Mark Scheme

Section A — Module 1

Question 1

(a) (6) — 2 marks per feature, awarded only where both cell types are contrasted; one-sided statements score 1.

• Nucleus: prokaryote has no true nucleus, DNA free in cytoplasm as a circular chromosome / eukaryote has DNA enclosed in a double membrane-bound nucleus.
• Organelles: prokaryote lacks membrane-bound organelles (no mitochondria, ER, Golgi) / eukaryote possesses them.
• Ribosomes: prokaryote 70S / eukaryote 80S in cytoplasm (accept 70S in mitochondria and chloroplasts).
• Cell wall: prokaryote wall of peptidoglycan / eukaryote wall of cellulose in plants, chitin in fungi, absent in animals.
• Size: typically 0.5–5 µm / 10–100 µm. Accept any three. Do not credit "prokaryotes are smaller" without a figure or comparative qualifier.

(b) (6) — 1 mark for naming each of four components, 1 mark each for two correctly stated functions (max 6; a candidate naming four with four functions still caps at 6, so credit the best-answered).

• Phospholipid bilayer — hydrophilic phosphate heads outward, hydrophobic fatty acid tails inward; forms a partially permeable barrier allowing small non-polar molecules through and blocking large polar/charged ones.
• Cholesterol — sits between phospholipids regulating fluidity; prevents excessive fluidity at high temperature and excessive rigidity at low temperature.
• Intrinsic (integral) proteins — channel and carrier proteins for facilitated diffusion and active transport of polar molecules and ions.
• Extrinsic (peripheral) proteins — structural support, enzymic activity, attachment to cytoskeleton.
• Glycoproteins / glycolipids — cell recognition, receptor sites for hormones, antigens, cell adhesion. Credit "fluid" (components move laterally) and "mosaic" (proteins scattered through the bilayer) if explained.

(c) (3)

• Percentage change = (4.30 − 5.00) / 5.00 × 100 = −14% (1 for correct method, 1 for correct value with sign or the word "decrease").
• At 0.40 mol dm⁻³ there is no net change in mass, therefore the water potential of the potato tissue is equal to that of the solution and there is no net osmotic movement of water (1). Accept "the solution is isotonic to the tissue". Do not credit "the tissue has no water potential".

Question 2

(a) (9) — 3 marks per polysaccharide: 1 for monomer/linkage, 1 for structural description, 1 for the function linked to that structure. A candidate who describes structure without relating it to function caps at 6.

• Starch — α-glucose; amylose is 1,4-linked and helical, amylopectin is 1,4- with 1,6- branches. Compact, coiled and insoluble, so it exerts no osmotic effect and is a good storage carbohydrate in plants; branching gives many ends for rapid hydrolysis.
• Glycogen — α-glucose, 1,4- and 1,6-linked but more highly branched than amylopectin. More branch ends means faster glucose mobilisation, suiting the higher metabolic rate of animals; stored in liver and muscle.
• Cellulose — β-glucose with alternate residues inverted, giving straight unbranched chains; hydrogen bonding between parallel chains forms microfibrils of high tensile strength, providing rigidity to the plant cell wall and resisting turgor pressure. The discriminating marks are the β-linkage/inversion point for cellulose and the branching comparison between starch and glycogen.

(b) (6) — Up to 3 marks for the sketch: axes labelled rate of reaction (y) against temperature/°C (x), curve rising from near zero, peaking at an optimum (accept 35–45 °C), then falling more steeply than it rose to reach zero. A symmetrical curve caps the sketch at 2. Up to 3 for explanation:

• Rising phase: increasing temperature increases kinetic energy of enzyme and substrate, so more frequent collisions and a greater proportion with energy exceeding the activation energy — more enzyme–substrate complexes per unit time (1–2).
• Optimum: temperature at which rate is maximal (1).
• Falling phase: above the optimum, increased vibration breaks hydrogen and ionic bonds maintaining tertiary structure; the active site changes shape and is no longer complementary to the substrate — the enzyme is denatured, and this is irreversible, which is why the fall is steeper than the rise (1–2). Do not credit "the enzyme is killed". "Denatured" must be linked to active-site shape for the final mark.

Section B — Module 2

Question 3

(a) (6) — 2 marks per point of comparison, requiring both sides.

• Divisions: mitosis one division / meiosis two successive divisions.
• Daughter cells: mitosis two diploid cells / meiosis four haploid cells; chromosome number maintained / halved.
• Variation: mitosis produces genetically identical clones / meiosis produces genetically distinct cells.

(b) (4) — 2 marks each for any two, requiring a mechanism not just a name.

• Crossing over in prophase I — non-sister chromatids of homologous pairs exchange segments at chiasmata, producing new combinations of alleles on a chromatid.
• Independent assortment in metaphase I — homologous pairs align at the equator in a random orientation, so maternal and paternal chromosomes are distributed independently; 2ⁿ possible combinations.
• Random fusion of gametes at fertilisation is acceptable only if the candidate notes it is a source of variation additional to meiosis itself.

(c) (5)

• Parental genotypes Rr (red) × rr (white) (1)
• Gametes: R and r from the heterozygote; r and r from the white parent — must be circled or clearly identified as gametes (1)
• Correct Punnett square or fusion lines showing all four combinations (1)
• Offspring genotypes Rr, Rr, rr, rr (1)
• Phenotypic ratio 1 red : 1 white (1) A correct ratio with no genetic diagram scores a maximum of 2 — the question asks for the diagram.

Question 4

(a) (5) — Definition (2): the process by which individuals with phenotypes better suited to the environment survive and reproduce more successfully, so the alleles conferring those phenotypes increase in frequency in the population over generations. Conditions (1 each, max 3): genetic variation exists within the population; that variation is heritable; organisms produce more offspring than can survive, creating competition / a selection pressure; differential survival and reproduction.

(b) (10) — Explanation up to 7, measures up to 3. Explanation (levels-marked; a top response makes the pre-existence point explicitly):

• Genetic variation for resistance existed in the population before the insecticide was applied, arising by random mutation; the insecticide did not cause resistance (2 — this is the principal discriminator).
• The insecticide acts as a selection pressure: susceptible individuals die, resistant individuals survive (1–2).
• Survivors reproduce and pass the resistance allele to offspring, so its frequency rises each generation (2).
• Reference to the data: the rise is slow at first (2% to 5%) because the allele is rare, then accelerates sharply (14% to 89% in three years) as resistant survivors come to dominate the breeding population; mosquitoes' short generation time allows this within six years (1–2). Candidates must quote figures from the table for the data marks. Measures (1 each, max 3, accept any sensible): rotate between insecticides with different modes of action; use insecticides only where necessary rather than continuously, at correct dose; integrate non-chemical control (removing standing-water breeding sites, biological control with larvivorous fish, bed nets); apply mixtures of insecticides simultaneously. Do not credit "the mosquitoes became immune" or any Lamarckian phrasing such as "they adapted to survive" — such answers cap at 4 however well written.

Section C — Module 3

Question 5

(a) (8) — 1 mark per point, max 8.

• Pollen grain lands on a stigma of the same species and adheres; chemical recognition permits germination.
• Pollen grain absorbs water and nutrients from the stigma surface.
• A pollen tube grows out through the style, controlled by the tube nucleus at its tip.
• The tube secretes hydrolytic enzymes digesting a path through the style tissue and grows down a chemical/nutrient gradient towards the ovary.
• The generative nucleus divides by mitosis to give two male gametes which travel down the tube.
• The tube enters the ovule through the micropyle and the tip breaks down, releasing both male gametes into the embryo sac.
• Double fertilisation: one male gamete fuses with the egg cell nucleus to form the diploid zygote.
• The second male gamete fuses with the two polar nuclei to form the triploid (3n) primary endosperm nucleus, which develops into the food store. The double fertilisation detail and the triploid endosperm are the marks that separate strong from average responses.

(b) (7) — 3 marks for three wind-pollinated adaptations, 3 for three insect-pollinated, 1 for overall comparative quality. Each adaptation must be structural, not behavioural.

• Wind: small, dull, green or absent petals (no need to attract); anthers large, loosely attached and exposed outside the flower so pollen is caught by air currents; stigmas large, branched and feathery to filter pollen from the air; pollen very light, smooth and produced in enormous quantity; no nectaries or scent.
• Insect: large, brightly coloured petals with honey guides; scent and nectaries to attract and reward insects; anthers enclosed within the flower and firm, positioned to brush the insect's body; stigma sticky and enclosed; pollen grains larger, fewer, sculptured or spiky to adhere to the insect. Award the seventh mark only where the answer is organised as a genuine comparison of matched features rather than two separate lists.

Question 6

(a) (8) — 2 marks per hormone.

• FSH — secreted by the anterior pituitary; stimulates development of a primary follicle in the ovary and stimulates the follicle to secrete oestrogen.
• Oestrogen — secreted by the developing follicle; causes proliferation and repair of the endometrium; at low levels it inhibits FSH by negative feedback, but the high pre-ovulatory peak exerts positive feedback triggering the LH surge.
• LH — secreted by the anterior pituitary; the mid-cycle surge triggers ovulation (around day 14) and stimulates conversion of the ruptured follicle into the corpus luteum.
• Progesterone — secreted by the corpus luteum; maintains and further thickens the vascularised endometrium ready for implantation; inhibits FSH and LH by negative feedback; its fall as the corpus luteum degenerates (if no implantation occurs) triggers menstruation. Credit the feedback relationships generously — candidates who list functions without any reference to feedback cap at 5.

(b) (7) — Mechanism up to 4:

• Contains synthetic oestrogen and progesterone (1).
• Maintained high levels exert negative feedback on the anterior pituitary and hypothalamus, inhibiting secretion of FSH and LH (1).
• Without FSH no follicle matures; without the LH surge ovulation does not occur (1).
• Additional effects: thickening of cervical mucus impeding sperm passage, and thinning of the endometrium making implantation unlikely (1). Considerations up to 3 (any two developed): reliability depends on being taken consistently at the same time daily, so effectiveness in practice is lower than in theory; it gives no protection against sexually transmitted infections, unlike barrier methods; possible side effects including raised risk of thrombosis, particularly in smokers and older users, and effects on blood pressure; interactions with some antibiotics and anticonvulsants; non-contraceptive benefits such as regulation of cycles and reduced menstrual pain; return of fertility on cessation. Accept ethical or religious considerations only if expressed in biological terms; the question asks for biological considerations.

Level descriptors (extended-response parts)

• Level 4 (top band): Accurate, detailed biological knowledge using correct terminology throughout; data quoted and manipulated where supplied; sketches correctly labelled and referred to in the prose; explanations form complete causal chains rather than lists; no Lamarckian or teleological phrasing.
• Level 3: Sound knowledge with mostly correct terminology; some development but chains of reasoning incomplete; data referred to only in general terms.
• Level 2: Largely descriptive; terminology imprecise; sketch absent or unlabelled; assertion in place of mechanism.
• Level 1: Isolated correct facts with no developed explanation.

Sample Answers with Examiner Commentary

Question 4(b) — Sample Answers

Grade I (Distinction) answer (extract)

"The critical point is that resistance was not created by the insecticide. Within the mosquito population of Year 1 there were already a small number of individuals — the 2% shown in the table — carrying an allele, arising originally by random mutation, that conferred resistance. That allele was present before spraying began and would have persisted at low frequency indefinitely without it, since in the absence of insecticide it confers no advantage and may even carry a metabolic cost.

Continuous application of the insecticide imposed a powerful selection pressure. Susceptible individuals were killed before reproducing; resistant individuals survived, bred, and passed the allele to their offspring. The frequency of the resistance allele therefore rose in each successive generation.

The shape of the change in the table reflects this precisely. Between Years 1 and 2 the rise is modest, from 2% to 5%, because the allele is initially rare and most matings involve two susceptible parents. Once resistant individuals form an appreciable fraction of the breeding population the increase becomes sharp — 14% to 38% to 67% — because a growing proportion of all offspring inherit the allele. By Year 6, at 89%, the population is close to fixation and the insecticide is effectively useless. That this occurred within six years reflects the very short generation time of mosquitoes, which allows many rounds of selection in a period that would represent only two or three generations in a long-lived species.

Two measures would slow this. First, rotating between insecticides with different modes of action, so that an individual resistant to one is not thereby protected against the next and the selection pressure for any single allele is never sustained. Second, integrating non-chemical control — eliminating standing-water breeding sites and using larvivorous fish — which reduces reliance on the chemical and so reduces the intensity of selection acting on the population."

Mark: 10/10. Examiner commentary: This is a model Level 4 response. The candidate opens with the pre-existence of the mutation, which is the single most heavily weighted point in this mark scheme and the one most often lost. Data is not merely mentioned but used: the candidate explains why the curve is slow then steep in terms of allele frequency in the breeding population, which is genuine analysis rather than description. The observation about generation time is beyond what the mark scheme requires and shows secure understanding. Both control measures are justified by reference to selection pressure rather than simply named. Full marks.

Grade III (Pass) answer (extract)

"The mosquitoes were sprayed with insecticide so they got used to it and became resistant. Over the years more and more of them were resistant, going from 2% up to 89% by Year 6, because they passed the resistance on to their offspring. To slow it down you could use a different insecticide or use less of it."

Mark: 4/10. Examiner commentary: There is credit here — the response correctly notes inheritance of the resistance and quotes the endpoints of the data, and both control measures are valid in outline. But the answer opens with "they got used to it and became resistant", which is Lamarckian: it says the insecticide caused the resistance rather than selecting for variation that already existed. The mark scheme caps such responses at 4 regardless of other merit, because this misconception is precisely what the question sets out to test. There is also no mention of selection pressure, differential survival, or mutation, and the data is quoted rather than explained — the candidate does not address why the increase accelerates. Rewriting the first sentence to say that a few resistant individuals already existed by mutation, and that spraying killed the rest, would roughly double this mark.

Question 2(b) — Sample Answers

Grade I (Distinction) answer (extract)

"[Sketch: y-axis labelled rate of reaction, x-axis labelled temperature / °C running 0 to 60; curve rises gently from near zero at 0 °C, steepening to a peak at approximately 40 °C, then falling away sharply to zero by about 55 °C — the fall notably steeper than the rise.]

Below the optimum, raising the temperature increases the kinetic energy of both enzyme and substrate molecules. They move faster, so collisions between them are more frequent, and a greater proportion of those collisions have energy exceeding the activation energy. More enzyme–substrate complexes form per unit time and the rate rises.

Above about 40 °C the rate falls, and it falls for an entirely different kind of reason. The increased vibration of the enzyme molecule breaks the hydrogen bonds and ionic interactions that hold its tertiary structure in shape. The active site is progressively distorted until it is no longer complementary to the substrate, so the substrate can no longer bind and the enzyme is denatured. Because this change to the tertiary structure is not reversible on cooling, denatured enzyme is lost permanently from the reaction — which is why the curve falls more steeply than it climbed rather than mirroring the rising phase."

Mark: 6/6. Examiner commentary: The sketch is correctly labelled on both axes and, importantly, is asymmetric — the steeper fall is drawn, not merely described. The rising phase is explained through collision frequency and activation energy, and the falling phase is taken all the way from vibration through bond breaking to active-site complementarity, which is the full causal chain. The closing sentence explicitly links the irreversibility of denaturation to the asymmetry of the curve, connecting the explanation back to the sketch. Full marks.

Grade II (Merit) answer (extract)

"[Sketch: bell-shaped curve, axes labelled, symmetrical about a peak at 40 °C.]

As temperature increases the molecules have more kinetic energy so they collide more often and the rate of reaction goes up. After the optimum temperature of 40 °C the enzyme becomes denatured so the rate goes down and eventually stops."

Mark: 4/6. Examiner commentary: Both phases are correctly identified and the optimum is stated, so this is secure Level 3. Two marks are lost. The sketch is symmetrical, whereas the fall should be visibly steeper than the rise — the mark scheme caps a symmetrical curve at 2 of the 3 sketch marks. And "the enzyme becomes denatured" is stated but not explained: there is no reference to bonds breaking, to the tertiary structure, or to the active site ceasing to be complementary to the substrate. Naming a process is not explaining it, and at CAPE the mechanism is where the marks sit.

How Unit 1 is assessed

Paper 02 carries 90 marks and contributes 50% of the Unit 1 external assessment mark, alongside Paper 01 (multiple choice, 30%) and the Internal Assessment (20%). All six questions are compulsory and the three Modules carry equal weight, so no Module can be neglected. Grades are reported Grade I – Grade VII, with Grade I the highest.

Three habits separate the top band. First, explain mechanisms rather than naming them — "the enzyme is denatured" and "the mosquitoes became resistant" are conclusions, and the marks lie in the causal chain that produces them. Second, use supplied data: where a table is given, the mark scheme reserves marks for quoting specific figures and accounting for the shape of the trend, not merely for restating the first and last values. Third, avoid teleological and Lamarckian phrasing — organisms do not adapt in order to survive, and populations do not become resistant because they were exposed; variation arises first by mutation and selection acts on it afterwards. This single distinction is examined somewhere on almost every Unit 1 paper.