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MATSEC AM Biology study materials
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MATSEC AM Biology practice questions

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Question 1Biology (AM syllabus content)

Which structural feature is shared by mitochondria and chloroplasts and is used as evidence for the endosymbiotic theory?

  1. A rigid peptidoglycan wall outside the membrane
  2. A single membrane continuous with the endoplasmic reticulum
  3. Linear chromosomes associated with histone proteins
  4. A double membrane envelope surrounding the organelle Correct answer.

Explanation

Both organelles are enclosed by two membranes, consistent with engulfment of a prokaryote by an ancestral host cell. Their DNA is circular and naked rather than histone-bound, and neither possesses peptidoglycan. Continuity with the endoplasmic reticulum is a feature of the nuclear envelope, not of these organelles.

Question 2Biology (AM syllabus content)

A cell is placed in a solution whose solute potential is more negative than that of the cytoplasm. Which immediate response is expected in a plant cell with a freely permeable membrane to water?

  1. Solutes are actively pumped out to equalise the potentials
  2. Water enters the cell and turgor pressure rises Correct answer.
  3. There is no net movement of water in either direction
  4. Water leaves the cell and the protoplast shrinks

Explanation

Water moves down its water potential gradient, from the less negative (higher) potential outside to the more negative potential inside. In a plant cell the incoming water raises turgor pressure, which opposes further entry. Solute pumping is not an immediate osmotic response.

Question 3Biology (AM syllabus content)

Which property of the phospholipid bilayer directly limits the free diffusion of large polar molecules such as glucose?

  1. The cholesterol content of the bilayer
  2. The hydrophilic phosphate head groups at the surface
  3. The presence of peripheral membrane proteins
  4. The hydrophobic fatty acid tails in the interior Correct answer.

Explanation

The hydrophobic core formed by the fatty acid tails repels polar and charged substances, so large polar molecules cannot pass freely. Peripheral proteins sit at the surface and do not form the diffusion barrier. Cholesterol modulates fluidity rather than creating the main permeability barrier.

Question 4Biology (AM syllabus content)

During which stage of mitosis do sister chromatids separate and move towards opposite poles of the cell?

  1. Telophase
  2. Metaphase
  3. Anaphase Correct answer.
  4. Prophase

Explanation

In anaphase the cohesin holding sister chromatids is cleaved and the chromatids are pulled poleward by shortening spindle microtubules. Prophase involves chromosome condensation, metaphase alignment at the equator, and telophase nuclear reformation.

Question 5Biology (AM syllabus content)

Which bond is formed between the amino group of one amino acid and the carboxyl group of another during translation?

  1. A hydrogen bond
  2. A glycosidic bond
  3. A phosphodiester bond
  4. A peptide bond Correct answer.

Explanation

The ribosome catalyses a condensation reaction between the amino group and carboxyl group, releasing water and forming a peptide bond. Glycosidic bonds join sugars, phosphodiester bonds join nucleotides, and hydrogen bonds are weaker interactions.

Question 6Biology (AM syllabus content)

In a dihybrid cross between two individuals heterozygous at both loci (AaBb x AaBb), assuming independent assortment, what proportion of offspring is expected to show both recessive phenotypes?

  1. 3/16
  2. 1/4
  3. 9/16
  4. 1/16 Correct answer.

Explanation

Each recessive phenotype has a probability of 1/4 from a heterozygous monohybrid cross. Multiplying the independent probabilities gives 1/4 x 1/4 = 1/16 for the double recessive combination.

Question 7Biology (AM syllabus content)

Which enzyme unwinds the DNA double helix ahead of the replication fork by breaking hydrogen bonds between base pairs?

  1. DNA helicase Correct answer.
  2. DNA polymerase III
  3. Primase
  4. DNA ligase

Explanation

Helicase separates the two strands by disrupting hydrogen bonds between complementary bases, creating the replication fork. Ligase joins Okazaki fragments, polymerase extends the new strand, and primase synthesises RNA primers.

Question 8Biology (AM syllabus content)

Which feature distinguishes a prokaryotic cell from a eukaryotic cell?

  1. Use of DNA as genetic material
  2. Presence of ribosomes in the cytoplasm
  3. Absence of a membrane-bound nucleus Correct answer.
  4. Presence of a plasma membrane

Explanation

Prokaryotes lack a nuclear envelope, so their circular DNA lies free in the cytoplasm. Ribosomes, a plasma membrane, and DNA as genetic material are common to both cell types and therefore do not distinguish them.

Question 9Biology (AM syllabus content)

A respirometer records oxygen uptake by germinating seeds. Which conclusion is best supported if the rate of carbon dioxide release exceeds the rate of oxygen uptake?

  1. Only carbohydrate is being respired aerobically
  2. The respiratory quotient is exactly 1.0
  3. Photosynthesis is occurring faster than respiration
  4. Anaerobic respiration is contributing to the carbon dioxide produced Correct answer.

Explanation

When carbon dioxide output exceeds oxygen consumption, some carbon dioxide must arise from a pathway that does not use oxygen, indicating anaerobic respiration alongside aerobic respiration. A respiratory quotient of 1.0 would mean equal volumes of oxygen used and carbon dioxide released. Photosynthesis would reduce net carbon dioxide release, not increase it.

Question 10Biology (AM syllabus content)

In the light-dependent reactions of photosynthesis, what is the immediate role of the proton gradient across the thylakoid membrane?

  1. It splits water molecules to release oxygen
  2. It directly reduces NADP+ to NADPH
  3. It fixes carbon dioxide onto ribulose bisphosphate
  4. It drives ATP synthesis by chemiosmosis through ATP synthase Correct answer.

Explanation

Protons accumulate in the thylakoid lumen, and their flow back into the stroma through ATP synthase drives ATP production by chemiosmosis. NADP+ reduction occurs at the end of the electron transport chain, water splitting generates the protons and electrons, and carbon fixation happens in the stroma during the light-independent reactions.

Question 11Biology (AM syllabus content)

Which statement best explains why the genetic code is described as degenerate?

  1. Each codon can specify more than one amino acid
  2. The code differs between species
  3. Most amino acids are specified by more than one codon Correct answer.
  4. Codons are read continuously without punctuation

Explanation

Degeneracy means several different codons can code for the same amino acid, which reduces the impact of some point mutations. A single codon specifying several amino acids would make translation ambiguous, and continuity and universality are separate properties of the code.

Question 12Biology (AM syllabus content)

In a population at Hardy-Weinberg equilibrium, the frequency of the recessive allele is 0.2. What proportion of the population is expected to be heterozygous?

  1. 0.64
  2. 0.32 Correct answer.
  3. 0.80
  4. 0.04

Explanation

The dominant allele frequency is 0.8, and heterozygote frequency is 2pq = 2 x 0.8 x 0.2 = 0.32. The homozygous recessive frequency q squared is 0.04, and the homozygous dominant frequency p squared is 0.64.

Question 13Biology (AM syllabus content)

Which event during meiosis is the principal source of new combinations of alleles on the same chromosome?

  1. Condensation of chromatin in prophase I
  2. Crossing over between non-sister chromatids in prophase I Correct answer.
  3. Separation of sister chromatids in anaphase II
  4. Separation of homologous chromosomes in anaphase I

Explanation

Crossing over exchanges equivalent segments between non-sister chromatids, creating recombinant chromosomes that carry new allele combinations. Chromosome and chromatid separation redistribute existing combinations but do not create new linkage arrangements, and condensation is purely structural.

Question 14Biology (AM syllabus content)

A mutation changes a single base in a coding region but the polypeptide produced is unchanged in length and amino acid sequence. Which type of mutation is this?

  1. A nonsense mutation
  2. A missense mutation
  3. A frameshift insertion
  4. A silent substitution Correct answer.

Explanation

A silent substitution changes a codon to another codon for the same amino acid because of code degeneracy, leaving the protein unchanged. Nonsense mutations create a stop codon, insertions shift the reading frame, and missense mutations alter an amino acid.

Question 15Biology (AM syllabus content)

Which feature of the xylem is most important for maintaining a continuous column of water under tension?

  1. Living companion cells that load sucrose
  2. Lignified walls with pits that allow lateral water movement Correct answer.
  3. Sieve plates with large pores
  4. A Casparian strip in the endodermis

Explanation

Lignified walls prevent collapse of vessels under tension and pits permit lateral movement while restricting air entry. Companion cells and sieve plates belong to the phloem, and the Casparian strip lies in the endodermis where it blocks apoplastic flow.

Question 16Biology (AM syllabus content)

Which change would most increase the rate of transpiration from a leaf, assuming stomata remain open?

  1. Closure of the stomatal pores
  2. An increase in atmospheric humidity
  3. A decrease in wind speed around the leaf
  4. A rise in temperature with a steep water vapour gradient Correct answer.

Explanation

Higher temperature increases evaporation from the leaf surface and steepens the diffusion gradient for water vapour, raising transpiration. High humidity and still air reduce the gradient, and closed stomata restrict water loss.

Question 17Biology (AM syllabus content)

In the cardiac cycle, what happens during ventricular systole?

  1. Both sets of valves are closed while the atria contract
  2. The atrioventricular valves close and the semilunar valves open Correct answer.
  3. The atrioventricular valves open and the semilunar valves close
  4. Both sets of valves are open simultaneously

Explanation

When ventricles contract, rising pressure forces the atrioventricular valves shut and pushes blood through the semilunar valves into the aorta and pulmonary artery. Valves open and close reciprocally, so they are never all open at once.

Question 18Biology (AM syllabus content)

A student measures the rate of an enzyme-catalysed reaction at increasing substrate concentrations and obtains a hyperbolic curve. Which interpretation is correct?

  1. The substrate is acting as a competitive inhibitor
  2. The rate approaches a maximum because active sites become saturated Correct answer.
  3. The enzyme is denaturing at high substrate concentration
  4. The enzyme concentration is increasing during the reaction

Explanation

At high substrate concentration all active sites are occupied, so the reaction reaches a maximum velocity limited by enzyme concentration. Saturation does not involve denaturation, and high substrate would outcompete rather than act as an inhibitor. Enzyme concentration is held constant in such experiments.

Question 19Biology (AM syllabus content)

Which feature of the mammalian gas exchange surface is most directly responsible for maintaining a steep diffusion gradient along the entire length of the alveolus?

  1. A large total surface area
  2. Continuous blood flow that carries oxygen away Correct answer.
  3. Surfactant secretion by alveolar cells
  4. A thin single-cell epithelium

Explanation

Blood flow removes oxygen and delivers carbon dioxide, keeping a concentration difference between alveolar air and blood along the whole capillary. Surface area and thinness increase the rate of diffusion but do not by themselves maintain the gradient, and surfactant reduces surface tension.

Question 20Biology (AM syllabus content)

Which of these is an example of innate, non-specific immunity in humans?

  1. Formation of memory B cells after infection
  2. Phagocytosis of bacteria by neutrophils at a wound site Correct answer.
  3. Production of antibodies by plasma cells
  4. Activation of cytotoxic T cells by infected cells

Explanation

Phagocytosis by neutrophils is a rapid, non-specific innate response that does not depend on recognising a particular antigen. Antibody production, memory cell formation, and cytotoxic T cell activity are all antigen-specific adaptive responses.

Question 21Biology (AM syllabus content)

In a food chain, only about 10% of the energy at one trophic level is transferred to the next. Which explanation best accounts for this low efficiency?

  1. Energy flows in both directions between trophic levels
  2. Consumers convert most ingested energy into new biomass
  3. Energy is created by producers and destroyed by consumers
  4. Most energy is lost as heat through respiration and in undigested material Correct answer.

Explanation

Energy is lost at each transfer because organisms respire, releasing heat, and because not all material is digested or consumed. Energy is neither created nor destroyed but degrades to heat, and it flows one way through ecosystems.

Question 22Biology (AM syllabus content)

A woman who is a carrier of an X-linked recessive disorder has children with an unaffected man. What is the probability that a son will be affected?

  1. 1/2 Correct answer.
  2. 0
  3. 1/1
  4. 1/4

Explanation

A son inherits his single X chromosome from his mother, and she has a 1/2 chance of passing on the allele carrying the mutation. Since males are hemizygous, one copy of the recessive allele is sufficient to produce the disorder.

Question 23Biology (AM syllabus content)

In an operon controlled by a repressible mechanism, such as the trp operon, what happens when the end product accumulates?

  1. It binds the repressor, enabling it to block transcription Correct answer.
  2. It removes the repressor from the operator
  3. It binds the operator and directly blocks RNA polymerase
  4. It activates the repressor gene to increase enzyme synthesis

Explanation

The end product acts as a co-repressor, binding the inactive repressor and enabling it to attach to the operator and block transcription. The product does not bind DNA directly, and accumulation of product reduces rather than increases enzyme synthesis.

Question 24Biology (AM syllabus content)

A population of insects develops resistance to an insecticide within a few years of repeated spraying. Which explanation is consistent with natural selection?

  1. The insecticide causes targeted mutations that confer resistance in exposed insects
  2. Resistance is acquired because insects learn to avoid the insecticide
  3. Pre-existing resistant alleles become more frequent because resistant individuals survive and reproduce Correct answer.
  4. Individual insects gradually strengthen their resistance during their lifetime

Explanation

Selection acts on variation already present in the population; resistant individuals survive and pass on their alleles, so allele frequencies shift over generations. The insecticide does not direct mutation, and acquired characteristics or learned avoidance are not inherited.

Question 25Biology (AM syllabus content)

In a cross between a heterozygous tall pea plant and a homozygous short plant, all offspring are tall. Which conclusion is justified?

  1. Tallness is a recessive trait in this cross
  2. Half of the offspring should be short by chance
  3. The short parent must also be heterozygous
  4. The data are inconsistent with simple dominance and suggest an error or a different inheritance pattern Correct answer.

Explanation

A cross between a heterozygote and a homozygous recessive should yield roughly equal numbers of tall and short offspring under simple dominance. Obtaining only tall offspring suggests a recording or labelling error or another genetic explanation. Shortness cannot be recessive if it fails to appear in this test cross.

Question 26Biology (AM syllabus content)

Which property of water allows insects such as pond skaters to move across the surface of a pond without breaking through?

  1. The high latent heat of vaporisation of water
  2. Surface tension arising from hydrogen bonding between water molecules Correct answer.
  3. High specific heat capacity of liquid water
  4. Low viscosity of water at room temperature

Explanation

Water molecules at an air-water interface are pulled inward and sideways by hydrogen bonds, creating a cohesive surface film that resists penetration by small objects. Cohesion between surface molecules is what generates this tension, and it is unrelated to heat capacity or viscosity.

Question 27Biology (AM syllabus content)

Which structural feature of a phospholipid makes it suitable as the main component of a cell-surface membrane bilayer?

  1. A hydrophilic phosphate head and a hydrophobic hydrocarbon tail Correct answer.
  2. A rigid ring structure that resists bending
  3. A single long-chain fatty acid esterified to glycerol
  4. A charged glycerol backbone with no fatty acid component

Explanation

Phospholipids are amphipathic: the phosphate-containing head interacts with water while the two hydrocarbon tails are repelled by it. In water this drives the tails together and the heads outward, forming a stable bilayer that is an effective barrier to polar solutes.

Question 28Biology (AM syllabus content)

A cell is placed in a solution with a lower water potential than its cytoplasm. What is the most likely immediate effect on an animal cell?

  1. Water enters the cell and the cell swells
  2. Water leaves the cell by osmosis and the cell shrinks Correct answer.
  3. The cell wall prevents any net movement of water
  4. Solute moves out of the cell until potentials equalise

Explanation

Water moves by osmosis from higher to lower water potential. The surrounding solution has the lower water potential, so net movement is out of the cell, reducing its volume and causing crenation or shrinkage.

Question 29Biology (AM syllabus content)

Which level of protein structure is held together mainly by hydrogen bonds between the carbonyl oxygen and amide hydrogen of the polypeptide backbone?

  1. Secondary structure such as an alpha helix Correct answer.
  2. Primary structure
  3. Quaternary structure
  4. Tertiary structure

Explanation

Secondary structure refers to local regular folding, such as alpha helices and beta pleated sheets, stabilised by hydrogen bonds between backbone C=O and N-H groups. These bonds do not involve the side chains, which distinguishes this level from tertiary interactions.

Question 30Biology (AM syllabus content)

During which phase of mitosis do sister chromatids separate and move to opposite poles of the cell?

  1. Anaphase Correct answer.
  2. Metaphase
  3. Prophase
  4. Telophase

Explanation

In anaphase the centromeres divide and the spindle fibres shorten, pulling the separated chromatids, now individual chromosomes, towards opposite poles. This is the defining event that distributes one copy of each chromosome to each future daughter cell.

Question 31Biology (AM syllabus content)

Which blood vessel carries oxygenated blood from the lungs to the heart?

  1. Vena cava
  2. Aorta
  3. Pulmonary artery
  4. Pulmonary vein Correct answer.

Explanation

The pulmonary veins return blood that has been oxygenated in the lungs to the left atrium of the heart. This is the standard exception to the rule that veins carry deoxygenated blood.

Question 32Biology (AM syllabus content)

Which process in the nitrogen cycle converts ammonium ions in the soil into nitrite and then nitrate ions?

  1. Nitrification by nitrifying bacteria Correct answer.
  2. Ammonification by decomposers
  3. Denitrification by denitrifying bacteria
  4. Nitrogen fixation by Rhizobium

Explanation

Nitrification is a two-step oxidation carried out by nitrifying bacteria such as Nitrosomonas and Nitrobacter, converting ammonium to nitrite and then nitrite to nitrate. Nitrate is the form most readily absorbed by plant roots.

Question 33Biology (AM syllabus content)

Which molecule is the final electron acceptor at the end of the electron transport chain in aerobic respiration?

  1. ATP synthase
  2. NAD+
  3. Pyruvate
  4. Oxygen Correct answer.

Explanation

Oxygen accepts electrons and protons at the end of the chain, forming water, which keeps electrons flowing through the carriers. Without this terminal acceptor the chain and associated proton pumping would halt.

Question 34Biology (AM syllabus content)

A plant is kept in continuous darkness for several days. Which immediate consequence for the Calvin cycle is expected?

  1. RuBP carboxylase is irreversibly denatured by the absence of light
  2. The supply of ATP and NADPH from the light-dependent reactions falls, so carbon fixation slows Correct answer.
  3. Carbon dioxide concentration in the leaf rises because stomata close permanently
  4. The rate of the Calvin cycle increases because photorespiration is suppressed

Explanation

The Calvin cycle depends on ATP and NADPH produced by the light-dependent reactions. In darkness these products are no longer generated, so reduction of glycerate-3-phosphate and regeneration of RuBP are limited, slowing carbon fixation.

Question 35Biology (AM syllabus content)

In a cross between a plant heterozygous for a single gene (Tt) and a plant homozygous recessive (tt), what proportion of the offspring is expected to show the dominant phenotype?

  1. One quarter
  2. One half Correct answer.
  3. Three quarters
  4. All offspring

Explanation

A Tt x tt cross produces gametes T and t from the heterozygote and only t from the homozygote. Half the offspring are Tt and show the dominant phenotype, while the other half are tt.

Question 36Biology (AM syllabus content)

A section of a DNA molecule has the template strand sequence 3'-TACGGATCC-5'. Assuming transcription proceeds antiparallel to this template, which mRNA sequence is produced?

  1. 5'-GUAGGCUAA-3'
  2. 5'-AUGCCUAGC-3'
  3. 5'-AUGCCUAGG-3' Correct answer.
  4. 5'-TACGGATCC-3'

Explanation

RNA polymerase reads the template 3' to 5' and builds mRNA 5' to 3', using complementary bases with uracil replacing thymine. Complementing T-A-C-G-G-A-T-C-C gives A-U-G-C-C-U-A-G-G.

Question 37Biology (AM syllabus content)

A mutation inserts one extra base pair early in a coding sequence. Which consequence is most likely?

  1. The reading frame is restored after the next codon
  2. A single amino acid is substituted with no other change to the protein
  3. The polypeptide is shortened by exactly one amino acid
  4. A frameshift alters every subsequent codon, usually producing a non-functional polypeptide Correct answer.

Explanation

Because the genetic code is read in non-overlapping triplets, adding one base shifts the reading frame for all codons downstream. The resulting amino acid sequence is completely different and usually includes a premature stop codon.

Question 38Biology (AM syllabus content)

In a mesophyll cell, which combination of features best adapts the chloroplast for efficient light absorption?

  1. Stacked thylakoid membranes bearing photosystems and a large surface area Correct answer.
  2. A starch grain occupying most of the stroma volume
  3. A single circular DNA molecule coding for the whole genome
  4. A double envelope membrane containing many carrier proteins

Explanation

Grana provide a large membrane surface area in a small volume, and the thylakoid membranes hold the photosystems and electron carriers needed for light absorption and the light-dependent reactions. The stacking maximises the capture of photons.

Question 39Biology (AM syllabus content)

Which feature of the small intestine most directly increases the surface area available for absorption of glucose?

  1. A low pH maintained by gastric acid
  2. Presence of goblet cells secreting mucus
  3. Villi and microvilli on the epithelial cells Correct answer.
  4. A thick layer of smooth muscle in the wall

Explanation

Finger-like villi project into the lumen and the epithelial cells bear microvilli, together multiplying the absorptive surface many times over. This large area allows rapid uptake of glucose by co-transport with sodium ions.

Question 40Biology (AM syllabus content)

A student finds that a plant stem bends towards a unilateral light source. Which explanation is consistent with the auxin theory of phototropism?

  1. Auxin is distributed evenly but light directly elongates cells on the shaded side
  2. Auxin accumulates on the shaded side and stimulates greater cell elongation there Correct answer.
  3. Auxin is destroyed on the shaded side, so cells there elongate less
  4. Auxin accumulates on the illuminated side and inhibits cell elongation there

Explanation

Lateral redistribution of auxin away from the light raises its concentration on the shaded side of the stem. Higher auxin promotes elongation of those cells, so the shaded side grows faster and the stem bends towards the light.

Question 41Biology (AM syllabus content)

In the mammalian kidney, which process occurs in the loop of Henle and contributes to the concentration of urine?

  1. Filtration of plasma proteins into the descending limb
  2. Active transport of sodium chloride out of the ascending limb, which is impermeable to water Correct answer.
  3. Secretion of urea into the distal convoluted tubule only
  4. Reabsorption of glucose by the descending limb

Explanation

The ascending limb actively pumps sodium and chloride into the medulla while remaining impermeable to water, so the medullary interstitial fluid becomes hypertonic. This gradient allows water to be withdrawn osmotically from the collecting duct under the influence of ADH.

Question 42Biology (AM syllabus content)

Two species of finch on the same island eat seeds of different sizes and have distinctly different beak depths. Which term best describes this pattern?

  1. Mutualism between the two species
  2. Competitive exclusion of one species
  3. Commensalism between the two species
  4. Resource partitioning reducing competition Correct answer.

Explanation

When species use different parts of an available resource range, they divide the resource and reduce direct competition, a pattern called resource partitioning. The divergence in beak depth is consistent with character displacement driven by competition.

Question 43Biology (AM syllabus content)

A DNA sample contains 28% adenine. What percentage of the bases is guanine?

  1. 22% Correct answer.
  2. 28%
  3. 44%
  4. 14%

Explanation

Chargaff's rules state that adenine equals thymine and guanine equals cytosine in double-stranded DNA. If adenine is 28%, thymine is also 28%, leaving 44% shared equally between guanine and cytosine, so guanine is 22%.

Question 44Biology (AM syllabus content)

Which statement describes the role of the enzyme DNA ligase in DNA replication?

  1. It adds complementary nucleotides to the growing strand
  2. It removes RNA primers and replaces them with DNA
  3. It separates the two parental strands by breaking hydrogen bonds
  4. It joins Okazaki fragments by forming phosphodiester bonds between adjacent nucleotides Correct answer.

Explanation

DNA ligase seals the nicks left between Okazaki fragments on the lagging strand, catalysing phosphodiester bonds between the sugar-phosphate backbones. This produces a continuous new strand.

Question 45Biology (AM syllabus content)

A respirometer is used to measure the rate of aerobic respiration in small invertebrates at two temperatures, 15 degrees Celsius and 25 degrees Celsius. Which result is expected and why?

  1. Oxygen uptake is the same at both temperatures because respiration is not enzyme-controlled
  2. Oxygen uptake is higher at 25 degrees Celsius because anaerobic respiration replaces aerobic respiration
  3. Oxygen uptake is higher at 15 degrees Celsius because gases dissolve better in cold water
  4. Oxygen uptake is higher at 25 degrees Celsius because enzyme-controlled reactions proceed faster Correct answer.

Explanation

Aerobic respiration is catalysed by enzymes, and within the physiological range a rise in temperature increases kinetic energy and the frequency of successful enzyme-substrate collisions. This raises the rate of oxygen consumption, provided the temperature remains below the optimum.

Question 46Biology (AM syllabus content)

A population of bacteria is exposed to an antibiotic. Which observation would provide the strongest evidence that resistance arose from a pre-existing mutation rather than induced by the antibiotic?

  1. Resistant colonies appear on a plate that was never exposed to the antibiotic Correct answer.
  2. Resistant colonies appear only after several days of exposure to the antibiotic
  3. The antibiotic causes the bacteria to increase their mutation rate
  4. Resistance is lost as soon as the antibiotic is removed

Explanation

Finding resistant cells in a population never exposed to the antibiotic shows that the resistance mutation existed before selection acted. The antibiotic then acts only as a selecting agent, favouring those already-resistant cells.

Question 47Biology (AM syllabus content)

In a species of mammal, a gene shows two alleles in a population. A survey finds 64% of individuals are homozygous dominant and 4% are homozygous recessive. Assuming Hardy-Weinberg equilibrium, what is the frequency of the recessive allele?

  1. 0.80
  2. 0.20 Correct answer.
  3. 0.04
  4. 0.32

Explanation

The frequency of homozygous recessive individuals equals q squared, so q squared is 0.04 and q is 0.20. The heterozygote frequency 2pq is then 0.32, and the homozygous dominant frequency p squared is 0.64, which matches the survey.

Question 48Biology (AM syllabus content)

A gene is subject to genomic imprinting. A child inherits a mutant allele from the father and a normal allele from the mother, and expresses the disease. Which interpretation is most consistent?

  1. The paternal allele is expressed because the maternal allele is silenced by imprinting Correct answer.
  2. The mutant allele is recessive, so both alleles must be mutant
  3. Imprinting means the mutant allele is always inactivated
  4. The disease arises from mitochondrial inheritance from the father

Explanation

Imprinting is parent-of-origin specific silencing, often by DNA methylation. If the maternal copy is imprinted and silenced, the child expresses only the paternal mutant allele, so disease appears even though the other allele is normal.

Question 49Biology (AM syllabus content)

In a study of a lake ecosystem, a sudden input of nitrate and phosphate triggers an algal bloom. Which sequence of consequences is most likely?

  1. Algal decay consumes oxygen, so dissolved oxygen falls and fish die Correct answer.
  2. Phosphate is absorbed by fish and removed from the lake
  3. Dissolved oxygen rises because photosynthesis increases, so fish thrive
  4. Nitrate is converted to nitrogen gas, which poisons fish

Explanation

Nutrient enrichment causes excessive algal growth, and when the algae die, aerobic decomposers multiply and consume oxygen. The resulting deoxygenation, together with shading of submerged plants, can kill fish and other aerobic organisms.

Question 50Biology (AM syllabus content)

A mutant strain of a flowering plant fails to produce functional phytochrome. Which response is most likely to be impaired?

  1. Water uptake by root hair cells
  2. Translocation of sucrose in the phloem
  3. Active transport of mineral ions into root hairs
  4. Germination and flowering responses to changes in day length and red light Correct answer.

Explanation

Phytochrome is the photoreceptor that detects red and far-red light and mediates responses such as seed germination and photoperiodic control of flowering. Lacking functional phytochrome, the plant cannot adjust these responses to day length or light quality.