CBSE Sample Papers for Class 11 Biology Set-5

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Time : 3 Hours
Maximum Marks : 70

General Instructions:

1. All questions are compulsory.
2. The question paper has five sections and 33 questions. All questions are compulsory.
3. Section – A has 16 questions of 1 mark each; Section – B has 5 questions of 2 marks each; Section – C has 7 questions of 3 marks each; Section- D has 2 case – based questions of 4 marks each; and Section – E 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 labeled diagrams should be drawn.

Section – A

Question 1.
A person breathes in some volume of air by forced inspiration after having a forced expiration. This quantity of air
taken in is: (1)
(A) Total lung capacity
(B) Tidal volume
(C) Vital capacity
(D) Inspiratory capacity
Answer:
Option (C) is correct.
Explanation: Vital capacity is the maximum volume of air a person can breathe in after a forced expiration. VC = ERV + TV + IRV here, EVR is Expiratory reserve volume, TV is Tidal volume, and IRV is Inspiratory Reserve volume.

Question 2.
Ethylene is used for (1)
(A) Retarding ripening of tomatoes
(B) Hastening of ripening of fruits
(C) Slowing down the ripening of apples
(D) Both (B) and (C)
Answer:
Option (B) is correct.
Explanation: Ethylene is gaseous hormone which is i also called fruit ripening hormone. It stimulate the biochemical changes for fruit ripening.

Question 3.
Match the following and choose the 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.

Group A	Group B
(a) Aleurone layer	(ii) Nutrition
(b) Parthenocarpic fruit	(i) Without fertilisation
(c) Ovule	(iv) Seed
(d) Endosperm	(iii) Double fertilisation

Question 4.
The role of ABA is antagonistic to: (1)
(A) Ethylene
(B) Gibberellic acid
(C) Cytokinin
(D) IAA
Answer:
Option (B) is correct.
Explanation: Gibberellic acid is antagonotstic to ABA because of some anti characteristics such as
(i) ABA inhibits growth of the plant While gibberellin promotes growth.
(ii) ABA promotes Leaf senescence while gibberrelin prevent leaf senescence.

Question 5.
Potential difference across resting membrane is negatively charged. This is due to differential distribution of the
following ions. (1)
(A) Na⁺ and K⁺ ions
(B) CO₃ and Cl^– ions
(C) Ca⁺⁺ and Mg⁺⁺ ions
(D) Ca⁴⁺and Cl^– ions.
Answer:
Option (A) is correct.
Explanation: In a neuron the potential difference across resting membrane maintained by sodium and potassium ions. The ion channels control the movement of sodium and potassium ions in both inner and outer side of a neuron.

Question 6.
Corpus luteum secretes a hormone called: (1)
(A) Prolactin
(B) Progesterone
(C) Aldosterone
(D) Testosterone
Answer:
Option (B) is correct.
Explanation: progesterone is secreted by ovaries as corpus leutem is part of an ovary so the hormone is progesterone.

Question 7.
A fully grown tadpole larva of a frog respires through: (1)
(A) Gills
(B) Skin
(C) Lungs
(D) Tail fin
Answer:
Option (A) is correct.
Explanation: Frog is an amphibian and shows metamorphosis . At tadpole stage its respire through gills in water.

Question 8.
Identify the phylum among the following, in which adults exhibit this type of symmetry but, their larvae exhibit
different symmetry. (1)

(A) Porifera
(B) Coelenterata
(C) Aschelminthes
(D) Echinodermata
Answer:
Option (D) is correct.
Explanation: Echinodermates adult have radial symmetry while their larvae has bilateral symmetry, as shown in the figure.

Question 9.
Which range of wavelength (in nm) is called photosynthetically active radiation (PAR)? (1)
(A) 100 – 390
(B) 390 – 430
(C) 400 – 700
(D) 760 – 100,00
Answer:
Option (C) is correct.
Explanation: Photo synthetically active radiation (PAR) is the light of wavelength 400-700 nm and this portion of light is utilised by the plants for photosynthesis.

Question 10.
Identify the ‘pacemaker’ of the heart from the given diagram.
(A) (iii)
(B) (iv)
(C) (i)
(D) (ii)

Answer:
Option (C) is correct.
Explanation: Label (a) shows the SA (sinoauricular node) which is also called pace maker of human heart, which is responsible to generate impulses.

Question 11.
Many physico-chemical processes are enzyme-mediated. Which of the following reactions is not enzyme-mediated in biological system? (1)

(A) Dissolving CO₂ in water
(B) Untwining the two strands of DNA
(C) Hydrolysis of sucrose
(D) Formation of peptide bond
Answer:
Option (A) is correct.
Explanation: CO₂ Dissolution is non-enzymatic reaction while to unwinding of two strands enzyme topoisomerase is used, to hydrolyse sucrose enzyme sucrase is used and to form peptide bonds enzyme peptidyl transferase is used.

Question 12.
Identify the given stage shown in the diagram : (1)
(A) Late prophase
(B) Early Prophase
(C) Interphase
(D) Metaphase.
Answer:
Option (A) is correct.
Explanation: The given diagram shows late prophase stage of cell cycle in which nuclear membrane get dissolved and chromosomes get more condensed and visible.

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(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.

Question 13.
Assertion (A): Bryophytes are successful land plants. (1)
Reason (R): They grow successfully on land but require water for the completion of their life cycle.
Answer:
Option (D) is correct.
Explanation: Bryophytes are amphibians of plant kingdom. They grow on moist habitats and require water for reproduction.

Question 14.
Assertion (A): Arthritis or inflammation of a joint makes the joint painful. (1)
Reason (R): Some toxic substances are deposited at the joint.
Answer:
Option (C) is correct.
Explanation .-Due to the deposition of uric acid in joints the movement of joint become painful

Question 15.
Assertion (A): Nephridia are present in earthworms. (1)
Reason (R): They help in respiration.
Answer:
Option (C) is correct.
Explanation: Nephridia is excretory organs present in earthworm.

Question 16.
Assertion (A): Mitochondria help in cellular respiration (1)
Reason (R): Mitochondria have enzymes for dark reaction.
Answer:
Option (C) is correct.
Explanation: Mitochondria help in cellular respiration by transferring energy from organic compounds to ATE Chloroplast helps in photosynthesis. The dark reaction takes place in the stroma of the chloroplast.

Section – B

Question 17.
‘Omnis cellula-e-cellula’. Who gave this statement and what does it mean ? (2)
Answer:
‘Rudolf Virchow’ gave this statement. It means “All cells come from pre-existing cells”. Every cell is born from previous cell, which was born from a previous cell. Life comes from life.’

Question 18.

(a) Name the type of joint represented in the given diagram. (2)
(b) What is the significance of the fluid present between this type of joint?
Answer:
(a) Synovial Joint.
(b) There is a membrane present called synovial membrane, which is composed of secretory epithelial cells which secrete a thick sticky fluid, of the consistency of the white of an egg called synovial fluid.
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.

Question 19.

(a) Name the type of bond highlighted in the given diagram (i) and (ii). (2)
(b) Define each type of bond highlighted in the given diagram (i) and (ii), respectively.
Answer:
(a) (i) Phosphodiester bond.
(ii) Peptide bond.

(b) (i) Peptide bond is a covalent bond that joins the two amino acids by -NH-CO linkage.
(ii) Phosphodiester bond is a strong covalent bond between phosphate and two sugar groups.

Question 20.
(a) What is placentation? (2)
(b) Identify the given type of placentation.
Answer:
(a) The number, position and arrangement or distribution of placenta inside an ovary is called placentation. It is of the following types:

1. Marginal
2. Parietal
3. Axile
4. Free central
5. Basal
6. Apical and
7. Superficial.

(b) Axile placentation.

Question 21.
Endoparasites are found inside the host body. They live in their tissues and internal organs and survive on their nutrition. (2)
(a) Mention the special structure, possessed by these parasites.
(b) How these structures help the host survive in those conditions.

OR
(a) Study the given table and
Fill up the spaces appropriately.

Phylum/Class	Excretory organ	Circulatory organ	Respiratory organ
Arthropoda			Lungs/Gills/Tracheal system
	Nephridia	Closed	Skin/Parapodia
	Metanephridia	Open
Amphibia		Closed	Lung

(b) Define the scientific term used for cold-blooded animals .
Answer:
(a) Hook and Suckers.

(b) Special characters present in endoparasites are as follows:

1. Presence of adhesive organ for attachment to the host. Fasciola hepatica possesses posterior sucker for 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 hermaphrodite.

OR
(a)

Phylum/ Class	Excretory organ	Circulatory organ	Respiratory organ
Arthropoda	Malpighian tubule	Open	Lungs/Gills/Tracheal system
Annelida	Nephridia	Closed	Skin/parapodia
Mollusca	Metane- phridia	Open	Gills/Mantle, Pulmonary sac
Amphibia	Kidneys	Closed	Lung

(b) Poikilotherms; their body temperature varies with the varying environment.

Section – C

Question 22.
(a) Label the given diagram of VS. of maize seed: (3)

(b) Write down the function of part (2.)
Answer:

(b) Aleurone layer is a sheath of special tissue. It contains proteins, which helps in the nourishment of developing embryo enclosed in the seed. It also secretes a few enzymes which help in the process of seed germination.

Question 23.
How diabetes insipidus is different from diabetes mellitus ? (3)
Answer:
Deficiency of ADH leads to diabetes insipidus, a condition marked by the output of huge amounts of urine and intense thirst. The name itself (diabetes =overflow; insipidus = tasteless) distinguishes it from diabetes mellitus (mel = honey), in which insulin deficiency causes large amounts of blood sugar to be lost in the urine.

Question 24.
The process of respiration requires oxygen. In plants, oxygen is taken in by stomata, lenticels and root hairs. During the process of respiration, oxygen is utilised and carbon dioxide and water is released along with energy molecules in form of ATP.
(a) What are the two crucial events in anaerobic respiration? Where does these take place? (3)
(b) Why anaerobic respiration yields much less energy than aerobic respiration?
(b) Name the part which helps in cross-bridge formation.
Answer:
(a) The two important events in aerobic respiration are:
(a) Complete oxidation of pyruvate in the matrix of mitochondria, all this happens by step¬wise removal of hydrogen atoms, and finally carbon dioxide is left.
(b) Electron removes as part of hydrogen atoms passed on to molecular oxygen and ATP formed simultaneously.

(b) Anaerobic respiration yields much less energy than aerobic respiration. The main reasons are:

• There is an incomplete breakdown of the respiratory substrate.
• One of the end products of anaerobic respiration is organic which still contains energy.
• NADH₂ produced during glycolysis is often re-utilized.
• Regeneration of NAD⁺ from NADH₂ does not produce ATP.
• Electron transport chain is absent.
• Oxygen is not used for accepting electrons and protons.

Question 25.
(a) Label the different components of actin filament in the diagram given below. (3)

(b) Name the part which helps in cross-bridge formation.
Answer:

(b) F-Actin.

Question 26.
Mention the initial fixation in C₄ pathway. (3)
Answer:
(i) Initial fixation of CO₂ occurs in mesophyll cells.

(ii) The primary acceptor of CO₂ is phosphoenol pyruvate or PEP. It combines with CO₂ in the presence of PEP carboxylase to form oxaloacetic acid or oxaloacetate.
PEP + CO₂ + H₂O → Oxaloacetic acid + H₃PO₄

(iii) Oxaloacetic acid is reduced to malic acid or transaminated to form aspartic acid.
Oxaloacetic acid + NADPH Dehydrogenase → Malic acid + H₃PO₄
Oxaloacetic acid + Alanine Transaminase → Aspartic acid + Pyruvic acid

Question 27.
Under high concentration of O₂, RuBP binds with O₂ and forms one molecule of phosphoglycerate and phosphoglycolate. (3)
(a) What is this process called?
(b) Why is this process called a wasteful process?
(c) Name three cell organelles involved in this process.
OR
The term glycolysis originated from the Greek words, glycos for sugar and lysis for splitting. It is a metabolic pathway given by G. Embden, Otto Meyerhoff, and J. Parna Spl
(a) (i) Where does glycolysis occur in the cell?
(ii) Mention the end product.
(b) Which intermediate compound is a connecting link between glycolysis and Kreb’s cycle?
Answer:
(a) Photorespiration

(b) Photorespiration is called a wasteful process because:

1. No energy is produced in this process.
2. Oxygen is consumed for nothing.
3. H₂O₂ is produced which is highly toxic.
4. The yield of photosynthesis is reduced to 50%.

(c) Chloroplast, Mitochondria and Peroxisomes
OR
(a)
(i) Glycolysis: Glycolysis means “the splitting of sugar”. A glucose molecule is converted into 2 molecules of pyruvic acid during glycolysis. It occurs in the cytoplasm of the cell.
(ii) End products are 2 pyruvate molecules and 2 ATP molecules.

(b) Acetyl CoA is a connecting link between glycolysis and kreb’s cycle.

Question 28.
Why bryophytes do not attain great heights? (3)
Answer:
Bryophytes seldom achieve great heights. The possible reasons are:

1. Absence of roots.
2. Materials are transported from cell to cell.
3. Absence of vascular tissues.
4. Absence of cuticle on the plant body.
5. Requirement of an external sheet of water for capillary conduction to all parts and transport of male gametes.

Section – D

Question 29.
Study the given structure of cyanobacteria and answer the following questions:
(a) Name the reserve food material in cyanobacteria.
OR
Name the type of substance which surrounds the colonies of blue-green algae.
(b) (i) Name the pigments present in cyanobacteria.
(ii) Name the type of reproduction through which blue green algae reproduces.
(c) Name the kingdom which includes cyanobacteria. (4)

Answer:
(a) Oil droplets.
OR
Gelatinous sheath

(b) (i) The pigments present in cyanobacteria are: Chlorophyll ‘a’, phycobilin, phycoerythrin and phycocyanin.
(ii) Blue-green algae multiply asexually by binary fission, fragmentation or by the formation of small segments called hormogonia.
• Typical sexual reproduction is absent.

(c) Kingdom Monera.

Question 30.

S. No.	Major pigments	Stored food	Cell wall	Flagellar number and position of insertion
1.	chlorophyll a, b	Starch	Cellulose	2 – 8, equal apical
2.	chlorophyll a, c, fucoxanthin		Cellulose and algin	2, unequal lateral
3.	chlorophyll a, d, phycoerythrin	Floridean starch		Absent

(a) Write the common names for the groups respective to their characteristics from the given table.
(b) Write the stored food material for group ‘2’.
(c) Write the cell wall composition for group ‘3’. (4)
OR
There are millions of living organisms on earth. All these living organisms differ in shape, size, colour, habitat and many other characteristics. To understand their origin, diversity, distribution and interrelationship, the scientists have devised mechanisms to classify all of them. Classification of living organisms help in revealing the relationship between various organisms. It also helps in making the study of organisms easy and organised.
(a) Name the scientist w’ho made a significant contribution of field of classification.
(b) All living organisms are linked to one another. Why?
(c) What is the important criteria in the present day classification?
Answer:
(a)
(i) Green Algae
(ii) Brown Algae
(iii) Red Algae

(b) Mannitol, laminarin.

(c) Cellulose, pectin, and sulphate esters.
OR
(a) Linnaeus made significant contribution in the field of classification.

(b) All living organisms are linked to one another because they share common genetic material but to varying degrees. Living organisms share common genetic material to some extent of varying degrees and hence they linked to one another.

(c) An important criteria in the present day classification is: Anatomical and physiological traits.

Section – E

Question 31.
On the basis of the given table answer the following questions: (5)

S. No.	Red muscle fibre	White muscle fibre
(i)	They are thin and smaller in size.	They are thick and larger in size.
(b)	They are red in colour.	They are white in colour.
(iii)	They contain numerous mitochondria.	They contain less number of mitochondria.
(iv)	They carry out slow and sustained contractions for a long period.	They carry out fast work for short duration.
(v)	They do not undergo fatigue.	They undergo early fatigue.

(a) Why red muscle fibre are red in colour?
(b) What is muscle fatigue? How is it caused?
(c) Give examples of both the muscle fibre.
OR
In the following diagram shows the myofibrils in different state of action. Answer the questions as follows:

(a) What is the state of myofibrils in (A) and (B) respectively?
(b) What are I band and A band also called as respectively?
(c) Which ion is responsible for contraction or relaxation of myofibrils
Answer:
(a) Red muscle fibres are red in colour due to large amounts of myoglobin content present in them.

(b) The reduction in muscle contraction force after prolonged stimulation is called muscle fatigue.
(i) A muscle is able to contract for a short time in the absence of oxygen. But it gets fatigue sooner because the metabolic products of glycolysis mainly lactic acid accumulate, in the absence of oxygen.
(ii) The accumulation of lactic acid leads to muscle fatigue.
(c)

Red muscle fibre	White muscle fibre
Example: Flight muscles of birds like kites, Extensor muscles of the back to maintain erect posture of the body	Example: Muscles of the eyeball and flight muscles of birds like sparrow.

OR
(a) (a) is relaxed state and (B) is contracted state of myofibril.
(b) I band is light band or isotropic band while A band is dark band or anisotropic band.
(c) Ca²⁺ ions are responsible for the contraction or relaxation of myofibril.

Question 32.
Eukaryotic cells are present in protista, plants, animals and fungi. Cytoplasm is divided into compartments due to
the presence of membrane bound organelles. (5)
(a) Name two cell-organelles that are double membrane bound.
(b) What are the characteristics of these two organelles?
(c) State their functions and draw labelled diagrams of both.
OR
The genetic material is contained in the chromosomes of the cell. The chromosome has different parts; arms, secondary constrictions, and centromere.
(a) What is a centromere?
(b) How does the position of centromere form the basis of classification of chromosomes?
(c) Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Answer:
(a) Mitochondria and chloroplast.

(b) Mitochondria:

1. Each mitochondrion is a double membrane- bound structure with the outer membrane and the inner membrane.
2. The outer membrane forms the continuous limiting boundary of the organelle whereas the inner membrane forms a number of infolding called the cristae. The cristae increase the surface area.
3. The matrix possesses a single circular DNA molecule, a few RNA molecules and 70S ribosomes.

Chloroplast:

1. The chloroplast are double membrane structures. They are divided into outer and inner membranes, further divided into two distinct regions: Grana and stroma.
2. Between the outer and inner membranes, intermembrane space is present.
3. The inside of the chloroplast is clearly marked into a colourless ground matrix called stroma.
4. A number of organised flattened membranous sacs called thylakoids are present in the stroma.
5. Thylakoids are arranged in stacks called grana. Besides these, there are flat membranous tubules called the stroma lamellae connecting the thylakoids of the different grana.
6. The stroma of the chloroplast contains small, double-stranded circular DNA and 70S ribosomes.

(c) Mitochondria – Function:
(i) Mitochondria are the sites of aerobic respiration and produce cellular energy in the form of ATP, hence they are called powerhouses of the cell.

Chloroplast- Functions:
(i) Chloroplasts are the centres of photosynthesis.
(ii) They are able to trap solar energy and convert it into chemical energy.

OR
(a) Centromere is the constriction present on the chromosomes, where chromatids are held together.

(b) 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. In this, 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 33.
Explain briefly the structural levels of proteins. (5)
OR
What are the methods of passage of substances across the cell membrane?
Answer:
Proteins are macromolecules having one or more polypeptide chains of amino acids. These are highly organised linear polymers of amino acids. Proteins have four structural levels – primary, secondary, tertiary and quaternary.

(i) Primary structure: The linear sequence of amino acids in a polypeptide chain is the primary structure of protein. The first amino acid is called N-terminus and the last amino acid is called C-terminus.

(ii) Secondary structure: The polypeptide chains are coiled or folded in the form of helix. The structure is maintained by a series of regularly spaced infra or inter molecular hydrogen bonds formed between the amino acids of the same or different polypeptide chains. Such a conformation is termed as secondary structure. There are 3 types of secondary structure, α – helix, β -pleated and collagen helix.

(iii) Tertiary structure: The helical polypeptide molecules further coiled to assume a complex but specific form called tertiary structure. They are so arranged as to hide non-polar amino acid inside and expose the polar side chains. These structure are maintained by several types of bonds such as hydrogen bonds, Van der Waal interaction, ionic bonds, disulphide bonds, etc., between the polypeptide chains.

(iv) Quaternary structure: Each polypeptide develops its own tertiary structure and functions as subunit of the protein. The different charged subunits pack together to give specific conformation. Large proteins such as haemoglobin have a quaternary structure. It has four chains, two α-chains and two β – chains.
OR
The passage of substances across the cell membrane occurs by three methods:
(i) Passive transport: It is a mode of membrane transport which occurs without the expenditure of cell energy. Passive transport occurs by diffusion or osmosis. Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient. Osmosis is a special type of diffusion in which only water molecules migrate across a semi- permeable membrane from the area of high concentration to the area of low concentration.

(ii) Active transport: It is the movement of materials across the membrane against the concentration gradient. This often leads to accumulation of substances at a higher concentration within the cell than outside. Energy is required for such a process, which is usually provided by ATP. There are certain proteins in the membrane, which acts as a carrier molecules or carrier proteins. It has a binding site for the substrate. This transports the carrier-bound substrate to the other side of the membrane.

(iii) Bulk transport: It includes exocytosis and endocytosis, which are the processes for active cellular extrusion and intake respectively of such materials that cannot pass through the unbroken plasma membrane. Both the processes involve movement and folding of the membrane, which depends on its fluidity and mobility.