Unit Exam 1 — 90 minutes

IGCSE Biology 0610 — Topics 14, 16, 17 and 18 — Section A: 20 multiple choice (20 marks) · Section B: structured (50 marks)

Duration90 minutes

Total Marks70

Questions20 MCQ + 5

Time Remaining90:00

Syllabus0610

Instructions

Answer all questions in both sections.
Section A: choose one option per question — marked automatically when you submit, with an explanation of the option you chose.
Section B: answer in the boxes; after submitting, mark yourself against each scheme and tick the marks you earned.
Show all working, especially in the genetics questions — a genetic diagram earns marks even when the final probability is wrong.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the four topics of your school Unit Assessment — Topic 14 Coordination and Response, Topic 16 Reproduction, Topic 17 Inheritance and Topic 18 Variation and Selection — and, like a real Cambridge paper, it mixes them inside single questions. It is deliberately set at Challenge level: expect to have to deduce, calculate and justify, not just recall. Section A is 20 multiple-choice questions (about 25 minutes); Section B is the 50-mark structured paper (about 55 minutes).

Section A — Multiple Choice

20 questions · 20 marks · about 25 minutes — Topics 14, 16, 17 and 18 mixed

Choose one option for each question. Section A is marked automatically when you submit the exam — and for every question you will be shown why the option you chose is right or wrong. An unanswered question scores 0.

1An impulse arrives at a synapse. Why can the message cross the synapse in one direction only?

Athe synaptic gap is too wide for any impulse to jump back across it Bonly the presynaptic side has vesicles of transmitter and only the postsynaptic side has receptors Cthe neurotransmitter is broken down and destroyed before it has any time to diffuse back across the gap Dthe impulse loses too much energy crossing the gap to make a return journey

2In a withdrawal reflex, which sequence of structures is correct?

Areceptor → sensory neurone → relay neurone → motor neurone → effector Breceptor → relay neurone → sensory neurone → motor neurone → effector Creceptor → motor neurone → relay neurone → sensory neurone → effector Deffector → sensory neurone → relay neurone → motor neurone → receptor

3A faint star is easier to see by looking slightly to one side of it. Why?

Athe pupil dilates further when the eyes point slightly away from an object Bthe lens becomes thinner and so focuses the dim starlight more strongly Cthe image falls on the fovea, where the cone cells are packed most densely Dthe image falls outside the fovea, onto rods, which respond in dim light

4A young shoot is lit from the left side only. What happens to the auxin, and to the shoot?

Aauxin gathers on the lit left side, that side elongates more, the shoot bends right Bauxin gathers on the lit left side, that side stops growing, the shoot bends left Cauxin gathers on the shaded right side, that side elongates more, the shoot bends left Dauxin gathers on the shaded right side, that side elongates more, the shoot bends right

5Which row shows three effects of adrenaline?

Aheart rate rises, pupils dilate, blood glucose concentration rises Bheart rate rises, pupils constrict, blood glucose concentration falls Cheart rate falls, pupils dilate, blood glucose concentration falls Dheart rate falls, pupils constrict, blood glucose concentration rises

6Which of these is fertilisation in a flowering plant?

Aa pollen grain being carried from the anther of one flower to the stigma of another flower Bthe growth of a pollen tube down through the style towards the ovary Cthe fusion of the male gamete nucleus with the egg cell nucleus in an ovule Da pollen grain landing on a mature, sticky and receptive stigma surface

7In the second half of the menstrual cycle, which hormone keeps the uterus lining maintained, and where is it made?

Aoestrogen, secreted by the ovary Bprogesterone, secreted by the ovary Cprogesterone, secreted by the uterus lining Doestrogen, secreted by the pituitary gland

8Which combination of features would you expect in a wind-pollinated flower?

Abright petals, nectar, and large amounts of small and smooth pollen Bsmall green petals and small amounts of large, spiky and sticky pollen Cbright showy petals, with the anthers and the sticky stigma enclosed deep inside the flower Dsmall dull petals, dangling anthers, and huge amounts of light, smooth pollen

9A human body cell contains 46 chromosomes. How many chromosomes are in a sperm cell, and in a zygote?

Asperm cell 23, zygote 23 Bsperm cell 23, zygote 46 Csperm cell 46, zygote 46 Dsperm cell 46, zygote 92

10Which statement about asexual reproduction is correct?

Atwo parents produce offspring that are genetically identical to each other Bone parent produces genetically different offspring by repeated mitosis Cone parent produces genetically identical offspring by the process of mitosis Done parent produces genetically identical offspring by the process of meiosis

11In pea plants the allele for tall (T) is dominant over short (t). Two Tt plants are crossed. What ratio of phenotypes is expected in the offspring?

A3 tall : 1 short, because Tt plants look exactly like TT plants B1 tall : 2 medium : 1 short, because Tt plants are half way in height C1 tall : 1 short, because the two alleles are passed on equally often Dall tall, because every offspring must inherit at least one T allele

12A tall pea plant could be TT or Tt. Which single cross best reveals its genotype?

Across it with a TT plant and count how many tall offspring appear Bcross it with another tall plant that has the same unknown genotype Cself-pollinate a short plant and compare its offspring with the tall plant Dcross it with a short (tt) plant and examine the offspring phenotypes

13One parent has blood group AB and the other has blood group O. Which blood groups are possible in their children?

Agroup AB or group O only, matching one or other of the parents Bgroup A or group B only Cgroups A, B, AB and O are all equally possible Dgroup AB only, because the A and B alleles are both codominant

14Red-green colour blindness is X-linked and recessive. Why are boys affected much more often than girls?

Athe allele is carried on the Y chromosome, which only boys have Bboys inherit the allele from their fathers, together with the Y Ca boy has only one X, so a single recessive allele is always expressed Dthe allele behaves as if dominant in boys but stays fully recessive in girls

15Which statement about meiosis is correct?

Ait makes four genetically different haploid cells from one diploid cell Bit makes two genetically identical diploid cells from one diploid cell Cit makes four genetically identical haploid cells from one diploid cell Dit makes two genetically different haploid cells that are used for growth

16Which feature of humans shows discontinuous variation?

Abody mass, which is affected by genes and also by diet Bheight, which is affected by many genes plus nutrition Chand span, which shows a complete continuous range of in-between values DABO blood group, which has four classes and nothing in between

17A population of bacteria became resistant to an antibiotic. What actually happened?

Aindividual bacteria changed themselves so they could survive the drug Bthe antibiotic itself caused the exact mutation that the bacteria needed to resist it Ca few bacteria already carried a resistance mutation and survived to breed Dthe bacteria learned to avoid the antibiotic and passed the habit down

18What is a mutation?

Aa change in an organism’s body caused directly by its environment Ba random, spontaneous change in the base sequence of DNA Ca deliberate change an organism makes to fit its surroundings Dthe new mixture of parental alleles made in sexual reproduction

19After soot darkened tree bark near factories, dark-winged moths became commoner than pale ones. What was the selection pressure?

Apredation by birds that hunt the resting moths by sight Bthe soot particles that settled on and darkened the tree bark Ca rise in the mutation rate of the gene for the wing colour Ddark moths choosing other dark-winged moths as their mates

20Which statement gives a correct difference between selective breeding and natural selection?

Aselective breeding produces offspring that are clones of the chosen parents Bselective breeding causes more mutations, so new alleles appear faster Cselective breeding favours features that aid survival; natural selection favours features useful to us Din selective breeding humans choose the parents; in natural selection the environment decides

Question 1 — Two Ways of Sending a Message

Total: 11 marks — nervous coordination, the eye, and hormones of the menstrual cycle

(a) [3]

An impulse travelling along a reflex arc must cross two synapses inside the spinal cord. Describe, in the correct order, what happens at a synapse, and explain why an impulse can only ever cross it in one direction.

Model Answer — 1(a)

when the impulse arrives, vesicles in the ending of the presynaptic neurone move to the membrane and release a neurotransmitter into the gap [1]

the neurotransmitter diffuses across the synaptic gap and binds to receptor protein molecules on the membrane of the next neurone, starting a new impulse there [1]

one direction only because only the presynaptic side has vesicles of neurotransmitter and only the postsynaptic membrane has the receptors [1]

⚠ If you missed marks here: The order is the mark. Vesicles release the chemical, the chemical diffuses, the chemical binds, and only then does a new impulse start. Candidates who write “the impulse jumps across the gap” score nothing — the impulse itself never crosses; a chemical carries the message. And the one-way explanation must name both halves of the asymmetry: vesicles on one side, receptors on the other. Naming only one is half an argument.

1 mark — vesicles in the presynaptic ending release a neurotransmitter into the gap 1 mark — neurotransmitter diffuses across and binds to receptor proteins on the postsynaptic membrane, starting a new impulse 1 mark — one-way because vesicles are only presynaptic and receptors only postsynaptic

(b) [3]

Tara looks up from her book to read a clock on the far wall of a dim room.
(i) Describe how her eyes change focus from the page to the distant clock. Name the structures involved. [2]
(ii) The room is dark and the clock face is faint. Explain why she can see the faint clock more clearly by looking slightly to one side of it rather than straight at it. [1]

Model Answer — 1(b)

(i) the ciliary muscles relax, so the suspensory ligaments are pulled taut [1]

(i) the taut ligaments pull the lens thinner / less curved, so light from the distant clock is refracted less and focuses on the retina [1]

(ii) rods work in dim light and are found outside the fovea; looking slightly to one side places the clock’s image on the rod-rich part of the retina instead of the cone-packed fovea, which needs bright light [1]

⚠ If you missed marks here: Accommodation answers collapse when the muscle and the ligament are swapped. Fix the logic once and keep it: the ciliary muscle is a ring — when it contracts the ring tightens and the ligaments go slack (near focus, fat lens); when it relaxes the ring widens and the ligaments go taut (distant focus, thin lens). If your answer had the muscles contracting for the far-away clock, that is the exact reversal examiners set this question to catch. Part (ii) is pure rods-versus-cones: dim light is rod territory, and rods live away from the fovea.

1 mark — ciliary muscles relax so the suspensory ligaments become taut 1 mark — lens pulled thinner / less curved, refracting light less for a distant object 1 mark — rods work in dim light and lie outside the fovea, so an off-centre image falls on the rods

Fig. 1.1 shows the concentrations of two hormones, labelled P and Q, in a woman’s blood across one 28-day menstrual cycle. Neither hormone has been named for you.

(i) One of the two hormones is progesterone. State which line shows progesterone and justify your choice using days and values read from Fig. 1.1. [2]
(ii) A third hormone, LH, is not shown. State the event marked by the dashed line at day 14 and the role of LH in causing it. [1]
(iii) State the role of progesterone between day 16 and day 26 of this cycle. [1]
(iv) Use the graph to explain why menstruation begins at the end of this cycle. [1]

Model Answer — 1(c)

(i) progesterone is the amber line labelled Q [1]

(i) justification with figures: the amber line stays near 5 units until day 14 and only rises after ovulation, peaking at about 85 units around day 21 — whereas the blue line (oestrogen) peaks at about 90 units at day 12–13, before day 14 [1]

(ii) day 14 is ovulation — a surge of LH triggers the release of the egg from the ovary [1]

(iii) between days 16 and 26 progesterone maintains the thickened lining of the uterus, keeping it ready for a fertilised egg to implant [1]

(iv) no fertilisation occurred, so progesterone falls steeply after about day 24 (from roughly 70 down towards 8 units); without progesterone the uterus lining is no longer maintained and breaks down — menstruation [1]

⚠ If you missed marks here: “Justify using the graph” means numbers, not adjectives. “Progesterone rises later” earns nothing that the grid did not already say — quote a day and a value: flat near 5 units until day 14, peak of about 85 at day 21. The deep pattern to memorise: oestrogen’s peak comes before ovulation (it repairs and rebuilds the lining), progesterone’s comes after (it maintains the lining). And menstruation is not caused by “the cycle ending” — it is caused by the fall of progesterone. Cause, then consequence.

1 mark — progesterone correctly identified as the amber line (the one labelled Q) 1 mark — justified with figures: flat near 5 units until day 14, rising only after ovulation to peak about 85 units near day 21, while the other hormone peaks before day 14 1 mark — day 14 is ovulation, triggered by a surge of LH 1 mark — progesterone maintains the thickened uterus lining ready for implantation 1 mark — progesterone falls steeply near the end of the cycle, so the lining breaks down and menstruation begins

Question 2 — What Plants Do Instead of Thinking

Total: 10 marks — flower structure, pollination, and the auxin experiments

(a) [4]

Table 2.1 compares two flowers, R and S.

feature	flower R	flower S
petals	small, green	large, brightly coloured
stigma	feathery, hangs outside the flower	flat and sticky, enclosed inside the flower
anthers	dangle outside on long filaments	held inside the flower
nectar	none	present

(i) State which flower is wind-pollinated and justify your choice using two features from Table 2.1, explaining how each feature suits that method. [2]
(ii) A student writes: “Once pollination has happened the flower is fertilised.” Explain the mistake, by defining both terms precisely. [2]

Model Answer — 2(a)

(i) the wind-pollinated flower is the one with green petals and no nectar (flower R): its feathery stigma hanging outside the flower gives a large surface to filter drifting pollen from the air [1]

(i) and its anthers dangling outside on long filaments let the wind shake the pollen out (accept: small dull petals / no nectar — nothing needs to attract insects) [1]

(ii) pollination is only the transfer of pollen from an anther to a stigma [1]

(ii) fertilisation happens later and separately: the male gamete nucleus from the pollen grain fuses with the egg cell nucleus inside the ovule — pollen can land on a stigma and never achieve this [1]

⚠ If you missed marks here: A feature alone is half a mark of thought — the mark lives in the because. “Feathery stigma” scores only when you say what the feathers are for: a big net hung in the airstream. For (ii), the two definitions must contain their key nouns: pollination = transfer, anther → stigma; fertilisation = fusion of nuclei, inside the ovule. Any answer using “the pollen fertilises the flower” without mentioning nuclei fusing has restated the student’s error, not corrected it.

1 mark — wind-pollinated flower correctly chosen with one feature explained (feathery hanging stigma filters airborne pollen) 1 mark — second feature explained (dangling anthers release pollen to wind / dull petals and no nectar as no insects needed) 1 mark — pollination defined as transfer of pollen from anther to stigma 1 mark — fertilisation defined as fusion of the male gamete nucleus with the egg nucleus in the ovule

(b) [5]

A student grew five sets of wheat coleoptiles (young shoots) and lit them from one side only. Table 2.2 shows the treatments and the results after 24 hours.

set	treatment	curvature towards light (degrees)	increase in length (mm)
A	left intact	21	8
B	tip cut off	0	1
C	tip covered with an opaque (black) cap	1	7
D	tip covered with a transparent cap	19	7
E	opaque tube around the base, tip left bare	20	8

(i) Using named sets from Table 2.2, state what the results show about where the growth substance is made and where the light is detected. [2]
(ii) Explain, in terms of auxin, why the intact shoots in set A curved 21° towards the light. [3]

Model Answer — 2(b)

(i) the decapitated set (B) grew barely at all (1 mm) and never curved — so the tip makes the growth substance; without the tip there is almost no elongation [1]

(i) the opaque-cap set (C) still grew (7 mm) but did not bend, while the transparent-cap set (D) bent 19° — comparing them shows light is detected by the tip, and the cap itself is not the cause (that is what the transparent cap controls for); the covered-base set (E) bending normally confirms the base does not detect light [1]

(ii) auxin is made in the tip and moves down the shoot to the region of elongation [1]

(ii) with light from one side, auxin becomes unequally distributed — it accumulates on the shaded side [1]

(ii) the extra auxin makes the cells on the shaded side elongate more, so that side grows longer and the shoot curves towards the light — positive phototropism [1]

⚠ If you missed marks here: Two classic losses. First, quoting a set without its numbers: the decapitated shoots are the evidence for production only because growth collapsed to 1 mm — say so. Second, and far more costly: writing that auxin “moves to the light side” or “makes the plant grow towards the light”. Auxin accumulates on the shaded side and that side grows longer — the shoot bends towards the light precisely because its dark side outgrows its lit side. If your arrow of cause points the other way, every following sentence is wrong with it.

1 mark — decapitated set (1 mm growth, no curve) shows the tip makes the growth substance 1 mark — opaque vs transparent cap comparison shows the tip detects the light (transparent cap = control for the cap itself) 1 mark — auxin made in the tip, moves down to the elongating region 1 mark — auxin becomes unequally distributed, accumulating on the shaded side 1 mark — cells on the shaded side elongate more, so the shoot bends towards the light

Many grasses can also reproduce asexually from runners. State one advantage to the plant of reproducing asexually in a stable, favourable environment.

Model Answer — 2(c)

any one: offspring are genetically identical to a parent already proven successful in that environment / reproduction is faster / no mate, pollinator or pollen transfer is needed [1]

⚠ If you missed marks here: The trap is answering with a disadvantage-shaped fact: “the offspring are all the same” is only an advantage if you finish the thought — the same as a parent that already thrives here. In a stable environment identical is optimal; in a changing one it is fatal. That single sentence is the whole asexual-versus-sexual argument, and examiners pay for the second half of it.

1 mark — one advantage: identical copies of an already-successful parent / faster / no mate or pollinator needed

Question 3 — Reading a Family Like a Puzzle

Total: 9 marks — pedigree deduction, probability, blood groups and meiosis

Cystic fibrosis is caused by a recessive allele, f. The normal allele is F. Fig. 3.1 shows a family. Shaded symbols are people who have cystic fibrosis; squares are males, circles are females.

(a) [3]

Deduce the genotypes of I-1, II-3 and III-1. For each person, give the reasoning that forces your answer — “could be either” is not a deduction, so show why each genotype is the only one possible.

Model Answer — 3(a)

I-1 is Ff — his daughter II-1 has cystic fibrosis (ff), so she received one f from each parent; I-1 is unaffected, so his other allele must be F [1]

II-3 is Ff — she is unaffected but her son III-2 is ff, and one of his two f alleles had to come from her [1]

III-1 is Ff, with certainty — her father II-4 has cystic fibrosis (ff) so she must have received an f from him; being unaffected, her other allele is F. She is a guaranteed carrier, unlike II-2, whose genotype cannot be pinned down [1]

⚠ If you missed marks here: The examiner’s favourite trick in a pedigree is the child of an affected parent: everyone such a parent produces must receive one recessive allele, so every unaffected child of an affected parent is a certain carrier — no probability about it. Compare that with the unaffected son in generation II, who could be FF or Ff and stays ambiguous forever. Knowing which individuals are deducible and which are not is the skill this question is testing; reasoning that says “probably Ff” for the guaranteed carrier loses the mark.

1 mark — the grandfather (I-1) shown to be Ff via his affected daughter 1 mark — the unaffected mother (II-3) shown to be Ff via her affected son 1 mark — the unaffected granddaughter (III-1) shown to be exactly Ff, because her affected father could only give her f

(b) [3]

II-3 and II-4 plan another child. Draw a genetic diagram (Punnett square) for this cross and use it to state:
(i) the probability the child will have cystic fibrosis [working + answer, 2]
(ii) the probability the child will be a daughter who has cystic fibrosis. [1]

Model Answer — 3(b)

gametes correct: F and f from the unaffected carrier mother; only f from the affected father [1]

(i) offspring 1 Ff : 1 ff, so the probability the child has cystic fibrosis = 1/2 (50%) [1]

(ii) P(has cystic fibrosis) × P(daughter) = 1/2 × 1/2 = 1/4 (25%) [1]

⚠ If you missed marks here: Part (ii) multiplies two independent events — the sex of the child has nothing to do with an autosomal allele, so the two halves simply multiply: 1/2 × 1/2 = 1/4. Candidates who answer 1/2 forgot the sex condition; candidates who answer 1/8 have smuggled in an extra halving that nothing in the question justifies. Also note the affected parent writes only one kind of gamete into the table — drawing four different gametes for two homozygous alleles is the most common Punnett error at this level.

1 mark — correct gametes: F and f from the carrier mother, f only from the affected father, in a Punnett square 1 mark — offspring 1 Ff : 1 ff read off, probability affected = 1/2 1 mark — probability of an affected daughter = 1/2 × 1/2 = 1/4

In a different family, a mother with blood group A and a father with blood group B have a first child with blood group O.
(i) Deduce the genotypes of both parents, using the symbols I^A, I^B and I^O. [1]
(ii) State all the blood groups possible for their next child. [1]

Model Answer — 3(c)

(i) the group O child is I^OI^O, so each parent must carry (and have passed on) an I^O: mother I^AI^O, father I^BI^O [1]

(ii) the cross I^AI^O × I^BI^O gives I^AI^B, I^AI^O, I^BI^O and I^OI^O — so the next child could be any of the four groups: AB, A, B or O [1]

⚠ If you missed marks here: This is the famous couple who can produce a child of every blood group — if you found only two or three groups, redo the square rather than trusting intuition. Remember the two layers of the system: I^A and I^B are codominant with each other (together they give group AB, not a blend), while both are dominant over I^O. “A is dominant over B” is the error that makes group AB impossible on paper — and it never is.

1 mark — parental genotypes deduced from the group O child: mother heterozygous A, father heterozygous B 1 mark — all four blood groups possible for the next child: AB, A, B and O

(d) [1]

II-1 and II-2 in Fig. 3.1 have the same two parents, yet one has cystic fibrosis and the other does not. Explain how meiosis makes this possible.

Model Answer — 3(d)

meiosis separates each pair of alleles so that a gamete receives a random one of the two — a parent who is Ff makes F gametes and f gametes; which gamete meets which at (random) fertilisation differs for each child, so the same parents produce genetically different offspring [1]

⚠ If you missed marks here: The word doing the work is random — random which allele enters a gamete, random which gametes fuse. An answer that only says “meiosis makes gametes” describes the process without explaining the difference between the siblings. Mitosis is the wrong answer twice over: it makes genetically identical cells, which is precisely what this family is not.

1 mark — meiosis puts a random one of each allele pair into each gamete (plus random fertilisation), so siblings inherit different combinations

Question 4 — The X Chromosome, and What Varies

Total: 9 marks — sex linkage, variation, and blood glucose

(a) [4]

Red-green colour blindness is caused by a recessive allele carried only on the X chromosome. Use the symbols X^N (normal vision) and Xⁿ (colour blindness).

A woman with normal colour vision, whose father was colour blind, has children with a man who has normal vision.
(i) Explain why the woman must be a carrier. [1]
(ii) Draw a genetic diagram for this cross in X notation, showing the parents’ genotypes, their gametes and the four possible offspring. [2]
(iii) Their first child is a boy. State the probability that he is colour blind. [1]

Model Answer — 4(a)

(i) her colour-blind father was XⁿY — the only X he could give his daughter was Xⁿ; she has normal vision, so her genotype is X^NXⁿ: a carrier by necessity [1]

(ii) gametes in X notation: X^N and Xⁿ from the mother; X^N and Y from the father [1]

(ii) four offspring correctly derived: X^NX^N (daughter, normal), X^NXⁿ (daughter, carrier), X^NY (son, normal), XⁿY (son, colour blind) [1]

(iii) the child is known to be a boy, so only the two sons count: one of the two is XⁿY → probability = 1/2 (not 1/4 — 1/4 is the chance for a child whose sex is unknown) [1]

⚠ If you missed marks here: Two traps in one part. First, notation: writing N and n floating free, without the X they ride on, loses the diagram mark — the whole point of sex linkage is that the allele travels on the X, and the Y has no copy at all. Second, the conditional probability: once the question tells you the child is a boy, the two daughter boxes of the square are off the table. Halve the world first, then count. Answering 1/4 to part (iii) means the condition was read but not used.

1 mark — carrier status forced: her colour-blind father could only give her his one X, carrying the colour-blind allele 1 mark — genetic diagram in X notation with correct gametes from each parent 1 mark — four offspring correct: normal daughter, carrier daughter, normal son, colour-blind son 1 mark — probability the first son is colour blind = 1/2, with the sons-only reasoning

(b) [3]

A class of 100 students recorded two features of themselves: height, plotted as a histogram with many narrow bars forming a smooth hump, and ABO blood group, plotted as exactly four separate bars (O, A, B, AB) with nothing in between.
(i) State which of the two features shows discontinuous variation and justify your answer from the shape of its chart. [1]
(ii) Explain the genetic difference between the causes of the two kinds of variation, and state the extra factor that affects one of them. [2]

Model Answer — 4(b)

(i) blood group is discontinuous — the chart shows only four distinct classes with no intermediates between the bars; every person falls exactly into one class [1]

(ii) discontinuous variation is controlled by a single gene (or a small number of genes), and the environment has no effect on it [1]

(ii) continuous variation, like height, is controlled by many genes acting together AND is also affected by the environment (for example diet), which is why it gives a complete range of values rather than classes [1]

⚠ If you missed marks here: The justification must come from the chart, as asked — “blood group is genetic” is true and scores nothing, because both features are genetic. The chart evidence is the gaps: classes with nothing between them. And in (ii) the examiners are listening for two exact ideas about continuous variation — many genes and environment as well. An answer giving genes-only for both kinds has described one cause twice, and the second mark goes unclaimed.

1 mark — blood group identified as discontinuous, justified by distinct classes with no intermediates 1 mark — discontinuous variation controlled by one or few genes, unaffected by environment 1 mark — continuous variation controlled by many genes plus the environment, giving a full range

During a long cross-country race, a runner’s blood glucose concentration begins to fall. Explain how her body returns the concentration towards normal without her eating anything. Name the hormone, its source and its effect.

Model Answer — 4(c)

the fall is detected by the pancreas, which secretes glucagon (not insulin — and not “glycogen”) into the blood [1]

glucagon causes the liver to break down its stored glycogen into glucose and release it into the blood, raising the concentration back towards the set point — negative feedback [1]

⚠ If you missed marks here: Three near-identical words, three different things: glucagon is the hormone, glycogen is the store, glucose is the sugar in the blood. One letter swapped and the sentence says the store signals itself. Keep the direction straight by anchoring to the situation: glucose low → glucagon → glycogen broken down. Insulin is the mirror image and belongs to the after-a-meal story, never to the runner.

1 mark — pancreas detects the fall and secretes glucagon 1 mark — glucagon makes the liver break glycogen down to glucose, released into the blood (negative feedback)

Question 5 — Selection, Natural and Otherwise

Total: 11 marks — antibiotic resistance, viruses, and selective breeding

(a) [6]

Table 5.1 shows the percentage of Staphylococcus aureus samples from one hospital that were resistant to the antibiotic meticillin.

year	2005	2010	2015	2020
samples resistant (%)	3	18	41	57

Explain the rise from 3% to 57% using natural selection. Your answer must give the full mechanism in the correct sequence — where the variation came from, what did the selecting, what happened to each kind of bacterium, and what changed in the population over the years.

Model Answer — 5(a)

the bacterial population varied: a few individuals already carried a resistance allele, which arose by chance mutation — before, and independently of, any exposure to the antibiotic [1]

meticillin acts as the selection pressure [1]

differential survival: the susceptible bacteria are killed; the few resistant ones survive [1]

the resistant survivors reproduce [1]

and pass the resistance allele to their offspring [1]

repeated over many generations, the proportion of the population carrying the allele rises — exactly the climb from 3% of samples in 2005 to 57% in 2020 [1]

⚠ If you missed marks here: The one sentence that destroys this answer is “the bacteria became resistant so that they could survive the antibiotic” — that is Lamarck’s error, organisms changing because they need to. Nothing in natural selection is on purpose. The mutation happened first, blindly, by chance; the antibiotic never creates or instructs resistance, it only selects among variants that already exist — it is a sieve, not a teacher. Every phrase like “in order to”, “adapted so that”, “learned to resist”, or “became immune” costs the variation mark and usually the selection mark with it. (“Immune” is doubly wrong — immunity is an animal’s defence response; bacteria are resistant.) The examiners also want the sequence: variation → selection pressure → differential survival → reproduction → inheritance → changed proportion. Six links; a missing link is a missing mark.

1 mark — variation already present: resistance allele arose by chance mutation before exposure 1 mark — the antibiotic named as the selection pressure 1 mark — differential survival: susceptible killed, resistant survive 1 mark — the resistant survivors reproduce 1 mark — resistance allele passed to offspring 1 mark — over generations the proportion carrying the allele rises, matching 3% to 57%

(b) [2]

A patient at the same hospital has an HIV infection. A friend suggests treating it with antibiotics.
(i) Explain why antibiotics cannot work against HIV. [1]
(ii) State one way the transmission of HIV can be reduced. [1]

Model Answer — 5(b)

(i) antibiotics work by attacking structures and processes of bacterial cells (such as cell-wall building); a virus is not a cell and has no metabolism of its own — it only replicates inside the host’s cells — so there is nothing for the antibiotic to act on [1]

(ii) any one: using condoms / never sharing needles / screening donated blood [1]

⚠ If you missed marks here: “HIV is a virus and antibiotics only kill bacteria” restates the question in different words — the mark needs the why: a virus has no cell wall, no metabolism, no machinery of its own for the drug to poison. It borrows the host’s. Any drug that could stop the replication would be attacking the patient’s own cells — which is exactly what makes viral disease hard to treat.

1 mark — viruses are not cells and have no own metabolism or cell wall, so antibiotics have no target 1 mark — one valid transmission control: condoms / not sharing needles / screening blood

A farmer wants to raise the average milk yield of her herd of cows.
(i) Describe how she could do this by selective breeding. [2]
(ii) State one way selective breeding differs from natural selection. [1]

Model Answer — 5(c)

(i) choose the cows with the highest milk yield (and a bull whose female relatives yield well) and cross them [1]

(i) from the offspring, again select the best yielders as parents and repeat over many generations — the repetition is what shifts the average [1]

(ii) in selective breeding humans choose which individuals breed, for features useful to people; in natural selection the environment does the selecting, for features that help survival and reproduction (also acceptable: selective breeding is usually much faster / can favour features that would be harmful in the wild) [1]

⚠ If you missed marks here: The second mark of (i) is the one candidates drop: a single cross is not selective breeding. The scheme wants select → cross → select the best offspring → repeat over many generations — without the repetition you have described buying a good cow, not breeding one. A subtle trap in the bull: he gives no milk himself, so he is judged by his mother’s and daughters’ yields — mentioning that is the mark of a strong answer.

1 mark — select the highest-yielding cows (and a bull from a high-yield line) and cross them 1 mark — select the best offspring as new parents and repeat over many generations 1 mark — one difference: humans choose for usefulness vs the environment selecting for survival

Self-Assessment

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