CBSE Sample Papers for Class 11 Biology Set 2 with Solutions

Students can access the CBSE Sample Papers for Class 11 Biology with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 11 Biology Set 2 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 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.
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.
Members of phycomycetes are found in: 1
(i) 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.
Answer:
Option (D) is correct.
Explanation:Members of the phycomycetes are found in the aquatic habitats, on decaying wood, moist and damp places and as obligate parasites on plants.

Question 2.
If the diploid number of a flowering plant is 36. What would be the chromosome number in its endosperm? 1
(A) 36
(B) 18
(C) 54
(D) 72
Answer:
Option (C) is correct.
Explanation: The end result of the triple fusion is endosperm : One male nuclei (n = 18) fused with diploid secondary nucleus (2n=36). Endosperm of flowering plants is a triploid structure. As 2n = 36, then n = 18, therefore 3n = 54.

Question 3.
Match the following terms with their appropriate example. 1

Column – I	Column – II
(a) Hypogynous flowers	(i) Sunflower
(b) Epigynous flowers	(ii) Nerium
(c) Perigynous flower	(iii) China rose
(d) Whorled phyllotaxy	(iv) Rose

(A) (a) – iv, (b) – ii, (c) – i, (d) – iii
(B) (a) – iii, (b) – i, (c) – iv, (d)- ii
(C) (a) – ii, (b) – i, (c) – iii, (d) – iv
(D) (a) – i, (b) – iii, (c) – iv, (d) – ii
Answer:
Option (B) is correct.
Explanation: Hypogynous flowers- The ovary is said to be superior. Example: China rose.
• Epigynous flowers- The ovary is said to be inferior.
Example: sunflower.
The ovary is half inferior e.g., Rose.
Whorled phyllotaxy- Leaves arise in whorl from one point. e.g., Nerium.

Question 4.
Which of the following statements is not true for plasma membrane? 1
(A) It is present in both plant and animal cell.
(B) Lipid is present as a bilayer in it,
(C) Proteins are present integrated as well as loosely associated with the lipid bilayer.
(D) Carbohydrate is found in it.
Answer:
Option (D) is correct.
Explanation: Carbohydrates are never found in the plasma membrane.

Question 5.
A homo-polymer has only one type of building block called monomer repeated ‘n’ number of times. A hetero-polymer has more than one type of monomer. Proteins are hetero-polymers made of: 1
(A) 20 types of monomers
(B) 40 types of monomers
(C) 3 types of monomers
(D) Only one type of monomer
Answer:
Option (A) is correct.
Explanation: The function and shape of a protein is affected by sequence of 20 types of amino acids, each having an amino group (NH₂), a carboxylic acid group (COOH), a hydrogen atom each linked to the carbon located next to the -COOH group, and a side chain R that varies from amino to amino.

Question 6.
A person suffers punctures in his chest cavity in an accident, without any damage to the lungs. Its effect could be: 1
(A) Reduced breathing rate
(B) Rapid increase in breathing rate
(C) No change in respiration
(D) Cessation of breathing
Answer:
Option (D) is correct.
Explanation: To breathe, there must be a pressure gradient in the lungs with regard to the atmosphere. For example, while inhaling, there must be a negative pressure and while exhaling, there must be a positive pressure gradient inside the lungs. Thus, if there is a puncture in chest cavity of a person, there would be very less or no pressure gradient in that person’s lungs. As a result, his/her breath would cease.

Question 7.
Mark the pair of substances among the following which is essential for coagulation of blood. 1
(A) Heparin and calcium ions
(B) Calcium ions and platelet factors
(C) Oxalates and citrates
(D) Platelet factors and heparin
Answer:
Option (B) is correct.
Explanation: Platelets and calcium ions are crucial components for blood coagulation. When there is an injury, the platelets release inactive platelet factors that are involved in activation of other factors involved in blood clot formation. Numerous clot-forming factors, including factor VIII, factor IX, factor V and factor XIII, are activated by calcium ions in the blood.

Question 8.
Identify the structure given below. 1

(A) Spirochete
(B) Spirillum
(C) Vibrio
(D) Coccobacillus
Answer:
Option (B) is correct.
Explanation: Bacteria that have a hard spiral (helical) structure are thick, long, and move with flagella. They measure 6 to 15 m in length and resemble a spiral shape. Hence, known as spirillum.

Question 9.
Filtration of the blood takes place at: 1
(A) PCT
(B) DCT
(C) Collecting ducts
(D) Malpighian body
Answer:
Option (D) is correct.
Explanation: Each nephron has two parts: the glomerulus and renal tubule. Glomerulus is a tuft of capillaries formed by afferent arteriole. The renal tubule begins with a double-walled cup¬like structure known as bowman’s capsule, which encloses glomerulus. Glomerulus along with the bowman’s capsule is called malpighian body or renal corpuscle. The tube proceeds further to form PCT, a loop of Henle and DCT The first step in urine formation is the filtration of blood, which takes place at malpighian body.

Question 10.
Identify the diagram given below. 1

(A) Pinnately compound
(B) Palmately compound leaf
(C) Trifoliolate leaf
(D) Ternate leaf
Answer:
Option (B) is correct.
Explanation: Palmately compound leaves are a type of compound leaves where the leaflets are attached at the tip of the petiole. Based on how the lamina is divided, leaves can be divided into two basic categories. Simple leaves are undivided and the incision does not reach to the midrib.

Question 11.
The function of our visceral organs is controlled by: 1
(A) Sympathetic and somatic neural system.
(B) Sympathetic and parasympathetic neural system.
(C) Central and somatic nervous system.
(D) None of the above.
Answer:
Option (B) is correct
Explanation: Autonomic nervous system includes sympathetic and parasympathetic nervous systems. This system regulates the body’s automatic processes, including breathing, digestion, urine, and heart rate. They have varying effects on the visceral organs.

Question 12.
The figuie given shows the conversion of a substrate into product by an enzyme. Which one of the four
options of reaction labelled as A, B, C and D are identified correctly? 1

1. Activation energy with enzyme.
2. Potential energy.
3. Activation energy without enzyme.
4. Transition energy.
(A) A – 1, B – 4, C – 2, D – 3
(B) A – 3, B – 1, C – 2, D – 4
(C) A – 2, B – 4, C – 3, D – 1
(D) A – 3, B – 2, C – 3, D – 4 .
Answer:
Option (C) is correct.
Explanation: In the given graph: A- Potential energy, B- Transition energy, C – Activation energy without enzyme and D- Activation energy with the enzyme.

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): The generic name begins with a capital letter while specific name begins with small letter.
Reason (R): Scientifically, mango is written as Mangifera indica. 1
Answer:
Option (B) is correct.
Explanation: Scientifically mango is written as Mangifera indica. The generic name begins with capital letter while a specific name begins with small letter.

Question 14.
Assertion (A): Apple is a true fruit. 1
Reason(R): In apple, thalamus and perianth take part in fruit formation.
Answer:
Option (D) is correct.
Explanation: Apple is a false fruit. It develops from parts other than the ovary.

Question 15.
Assertion (A): Chloroplasts are found in all the eukaryotic cells. 1
Reason (R): They perform the function of photosynthesis as they trap energy.
Answer:
Option (D) is correct.
Explanation: Chloroplast is found in all green cells which perform the function of photosynthesis. It traps energy.

Question 16.
Assertion (A): Uremia is an excretory disorder. 1
Reason (R): The accumulation of urea in blood is known as uremia.
Answer:
Option (A) is correct.
Explanation: Uremia is an excretory disorder. It is the accumulation of urea in blood due to malfunctioning of kidney.

Section – B

Question 17.
Explain the role of neural system in regulation of respiration. 2
Answer:
Neural system plays an important role in the regulation of respiration. There is a specialised centre present in the medulla oblongata region of the brain called respiratory rhythm centre that regulates the respiration. The function of respiratory rhythm centre is controlled by another centre present in the pons varolli called pneumotaxic centre. Neural signals from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.

Question 18.
Explain the given two types of anther lobes. 2

(a) Dithecous.
(b) Monothecous.
Answer:
(a) Dithecous: Dithecous anthers are those anthers that have two lobes that are joined by connective tissue. They are two-chambered and contain four pollen sacs. Such anthers can be seen in Solanum.
(b) Monothecous: Monothecous anthers are those anthers that have only one lobe. They are one chambered and contain two pollen sacs. Such anthers can be seen in Hibiscus.

Question 19.
Study the given flow chart and define the process. Where does it occur ? Explain. 2

Answer:
Glycolysis is the breakdown of glucose or similar hexose sugar into two molecules of pyruvic acid through a series of enzyme-mediated reactions releasing some energy (ATP) and reducing power (NADH). It occurs in cytosol or cytoplasm.

Question 20.
Study the given diagram of mitochondria. 2

(a) Why mitochondria is called powerhouse of the cell?
(b) What is the main function of mitochondria?
Answer:
(a) The fact that mitochondria play a crucial role in generating energy in the form of adenosine triphosphate (ATP) has led to the nickname “powerhouse of the cell” being applied to them. The fundamental energy unit of cells is ATP, which powers a variety of cellular functions. Cellular respiration, which is carried out by mitochondria, entails the breakdown of organic molecules like glucose to produce ATP

(b)

1. Mitochondria have an inner and outer membrane, The inner membrane is highly folded into structures called cristae, which increase its surface area and provide more space for chemical reactions to occur.
2. In addition to producing energy, mitochondria also participate in other cellular functions. They involved in apoptosis (programmed cell death), calcium ion homeostasis, and cell signalling.
3. Furthermore, mitochondria can autonomously replicate within the cell and have their own DNA, known as mitochondrial DNA (mtDNA). (Any two)

Question 21.
Observe the information provided in the table to answer the following questions.

(a) What has the International Code of Botanical Nomenclature ICBN provided? 2
(b) Briefly explain the taxonomic hierarchy with the help of an example.
OR
A plant species shows several morphological variations in response to altitudinal gradient. When grown under similar conditions of growth, the morphological variations disappear and all the variants have common morphology. What are these variants called?
Answer:
(a) To make it easier for botanists to study plants, the ICBN has established a number of norms and guidelines. Through accurate identification and nomenclature, it aids in the proper placing of any newly found organism.

(b) The taxonomic hierarchy, which is used to categorize any plant, is provided below.
Kingdom-Plantae.
Division-phyta.
Class-ae.
Order-ales.
Family-eae/ceae.
Genus: An organism’s first name is typically written in italics and includes a Latin word. The second word of the scientific term, “species,” is similarly italicized.
OR
These morphological variants are called biotypes. Biotypes are group of genetically similar plants which show similarity when grown in the same environmental and geographical regions. The same environment provides them the similar abiotic factors like soil, pH, temperature, etc. When such plants are grown in two different geographical regions, they are exposed to different abiotic characters. This in turn, affects their growth and development and brings changes in their external morphological features, but their genetic constitution remains the same.

Section – C

Question 22.
The figure given below shows structure of Laminaria. 3
(a) What is the function of frond, stipe, holdfast in Laminaria?
(b) Name the class of algae to which Laminaria be1ong.

Answer:
(a) Laminaria belongs to the phaeophyceae family, which includes organisms that are frequently referred to as brown algae. Holdfast, stripe, and frond are the three distinct sections of the plant body. Frond can be described as a leaf-like photosynthetic organ, holdfast is used for attachment to the substratum and the stipe acts as a stalk.

(b) Phaeophyceae is the class of algae to which laminaria belongs. Phaeophyceae or brown algae are the large group of marine multicellular algae. They play a significant role in aquatic habitats.

Question 23.
What are the two crucial events in aerobic respiration? Where does these take place? 3
Answer:
The two important events in aerobic respiration are:

1. The complete oxidation of pyruvate by the step-wise removal of all the hydrogen atoms with the release of three molecules of carbon dioxide.
2. The passing on of the electrons removed as part of the hydrogen atoms to molecular O₂ along with the synthesis of ATP.
   The first step takes place in the matrix of the mitochondria while the second step occurs in the inner membrane of mitochondria.

Question 24.
(a) What is a fruit? Explain with the help of diagram. 3
(b) Describe various zones of fruit by taking an example of succulent fruit.
Answer:
(a) Fruit: It is regarded as a ripened ovary developed after fertilisation.

(b) Various zones of fruit are:

1. A succulent fruit like mango consists of pericarp and the seed.
2. The pericarp is divided into three zones: epicarp, mesocarp and the endocarp.
3. In mango, the outer skin is epicarp.
4. The sweet, pulpy and edible portion is mesocarp.
5. Endocarp is the innermost hard zone which encloses the seed.

Question 25.
(a) What are the names given to A and B. Explain. 3

(b) Write about monadeiphous stamens?
Answer:
(a) The name are given to A is syngenesious stamens and B is synandrous stamens.
In syngenesious stamens, anthers of all the stamens are fused. Fused anthers make a tube around the style. Filaments of all the stamens are free.

In synandrous stamens, anthers and filaments of all the stamens are fused. Fused anthers and filaments form a compact mass. No part of stamen is free.

(b) In Monadelphous stamens, all the stamens of a flower are fused in one bundle by their filaments only. Staminal tube is formed 2g around the base of the pistil which may extend to cover the style. In this, anthers and upper part of the filaments are free.

Question 26.
Explain the significance of citric acid cycle. 3
Answer:
Significance of citric acid cycle:
(i) It explains the process of breaking of pyruvate into CO2 and water. It is the major pathway of generation of ATP.

(ii) More energy is released (24 ATP) in this process, as compared to glycolysis.

(iii) In Kreb’s cycle, carbon skeletons are obtained for use in growth and maintenance of the cell. Many intermediate compounds are formed. They are used in the synthesis of other biomolecules like amino acids, nucleotides, chlorophyll, cytochromes and fats. Succinyl CoA is the starting molecule for synthesis of chlorophyll. Amino acids are synthesised from α-ketoglutaric acid, pyruvic acid, and oxaloacetic acid.

Question 27.
Describe the following disorders of the muscular and skeletal system. 3
(a) Myasthenia gravis
(b) Tetany
(c) Muscular dystrophy
OR
Nerve impulse is any signal initiated by the stimulus at the receptor. Nerve cells are excitable.
(a) (i) Describe conduction in nerve impulse of the resting phase.
(ii) Action potential.
(b) Explain about recovery phase with the help of diagram
Answer:
(a) Myasthenia gravis: It is an autoimmune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.

(b) Tetany: Rapid spasms (wild contractions) in muscle due to lesser Ca⁺⁺ in body fluid is tetany.

(c) Muscular dystrophy: It is an inborn abnormality associated with progressive degeneration of skeletal muscle.
OR
(a) (i) Resting phase: During this phase, there is more Na⁺ ions on outside the axon membrane in ECF than inside of the axon membrane. K⁺ ions that move out and the membrane is positively charged outside and negatively charged inside. Thus, membrane is in resting potential (-80 mV).

(ii) Action Potential: It is another name of nerve impulse with resting potential. When the stimulus is applied, the axon membrane is negatively charged outside and positively charged inside. It is called depolarised or an action potential ( + 30 mV). The electrogenic pump expels 3 Na⁺ for every 2 K⁺ imported. Na⁺ K⁺ ions transmembrane pumps are very important in membrane potential.

(b) Recovery phase: During this phase, depolarised point becomes repolarised and Na+ ions move to outside while K+ ions move inside.

Question 28.
Name the hormones of posterior pituitary. Give reason why are these called neurohormones. 3
Answer:
Oxytocin and vasopressin are two hormones of posterior pituitary.
These hormones are called neurohormones because they are actually synthesised by the hypothalamus and are transported axonally to neurohypophysis.

Section – D

Question 29.
The structure below shows Fluid mosaic model. According to this model, there is a central lipid bilayer of phospholipids with their polar head group towards the outside and the non-polar tails pointing inwards. Some proteins are embedded in the lipid layer and are called integral or intrinsic proteins as they cannot be separated from the membrane. There are large globular integral proteins which project beyond the lipid layer on both the sides. Superficially attached proteins are called extrinsic or peripheral proteins, which can be easily removed.
Some membrane lipids and integral proteins remains bound to oligosaccharide which project into the extracellular fluid. 4

(a) Describe the constituent of cell membrane?
(b) The passage of substances across the cell membrane occurs by.
(c) Why is membrane fluidity important?
OR
What is the fluid mosaic model of the plasma membrane? Who proposed it?
Answer:
(a) The biological membranes are made up of lipids and proteins. The various lipids present in the membrane are phospholipids, cerebrosides, gangliosides, cholesterol, glycolipids, etc. Proline is generally absent in membrane proteins.

(b) Mechanism by which molecules can cross the plasma membrane is passive diffusion Passive transport mechanism uses no energy while active transplant require energy to get it done.

(c) Membrane fluidity is important for maintaining the integrity and functionality of cell membranes, facilitating the movement of molecules and ions across the membrane, supporting protein function, and enabling cellular processes such as signal transduction.
OR
According to the fluid mosaic model, the membrane is made up of a lipid bilayer and integral and peripheral proteins in mosaic pattern. This model was proposed by Singer and Nicolson.

Question 30.
Observe the graph given below.
In tropical rainforests, 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 comes 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. 4

(a) Name the accessory pigments?
(b) Which light range is most effective in photosynthesis?
(c) What is the range of wavelength (in nm),which is called photosynthetically active radiation (PAR)?
OR
Why is the rate of photosynthesis higher in the red and blue regions of the spectrum of light?
Answer:
(a) Carotenoids and xanthophylls are accessory pigments.
(b) Red light is most effective for photosynthesis.
(c) The range of wavelength (in nm) between 400 – 700 nm, is called photosynthetically active radiation (PAR).
OR
The blue and red regions of spectrum show higher rate of photosynthesis. It is because, red and blue light have maximum energy which is absorbed by chlorophyll pigment and then gets excited to initiate the process of photosynthesis. Also, its wavelength are 400¬700 nm i.e., between the photosynthetic active regions.

Section – E

Question 31.
(a) Describe the uitrastructure of mitochondria? 5
(b) Write any two functions of golgi apparatus?
(c) Draw a well labeled diagram of golgi apparatus?
OR
(a) Name three events which are responsible for gametic consequences during meiosis?
(b) Write any two significance of meiosis.
(c) What do you understand by synapsis? In which stage of prophase-I you observe it.
Answer:
(a) 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.

1. The outer membrane is smooth and is permeable to small molecules.
2. The inner membrane is infolded into the matrix as incomplete partitions called cristae.
3. The cristae bear numerous small tennis rackets like particles called elementary particles, F₀ – F₁ particles or
   oxysomes. Each oxysome has a head, a stalk and a base.
4. The enzymes of electron transport are located in the inner membrane in contact with elementary particles.
5.  The F₀ – F₁ combination function as ATP synthetase.
6. The matrix possesses single circular DNA molecule, a few RNA molecules, 70 S type ribosomes. 2

(b) The functions of golgi apparatus are-

1. Cell wall formation in plants: In plants, dictyosomes are known to synthesise pectin and some carbohydrates necessary for the formation of the cell walls.
2. Acrosome formation: During spermatogenesis, golgi apparatus forms the acrosome.

(c) Ultrastructure of golgi apparatus-

OR
(a) Three events responsible for gametic consequences during meiosis are:

1. The pairing of the homologous chromosomes.
2. The process of crossing over and recombination.
3. The segregation of homologous chromosomes.

(b) Formation of gametes, genetic variability, maintenance of chromosomal number.

(c) Pairing of homologous chromosomes is synapsis. It is observed during the zygotene stage.

Question 32.
Discuss the role of Ca²⁺ ions in muscle contraction. 5
OR
Describe the microscopic structure of a neuron with the help of a diagram.
Answer:

1. The sarcoplasmic reticulum releases the stored Ca⁺⁺ which binds with the specific sites present on the troponin component of the thin filament.
2. As a result, conformational change occurs in the troponin molecule and the active sites present on the F-actin molecules are exposed.
3. These sites are specific to myosin head which exhibits Mg⁺⁺ dependent ATPase activity.
4. During the relaxation of the muscle, the Ca⁺⁺ is pumped back into the sarcoplasmic reticulum.
5. As a result, the troponin component becomes free to mask the active sites for myosin head.
6. The cross-bridge breaks and thin filament occupies its normal position and the muscle relaxes.
7. OR
   A neuron is composed of three main parts, namely cell body or cyton, dendrite and axon.
   (i) Cyton: It contains a well-defined nucleus, surrounded by granular cytoplasm called neuroplasm and all the cell organelles along with, nissl’s granules. Only centrosome is absent because nerve cells have lost the ability to divide.

(ii) Dendrites: These are branched cytoplasmic projections of cyton known as the dendrons which receives nerve impulses.

(iii) Axon: It is a very long process and projects from a part of a cyton axon hillock. The lower end of the axon is branched and each branch terminates as a bulb like structures called nerve endings which have swollen synaptic knobs. There knobs possess synaptic vesicles, which store neurotransmitters. The axons transmit nerve impulses away from the cell body to a synapse or to a neuromuscular junction.

Question 33.
Certain aspects are provided in this table on the basis of major divisions of classifications. Analyse the aspects,
complete the table given below. And explain briefly the major divisions of classifications. 5

Common name	Man	Housefly
Biological name
Genus
Family
Order
Class
Phylum/Division

OR
The five-kingdom classification was proposed by R.H. Whittaker, in 1969. The five-kingdom classification are Monera, Protista, Fungi, Plantae, Animalia. There are certain characteristics given below in the table. Answer all the following characteristics.

Answer:

Common name	Man	Housefly
Biological name	Homo sapiens	Musca domestica
Genus	Homo	Musca
Family	Hominidae	Muscidae
Order	Primata	Diptera
Class	Mammalia	Insecta
Phylum/Division	Chordata	Arthropoda

Major division of classification.

1. Kingdom: It is the highest category in the classification. Example, Plantae and Animalia.
2. Phylum: A group of closely related classes having certain common characters.
3. Class: A group of closely related orders having certain common characters.
4. Order: A group of closely related families having certain common characters.
5. Family: A group of closely related genera having certain common characters.
6. Genus: A group of closely related species having certain common characters.
7. Species: Individuals having certain common character plus some characters of their own.

OR