CBSE Sample Papers for Class 11 Biology Set 2 with Solutions

students must start practicing the questions from CBSE Sample Papers for Class 11 Biology with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Biology Set 2 with Solutions

Time Allowed : 3 hours
Maximum Marks: 70

General Instructions :

1. All questions are compulsory.
2. The question paper has four sections: Section A, Section B, Section C and Section D. There are 27 questions in the question paper.
3. Sedion-A has 5 questions ofl mark each. Section-B has 7 questions of 2 marks each. Section-C has 12 questions of 3 marks each and Section-D has 3 questions of 5 marks each.
4. 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.
5. Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

Question 1.
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
Answer:
(D) Crossing over
Explanation:
Crossing over does not occur in mitosis as there is no pairing of homologous chromosomes seen.

Question 2.
Which type of neurons are found in eyes?
(A) Multipolar.
(B) Bipolar.
(C) Unipolar.
(D) None of these.
Answer:
(B) Bipolar.

Question 3.
Gibberellins were named after a fungus called as……….
(A) Gibberella gaditijirrii
(B) Gibberella fujikuroi
(C) Gibberella Africana
(D) Gibberella acuminate
Answer:
(B) Gibberella fujikuroi

Question 4.
Directions (Qs. 4 & 5): In the below questions, the Assertions (A) and Reasons (R) have been put forward. Read both the statements and choose the correct option from the following:
Assertion (A): Chloride shift occurs from plasma to RBCs (erythrocytes) in human body.
Reason (R): It maintains ionic balance and electrochemical neutrality.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Answer:
(A) Both A and R are true and R is the correct explanation of A.
Explanation:
Chloride shift is the diffusion of Cl- ions from blood plasma into the RBCs is known as chloride shift. It occurs from plasma to RBCs (erythrocytes) in the human body to maintain ionic balance and electrochemical neutrality.

Question 5.
Assertion (A): Mitochondria help in cellular respiration.
Reason (R): Mitochondria have enzymes for dark reaction.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Answer:
(C) A is true but R is false.
Explanation:
Mitochondria help in cellular respiration by transferring energy from organic compounds to ATP. Chloroplast helps in photosynthesis. The dark reaction takes place in the stroma of the chloroplast.

Section – B

Question 6.
Distinguish between a genus and a species.
Answer:

Genus	Species
(i) It is the first highest category above the species level.	It is the basic unit of taxonomy.
(ii) It is a group of species which are closely related.	It is a dynamic genetically distinct group of organisms.

Question 7.
Which features make reptiles successful on land?
Answer:
Four features that make reptiles as true land animals are:

1. The amnion (embryonic membrane) encloses the embryo and provides it with a watery environment during development, therefore, the embryo does not need watery environment.
2. Internal fertilisation.
3. Shell around the egg to check desiccation.
4. Horny scales on the body of reptiles check for loss of water.

Question 8.
Describe the role of hemoglobin in the transport of respiratory gases.
Answer:
Role of hemoglobin in the transport of respiratory gases are:

1. Oxygen binds with hemoglobin to form oxyhemoglobin. 1 molecule of hemoglobin carries 4 molecules of oxygen.
2. CO₂ binds with hemoglobin to form carbamino hemoglobin. CO₂ is carried as carbamino hemoglobin by blood.

Question 9.
What is cholesterol? What diseases occur due to this?
Answer:
Cholesterol (C₂₇H₄₅OH) occurs in both free and combined form when it is esterified with a fatty acid. Cholesterol and its ester are insoluble in water. When its level rises in the blood, it tends to get deposited in the walls of arteries, known as atherosclerosis.

Question 10.
Describe the ultrastructure of mitochondria with a diagram.
Answer:
Ultrastructure:

1.  A mitochondrion is enclosed by a double membrane envelope, outer and inner membrane, which are separated by a narrow fluid-filled space called the outer compartment or peri-mitochondrial space.
2. The outer membrane is smooth and is permeable to small molecules.
3. The inner membrane is infolded into the matrix as incomplete partitions called cristae.
4. The cristae bear numerous small tennis racket-like particles called elementary particles, F₀ – F_x particles or oxysomes. Each oxysome has a head, a stalk and a base.
5. The enzymes of electron transport are located in the inner membrane in contact with elementary particles.
6. The F₀ – F_x combination functions as ATP synthetase.
7. The matrix posess possesses single circular DNA molecule, a few RNA molecules, 70S type ribosomes.
8. Mitochondria was first observed by Kolliker and the name mitochondria was given by C.Benda.

Question 11.
What are archaebacteria?
Answer:
(i) Some bacteria can survive in extreme environmental conditions like high temperature, high salt concentration, in absence of oxygen or in presence of oxygen, in high acidic or alkaline pH. Such bacteria are called archaebacteria.

(ii) Their cell wall consists of non-cellulosic polysaccharides or protein and lacks peptidoglycan. This allows them to survive in extreme conditions.

Question 12.
Mention the importance of trichomes and root hair for a plant.
Answer:
Trichomes are multicellular epidermal hairs on the stem, seeds and fruits. They help in protection, dispersal of seed and fruits and reduction of water loss. Root hair helps in absorbing water and mineral from soil.

Section – C

Question 13.
Briefly write the significance of mitosis in a multicellular organism.
Answer:
Significance of mitosis:
It is the only mode of multiplication in unicellular organisms. It is the process by which growth takes place in animals and plants by constantly adding more and more cells. Somatic cells are formed by mitosis. It maintains the genetic continuity and equality. This helps in proper coordination among different cells. It also plays a role in repair by growth, for example in wound healing, regeneration of damaged parts (as in the tail of lizard), and replacement of cells lost during normal wear and tear (as the surface cells of the skin or the red blood cells).

Question 14.
Compare the following:
(i) Central neural system (CNS) and Peripheral neural system (PNS)
(ii) Resting potential and action potential
Answer:
(i) The meristematic cells can be identified by the following features:

Central neural system	Action potential
(i) It is the first highest category above the species level.	It is the potential difference across nerve fiber when there is conduction of nerve impulse.
(ii) It is a group of species which are closely related.	In action potential the interior of axon is electropositive and the exterior is electro-negative.

(ii) Differences between Resting potential and Action potential.

Resting potential	Action potential
(i) It is the potential difference across the nerve fibre when there is no conduction of nerve impulse.	It is the potential difference across nerve fibre when there is conduction of nerve impulse.
(ii) In resting potential, the interior of the axon is electronegative and the exterior is electropositive.	In action potential the interior of axon is electropositive and the exterior is electro-negative.

Question 15.
Describe the structure of synovial joint.
Answer:

1. There is a membrane present called synovial membrane.
2. This membrane is composed of secretory epithelial cells which secrete a thick sticky fluid of the consistency of the white of an egg called synovial fluid.
3. It acts as a lubricant to the joint, provides nutrient materials for the structures within the joint cavity and helps to maintain the stability of the joint.
4. Bone ends are covered by articular cartilage.
5. Ligaments join the bones and tendons connect the bones with muscles.
6. Little sacs of synovial fluid or bursae are found in some joints.

Question 16.
What are the various types of nitrogenous bases found in DNA? Name the type of bond seen between
(i) two nitrogen bases of DNA.
(ii) phosphate and hydroxyl group of sugar of DNA.
Answer:
Types of nitrogenous bases found in DNA are:
(a) Adenine: Abbreviated A’, has a 2-ring structure, so that it is a purine.
(b) Thymine: Abbreviated ‘T’, is a pyrimidine, which means it has a 1-ring structure. It is only present in DNA, where it pairs with adenine.
(c) Guanine: Abbreviated ‘G’, is part of DNA, where it bonds with cytosine.
(d) Cytosine: abbreviated ‘C’, is part of DNA and bonds with guanine. It has one ring, so it’s a pyrimidine.

• Hydrogen bond
• Phosphodiester bond.

Question 17.
Endoparasites are found inside the host body. Mention the special structure, possessed by these and which enables them to survive in those conditions.
Answer:
Special characters present in by endoparasites are as follows:

1. Presence of adhesive organ for attachment to the host. Fasciola hepatica possess posterior sucker for the attachment. Taenia solium possess hook and suckers for the attachment within the host.
2. Presence of thick integument which is resistant to host s digestive enzymes and antitoxin.
3. Absence of locomotory organs.
4. Lack of digestive organs because digested and semi-digested food of the host is directly absorbed through their body surface.
5. They are generally hermaphrodites.

Question 18.
Plastids are the cell organelles which are found only in plant cells. Name any two types of plastids and mention the function of each.
Answer:
The plastids are flat, circular cytoplasmic cell organelles. Plastids are found only in plant cells and certain protists. Two types of plastids are:
(i) Chromoplasts: These are the coloured plastids which have the pigment other than the chlorophyll. These are found in the floral leaves and fruits imparting them different colours for attracting insects and animals for pollination.

(ii) Chloroplasts: These are the green plastids found in the green parts of the plants. These plastids are concerned with the manufacture of carbohydrates by photosynthesis.

Question 19.
What is the significance of hepatic portal system in the circulatory system?
Answer:
Hepatic portal system is a system which includes the hepatic portal vein that carries blood from alimentary canal and associated glands like pancreas, spleen etc., to the liver (which acts as an intermediate organ) before returning to the heart.

Significance of hepatic portal system is:
(i) Blood that comes from alimentary canal contains glucose, amino acids and other nutrients. Liver stores the excess of glucose as glycogen for later use during starvation when blood passes through hepatic portal system.

(ii) Harmful nitrogenous wastes like ammonia is converted into urea which is later removed by kidney.

(iii) Liver produce proteins which are transported through blood circulation.

Question 20.
What is abscisic acid and why it is called as stress hormone?
OR
Discuss the role of plant growth regulators in agriculture.
Answer:
Abscisic acid is a mildly acidic growth hormone, which functions as a general growth inhibitor by  counteracting other hormones or reactions mediated by them. Abscisic acid is also called as stress hormone because the synthesis of abscisic acid is stimulated by drought, water logging and other adverse  environmental conditions. It is produced in many parts of the plants but more abundantly inside the chloroplasts of green cells.
OR
(i) Parthenocarpy: The knowledge of the fact that application of auxins and gibberellins on unpollinated flowers results in the production of seedless fruits has been a great horticultural importance. It has been of great significance in such fruits where the number of seeds is very high and the seeds are of light food value (e.g., tomato, brinjal, guava, papaya, watermelon, orange) or the seeds are large in size (e.g., Litchi and Mango).

(ii) Flower thinning: This problem refers to fall of unpollinated or unfertilised flowers from certain fruit trees e.g., in pineapple and litchi. This problem has been solved to a great extent by spraying the plants with 2, 4-D and Naphthalene Acetic Acid (NAA) both of which are synthetic auxins.

(iii) Fruit ripening: Ethylene plays an important role in ripening of various fruits like oranges, lemons, grapes, banana, etc.

Question 21.
The gametophyte is a dominant phase in the life cycle of a bryophyte. Explain.
OR
Distinguish between intracellular and extracellular digestion?
Answer:
The gametophyte is a dominant phase in the life cycle of a bryophyte. It is more conspicuous, long-living, independent, green and freely branched whereas, the sporophytic phase is short-lived and dependent upon the gametophyte. The main plant body is haploid and bears sex organs i.e., antheridium and archegonium. Antheridium produces a large number of flagellated male gametes called antherozoids and archegonium is flask-shaped with tubular neck and swollen venter.

The gametophytic plant body of bryophytes is thalloid in liverworts whereas foliose in mosses. In liverworts, the thallus is differentiated into a dorsal photosynthetic and ventral storage region. In mosses, the gametophyte has two growth stages i.e., protonema stage and leafy stage or gametophore.
OR
Differences between intracellular and extracellular digestion are:

Intracellular digestion	Extracellular digestion
(i) In this, the digestive enzymes are poured into the food vacuoles, where digestion of food takes place	In this, the digestive enzymes are poured into the cavity for digestion to take place.
(ii) It occurs within the cells. e.g., Protozoans and Sponges	It occurs outside the cell in a cavity. e.g., Higher invertebrates and all vertebrates.

Question 22.
State the difference between protoxylem and metaxylem.
Answer:
Difference between protoxylem and metaxylem

Protoxylem	Metaxylem
(i) Cells are mostly thin-walled, rarely lignified	Cells are thick-walled.
(ii) Parenchyma cells are more in protoxylem.	Parenchyma cells are less in number
(iii) Formed much earlier in primary plant body.	Formed much later in primary plant body.

Question 23.
How many ATP are produced during cellular respiration during-Glycolysis, Oxidative decarboxylation and Kreb’s cycle? What is the net production of ATP?
Answer:
Complete aerobic breakdown of one molecule of hexose results in the release of ATP molecule.
(i) Glycolysis: Two molecules of NADPH are released during glycolysis. The glycolysis also yields 4 ATP molecules out of which 2 ATP molecules are consumed. Thus, the glycoysis contributes a total of 2 ATP molecule.

(ii) 2 molecules of NADH are produced in glycolysis and each molecule of NADH is equivalent to 2 ATT, so the net gain of ATP is 6 ATP

(iii) Oxidative decarboxylation: Two molecules of NADH are released during the oxidative decarboxylation and synthesis of acetyl CoA from pyruvic acid. Thus, number of ATP molecules will be 2 × 3 = 6. As each NADH is equivalent to 2 ATP in cytoplasm and 3 ATP in mitochondria. and oxidative decarboxylation occurs in mitochondria.

Kreb’s cycle:
Kreb’s cycle releases 6 molecules of NADH and 2 molecules of FADH2. The Kreb’s cycle also releases 2 ATP molecules, thus total number of ATP molecules produced during Kreb’s cycle are 2 + 2 × 2 + 6 × 3 = 24 molecules. Net production of ATP molecules is 36. i.e., 6+6+24 = 36

Question 24.
Differentiate between red algae and brown algae:
Answer:
Differences between red algae and brown algae:

Red Algae	Brown Algae
(i) It belongs to Rhodophyceae.	It belongs to the class phaeophyceae.
(ii) Phycoerythrin, phycocyanin and phycobilins pigments are present with chlorophyll ‘a’	Chlorophyll a, c and fucoxanthin are the major pigments.
(iii) Reserve food material is floridean starch.	Reserve food is laminarin and mannitol.
(iv) Unicellular and microscopic, only a few are filamentous and heterotrichous.	Unicellular, multicellular may be motile or flagellated.
Examples: Gelidium, Porphyra, Chondrus.	Examples: Laminaria, Sargassum & Fucus.

Section – D

Question 25.
Read the following and answer any four questions from (i) to (v) given below: 5×1=5
Adaptation for photosynthesis in tropical rain forests, the canopy is thick and shorter plants growing below it, receive filtered light. Shorter plants growing below the thick canopy are adapted to carry out the process of photosynthesis in low light intensity, which come down as filtered light. For this, the shade plants possess accessory pigments that can absorb green wavelength and then hand over the energy to chlorophyll a molecule for its photoconversion.

(i) The accessory pigments are …………
(A) lycopene and carotenoids
(B) xanthophylls and lycopene
(C) carotenoids and xanthophylls
(D) carotenoids, xanthophylls and lycopene
Answer:
(C) carotenoids and xanthophylls
Explanation:
Carotenoids and xanthophylls are accessory pigments.

(ii) range of wavelength (in nm) is called photosynthetically active radiation (PAR).
(A) 100-390
(B) 390 – 430
(C) 400 – 700
(D) 760-1000
Answer:
(C) 400 – 700

(iii) light is least effective in photosynthesis?
(A) Purple
(B) Green
(C) Red
(D) Yellow
Answer:
(B) Green
Explanation:
Carotenoids and xanthophylls are accessory pigments.

(iv) The accessory pigments not only absorb energy and transfer it to chlorophyll ‘a’ but also protect the
chlorophyll molecule from
(A) photo-oxidation
(B) photo-reduction
(C) photo-radiation
(D) photo-oxidation and photo-reduction
Answer:
(A) photo-oxidation

(v) If these plants lack chlorophyll ‘a1 and have a high concentration of chlorophyll b’, then this plant would:
(I) undergo photosynthesis.
(II) not undergo photosynthesis.
(III) gradually dies.
(IV) not show any change.
Choose from below the correct alternative.
(A) Only (I) is true.
(B) (I) and (IV) are true.
(C) (III) and (II) are true.
(D) (I) and (III) are true.
Answer:
(C) (III) and (II) are true.

Question 26.
What is blood dotting? Describe the process.
OR
Explain sliding filament theory of muscle contraction with neat sketches.
Answer:
When an injury is caused to a blood vessel bleeding starts which is stopped by a process called blood clotting or blood coagulation. This process can be described under three major steps.
(i) First step:
(a) At the site of an injury, the blood platelets disintegrate and release platelet factor-3 (= Platelet thromboplastin).
(b) Injured tissues also release thromboplastin.
(c) These two factors combine with calcium ions (Ca⁺⁺) and certain proteins in the blood plasma to form an enzyme called prothrombinase.

(ii) Second step:
(a) The prothrombinase inactivates heparin (or anti prothrombin anticoagulant) in the presence of calcium.
(b) Prothrombinase catalyses breakdown of prothrombin (inactive plasma protein) into an active protein thrombin.

(iii) Third step:
(a) Fibrins are formed by the conversion of inactive fibrinogens in the plasma by the enzyme thrombin.
(b) The fibres of fibrin form a dense network upon the wound and trap blood corpuscles to form a clot.
(c) The clot seals the wound and stops bleeding.

OR
Each muscle fibre has an alternate light and dark band, which contains a special contractile protein, called actin and myosin respectively. There is an elastic fibre called Z line that bisects each I-band.
The thin filament is firmly anchored to the Z line. The central part of the thick filament that is not overlapped by the thin filament is known as the H-zone. During muscle contraction, the myosin heads or cross bridges come in close contact with the thin filaments.

As a result, the thin filaments are pulled towards the middle of the sarcomere. The Z line attached to the actin filaments is also pulled leading to the shortening of the sarcomere. Hence, the length of the A-band remains constant as its original length and the I-band shortens and the H-zone disappears. Similar action in all the sarcomeres results in shortening of whole myofibril, and thereby the whole muscle fibre and muscle.

Question 27.
Describe the male reproductive organs in the frog with labeled sketch.
OR
The arrangement of veins and veinlets in the lamina of a leaf is called venation. The veins are not only the conducting channels for water, minerals and organic food, they also provide firmness to the lamina and keep it expanded. They give rise to lateral veins, which traverse the entire lamina. Venation is of two main types: Reticulate and parallel. When the veinlets form a network, the venation is termed as reticulate venation, for e.g., leaves of dicot plants. When the veins arising from mid rib or main veins, run parallel to each other towards the margin or the apex of the lamina, venation is termed as parallel venation, present in the leaves of monocot plants.

Parallel venation is of two sub-types: Pinnate of palmate type. When lamina has a single prominent vein or midrib running from the base to the apex of lamina, e.g., Banana, Canna, etc., it is called pinnate when lamina has several principal veins arising from the base and running towards the apex or margin of the lamina, it is called palmate type. 5×1=5
Answer:
Male reproductive organs consist of a pair of the yellowish testis that is attached to the upper part of the kidneys by a double fold of peritoneum called mesorchium.

(i) Venation is a term used to describe the pattern of arrangement of:
(A) Floral organs.
(B) Flower in inflorescence.
(C) Veins and veinlets in a lamina.
(D) All of them.
Answer:
(C) Veins and veinlets in a lamina.
Explanation:
Venation is the distribution of veins in lamina. They contain xylem and phloem in them and therefore supply water, nutrient and food and support to the lamina.

(ii) If leaf has more than one prominent vein it is said to be:
(A) Unicostate.
(B) Multicostate.
(C) Pinnate.
(D) Palmate.
Answer:
(B) Multicostate.

(iii) Parallel venation is not a characteristic in:
(A) Hibiscus.
(B) Solatium.
(C) Rice.
(D) Maize.
Answer:
(A) Hibiscus.
Explanation:
Parallel venation is characteristics of all monocots with only a few exceptions.

(iv) Reticulate venation is found is:
(A) Canna.
(B) Mango.
(C) Musa.
(D) Rice.
Answer:
(B) Mango.
Explanation:
Mango is a dicot plant. Reticulate venation is characteristic of all dicotyledonous plants with a few exceptions.

(v) Smilax leaves show …………… venation while leaves of ……………… show parallel venation:
(A) Parallel, Smilax
(B) Musa, Parallel
(C) Reticulate, Solarium
(D) None of these
Answer:
(C) Reticulate, Solarium
Explanation:
Smilax leaves show reticulate venation. Leaves of Solanum show parallel venation.