: Centromere

Q8. The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over?

Ans: Sister chromatids of homologous chromosome.

Q9. If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone?

Q10. An anther has 1200 pollen grains. How many pollen mother cells (pmc) must have been there to produce them?
Ans: 4 pollen grains are produced by 1 pmc
1200 pollen grains are produced by = 1200/4
= 300 pmc

Q11. At what stage of cell cycle does DNA synthesis take place?
Ans: S-phase (interphase)

Q12. It is said that the one cycle of cell division in human cells (eukaryotic cells) takes 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Ans: It is significant to note that in the 24 hour average duration of cell cycle of a human cell, cell division proper lasts for only about an hour. The interphase lasts more than 95% of the duration of cell cycle.

Q13. It is observed that heart cells do not exhibit cell division. Such cells do not
divide further and exit phase to enter an inactive stage called of cell cycle. Fill in the blanks.
Ans: It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit Gt phase to enter an inactive stage called G0 of cell cycle.

Q14. In which phase of meiosis are the following formed? Choose the answers from hint points given below.
a. Synaptonemal complex
b. Recombination nodules
c. Appearance/activation of enzyme recombinase
d. Termination of chiasmata
e. Interkinesis
f. Formation of dyad of cells
[Hints: (1) Zygotene, (2) Pachytene, (3) Pachytene, (4) Diakinesis, (5) After Telophase-I /before Meosis-II, (6) Telophase-I /after Meiosis-I]
Ans: a. Synaptonemal complex: zygotene
b. Recombination nodules: pachytene
c. Appearance/activation of enzyme recombinase: pachytene
d. Termination of chiasmata: diakinesis
e. Interkinesis: after Telophase-I /before Meosis-II
f. Formation of dyad of cells: Telophase-I /after Meiosis-I.

Short Answer Type Questions
Q1. State the role of centrioles other than spindle formation.
Ans: The centrioles form the basal body of cilia or flagella.

Q2. Mitochondria and plastids have their own DNA (genetic material). What is known about their fate during nuclear division like mitosis?
Ans: At the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.

Q3. Label the diagram and also determine the stage at which this structure is visible.

Ans: This is a transition to metaphase

Q4. A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number (TV) during metaphase? What would be the DNA content (Q during anaphase?
Ans: Chromosome number (N) during metaphase = 32 (N)
DNA content (C) during anaphase = 2C

Q5. While examining the mitotic stage in a tissue, one finds some cells with 16
chromosomes and some with 32 chromosomes. What possible reasons could you assign to this difference in chromosome number? Do you think cells with 16 chromosomes could have arisen from cells with 32 chromosomes or vice versa?
Ans: Cells with 16 chromosomes are produced by meiosis while that with 32 chromosomes are produced by mitosis.
• Cells with 16 chromosomes could have arisen from cells with 32 chromosomes through meiosis.
• Cells with 32 chromosomes could have arisen from cells with 16 chromosomes through fertilisation or syngamy.

Q6. The following events occur during the various phases of the cell cycle. Name the phase against each of the events.
a. Disintegration of nuclear membrane ________
b. Appearance of nucleolus ________
c. Division of centromere ________
d. Replication of DNA ________
Ans: a. Disintegration of nuclear membrane: Late prophase
b. Appearance of nucleolus: Telophase
c. Division of centromere: Anaphase
d. Replication of DNA: S-phase

Q7. Mitosis results in producing two cells which are similar to each other. What would be the consequence if each of the following irregularities occur during mitosis?
a. Nuclear membrane fails to disintegrate
b. Duplication of DNA does not occur
c. Centromeres do not divid
d. Cytokinesis does not occu;
Ans: a. Nuclear membrane fails to disintegrate: In this condition, mitosis
takes place within nucleus. This is called endoduplication.
b. Duplication of DNA does not occur: There will be no mitosis
c. Centromeres do not divide: Polyploidy appears
d. Cytokinesis does not occur: In some organisms karyokinesis is not followed by cytokinesis as a result of which multinucleate condition arises leading to the formation of syncytium (e.g., liquid endosperm in coconut)

Q8. Both unicellular and multicellular organisms undergo mitosis. What are the differences, if any, observed in the process between the two?
Ans:
• The growth of multicellular organisms is due to mitosis.
• The reproduction of unicellular organisms is due to mitosis.

Q9. Name the pathological condition when uncontrolled cell division occurs.
Ans: Cancer

Q10. Two key events take place, during S phase in animal cells, DNA replication and duplication of centriole. In which parts of the cell do events occur?
Ans: In animal cells, during the S phase, DNA replication begins in the nucleus, and the centriole duplicates in the cytoplasm.

Q11. Comment on the statement—Meiosis enables the conservation of specific chromosome number of each species even though the process per se, results in a reduction of chromosome number.
Ans: Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms, even though the process, per se, paradoxically, results in reduction of chromosome number by half. But fertilisation restores the chromosome number.

Q12. Name a cell that is found arrested in diplotene stage for months and years. Comment in 2-3 lines how it completes cell cycle?
Ans: In oocytes of some vertebrates, diplotene can last for months or years.
• Lampbrush chromosomes or diplotene chromosome are found in diplotene
stage of most animal oocytes of frog or amphibians.
• Lampbrush chromosomes are observed in meiotic prophase. These chromosomes become normal after growth and thus completing the cell cycle.

Q13. How does cytokinesis in plant cells differ from that in animal cells?
Ans: In an animal cell, cytokinesis is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.
Plant cells however, are enclosed by a relatively inexte’nsible cell wall, therefore they undergo cytokinesis by a different mechanism. In plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of the new cell wall begins with the formation of a simple precursor, called the cell-plate that represents the middle lamella between the walls of two adjacent cells.

Long Answer Type Questions
Q1. Comment on the statement— Telophase is reverse of prophase.
Ans: Prophase is marked by the initiation of condensation of chromosomal material. The chromosomal material becomes untangled during the process of chromatin condensation. At the beginning of the final stage of mitosis, i.e. telophase, the chromosomes that have reached their respective poles decondense and lose their individuality.
Cells at the end of prophase, when viewed under the microscope, do not show golgi complexes, endoplasmic reticulum, nucleolus and the nuclear envelope. In the telophase stage nuclear envelope assembles around the chromosome clusters. Nucleolus, golgi complex and ER reform.

Q2. What are the various stages of meiotic prophase-I? Enumerate the chromosomal events during each stage?
Ans: Meiosis-I:
Prophase-I: Prophase of the first meiotic division is typically longer and more complex when compared to the prophase of mitosis. It has been further subdivided into the following five phases based on chromosomal behaviour, i.e. Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis. During leptotene stage, the chromosomes become gradually visible under the light microscope.
• The compaction of chromosomes continues throughout leptotene. This is followed by the second stage of prophase-I called zygotene. During this stage chromosomes start pairing together and this process of association is called synapsis. Such paired chromosomes are called homologous chromosomes. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex.
• The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. However, these are more clearly visible at the next stage. The first two stages of prophase-I are relatively short-lived compared to the next stage that is pachytene. During this stage bivalent
. chromosomes now clearly appears as tetrads. This stage is characterised by the appearance of recombination nodules, the sites at which crossing over occurs between non-sister chromatids of the homologous chromosomes. Crossing over is the exchange of genetic material between two homologous chromosomes.
• Crossing over is also an enzyme-mediated process and the enzyme involved is called recombinase. Crossing over leads to the recombination of genetic material on the two chromosomes. Recombination between homologous chromosomes is completed by the end of pachytene, leaving the chromosomes linked at the sites of crossing over.
• The beginning of diploteneis recognised by the dissolution of the synaptonemal complex and the tendency of the recombined homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers. These X-shaped structures, are called chiasmata. In oocytes of some vertebrates, diplotene can last for months or years.
• The final stage of meiotic prophase-I is diakinesis. This is marked by terminalisation of chiasmata. During this phase the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. By the end of diakinesis, the nucleolus disappears and the nuclear envelope also breaks down.

Q3. Differentiate between the events of mitosis and meiosis

	Mitosis		Meiosis
1.	Take place in the somatic cells of the body.	1.	Take place in the germ cells
2.	Occurs in both sexually as well as asexually reproducing organisms.	2.	Occurs only in sexually reproducing organisms.
3.	Mitosis involves only one cycle of nuclear and cell division.	3.	Meiosis          involves         two sequential cycles of nuclear and cell division called meiosis-I and meiosis-II.
4.	The DNA replicates once for one cell division.	4.	The DNA replicates once for two cell divisions.
5.	The prophase is shorter.	5.	Prophase is typically longer
6.	Prophase is comparatively simple.	6.	Prophase of the first meiotic division is more complex when compared to prophase of mitosis.
7.	The cell divides only once and the chromosomes also divide only once. .	7.	There are two cell divisions but the chromosomes divide only once.
8.	Mitosis does not involves pairing                     of        homologous chromosomes                       and recombination between them.	8.	Meiosis involves pairing of homologous chromosomes and recombination between them.
9.	Two cells are formed at the end of mitosis.	9.	Four haploid cells are formed at the end of meiosis.

Q4. Write brief note on the following:
a. Synaptonemal complex
b. Metaphase plate
Ans: a. Synaptonemal complex: During zygotene stage chromosomes start pairing together and this process of association is called synapsis. Such paired chromosomes are called homologous chromosomes. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex. The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. However, these are more clearly visible at the next stage.
b. Metaphase plate: At this stage, metaphase chromosome is made up of two sister chromatids, which are held together by the centromere. Small disc-shaped structures at the surface of the centromeres are called kinetochores. These structures serve as the sites of attachment of spindle fibres (formed by the spindle fibres) to the chromosomes that are moved into position at the centre of the cell. Hence, the metaphase is characterised by all the chromosomes coming to lie at the equator with one chromatid of each chromosome connected by its kinetochore to spindle fibres from one pole and its sister chromatid connected by its kinetochore to spindle fibres from the opposite pole. The plane of alignment of the chromosomes at metaphase is referred to as the metaphase plate.

Q5. Write briefly the significance of mitosis and meiosis in multicellular organism.
Ans: Significance of Mitosis:
Mitosis or the equational division is usually restricted to the diploid cells only. However, in some lower plants mitosis usually results in the production of diploid daughter cells with identical genetic complement. The growth of multicellular organisms is due to mitosis. A very significant contribution of mitosis is cell repair. The cells of the upper layer of the epidermis, cells of the lining of the gut, and blood cells are being constantly replaced. Mitotic divisions in the meristematic tissues —the apical and the lateral cambium, result in a continuous growth of plants throughout their life.
Significance of Meiosis:
Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms, even though the process, per se, paradoxically, results in reduction of chromosome number by half. It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Q6. An organism has two pair of chromosomes (i.e., chromosome number = 4). Diagrammatically represent the chromosomal arrangement during different phases of meiosis-II.

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We hope the NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 10 Cell Cycle and Cell Division, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 6 Anatomy of Flowering Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 6 Anatomy of Flowering Plants.

NCERT Exemplar Class 11 Biology Chapter 6 Anatomy of Flowering Plants

Multiple Choice Questions

Q1. A transverse section of stem is stained first with safranin and then with fast green following the usual schedule of double staining for the preparation of a permanent slide. What would be the colour of the stained xylem and phloem?
(a) Red and green (b) Green and red
(c) Orange and yellow (d) Purple and orange
Ans: (a) A transverse section of stem is stained first with safranin and then with fast green following the usual schedule of double staining for the preparation of a permanent slide. Red and green colour of the stained xylem and phloem appear.

Q2. Match the following and choose the correct option from below.

A.	Meristem	(i)	Photosynthesis, storage
B.	Parenchyma	(ii)	Mechanical support
C.	Collenchyma	(iii)	Actively dividing cells
D.	Sclerenchyma	(iv)	Stomata
E.	Epidermal tissue	(v)	Sclereids

Options:
(a) A—(i), B—(iii), C—(v), D—(ii), E—(iv)
(b) A—(iii), B—(i), C—(ii), D—(v), E—(iv)
(c) A—(ii), B—(iv), C—(v), D—(i), E—(iii)
(d) A—(v), B—(iv), C—(iii), D—(ii), E—(i)

Ans:

A.	Meristem	(iii)	Actively dividing cells                    •
B.	Parenchyma	(i)	Photosynthesis, storage
C.	Collenchyma	(ii)	Mechanical support
D.	Sclerenchyma	(v)	Sclereids
E.	Epidermal tissue	(iv)	Stomata

Q3. Match the following and choose the correct option from below.

A.	Cuticle	(i)	Guard cells
B.	Bulliform cells	(ii)	Single layer
C.	Stomata	(iii)	Waxy layer
D.	Epidermis	(iv)	Empty colourless cell

Options:
(a) A—(iii), B—(iv), C—(i), D—(ii)
(b) A—(i), B—(ii), C—(iii), D—(iv)
(c) A—(iii), B—(ii), C—(iv), D—(i)
(d) A—(iii), B—(ii), C—(i), D—(iv)

Ans. (a)

A.	Cuticle	(iii)	Waxy layer
B.	Bulliform cells	(iv)	Empty colourless cell
C.	Stomata	(i)	Guard cells
D.	Epidermis	(ii)	Single layer

Q4. Identify the simple tissue from among the following.
(a) Parenchyma (b) Xylem (c) Epidermis (d) Phloem
Ans: (a) A simple tissue is made of only one type of cells so the origin of simple tissue is homogenous. Parenchyma, collenchyma and sclerenchyma comes under simple tissue.

Q5. Cells of this tissue are living and show angular wall thickening. They also provide mechanical support. The tissue is
(a) Xylem (b) Sclerenchyma fc) Collenchyma (d) Epidermis
Ans: (c) Collenchyma occurs only in dicots. Collenchyma is living mechanical tissue having cellulosic cell wall. Collenchyma is found in climbing stems like Cucurbita. In Cucurbita lacunate or angular collenchyma is present.

Q6. Epiblema of roots is equivalent to
(a) Pericycle (b) Endodermis (c) Epidermis (d) Stele
Ans: (c) The outermost (piliferous) layer is epidermis (epiblema) or rhizodermis.

Q7. A conjoint and open vascular bundle will be observed in the transverse section of
(a) Monocot root (b) Monocot stem (c) Dicot root (d) Dicot stem
Ans: (d) Each vascular bundle is conjoint, collateral, open, and with endarch in protoxylem dicot stem.

Q8. Interfascicular cambium and cork cambium are formed due to
(a) Cell division (b) Cell differentiation
(c) Cell dedifferentiation (d) Redifferentiation
Ans: (c) Interfascicular cambium and cork cambium are formed due to cell dedifferentiation. .

Q9. Phellogen and phellem respectively denote
(a) Cork and cork cambium (b) Cork cambium and cork
(c) Secondary cortex and cork (d) Cork and secondary cortex
Ans: (b) Phellogen and phellem respectively denote cork cambium and cork.

Q10. In which of the following pairs of parts of a flowering plants is epidermis absent?
(a) Root tip and shoot tip (b) Shoot bud and floral bud
(c) Ovule and seed (d) Petiole and pedicel
Ans. (a) In root tip and shoot tip, epidermis is absent. .

Q11. How many shoot apical meristems are likely to be present in a twig of a plant possessing, 4 branches and 26 leaves?
(a) 26 (b) L (c) 5 (d) 30 (e) 4
Ans: (c) Five shoot apical meristems are likely to be present in a twig of a plant possessing 4 branches and 26 leaves.

Q12. A piece of wood having no vessels (trachea) must be belonging to
(a) Teak (b) Mango (c) Pine (d) Palm
Ans: (c) The presence of vessels is a characteristic feature of angiosperms.

Q13. A plant tissue, when stained, showed the presence of hemicellulose and pectin in cell wall of its cells. The tissue represents
(a) Collenchyma (b) Sclerenchyma (c) Xylem (d) Meristem
Ans: (a) A plant tissue, when stained, showed the presence of hemicellulose and pectin in cell wall of its cells. The tissue represents collenchyma.

Q14. In conifers fibres are likely to be absent in
(a) Secondary phloem (b) Secondary xylem
(c) Primary phloem (d) Leaves
Ans: (b) In conifers fibres are likely to be absent in secondary xylem.

Q15. When we peel the skin of a potato tuber, we remove
(a) Periderm (b) Epidermis (c) Cuticle (d) Sapwood
Ans: (a) When we peel the skin of a potato tuber, we remove periderm.

Q16. A vesselless piece of stem possessing prominent sieve tubes would belong to
(a) Pinus (b) Eucalyptus
(c) Grass (d) Trochodendron
Ans: (d) A vesselless piece of stem possessing prominent sieve tubes would belong to Trochodendron.

Q17. Which one of the following cell types always divides by anticlinal cell division?
(a) Fusiform initial cells (b) Root cap
(c) Protoderm (d) Phellogen
Ans: (c) Protoderm cell always divides by anticlinal cell division.

Q18. What is the fate of primary xylem in a dicot root showing extensive secondary growth?
(a) It is retained in the centre of the axis.
(b) It gets crushed.
(c) May or may not get crushed.
(d) It gets surrounded by primary phloem.
Ans: (a) The fate of primary xylem in a dicot root showing extensive secondary growth because it is retained in the centre of the axis.

Very Short Answer Type Questions

Q1. Product of photosynthesis is transported from the leaves to various parts of the plants and stored in some cell before being utilised. What are the cells/ tissues that store them?
Ans: Parenchyma –

Q2. Protoxylem is the first formed xylem. If the protoxylem lies next to phloem what kind of arrangement of xylem would you call it?
Ans: Exarch

Q3. What is the function of phloem parenchyma?
Ans: The phloem parenchyma stores food material and other substances like resins, latex and mucilage.

Q4. What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots?
Ans: Cuticle is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots.

Q5. What is the epidermal cell modification in plants which prevents water loss?
Ans: In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells. These are called bulliform cells or motor cells. Bulliform cells help in folding and unfolding of grass leaves.
When the bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards (inrolling) to minimise water loss (transpiration).

Q6. What part of the plant would show the following?
a. Radial vascular bundle
b. Polyarch xylem
c. Well developed pith
Ans: a. Radial vascular bundle: Root
b. Polyarch xylem: Monocot root
c. Well developed pith: Dicot stem and monocot root

Q7. What are the cells that make the leaves curl in plants during water stress?
Ans: Bulliform/motor cells

Q8. What constitutes the cambial ring?
Ans: Interfascicular cambium + intrafascicular cambium .

Q9. Give one basic functional difference between phellogen and phelloderm.
Ans. Phellogen is a couple of layers thick. It is made of narrow, thin-walled and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm.

Q10. Arrange the following in the sequence you would find them in a plant starting from the periphery—phellem, phellogen, phelloderm.
Ans: Phellem -» phellogen -» phelloderm.

Q11. If one debarks a tree, what parts of the plant is being removed?
Ans: On debarking a tree, all tissue exterior to vascular cambium gets removed (i.e., periderm and 2° phloem).

Q12. The cross-section of a plant material showed the following features when viewed under the microscope.
a. The vascular bundles were radially arranged.
b. Four xylem strands with exarch condition of protoxylem.
To which organ should it be assigned?
Ans: Dicot root

Q13. What do hard wood and soft wood stand for?
Ans: Gymnosperm wood is called softwood due to absence of vessels (therefore it is called non-porous wood). Dicot angiospermic wood is called hardwood because it possess abundant vessels.

Short Answer Type Questions

Q1. While eating peach or pear it is usually seen that some stone like structures get entangled in the teeth, what are these stone like structures called?
Ans: The structures that get entangled in the teeth while eating fruits like peach and pear are actually the stone cells or brachysclereids which are unbranched, short and isodiametric type of sclereids. These stone cells usually occur in groups and provide grit or stone like hardness that get entangled in the spaces between teeth.

Q2. What is the commercial source of cork? How is it formed in the plant?
Ans: Commercial source of cork is oak (Quercus suber). As the stem continues to
increase in girth due to the activity of vascular cambium, the outer cortical and epidermis layers get broken and need to be replaced to provide new protective cell layers. Hence, sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled and
nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem.

Q3. Below is a list of plant fibres. From which part of the plant these are obtained
a. Coir b. Hemp
c. Cotton d. Jute
Ans: a. Coir—Mesocarp of coconut fruit (drupe)
b. Hemp—Phloem or bast fibre
c. Cotton—Epidermal hair of seed
d. Jute—Phloem or bast fibre

Q4. What are the characteristic differences found in the vascular tissue of gymnosperms and angiosperms?
Ans.
• Xylem functions as a conducting tissue for water and minerals from roots to the stem and leaves. It also provides mechanical strength to the plant parts. It is composed of four different kinds of elements, namely, tracheids, vessels, xylem fibres and xylem parenchyrtia. Gymnosperms lack vessels in their xylem.
• Phloem transports food materials, usually from leaves to other parts of the plant. Phloem in angiosperms is composed of sieve tube elements, companion cells, phloem parenchyma and phloem fibres. Gymnosperms have albuminous cells and sieve cells. They lack sieve tubes and companion cells.

Q5. Epidermal cells are often modified to perform specialized functions in plants. Name some of them and function they perform.
Ans:
• Root hairs: The root hairs are unicellular elongations of the epidermal cells and help absorb water and minerals from the soil.
• Stem hairs or trichomes: On the stem the epidermal hairs are called trichomes. The trichomes in the shoot system are usually multicellular.
They may be branched or unbranched and soft or stiff. They may even be secretory. The trichomes help in preventing water loss due to transpiration.
• Bulliform cells: In grasses, certain adaxial epidermal cells along the veins modify themselves into large, empty, colourless cells. These are called
bulliform cells. When the bulliform cells in the leaves have absorbed water and are turgid, the leaf surface is exposed. When they are flaccid due to water stress, they make the leaves curl inwards to minimise water loss.

Q6. The lawn grass (Cyandon dactylon) needs to be mowed frequently to prevent its overgrowth. Which tissue is responsible for its rapid growth?
Ans: Intercalary meristem: The meristem which occurs between mature tissues is known as intercalary meristem. They occur in grasses and regenerate parts removed by the grazing herbivores.

Q7. Plants require water for their survival. But when watered excessively, plants die. Discuss.
Ans: Irrigation without proper drainage of water leads to waterlogging in the soil. Besides affecting the crops, waterlogging draws salt to the surface of the soil. The salt then is deposited as a thin crust on the land surface or starts collecting at the roots of the plants. This increased salt content is inimical to the growth of crops and is extremely damaging to agriculture and plants may die.

Q8. A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What is the significance of these rings?
Ans: The activity of cambium is under the control of many physiological and environmental factors. In temperate regions, the climatic conditions are not uniform through the year. In the spring season, cambium is very active and produces a large number of xylary elements having vessels with wider cavities. The wood formed during this season is called spring wood or early wood. In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels, and this wood is called autumn wood or late wood.
The spring wood is lighter in colour and has a lower density whereas the autumn wood is darker and has a higher density. The two kinds of woods that appear as alternate concentric rings, constitute an annual ring. Annual rings seen in a cut stem give an estimate of the age of the tree.

Q9. Trunks of some of the aged tree species appear to be composed of several fused trunks. Is it a physiological or anatomical abnormality? Explain in detail.
Ans: Trunks of some of the aged tree species appear to be composed of several fused trunks. It is an anatomical abnormality. It is due to the abnormal secondary growth.

Q10. What is the difference between lenticels and stomata?
Ans: At certain regions, the phellogen cuts off closely arranged parenchymatous cells on the outer side instead of cork cells. These parenchymatous cells soon rupture the epidermis, forming a lensshaped openings called lenticels.
• Lenticels permit the exchange of gases between the outer atmosphere and the internal tissue of the stem. These occur in most woody trees. Lenticels remains open permanently.
• Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two beanshaped cells known as guard cells which enclose stomatal pore. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells (away from the stomatal pore) are thin and the inner walls (towards the stomatal pore) are highly thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata.

Q11. Write the precise function of:
a. Sieve tube
b. Interfasicular cambium
c. Collenchyma
d. Aerenchyma
Ans: a. Sieve tube—Conduction of food
b. Interfasicular cambium—Formation of vascular cambium
c. Collenchyma—Mechanical support
d. Aerenchyma—Provides buoyancy

Q12. The stomatal pore is guarded by two kidney shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Ans: Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two beanshaped cells known as guard cells which enclose stomatal pore. The guard cells possess chloroplasts which are absent in other epidermal cells. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary ceils.

 Q13. Point out the differences in the anatomy of, leaf of peepal (Ficus religiosa) and maize {Zea mays). Draw the diagrams and label the differences.
Ans: Leaf of peepal (Ficus religiosa) is an example of dorsiventral leaf (dicot) and maize (Zea mays) is an example of isobilateral leaf (monocot).

Dicot leaves                      ‘		Monocot leaves
1.	Dorsiventral	1.	Isobilateral
2.	Stomata usually present in lower surface.	2.	Surface equally present on both surfaces.
3.	Stomata have bean shaped guard cells.	3,	Stomata have dumb bell shaped guard cells.
4.	Mesophyll is differentiated in palisade and spongy parenchyma.	4.	Mesophyll is undifferentiated.
5.	Bulliform cells are absent.	5.	Bulliform cells are present.

Q14. Palm is a monocotyledonous plant, yet it increases in girth. Why and how?
Ans: A palm tree is a monocotyledonous plant and like all other monocots the stems do not have primary cambium in the vascular bundles. However, with age, the tree grows in diameter, though slowly, as a result of the growth of the ground tissue. A secondary cambium may be formed in the hypodermal region of the stem. The later forms the conjunctive tissue and patches of meristematic cell. The activity of the meristematic cells result in the formation of secondary vascular bundles.

Long Answer Type Questions

Q1. The arrangement of ovules within the ovary is known as placentation. What does the term placenta refer to? Draw various types of placentations in the flower as seen in T.S. and V.S.
Ans: Placentation: The arrangement of ovules within the ovary is known as placentation. The placentation are of different types namely, marginal, axile, parietal, basal, central and free central. In marginal placentation, the placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows, as in pea. When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, as in china rose, tomato and lemon. In parietal placentation, the ovules develop on the inner wall of the ovary or on peripheral part. Ovary is one- chambered but it becomes two-chambered due to the formation of the false septum, e.g., mustard and Argemone. When the ovules are borne on central axis and septa are absent, as in Dianthus and Primrose the placentation is called free central. In basal placentation, the placenta develops at the base of ovary and a single ovule is attached to it, as in sunflower, marigold.

Q2. Deciduous plants shed their leaves during hot summer or in autumn. This process of shedding of leaves is called abscission. Apart from physiological changes what anatomical mechanism is involved in the abscission of leaves.
Ans: Anatomical mechanism involved in the abscission of leaves:
(1) Structural: In deciduous trees, an abscission zone, also called a separation zone, is formed at the base of the petiole. It is composed of a top layer that has cells with weak walls, and a bottom layer that expands in the autumn, breaking the weak walls of the cells in the top layer. This allows the leaf to be shed.
(2) The loss of chlorophyll may also contribute to the abscission process.
(3) Hormonal: Abscisic acid to be the hormone that stimulates abscission (for which the hormone was named).

Q3. Is Pinus an evergreen tree? Comment.
Ans: Yes, Pinus trees are evergreen, coniferous resinous trees. The leaves in gymnosperms are well-adapted to withstand extremes of temperature, humidity and wind. In conifers, the needle-like leaves reduce the surface area. Their thick cuticle and sunken stomata also help to reduce water loss. Unlike deciduous plants, Pinus do not shed their leaves so it is an evergreen tree.

Q4. Assume that a pencil box held in your hand, represents a plant cell. In how many possible planes can it be cut? Indicate these cuts with the help of line drawings.
Ans: When any plane passing through the central axis of the plant cell divides the cell into two identical halves, it is called radial symmetry. If the plant cell
can be divided into identical left and right halves in only one vertical plane, exhibit bilateral symmetry.

Q5. Each of the following terms has some anatomical significance. What do these terms mean? Explain with the help of line diagrams.
a. Plasmadesmoses/Plasmodesmata
b. Middle lamella
c. Secondary wall
Ans: a. Plasmodesmata are the protoplasmic strands between adjacent plant cells. Plasmodesmata connections help in movement of substances between cells.
b. The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together. Middle lamella is made up of calcium pectate (mainly) and magnesium pectate.During ripening a fruit becomes soft and pulpy due to dissolution of middle lamella.
c. The cell wall of a young plant cell, the primary wall is capable of growth, which gradually diminishes as the cell matures and the secondary wall is formed on the inner (towards membrane) side of the cell. Secondary wall may have the deposition of lignin. Secondary wall help in the differentiation of the cell.

Q6. Distinguish between the following:
(a) Exarch and endarch condition of protoxylem
(b) Stele and vascular bundle
(c) Protoxylem and metaxylem
(d) Interfasicular cambium and intrafasicular cambium
(e) Open and closed vascular bundles
(f) Stem hair and root hair. .
Ans: (a) Exarch and endarch condition of protoxylem: In stems, the protoxylem lies towards the centre (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called endarch. In roots, the protoxylem lies towards periphery and metaxylem lies towards the centre. Such arrangement of primary xylem is called exarch.
(b) Stele and vascular bundle: All tissues on the innerside of the endodermis such as pericycle, vascular bundles and pith constitute the stele. The vascular system consists of complex tissues, the phloem and the xylem. The xylem and phloem together constitute vascular bundles.
(c) Protoxylem and metaxylem: Primary xylem is of two types— protoxylem and metaxylem. The first formed primary xylem elements are called protoxylem and the later formed primary xylem is called
metaxylem.
(d) Interfasicular cambium and intrafasicular cambium; In dicot stems, the cells of cambium present between primary xylem and primary phloem is the intrafascicular cambium. The cells of medullary rays, adjoining these intrafascicular cambium become meristematic and form
the interfascicular cambium.
(e) Open and closed vascular bundles: In dicotyledonous stems, cambium is present between phloem and xylem. Such vascular bundles because of the presence of cambium possess the ability to form secondary xylem and phloem tissues, and hence are called open vascular bundles. In the monocotyledons, the vascular bundles have no cambium present in them. Hence, since they do not form secondary tissues they are referred to as closed.
(f) Stem hair and root hair: The root hairs are unicellular elongations of the epidermal cells and help absorb water and minerals from the soil. On the stem the epidermal hairs are called trichomes or stem hairs. The trichomes in the shoot system are usually multicellular. They may be branched or unbranched and soft or stiff. They may even be secretory. The trichomes help in preventing water loss due to transpiration.

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NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption.

NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption

Multiple Choice Questions

Q1. Select what is not true of intestinal villi among the following
(a) They possess microvilli
(b) They increase the surface area
(c) They are supplied with capillaries and the lacteal vessels (d_) They only participate in digestion of fats
Ans: (d) They only participate in digestion of fats.

Q2. Hepato-pancreatic duct opens into the duodenum and carries
(a) Bile (b) Pancreatic juice
(c) Both bile and pancreatic juice (d) Saliva
Ans: (c) Hepato-pancreatic duct opens into the duodenum and carries both bile and pancreatic juice.

Q3. One of the following is not a common disorder associated with digestive system
(a) Tetanus (b) Diarrhoea (c) Jaundice (d) Dysentery
Ans: (a) Tetanus is not a common disorder associated with digestive system.

Q4. A gland not associated with the alimentary canal is
(a) Pancreas (b) Adrenal
(c) Liver (d) Salivary glands
Ans: (b) A gland not associated with the alimentary canal is adrenal (this is endocrine gland).

Q5. Match the two columns and select the correct among options given

	Column I		Column II
A.	Biomacromolecules	(i)	Alimentary canal and of food associated gland
B.	Human digestive system	(ii)	Embedded in jawbones
C.	Stomach	(iii)	Outer wall of visceral organs

D.	Thecodont	(iv)	Converted into simple substances
E.	Serosa	(v)	J-shaped bag like structure

 

Options:
(a) A—(ii), B—(i), C—(v), D—(iii), E—(iv)
(b) A—(iv), B—(i), C—(v), D—(ii), E—(iii)
(c) A—(i), B—(ii), C—(iii), D—(iv), E—(v)
(d) A—(i), B—(iii), C—(ii), D—(iv), E-(v)

Ans. (b)

A.	Biomacromolecules	(iv)	Converted into simple substances
B.	Human digestive	(i)	Alimentary canal and of food associated
	system		gland
C.	Stomach	(v)	J-shaped bag like structure
D.	Thecodont	(ii)	Embedded in jawbones
E.	Serosa	(iii)	Outer wall of visceral organs

6. Match the two columns and select the right one among options given

Column I		Column II
A.	Duodenum	(i)	A cartilagenous flap
B.	Epiglottis	(ii)	Small blind sac
C.	Glottis	(Hi)	‘C’ shaped structure emerging from the stomach
D.	Caecum	(iv)	Opening of wind pipe

(a) A—(i), B—(ii), C—(iii), D—(iv)
(b) A—(iv), B—(iii), C—(ii), D—(i)
(c) A—(iii), B—(i), C—(iv), D—(ii)
(d) A—(ii), B—(iv), C—(i), D—(iii)

A.	Duodenum	(iii)	‘C’ shaped structure emerging from the stomach
B.	Epiglottis	(i)	A cartilagenous flap
C.	Glottis	(iv)	Opening of wind pipe
D.	Caecum	(ii)	Small blind sac

Q7. Match the enzymes with their respective substrates choose the right one among options given.

Column I		Column II
A.	Lipase	(i)	Dipeptides
B.	Nuclease	(ii)	Fats
C.	Carboxypeptidase	(iii)	Nucleic acids
D.	Dipeptidases	(iv)	Proteins, peptones and proteoses

Options:
(a) A—(ii), B—(iii), C—(i), D—(iv)
(b) A—(iii), B—(iv), C—(ii), D—(i)
(c) A—(iii), B—(i), C—(iv), D—(ii)
(d) A—(ii), B—(iii), C—(iv), D—(i)

 

A.	Lipase	(ii)	Fats
B.	Nuclease	(iii)	Nucleic acids
C.	Carboxypeptidase	(iv)	Proteins, peptones and proteoses
D.	Dipeptidases	(i)	Dipeptides

Ans. (d)

Q9. Liver is the largest gland and is associated with functions, choose one which is not correct
(a) Metabolism of carbohydrate
(b) Digestion of fat
(c) Formation of bile
(d) Secretion of hormone called gastrin
Ans: (d) Liver is the largest gland and is associated with functions metabolism of carbohydrate, digestion of fat and formation of bile.

Q10. Mark the right statement among the following
(a) Trypsinogen is an inactive enzyme
(b) ’ Trypsinogen is secreted by intestinal mucosa
(c) Enterokinase is secreted by pancreas
(d) Bile contains trypsin
Ans: (a)

Very Short Answer Type Questions

Q1. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall. What do we call the food then?
Ans: Chyme

Q2. Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase, activates it. Which tissue/cells secrete this enzyme?/ How is it activated?
Ans: Intestinal mucosa

Q3. In which part of alimentary canal does absorption of water, simple sugars and alcohol takes place?
Ans: Stomach

Q4. Name the enzymes involved in the breakdown of nucleotides into sugars and bases.
Ans: Nucleotidases and Nucleosidases

Q5. Define digestion in one sentence.
Ans: The process of conversion of complex food substances in the digestive system to simple absorbable forms is called digestion.

Q6. What do we call the type of teeth attachment to j aw bones in which each tooth is embedded in a socket of jaws bones?
Ans: Thecodont

Q7. Stomach is located in upper left portion of the abdominal cavity and has three major parts. Name these three parts.
Ans: Cardiac, fundic and pyloric

Q8. Does gall bladder make bile?
Ans: No, it only stores.

Q9. Correct the following statements by deleting one of entries (given in bold).
a. Goblet cells are located in the intestinal mucosal epithelium and secrete chymotrypsin/mucus.
b. Fats are broken down into di- and monoglycerides with the help of amylase/lipases.
c. Gastric glands of stomach mucosa have oxyntic cell/chief cells which secrete HCl.
d. Saliva contains enzymes that digest starch/protein.
Ans: a. Goblet cells are located in the intestinal mucosal epithelium and secrete _ mucus.
b. Fats are broken down into di- and monoglycerides with the help of lipases.
c. Gastric glands of stomach mucosa have oxyntic cell which secrete HC1.
d. Saliva contains enzymes that digest starch.

Short Answer Type Questions
Q1. What is pancreas? Mention the major secretions of pancreas that are helpful in digestion.
Ans: Pancreas is a gland having exocrine and endocrine portions involved in secreting digestive enzymes as well as hormones. Major secretions of pancreas involved in digestion are inactive enzymes listed below:
a. Trypsinogen
b. Chymotrypsinogen
c. Procarboxypeptidases
d.Amylases
e. Lipases
f. Nucleases

Q2. Name the part of the alimentary canal where major absorption of digested food takes place. What are the absorbed forms of different kinds of food materials? ‘
Ans: Small intestine is the part of alimentary canal where digested food materials are mainly absorbed. Amino acids (proteins), monosaccharides like glucose, fructose, galactose, etc. (carbohydrate) and fatty acids and glycerol (fats) are different absorbable forms of food materials.

Q3. List the organs of human alimentary canal and name the major digestive glands with their location.
Ans: The alimentary canal begins with an anterior opening the mouth and then buccal cavity, pharynx, oesophagus, stomach, small intestine, large intestine and it opens out posteriorly through the anus.
The digestive glands associated with the alimentary canal include the salivary glands, the liver and the pancreas. Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub¬mandibular (lower jaw) and the sublinguals (below the tongue). Liver is situated in the abdominal cavity, just below the diaphragm. The pancreas is situated between the limbs of the ‘C’-shaped duodenum.

Q4. What is the role of gall bladder? What may happen if it stops functioning or is removed?
Ans: The bile secreted by the hepatic cells passes through the hepatic ducts and is stored and concentrated in a thin muscular sac called the gall bladder. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases. If it stops functioning or is removed then digestion of fat will be affected.

Q5. Correct the statement given below by the right option shown in the bracket against them.
a. Absorption of amino acids and glycerol takes place in the (small intestine/ large intestine).
b. ‘The faeces in the rectum initiate a reflex causing an urge for its removal, (neural /hormonal)
c. Skin and eyes turn yellow in infection, (liver/stomach)
d. Rennin is a proteolytic enzyme found in gastric juice in (infants/adults).
e. Pancreatic juice and bile are released through (intestinepancreatic/ hepato-pancreatic duct).
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in region of small intestine, (jejunum/duodenum)
Ans: a. Absorption of amino acids and glycerol takes place in the small intestine.
b. The faeces in the rectum initiate a neural reflex causing an urge for its removal.
c. Skin and eyes turn yellow in infection of liver. .
d. Rennin is a proteolytic enzyme found in gastric juice in infants.
e. Pancreatic juice and bile are released through hepato-pancreatic duct.
f. Dipeptides, disaccharides and glycerides are broken down into simple substances in duodenum region of small intestine.

Q6. What are three major types of cells found in the gastric glands? Name their secretions.
Ans: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).
Q7. How is the intestinal mucosa’protectcd from the acidic food entering from stomach?
Ans: The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from excoriation by the highly concentrated hydrochloric acid.

Q8. How are the activities of gastro-intestinal tract regulated?
Ans: The activities of the gastro-intestinal tract are under neural and hormonal control for proper coordination of different parts. The sight, smell and/or the presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also, similarly, stimulated by neural signals. The muscular activities of different parts of the alimentary canal can also be moderated by neural mechanisms, both local and through CNS. Hormonal control of the secretion of digestive juices is carried out by local hormones produced by the gastric and intestinal mucosa.

Q9. Distinguish between constipation and indigestion. Mention their major causes.
Ans: Constipation: In constipation, the faeces are retained within the rectum as the bowel movements occur irregularly.
Indigestion: In this condition, the food is not properly digested leading to a feeling of fullness. The causes of indigestion are inadequate enzyme secretion, anxiety, food poisoning, over eating, and spicy food.

10. Describe the enzymatic action on fats in the duodenum.
Ans: In the duodenum fats are broken down by pancreatic lipases with the help of bile into di- and monoglycerides.

Long Answer Type Questions
Q1. A person had roti and dal for his lunch. Trace the changes in those during its passage through the alimentary canal.
Ans: The process of digestion is accomplished by mechanical and chemical processes. The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About 30% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide – maltose.
• The stomach stores the food for 4-5 hours. The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and is called the chyme. The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteoses and peptones (peptides).
• The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
Proteins proteoses and peptones (partially hydrolysed proteins) in the chyme reaching the intestine are acted upon by the proteolytic enzymes of pancretic juice as given below.

Q2. What are the various enzymatic types of glandular secretions in our gut helping digestion of food? What is the nature of end products obtained after complete digestion of food?
Ans: Enzymatic types of glandular secretions in our gut:
a. Salivary glands: Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub-mandibular (lower jaw) and the sublinguals (below the tongue). These glands situated just outside the buccal cavity secrete salivary juice into the buccal cavity.
b. Gastric glands: The mucosa of stomach has gastric glands. Gastric glands have three major types of cells namely
(i) mucus neck cells which secrete mucus;
(ii) peptic or chief cells which secrete the proenzyme pepsinogen; and
(iii) parietal or oxyntic cells which secrete HC1 and intrinsic factor (factor essential for absorption of vitamin B12).
c. The bile, pancreatic juice and the intestinal juice are the secretions
released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes—trypsinogen, chymotrypsinogen, procarboxypeptidases,
amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes.

Q3. Discuss mechanisms of absorption.
Ans: Mechanisms of absorption: Absorption is the process by which the end products of digestion pass through the intestinal mucosa into the blood or lymph. It is carried out by passive, active or facilitated transport mechanisms. Small amounts of monosaccharides like glucose, amino acids and some electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon the concentration gradients. However, some substances like glucose and amino acids are absorbed with the help of carrier proteins. This mechanism is called the facilitated transport.
Transport of water depends upon the osmotic gradient. Active transport occurs against the concentration gradient and hence requires energy. Various nutrients like amino acids, monosaccharides like glucose, electrolytes like Na+ are absorbed into the blood by this mechanism.

Q4. Discuss the role of hepato-pancrdatic complex in digestion of carbohydrate, protein and fat components of food.
Ans: The bile, pancreatic juice and the intestinal juice are the secretions released into the small intestine. Pancreatic juice and bile are released through the hepato-pancreatic duct. The pancreatic juice contains inactive enzymes— trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes. Bile helps in emulsification of fats, i.e., breaking down of the fats into very small micelles. Bile also activates lipases.
Proteins proteoses and peptones (partically hydrolysed proteins) in the chime reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice as given below:

Q5. Explain the process of digestion in the buccal cavity with a note on the arrangement of teeth.
Ans: The process of digestion in the buccal cavity: The buccal cavity performs two major functions, mastication of food and facilitation of swallowing. The teeth and the tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into a bolus. The bolus is then conveyed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus by successive waves of muscular contractions called peristalsis. The gastro-oesophageal sphincter controls the passage of food into the stomach. The saliva secreted into the oral cavity contains electrolytes and enzymes, salivary amylase and lysozyme. The chemical process of digestion is initiated in the oral cavity by the hydrolytic action of the carbohydrate splitting enzyme, the salivary amylase. About% of starch is hydrolysed here by this enzyme (optimum pH 6.8) into a disaccharide—maltose. Lysozyme present in saliva acts as an antibacterial agent that prevents infections.

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We hope the NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 16 Digestion and Absorption, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 3 Plant Kingdom are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 3 Plant Kingdom.

NCERT Exemplar Class 11 Biology Chapter 3 Plant Kingdom

Multiple Choice Questions

Q1. Cyanobacteria are classified under
(a) Protista (b) Plantae (c) Monera (d) Algae
Ans: (c) Cyanobacteria are classified under Kingdom Monera.
• Protista— unicellular eukaryotes
• Plantae, all members of Kingdom Plantae are eukaryotic chloroplast ‘chlorophyll containing organisms commonly called plants. These are autotrophic/holophytic.

Q2. Fusion of two motile gametes which are dissimilar in size is termed as
(a) Oogamy (b) Isogamy (c) Anisogamy (d) Zoogamy
Ans: (c) Fusion of two motile gametes which are dissimilar in size is termed as anisogamy.

Q3. Holdfast, stipe and frond constitute the plant body in case of
(a) Rhodophyceae (b) Chlorophyceae
(c) Phaeophyceae (d) All of the above
Ans: (c) The plant body of phaeophyceae is usually attached to the substratum by a holdfast, and has a stalk, the stipe and leaf like photosynthetic organ—the frond.

Q4. A plant shows thallus level of organization. It shows rhizoids and is haploid. It needs water to complete its life cycle because the male gametes are motile. It may belong to
(a) Pteridophytes (b) Gymnosperms
(c) Monocots (d) Bryophytes
Ans: (d) A plant shows thallus level of organization. It shows rhizoids and is haploid. It needs water to complete its life cycle because the male gametes are motile. It may belong to bryophytes.

Q5. A prothallus is ‘ ‘
(a) A structure in pteridophytes formed before the thallus develops
(b) A sporophytic free living structure formed in pteridophytes
(c) A gametophyte free living structure formed in pteridophytes
(d) A primitive structure formed after fertilization in pteridophytes
Ans. (c) In pteridophytes, meiosis or R/D occurs at the time of spore formation. The spores germinate to give rise to inconspicuous, small but multicellular, free-living, mostly photosynthetic thalloid gametophytes called prothallus. Prothallus tepresents the gametophytic phase in pteridophytes.

Q6. Plants of this group are diploid and well adapted to extreme conditions. They grow bearing sporophylls in compact structures called cones. The group in reference is
(a) Monocots (b) Dicots
(c) Pteridophytes (d) Gymnosperms
Ans: (d) Plants of this group are diploid and well adapted to extreme conditions. They grow bearing sporophylls in compact structures called cones. The group in reference is gymnosperms.

Q7. The embryo sac of an Angiosperm is made up of
(a) 8 cells .(b) 7 cells and 8 nuclei
(c) 8 nuclei (d) 7 cells and 7 nuclei
Ans: (b) The embryo sac of an Angiosperm is made up of 7 cells and 8 nuclei.

Q8. If the diploid number of a flowering plant is 36, what would be the chromosome number in its endosperm?
(a) 36 (b) 18 (c) 54 ‘ (d) 72
Ans: (c) Diploid number (2«) of a flowering plant is 36.
The chromosome number in its endosperm 3n = 54.

Q9. Protonema is
(a) Haploid and is found in mosses
(b) Diploid and is found in liverworts
(c) Diploid and is found in pteridophytes
(d) Haploid and is found in pteridophytes
Ans: (a) The predominant stage of the life cycle of a moss is the gametophyte which consists of two stages. The first stage is the protonema stage (juvenile stage) and the second stage is the leafy stage. Moss protonema resembles to multicellular green algae in structure. Moss plant develops from protonema.

Q10. The giant Redwood tree (Sequoia sempervirens) is a/an .
(a) Angiosperm (b) Free fern
(c) Pterdophyte (d) Gymnosperm
Ans: (d) One of the gymnosperms, the giant redwood tree Sequoia is one of the tallest tree species.

Very Short Answer Type Questions

Q1. Food is stored as Floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
Ans: Mannitol is the reserve food material of brown algae or phaeophyceae.

Q2. Give an example of plants with
a. Haplontic life cycle
b. Diplontic life cycle
c. Haplo-diplontic life cycle
Ans: a. Haplontic life cycle—Volvox, Spirogyra and some species of Chlamydomonas
b. Diplontic life cycle—AH seed-bearing plants, i.e. (gymnosperms and angiosperms)
c. Haplo-diplontic life cycle—Bryophyfes and Pteridophytes

Q3. The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorption. What is the equivalent of roots in the less developed lower plants?
Ans: In lower plants like algae, holdfast is present and in bryophytes, rhizoids are present instead of roots.

Q4. Most algal genera show haplontic life style. Name an alga which is
a. Haplo-diplontic
b. Diplontic
Ans: a. Haplo-diplontic—Ectocarpus, Polysiphonia and Kelps b. Diplontic—Fucus

Q5. In Bryophytes male and female sex organs are called _______ and __________
Ans: In Bryophytes male sex organ is called antheridium and female sex organ is called archegonium.

Short Answer Type Questions

Q1. Why are bryophytes called the amphibians of the plant kingdom?
Ans: Bryophytes are also called amphibians of the plant kingdom because these . plants can live in soil but are dependent on water for sexual reproduction.

Q2. The male and female reproductive organs of several pteridophytes and gymnosperms are comparable to floral structures of angiosperms. Make an attempt to compare the various reproductive parts of pteridophytes and gymnosperms with reproductive structures of angiosperms

Ans.

	Reproductive parts of pteridophytes and gymnosperms	Reproductive structures of angiosperms
(0	Strobili/cone	Flower
(ii)	Microsporophyll	Stamen
(iii)	Megasporophyll	Pistil/Carpel
(iv)	Microsporangium	Anther
(v)	Megasporangium	Ovule

 

Q3. Heterospory, i.e. formation of two types of spores—microspores and megaspores is a characteristic feature in the life cycle of a few members of pteridophytes and all spermatophytes. Do you think heterospory has some evolutionary significance in plant kingdom?
Ans: In majority of the pteridophytes all the spores are of similar kinds, such plants are called hom’osporous. Genera like Selaginella, Salvirtia, Marsilea and Azolla which produce two kinds of spores, macro (large) and micro (small) spores are known as heterosporous. The megaspores and microspores germinate and give rise to female and male gametophytes, respectively.
The female gametophytes in these plants are retained on the parent sporophytes for variable periods. The development of the zygotes into young embryos take place within the female gametophytes. This event is a precursor to the seed habit considered an important step in evolution.

Q4. How far does Selaginella one of the few living members of lycopodiales (pteridophytes) fall short of seed habit?
Ans: Selaginella produce two kinds of spores, macro (large) and micro (small) spores. The megaspores and microspores germinate and give rise to female and male gametophytes, respectively. But Selaginella falls short of seed habit due to lack of integument around the megasporangium.

Q5. Each plant or group of plants has some phylogenetic significance in relation to evolution: Cycas, one of the few living members of gymnosperms is called as the ‘relic of past’. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?
Ans: Cycas, one of the few living members of gymnosperms is called as the ‘relic of past’ because it shows many characteristics which are similar to pteridophytes, like, flagellated antherozoids, circinate ptyxis, megasporophyll is leaf like, presence of archegonia, etc.

Q6. The heterosporous pteridophytes show certain characteristics, which are precursor to the seed habit in gymnosperms. Explain.
Ans: In majority of the pteridophytes all the spores are of similar kinds, such plants are called homosporous. Genera like Selaginella, Salvinia, Marsilea and Azolla which produce two kinds of spores, macro (large) and micro (small) spores, are known as heterosporous. The megaspores and microspores germinate and give rise to female and male gametophytes, respectively.
The female gametophytes in these plants are retained on the parent sporophytes for variable periods. The development of the zygotes into young embryos take place within the female gametophytes. This event is a precursor to the seed habit considered an important step in evolution.

Q7. Comment on the life cycle and nature of a fern prothallus.
Ans: The diploid sporophyte is represented by a dominant, independent, photosynthetic, vascular plant body. It alternates with multicellular, saprophytic/autotrophic, independent but short-lived haploid gametophyte . called prothallus. Such a pattern is known as haplo-diplontic life cycle. All
pteridophytes exhibit this pattern.
These gametophytes require cool, damp, shady places to grow. Because of this specific restricted requirement and the need of water for fertilisation, the spread of living pteridophytes is limited and restricted to narrow geographical regions. The gametophytes (prothallus) bear male and female sex organs ‘ called antheridia and archegonia, respectively.
Water is required for transfer of antherozoids—the male gametes released from the antheridia, to the mouth of archegonium. Fusion of male gamete … with the egg present in the archegonium result in the fonnation of zygote.
• Zygote thereafter produces a multicellular well-differentiated sporophyte which is the dominant phase of the pteridophytes.

Q8. How arc the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Ans: Male and female gametophytes of pteridophytes are free living while in gymnosperms male and female gametophyte do not have free-living * existence. They remain within the sporangia retained on sporophytes

Pteridophytes		Gymnosperms
(0	Flagellated male gamete	(a)	Non-flagellated male gamete
(ii)	Water is essential for fertilisation	(b)	Water is not essential
(iii)	Pollen tubes are not formed	(c)	Pollen tubes are formed
(iv)	Archegonia with neck canal cells	(d)	Neck canal cells are absent

Q9. In which plant will you look for mycorrhiza and corolloid roots? Also explain w’hat these terms mean.
Ans: Roots in some genera have fungal association in the form of mycorrhiza (Finns), while in some others (Cvms) small specialised roots called coralloid roots are associated with N2-fixing cyanobacteria

Long Answer Type Questions

Q1. Gametophyte is a dominant phase in the life cycle of a bryophyte. Explain.
Ans: The main plant body of the biyophyte is haploid. It produces gametes, hence is called a gametophyte. The sex organs in bryophytes are multicellular.
The male sex organ is called antheridium. They produce biflagellate antherozoids or biciliated sperms. The female sex organ called archegonium is flask-shaped and produces a single egg. The antherozoids are released into water where they come in contact with archegonium. An antherozoid fuses with the egg to produce the zygote. Zygote do not undergo reduction division immediately. They produce a multicellular body called a sporophyte.

Q2. With the help of a schematic diagram, describe the haplo–diptontic life cycle pattern of a plant group.
Ans: In a sexually reproducing plant there is an alternation of generation between a haploid and a diploid phase of plant bodies. The haploid plant body is termed gametophyte while the diploid plant body is called sporophyte. The gametophyte produces gametes by mitosis while the haploid spores are produced by sporophyte following meiosis (reduction division). Two gamete fuse together to produce a zygote which develops into the diploid sporophyte.

In a haplodiplontic life cycle pattern, such as in bryophyta or pteridophyta both the phases of life are multicellular. However, in bryophytes, the gametophytes are small, photosynthetic, independent and represent dominant phase. The partly or totally dependent sporophyte is physically attached to the gametophyte. The (n) spores dispersed by sporophyte germinate into individual gametophytic plants. However, in pteridophytes the 2n (diploid) phase is dominant, well organized, independent while the n phase though also free-living and independent is short lived and photosynthetic. In both of these groups of plants the mobile male gametes, antherozoid produced by sex organ antheridium, travel to archegonium (bearing an egg cell) via the medium of water. Egg cell is non-motile hence the reproduction is oogamous.

Q3. Lichen is usually cited as an example of‘symbiosis’ in plants where an algal and a fungal species live together for their mutual benefit. Which of the following will happen if algal and fungal partners are separated from each other?
a. Both will survive and grow normally and independent from each other.
b. Both will die.
c. Algal component will survive while the fungal component will die.
d. Fungal component will survive while algal partner will die.
Based on your answer how do you justify this association as symbiosis.
Ans: Lichen is usually cited as an example of symbiosis in biology where in a fungal and an algal species live together for mutual benefit. The algal component synthesizes the food through photosynthesis which is utilized by the fungal species for its survival. The fungal component in return provides shelter and waste products that are consumed by algal species.
Experiments though have shown that algal component can grow independently when separated from fungal species. But same is not true with the fungal component which dies when separated from algal component. This association is, therefore, a typical case of master-slave relationship where fungus (master) has trapped the algal components (slave) for its own survival while giving nothing in return to it. Some authors consider this association as controlled parasitism or helotism due to the fact that sometimes the fungus sends its haustoria into the algal cells to derive nourishment.

Q4. Explain why sexual reproduction in angiosperms is said to take place through double fertilization and triple fusion. Also draw a labelled diagram of embryo sac to explain the phenomena.
Ans: After entering one of the synergids, the pollen tube release the two male gametes into the cytoplasm of the synergid.

Q5. Draw labelled diagrams of
a. Female and male thallus of a liverwort.
b. Gametophyte and sporophyte of Funaria.
c. Alternation of generation in Angiosperm.

Ans:  a. Female and male thallus of a liverwort.

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We hope the NCERT Exemplar Class 11 Biology Chapter 3 Plant Kingdom help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 3 Plant Kingdom, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants.

NCERT Exemplar Class 11 Biology Chapter 11 Transport in Plants

Multiple Choice Questions

Q1. Which of the following statements does not apply to reverse osmosis?
(a) It is used for water purification.
(b) In this technique, pressure greater than osmotic pressure is applied to the system.
(c) It is a passive process.
(d) It is an active process.
Ans: (c) If pressure greater than the osmotic pressure is applied to the higher concentration, the direction of water flow through the membrane can be reverse. This is called reverse osmosis. Reverse osmosis occurs when water is moved across the membrane against the concentration gradient, from lower concentration to higher concentration. Reverse osmosis is an active process.

Q2. Which one of the following will not directly affect transpiration?
(a) Temperature
(b) Light
(c) Wind speed
(d) Chlorophyll content of leaves
Ans: (d) The chlorophyll content of leaves will not directly affect transpiration, while temperature, light and wind speed directly affect the transpiration.

Q3. The lower surface of leaf will have more number of stomata in a
(a) Dorsiventral leaf
(b) Isobilateral leaf
(c) Both (a) and (b)
(d) None of the above
Ans: (a) Usually, the lower surface of a dorsiventral (dicotyledonous) leaf has a greater number of stomata. On the upper surface, stomata may be even absent sometimes.

Q4. The form of sugar transported through phloem is
(a) Glucose (b) Fructose (c) Sucrose (d) Ribose
Ans: (c) Food, primarily sucrose, is transported by the vascular tissue phloem from a source to a sink.

Q5. The process of guttation takes place
(a) when the root pressure is high and the rate of transpiration is low
(b) when the root pressure is low and the rate of transpiration is high
(c) when the root pressure equals the rate of transpiration
(d) when the root pressure as well as rate of transpiration are high.
Ans: (a) The effect of root pressure is observable at night as well as early morning when evaporation is low. Excess water gets collected in the form of droplets around special openings of veins near the tip of grass blades and leaves of many herbaceous parts of plants such as Tropaeolum, Balsam and grasses. Such water loss in its liquid phase is known as guttation.

Q6. Which of the following is an example of imbibition?
(a) Uptake of water by root hair (b) Exchange of gases in stomata (c) Swelling of seed when put in soil (d) Opening of stomata
Ans: (c) Imbibition is a special type of diffusion. A classic example of imbibition is absorption of water by seeds and dry wood.

Q7. When a plant undergoes senescence, the nutrients may be
(a) accumulated
(b) bound to cell wall
(c) translocated
(d) None of the above
Ans: (c) Mineral ions are frequently remobilized (translocation), particularly from older senescing parts. Before the leaf fall in deciduous plants, minerals are translocated to other parts.

Q8. Water potential of pure water at standard temperature is equal to
(a) 10
(b) 20
(c) Zero
(d) None of these
Ans: (c) The water potential of pure water at standard temperature is equal to zero.

Q9. Choose the correct option. Mycorrhiza is a symbiotic association of fungus with root system which helps in A. Absorption of water B. Mineral nutrition
C. Translocation D. Gaseous exchange
(a) Only A
(b) Only B –
(c) Both A and B
(d) Both B and C
Ans: (c) Mycorrhiza is a symbiotic association of fungus with root system which helps in absorption of water and mineral nutrition.

Q10. Based on the figure given below, which of the following statements is not correct?

(a) Movement of solvent molecules will take place from chamber AtoB
(b) Movement of solute will take place from A to B
(c) Presence of a semipermeable is a prerequisite for this process to occur
(d) The direction and rate of osmosis depends on both the pressure gradient and concentration gradient.
Ans: (b) The movement of solute will take place from B to A

Q11. Match the following and choose the correct option.

A.	Leaves	(i)	Anti-transpirant
B.	Seed	(ii)	Transpiration
C.	Roots	(iii)	Negative osmotic potential
D.	Aspirin	(iv)	Imbibition ‘
E.	Plasmolyzed cell	(v)	Absorption

Options:
(a) A—(ii), B—(iv), C—(v), D—(i), E—(iii)
(b) A—(iii), B—(ii), C—(iv), D—(i), E—(v)
(c) A—(i), B—(ii), C—(iii), D—(iv), E—(v)
(d) A—(v), B—(iv), C—(iii), D—(ii), E—(i)
Ans: (a)

A.	Leaves	(ii)	Transpiration
B.	Seed	(iv)	Imbibition
e.	Roots	(v)	Absorption
D.	Aspirin	(i)	Anti-transpirant
E.	Plasmolyzed cell	(ii)	Negative osmotic potential

 

Q12. Mark the mismatched pair.
(a) Amyloplast—Store protein granule
(b) Elaioplast—Store oils or fats
(c) Chloroplasts—Contain chlorophyll pigments
(d) Chromoplasts—Contain coloured pigments other than chlorophyll
Ans: (a) Aleuroplasts—Store proteins
Amyloplast—Store carbohydrate (starch)
Very Short Answer Type Questions .
Q1. Smaller, lipid soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain transporters which are chemically ________.
Ans: Protein

Q2. In a passive transport across a membrane, when two protein molecules move in opposite direction and independent of each other, it is called as ________
Ans: Antiport

Q3. Osmosis is a special kind of’diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both ________
Ans: Pressure and concentration gradient

Q4. A flowering plant is planted in an earthen pot and irrigated. Urea is added to make the plant grow faster, but after some time the plant dies. This may be due to ________.
Ans: Exosmosis

Q5. Absorption of water from soil by dry seeds increases the ________ thus helping seedlings to come out of soil.
Ans: Pressure

Q6. Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. The most important physiological phenomenon which is responsible for the upward movement of water is _________
Ans: Transpiration pull

Q7. The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because _________ .
Ans: The cell wall is freely permeable to water and substances in solutions but membrane is selectively permeable.

Q8. The C4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only _________ as much water as C3 plants for the same amount of C02 fixed.
Ans: Half

Q9. Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Explain.
Ans: The direction of movement in the phloem can be upwards or downwards, i.e. bi-directional. This contrasts with that of the xylem where the movement is always unidirectional, i.e. upwards. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction, as it is a source of sugar and works as a sink to use, store or remove the sugar

Q10. Identify the process occurring in I, II and III.

Ans: I—Uniport facilitated diffusion
II— Antiport facilitated diffusion
III— Symport facilitated diffusion

Q11. Given below is a table. Fill in the gaps.

	Property	Simple diffusion	Facilitated transport	Active transport
i.	Highly selective		Yes
ii.	Uphill transport			Yes
iii.	Requires ATP

Ans:

	Property	Simple diffusion	Facilitated transport	Active transport
i.	Highly selective	No	Yes	Yes
ii.	Uphill transport	No	No ,	Yes
iii.	Requires ATP	No	No	Yes

Q12. Define water potential and solute potential.
Ans: Water potential is considered as the potential energy of water. It is also taken as a measure of the difference between the potential energy in a given sample of wafer and pure water.
If a solute is dissolved in pure water, the solution is considered having fewer free water and hence the concentration of water decreases. Thus, all solutions have lower water potential than pure water. The magnitude of lowering in water potential due to dissolution of solute is called solute potential.

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