CBSE Sample Papers for Class 11 Biology Set 10 with Solutions

students must start practicing the questions from CBSE Sample Papers for Class 11 Biology with Solutions Set 10 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Biology Set 10 with Solutions

Time Allowed : 3 hours
Maximum Marks: 70

General Instructions :

All questions are compulsory.
The question paper has four sections: Section A, Section B, Section C and Section D. There are 27 questions in the question paper.
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.
There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

Question 1.
Assertion (A): Systematics is the branch of biology that deals with the identification, naming and classification of
organisms into groups.
Reason (R): The aim of classification is to group organisms.
(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:
(B) Both A and R are true but R is NOT the correct explanation of A.
Explanation:
Systematics is the branch of biology that deals with identification, naming and classification of the organisms into groups. The aim of classification is to group the organisms.

Question 2.
Assertion (A): Meiosis is also known as ‘Reduction Division’.
Reason (R): The chromosomes replicate and get equally distributed both quantitatively and qualitatively into two daughter cells.
(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:
(D) A is false but R is true.
Explanation: Meiosis occurs during gametogenesis in plants and animals, This results in the production of haploid gametes. Meiosis I results in the number of chromosomes and crossing over during pachytene and Meiosis I results in the introduction of variation.

Question 3.
Modern classification is based on:
(A) Fossils
(B) Phycology
(C) Phylogeny
(D) Morphology
Answer:
(C) Phylogeny

Question 4.
Pyruvic acid gets reduced to lactic acid by enzyme/enzymes
(A) alcohol dehydrogenase.
(B) pyruvate dehydrogenase.
(C) lactate and pyruvate dehydrogenase.
(D) alcohol and pyruvate dehydrogenase.
Answer:
(D) alcohol and pyruvate dehydrogenase.

Question 5.
The part of the pancreas that secretes insulin is:
(A) Glomerulus
(B) Bowman’s capsule
(C) Islets of Langerhans
(D) Loop of Henle
Answer:
(C) Islets of Langerhans

Section – B

Question 6.
“There exists a clear division of labour within the chloroplast”. Comment.
Answer:
There is a clear division of labour within the chloroplast. The membrane system is responsible for the light reaction (trapping light energy and synthesis of ATP and (NADPH) while the dark reaction i.e., enzymatic reactions for the reduction of carbon dioxide into sugars using ATP and NADPH takes place in the stroma region of chloroplast.

Question 7.
What are the unique features of sponges?
Answer:
The sponges have pores all over the body, cellular level of organisation, a canal system for the passage of water current, choanocytes lining the main cavity (spongocoel) or certain canal (radial canal), skeleton made up of spicules and spongin fibres and they are devoid of mouth and digestive cavity.

Question 8.
Explain the process of ATP formation during aerobic respiration (in mitochondria).
Answer:
The synthesis of ATP from ADP and inorganic phosphate using energy from proton gradient is called oxidative phosphorylation. This takes place in elementary particles present on the inner membrane of cristae of mitochondria. The process in mitochondria is catalysed by ATP synthase. This complex has two major components-F0 and Fr F0 acts as channel for proton and F( acts as an ATP synthase.

Question 9.
From which cells do platelets originate? What is their life span? How do they act when blood vessels get injured?
Answer:
Platelets are produced from megakaryocytes present in the bone marrow. The life span of platelets is 3 to 7 days only. When an injury is caused, the blood platelets disintegrate and release certain chemicals called the platelet factors which help in the clotting of blood.

Question 10.
How are secondary vascular tissues formed in dicot roots?
Answer:

The primary dicot roots lack cambium,
The vascular cambium appears later as a secondary meristem, which develops partially from conjunctive parenchyma and partially from pericycle.
The parenchyma cells lying beneath the phloem bundles become meristematic to form cambium strips,
The cells of the pericycle just outside the protoxylem also become meristematic and joins with the cambium strips to form a wavy band of cambium ring
This cambium ring form secondary xylem on the inner side and secondary phloem on the outer side.
The cambium is always more active towards the inner side during the formation of secondary vascular tissues.

Question 11.
What is a fruit? Describe various zones of fruit by taking an example of succulent fruit.
Answer:
Various zones of fruit:

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

Question 12.
Mention the bones which form limbs.
Answer:
The bones Which make up arm are: 1 humerus, 1 radius, 1 ulna, 8 carpal bones, 5 metacarpal, 5 digits (14 phalanges). The bones which make up leg are: 1 femur, 1 tibia, 1 fibula, 1 patella (knee cap), 7 tarsal bones, 5 metatarsals, 5 digits (14 phalanges).

Section – C

Question 13.
What are hormogonia? Give one example of cyanobacteria which reproduce by binary fission.
Answer:

Hormogonia are motile filaments of cells formed by some cyanobacteria.
They are formed during asexual reproduction in unicellular, filamentous cyanobacteria.
The thick-walled hormogonium is referred to as hormocysts. Hormocysts are helpful in reproduction.
Unicellular cyanobacteria which reproduce by binary fission is Chroococcus.

Question 14.
Explain different types of venations.
Answer:
Venation: 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 types: Reticulate and parallel.

Reticulate Venation: When the veinlets form a network, the venation is termed as reticulate venation, for e.g, leaves of dicot plants.
Parallel Venation: 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,. pPresent in the leaves of monocot

plants. Parallel venation is of two sub-types:
(a) Pinnate or Unicostate parallel venation: The lamina has a single prominent vein or midrib running from the base to the apex of lamina, e.g., Banana, Canna, etc.

(b) Palmate or multicostate parallel venation: The lamina has several principal veins arising from the base and running towards the apex or margin of the lamina.

Question 15.
Describe the gametophytic phase exhibited by moss.
Answer:
The predominant stage in the life cycle of moss is the gametophyte which consists of two stages:
(a) The first stage is the protonema stage, which develops directly from a spore; it is a creeping, green, branched and filamentous stage.

(b) The second stage is the leafy stage, which develops from the secondary protonema as a lateral bud. The leafy stage consists of upright, slender axes bearing spirally arranged leaves and bears the sex organs.

Question 16.
What is the significance of the juxta glomerular apparatus (JGA) in kidney function?
Answer:
Significance of juxta glomerular apparatus:

The aldosterone is a hormone secreted by the adrenal gland. It controls the level of sodium in the blood.
Reabsorption is controlled by a hormone, renin produced in the kidneys. The function of the renin is the conversion of angiotensinogen into angiotensin II, which is produced in the liver.
This angiotensin stimulates the adrenal cortex to secrete aldosterone. Aldosterone induces the distal nephron to reabsorb more water and Na⁺.

Question 17.
Name the three types of respiration in the frog. How does a frog respire during hibernation?
Answer:
The three types of respiration in a frog are:

Cutaneous respiration,
Buccal respiration,
Pulmonary respiration.

The skin of the frog is moist, slimy and highly vascularised which is especially useful for respiration in hibernation and aestivation. The oxygen of the atmosphere enters the thin film of skin moisture where it goes to the blood capillaries of the skin. The oxygen mixes with the blood and passes to the different organs of the body. The carbon dioxide formed in the body organs is taken up by the blood capillaries to the skin from where it diffuses out into the air.

Question 18.
What are the characteristics of enzymes?
Answer:
Enzymes-characteristics:
(i) Approximately all the enzymes are proteinaceous in nature. An enzyme/protein has a primary structure (amino add sequence of protein). Enzymes like protein has secondary and tertiary structure.

(ii) The backbone of protein chain folds upon itself, chain criss-crosses itself and so several crevices or pockets are made out. One pocket is known as ‘active site’. So, the enzymes, through their active site, catalyse reactions at a high rate.

(iii) The enzymes get denatured at a high temperature (above 40°C). The enzymes isolated from organisms who live under extremely high temperatures e.g., hot vents and sulphur springs, are stable and retain their catalytic power at high temperatures (up to 80°-90°C).

Question 19.
(a) What is meant by cytokinesis?
(b) How cytokinesis in plants differ from that in animals?
(c) With reference to the cell types undergoing mitosis, differentiate between animal cells and plant cells.
Answer:
(a) Cytokinesis is the cytoplasmic division of a cell at the end of mitosis or meiosis, which bring about the separation into two daughter cells.

(b) In animal cell, division of cytoplasm takes place by deavage while in plant cell, division of cytoplasm takes place by cell-plate formation.

(c) In plants, a new cell wall is fashioned between the new daughter cells, while in animal cells, the cell membrane constricts to pinch the parent cell into daughter cells.

Question 20.
Mention the similarities between mitochondria and chloroplasts.
Answer:
Similarities between mitochondria and chloroplasts:

Presence of double membrane envelope.
Formation of involutions from the inner membrane.
Both are semi-autonomous.
DNA is naked in both and both possess 70S ribosomes.
Take part in energy transduction.
Both of them produce ATE

Question 21.
Why is ABA known as ‘stress hormone’? Mention any two functions of this hormone. How are they antagonistic to
gibberellins?
Answer:
Absdsic add is also called as stress hormone because the synthesis of absdsic add 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 chloroplast of green cells. Absdsic add owes its name to its role in the absdssion of plant leaves. In preparation for winter, ABA is produced in terminal buds. This slows plant growth and directs leaf primordia to develop scales to protect the dormant buds during the cold season. They are antagonistic to gibberellins as gibberellins promotes stem elongation while absdsic acid acts as growth inhibitor.

Question 22.
Describe briefly the regulation of kidney function by the following:
(i) ADH
(ii) ANF
Answer:
Regulation of Kidney:
(i) ADH: Antidiuretic hormone (ADH) is secreted by hypothalamus of the brain and released into the blood from the posterior Jobe of the pituitary gland. The release of ADH is triggered when osmoreceptors in the hypothalamus detect an increase in the osmolarity of the blood above a set point of 300 osmo mL^-1 In this situation, the osmoreceptor cells also promote thirst. It increases the reabsorption of water in the DCT and collecting duct.

(ii) ANF: Atrial natriuretic factor is another hormone which opposes the regulation by RAAS. The walls of the artety of the heart release ANF in response to an increase in blood volume and pressure. ANF inhibits the release of renin from the 1G cells and thereby inhibits NaC1 reabsorption by the collecting duct and reduces aldosterone release from the adrenal gland. Thus, ADH, RAAS and ANF regulate the function of the kidneys. As a result, they control body fluid osinolatit salt concentration, blood pressure and blood volume.

Question 23.
Why do we call our heart myogenic?
Answer:
The human heart is myogenic as the contraction is initiated by a specially modified heart muscle known as the sino- atrial node (SA node). It is located in the right atrium. The SA node has the inherent power of generating a wave of contraction and controlling the heartbeat. Hence, it is known as the pacemaker.

Question 24.
Describe the role of red blood cells in the transport of oxygen and carbon dioxide by blood.
Answer:
Role of red blood cells in the transport of oxygen:
(i) Red blood cells are known as erythrocytes. They carry oxygen in the form of an unstable, easily dissociable chemical combination with the haemoglobin.

(ii) O2 diffuse into RBCs and combines usually with haemoglobin to form oxyhaemoglobin.
Hb₄ + O2 ⇔ Hb₄O₈

(iii) In tissues oxyhaemoglobin dissociates into oxygen and deoxyhemoglobin at low partial pressure.

(iv) The blood carries its full load to O₂ until it enters the capillaries in tissues to the lungs.

Role of red blood cells in the transport of CO₂:
(i) Carbon dioxide is carried by RBCs. About 30% of CO₂ is carried by RBC’s from tissues to the lungs.

(ii) COz on entering the red blood cell loosely combines with the amino group of reduced Hb and forms carbaminohaemoglobin.

CO₂ + Hb ⇔ HbCO₂
(Carbaminohaemoglobin)

Section – D

Question 25.
Read the following to answer any four questions 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 ceEs. These spores germinate to give rise to gametophytes. The gametophytes bear male and female sex organs. Water is required for 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 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 zygote develops 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
(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 ¡s 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 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) hornosporous and heterosporous
(D) megasporous.
Answer:
(C) hornosporous 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 ptendophytes, the male and female sex organs 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 all type of pteridophytes, all the spores are of similar kinds. (True/False)
Answer:
False.
Explanation:
Some genera also produce two kinds of spores, macro (large) and micro (small) spores.

Question 26.
Describe the structure of cell wall.
OR
Describe the sub-phases of prophase I of meiosis I.
Answer:
The cell wall is differentiated into three parts i.e., middle lamella, primary wall and secondary wall.

(i) Middle lamella:
(a) It is a thin amorphous cementing layer between two adjacent cells.
(b) It is the first Layer, which is deposited at the time of cytokinesis.
(c) It is composed of calcium and magnesium pectate.

(ii) Primary wall:
(a) It is thin, elastic and extensible in growing cells.
(b) It grows by addition of wall material within the existing one. Such growth is termed as intussusception.
(c) This wall consist consists of a loose network of cellulose microfibrils.
(d) These microfibrils in fungi are made up of a polymer of acetyl glucosamine. (y) The matrix of wall is chiefly composed of water, hemicellulose, pectins and glycoproteins.

(iii) Secondary wall:
(a) After maturity, a thick secondary wall is laid inner to the primary wall by accretion or deposition of materials. It is found in many layers.
(b) A number of different materials may be deposited in the secondary wall, e.g., lignin and suberin.

OR
Prophase I of meiosis I:
It is more complicated and prolonged as compared to the similar stage of mitosis. It may be divided into the five sub-phases.
(i) Leptotene or leptonema:
(a) It is the first stage of meiotic division following the interphase.
(b) The chromatin fibres of interphase nucleus gets shortened and elongated, chromosomes become clear.

(ii) Zygotenc or zygonema:
The homologous chromosomes come to Lie side by side and get attadwd laterally due to the development of nudeoprotein between them. This pairing of homologous chromosomes is called synapsis or synthesis. On account of synapsis, chromosomes form pairs or bivalent. The number of bivalents is half the
number of the total chromosomes.

(iii) Pachytene or pachynema:
(a) The paired chromosomes or bivalent become short and thick. The chromatids of each paired chromosome slightly separate and become visible. Thus, each bivalent or chromosome pair is made up of actually four chromatids, two of each chromosome,

(b) Each group of four chromatids is referred to as tetrad. During pachytene, the exchange of the corresponding segments of non-sister chromatids of homologous chromosomes, occurs. This process is called crossing over,

(iv) Diplotene or Diplonema:
(a) The nucleoprotein fusion complex of the synapsed chromosomes dissolves.
(b) The homologous chromosomes start separating except in the region of crossing over.
(c) The points of attachment between the homologous chromosomes are called chiasmata.

(v) Diakinesis: This is marked by terminalisation of chiasmata. During this phase, the chromosomes are fully condensed and the meiotic spindle is assembled to prepare the homologous chromosomes for separation. The nucleolus/ nucleoli disappear and nudear envelope breaks down into vesicles.

Question 27.
Read the following and answer any four questions from (i) to (v) given below: 5×1=5
The current and projected changes in abiotic stresses such as heat, cold, drought and salinity adversely affect the plant growth that ultimately limits productivity and is the leading cause of crop losses worldwide. During stress conditions, plants reduce the growth and development process, which ultimately affects the yield. In stress conditions, plants develop various stress mechanisms to face the magnitude of stress challenges.

Therefore, many strategies have been used to produce abiotic stress tolerance crop plants, among them, absdsic acid (ABA) phytohormone engineering could be one of the methods of choice. ABA regulates various physiological processes ranging from stomatal opening to protein storage and provides adaptation to many stresses like drought, salt, and cold stresses. ABA acts as the signalling mediator for regulating the adaptive response of plants to different environmental stress conditions.

(i) In the given paragraph which phytohormone is a derivative of carotenoids?
Answer:

(ii) Why Abscisic acid is called stress hormone of tree?
Answer:
Absrisic add helps the plant to survive during unfavourable conditions of the environment. It is observed that when there is a defitiency of water in leaves, ABA concentration increases and it promotes closure of stomata hence it is called stress hormone.

(iii) How ABA is antagonistic to Gibberellic acid?
Answer:
ABA is a plant growth inhibitor. It induces dormancy and promotes senescence. These effects are opposite to that of gibberellins, hence ABA acts as an antagonist to gibberellins.

(iv) State some factors for development of stress in plant.
Answer:
Drought, salt, heat and cold weather are some factors of plant stress.

(v) State some effects of ABA.
Answer:
Absdsic acid stimulates the growth and development of the root system including adventitious roots of the hypocotyl, the formation and growth of the lateral shoots of the cotyledonary node etc.

OR
Under high concentration of O₂, RuBP binds with O₂ and forms one molecule of phosphoglycerate and phosphoglycolate. 1+2+2

(i) What is this process called?
Answer:
Photorespiration

(ii) Why is this process called a wasteful process?
Answer:
Photorespiration is called wasteful process because :
(a) No energy is produced in this process.
(b) Oxygen is consumed for nothing.
(c) H₂O₂ is produced which is highly toxic.
(d) The yield of photosynthesis is reduced to 50%.

(iii) Briefly explain the pathway by which some plants ensure that the RuBisCO functions as a carboxylase, minimizing the oxygenase activity?
Answer:
In C₄ plants e.g., maize, sugar cane, grasses etc. light reaction occurs in mesophyll cells and RuBisCO is found in bundle sheath cells in which CO₂ fixation occurs. Thus, in C₄ plants RuBisCO remains protected from sunlight and is also protected from oxygenation, because in bundle sheath cells only dark reactions occurs. In these plants the phosphoenol pyruvate (PEP), a 3-carbon compound acts as CO₂ acceptor. The PEP combines with CO₂ to form a 4-carbon compound oxaloacetric acid (OAA) which is soon converted into malic acid, a 4-carbon compound. The malic acid releases CO₂ which is transferred to bundle sheath cells where it is fixed through Calvin cycle. The C₄ plants have a characteristic Kranz anatomy of leaf. There is no photorespiration in these plants.