CBSE Sample Papers for Class 11 Biology Set 3 with Solutions

students must start practicing the questions from CBSE Sample Papers for Class 11 Biology with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Biology Set 3 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.
CO₂ dissociates from carbamino-hemoglobin when:
(A) pCO₂ is high & pO₂ is low
(B) p02 is high and pCO₂ is low
(C) pCO₂ and p02 are equal
(D) None of the above
Answer:
(C) pCO₂ and pO₂ are equal

Question 2.
The condition of accumulation of urea in the blood is termed as:
(A) Renal calculi
(B) Glomerulonephritis
(C) Uremia
(D) Ketonuria
Answer:
(B) Glomerulonephritis

Question 3.
ATPs of the muscle is located in:
(A) Actinin.
(B) Troponin.
(C) Myosin.
(D) Actin.
Answer:
(D) Actin.
Explanation:
The nucleolus is larger in actively working cells as it is involved in actively synthesizing RNA which ultimately results in protein formation.

Direction (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:

Question 4.
Assertion (A): Nucleolus is smaller in actively working cells.
Reason (R): It is involved in actively synthesising DNA which ultimately results in protein formation.
(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 and R is true.
Answer:
(C) A is true but R is false.

Question 5.
Assertion (A): Photorespiration takes place in C₄ plants.
Reason (R): C₄ plants have a mechanism that increases the concentration of CO₂ at the enzyme site.
(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 and R is true.
Answer:
(D) A is false and R is true.
Explanation:
In C₄ plants, photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site. This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂. This results in increasing the intracellular concentration of CO₂. In turn, this ensures that the RuBisCO functions as a carboxylase minimising the oxygenase activity.

Section – B

Question 6.
(a) What do you mean by Plasmodium?
(b) Name the pigments present in cyanobacteria.
OR
How important is the presence of an air bladder in Pisces?
Answer:
(a) A multinucleate mass of protoplasm of slime moulds which creep like Amoeba is known as Plasmodium.
(b) The pigments present in cyanobacteria are chlorophyll a, phycobilin, phycoerythrin and phycocyanin.

OR

1. A gas-filled swim bladder or air sac is always present above the alimentary canal which acts as buoyancy regulator. It enables the fish to stay at a particular depth without expending energy in swimming.
2. In some bony fishes, the swim bladder is used as a lung for breathing air.

Question 7.
What do you call the conduction of impulse along a myelinated nerve fiber?
Answer:

1. Along a myelinated nerve fibre, the conduction of impulse is called saltatory conduction.
2. This is so because the ionic changes and consequent Depolarisation take place only at the nodes of Ranvier free from myelin sheath leading to the jumping of action potential from one node to the next.

Question 8.
Briefly describe the types of joints.
Answer:

1. Fibrous or immovable joints: There is no movement in such joints. There is white fibrous tissue between the ends of the bones. e.g., the joints of skull, teeth etc.
2. Cartilaginous or slightly movable joints: It is an articulation between bones in which motion is limited due to either fibrous tissue or cartilage.
3. Synovial or freely movable joints: It is a freely movable joint characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bone.

Question 9.
Differentiate between Cryptogams and Phanerogams?
Answer:
Difference between Cryptogams and Phanerogams:

Cryptogams	Phanerogams
(i) These are seedless plants.	Phanerogam is seeded plants.
(ii) It has three divisions i.e., thallophytes, bryophytes and pteridophytes.	There is only one division of spermatophyte.
(iii) An ovule is not formed	An ovule is present.

Question 10.
What is the epidermal cell modification in plants which prevents water loss?
Answer:
Bulliform cells are modified epidermal cells that checks the water loss. Under stressed conditions, they help in shutting down stomata and thus reduce water loss through transpiration.

Question 11.
Even though a very few cells in a C₄ plant carry out the biosynthetic – Calvin pathway, yet they are highly
productive. Can you discuss why?
Answer:
C₄ plants are highly productive as they have a mechanism for increasing the concentration of CO₂. The increase in CO₂ ensures that the enzyme RuBisCO does not act as an oxygenase, but as carboxylase. This prevents photorespiration and increases the rate of photosynthesis

Question 12.
How frog is useful to mankind?
Answer:
(a) Frogs eat up insects and thus, protect the crop.
(b) Frogs are largely used as experimental material for teaching and research.
(c) The muscular legs of the frog are largely used as food by man in some parts of India and many other countries.

Section – C

Question 13.
What are the different parts of forebrain?
Answer:
Parts of forebrain are:
(i) Olfactory lobes: Each lobe consists of two parts, an anterior olfactory bulb and a posterior olfactory tract. They are fully covered by the cerebral hemispheres.

(ii) Cerebrum: The cerebrum is the largest part of the human brain. It consists of left and right hemispheres connected by a large bundle of myelinated fibres, the corpus callosum and other smaller fibre bundles. The outer portion of cerebrum is called the cerebral cortex that makes up the grey matter of the cerebrum. There are three kinds of cortex: sensory, motor and associative. Beneath the grey matter, the large concentration of medullated nerve fibres gives this tissue an opaque white appearance, collectively called white matter. Each cerebral hemisphere is divided into four lobes: frontal, parietal, temporal and occipital lobe.

(iii) Diencephalon: Its main parts are epithalamus, thalamus and hypothalamus. Epithalamus is thin and not formed of nervous tissue. Hypothalamus is relatively small but is highly vascular. The thalamus lies superior to mid brain and is composed primarily of grey matter.

Question 14.
Describe the various types of placentation found in flowering plants.
Answer:

• Marginal: Ovary one chambered and ovules lie along the margin of the ovary. e.g., Gram and pea.
• Parietal: Ovary one chambered and ovules lie at the level of the fusion of two carpels. e.g., Mustard.
• Axile: Ovary many chambered and the ovules are attached to the central column. e.g., Onion and lemon.
• Free central: Ovary one-chambered. The center bears many ovules. e.g., Dianthus and Primula.
• Basal: Ovary one chambered and ovules develop on the thalamus. e.g., Sunflower.
• Superficial: Ovary is multilocular and syncarpous. Ovules develop on the minor surface of the ovary. e.g., Nymphea.

Question 15.
Write a note on the economic importance of algae and gymnosperms.
Answer:
Economic importance of Algae:

1. Agar is obtained commercially from Gelidium and Gracilaria which is widely used in the preparation of jellies, puddings, cream etc.
2. Some brown algae are menace for shipping.
3. Some species of Porphyra, Laminaria and Sargassum are used as food.
4. An antibiotic chlorellin is extracted from Chlorella.

Economic importance of Gymnosperms:

1. Conifers provide an enormous amount of softwood for construction, plywood, paper industry, etc.
2. Seeds of Pinus (Chilgoza) are edible.
3. Saw dust of conifers is used in making linoleum and plastics.
4. Ephedrine drug is obtained from Ephedra used in asthma.

Question 16.
What is the difference between nucleotide and nucleoside? Give one example of each.
Answer:
Difference between Nucleotide and Nucleoside:

Nucleotide	Nucleoside
(i) Nucleotide is made up of base, sugar and phosphoric acid.	Nitrogenous base and sugar form a nucleoside.
(ii) Nucleotide of RNA is called ribonucleo tide and nucleotide of DNA is called deoxyribonucleotide.	A nucleoside with ribose sugar is called a riboside or ribonucleoside. A nucleoside with deoxyribose sugar is called deoxyribonucleoside.
(iii) Example: Adenylic acid, guanylic acid, cytidylic acid, thymidylic acid, uridylic acid, AMP.	Example: Adenosine, guanosine, cytidine, thymidine and uridine.

Question 17.
Describe the role of DNA in the life of an organism.
OR
What is the necessity of meiosis II in meiosis in spite of the reduction division of meiosis I?
Answer:
Role of DNA in the life of an organism:

1. DNA is known as the master molecule of the organism. It is the vehicle of the heredity and store house of all information’s required for the growth, differentiation etc.
2. This information in deoxyribose nucleic acid is contained in the form of a blueprint.
3. It gives message to the cell for duplication and transcription. This message is used for protein synthesis by the cells.

OR
At the end of Meiosis-I, two groups of chromosomes are produced, with each group having half the number of chromosomes present in the parent nucleus; therefore meiosis-I is called reductional division. But each chromosome is still a dyad i.e., formed of two sister chromatids joined at the common centromere. Though the chromosome number has been halved, each chromosome still has double (2N) amount of DNA. So, there is need for meiosis-II during which sister chromatids of each chromosome separate into different cells. Haploidy occurs even in DNA amount. The main purpose of meiosis-II is to separate chromosomes formed during crossing over in meiosis-I.

Question 18.
Explain the following in short:
(i) Nephridium
(ii) Parapodia
(iii) Mantle
Answer:
Nephridium:
These are delicate, coiled, respiratory units of Annelids. They collect waste matter from the body cavity and discharge the same into the alimentary canal or outside the body.

Parapodia:
These are paired jointed, leaf -like locomotory appendages of polychaetes (a group of annelids). It helps in locomotion in water and on land.

Mantle:
It is the outer, soft fold that covers the visceral hump of molluscs. It also secretes the outer shell in them.

Question 19.
What is activation energy?
Answer:
Activation energy: It is the energy required to initiate a chemical or biochemical reaction.
Activation energy overcomes the energy barrier of the reactants which occurs amongst the reactants due to

• the presence of electrons over their surface.
• absence of precise and forceful collisions, essential for bringing the reactive sites of the chemical together.

Question 20.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Answer:
By seeing the V S. of leaves, one of the C₃ plant and the other of aC₄ plant.

1. The C₄ plant leaves have Kranz anatomy.
2. The bundle sheath cells make several layers around the vascular bundles.
3. They possess a large number of chloroplasts, thick cell wall impervious to gaseous exchange and there are no intercellular spaces e.g., maize and Sorghum leaves.
4. The chloroplasts in C₄ leaves are dimorphic.
5. The C₃ plants do not have bundle sheath.
6. They possess one type of chloroplasts.

Question 21.
How is oxygen transported in the blood and released in the tissues?
Answer:

1. Blood carries oxygen from the lungs to the heart and from the heart to various body parts.
2. About 3 % of oxygen in the blood is dissolved in the plasma which carries oxygen to the body cells.
3. About 97% of oxygen is carried in combination with hemoglobin of the erythrocytes.
4. The haem portion of hemoglobin contains four atoms of iron, each capable of combining with a molecule of oxygen.
5. Oxygen and hemoglobin combine in an easily reversible reaction to form oxyhemoglobin (HbO₂).
6. Under the high partial pressure, oxygen easily binds with hemoglobin in the pulmonary (lung) blood capillaries.
7. When this oxygenated blood reaches the different tissues, the partial pressure of oxygen declines and the bonds holding oxygen to hemoglobin become unstable.
8.  As a result, oxygen is released from the blood capillaries.

Question 22.
Seeds of some plants germinate immediately after shedding from the plants while in other plants they require a
period of rest before germination. The latter phenomenon is called dormancy. Give the reasons for seed dormancy and some methods to break it.
Answer:
Reasons for seed dormancy are:

1. Impermeability of seed coat to water, e.g., Chenopodium, Trigonella, Meliotus, etc.
2. Impermeability of seed coat to oxygen. e.g., Brassica alba, Pyrus malus, etc.
3. Hard seed coat which does not allow proper growth of developing embryo. e.g., Mustard
4. Some plants produce certain chemical substances like germination inhibitors that inhibit the germination of their own seeds. e.g., Tomato.
5. Some seeds contain an imperfectly developed immature embryo.

Methods of breaking dormancy are:

1. Mechanical scarification weakens the tough and impermeable seed coat.
2. Treating seeds with dilute acids, fats solvents, etc., also weakens the tough seed coat.
3. Inactivation of growth inhibitors by chilling treatment. In this, seeds are treated with alternate low and high temperatures.
4. Permeability of seed coats can be increased when they are kept in water at high temperatures for some time.

Question 23.
Why C₄ plants are preferred in tropical region.
OR
Why does photorespiration not take place in C₄ plants?
Answer:
C₄ plants preferred in tropical region because:
(a) They consume 30 ATPs (12 ATPs more than C₃ plants) to produce one molecule of glucose favoured in tropical region.

(b) In these plants, photorespiration is the mechanism not to lose the photosynthetic carbon.

(c) In the process of photorespiration, RuBP is catabolized to a 3-carbon atom compound instead of combining with CO₂.

(d) More than 50% CO₂ fixed by photosynthesis is lost in photorespiration.

(e) Photorespiration acts to undo the work of photosynthesis as no energy rich compound is produced during this process.

(f) Thus, C₄ plants are better photosynthesizers than C₃ plants and C₄ pathway is of adaptive advantage in the tropical region and thus these plants are preferred.

OR
In C₄ plants, photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site. This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂: This results in increasing the intracellular concentration of CO₂. In turn, this ensures that the RuBisCO functions as a carboxylase, minimizing the oxygenase activity.

Question 24.
What is the significance of Pentose Phosphate pathway?
Answer:
The significance of Pentose phosphate pathway is:

1. PPP constitutes an alternate pathway for the breakdown of carbohydrates in respiration.
2. It produces ribose-5-phosphate, which is used in the synthesis of nucleic acid.
3. Erythrose-4-phosphate produced in PPP is required for the synthesis of lignin, anthocyanin, IAA and a number of other compounds.

Section – D

Question 25.
Draw a standard ECG and explain the different segments in it.
Answer:
ECG is a graphic record of the electric current produced by the excitation of the cardiac muscles. The instrument used to record the change is an electrocardiograph.

Components:
(i) A normal electrocardiogram (ECG) is composed of a P-wave, a QRS wave (complex) and a T-wave.
(ii) The letters are arbitrarily selected and do not stand for any particular words
(iii) The P wave is a small upward wave that indicates the depolarisation of the atria. It is caused by the activation of SA node.

(iv) The QRS wave (complex) begins after a fraction of second of the P wave. It represents ventricular depolarisation (ventricular contraction). It is caused by the impulses of the contraction from AV node through the bundle of His and Purkinje fibres and marks the beginning of the systole.

(v) The T-wave is dome-shaped which indicates ventricular repolarisation (ventricular relaxation). The potential generated by the recovery of the ventricle from the depolarisation state is called the repolarisation wave.

(vi) Normal P-R interval is 0.12 to 0.2 sec. The normal QRS complex duration is 0.12 sec. Normal Q-T interval is 0.4 sec. The importance of ECG is that it gives accurate information about the heart. Therefore, ECG is of great diagnostic value in cardiac diseases.

Question 26.
Read the following to answer the questions from (i) to (v) given below: 5×1=5
Mitosis takes place both in somatic and reproductive cells of plants and animals. In multicellular organisms, mitosis produces more cells for growth and repair. Mitosis division is responsible for the growth and development of a single-celled zygote into a multicellular organism. Mitotic division helps in maintaining the proper size.

Mitosis also helps in restoring wear and tear in body tissues, replacing damaged or lost part, healing wounds and regeneration of detached parts. Mitosis is a method of multiplication of unicellular organisms. It produces diploid daughter cells with identical genetic complements (both quantitatively and qualitatively) as in the parent cell. Mitosis is a continuous process and it is divided into four phases viz: prophase, metaphase, anaphase and telophase.

(i) The correct statement for the significance of mitosis is:
(A) Equal distribution of chromosomes.
(B) Chromosome number gets reduced to its half.
(C) It is a method of multiplication of unicellular organisms.
(D) All of the above
Answer:
(C) It is a method of multiplication of unicellular organisms.
Explanation:
Mitosis is an equational division and is a means of multiplication in unicellular organisms.

(ii) In mitotic cell division, the:
(A) Amount of DNA is the daughter cells will be equal to the parent cell.
(B) Size will be half of the parent cell.
(C) DNA will be double of the parent cell.
(D) Both (A) and (B)
Answer:
(A) Amount of DNA is the daughter cells will be equal to the parent cell.
Explanation:
Daughter cells formed have the same genetic constitution as the qualitatively and quantitatively and as the parent cell.

(iii) Mitosis is characterised by:
(A) Reduction division.
(B) Equational division
(C) Both reduction and equational division
(D) None of the above
Answer:
(B) Equational division
Explanation:
Mitosis is an equational division. Daughter cells formed are genetically identical and similar to the mother cell. The number of chromosomes remain same in parental and progeny cell.

(iv) 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:
There is no crossing over in mitosis because homologous chromosomes do not pair up (Synapsis), therefore there is no chance for crossing over.

(v) In this question the Assertions (A) and Reasons (R) have been put forward. Read both the statements and choose the correct option from the following:
Assertion (A): Mitosis is means of multiplication in unicellular organisms.
Reason (R): Mitosis in multicellular organisms bring about growth and repair.
(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 and R is true.
Answer:
(B) Both A and R are true, but R is not the correct explanation of A.
Explanation:
Mitosis brings about growth and repair in multicellular organism but a means of multiplication in unicellular organisms.

Question 27.
Read the following to answer the questions with explanation from (i) to (v) given below: 5×1=5
In pteridophytes, the sporophyte produces spores through meiosis. The spores are produced by the sporangia in the spore mother cells. These spores germinate to give rise to gametophytes. The gametophytes bear male and female sex organs. Water is required for the transfer of antherozoids – the male gametes released from the male sex organ to the mouth of the female sex organ. Fusion of male gamete with the egg results in the formation of zygote. Zygote thereafter produces a multicellular well- differentiated sporophyte.

In the majority of the pteridophytes, all the spores- are of similar kinds. However, some genera also produce two kinds of spores, macro (large) and micro (small) spores. The megaspores and microspores germinate and give rise to female and male gametophytes, respectively. In these plants, the female gametophytes are retained on the parent sporophytes for variable periods. The development of zygotes into young embryos within the female gametophytes. This event is a precursor to the seed habit considered an important step in evolution.

(i) The gametophytes are called as.
(A) thalli
(B) prothallus
(C) strobilus
(D) liverworts
Answer:
(B) prothallus
Explanation:
It bears sex organs and represents the gametophyte phase in the life cycle of pteridophytes.

(ii) In pteridophytes, the dominant form of life is the
(A) gametophyte
(B) haploid
(C) sporophyte
(D) triploid
Answer:
(C) sporophyte
Explanation:
In pteridophytes, the dominant phase is sporophyte which represents main plant body is differentiated into root, stem and leaves. Gametophyte is small and short-lived. Life cycle is diplohaplontic type.

(iii) Pteridophytes are plants.
(A) homosporous
(B) heterosporous
(C) homosporous and heterosporous
(D) mega sporous
Answer:
(C) homosporous and heterosporous
Explanation:
If all the spores are of similar kind, the condition is homosporous whereas if they are of two types it is called heterosporous.

(iv) In pteridophytes, the male and female sex organ of the gametophytes are called , respectively.
(A) anthers and archegonia
(B) archegonia and antheridia
(C) archegonia and anthers
(D) antheridia and archegonia
Answer:
(D) antheridia and archegonia
Explanation:
Antheridia produces male gamete and archegonia produces egg or female gametes.

(v) In this question the Assertions (A) and Reasons (R) have been put forward. Read both the statements and choose the correct option from the following:
Assertion (A): In pteridophytes, sporophytic and gametophytic phases are independent.
Reason (R): The sporophytic plant bears sporangia on the lateral side of leaf-like appendages called sporophylls.
(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:
Sporangia are not borne on the lateral side of sporophylls. They are produced adaxially or on the ventral