students must start practicing the questions from CBSE Sample Papers for Class 11 Biology with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Biology Set 7 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.
Assertion: Plant hormones are called phytohormones.
Reason: They increase the rate of reaction, accelerate growth and other changes in plants.
(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:
The phytohormones either increase or decrease the metabolic activities to regulate the growth, development and physiological process in plants.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

Question 2.
Assertion: Energy is released during the oxidation of compounds in respiration.
Reason: Energy is stored as bond energy of ATP.
(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:
Energy is stored as bond energy of ATE This bond is broken to release energy when required.

Question 3.
In breathing movements, air volume can be estimated by:
(A) Stethoscope
(B) Hygrometer
(C) Sphygmomanometer
(D) Spirometer.
Answer:
(D) Spirometer.

Question 4.
A bivalent of meiosis-I consists of:
(A) Two chromatids and one centromere.
(B) Two chromatids and two centromeres.
(C) Four chromatids and two centromeres.
(D) Four chromatids and four centromeres
Answer:
(C) Four chromatids and two centromeres.

Question 5.
A primary protein should normally have:
(A) Two ends
(B) One end
(C) Three ends
(D) No ends
Answer:
(A) Two ends
Explanation:
Primary protein is simply an amino acid sequence that has two ends, the carboxyl amino terminals.

Section -B

Question 6.
Name the four different forms of lysosomes.
Answer:
Four different forms of Lysosomes are:

  1. Primary lysosome,
  2. Secondary lysosome,
  3. Autophagosomes,
  4. Residual bodies.

Question 7.
Define and explain the “Living State”.
Answer:
Living State: The living state is a non-equilibrium steady state to be able to perform work.

  1. The living process is a constant effort to prevent falling into equilibrium and is achieved by energy input.
  2. The metabolism provides a mechanism for the production of energy.
  3. The living state and metabolism are synonymous, without metabolism there cannot exist a living state.
  4. A living organism works continuously so it cannot afford to reach equilibrium because a system at equilibrium cannot perform the work

Question 8.
Name a substance that prevent the blood coagulation in uninjured blood vessels? How do they act?
Answer:

  1. The anticoagulant heparin prevents the blood form clotting.
  2. The injured platelets don’t form thromboplastin.

Question 9.
Mango and coconut are ‘drupe’ type of fruits. In mango fleshy mesocarp is edible. What is the edible part of
coconut? What does milk of tender coconut represent?
Answer:
The edible part of coconut is endosperm. The milk of tender coconut represents nuclear endosperm in liquid form. Later, this endosperm gets deposited along the walls of endocarp and forms edible flesh.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

Question 10.
What is bioluminescence? Give an example.
OR
Write a short note on slime moulds.
Answer:
Bioluminescence is the production and emission of light by a living organism. It is a form of chemiluminescence. Example: Nodiluca, Gonyaulax.
OR

  1. Slime moulds are both plant and animal like.
  2. They are plant like in the production of spores during reproduction and animal like in the mode of nutrition.
  3. Their somatic structures consist of wall-less, multinucleate mobile mass of protoplasm called plasmodium.
  4. They absorb nutrients directly from the substratum.
  5. The reproductive stage consists of sporangia and spores which are formed after meiosis

Question 11.
What do you mean by metagenesis? Give one example of an animal that shows metagenesis.
OR
What do you understand by metameric segmentation?
Answer:
Certain cnidarians exhibit both body forms (polyp and medusa), exhibit alternation of generation, i.e., polyps produce medusae asexually and medusae form the polyps sexually. This phenomenon is called metagenesis. Example: Obelia.
OR
Metameric segmentation is also known as true segmentation. It is a type of segmentation where external divisions correspond to internal divisions.

  1. The body is often divided both externally and internally into a number of segments, e.g., annelids.
  2. Segmentation is mostly external in arthropods and mainly internal in humans (vertebrae, body muscles, blood vessels, nerves).

Question 12.
Fungi are cosmopolitan, write the role of fungi in your daily life.
Answer:
Fungi are cosmopolitan. They are present in air, water, soil, over and inside plants and animals. They have both beneficial and harmful roles. For example:

  1. Few fungi like Mushrooms are used as nutritious and delicious food.
  2. Saprophytic fungi lives upon dead and decaying organic matter and break complex substances into simpler forms, which are then absorbed by plants as nutrients.
  3. Some fungi like Mucor, and Rhizopus possess soil binding properties and make the soil fertile for cultivation.
  4. Yeast has the property of fermentation, thus used in alcohol and dough preparation.

Section – C

Question 13.
(a) What do you understand by the secondary structure of a protein? Explain.
(b) What are amphipathic lipids?
Answer:

  1. Secondary structure is a three-dimensional form of a polypeptide chain. Many secondary structures in proteins possess two forms i.e., a-helix and P-pleated secondary structure.
  2. Lipid molecules that possess both hydrophilic and hydrophobic properties are called amphipathic lipids.

Question 14.
In a slide showing different types of cells can you identify which type of the cell may be meristematic and the one which is incapable of dividing and how?
Answer:
The meristematic cells can be identified by the following features:

  1. Cells have a cellulose cell wall and dense cytoplasm with a large nucleus.
  2. Plasmodesmata connections are more numerous among meristematic cells.
  3. Cell division i.e., mitosis and its various stages are distinctly visible.

Question 15.
The formation of the enzyme-substrate complex (ES) is the first step in catalyzed reactions. Describe the other steps till the formation of the product.
Answer:
Each enzyme molecule has an active site for specific binding of substrate molecules. The shape of the active sites of enzymes is exactly complementary to the shape of the substrate. The enzyme molecule operates by chemically binding with the substrate molecule, to form an enzyme-substrate complex. The enzyme’s tertiary structure consists of a unique pocket or site on which the substrate molecules can become attached and interact subsequently. This brings about an interaction between the specific active sites in the enzyme molecule and the reactive sites in the substrate molecule. The enzyme now breaks down the substrate into products.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

The products initially remain attached to the enzyme for a short while forming an enzyme product complex.
The products get released from the enzyme molecule subsequently. The enzyme is now ready to receive another substrate molecule again. Thus, the same enzyme can be used again and again.
S [substrate] + E [Enzyme] ⇔ ES [Enzyme Substrate complete]
EP [Enzyme Substrate complex] → E [Enzyme] + P [Product.]

Question 16.
What is the economic importance of plants of the family Solanaceae?
Answer:

  1. Potato (Solatium tuberosum) and tomato are the most common articles for cooking.
  2. Tobacco comes from the dried leaves of Nicotiana tabacum.
  3. Atropa bellodona yield belladona for relieving pain, cough, excessive respiration

Question 17.
Write the significance of citric add cycle.
Answer:
Significance of citric acid cycle:

  1. It explains the process of breaking of pyruvate into CO2 and water. It is the major pathway of generation of ATP
  2. More energy is released (24 ATP) in this process, as compared to glycolysis.
  3. in Kreb’s cyde, carbon skeletons are obtained for use in the growth and maintenance of the cell.

Many intermediate compounds are formed. They are used in the synthesis of other biomolecules like amino adds, nucleotides, chlorophyll, cytochromes and fats. Succinyl CO2 A is the starting molecule for synthesis of chlorophyll. Amino acids are synthesised from a-ketoglutaric acid, pyruvic acid, and oxaioacetic acid.
Pyruvic acid + 4NAD+ + FAD + 2H2 O + ADP +CBSE Sample Papers for Class 11 Biology Set 7 with Solutions 2 3CO2 + 4NADH +4W + FADH2 + ATP

Question 18.
Briefly describe the cell structure of blue-green algae.
Answer:

  1. Cyanobacteria or blue-green algae are gram (+) photosynthetic prokaryotes that perform oxygenic
    photosynthesis.
  2. Cell structure is typically prokaryotic one envelope organization with peptidoglycan wall, naked DNA, 70S ribosomes and devoid of membrane-bound cell organelles.
  3. The outer part of the protoplast contains a number of photosynthetic thylakoids. It is called as chromoplast.
  4. Their membranes contain chlorophyll-a, carotenes and xanthophylls. Chlorophyll-b is absent.
  5. DNA lies coiled generally in the central part of the cytoplasm known as centroplasm.
  6. Small circular DNA segments may also occur in nudeoid, known as plasmid or transposons.

Question 19.
Explain the stage diakinesis of meiosis I division.
Answer:
Diakinesis of meiosis I:

  1. The bivalents become condensed to the maximum. These are evenly distributed throughout nucleus.
  2. Nucleolus disappears.
  3. The centromeres of bivalents repel each other and the chiasmata move laterally.
  4. As a result, the paired chromosomes are united only at the ends. This process is known as the “terminalisation”.
  5.  In big-sized chromosomes with many chiasmata, terminalisation is never complete.
  6. They remain joined together until metaphase-I. Nuclear membrane is broken into pieces and disappears then.
  7. Bivalents lie evenly distributed in cytoplasm without overlapping among them.
  8. For counting the chromosome in sexually reproducing organisms, it seems the best stage of cell division.

Question 20.
Describe the three disorders of the muscular and skeletal system.
Answer:
The three disorders are:

  1. Myasthenia gravis: It is an autoimmune disorder affecting neuromuscular junction leading to fatigue, weakening and paralysis of skeletal muscle.
  2. Tetany: Rapid spasms (wild contractions) in muscle due to lesser Ca++ in body fluid is tetany.
  3. Muscular dystrophy: It is an inborn abnormality associated with progressive degeneration of skeletal muscle.

Question 21.

  1. Draw the labelled diagram of a nephron.
  2. What is the role of Henle’s loop?

Answer:
CBSE Sample Papers for Class 11 Biology Set 7 with Solutions 1
(ii) Loop of Henle plays a significant role in the maintenance of high osmolalrity of medullary interstitial fluid.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

Question 22.
Explain the auto-regulatory mechanism of GFR.
OR
Explain, why a hemodialysis unit is called an artificial kidney.
Answer:
The juxtaglomerular apparatus is a microscopic structure located between the vascular pole of the renal corpuscle and the returning distal convoluted tubule of the same nephron. It plays a major role in regulating renal blood flow and glomerular filtration rate (GFR). When there is a fall in the glomerular filtration rate, it activates the juxtaglomerular cells to release renin.
OR
A haemo dialysing unit is called artificial kidney because it performs the same functions as that of a kidney. It helps in removing the nitrogenous wastes from the blood without losing plasma proteins. This method is a boon for people suffering from uremia. Uremia is the accumulation of urea in the blood. In this process, the dialysing unit contains a coiled cellophane tube surrounded by dialysing fluid having the same composition of plasma except for the nitrogenous wastes. Blood drained from a convenient artery is pumped into dialysing unit after adding an anticoagulant like heparin.

The porous cellophane membrane of the tube allows the passage of molecules based on concentration gradient. As nitrogenous wastes are absent in dialysing fluid, these substances freely move out, thereby clearing the blood. The cleared blood is pumped back to the body through a vein after adding anti-heparin to it. This stimulates the glomerular blood flow, thereby bringing the GFR back to normal. Renin brings the GFR back to normal by the activation of renin – angiotensin mechanism.

Question 23.

  1. How do fungi form partnership with most plants?
  2. Explain the following in terms of mycorrhizal assodation: Nourishment, Absorption of water and Growth promotion.

OR
What is a bacteriophage? Also describe its structure.
Answer:
(i) Some fungi have evolved essential relationships with the roots of many living plants. This partnership is called mycorrhizal symbiosis.

(ii) Nourishment: In orchids, the fungi absorb nourishment from outside, transport it over to the germinating seed. Absorption of water: Fungal association increases the water availability to the root. Growth promotion: Fungus produces growth-promoting hormones and provides it to the plant.
OR
Bacteriophage: Viruses infecting bacteria are known as bacteriophage.

  1. These are obligate parasites that occur in soil, sewage water, and fruits.
  2. The phage possesses a tail and a head.
  3. The viral DNA is a thread-like double-stranded macromolecule.
  4. There are 4 segments of the tail.
  5. Bacteriophage are of two types:
    • Lytic and
    • Lysogenic phages e.g., phage (Lambda), T2 phage etc.

Question 24.
Give distinguishing features of Phylum Porifera.
OR
Write a brief note on reproduction of Platyhelminthes.
Answer:
Phylum Porifera includes the most primitive and simplest animals. They are commonly called sponges.

  1. They show the cellular level of organisation. The cells are almost independent, cooperating very little with one another in their function.
  2. Sponges are neither diploblastic nor triploblastic because they have cellular level of organisation.
  3. The sponges have porous body viz. provided with pores. The pores are of two types: inhalant pores are called ostia (e.g., ostium) and exhalent pores are known as oscula (e.g., osculum).
  4. The animals possess a large cavity called spongocoel or paragastric cavity. It opens to the outside by a terminal opening called osculum.
  5. One of the most important features of sponges is the presence of an interconnected network of channels called canal system. Water enters the spongocoel through ostia and goes outside through osculum.
  6. The sponges possess an internal skeleton of calcareous or siliceous spicules or irregular proteinaceous spongin fibers or both located in the mesenchyme.
  7. Asexual reproduction occurs by budding or special cells mass called gemmules or internal buds. Sexual reproduction involves the formation of ova and spermatozoa.

OR

  1. Flatworms are generally hermaphrodites. However, the anatomy of reproduction system encourages cross-fertilization.
  2. Gonads have ducts that leads to the copulatory organs.
  3. They are viviparous.
  4. Life history often includes one or more larva stages.
  5. Most of the parasitic forms involve more than one host to complete their life cycle.
  6. Asexual reproduction by transverse fission occurs in some forms.
  7. They also have the power of regeneration.

Section – D

Question 25.
What are the products of light reaction?
Name the first stable product of C3 plants. Explain different steps of Calvin cycle.
OR
Over-secretion of GH stimulates abnormal growth of the body leading to gigantism and low secretion of GH results in stunted growth resulting in pituitary dwarfism. Excess secretion of growth hormone in adults espedally in middle age can result in severe disfigurement (especially of the face) called Acromegaly, which may lead to serious complications, and premature death if unchecked. The disease is hard to diagnose in the early stages and often goes undetected for many years, until changes in external features become noticeable.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

Prolactin regulates the growth of the mammary glands and the formation of milk in them. TSH stimulates the synthesis and secretion of thyroid hormones from the thyroid gland. ACTH stimulates the synthesis and secretion of steroid hormones called glucocorticoids from the adrenal cortex. LH and FSH stimulate gonadal activity and hence are called gonadotropins. In males, LH stimulates the synthesis and secretion of hormones called androgens from testis. In males, FSH and androgens regulate spermatogenesis.
Answer:
ATP and NADPH, the product of light reaction are used in synthesis of food. The firstCO2 fixation product in C3 Plants is 3-phosphoglyceric acid. The CO2 acceptor is RuBE Calvin cycle has three stages :
(i) Carboxylation: CO2 + RuBP → 2 molecule of PGA.

(ii) Reduction: Glucose is formed at the expense of ATP and NADPH. It involves 2 ATP for phosphorylation and 2 NADPH for reduction per CO2 molecule fixed.

(iii) Regeneration: RuBP has formed again. 6 turns of Calvin cycle and 18 ATP molecules are required to synthesise one molecule of glucose.
6CO2 + 6RuBP + 18ATP + 12NADPH → C6 H12 C6 + 6RuBP + 18ADP + 18Pi + 12NADR

(i) Dwarfism is caused by secretion of growth hormone.
Answer:
(i) Low.

(ii) How Acromegaly is different from Gigantism?
Answer:
Gigantism is the abnormal growth of the body from the initial stage of growth due to over secretion of growth hormone and Acromegaly is caused by excess secretion of growth hormone at middle age which causes disfigurement.

(iii) ACTH stimulates the synthesis and storage of peptide hormones called glucocorticoids from the adrenal medulla. [Correct the statement if it is wrong]
Answer:
ACTH stimulates the synthesis and SECRETION of STEROID hormones called glucocorticoids from the adrenal CORTEX.

(iv) Spermatogenesis is controlled by this hormone.
Answer:
FSH [follicle stimulating hormone ] and androgens regulate spermatogenesis.

(v) Androgen secretion from the testis is controlled by which hormone?
Answer:
LH [ luteinizing hormone].

Question 26.
Trace the events in a muscle fibre from the time it receives the impulse through the neuromuscular junction up to the contractile response.
Answer:
Events in Muscle fibre:

  1. A nerve impulse arriving at the neuromuscular junction stimulates contractile response.
  2. Due to the depolarisation of the surface of sarcomeres, it spreads quickly.
  3. Neurotransmitter is released at the neuromuscular junction.
  4. It enters into the sacromere through membrane channel. Na+.moves inside the sarcomere.
  5.  It is called inflow of Na+.
  6. Action potential is generated in the sarcomere. Action potential travels to the full length of muscle fiber.
  7. The sacroplasmic reticulum, then, releases the Ca++ which is stored here.
  8. It binds to the specific sites found in the troponin of the thin actin filament.
  9. Change takes place in troponin and active sites of F-actin are exposed to myosin head.
  10. Myosin head shows Mg++ dependent ATPase activity.
  11. During relaxation of muscle, Ca++ is pumped back into the sarcoplasmic reticulum.
  12. Consequently, the troponin component is freed to inhibit the active sites for myosin head. Cross bridges are broken.
  13. The thin filaments assume their normal position.
  14. The muscle fiber then comes in relaxed state

Question 27.
Read the following to answer any four questions from (i) to (v) given below: (5×1=5)
The anatomy of monocot root is similar to the dicot root in many respects. It has epidermis, cortex, endodermis, pericycle, vascular bundles and pith. As compared to dicot root which have fewer xylem bundles there we usually more than six (polyarch) xylem bundles in the monocot root.

(i) Outer layer of monocot root is?
Answer:
Epidermis is the outer layer of monocot root.

(ii) The parenchymatous cells which lie between the xylem and Phloem are?
Answer:
Conjunctive tissue.

(iii) What is stele?
Answer:
All tissue on the inner side of endodermis such as pericycle, vascular bundles and pith constitute the stele.

(iv) Differentiate monocot and dicot roots. On the basis of xylem bundles.
Answer:
Dicot roots have fewer xylem bundles, there are more than six (Polyarch) xylem bundles in the monocot root.

CBSE Sample Papers for Class 11 Biology Set 7 with Solutions

(v) Do monocot roots undergo secondary growth?
Answer:
Monocot roots do not undergo Secondary growth.

OR
The important functions of roots are; fixation of plant in the soil i.e., ground, absorption of nutrients and water from soil and conduction of absorbed materials from soil to aerial parts of the plant. In addition to the above functions, some adventitious roots perform different function i.e., in Cuscuta (a parasitic plant) they absorb food from the host’s body; in banyan; the prop roots provide support to the plant, in maize, Rhizophora, they support the plant; in Tinospora the green roots perform the function of photosynthesis; in some plants they get swollen and perform as storage organs for the plant; other perform the function of vegetative reproduction. Some roots perform the functions of storage of food reproduction, climbing, giving support to plants.

(i) Roots developed from parts other than radicle or called:
(A) Tap roots.
(B) Fibrous roots.
(C) Adventitious roots.
(D) Nodular roots.
Answer:
(C) Adventitious roots.
Explanation:
Roots which develop from parts other than radicle or which arise from an organ other than the root usually a stem of sometime leaf are called adventitious roots. Root arising from radicle is the tap root.

(ii) Tuberous roots are found in
(A) Solatium tuberosum
(B) Ipomoea batata
(C) Momordica Charantia
(D) Piper betel
Answer:
(B) Ipomoea batata
Explanation:
These are the adventitious roots e.g., sweet potato which get swollen to store food.

(iii) Tap roots are seen in.
Answer:
Tap roots are seen in mustard plant.

(iv) What is the function of prop roots?
Answer:
The prop roots provide support to the plant.

(v) What are the functions of Tinospora root?
Answer:
Tinospora roots helps in climbing and also perform photosynthesis as their colour is green.