This includes questions from Class 11 Biology chapters.
Q 1: Which taxonomical category is not used for plants?
(1) Phylum
(2) Class
(3) Division
(4) Order
✅ (1) Phylum
▶️ In plants, the term ‘Division’ is used instead of Phylum.
▶️ Order: It is an assemblage of related families. E.g. Families Convolvulaceae & Solanaceae belong to Order Polymoniales.
▶️ Class: It is an assemblage of related orders. E.g. Orders Primata, Carnivora etc. belong to class Mammalia.
▶️ Phylum (Division in case of plants): It is an assemblage of related classes. E.g. Classes Amphibia, Reptilia, Aves, Mammalia etc. come under phylum Chordata.
Q 2: Select the wrong statement related to diatoms.
(1) They are the chief ‘producers’ in the oceans.
(2) They have siliceous cell walls forming two thin overlapping shells.
(3) They are photosynthetic dinoflagellates.
(4) Diatomaceous earth is the cell wall deposit of diatoms over billions of years in their habitat.
✅ (3) They are photosynthetic dinoflagellates.
▶️ Diatoms belong to Chrysophyta (a phylum of Kingdom Protista).
▶️ They have siliceous cell walls forming two thin overlapping shells, which fit together as in a soap box.
▶️ The cell wall deposit of diatoms over billions of years in their habitat is known as ‘diatomaceous earth’. This is used in polishing, filtration of oils and syrups.
Q 3: Of the following organisms, cell wall is present in
(1) Chlamydomonas and Paramecium
(2) Amoeba and Chlorella
(3) Euglena and Chlamydomonas
(4) Chlamydomonas and Chlorella
✅ (4) Chlamydomonas and Chlorella
▶️ Chlamydomonas and Chlorella belong to Algae.
▶️ Paramoecium and Amoeba belong to Protozoa (a phylum under Kingdom Protista). It has no cell wall.
▶️ Euglena belongs to Euglenoid (a phylum under Kingdom Protista). Instead of a cell wall, they have a protein rich layer called pellicle. It makes their body flexible.
Q 4: Examples for filamentous form of algae are
(1) Ulothrix and Spirogyra
(2) Ulothrix and Chlamydomonas
(3) Volvox and Chlamydomonas
(4) Ulothrix and Volvox
✅ (1) Ulothrix and Spirogyra
▶️ The form and size of algae is highly variable.
▶️ Microscopic unicellular forms: E.g. Chlamydomonas.
▶️ Colonial forms: E.g. Volvox.
▶️ Filamentous forms: E.g. Ulothrix and Spirogyra.
Q 5: Algin and Carrageen are Hydrocolloids (water holding substances) produced from
(1) Red algae and Brown algae respectively
(2) Green algae and Red algae respectively
(3) Brown algae and Red algae respectively
(4) Red algae and Green algae respectively
✅ (3) Brown algae and Red algae respectively
▶️ Some marine brown & red algae produce hydrocolloids (water holding substances). E.g. algin (brown algae) and carrageen (red algae). These are used commercially.
▶️ Agar (from Gelidium & Gracilaria) is used to grow microbes and in ice-creams and jellies.
▶️ Protein-rich unicellular algae like Chlorella & Spirulina are used as food supplements by space travelers.
Q 6: The following figure represents 
(1) Sycon which belongs to Porifera
(2) Sycon which belongs to Cnidaria
(3) Hydra which belongs to Porifera
(4) Hydra which belongs to Cnidaria
✅ (1) Sycon which belongs to Porifera
Unique features of Porifera (Sponges):
▶️ Cellular grade of organisation.
▶️ Water canal (water transport) system.
▶️ Millions of ostia (pores).
▶️ Spongocoel & canals are lined with choanocytes (collar cells).
▶️ Body is supported by spicules and spongin fibres.
▶️ E.g., Sycon (Scypha), Spongilla (fresh water sponge), Euspongia (Bath sponge)
Q 7: The phyllotaxy A, B & C respectively represent 
(1) A= Alternate B= Opposite, C= Whorled
(2) A= Opposite, B= Alternate, C= Whorled
(3) A= Whorled, B= Opposite, C= Alternate
(4) A= Opposite, B= Whorled, C= Alternate
✅ (1) A= Alternate B= Opposite, C= Whorled
▶️ Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. 3 types:
1. Alternate: A single leaf arises at each node in alternate manner. E.g. China rose, mustard & sun flower.
2. Opposite: A pair of leaves arise at each node and lie opposite to each other. E.g. Calotropis and guava.
3. Whorled: More than two leaves arise at a node and form a whorl. E.g. Alstonia.
Q 8: Casparian strip is mainly found in
(1) Epidermis of roots
(2) Endodermis of stem
(3) Endodermis of roots
(4) Epidermis of stem
✅ (3) Endodermis of roots
TS of the sunflower root shows the following parts:
▶️ Epidermis: Outermost layer. Unicellular root hairs present.
▶️ Cortex: Several layers of parenchyma cells with intercellular spaces.
▶️ Endodermis: Innermost single layer of the cortex. The tangential and radial walls of the endodermal cells have a deposition of suberin (water impermeable, waxy material) in the form of casparian strips.
▶️ Stele: All tissues on the inner side of the endodermis together constitute stele.
Q 9: A: Polar molecules require membrane carrier protein for transport.
B: Polar molecules alone cannot pass through the non-polar lipid bilayer.
B: Polar molecules alone cannot pass through the non-polar lipid bilayer.
(1) A and B are correct and B is correct explanation for A.
(2) A and B are correct and B is not correct explanation for A.
(3) A is correct but B is wrong.
(4) Both A and B are wrong.
✅ (1) A and B are correct and B is correct explanation for A.
Types of Transport
▶️ 1.Passive transport: Movement of molecules across the membrane from higher concentration to the lower without the expenditure of energy. It is 2 types:
a. Simple diffusion: Movement of neutral solutes across the membrane.
b. Osmosis: Movement of water by diffusion across the membrane.
Polar molecules cannot pass through the non-polar lipid bilayer. So they require membrane carrier protein for transport.
▶️ 2. Active transport: Movement of molecules across the membrane from lower to the higher concentration with the expenditure of energy.
Q 10: General formula of fatty acids is
(1) R-COOH where R is any carbohydrate molecule
(2) R-CHO where R is long chain hydrocarbon
(3) R-CHO where R is any carbohydrate molecule
(4) R-COOH where R is long chain hydrocarbon
✅ (4) R-COOH where R is long chain hydrocarbon
▶️ Fatty acids are lipids with a hydrocarbon chain (R- group) ending in –COOH group. i.e. R-COOH.
▶️ E.g. Palmitic acid has 16 carbons (CH3 - (CH(2)14 - COOH or C15H31-COOH) and Arachidonic acid has 20 Carbons.
▶️ Fatty acids are 2 types:
1. Saturated fatty acids: They have no double or triple bonds between carbon atoms. E.g. Palmitic acid, Stearic acid (C17H35COOH) etc.
2. Unsaturated Fatty acids: They have one or more C=C bonds. E.g. Oleic acid (C17H33COOH), Arachidonic acid (C19H31COOH) etc.
Q 11: The correct sequence of stages in karyokinesis is
(1) Metaphase, Prophase, Telophase & Anaphase
(2) Prophase, Metaphase, Anaphase & Telophase
(3) Telophase, Prophase, Anaphase & Metaphase
(4) Anaphase, Prophase, Telophase & Metaphase
✅ (2) Prophase, Metaphase, Anaphase & Telophase
Key events of karyokinesis of mitosis
▶️ Prophase: The nucleolus and nuclear membrane vanish. Chromosomes condense and become visible.
▶️ Metaphase: Chromosomes align at the metaphase plate (the equatorial plane), and spindle fibers connect the chromosomes to the centromeres.
▶️ Anaphase: The sister chromatids separate at the centromere and move to opposite poles.
▶️ Telophase: The chromosomes reach the poles and begin to decondense. The nuclear membrane and nucleolus reappear.
Q 12: Assertion: Chloroplasts are present in the walls of mesophyll cells of leaves.
Reason: It helps to get optimum quantity of incident light.
Reason: It helps to get optimum quantity of incident light.
(1) Both assertion and reason are true and reason is the correct explanation of assertion.
(2) Both assertion and reason are true but reason is not the correct explanation of assertion.
(3) Assertion is true but reason is false.
(4) Both assertion and reason are false.
✅ (1) Both assertion and reason are true and reason is the correct explanation of assertion.
▶️ Chloroplast contains a membranous system. It consists of grana, stroma lamellae & matrix stroma.
▶️ Each granum is a group of membrane-bound sacs called thylakoids (lamellae). They contain leaf pigments.
▶️ The membrane system traps light energy and synthesize ATP and NADPH. It is called light reactions.
▶️ In stroma, enzymatic reactions synthesize sugar, which in turn forms starch. It is called dark reactions (carbon reactions).
Q 13: In anaerobes, the only process in respiration is
(1) Krebs’ cycle
(2) EMP pathway
(3) Lactic acid fermentation
(4) Oxidative phosphorylation
✅ (2) EMP pathway
▶️ In anaerobic organisms, the Embden-Meyerhof-Parnas (EMP) pathway, also known as glycolysis, is the primary pathway for energy production. This process breaks down glucose into pyruvate, producing ATP and NADH in the absence of oxygen. Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms.
▶️ Krebs’ cycle and Oxidative phosphorylation are parts of aerobic respiration, which requires oxygen.
▶️ Lactic acid fermentation is a specific type of anaerobic respiration that occurs in some organisms, but it’s not the only process in anaerobic respiration.
Q 14: Complete A, B, C and D in the following diagram. 
(1) A= plasmatic growth, B= cell division, C= Maturation, D= Senescence
(2) A= cell division, B= plasmatic growth, C= Maturation, D= Senescence
(3) A= cell division, B= Maturation, C= plasmatic growth, D= Senescence
(4) A= plasmatic growth, B= Maturation, C= cell division, D= Senescence
✅ (2) A= cell division, B= plasmatic growth, C= Maturation, D= Senescence
▶️ In plants, development is a process that includes all changes in the life cycle of an organism from seed germination to senescence.
▶️ It is the sum of growth and differentiation.
Q 15: The factor which does not affect the rate of alveolar diffusion is
(1) Solubility of gases
(2) Thickness of membranes
(3) Pressure gradients
(4) Reactivity of the gases
✅ (4) Reactivity of the gases
Factors affecting rate of diffusion of gases across the alveolar membrane:
▶️ Solubility of gases: The rate of diffusion of a substance is directly proportional to its solubility.
▶️ Thickness of membranes: The rate of diffusion is inversely proportional to the thickness of the membrane.
▶️ Pressure gradient: Rate of diffusion of a substance across a membrane is directly proportional to the concentration gradient (pressure gradient for gases).
▶️ Surface area of respiratory membrane: Alveoli increase the surface area of lungs. It increases the gas exchange.
Q 16: Number of heart chambers present in fish, amphibians, reptiles, birds and mammals respectively is
(1) 2, 3, 4, 4, 4
(2) 2, 2, 3, 4, 4
(3) 2, 3, 3, 3, 4
(4) 2, 3, 3, 4, 4
✅ (4) 2, 3, 3, 4, 4
▶️ All vertebrates have a muscular chambered heart.
▶️ Fishes: 2-chambered heart (an atrium + a ventricle).
▶️ Amphibians: 3-chambered heart (2 atria + a ventricle).
▶️ Reptiles (except crocodiles): 3-chambered heart (2 atria + a ventricle). Ventricle is incompletely partitioned.
▶️ Crocodiles, birds & mammals: 4-chambered heart.
Q 17: The vessel that takes blood away from the glomerulus is
(1) Renal artery
(2) Renal vein
(3) Efferent arteriole
(4) Afferent arteriole
✅ (3) Efferent arteriole
▶️ Renal artery: It carries blood into kidney.
▶️ Renal vein: It carries blood away from kidney.
▶️ Afferent arteriole: A fine branch of renal artery which carries blood into glomerulus.
▶️ Efferent arteriole: It carries blood away from the glomerulus.
Q 18: If sliding filament theory is true, choose the correct statement.
(1) The sarcomere length remains unchanged.
(2) The distance between two adjacent Z lines remains unchanged.
(3) The A-band length remains unchanged.
(4) During fibre contraction both A band and I band shortens.
✅ (3) The A-band length remains unchanged.
According to the sliding filament theory, during muscle contraction, the thin filaments slide past the thick filaments, causing the sarcomere to shorten.
This results in changes to several components of the sarcomere:
▶️ The sarcomere length decreases.
▶️ The distance between two adjacent Z lines decreases.
▶️ The A-band length remains unchanged. This is because the A-band is made up of the entire length of the thick (myosin) filaments, and these do not change length during contraction.
▶️ During fibre contraction, the I band shortens but the A band does not.
Q 19: During the propagation of a nerve impulse, the action potential results from the movement of
(1) K+ ions from extracellular fluid to intracellular fluid
(2) Na+ ions from intracellular fluid to extracellular fluid
(3) K+ ions from intracellular fluid to extracellular fluid
(4) Na+ ions from extracellular fluid to intracellular fluid
✅ (4) Na+ ions from extracellular fluid to intracellular fluid
During the propagation of a nerve impulse, the action potential is generated by the movement of ions across the neuron’s membrane.
This process involves two key steps:
▶️ Depolarization: During this, Na+ ions move from the extracellular fluid to the intracellular fluid, causing the inside of the neuron to become more positive.
▶️ Repolarization: After the action potential, the neuron returns to its resting state. This is called repolarization. During this, K+ ions move from the intracellular fluid to the extracellular fluid, restoring the negative charge inside the neuron.
Q 20: If pituitary is surgically removed, blood level of sodium falls and that of potassium rises up due to
(1) Fact that LTH from pituitary is no longer available
(2) Atrophy of adrenal medulla
(3) Atrophy of adrenal cortex
(4) Fact that oxytocin from pituitary is no longer available
✅ (3) Atrophy of adrenal cortex
▶️ The surgical removal of the pituitary gland leads to a decrease in the blood level of sodium and an increase in the blood level of potassium. This is due to atrophy of the adrenal cortex, which results in a deficiency of aldosterone.
▶️ Aldosterone is a hormone that helps regulate the balance of sodium and potassium in the body. When aldosterone levels decrease, the body may excrete more sodium and retain more potassium.