A-Level Biology Practice Test 16 – 60 Questions with Answers (Cambridge 9700)
This test works through the whole A-Level syllabus at full depth, from cell ultrastructure to genetic technology, so it is a good check of how exam-ready your understanding really is.
Cell structure
Q1. Which feature of a prokaryotic cell distinguishes it most clearly from a eukaryotic cell?
- (A) Absence of membrane-bound organelles and a true nucleus
- (B) Presence of a cell wall made of cellulose
- (C) Presence of mitochondria for aerobic respiration
- (D) Presence of a large permanent vacuole
Show Answer
Answer: A – Prokaryotes lack a nucleus and membrane-bound organelles, whereas eukaryotes possess both.
Q2. The rough endoplasmic reticulum is closely associated with protein synthesis because it is studded with which structures?
- (A) Lysosomes
- (B) Ribosomes
- (C) Peroxisomes
- (D) Centrioles
Show Answer
Answer: B – Ribosomes on the rough endoplasmic reticulum synthesise proteins that are then folded and transported within the system.
Q3. Which organelle modifies, packages and secretes proteins, forming vesicles that carry them to the cell surface?
- (A) Nucleolus
- (B) Smooth endoplasmic reticulum
- (C) Golgi apparatus
- (D) Chloroplast
Show Answer
Answer: C – The Golgi apparatus processes proteins from the endoplasmic reticulum and packages them into secretory vesicles.
Biological molecules
Q4. Which bond is formed when two monosaccharides join in a condensation reaction?
- (A) Peptide bond
- (B) Glycosidic bond
- (C) Hydrogen bond
- (D) Ester bond
Show Answer
Answer: B – A glycosidic bond links monosaccharides, with water released during condensation.
Q5. Why is glycogen a more suitable storage polysaccharide in animals than glucose alone?
- (A) It is soluble and raises the water potential of liver cells
- (B) It is transported freely in the blood plasma
- (C) It is insoluble and therefore has little osmotic effect, yet is rapidly hydrolysed when glucose is needed
- (D) It cannot be broken down by enzymes in animal cells
Show Answer
Answer: C – Glycogen is compact and insoluble, so it stores much glucose without disturbing the osmotic balance of the cell.
Q6. A triglyceride is formed from glycerol and three fatty acids. Which type of bond links each fatty acid to the glycerol?
- (A) Glycosidic bond
- (B) Peptide bond
- (C) Hydrogen bond
- (D) Ester bond
Show Answer
Answer: D – Ester bonds form by condensation between the hydroxyl groups of glycerol and the carboxyl groups of fatty acids.
Enzymes
Q7. Which statement best explains why a competitive inhibitor reduces the rate of an enzyme-catalysed reaction at low substrate concentration?
- (A) It denatures the enzyme by breaking its tertiary structure
- (B) It binds to an allosteric site and changes the shape of the active site
- (C) It occupies the active site, so fewer substrate molecules can bind
- (D) It increases the activation energy of the reaction permanently
Show Answer
Answer: C – A competitive inhibitor resembles the substrate and competes for the active site, an effect that high substrate concentration can overcome.
Q8. A non-competitive inhibitor differs from a competitive inhibitor in that it does what?
- (A) Binds only when substrate concentration is very high
- (B) Binds reversibly to the active site of the enzyme
- (C) Raises the maximum rate of reaction if more substrate is added
- (D) Binds to a site other than the active site and alters the active site shape
Show Answer
Answer: D – Non-competitive inhibitors bind away from the active site, so adding substrate cannot restore the maximum rate.
Q9. Immobilised enzymes are widely used in industry because they offer which advantage over free enzymes?
- (A) They can be recovered and reused, and they do not contaminate the product
- (B) They always work faster than the same enzyme in solution
- (C) They are unaffected by temperature or pH changes
- (D) They no longer require a specific substrate to function
Show Answer
Answer: A – Immobilisation keeps the enzyme separate from the product stream, allowing reuse and a purer product.
Cell membranes and transport
Q10. In the fluid mosaic model, which component is mainly responsible for the fluidity of the membrane?
- (A) Glycoproteins anchored to the cytoskeleton
- (B) Cholesterol molecules packed between phospholipids only at low temperature
- (C) Channel proteins spanning the whole bilayer
- (D) Phospholipid molecules moving laterally within each layer
Show Answer
Answer: D – Lateral movement of phospholipids gives the membrane its fluid character, while proteins are embedded within it.
Q11. Facilitated diffusion differs from simple diffusion in that it requires what?
- (A) A transport protein in the membrane
- (B) ATP hydrolysed directly by the transported molecule
- (C) Movement against a concentration gradient
- (D) Endocytosis of the surrounding fluid
Show Answer
Answer: A – Facilitated diffusion is passive but uses channel or carrier proteins, so it is selective and saturable.
Q12. Which process moves sodium ions out of a cell against their concentration gradient using energy from ATP?
- (A) Osmosis through aquaporins
- (B) Active transport by a sodium-potassium pump
- (C) Facilitated diffusion through a channel
- (D) Simple diffusion across the lipid bilayer
Show Answer
Answer: B – Active transport uses ATP to pump ions against their gradient, maintaining the steep gradients used in nerve and kidney function.
The mitotic cell cycle
Q13. During which stage of the cell cycle does DNA replication occur?
- (A) S phase of interphase
- (B) Prophase of mitosis
- (C) Cytokinesis
- (D) Metaphase of mitosis
Show Answer
Answer: A – DNA is copied during the S phase, so each chromosome entering mitosis consists of two identical sister chromatids.
Q14. What is the key event of anaphase in mitosis?
- (A) Chromosomes condense and the nuclear envelope breaks down
- (B) Sister chromatids separate and move to opposite poles
- (C) Chromosomes align along the equator of the cell
- (D) The nuclear envelope reforms around each set of chromosomes
Show Answer
Answer: B – Spindle fibres shorten in anaphase, pulling sister chromatids apart so each daughter cell receives an identical set.
Q15. Uncontrolled mitosis is dangerous because it can lead to which outcome?
- (A) Loss of all cell differentiation in every tissue
- (B) Failure of meiosis in reproductive organs only
- (C) Formation of a tumour of genetically identical cells
- (D) Immediate death of the affected cell by apoptosis
Show Answer
Answer: C – When cell-cycle checkpoints fail, repeated mitosis produces a mass of cells that may invade surrounding tissue.
Nucleic acids and protein synthesis
Q16. Which feature of DNA structure allows it to be copied accurately during semi-conservative replication?
- (A) The sugar-phosphate backbone is identical in both strands
- (B) Complementary base pairing between the two strands
- (C) The presence of uracil in place of thymine
- (D) The double helix is held together only by covalent bonds
Show Answer
Answer: B – Each strand acts as a template because adenine pairs with thymine and cytosine with guanine.
Q17. In protein synthesis, the role of tRNA is best described as what?
- (A) Carrying the genetic code from the nucleus to the cytoplasm
- (B) Forming the structural framework of the ribosome
- (C) Carrying a specific amino acid to the ribosome according to its anticodon
- (D) Splicing introns out of pre-mRNA in the nucleus
Show Answer
Answer: C – Each tRNA has an anticodon complementary to an mRNA codon and carries the corresponding amino acid.
Q18. A mutation that changes a single base in a codon but leaves the amino acid unchanged is best described as which type?
- (A) A frameshift mutation
- (B) A nonsense mutation
- (C) A deletion mutation
- (D) A silent mutation
Show Answer
Answer: D – The genetic code is degenerate, so some base changes produce a codon for the same amino acid.
Transport in plants
Q19. Which tissue transports assimilates such as sucrose from sources to sinks, and which feature supports that role?
- (A) Xylem, with lignified walls that withstand tension
- (B) Cambium, with rapidly dividing meristematic cells
- (C) Phloem, with sieve tube elements lacking nuclei but supported by companion cells
- (D) Epidermis, with a waxy cuticle that limits water loss
Show Answer
Answer: C – Sieve tube elements form a continuous tube for translocation, while companion cells carry out their metabolic functions.
Q20. Transpiration pull depends mainly on which property of water?
- (A) Water is a good solvent for mineral ions
- (B) Water has a low specific heat capacity
- (C) Water molecules are small enough to pass through cell walls
- (D) Cohesion between water molecules and adhesion to xylem walls
Show Answer
Answer: D – Cohesion holds the water column together and adhesion to the walls prevents it breaking under tension.
Q21. At a sink in mass flow, sucrose is actively unloaded from phloem sieve tubes. What happens to water in the sieve tubes as a result, and why?
- (A) Water potential in the sieve tubes rises as sucrose is removed, so water leaves by osmosis and hydrostatic pressure falls
- (B) Water evaporates from leaf stomata, pulling sap upward
- (C) Root pressure pushes sap from the roots into the leaves
- (D) Gravity acting on sap in vertical sieve tubes drives the flow
Show Answer
Answer: A – Unloading lowers solute concentration at the sink, water follows by osmosis out of the sieve tube, and the resulting pressure difference drives sap from source to sink.
Transport in mammals
Q22. Why is the left ventricle wall much thicker than the right ventricle wall?
- (A) It must prevent backflow into the left atrium
- (B) It contracts less frequently than the right ventricle
- (C) It receives blood directly from the pulmonary vein
- (D) It must generate enough pressure to pump blood around the whole body
Show Answer
Answer: D – The systemic circulation is far longer and higher-resistance than the pulmonary circulation, so the left ventricle needs more muscle.
Q23. Which sequence correctly describes the mammalian double circulation, starting at the right ventricle?
- (A) Pulmonary artery, lungs, pulmonary vein, left atrium, left ventricle, body, vena cava, right atrium
- (B) Aorta, body, pulmonary vein, left atrium, lungs, right atrium, vena cava
- (C) Pulmonary vein, lungs, aorta, body, vena cava, left atrium, right ventricle
- (D) Vena cava, right atrium, aorta, body, pulmonary artery, lungs, left ventricle
Show Answer
Answer: A – Blood passes through the lungs before returning to the left side of the heart and then being pumped to the body.
Q24. Tissue fluid formation at the arteriole end of a capillary is driven mainly by which pressure difference?
- (A) Lower solute potential in the tissue fluid than in the blood
- (B) Higher hydrostatic pressure in the capillary than the opposing solute potential gradient
- (C) Active transport of plasma proteins out of the capillary
- (D) Lower hydrostatic pressure in the capillary than in the surrounding tissue
Show Answer
Answer: B – High hydrostatic pressure forces fluid out, while water potential from plasma proteins opposes it and dominates at the venule end.
Q25. Haemoglobin unloads oxygen more readily in actively respiring tissue because of which combined effect?
- (A) Higher carbon dioxide concentration and lower pH shifting the dissociation curve to the right
- (B) Higher oxygen concentration and higher pH shifting the curve to the left
- (C) Lower temperature and lower carbon dioxide concentration in the tissue
- (D) Direct binding of carbon dioxide to the haem group of haemoglobin
Show Answer
Answer: A – The Bohr effect means that in respiring tissue, haemoglobin affinity for oxygen falls and oxygen is released where it is needed.
Gas exchange
Q26. Which feature of mammalian alveoli most increases the rate of gas exchange?
- (A) A very large total surface area and a thin diffusion barrier
- (B) A thick layer of mucus trapping inhaled particles
- (C) Cartilage rings keeping the airways permanently open
- (D) Ciliated cells moving mucus toward the pharynx
Show Answer
Answer: A – Millions of alveoli give a huge surface, and their walls are one cell thick with a dense capillary network.
Q27. In fish, countercurrent flow across the gills is advantageous because it does what?
- (A) Prevents water from entering the mouth
- (B) Maintains a concentration gradient along the whole length of the lamella
- (C) Allows blood to flow in the same direction as water for faster exchange
- (D) Increases the oxygen content of water before it reaches the gills
Show Answer
Answer: B – Blood and water flow in opposite directions, so oxygen continues to diffuse into blood along the entire lamella.
Q28. During inspiration in mammals, which sequence of events occurs?
- (A) Diaphragm relaxes, intercostal muscles relax, thorax volume falls, pressure rises
- (B) Diaphragm contracts, internal intercostal muscles contract, ribs move down and in
- (C) Diaphragm contracts and flattens, external intercostal muscles contract, thorax volume rises, pressure falls
- (D) Diaphragm flattens, thorax volume falls, air is forced out of the lungs
Show Answer
Answer: C – Contraction of the diaphragm and external intercostal muscles enlarges the thorax, lowers pressure, and draws air in.
Q29. Insects use a tracheal system for gas exchange. Which feature of tracheoles makes them effective for direct delivery to tissues?
- (A) They are lined with cartilage that holds them open
- (B) They contain blood that carries oxygen to muscles
- (C) They open and close using a diaphragm muscle
- (D) Their walls are very thin and they penetrate deep into tissues
Show Answer
Answer: D – Thin-walled tracheoles end close to respiring cells, giving a short diffusion distance without a blood transport step.
Infectious diseases
Q30. Which statement about the transmission of cholera is correct?
- (A) It is spread mainly by airborne droplets from coughing
- (B) It is spread through water or food contaminated with faeces containing Vibrio cholerae
- (C) It is spread only by direct skin contact with an infected person
- (D) It is spread by the bite of an infected mosquito
Show Answer
Answer: B – Cholera is a water-borne disease, so clean water supply and sanitation are the key controls.
Q31. The human immunodeficiency virus (HIV) causes AIDS by progressively destroying which cells of the immune system?
- (A) B lymphocytes that produce antibodies
- (B) Neutrophils that carry out phagocytosis
- (C) Helper T lymphocytes (CD4 cells) that coordinate the immune response
- (D) Platelets involved in blood clotting
Show Answer
Answer: C – HIV infects helper T cells, so as their numbers fall the whole specific immune response weakens.
Q32. Why does antibiotic resistance spread rapidly in a bacterial population exposed to an antibiotic?
- (A) Antibiotics cause new mutations in every bacterium they contact
- (B) Bacteria learn to recognise and avoid the antibiotic molecule
- (C) Antibiotics weaken the immune system so bacteria multiply unchecked
- (D) Resistant mutants survive and reproduce, passing resistance genes to offspring and sometimes to other bacteria
Show Answer
Answer: D – Selection favours pre-existing resistant variants, and horizontal gene transfer can spread resistance further.
Q33. Which measure is most effective for controlling malaria in a region where the mosquito vector is present?
- (A) Quarantining all infected people in hospitals
- (B) Reducing mosquito breeding sites, using insecticide-treated nets, and treating cases promptly
- (C) Vaccinating the whole population against the Plasmodium parasite with a single lifelong dose
- (D) Adding antibiotics to the local drinking water supply
Show Answer
Answer: B – Malaria control combines vector control with rapid diagnosis and treatment, since no single measure alone is sufficient.
Immunity
Q34. In the primary immune response, which sequence occurs after a pathogen enters the body?
- (A) Memory cells are activated immediately and produce large amounts of antibody within hours
- (B) Antibodies are produced before any phagocyte encounters the pathogen
- (C) Phagocytes engulf the pathogen, present antigens, and specific B and T lymphocytes are activated and divide
- (D) Plasma cells form first and only later stimulate phagocytosis
Show Answer
Answer: C – Antigen presentation triggers clonal selection, producing plasma cells that secrete antibody and memory cells for future encounters.
Q35. What distinguishes the secondary immune response from the primary response?
- (A) It involves only the non-specific defences of the skin and mucus
- (B) It produces no memory cells at all
- (C) It relies entirely on antibiotics rather than antibodies
- (D) It is faster and stronger because memory cells are already present
Show Answer
Answer: D – Memory cells from the first encounter divide rapidly on re-exposure, so antibody levels rise sooner and higher.
Q36. A vaccine gives long-lasting protection because it does what?
- (A) Introduces memory cells and triggers the formation of memory cells without causing the full disease
- (B) Kills all pathogens already present in the body permanently
- (C) Replaces the need for phagocytes in future infections
- (D) Provides ready-made antibodies that remain in the blood for life
Show Answer
Answer: A – Vaccines present antigens safely, stimulating a primary response that leaves memory cells ready for real infection.
Energy and respiration
Q37. In aerobic respiration, most ATP is produced during which stage and by what mechanism?
- (A) Glycolysis, by substrate-level phosphorylation in the cytoplasm
- (B) Link reaction, by direct transfer of phosphate to ADP
- (C) Krebs cycle, by photophosphorylation in the mitochondria
- (D) Oxidative phosphorylation, by chemiosmosis as protons flow through ATP synthase
Show Answer
Answer: D – Electron transport pumps protons into the intermembrane space, and their return through ATP synthase drives most ATP synthesis.
Q38. Anaerobic respiration in mammalian muscle produces lactate. Why is lactate formation necessary?
- (A) It regenerates NAD from reduced NAD, allowing glycolysis to continue
- (B) It produces more ATP than aerobic respiration per glucose molecule
- (C) It removes all carbon dioxide from the muscle cell
- (D) It directly supplies oxygen to the mitochondria
Show Answer
Answer: A – Without oxygen as the final electron acceptor, pyruvate accepts hydrogen from reduced NAD so glycolysis can keep running.
Q39. Which molecule is the final electron acceptor in the electron transport chain of aerobic respiration?
- (A) Carbon dioxide
- (B) Oxygen
- (C) NAD
- (D) Glucose
Show Answer
Answer: B – Oxygen accepts electrons and hydrogen ions at the end of the chain, forming water.
Photosynthesis
Q40. In the light-dependent stage, photolysis of water provides which products for the rest of photosynthesis?
- (A) Electrons, protons and oxygen
- (B) Glucose and carbon dioxide
- (C) ATP and glycerate 3-phosphate only
- (D) NADP and ADP only
Show Answer
Answer: A – Splitting water replaces electrons lost by chlorophyll, supplies protons for chemiosmosis, and releases oxygen.
Q41. Cyclic photophosphorylation differs from non-cyclic photophosphorylation in that it produces what?
- (A) Oxygen and reduced NADP in equal amounts
- (B) ATP only, with no reduced NADP and no oxygen released
- (C) Glucose directly from carbon dioxide
- (D) Reduced NADP only, with no ATP formed
Show Answer
Answer: B – In cyclic flow, electrons return to photosystem I, pumping protons for ATP synthesis without photolysis or NADP reduction.
Q42. In the Calvin cycle, what is the role of reduced NADP and ATP from the light-dependent stage?
- (A) They split water to release oxygen
- (B) They fix carbon dioxide directly to ribulose bisphosphate
- (C) They reduce glycerate 3-phosphate to triose phosphate
- (D) They regenerate chlorophyll in the photosystems
Show Answer
Answer: C – ATP supplies energy and reduced NADP supplies hydrogen for reducing glycerate 3-phosphate to triose phosphate.
Homeostasis
Q43. Which sequence correctly describes negative feedback control of blood glucose after a meal?
- (A) Blood glucose falls, glucagon is released, liver glycogen is broken down
- (B) Blood glucose rises, insulin is released, liver and muscle cells take up glucose and store glycogen
- (C) Blood glucose rises, glucagon is released, glucose is converted to glycogen in the liver
- (D) Blood glucose falls, insulin is released, fat stores are mobilised
Show Answer
Answer: B – Insulin lowers blood glucose, and as levels return to normal, insulin secretion falls, completing the negative feedback loop.
Q44. In the kidney, ultrafiltration occurs in the glomerulus because of which combination of factors?
- (A) Active transport of plasma proteins through the basement membrane
- (B) Low blood pressure and a wide efferent arteriole
- (C) High hydrostatic pressure from a narrow efferent arteriole and a large filtration surface
- (D) Reabsorption of water in the collecting duct
Show Answer
Answer: C – The efferent arteriole is narrower than the afferent, building pressure that forces small molecules through the filtration membrane.
Q45. Antidiuretic hormone (ADH) increases water reabsorption in the kidney by acting on which structures?
- (A) Glomerular capillaries, increasing the filtration rate
- (B) Proximal convoluted tubule cells, increasing glucose reabsorption
- (C) Loop of Henle, preventing all sodium reabsorption
- (D) Collecting duct walls, inserting aquaporins that make them more permeable to water
Show Answer
Answer: D – ADH causes aquaporin channels to be inserted into collecting duct cells, so more water is reabsorbed into the blood.
Control and coordination
Q46. At a cholinergic synapse, an action potential arrives at the presynaptic neurone. Why is calcium influx into the synaptic knob essential for transmission?
- (A) Acetylcholinesterase is released from the postsynaptic membrane
- (B) The presynaptic membrane becomes impermeable to calcium ions
- (C) Calcium ions trigger synaptic vesicles to fuse with the presynaptic membrane and release acetylcholine
- (D) Vesicles in the postsynaptic cell release more acetylcholine
Show Answer
Answer: C – Calcium entry causes vesicle fusion and transmitter release into the synaptic cleft, the essential step in chemical transmission.
Q47. Which statement about the refractory period of a neurone is correct?
- (A) It allows action potentials to travel in both directions along the axon
- (B) It is the time when the resting potential is maintained by the sodium-potassium pump alone
- (C) It increases the speed of impulse transmission in myelinated neurones
- (D) It prevents a new action potential being generated immediately, ensuring impulses travel one way
Show Answer
Answer: D – During the refractory period, sodium channels are inactivated, so the impulse cannot reverse and discrete impulses remain separate.
Q48. In the eye, accommodation for a near object involves which changes?
- (A) Ciliary muscles contract, suspensory ligaments slacken, lens becomes more convex
- (B) Ciliary muscles relax, suspensory ligaments tighten, lens becomes flatter
- (C) Radial muscles of the iris contract and the pupil dilates
- (D) Suspensory ligaments contract and pull the lens into a flat shape
Show Answer
Answer: A – Contracted ciliary muscles release tension on the lens, which rounds up to refract light from a near object more strongly.
Inheritance
Q49. In a dihybrid cross between two heterozygotes (AaBb x AaBb), what phenotypic ratio is expected if the genes assort independently?
- (A) 3:1
- (B) 1:2:1
- (C) 1:1:1:1
- (D) 9:3:3:1
Show Answer
Answer: D – Independent assortment of two heterozygous gene pairs produces four phenotypes in the ratio 9:3:3:1.
Q50. A woman who is a carrier for haemophilia (X-linked recessive) has children with an unaffected man. What is the probability that a son will have haemophilia?
- (A) 1 in 2 (50 percent)
- (B) 1 in 4 (25 percent)
- (C) 0 percent, because the father is unaffected
- (D) 100 percent, because the mother is a carrier
Show Answer
Answer: A – Sons receive their X chromosome from their mother, so half inherit the haemophilia allele and are affected.
Q51. Codominance is best illustrated by which example?
- (A) A pink flower produced by crossing red and white parents
- (B) AB blood group, where both A and B antigens are expressed on red blood cells
- (C) A recessive allele masked in the heterozygous phenotype
- (D) Continuous variation in human height
Show Answer
Answer: B – In codominance both alleles are fully expressed in the heterozygote, unlike incomplete dominance where the phenotype is intermediate.
Selection and evolution
Q52. Which statement best describes natural selection?
- (A) Individuals with alleles that increase survival and reproduction become more common in the population over generations
- (B) Individuals change their genes during their lifetime to suit the environment
- (C) All individuals in a population acquire identical advantageous mutations
- (D) The environment directly creates the mutations that a species needs
Show Answer
Answer: A – Selection acts on existing variation, changing allele frequencies over generations without creating mutations on demand.
Q53. Directional selection differs from stabilising selection in that it does what?
- (A) Favours the average phenotype and reduces variation
- (B) Favours one extreme phenotype, shifting the population mean over time
- (C) Favours both extremes at the expense of the average
- (D) Acts only on discontinuous variation
Show Answer
Answer: B – Directional selection shifts the mean, for example toward smaller size when food becomes scarce, whereas stabilising selection preserves the mean.
Q54. The Hardy-Weinberg principle can be used to estimate allele frequencies in a population provided which condition holds?
- (A) Mutation and migration are frequent
- (B) The population is very small and isolated
- (C) Mating is random and there is no selection, mutation, migration or genetic drift
- (D) Only dominant alleles are present in the population
Show Answer
Answer: C – Hardy-Weinberg equilibrium assumes a large population with random mating and no selection, mutation, migration or drift.
Classification, biodiversity and conservation
Q55. Which hierarchy correctly lists taxonomic groups from largest to smallest?
- (A) Species, genus, family, order, class, phylum, kingdom, domain
- (B) Domain, kingdom, phylum, class, order, family, genus, species
- (C) Kingdom, domain, class, phylum, family, order, genus, species
- (D) Domain, phylum, kingdom, order, class, genus, family, species
Show Answer
Answer: B – The hierarchy runs from domain down through kingdom, phylum, class, order, family and genus to species.
Q56. The Simpson index of diversity takes account of which two factors?
- (A) The total area of the habitat and the age of the community
- (B) The number of trophic levels and the biomass of producers
- (C) Species richness and the evenness of abundance across species
- (D) Genetic variation within a single species only
Show Answer
Answer: C – Simpson’s index combines how many species are present with how evenly individuals are distributed among them.
Q57. Why is in situ conservation generally preferred to ex situ conservation for maintaining a species?
- (A) It is always cheaper and requires no management
- (B) It removes all predators and competitors from the habitat
- (C) It allows unlimited captive breeding without genetic problems
- (D) It preserves the species in its natural habitat, maintaining ecological interactions and natural selection
Show Answer
Answer: D – In situ conservation keeps species within their ecosystems, so adaptations and interactions continue to evolve.
Genetic technology
Q58. In recombinant DNA technology, what is the role of a restriction enzyme?
- (A) It joins DNA fragments together with phosphodiester bonds
- (B) It copies a target DNA sequence many times in a thermal cycler
- (C) It cuts DNA at specific recognition sequences, producing fragments that can be joined into a vector
- (D) It translates mRNA into a protein product
Show Answer
Answer: C – Restriction enzymes cut at defined sequences, often leaving sticky ends that match those of the vector.
Q59. In the polymerase chain reaction (PCR), what is the purpose of the high-temperature step in each cycle?
- (A) To activate DNA ligase
- (B) To allow primers to anneal to the template strands
- (C) To synthesise new DNA strands from primers
- (D) To denature double-stranded DNA into separate template strands
Show Answer
Answer: D – Heating to around 95 degrees Celsius breaks hydrogen bonds between strands, exposing templates for primer binding.
Q60. Genetically modified bacteria can produce human insulin because of which fact?
- (A) The genetic code is universal, so bacterial ribosomes can translate human insulin mRNA correctly
- (B) Bacteria naturally contain the human insulin gene in their plasmids
- (C) Human insulin is identical in structure to bacterial enzymes
- (D) Bacteria can absorb insulin directly from their growth medium
Show Answer
Answer: A – Because codons mean the same amino acids in almost all organisms, a transferred human gene can be expressed in bacteria.
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