This includes questions from Class 11 Biology chapters.
Q 1: What is true about Mangifera indica Linn.?
(1) Mangifera is the species name
(2) Linn is the generic name
(3) Mangifera is the name of author
(4) Indica is the specific epithet
✅ (4) Indica is the specific epithet.
Universal rules of Binomial nomenclature
▶️ Scientific names are in Latin and written in italics.
▶️ The first word is genus name (Generic name) and second word is the species name (specific epithet).
▶️ The Genus name starts with capital letter and the species name starts with small letter.
▶️ Name of the author (in abbreviated form) appears at the end of the biological name.
▶️ E.g., In Mangifera indica Linn. Mangifera is Genus name, indica is species name and Linn. is Author’s name.
Q 2: Which of the following is not a character of Protista?
(1) Protists are prokaryotes
(2) Some protists have cell walls
(3) Mode of nutrition is autotrophic & heterotrophic
(4) Membrane bound organelles are present in cells
✅ (1) Protists are prokaryotes.
▶️ Protists are eukaryotes, which means they have a well-defined nucleus and other membrane-bound organelles.
▶️ Only Kingdom Monera belongs to prokaryotes.
▶️ Protist groups such as Chrysophytes and Dinoflagellates have cell wall.
Q 3: Agar is obtained from
(1) Ulothrix, Chara & Porphyra
(2) Sargassum & Volvox
(3) Chlamydomonas & Volvox
(4) Gelidium & Gracilaria
✅ (4) Gelidium & Gracilaria.
▶️ Agar is used to grow microbes and in ice-creams and jellies.
▶️ Many marine algae are used as food. E.g. Porphyra, Laminaria and Sargassum.
▶️ Protein-rich unicellular algae like Chlorella & Spirulina are used as food supplements by space travelers.
Q 4: Polymorphism and alternation of generation are exhibited in some animals coming under
(1) Arthropoda
(2) Annelida
(3) Cnidaria
(4) Echinodermata
✅ (3) Cnidaria
▶️ Polymorphism refers to the occurrence of different forms within a population of a species. E.g., Polyp & Medusa are 2 types of body forms in cnidarians.
▶️ Alternation of generation (Metagenesis): The phenomenon in which polyps produce medusae asexually and medusae form the polyps sexually. E.g. Obelia.
Q 5: Which one of the following statements is false regarding Cymose inflorescence?
(1) Main axis terminates in a flower.
(2) Growth is limited.
(3) Flowers are borne in a basipetal order.
(4) Flowers are borne laterally.
✅ (4) Flowers are borne laterally.
Based on whether the apex gets converted into a flower or continues to grow, inflorescences are 2 types:
▶️ Racemose: In this, the main axis continues to grow. Flowers are borne laterally in an acropetal succession.
▶️ Cymose: In this, main axis terminates in a flower, hence is limited in growth. Flowers are borne in a basipetal order.
Q 6: A typical flower has four kinds of whorls arranged on the thalamus. They are
(1) Calyx, bracts, pedicel & gynoecium
(2) Pedicel, bracts, androecium & gynoecium
(3) Calyx, corolla, androecium & gynoecium
(4) Calyx, corolla, pedicel & bracts
✅ (3) Calyx, corolla, androecium & gynoecium.
▶️ Calyx & corolla are accessory organs, while androecium and gynoecium are reproductive organs.
▶️ In flowers like lily, the calyx and corolla are not distinct. It is called perianth.
▶️ When a flower has both androecium and gynoecium, it is bisexual. A flower having either only androecium or only gynoecium is unisexual.
Q 7: In stomata, a few epidermal cells, near the guard cells become specialized in their shape and size. They are called
(1) Mesophyll
(2) Subsidiary cells
(3) Bulliform cells
(4) Bundle sheath cells
✅ (2) Subsidiary cells.
▶️ Mesophyll: The tissue between upper and lower epidermis. It is made up of parenchyma. They contain chloroplasts for photosynthesis.
▶️ Bulliform cells: These are specialized large, empty, and colorless cells found in grasses, specifically arising from certain adaxial epidermal cells along the veins.
▶️ Bundle sheath cells: These are specialized cells that surround the vascular bundles in the leaves.
Q 8: Bidder’s canal is related with
(1) Reproductive system of male frog
(2) Reproductive system of female frog
(3) Excretory system of male frog
(4) Excretory system of female frog
✅ (1) Reproductive system of male frog.
▶️ Reproductive organs of male frog consist of a pair of yellowish ovoid testes. They are adhered to the upper part of kidneys by a double fold of peritoneum (Mesorchium).
▶️ Bidder’s canal is located in the male frog’s kidney. But it is not part of excretory system as it has no role in excretion. It receives sperms from the vasa efferentia and helps in sperm passage.
Q 9: Which of the following is not prokaryotic?
(1) Bacteria
(2) PPLO
(3) Amoeba
(4) Blue green algae
✅ (3) Amoeba
▶️ Amoeba is a eukaryotic which has membrane-bound organelles, including a nucleus.
▶️ Bacteria, PPLO (Pleuro pneumonia-like organisms and blue-green algae are examples of prokaryotic organisms, as they consist of prokaryotic cells without a true nucleus or membrane-bound organelles.
Q 10: Nucleic acids are the polymer of
(1) Nucleosomes linked by peptide bond
(2) Nucleotides linked by phosphodiester bond
(3) Nucleosides linked by N-glycosidic bond
(4) Nucleosides linked by phosphodiester bond
✅ (2) Nucleotides linked by phosphodiester bond.
▶️ Nucleosomes are the basic unit of DNA packaging in eukaryotes. A nucleosome is composed of a segment of DNA wound around eight histone proteins (histone octamer).
▶️ A peptide bond is a covalent chemical bond formed between two amino acids.
▶️ A nucleoside consists of a nitrogen base and sugar.
▶️ A nucleotide consists of a nitrogen base, a sugar and a phosphate group. Nucleotides are linked together by phosphodiester bonds to form a polynucleotide chain.
Q 11: The longest and shortest phases in mitosis is
(1) Prophase and Anaphase respectively
(2) Metaphase and Telophase respectively
(3) Anaphase and Prophase respectively
(4) Telophase and Metaphase respectively
✅ (1) Prophase and Anaphase respectively.
▶️ During prophase, chromatin condenses into chromosomes, and the nuclear envelope breaks down.
▶️ During metaphase, all the chromosomes align at the metaphase plate (the equatorial plane in the middle of the cell).
▶️ During anaphase, the sister chromatids separate and move to opposite poles of the cell.
▶️ During telophase, the separated chromosomes reach the opposite ends of the cell, and new nuclear envelopes form around each set of chromosomes.
Q 12: In Purple & green Sulphur bacteria, oxygen is not evolved during photosynthesis because hydrogen donor is
(1) Water (H2O)
(2) Hydrogen sulphide (H2S)
(3) Sulfur dioxide (SO2)
(4) Methane (CH4)
✅ (2) Hydrogen sulphide (H2S).
▶️ In plant photosynthesis H2O is the hydrogen donor and oxygen is released.
▶️ In Purple & green sulphur bacteria, H2S is the hydrogen donor and elemental sulphur is released.
Q 13: In EMP pathway, ATP molecules are formed when
(1) PGAL becomes BPGA and BPGA becomes PGA
(2) PGA becomes BPGA and PA becomes PEP
(3) PGAL becomes BPGA and PEP becomes PA
(4) BPGA becomes PGA and PEP becomes PA
✅ (4) BPGA becomes PGA and PEP becomes PA.
▶️ In Glycolysis (EMP pathway), ATP is utilized at 2 steps:
1. When glucose converts to glucose 6-phosphate.
2. When fructose 6-phosphate converts to fructose 1, 6-diphosphate.
▶️ ATP molecules are formed at 2 steps:
1. When 1, 3-bisphosphoglycerate (BPGA) becomes 3-phosphoglyceric acid (PGA).
2. When Phosphoenol pyruvate (PEP) converts to pyruvic acid (PA).
Q 14: Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called
(1) Plasticity
(2) Phototropism
(3) Apical dominance
(4) Differentiation
✅ (1) Plasticity
▶️ Phototropism: It is the movement or growth of an organism in response to light. In plants, it often involves the bending of stems or leaves toward a light source.
▶️ Apical Dominance: It is a phenomenon where the growth of a plant is primarily controlled by the apical bud (located at the tip of the stem), inhibiting the growth of lateral buds lower down the stem.
▶️ Differentiation: It is the process by which cells become specialized into different cell types with specific functions during the development of an organism. In plants, it leads to the formation of various tissues and structures.
Q 15: After forceful inspiration, the amount of air that can be breathed out by maximum forced expiration is equal to
(1) IRV + RV + ERV
(2) IRV + TV + ERV
(3) IRV + ERV + TV + RV
(4) TV + RV + ERV
✅ (2) IRV + TV + ERV
▶️ After forceful inspiration, the amount of air that can be breathed out by maximum forced expiration is equal to vital capacity (VC). i.e., VC = IRV + TV + ERV.
▶️ Inspiratory Reserve Volume (IRV): The additional volume of air that can be forcefully inhaled after a normal inspiration.
▶️ Tidal Volume (TV): The volume of air inhaled or exhaled during normal respiration.
▶️ Expiratory Reserve Volume (ERV): The additional volume of air that can be forcefully exhaled after a normal expiration.
▶️ Residual Volume (RV): The volume of air remaining in the lungs after maximum expiration.
Q 16: Which of the following matches correctly?
(1) Inferior vena cava – Receives deoxygenated blood from head and body
(2) Superior vena cava – Receives deoxygenated blood from the lower body and organs
(3) Pulmonary artery – Carries deoxygenated blood to the lungs
(4) Hepatic artery – Carries deoxygenated blood to the gut
✅ (3) Pulmonary artery – Carries deoxygenated blood to the lungs.
▶️ Inferior vena cava – Receives deoxygenated blood from lower body and organs.
▶️ Superior vena cava – Receives deoxygenated blood from the head and upper body.
▶️ Pulmonary vein – Carries oxygenated blood from the lungs to the heart.
▶️ Aorta – Carries oxygenated blood from the heart.
Q 17: Malpighian body is
(1) Glomerulus + Proximal convoluted tubule
(2) Glomerulus + Distal convoluted tubule
(3) Glomerulus + Bowman’s capsule
(4) Proximal convoluted tubule + Distal convoluted tubule
✅ (3) Glomerulus + Bowman’s capsule.
▶️ The Malpighian body, also known as the renal corpuscle, is composed of the Glomerulus + Bowman’s capsule.
▶️ The glomerulus is a tuft of capillaries, and Bowman’s capsule is a cup-like structure that encloses the glomerulus. This structure is part of the nephron in the kidney and plays a crucial role in the filtration of blood. This helps to remove waste products from blood and form urine.
Q 18: Identify A, B & C in the given diagram. 
(1) A: Tropomyosin, B: F actin, C: Troponin
(2) A: Myosin, B: F actin, C: Troponin
(3) A: Troponin, B: Myosin, C: F actin
(4) A: Troponin, B: Tropomyosin, C: F actin
✅ (4) A: Troponin, B: Tropomyosin, C: F actin.
▶️ In muscle fibre, there are two fibrous contractile proteins- thin Actin filament and thick Myosin filament.
▶️ An actin filament is made of 2 filamentous (F) actins which form double helix. F-actin is a polymer of monomeric Globular (G) actins.
▶️ Actin contains 2 other proteins called tropomyosin & troponin.
▶️ Each myosin filament is a polymer of many monomeric proteins called Meromyosins.
Q 19: Receptor sites for neurotransmitters are present on
(1) Pre-synaptic membrane
(2) Tips of axons
(3) Post-synaptic membrane
(4) Membranes of synaptic vesicles
✅ (3) Post-synaptic membrane.
▶️ In Chemical synapses, there is a fluid filled space (synaptic cleft) between the presynaptic neuron & postsynaptic neuron.
▶️ The presynaptic regions have swellings called Synaptic knob. They contain synaptic vesicles filled with neurotransmitters (acetylcholine or adrenaline).
Q 20: If ADH level of blood is less, then
(1) Volume of urine increases
(2) Glucose level of urine increases
(3) Volume of urine decreases
(4) Glucose level urine decreases
✅ (1) Volume of urine increases.
▶️ Antidiuretic Hormone (ADH or Vasopressin) plays a crucial role in regulating the body's water balance.
▶️ When ADH levels are low, the kidneys reduce their reabsorption of water, leading to increased urine production. This mechanism helps the body eliminate excess water.
▶️ The primary function of ADH is to control the concentration of urine and maintain water homeostasis in the body.