students must start practicing the questions from CBSE Sample Papers for Class 11 Biology with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Biology Set 5 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.
The taxonomic unit ‘Phylum’ in the classification of animals is equivalent to which hierarchical level in
(A) Class
(B) Order
(C) Division
(D) Family
Answer:
(C) Division

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Question 2.
The central hub is connected to the triplets through:
(A) Radial spokes.
(B) Lateral spokes.
(C) Transverse spokes.
(D) Pin head.
Direction (Qs. 3 & 4): 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:
Answer:
(A) Radial spokes.

Question 3.
Assertion (A): Vital capacity includes ERV, TV and IRV
Reason (R): It is the maximum volume of air a person can breathe out after a forced inspiration.
(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:
(A) Both A and R are true and R is the correct explanation of A.
Explanation:
Vital capacity is the maximum volume of air a person can breathe in after forced expiration. This includes ERV, TV and IRV

Question 4.
Assertion (A): The functional unit of contraction of muscle fibre is called sarcomere.
Reason (R): In the muscle fibre, the sarcomere is separated by an M line.
(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 functional unit of sarcomere is separated by Z line.

Question 5.
All living organisms are linked to one another because:
(A) They have common genetic material of the same type.
(B) They share common genetic material but to varying degrees.
(C) They have common cellular organization.
(D) All of the above.
Answer:
(B) They share common genetic material but to varying degrees.
Explanation:
Living organisms share common genetic material to some extent of varying degrees and hence they linked to one another.

Section – B

Question 6.
What is oxidative decarboxylation? What is its significance for Kreb’s cycle?
Answer:
Pyruvic acid in molecules enter the mitochondria where each of them is converted into two carbon atoms-acetic acid. One carbon is released as CO2. The removal of carbon dioxide from pyruvic acid is called decarboxylation. Oxidative decarboxylation plays a very significant role in Kreb’s cycle. This step is called link reaction or transition reaction or gateway step as it links glycolysis with Kreb’s cycle.

Question 7.
The transverse section of a plant material shows the following features:
Vascular bundles are conjoint, closed and scattered and are surrounded by the sclerenchymatous bundle sheath. What will you identify it as? Also, write any other four features of this specimen.
Answer:
It is a monocot stem. It is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundles is surrounded by sclerenchymatous

bundle-sheath cells. Other four features of monocot stem are:

  1. Vascular bundles are scattered in the ground tissue.
  2. Y-shaped xylem with endarch condition.
  3. Cambium is absent.
  4. Phloem parenchyma and medullary rays are absent in monocot stems.

Question 8.
What is pericardium and how it is helpful to humans?
Answer:
It is two layered sac consisting of outer parietal pericardium and inner visceral pericardium. In between the two layers, a space called the pericardial cavity is present which is filled with a pericardial fluid.

Significance:
The pericardium protects the heart from shocks and mechanical injuries and also allows free movement of the heart.

Question 9.
What are corals and coral reefs?
OR
In which plant will you look for mycorrhiza and corolloid roots? Also explain what these terms mean.
Answer:
Corals are minute cnidarians, which secrete deposits of lime and calcium carbonate. Coral reefs: In many coral species, new living corals develop over the old dead corals giving rise to coral reefs and island. The coral reefs provide a comfortable living place for a variety of animal species. They form stable marine ecosystems.
OR
Coralloid roots are a feature of Cycas plants. These are specialized roots, which grow on the surface of soil. These are irregular, dichotomously branched coral- like roots which do not possess root hair and root caps. Coralloid roots have a symbiotic relationship with blue-green algae like Nostoc and Anabaena species. Mycorrhiza is a symbiotic relationship between fungus and the roots of the vascular plants. Fungus colonised the roots of the host either intra or inter-cellular. It helps in the nutrient absorption from soil for the plant. It is commonly seen in plants like Pinus, Cedrus, etc.

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Question 10.
What is urine? Give the composition of urine.
Answer:
Urine is a pale yellow-coloured and transparent fluid.
(i) The pale yellow colour is due to pigment urochrome produced by the breakdown of haemoglobin from worn-out RBCs.

(ii) Normal urine contains about 95% water, 2% electrolytes (ions of salts, mainly chlorides, sulphates, bicarbonates of sodium, potassium etc.), 2.6% urea, 0.03% uric acid and traces of creatinine, ammonia, some pigments, hormones, allantoin and hippuric acid.

Question 11.
What is the advantage of having more than one pigment molecule in a photo centre?
Answer:

  1. The light reaction depends upon the amount of solar energy trapped by the pigment molecule.
  2. Energy trapped by a single pigment molecule is not enough to start the initial reactions which may occur in light.
  3. A number of pigment molecules capture light over a large surface area, and later on pass it on to photo centre.
  4. The presence of a number of pigment molecules would provide protection to the chlorophyll molecule against photo-oxidation

Question 12.
How do you distinguish a simple leaf from a compound leaf?
Answer:
Differences between simple and compound leaf:

Simple leaf Compound leaf
(i) Lamina is not divided into distinct lobes or leaflets i.e., it has a single lamina Lamina is incised into two or more distinct leaflets. The leaflets are borne either at the tip of petioles or on the sides of rachis. Leaflets do not bear any axillary bud.
(ii) Axillary bud is present in the axil of simple leaf Axillary bud is present in the axil of whole leaf.
(iii) Simple leaves are in acropetal succession on stem. Leaflets of a compound leaf are not in acropetal succession on rachis
(iv) Base of leaf may have stipules. Stipules may be present at the base of compound leaf.

Section – C

Question 13.
List any four salient features of the Watson-Crick model of DNA.
Answer:
Four salient features of Watson and Crick model are:

  1. DNA molecule consists of two polynucleotide chains, which are coiled like a rope in helical fashion. So the two strands form a ‘double helix’.
  2. The two strands of polynucleotides are antiparallel i.e., run in the opposite direction. The backbone is formed by the sugar-phosphate chain with the nitrogenous bases projected inside.
  3. Each DNA helix has alternate minor and major grooves having width of 12 A and 22 A respectively.
  4. The distance between two consecutive spirals is 34 A. Thus, between two consecutive spirals, 10 nucleotides can be adjusted.

Question 14.
Explain sexual dimorphism exhibited in frog.
OR

  • Name the larval stage in the life history of the frog.
  • Which is the most common Indian species of frog?
  • Why skin of the frog is moist and slippery?

Answer:
Frog exhibits sexual dimorphism. This means that the male and female frog can be distinguished by their external features. The male frog possesses vocal sacs which are most developed during the breeding season and also a copulatory pad on the first digit of the fore limbs which are absent in female frogs. Learn the concept of sexual dimorphism in frog carefully with proper explanation.
OR

  1. Tadpole
  2. Rana tigrina
  3. The skin of the frog is moist and slippery to perform cutaneous respiration as well as to slip away from the grip of predators.

Question 15.
Define plant hormones. In how many groups does plant growth regulators are classified?
Answer:
A plant hormone is a chemical substance produced naturally in plants which is translocated to another region for regulating one or more physiological reaction, when present in low concentration. Based on their functions, plant growth regulators are broadly divided into two groups:

  1. Growth Promoters: Such plant growth regulators are involved in growth-promoting activities such as cell division, cell enlargement, pattern formation etc. These include auxins, gibberellins and cytokinins.
  2. Growth Inhibitors: Such PGRs are involved in various growth-inhibiting activities such as promotion of dormancy and abscission. These include abscisic acid and ethylene.

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Question 16.
What are the differences between the roots of aquatic plants and terrestrial plants?
Answer:
Differences between the roots of aquatic plants and terrestrial plants are:

Simple leaf Compound leaf
(i) Roots may be absent e.g., Wolffia. If roots are present, they are not well-developed (i) Roots are well developed with root cap and root hair.
(ii) Vascular bundles are poorly developed. (ii) Vascular bundles are well developed.
(iii) Modified roots carry out the function of photosynthesis, food storage and exchange of gases. (iii) Roots help in anchorage and absorption of nutrients from soil.

Question 17.
Multicellular organisms have division of labour. Explain.
Answer:

  1. The division of labour in multicellular organisms is best represented by the presence of several types of organ system in the body.
  2. The cells in multicellular organisms originate by the division of single-celled zygote. These newly formed cells are specialised to perform specific function and a division of labour is established between these cells which co-exist in the body of the multicellular organisms.
  3. Division of labour provides better co-ordination and organisation to multicellular organisms. Thus, specialised structure performs different functions in the body.

Question 18.
Briefly describe the followings:

  1. Erythroblastosis foetalis.
  2. Angina pectoris
  3.  Atherosclerosis

Answer:
(i) Erythroblastosis foetalis: It is an Rh incompatibility between the Rh-ve blood of a pregnant mother and Rh+ve blood of the foetus. Rh antigens do not get mixed with maternal blood in first pregnancy because placenta separates the two bloods. But at the time of first delivery, there is a possibility of exposure of the maternal blood to small amounts of the Rh+ve blood from the foetus. This induces the formation of Rh antibodies in maternal blood. In case of her subsequent pregnancies, the Rh antibodies from the mother leak into the blood of the foetus (Rh +ve) and destroy the foetal RBCs. This is fatal to the foetus or cause severe anaemia and jaundice to the baby. This condition is called erythroblastosis foetalis.

(ii) Angina pectoris: Angina is an acute chest pain due to oxygen deficiency in heart muscles. It occurs due to improper blood flow. It is common among middle-aged and elderly people.

(iii) Atherosclerosis: It is a disorder, in which the deposition of calcium, fat, cholesterol and fibrous tissue occurs in the coronary arteries which makes the lumen of arteries narrower and thereby affecting the blood supply.

Question 19.
(a) What are green glands?
(b) What is radula?
(c) What are flame cells?
Answer:
(a) Excretory organs of phylum Arthropoda are called green glands.
(b) Radula is a teeth like structure found in the buccal cavity of snail.
(c) Flame cells are the excretory organs of Platyhelminthes and other related animals which possess flickering cilia or flagella for driving the excretory products. Both are excretory organs. Flame cells are found in phylum Platyhelminthes and green glands in phylum Arthropoda.

Question 20.
How does the blood flow through the heart during the different phases of the cardiac cycle?
Answer:
The action of the human heart consists of a series of events, which follow one another with great rapidity.
Main events in cardiac cycle are:

  1. Auricular systole,
  2. Ventricular systole and
  3. Joint diastole.

Thus, we can say that the contraction of the heart and its relaxation constitute cardiac cycle or heart beat. The duration of the cardiac cycle varies inversely with the cardiac rate. The heartbeat is 72 times per minute in man. A Single heartbeat lasts for 0.8 of a seconds. In such heart beats, blood comes to the heart and then is propelled from auricles to ventricles and then to arteries. The number of heartbeats per minute is very much influenced by age, size, sex, and temperature.

Question 21.
With respect to rib cage, explain the following:
(i) Bicephalic ribs
(ii) True ribs
(iii) Floating ribs
OR
What is the source of energy for muscle contraction?
Answer:

  1. Bicephalic ribs: Each ribs has two articulating surfaces on its dorsal end, hence, known as bicephalic ribs.
  2. True ribs: First 7 pairs of ribs are called true ribs. They are attached dorsally to the thoracic vertebrae and ventrally to the sternum with the help of hyaline cartilage.
  3. Floating ribs: Last two pairs of ribs are called floating ribs. They are not connected ventrally to the sternum, therefore, called as floating ribs.

OR
ATP (Adenosine Triphosphate) is the source of energy for muscle contraction. The head of myosin molecule contains an enzyme called myosin ATPase. This enzyme in the presence of calcium and magnesium ions, breaks down into ADP and inorganic phosphate, releasing energy in the head of myosin.
ATP → ADP + Pi + Energy
Energy from ATP causes energized myosin to form a cross bridge by binding to actin. These energised cross bridges move, causing thin myofilaments to slide along the thick myofilaments, thus initiating muscle contraction.

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

Question 22.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
OR
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:

  1. Tough cuticle protects the arthropods against predators and forms jointed exoskeleton for muscles attachment. This has enabled the arthropods to survive on land in almost any environment.
  2. Bilaterally symmetrical, segmented, triploblastic, coelomate animals, found almost everywhere in all habitats.
  3. Body enclosed by chitinous cuticle.
  4. They have joined appendages.
  5. They have mouth parts of various types.
  6. Trachea/book lungs or book gills for respiration.

OR
The number of eggs produced by an oviparous mother will be more than the viviparous mother. This is because, in oviparous animals, the development of a young one takes place outside the mother’s body. As a result, their eggs are more prone to environmental conditions and predators. Therefore, to overcome the loss, more eggs are produced by mothers so that even under harsh environment conditions, some eggs might be able to survive and produce young ones. On the other hand, a viviparous mother gives rise to fewer young ones because there are less chances of their death, as the development of young ones takes place in safe conditions inside the body of the mother.

Question 23.
What hormones are secreted by the posterior pituitary gland? What function does each serve? Where are these hormones actually produced? How are these hormones transported to the region from which they are released? 3
Answer:
(i) Oxytocin: It is released into the blood when hypothalamic neurons are stimulated by the widening of uterus at the time of delivery or by the sucking of breasts by an infant. It induces contraction of smooth muscles of the uterus during the birth of a young one and myoepithelial cells of mammary glands to cause the release of milk during sucking by an infant. Therefore, it is also known as ‘birth hormones’ and or ‘milk-ejecting hormones’.

(ii) Vasopressin: It is also called antidiuretic hormone (ADH). It decreases the loss of water in urine by increasing reabsorption of water in distal convoluted tubules, collecting tubules and collecting ducts of kidneys. It is transported as neurophysin – proteins bound secretory granules down the nerve fibre.

Question 24.
(i) What is G1 phase in cell cycle?
(ii) How does cytokinesis differ in plant and animal cells?
Answer:
(i) G1 phase: (a) In this phase, the cell is metabolically active and continuously grows but does not replicate its DNA. (b) RNA and proteins are synthesised and the cell grows in size.
(ii) Cytokinesis in plant cells Vs Cytokinesis in animal cells.

Cytokinesis in plant cells Cytokinesis in animals cells
(i) In these, cell wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. In these cells, cytokinesis is achieved by the ap-pearance of a furrow in the plasma membrane.
(ii) Formation of new cell wall begins with the formation of a precursor called cell plate. The furrow gradually deepens and ultimately join in the centre dividing the cytoplasm into two.

Section – D

Question 25.
Explain briefly the following terms with suitable examples: 5×1=5
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Sporophyll
(v) Isogamy
Answer:
(i) Protonema: It is a juvenile stage of a moss that develops from the germinating meiospores. When fully grown, it comprises slender, green, branching system of filaments called protonema.

(ii) Antheridium: Male sex organs in bryophytes and pteridophytes are known as antheridium. They are jacketed and multicellular which produces a number of flagellated male gametes called sperms or antherozoids.

(iii) Archegonium: The female sex organs of bryophytes, pteridophytes and gymnosperms are called as archegonia. It is a flask shaped structure with a tubular neck and a swollen venter. The single-layered wall of neck has 5-6 rows of the cell. It encloses a few sterile neck canal cells. The walls of the venter encloses a few sterile venter canal cell and a fertile egg or oosphere.

(iv) Sporophylls: Spores bearing leaves of ferns are called as sporophylls. Sporangia occur on the leaves on clusters called sori, so they are also known as fertile leaves. A sorus is covered by a flap-like outgrowth from its surface or turned margins of sporophylls. They are of two types: microsporophyll which bear microsporangia and megasporophyll which bear megasporangia.

(v) Isogamy: In isogamy, the fusing gametes are morphologically and physiologically similar. They may be flagellate or non-flagellate.

Question 26.
Read the following to answer the questions with explanation from (i) to (v) given below: 5×1=5
Fluid mosaic model was proposed by Singer and Nicolson. 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 embedded in the lipid layer are called integral or intrinsic proteins and they cannot be separated from the membrane. There are large globular integral proteins which project beyond the lipid layer on both sides. Superficially attached proteins are called extrinsic or peripheral proteins and can be easily removed. Some membrane lipids and integral proteins remain bound to oligosaccharide which projects into the extracellular fluid.

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

(i) The structure of membrane is:
(A) Lipid rafts were predicted by early models of cell membrane structure thalli
(B) Primarily made of cholesterol molecules.
(C) Glycoprotein on the cell surface is necessary for immune recognition.
(D) All of the above.
Answer:
(B) Primarily made of cholesterol molecules.
Explanation:
Glycosphingolipids, cholesterol glycoprotein lipid micro-domains are together called as lipid rafts.

(ii) The passage of substances across the cell membrane occurs by:
(A) Active transport
(B) Passive transport
(C) Bulk transport
(D) All of the above
Answer:
(D) All of the above
Explanation:
These are the mechanisms through which molecules can be moved across the membrane. Passive transport mechanism uses no energy while the active transplant requires energy to get it done.

(iii) Who proposed the fluid mosaic model of plasma membrane?
(A) homosporous
(B) Schleiden and Schwann
(C) Singer and Nicolson
(D) Robert brown
Answer:
(C) Singer and Nicolson
Explanation:
This mode was proposed by Singer and Nicolson. According to fluid mosaic model, the membrane is made up of lipid bilayer and integral and peripheral proteins in mosaic pattern.

(iv) Which of the following statements is not true for plasma membrane?
(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:
(D) Carbohydrate is found in it.
Explanation: Plasma membrane is of lipoprotein nature, carbohydrates are never found in it.

(v) Following proteins cannot be separated from the membrane?
(A) All types of protein.
(B) Intrinsic protein.
(C) Peripheral protein.
(D) None of the above.
Answer:
(B) Intrinsic protein.
Explanation: Some proteins are embedded in the lipid layer are called integral or intrinsic proteins and they cannot be separated from the membrane.

Question 27.
Read the following and answer any four questions from (i) to (v) given below: 5×1=5
The synthesis of energy-rich ATP molecules with the help of energy liberated during oxidation of reduced co-enzymes produced in respiration. It involves the enzymes required for ATP synthesis is called ATP synthase. The enzyme is activated with the passage of protons through the channel due to the proton gradient.
CBSE Sample Papers for Class 11 Biology Set 5 with Solutions 1

(i) The synthesis of ATP molecules is called
Answer:
Oxidative phosphorylation

(ii) How the protons are available for ATP formation?
Answer:
The proton available for ATP formation comes from the electron transport chain.

(iii) What are the reduced co-enzymes produced in the respiration?
Answer:
(iii) NADH2 and FADH2

(iv) In which part, enzyme ATP synthase is present?
(A) F0 particle
(B) F1 particle
(C) Mitochondrial matrix
(D) Inner mitochondrial membrane
Answer:
(C) Mitochondrial matrix

CBSE Sample Papers for Class 11 Biology Set 5 with Solutions

(v) Direction: In the following question, the Assertions (A) and Reason (R) have been put forward.
Read both the statements and choose the correct option from the following:
Assertion (A): 2H+ pass throughF0-F1 particle that forms a molecule of ATP from ADP and Pi.
Reason (R): 2H+ pass through F0-F1 from the lower proton gradient in the inter-membrane space to the higher proton gradient in the matrix.
(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:
2H+ pass through F0-F1 particles from the higher proton gradient in the intermembrane space to the lower proton gradient in the matrix.