Students can access the CBSE Sample Papers for Class 11 Biology with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Biology Set 3 with Solutions
Time : 3 Hours
Maximum Marks : 70
General Instructions:
- All questions are compulsory.
- The question paper has five sections and 33 questions. All questions are compulsory.
- Section-A has 16 questions ofl mark each; Section-B has 5 questions of 2 marks each; Section- C has 7 questions of 3 marks each; Section- Dhas 2 case-based questions of 4 marks each; and Section-E has 3 questions of 5 marks each.
- There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
- Wherever necessary, neat and properly labeled diagrams should be drawn.
Section – A
Question 1.
Fusion of two gametes which are dissimilar in size is termed as: 1
(A) Oogamy
(B) Isogamy
(C) Anisogamy
(D) Zoogamy
Answer:
Option (C) is correct.
Explanation: The fusion of two gametes with different sizes is known as anisogamy. The fusion of gametes that are similar in size is known as isogamy.
Question 2.
Which of the following pairs of animals has non-glandular skin? 1
(A) Snake and frog
(B) Chameleon and turtle
(C) Frog and pigeon
(D) Crocodile and tiger
Answer:
Option (B) is correct.
Explanation: The chameleon and turtle are members of the class Reptilia and have scales on their dry, non-glandular skin. The frog, pigeon, and tiger have modified skin in response to ecological adaptations.
Question 3.
Match the following and choose correct option. 1
Group A | Group B |
(a) Aleurone layer | (i) Without fertilisation |
(b) Parthenocarpic fruit | (ii) Nutrition |
(c) Ovule | (iii) Double fertilisation |
(d) Endosperm | (iv) Seed |
(A) (a) – (i), (b) – (ii), (c) – (iii), (d) – (iv)
(B) (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)
(C) (a) – (iv), (b) – (ii), (c) – (i), (d) – (iii)
(D) (a) – (ii), (b) – (iv), (c) – (i), (d) – (iii)
Answer:
Option (B) is correct.
Explanation: A layer of protein storage cells called the aleurone layer is found in grain and maize seeds and is a type of preserved food.
• A parthenocarpic fruit is one in which the ovary is induced to produce fruit without fertilizationas in grapes with fewer seeds and watermelon.
Ovule: Ovule develops into seed.
Endosperm: During double fertilisation process, PEN matures into endosperm, which feeds the growing embryo with nutrients.
Question 4.
Meiosis occurs in organisms during: 1
(A) Sexual reproduction
(B) Vegetative reproduction
(C) Both sexual and vegetative reproduction
(D) None of the above
Answer:
Option (A) is correct.
Explanation: Meiosis occurs in organisms during the sexual reproduction. During the process, the gametes fuse to form a single cell. Meiosis reduces the ploidy by halving it in order to preserve the same number of chromosomes in the new zygote as in the parent cell.
Question 5.
The enzyme that is not found in a C3 plant is: 1
(A) RuBP carboxylase
(B) PEP carboxylase
(C) NADP reductase
(D) ATP synthase
Answer:
Option (B) is correct.
Explanation: The mesophyll cells of C4 plants contain phosphoenolpyruvate (PEP) carboxylase, which catalyzes the conversion of PEP to oxaloacetate.
Question 6.
Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition, it form: 1
(A) Lactic acid
(B) CO2 + H2O
(C) Acetyl CoA + CO2
(D) Ethanol + CO2
Answer:
Option (C) is correct.
Explanation: Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition, it forms acetyl CoA + CO2.
Question 7.
When CO2 is added to PEP the first stable product synthesised is: 1
(A) Pyruvate
(B) Glyceraldehyde-3-phosphate
(C) Phosphoglycerate
(D) Oxaloacetate
Answer:
Option (D) is correct.
Explanation: Mesophyll cells in C4 plants are where the initial fixation of CO2 or carboxylation takes place. PEP carboxylase, also known as PEP case, is an enzyme that is present in the chloroplasts of mesophyll cells and is used to fix CO2 first. Phosphoenol pyruvate, or PEP, is the main acceptor of CO2. In the presence of PEP carboxylase. It reacts with CO2 to create oxaloacetic acid or oxaloacetate (OAA). So, “oxaloacetate” is the right response.
Question 8.
In the following flow chart, replace the symbols a,b,c and d with appropriate terms. 1
(i) Glyceraldehyde 3 phosphate Glucose
(ii) Phosphoenol pyruvic acid
(iii) Ethanol
(iv) Lactic acid
(A) (a) – (i), (b) – (ii), (c) – (iii), (d) – (iv).
(B) (a) – (ii), (b) – (i), (c) – (iii), (d) – (iv).
(C) (a) – (iv), (b) – (i), (c) – (iii), (d) – (ii).
(D) (a) – (iii), (b) – (i), (c) – (ii), (d) – (iv).
Answer:
Option (A) is correct.
Explanation:
Question 9.
Which one of the following types of cells lack nucleus? 1
(A) Erythrocytes
(B) Neutrophils
(C) Eosinophils
(D) Monocytes
Answer:
Option (A) is correct.
Explanation: Erythrocytes in humans and other mammals lack nucleus and the majority of cell organelles, which gives them maximum space to contain haemoglobin and allows them to transport oxygen gas to their full potential. Additionally, the absence of a nucleus gives RBCs their distinctive biconcave shape, which enhances the cell’s surface area and allows the storage of more haemoglobin. Erythrocytes with greater surface area promotes a more efficient exchange of oxygen and carbon dioxide.
Question 10.
In the diagram given below, replace the symbols i,ii,iii and iv with appropriate terms. 1
(1) Diaphragm
(2) Trachea
(3) Alveoli
(4) Pleural membranes
(A) (1) – (i), (2) – (iv), (3) – (ii),(4) – (iii).
(B) (1) – (iv), (2) – (i), (3) – (iii), (4) – (ii).
(C) (1) – (iii), (2) – (ii), (3) – (iv), (4) – (i).
(D) (1) – (iv), (2) – (iii), (3) – (ii), (4) – (i).
Answer:
Option (B) is correct.
Explanation:
Question 11.
Which one of the following statements is incorrect? 1
(A) The medullary zone of the 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 the 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 the kidney.
Answer:
Option (B) is correct.
Explanation: The renal columns are connective tissue extensions that radiate downward from the cortex into the medulla to divide the renal pyramids and renal papillae, which are the most distinguishing features of the medulla.
Question 12.
The following conclusion is drawn from the graph given below. 1
(i) Anaerobic respiration provides more energy.
(ii) Aerobic respiration provides more energy.
(iii) In aerobic respiration, organic food is completely oxidised.
(iv) In anaerobic respiration, organic food is completely oxidised.
Choose from below the correct alternative.
(A) Only (i) is true.
(B) (i) and (iv) are true.
(C) (ii) and (iii) are true.
(D) (i) and (iii) are true.
Answer:
Option (C) is correct.
Explanation: By completely oxidising organic compounds in the presence of oxygen, aerobic respiration produces CO2, water, and a significant quantity of energy in the form of ATP’s high-energy bonds. Hence, aerobic respiration provides more energy and in aerobic respiration, organic food is completely oxidised.
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer the questions selecting the appropriate option given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Question 13.
Assertion (A): Taxon and category are the same things. 1
Reason (R): Category shows hierarchical classification of an organism.
Answer:
Option (D) is correct.
Explanation: Taxon and category are different things. A category is a rank or level in the hierarchical classification of an organism while taxon is a unit in classification which may represent any level of grouping of organisms based on certain common characteristics.
Question 14.
Assertion (A): Nucleolus is smaller in actively working cell. 1
Reason (R): It is involved in actively synthesising DNA that ultimately results in protein formation.
Answer:
Option (D) is correct.
Explanation: Nucleolus is larger in actively working cells as it is involved in actively synthesising RNA that ultimately results in protein formation.
Question 15.
Assertion (A): Conduction of impulses is faster in non-myelinated nerve fibre. 1
Reason (R): In myelinated nerve, the action potential is conducted from one node to another.
Answer:
Option (D) is correct.
Explanatiomln a myelinated neuron fibre, conduction of impulses is faster as the action potential is conducted from one node to another.
Question 16.
Assertion (A): The statement “Omnis cellula-e-cellula” was given by Schleiden. 1
Reason (R): It means “All cells come from pre-existing cells.
Answer:
Option (D) is correct.
Explanation: Rudolf Virchow gave the statement “Omnis cellula-e-cellula”. It means that all cells come from pre-existing cells.
Section – B
Question 17.
What do you mean by taxonomic hierarchy? Give classification of mango up to species. 2
Answer:
The taxonomic hierarchy is a systematic framework in classification in which taxonomic groups are arranged in a definite order, from higher to lower categories. Each category is considered as a taxonomic unit and represents a taxon.
Classification of Mango:
Kingdom: Plantae.
Division: Angiospermae.
Class: Dicotyledonae.
Order: Sapindales.
Family: Anacardiaceae.
Genus: Mangifera.
Species: indica.
Question 18.
The diagram given below shows a typical angiosperm flower. 2
(a) Observe and labelled the given diagram properly.
(b) A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.
Answer:
(b) Floral whorls in an angiospermic flower from outside to inside are as follows-
Calyx, corolla, androecium and gynoecium
Question 19.
The given graph indicate activation energy. Describe the activation energy? 2
Answer:
Any chemical reaction includes formation of transition state between reactant and product. Formation of transition state requires high energy content of reactant molecules for alignment of reacting groups, formation of transient unstable charges, rearrangements of chemical bonds, etc. The energy required for formation of transition state is termed as activation energy. Enzymes enhance the rate of reaction by lowering down the activation energy of transition state. ln given graph, energy of substrate is higher than that of product reflecting the endothermic nature of reaction.
Question 20
The image below shows the result of synapsis and bivalent. 2
(a) Describe synapsis and where does synapsis occur?
(b) Describe bivalent and where does bivalent occur?
Answer:
(a) Synapsis- It is the pairing of homologous chromosomes during cell division. It occurs in zygotene stage of prophase I.
(b) Bivalent- The complex formed by a pair of synapsed homologous chromosomes is called a bivalent. lt is also known as a tetrad. lt occurs in the zygotene stage of prophase I.
Question 21.
Blood of human beings differ in certain aspects although it appears same in all individuals. Two main types of grouping are ABO and Rh. ABO grouping is based on presence or absence of two surface antigens on RBC, antigen A and antigen B. The plasma of an individual also contains two antibodies produced in response of antigens. 2
(a) Complete the given table by filling the space provided.
Blood Group | Antigens on RBCs | Antibodies in Plasma | Donor’s Group |
A | A | anti-B | A, O |
B | B | anti-A | B, O |
AB | A, B | nil | AB, A, B, O |
O | Nil | anti-A, B | O |
(b) Give the names of the blood groups which are universal donor and universal acceptor. Give a reason for your observations.
OR
From which cells do platelets originate? What is their life span? How do they act when blood vessels injured ?
Answer:
(a)
TA
(b) Universal donor blood group is O and universal acceptor blood group is AB.
OR
• Platelets are produced from megakaryocytes present in the bone marrow.
The life span of platelets is 3 to 7 days only. When an injury is caused the blood platelets disintegrate and release certain chemical called the platelet factors which help in the clotting of blood.
Section – C
Question 22.
Observe the figure given below X and Y and answer the following questions. 3
(a) Why bryophytes have thallus-like plant body?
(b) Identify the X and Y ? And also describe the liverworts and mosses ?
Answer:
(a) X is the liverwort- female thallus and Y is the liverwort- male thallus.
(i) Liverworts- The liverworts grow in moist and shady places like banks of water bodies, marshy ground, damp soil, bark of trees, etc. Their plant body is thalloid which is dorsiventral and closely appressed to the substrate. Some of the liverworts are leafy. Such forms have tiny leaf-like appendages in two rows on the stem-like structures.
(ii) Mosses- The life cycle of a moss involve both gametophytic phase and sporophytic phase, but the gametophytic phase is predominant one. Gametophytic phase consists of two stages – the first stage is the protonema stage and the second stage is leafy stage.
(b) Since bryophytes posses root like, stem like, and leaf like structures but they are not true forms of root, stem and leaf, so their body is called thallus.
Question 23.
“Green algae are ancestors of land plants”. Comment upon the statement. 3
Answer:
The various evidences which favour the chlorophycean (green algae) origin of land plants are-
- Both green algae and land plants possess the same type of chlorophylls, a and b.
- The carotenoid pigments are similar in the two group.
- Cell wall contain similar cellulose and pectic compounds in both the groups.
- Starch is the common storage carbohydrate in the two group. It is made of both amylose and amylopectin fractions.
- The flagella are similar in the motile forms of the two. (Any three)
Question 24.
Placentation refers to the arrangement of ovules within the ovary of a flower. 3
Differentiate between:
(a) Marginal and axile placentation
(b) Marginal and parietal placentation.
Answer:
(a) (Any three)
S. No | Marginal placentation | Axile placentation |
(i) | Ovary is always unilocular. | Ovary is two or more locular. |
(ii) | Ovary is simple or monocarpellary. | Ovary is compound and syncarpous. |
(iii) | Ovules is attached to wall of ovary. | Ovules is attached to central/axile column. |
(iv) | The ovule is found in a single file. | The number of ovule files are based on the number of fusing carpels or septa. |
(b)
S. No | Marginal placentation | Parietal placentation |
(i) | Found in monocarpellary or simple ovary. | Occurs in compound or syncarpous ovary. |
(ii) | A single longitudinal placenta or file of ovules are attached to the ovary wall. | Two or more longitudinal placentae or files of ovules are attached to the ovary wall. |
(iii) | Ovary is always unilocular. | Parietal placentation |
Question 25.
Study the following graph showing the effect of substrate concentration on the rate of enzyme activity and answer the questions that follow. 3
(a) Labelled the graph properly. What does A, B and C represent in the graph?
(b) Why is there no further increase in the velocity of enzyme action with the addition of substrate?
Answer:
(b) The reaction ultimately reaches a maximum velocity which is not exceeded by any further rise in the concentration of the substrate. Reason- The enzyme molecules are fewer than substrate molecules and after saturation of these molecules, there are no free enzymes to bind with the additional substrate molecules.
Question 26.
Explain the process of ATP formation during aerobic respiration (in mitochondria). 3
Answer:
The synthesis of ATP from ADP and inorganic phosphate using energy from a proton gradient is called oxidative phosphorylation. This takes place in elementary particles present on the inner membrane of cristae of mitochondria. This process in mitochondria is catalysed by ATP synthase. This complex has two major components- F0 and F1. F0 acts as channel for proton and F1 acts as on ATP synthase.
Question 27.
Explain why the following things happen? 3
(a) The more oxygen is released from oxyhaemoglobin in a more active tissue than in a less active one.
(b) Oxygenation of blood promotes the release of carbon dioxide from the blood in the lungs.
(c) Oxygen leaves the blood from tissue capillaries, but carbon dioxide enters the blood in tissue capillaries.
OR
Blood is specialised kind of living fluid connective tissue of opaque red colour of alkaline nature and salty in taste. It specific gravity is 1.050-1.060. The blood contains a fluid part – the plasma and the solid part the corpuscles.
(a) (i) What is the definition of plasma?
(ii) Explain the serum.
(b) Does it make any difference to have the haemoglobin in the corpuscles rather than in plasma? Explain.
Answer:
(a) The more oxygen is released from oxyhaemoglobin in a more active tissue because its partial oxygen pressure is lower than the least active tissue. The lower pO2 in the active tissue causes the dissociation of oxyhaemoglobin to release sufficient oxygen required by the tissues.
(b) The CO2 from the tissue is carried in the blood in three different forms; bicarbonate in plasma and erythrocytes, carbamino-haemoglobin in erythrocytes and small amounts of dissolved carbon dioxide in plasma. On reaching the lungs, blood is oxygenated. It donates H+ which joins bicarbonate (HCO3–) to form carbonic acid. This carbonic acid cleaves into H2O and CO2 by carbonic anhydrase. In this way, CO2 is released from the carbaminohaemoglobin. The oxygen affinity of haemoglobin also gets enhanced with the fall in the blood pCO2resulting from the elimination of CO2 from the blood in the lungs.
(c) The blood in the tissue capillaries contains higher pO2 than the tissue fluid. So, oxygen is released from oxyhaemoglobin and diffused from the capillary blood to the tissue fluid and finally to the cells to the tissue fluid to raise its pCO2 than the capillary blood. This enables carbon dioxide to diffuse from the tissue fluid to the capillary blood.
OR
(a) (i) Plasma is a pale straw-coloured fluid occupying about one half of the total blood volume. It has about 90 – 92 percent water and 8 to 10 percent proteins.
(ii) Serum is the name given to blood plasma which has its protein fibrinogen removed. In this form the plasma cannot clot, so it can be stored in hospital blood banks for transfusions in emergencies.
(b) There is great difference in respects of efficiency of carrying the oxygen from respiratory organs to the body tissue.
Haemoglobin in corpuscles | Haemoglobin in plasma |
As RBC are small, rounded and in more in quantity, so haemoglobin is exposed with large combined surface area to absorb O2. | In plasma, the exposed surface area for haemoglobin is very limited in comparison to the RBC, the result is the absorption of O2 in less amount. |
When RBC pass through the small capillaries of respiratory organs, one by one, they have ample time and surface, to absorb oxygen. | The haemoglobin dissolved in plasma has lesser time as it pass quickly, (being liquid) through the wall of capillaries. |
Question 28.
Describe glomerular filtration in human nephrons. 3
Answer:
The glomerular capillaries are narrower than the afferent renal arterioles, therefore the blood pressure in the glomerular capillaries becomes very high and there is a continuous process of ultrafiltration through the semi-permeable glomerular capillaries.
Filtration of blood occurs through three layers.
- The endothelium of glomerular blood vessels.
- The epithelium of Bowman’s capsule and
- The basement membrane between the two layers. The epithelial cells of Bowman’s capsule called podocytes are arranged intricately so as to leave minute spaces called filtration slits. The glomerular filtrate and blood plasma are similar except that glomerular filtrate does not have proteins and fats.
Section – D
Question 29.
Mycorrhiza is a symbiotic association between fungi and roots of higher plants. 4
(a) How do fungi form partnership with most plants?
OR
(b) Give the scientific name of a species of fungus which:
(i) produces a plant disease?
(ii) is used in the manufacture of ethanol?
(c) Explain the following in terms of mycorrhizal association: Nourishment, absorption of water, and growth promotion.
Answer:
(a) Some fungi have evolved essential relationships with the roots of many living plants. This partnership is called mycorrhizal symbiosis. In this association fungal hyphae are found in the roots of higher plants. Both members are benefitted from this association, that’s why it is called symbiosis or mutualism.
OR
(b) (i) Phytophthora infestans causes a disease known as late blight of potato.
(ii) Saccharomyces cerevisiae is used in the production of ethanol.
(c) Nourishment: In orchids, the fungi absorb nourishment from outside, transports it over to the germinating seed.
Absorption of water: Fungal association increases the water availability to the root.
Growth promotion: Fungus produces growth- promoting hormones and provides it to the plant.
Question 30.
The important functions of roots are: fixation of plant in the soil i.e., ground, absorption of nutrients and water from soil and conduction of absorbed materials from soil to aerial parts of the plant. In addition to the above functions, some adventitious roots perform different function i.e., in Cuscuta (a parasitic plant) they absorb food from the host’s body; in banyan; the prop roots provide support to the plant, in maize, Rhizophora, they support the plant; in Tinospora, the green roots perform the function of photosynthesis; in some plants they get swollen and perform as storage organs for the plant; other perform the function of vegetative reproduction. Some roots perform the functions of storage of food reproduction, climbing, giving support to the plant. 4
Observe the structure given below and answer the following questions.
(a) Identify he type of modified root.
(b) Roots developed from parts other than radicle are called …………… .
(c) Tap roots are seen in ………………. .
OR
Morphology is the branch of science concerned with the study of organisms, structure, characteristics, and forms. The flowering plants have a wide range of structural diversity that shares several common characteristics. The wide range in the structure of higher plants will never fail to fascinate even though the angiosperms show such a large diversity in external structure or morphology. They are all characterised by the presence of roots, stems, leaves, flowers, and fruits. These five similar traits may be found in the morphology of all angiosperms. The plants may or may not have flowers and fruits. Flowering plants known as angiosperms are plants that produce flowers. The flower is the reproductive unit in the angiosperms. The flower has four different kinds of whorls arranged successfully on the end of the pedicle called the thalamus. These are calyx, corolla, androecium, and gynoecium. The calyx and corolla are accessory organs while androecium and gynoecium are the reproductive organs. The manner in which sepals or petals are organised in a floral bud with reference to other parts of the same whorl is called aestivation.
Teacher showed the chart of aestivation to students and asked them to name the types A and B Identify (A) and (B)? How will you distinguish A and B ?
Answer:
(a) The given figure is of fasciculated root, which is the modified adventitious root that stores food material.
(b) Roots which develop from parts other than radicle or which arise from an organ other than the root usually a stem or some time leaf are called adventitious roots. Root arising from the radicle is the tap root.
(c) Normally, roots grow beneath the ground. But in some plants, the stem is modified into a tuber like structure for the storage of reserve food material. These stem tubers develop and grow under the ground. Potato is a stem because it has leaves, node, buds, etc.
OR
(a) – Valvate aestivation
(b) – Twisted aestivation
Valvate aestivation: Margins of sepals or petals remain either in contact or lie close to each other but do not overlap, e.g., Calyx of Datura, Calotropis.
Twisted aestivation: Margins of each sepal or petal is directed inwards and is overlapped. While the other margin is directed outwards and overlap the margin of adjacent, e.g., Corolla of China rose, cotton, etc.
Section – E
Question 31.
Ribosomes are granular organelles which are not enclosed by any membrane. And these are found either in a free state in the cytoplasm or attached to the endoplasmic reticulum. 5
(a) What are plastids?
(b) (i) What are the three structural elements of endoplasmic reticulum ?
(ii) Explain cisternae, vesicles and tubules with diagram.
(c) How many types of plastids are classified? Explain all types.
OR
(a) Describe the structure of nucleus with the help of labelled diagrams.
(b) Describe the structure of centrosome with the help of labelled diagrams.
(c) Centromere is the constriction present on the chromosomes, where chromatids are held together.
(i) How does the position of the centromere form the basis of classification of chromosomes?
(ii) Support the above answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
(a) Plastids are double-membrane organelles which are found in the cells of plants and algae. Plastids are responsible for manufacturing and storing of food. These often contain pigments that are used in photosynthesis and different types of pigments that can change the colour of the cell.
(i) There are three structural elements of ER 1. cisternae, 2. vesicles, 3. tubules
(ii) Cisternae- these are the broad flat membrane bound spaces arranged parallel to each other.
vesicles- these are spherical bodies inside it.
Tubules- These are narrow and branched tubular spaces which synthesis steroids.
(c) Depending upon their colour, plastids are of three types.
(i) Leucoplast: They are colourless plastids of varied shapes and sizes with stored nutrients. Leucoplast is of three types: starch containing amyloplast, fats- storing elaioplast and proteins bearing proteinoplast. Grana and photosynthetic pigments are absent in leucoplast.
(ii) Chromoplast: These are coloured plastids with reddish and yellow colours because of the presence of carotenoid pigments. The pigments are often found in crystallised state so that the shape of the plastids can be like needles, spindles or irregular. These provide colour to many flowers for attracting pollinating insects. They are also the site of synthesis of membrane lipids.
(iii) Chloroplast: They are green plastids which possess photosynthetic pigments, chlorophyll and carotenoids, and take part in the synthesis of food from inorganic raw materials in the presence of radiation energy.
OR
(a) Structure of a nucleus is as follows-
(b)
- Centrosome is an organelle usually containing two cylindrical structures called centrioles.
- They are surrounded by amorphous pericentriolar materials.
- Both the centrioles in a centrosome lie perpendicular to each other in which each has an organisation like the cart wheel.
- They are made up of nine evenly-spaced peripheral fibrils of tubulin.
- Each of the peripheral fibril is a triplet. The adjacent triplets are also linked.
- The central part of the centriole is also proteinaceous and called hub, which is connected with tubules of the peripheral triplets by radial spokes made of protein.
- The centrioles form the basal body of the cilia or flagella, and the spindle fibres that give rise to spindle apparatus during cell division in animal cells.
(c) (i) Chromosomes are divided into four types based on the position of centromere.
- Metacentric chromosome: The chromosome in which the centromere is present in the middle and divides the chromosome into two equal arms is known as a metacentric chromosome.
- Sub-metacentric chromosome: The chromosome in which the centromere is slightly away from the middle region is known as a sub-metacentric chromosome. In this, one arm is slightly longer than the other.
- Acrocentric chromosome: The chromosome in which the centromere is located close to one of the terminal ends is known as an acrocentric chromosome. One arm is extremely long and the other is extremely short.
- Telocentric chromosome: The chromosome in which the centromere is located at one of the terminal ends is known as a telocentric chromosome.
Question 32.
Explain the chemiosmotic hypothesis briefly? 5
OR
Why is the RuBisCO enzyme more appropriately called RuBP carboxylase-oxygenase and what important role does it play in photosynthesis?
Answer:
This hypothesis was proposed by Mitchell in 1961. ATP synthesis is linked to the development of proton gradient across the membrane of thylakoids and mitochondria. The process involved in the development of proton gradient across the membrane are:
(i) Splitting of water molecules occurs inside the thylakoid to produce hydrogen ion or proton.
(ii) As electron passes through the photosystem, protons are transported across the membrane because primary acceptor of electron is located towards the outer side of the membrane.
(iii) The NADP reductase enzyme is located on the stroma side of the membrane. Electrons come out from the acceptor of electrons of PS I. Protons are necessary for reduction of NADP+ to NADPH + H+. These protons are also removed from the stroma. This creates proton gradient across the thylakoid’s membrane along with pH in the lumen.
(iv) Gradient is broken down due to the movement of protons across the membrane to the stroma through trans-membrane channel of F0 of ATPase. One part of this enzyme is embedded in membrane to form trans-membrane channel. The other portion is called F1 that protrudes on the outer surface of thylakoid membrane which makes the energy-packed ATP.
(v) ATP and NADPH produced due to the movement of electron is used immediately to fix CO2 to form sugar.
OR
RuBisCO or RuBP carboxylase-oxygenase has dual nature. It has an affinity for both carbon dioxide and oxygen. But it has more affinity for CO2 than O2. Thus,
it is concentrations of the two, which determines which of the two will bind to the enzyme. In normal conditions, when CO2 and O2 concentration are normal, it acts as carboxylase. As carboxylase, it fixes CO2 by combining with ribulose bisphosphate and C3 cycle operates normally, producing glucose molecules as by-product of photosynthesis. However, when O2 concentration goes up and CO2 goes down, it acts as oxygenase. The process of photorespiration starts where RuBP binds with CO2 to form phosphoglycolate and phosphoglyceric acid. C4 plants have the mechanisms to increase the concentration of CO2 at enzyme site and thereby increases the intracellular concentration of CO2. Thus, here RuBisCo acts as carboxylase, minimising the effect of oxygenase.
Question 33.
Certain characteristics are provided in this table. Explain each characteristics briefly. 5
Aspects | Red Muscle | White Muscle |
Structure | ||
Color | ||
Function | ||
Fatigue resistance | ||
Energy metabolism | ||
Contraction speed | ||
Distribution | ||
Diameter | ||
Example |
OR
Certain characteristics are provided in this table. Explain each characteristics briefly.
Feature | Ammonotelism | Ureotelism | Uricotelism |
Main excretory product | |||
Toxicity | |||
Solubility | |||
Water loss | |||
Habitats | |||
Energy cost | |||
Excretion | |||
Advantage | |||
Disadvantage | |||
Examples |
Answer:
Aspects | Red muscle | White muscle |
Structure | Rich blood supply, numerous capillaries | Lower blood supply, fewer capillaries |
Color | Red appearance due to high myoglobin concentration | Pale appearance due to lower myoglobin concentration |
Function | Endurance, sustained contractions | Short bursts of intense power, strength |
Fatigue resistance | High | Low |
Energy metabolism | Aerobic metabolism (oxidative phosphorylation) | Anaerobic metabolism (glycogen breakdown) |
Contraction speed | Slower | Faster |
Distribution | Postural muscles | Muscles involved in rapid, powerful movements |
Diameter | Smaller diameter | Larger diameter |
Example | Extensor muscle of the human back. | Eye ball muscles. |
OR
Feature | Ammonotelism | Ureotelism | Uricotelism |
Main excretory Product | Ammonia | Urea | Uric Acid |
Toxicity | Highly toxic | Moderately toxic | Non-toxic |
Solubility | Highly soluble | Highly soluble | Insoluble |
Water loss | High | Moderate | Low |
Habitats | Aquatic animals, most invertebrates | Terrestrial animals, some aquatic animals | Terrestrial animals, reptiles, birds, insects |
Energy cost | Low | Moderate | High |
Excretion | Through gills, body surface, or specialised excretory organs | Primarily through kidneys | Primarily through Malpighian tubules in insects |
Advantage | Efficient in water; minimal energy expenditure | Moderate energy cost; conserves water | Efficient in arid environments; minimal water loss |
Disadvantage | High water loss; requires large amounts of water | Moderate energy cost; requires less water | High energy cost; requires more water |
Examples | Most fishes, aquatic invertebrates, larvae of amphibians | Mammals, adult amphibians, sharks, some bony fishes | Reptiles, birds, insects, land snails |