Students can access the CBSE Sample Papers for Class 11 Biology with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Biology Set 1 with Solutions
Time : 3 Hours
Maximum Marks : 70
General Instructions:
- All questions are compulsory.
- The question paper has five sections and 33 questions. All questions are compulsory.
- Section-A has 16 questions ofl mark each; Section-B has 5 questions of 2 marks each; Section- C has 7 questions of 3 marks each; Section- Dhas 2 case-based questions of 4 marks each; and Section-E has 3 questions of 5 marks each.
- 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 labeled diagrams should be drawn.
Section – A
Question 1.
Resting membrane potential is maintained by: 1
(A) Hormones
(B) Neurotransmitters
(C) Ion pumps
(D) None of the above
Answer:
Option (C) is correct.
Explanation: Different types of ion channels, such as the sodium-potassium pump and leak channels, maintain resting membrane potentials.
Question 2.
Contagium vivumfluidum was proposed by: 1
(A) D.J. Ivanowsky.
(B) M.W Beijerinck.
(C) Stanley.
(D) Robert Hooke.
Answer:
Option (B) is correct.
Explanation: Martinus Beijernick first
conceptualised a virus using the phrase “contagium vivum fluidum” (‘soluble living germ’).
Question 3.
Match the following (column I with column II). 1
Column I | Column II |
(a) Chlamydomonas | (i) Moss |
(b) Cycas | (ii) Pteridophyte |
(c) Selaginella | (iii) Algae |
(d) Sphagnum | (iv) Gymnosperm |
(A) (a) – (iii), (b) – (iv), (c) – (ii), (d) – (i).
(B) (a) – (iii), (b) – (iv), (c) – (i), (d) – (ii).
(C) (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii).
(D) (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i).
Answer:
Option (A) is correct.
Explanation: The correct matching pair is:
Column I | Column II |
(a) Chlamydomonas | (iii) Algae |
(b) Cycas | (iv) Gymnosperm |
(c) Selaginella | (ii) Pteridophyte |
(d) Sphagnum | (i) Moss |
(a) Chlamydomonas is a genus of biflagellated single- celled green algae that can be found in soil, ponds, and ditches (family Chlamydomonadaceae).
(b) Cycas belongs to gymnosperm order cycadales.
(c) An example of pteridophyte is Selaginella. It is sometimes known as club moss or spike moss.
(d) Sphagnum, commonly known as sphagnum moss, is a genus of about 380 recognised species of mosses.
Question 4.
What type of movable joint is present between the atlas and axis? 1
(A) Pivot
(B) Saddle
(C) Hinge
(D) Gliding
Answer:
Option (A) is correct.
Explanation: Pivot joints allow for one-way
movement. It is present between first (atlas) and second (axis) vertebral bone.
Question 5.
Epiblema of the root is equivalent to: 1
(A) Pericycle.
(B) Endodermis.
(C) Epidermis.
(D) Stele.
Answer:
Option (C) is correct.
Explanation: The epidermis is the outermost layer of cells on all plant organs, including the stem, root, leaf, flower, fruit, and seed. It is developed from the protoderm. The waxy cuticle layer of the epidermis serves as a defence against mechanical injury, water loss, and infection. Therefore, the epiblema of the root is equivalent to the epidermis.
Question 6.
The pH of human urine is approximately: 1
(A) 6.5
(B) 7
(C) 6
(D) 7.5
Answer:
Option (C) is correct.
Explanation: Human urine has a pH of 6 on average. The pH can, however, range from 4 to 8. The clinical identification of numerous metabolic and renal problems is aided by urine analysis.
Question 7.
PEP is primary CO2 acceptor in: 1
(A) C4 plants.
(B) C3 plants.
(C) C2 plants.
(D) Both C3 and C4 plants.
Answer:
Option (A) is correct.
Explanation: In the mesophyll cells of C4 plants, phosphoenolpyruvate (PEP), a 3-carbon molecule, serves as the primary CO2 acceptor, resulting in the generation of OAA (oxaloacetic acid), a C4 acid.
Question 8.
Identify the type of placentation in the diagram given below: 1
(A) Marginal placentation
(B) Axile placentation
(C) Parietal placentation
(D) Free central placentation
Answer:
Option (B) is correct.
Explanation: The given diagram represents axile placentation. In this type, placenta bears ovules developing from the central axis of a compound ovary corresponding to the fused margins of carpels. e.g., Citrus, Solanum.
Question 9.
The second heart sound (dubb) is associated with the closure of: 1
(A) Tricuspid valves.
(B) Semilunar valves.
(C) Bicuspid valves.
(D) Both (A) and (C)
Answer:
Option (B) is correct.
Explanation: The second sound of the heart is a high-pitched “DUB” caused by the semilunar valves. It results from vibrations which is due to closure of aortic and pulmonary valves.
Question 10.
Identify the structure given below: 1
(A) Maltose
(B) Lactose
(C) Sucrose
(D) Glucose
Answer:
Option (C) is correct.
Explanation: The given structure is of sucrose. It is a disaccharide, a molecule composed of two monosaccharides: glucose and fructose. It is a non¬reducing disaccharide with chemical formula of C12H22O11.
Question 11.
The oxygen – haemoglobin dissociation curve will show a right shift in case of: 1
(A) High pCO2.
(B) High pO2.
(C) Low pCO2.
(D) Less H+ concentration
Answer:
Option (A) is correct.
Explanation: An increase in [pCO2 ], [H+], and temperature, can lead to the right shift of the oxygen-haemoglobin dissociation curve.
Question 12.
Which of the following describes the given graph correctly? 1
(A) Endothermic reaction with energy A in the presence of enzyme and B in the absence of enzyme.
(B) Exothermic reaction with energy A in the presence of enzyme and B in the absence of enzyme.
(C) Endothermic reaction with energy A in the absence of enzyme and B in the presence of enzyme.
(D) Exothermic reaction with energy A in the absence of enzyme and B in the presence of the enzyme.
Answer:
Option (B) is correct.
Explanation: Exothermic reaction with energy A when an enzyme is present and B when an enzyme is absent.
Question No. 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer the questions selecting the appropriate option given below.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.
Question 13.
Assertion (A): Angina is an acute chest pain. 1
Reason (R): It involves oxygen deficiency in heart muscles.
Answer:
Option (A) is correct.
Explanation: Angina is an acute chest pain due to oxygen deficiency in heart muscles. It occurs due to improper blood flow.
Question 14.
Assertion (A): Species are static units in classification. 1
Reason (R): Species change over time.
Answer:
Option (D) is correct.
Explanation: Species are dynamic, ever-changing and genetically distinct group of organisms.
Question 15.
Assertion (A): In respiration process, oxygen is utilised while carbon dioxide, water and energy are obtained as
products. 1
Reason (R): The combustion of glucose produces CO2 and H2O and energy in the form of ATP.
Answer:
Option (C) is correct.
Explanation: Respiration is the breakdown of glucose that utilise O2 to produce CO2 and H2O and energy in the form of ATP. The complete combustion of glucose, which produces CO2 and H2O as end products, yields energy most of which is given out as heat.
Question 16.
Assertion (A): Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory
system. 1
Reason (R): Each bronchiole terminates into an irregular walled, vascularised bag-like structure called bronchi.
Answer:
Option (C) is correct.
Explanation: The alveoli have very thin walls consisting of squamous epithelium. The alveolar wall is also provided with an extensive network of blood. But other parts are lined by thick lining which does not allow exchange of gases. Each bronchiole terminates into an irregular walled, vascularised bag like structure called alveoli.
Section – B
Question 17.
What is two kingdom classification? Give its drawback. 2
Answer:
Carolus Linnaeus anciently divided all living organisms into two kingdoms: Plantae and Animalia.
The drawbacks of this classification are:
- First formed animals were neither plants nor animals.
- Fungi differ in structure, physiology and reproductively from plants.
- At lower level of organization, there are several instances where the distinction of plant and animal disappears.
Question 18.
In the diagrammatic representation of the heart given below, mark and label: 2
(a) SAN and AVN
(b) AV bundles-bundle of His and Purkinje fibres.
Answer:
Question 19.
Enzymes are proteins. Proteins are long chains of amino acids linked to each other by peptide bonds. Amino acids have many functional groups in their structure. These functional groups are, many of them at least, ionisable. As they are weak acids and bases in chemical nature, this ionisation is influenced by pH of the solution. For many enzymes, activity is influenced by surrounding pH. This is depicted in the curve below. Explain the graph briefly. 2
Answer:
Enzyme generally shows their highest activity at a particular pH called optimum pH. A rise or fall in pH reduces enzyme activity. The graph given represents the maximum enzyme activity at the optimum pH.
Question 20.
Given below is a diagram showing ATP synthesis during aerobic respiration. Replace the symbols A, B, C, D and E with appropriate terms given below:
F1 particle, Pi, 2H+, Inner mitochondrial membrane, ATP FO particle, ADP. 2
Answer:
Question 21.
Fill up the blank spaces (a), (b), (c) and (d) in the table given below. 2
Type of flower | Position of calyx, corolla and androecium in respect of the ovary on the thalamus | Type of ovary |
Hypogynous | (a) …………………. | Superior |
Perigynous | On the rim of the thalamus almost on the same level of the ovary. | (b) ………………….. |
(c) …………………. | (d) ………………….. | Inferior |
OR
Plants require water for their survival but when watered excessively plants die. Discuss.
Answer:
(a) Floral parts are situated just below the thalamus.
(b) Half inferior.
(c) Epigynous.
(d) Floral parts are present around the ovary.
OR
Excessive watering drives away air from the soil. Root absorption, root growth and root survival are dependent upon aerobic respiration. In the absence of air or oxygen, the roots fail to absorb water despite its excess presence in the soil. Therefore, the plants die.
Section – C
Question 22.
List any five functions of nucleic acids? 3
Answer:
(a) Primary structure of protein: The first level of structural organisation in a protein is the order/ sequence of amino acids which comprise the polypeptide chain. The primary structure is formed by covalent peptide bonds between the amine and carboxyl groups of adjacent amino acids. Primary structure controls all subsequent levels of protein organisation because it determines the nature of the interactions between R groups of different amino acids.
(b) Secondary structure of protein: In this, the amino acids in the polypeptide chain forms highly, regular shapes through the hydrogen bond between the carbonyl oxygen and the neighbouring amine hydrogen of the main chain. a-helix and P-pleated secondary structure are two most common substructures formed by proteins.
Question 23.
Explain sexual dimorphism exhibited in frog. 3
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 frog.
Question 24.
Distinguish between the following. 3
(a) Dicot flower and Monocot flower.
(b) Gamopetalous and Polypetalous.
Answer:
(a)
S. No | Dicot Flower | Monocot Flower |
(i) | Sepals and petals are present. | Perianth is present. |
(ii) | Floral leaves are in the multiple of 2 or 5. | Floral leaves are in the multiple of three. |
(b)
Gamopetalous | Polypetalous |
Flowers in which petals are partially or completely fused are termed as gamopetalous flowers. | Flowers in which petals are free are polypetalous flowers. |
e.g., Petunia, Tobacco. | e.g., Hibiscus rosa sinensis (China rose). |
Question 25.
Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions. 3
(a) At which point/s (A, B or C) in the curve is light a limiting factor?
(b) What do C and D represent on the curve?
Answer:
(a) At higher light intensities, the rate of photosynthesis becomes limiting. Light is limiting in A and 50 % of B.
(b) C is a saturation of light. Here, the rate of photosynthesis is not increased by increasing the intensity of light. (D) represents the intensity of light at which the same factor may be limiting.
Question 26.
Explain in brief the structure of prothallus of fern? 3
Answer:
Prothallus of ferns –
- It is a heart-shaped structure. The sex organs are present on the lower surface of the prothallus below the apical notch.
- Sex organs are antheridia and archegonia.
- Prothallus is produced from the meiospore gametophyte of fern.
- Below the sex organs are rhizoids.
- Archegonia are flask shaped but antheridia are globose.
- Male and female gametes are produced in antheridia and archegonia.
Question 27.
Differentiate between: 3
(a) S-phase and G2 phase.
(b) G1 phase and G2 phase.
OR
List any five functions of nucleic acids?
Answer:
(a)
S.No. | S-phase | G2 phase |
(i) | It occurs in between the G1 phase and G2 phase. | It occurs in between the S phase and mitotic phase. |
(ii) | Replication of DNA occurs. | It is the phase of rapid cell growth and protein synthesis. |
(iii) | It is the second sub -phase of interphase. It lasts for 6 – 8 hours. | It is the last sub phase of interphase that directly leads to divisional phase. It lasts for 2 – 5 hours. |
(b)
S.No. | G1 phase | G2 phase |
(i) | It is called the first growth period. | It is post growth phase. |
(ii) | Its duration is variable. | It lasts for 2 – 5 hour. |
(iii) | Cells grow in size. | Cell prepares to go into the mitotic phase. |
OR
Functions of nucleic acid:
- DNA is a genetic material which carries all the hereditary information.
- DNA also transfer genetic information from one generation to other.
- RNAs are involved in the expression of genetic code of DNA by forming specific protein.
- Some RNAs act as enzyme.
- RNA helps in protein synthesis.
Question 28.
Some symbiotic organism are very good pollution indicators and are composed of a chlorophyllous and a non- chlorophyllous members. Name and describe them. 3
Answer:
Lichens are symbiotic association between algal components (phycobiont) and fungal components (mycobiont).
Algae provide food while fungi provide shelter and absorb nutrients and water for alga.
They are good pollution indicator as they do not grow in polluted area.
Section – D
Question 29.
Gametophyte is a dominant phase in the life cycle of a bryophyte. It is more conspicuous, long living, independent, green and freely branched whereas, the sporophytic phase is short-lived and dependent upon the gametophyte. The main plant body is haploid and bears sex organs i.e., antheridium and archegonium. Antheridium produces a large number of flagellated male gametes called antherozoids and archegonium is flask shaped with tubular neck and swollen venter. The gametopnytic plant body of bryophytes is thalloid in liverworts whereas foliose in mosses. In liverworts, the thallus is differentiated into a dorsal photosynthetic and ventral storage region. In mosses, the gametophyte has two growth stages, protonema stage and leafy stage or gametophore. 4
(a) What is bryophyte?
(b) In a bryophyte, the cells of X are in haploid and cells of Y are in diploid condition. Identify X and Y.
(c) State any two differences between antheridia and archegonia.
OR
Explain the gametophyte of bryophyta?
Answer:
(a) Bryophytes are non-vascular, small plants which include hornworts, liverworts, and mosses.
(b) Cells of gametophyte are haploid and cells of sporophyte are diploid.
(c) Antheridia: (i) It is the male reproductive organ.
(ii) It is racket shaped.
Archegonia: (i) It is the female reproductive organ.
(ii) It is usually flask shaped.
OR
Bryophyta possesses both gametophyte (n) and sporophyte (2n). Gametes are produced by the gametophyte. The sex organs are multicellular. Male sex organ is called archegonium which produces antherozoids. Antherozoids possess two flagella and are motile. A flask-shaped structure called archegonium is the female sex organ. It produces only one egg.
Question 30.
In the following diagram shows the myofibrils in different state of action. Answer any four questions from (a) to (c)
as follows. 4
(a) What is the state of myofibrils in (i) and (ii) respectively?
(b) What are I band and A band also called as respectively?
(c) Which ion is responsible for contraction or relaxation of myofibrils?
(d) What does the H-zone of a sarcomere in a myofibril contain?
OR
Why microfibrils have striated appearance?
Answer:
(a) (i) It is relaxed state of myofibril.
(ii) It is contracted state of myofibril.
(b) I band is light band or isotropic band while A band is a dark band or anisotropic band.
(c) Ca2+ ions are responsible for the contraction or relaxation of myofibril.
(d) H-zone of a sarcomere contains thick myofilament. It is the center part of A-band.
OR
Myofibril has striated appearance is before due to the distribution pattern of two important proteins – Actin and myosin. The light bands contain actin and is called I-band or isotropic band, whereas the dark band called A’ or anisotropic band.
Section – E
Question 31.
Explain any two disorders of bones? 5
OR
Describe the structure of a human kidney with the help of a labelled diagram.
Answer:
Arthritis and osteoporosis are two diseases of bones in humans.
(a) Arthritis: Inflammation of the joints is the main cause of it. It is of many types like rheumatoid arthritis, osteoarthritis and gouty arthritis.
- The rheumatoid arthritis is diagnosed by the presence of rheumatoid factor. The joints becomes immovable. It is cured by heat treatment and physiotherapy.
- The osteoarthritis is a degenerative joint disorder. It is characterised by the degeneration of the articular cartilage and proliferation of new bones.
- The gouty arthritis is called gout. This is caused due to excessive formation of uric acid which gets deposited at the joints.’
(b) Osteoporosis:
-
- It is age-dependent systemic disorder.
- It is characterised by low bone mass, micro architectural deterioration of the bone.
- It is also characterised to fracture.
- It can occur in pregnant women.
- Thyrocalcitonin, parathyroid and sex- hormones imbalance is the main cause of this disease.
- Major causative factors are the deficiency of vitamin D and calcium deficiency in the body.
OR
Structure of human kidney.
-
- Kidneys are reddish brown, bean-shaped structures situated between the levels of last thoracic and 3rd lumbar vertebra.
- Each kidney is about 10 – 12 cm in length, 5 – 7 cm in width, and 2-3 cm in thickness.
- In adult males, the average weight of kidney is 150 gm, and in the adult female is about 135 gm.
- Each kidney is enclosed in a tough, three¬layered fibrous capsule.
- On the concave side of the kidney, there is an opening called hilum or hilus through which blood vessels, nerves, lymphatic ducts and ureter enter the kidney.
- Hilum leads to funnel shaped cavity called renal pelvis with projections called calyces.
- Each kidney has an outer dark region called cortex and inner lighter region called medulla.
- The medulla is divided into a number of conical projections called renal pyramids (medullary pyramids) projecting into the minor calyces. Minor calyces lead into major calyces.
- The major calyces open into a funnel shaped structure called renal pelvis which in turn leads into the ureter.
- Between the medullary pyramids, the substance of cortex extends into the medulla as renal columns called Columns of Bertini.
- Each kidney has nearly one million tubular nephrons.
Question 32.
Explain the following methods of the passage of substances across the cell membrane ? 5
(a) Passive transport.
(b) Active transport.
(c) Bulk transport.
OR
(a) Cell is the basic unit of life. Discuss in brief.
(b) What are the characteristics of prokaryotic cells?
(c) (i) What is a mesosome in a prokaryotic cell?
(ii) Mention the functions that it performs.
Answer:
The passage of substances across the cell membrane occurs by three methods.
(a) Passive transport: It is a mode of membrane transport which occurs without the expenditure of cell energy. Passive transport occurs by diffusion or osmosis. Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient. Osmosis is a special type of diffusion in which only water molecules migrate across a semipermeable membrane from the area of high concentration to the area of low concentration.
(b) Active transport: It is the movement of materials across the membrane against the concentration gradient. This often leads to the accumulation of substances at a higher concentration within the cell than outside. Energy is required for such a process, which ATP usually provides. There are certain proteins in the membrane, which acts as a carrier molecules or carrier proteins. It has a binding site for the substrate. This transports the carrier-bound substrate to other side of the membrane.
(c) Bulk transport: It includes exocytosis and endocytosis, which are the processes for active cellular extrusion and intake respectively of such materials that cannot pass through the unbroken plasma membrane. Both the processes involve movement and folding of the membrane, which depends on its fluidity and mobility.
OR
(a) Cell is the basic unit of life.
(i) All life begins as a single cell.
(ii) All life activities of an organism are present in miniature form in each and every cell of its body.
(b) (i) The outer covering of the protoplasm of bacterial cell is called as cell envelope, which comprises three components: Glycocalyx, cell wall and cell membrane.
(ii) Cytoplasm is granular, crystallo-colloid complex that forms the protoplasm.
(iii) The genetic material of prokaryotes is called as nucleoid. Nucleoid consists of a single circular strand of DNA duplex.
(iv) Some additional rings of extrachromosomal DNA called plasmids are present which can replicate independently.
(v) Bacteria bear flagella for locomotion. Bacterial flagellum is made up of 3 parts: basal body, hook and filament.
(c)
(i) Mesosomes are extensions of plasma membrane into the cell in the form of vesicles, tubules and lamellae.
(ii) The functions of a mesosome in a prokaryotic cell are:
- They help in the synthesis of cell wall and replication of DNA.
- The also help in respiration, and secretion processes to increase the surface area of the plasma membrane to carry out various enzymatic activities.
Question 33.
Given below is a certain name of different environment. Analyse and suggest about the adaptation that plant have
developed to optimise photosynthesis in extreme environment along with the suitable examples. 5
Extreme Environment | Adaptation | Example |
High altitude | ||
Desert | ||
Arctic tundra | ||
Waterlogged soils | ||
High salinity | ||
Extreme heat | ||
Low light | ||
Nutrient-poor soils |
OR
Certain aspects are provided in this table. Analyse the aspects and explain briefly according to the questions given below.
Aspects | C3 Pathway | C4 Pathway |
Leaf anatomy | ||
Initial carbon fixation | ||
First stable product | ||
Location of process | ||
Spatial separation | ||
Role of RuBisCO | ||
Photorespiration | ||
Water use efficiency | ||
Temperature sensitivity | ||
CO2 compensation point | ||
Examples |
Answer:
Extreme Environment | Adaptation | Example |
High altitude | Increased chlorophyll content to maximize light absorption in low light conditions. | Alpine plants have higher chlorophyll concentrations. |
Desert | Reduced leaf surface area to minimize water loss through transpiration. | Cacti have spines instead of leaves. |
Arctic tundra | Dark pigmentation on leaves to enhance light absorption and heat absorption. | Arctic willows have dark-colored leaves. |
Waterlogged soils | Aerenchyma tissue in roots and stems to facilitate oxygen transport to submerged tissues. | Rice plants have specialized air channels. |
High salinity | Salt excretion or accumulation mechanisms to maintain osmotic balance. | Mangroves excrete salt through specialized glands. |
Extreme heat | Crassulacean Acid Metabolism (CAM) or C4 photosynthesis to reduce water loss through stomata. | Agave plants use CAM photosynthesis. |
Low light | Elongated and thinner leaves to increase light capture and penetration. | Shade-tolerant plants, like ferns, have such adaptations. |
Nutrient-poor soils | Mycorrhizal associations with fungi to enhance nutrient uptake. | Orchids often form mycorrhizal associations. |
OR
Aspects | C3 Pathway | C4 Pathway |
Leaf anatomy | Normal leaf anatomy | Kranz anatomy (distinct bundle sheath cells) |
Initial carbon fixation | Carbon dioxide combines with RuBP | Carbon dioxide combines with PEPCase |
First stable product | 3-phosphoglycerate (PGA) | 4-carbon compound (oxaloacetate or malate) |
Location of process | Mesophyll cells | Mesophyll cells and bundle sheath cells |
Spatial separation | None | Carbon fixation in mesophyll, Calvin cycle in bundle sheath |
Role of RuBisCO | Responsible for both carboxylation and oxygenation reactions | Mainly carboxylation, minimal oxygenation |
Photorespiration | More pronounced due to oxygenation by RuBisCO | Reduced due to minimized oxygenation |
Water use efficiency | Lower water-use efficiency | Higher water-use efficiency |
Temperature sensitivity | Sensitive to high temperature | Better adaptation to high temperature |
CO2 compensation point | Lower CO2 compensation point | Higher CO2 compensation point |
Examples | Rice, wheat, soybeans, most plants | Corn, sugarcane, switchgrass, many grasses |