Unit Exam 2 — 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.

1Where does most of the bending (refraction) of light entering the eye take place?

Aat the lens, because its shape can be adjusted by the ciliary muscles Bat the pupil, because its diameter controls how much light enters Cat the curved cornea, where light first passes into the eye Dat the retina, where the light is turned into nerve impulses

2Tara looks up from the far wall to a book close to her face. Which row describes her eyes focusing on the book?

Aciliary muscles contract, suspensory ligaments slacken, lens becomes thicker Bciliary muscles contract, suspensory ligaments tighten, lens becomes thinner Cciliary muscles relax, suspensory ligaments slacken, lens becomes thicker Dciliary muscles relax, suspensory ligaments tighten, lens becomes thinner

3Tara steps out of a dark room into bright sunshine. What happens in the iris of each eye?

Acircular muscles relax and radial muscles contract, widening the pupil Bcircular muscles contract and radial muscles relax, narrowing the pupil Cthe circular and the radial muscles both contract, holding the pupil rigid Dthe circular and the radial muscles both relax, letting the pupil drift wider

4A seedling is laid on its side in complete darkness. Why does its shoot bend and then grow upwards?

Athe shoot grows up towards light coming from above the seedling Bauxin gathers on the upper side and the upper cells elongate more Cauxin spreads evenly, so the shoot simply keeps growing straight on Dauxin gathers on the lower side and the lower cells elongate more

5Which row correctly compares nervous and hormonal communication?

Anervous: chemical, slower, long-lasting — hormonal: electrical, faster, brief Bnervous: electrical, fast, short-lived — hormonal: chemical in blood, slower, longer-lasting Cnervous: electrical, fast, long-lasting — hormonal: chemical, slower, brief Dnervous: chemical in blood, fast, brief — hormonal: electrical, slow, lasting

6At the start of the menstrual cycle, which hormone from the pituitary gland stimulates an egg to start maturing?

AFSH, which makes a follicle and its egg develop in the ovary BLH, which surges in the middle of the cycle at about day 14 Coestrogen, which rebuilds the lining of the wall of the uterus Dprogesterone, which keeps the lining of the uterus in place

7Where does fertilisation normally happen in humans?

Ain the uterus, where the embryo will later implant and develop Bin the ovary, inside the follicle where the egg cell matured Cat the cervix, at the point where sperm first enter the uterus Din the oviduct, shortly after the egg cell has been released

8Which statement about the placenta is correct?

Athe mother’s blood flows through the fetus, delivering oxygen directly Bthe placenta blocks all pathogens, so no virus can ever reach the fetus Cthe mother’s and the fetus’s blood come close together but never mix Dthe umbilical vein carries deoxygenated blood away from the fetus

9Which set of conditions does a germinating seed need?

Awater, carbon dioxide and bright continuous sunlight Bwater, oxygen and a suitable warm temperature Coxygen, sunlight and a good supply of mineral ions Dwater, warmth and some light for early photosynthesis

10Compared with self-pollination, what is the main advantage of cross-pollination?

Ait needs no pollinators, because the pollen stays within one flower Bit guarantees that every ovule in the flower will be fertilised Cit produces offspring genetically identical to the parent plant Dit increases variation, helping the population survive change

11For a gene with alleles T and t, which of these is a genotype?

Atall — the visible feature that the organism displays BT — the dominant allele written on its own CTt — the pair of alleles that the organism carries Dtallness varying with the plant’s growing conditions

12In cattle the alleles for red coat (C^R) and white coat (C^W) are codominant; C^RC^W cattle are roan. Two roan cattle are crossed. What is the expected ratio in the calves?

A1 red : 2 roan : 1 white B3 roan : 1 white, with roan acting as the dominant coat C1 red : 1 white, with no roan calves appearing at all Dall roan calves, exactly like their two roan parents

13Which statement explains why about half of all babies are boys?

Ahalf of all egg cells carry an X and half carry a Y chromosome BXY zygotes divide faster, which keeps the two sexes balanced Cthe mother’s age decides which sex chromosome the baby gets Dhalf of the sperm carry an X and half carry a Y chromosome

14Colour blindness is X-linked recessive (X^N normal, Xⁿ colour blind). A carrier woman (X^NXⁿ) and a man with normal vision (X^NY) have children. Which child is NOT possible?

Aa son who is colour blind Ba daughter who is colour blind Ca daughter who carries the allele Da son who has normal colour vision

15A dominant allele is best described as one that ...

Ais the commonest allele of that gene in the population’s gene pool Bmakes an organism fitter and therefore more likely to survive Cis expressed in the phenotype when only one copy is present Dis expressed only when two copies of it are inherited together

16Which process creates completely NEW alleles?

Amutation, a random change in the DNA of a gene Bmeiosis, which shuffles the alleles into the gametes Cfertilisation, which combines alleles from two parents Dmitosis, which copies every chromosome set exactly

17Which is the best description of an adaptive feature?

Aany feature an organism perfects during its lifetime by practice Ba feature that a parent acquires and then passes to its offspring Cany feature that makes an organism larger or stronger than rivals Dan inherited feature that helps survival and reproduction in its environment

18Which sequence describes natural selection correctly?

Aa selection pressure appears; useful mutations then occur in response; the survivors change their alleles Bvariation already exists; a selection pressure acts; the better adapted survive and reproduce; their alleles become commoner Corganisms detect the danger; they alter their genes to cope; the altered genes pass on to the offspring Dvariation already exists; the biggest and strongest always survive; the survivors become genetically identical

19Why can many generations of selective breeding leave a crop badly placed against a NEW disease?

Athe breeding raises the mutation rate, so harmful alleles slowly build up Bthe selected plants are always weaker, since yield and health cannot both be bred Cthe breeding reduces variation, so few plants carry an allele that might resist it Dthe crop’s genes become modified, and modified genes fail as diseases change

20Bacteria can evolve antibiotic resistance within a few years. Which property makes evolution so FAST in bacteria?

Atheir very short generation time lets selection act over many generations quickly Bthey choose which of their mutations to keep and which to discard Cantibiotics make the bacteria copy their DNA with far more errors Dthey swap body cells with one another, spreading features rapidly

Question 1 — Signals at Two Speeds

Total: 10 marks — the reflex arc, the iris, and hormones of the menstrual cycle

(a) [3]

Tara picks up a hot pan by mistake and drops it before she feels any pain.
(i) Write out the full pathway of this reflex, from the structure that detects the stimulus to the structure that responds. [1]
(ii) Explain why the response happens before she feels pain, and why a reflex like this does not have to be learned. [2]

Model Answer — 1(a)

pathway: (pain/heat) receptor in the skin → sensory neurone → relay neurone in the spinal cord → motor neurone → effector (arm muscle contracts, dropping the pan) [1]

the arc runs through the spinal cord with only two synapses — the muscle is triggered before the information has travelled on up to the conscious brain, which is why the pain arrives after the pan has already been dropped [1]

reflexes are innate, automatic protective responses — no conscious decision is involved, so nothing has to be learned [1]

⚠ If you missed marks here: The commonest error is writing “the brain decides very quickly” — the whole point of a spinal reflex is that the response does NOT wait for the brain to decide anything. The pathway mark needs all five structures in order; dropping the relay neurone, or swapping sensory and motor, loses it.

1 mark — full pathway in order: receptor, sensory neurone, relay neurone (spinal cord), motor neurone, effector 1 mark — response triggered in the spinal cord before information reaches the conscious brain — so pain is felt after the drop 1 mark — reflex is automatic/innate, needing no conscious decision or learning

(b) [3]

Tara then walks out of the dark kitchen into bright sunshine. Name the two sets of muscles in her iris, state what each one does in the sunshine, and explain how this response protects the eye.

Model Answer — 1(b)

in bright light the circular muscles contract [1]

and the radial muscles relax — the two sets are antagonistic, so they can never usefully contract together [1]

the pupil narrows, so less light enters and the retina is protected from damage by too much light [1]

⚠ If you missed marks here: Both sets of muscles must be named AND given the correct action — “the iris contracts” names no muscle and scores nothing. If you wrote circular relax / radial contract, that is the dim-light response: you have described her walking back into the kitchen.

1 mark — circular muscles contract in bright light 1 mark — radial muscles relax — the two sets act antagonistically 1 mark — narrower pupil admits less light, protecting the retina

Fig. 1.1 shows the concentrations of two hormones, labelled X and Y, in a woman’s blood during one 28-day menstrual cycle. Neither hormone has been named for you.
(i) Identify X and Y, justifying each answer with figures read from the graph. [2]
(ii) Name the event that hormone Y’s surge causes, and state on which day it happens here. [1]
(iii) State the role of hormone X in the first half of the cycle. [1]

Model Answer — 1(c)

X is oestrogen: it climbs from about day 5 and peaks at about 80 units at about day 12 — before the middle of the cycle, which progesterone never does [1]

Y is LH: it stays near 5 units for most of the cycle, then shows a single sharp surge peaking at about 90 units on day 14 [1]

the LH surge triggers ovulation — release of the egg from the ovary — at about day 14 [1]

in the first half of the cycle oestrogen repairs and thickens the lining of the uterus after menstruation [1]

⚠ If you missed marks here: The tell is WHERE each curve peaks. A hormone peaking before day 14 is oestrogen; progesterone’s peak comes around day 21, in the second half — if you called X progesterone, that is the classic oestrogen/progesterone swap. Answers with no figures (“X rises then falls”) lose the justification marks: quote the numbers, e.g. “80 units at day 12”.

1 mark — X identified as oestrogen with figures: peak about 80 units at about day 12, before mid-cycle 1 mark — Y identified as LH with figures: sharp surge to about 90 units at day 14 1 mark — Y’s surge causes ovulation, at about day 14 here 1 mark — X’s first-half role: repairs/thickens the uterus lining after menstruation

Question 2 — From Pollen Grain to Seedling

Total: 10 marks — fertilisation in plants, germination, and a response to gravity

(a) [3]

The flowers of a bean plant are insect-pollinated. A bee has just left pollen grains on a stigma. Describe what must happen next for an ovule to become a seed. Use the words pollen tube, male gamete nucleus, ovule and fertilisation in your answer.

Model Answer — 2(a)

the pollen grain grows a pollen tube down through the style towards the ovary [1]

the male gamete nucleus travels down the tube and fuses with the egg cell nucleus inside the ovule — this fusion is fertilisation [1]

the fertilised ovule then develops into the seed (and the ovary around it into the fruit) [1]

⚠ If you missed marks here: Pollination has already happened — describing the bee again earns nothing. The middle mark needs the fusion of NUCLEI: “the pollen reaches the ovule” is transport, not fertilisation. And it is the ovule that becomes the seed — ovule→fruit or ovary→seed both lose the last mark.

1 mark — pollen tube grows down through the style to the ovary 1 mark — male gamete nucleus fuses with the egg cell nucleus in the ovule — fertilisation 1 mark — fertilised ovule develops into the seed (ovary into the fruit)

(b) [4]

A student set up four tubes of cress seeds on cotton wool, shown in Table 2.1.
(i) Using the tubes in pairs, state the three conditions cress seeds need in order to germinate, giving the evidence for each from the table. [2]
(ii) Explain why the germinating seeds need water, and why they need oxygen. [2]

tube	conditions	result after 5 days
1	moist cotton wool, air, 20 °C	all seeds germinated
2	dry cotton wool, air, 20 °C	no germination
3	moist cotton wool, nitrogen instead of air, 20 °C	no germination
4	moist cotton wool, air, 2 °C	no germination

Model Answer — 2(b)

water (tube 2 vs 1), oxygen (tube 3 vs 1) and a suitable/warm temperature (tube 4 vs 1) — each pair changes only ONE condition, with tube 1 as the comparison [2: all three conditions with their evidence 2, any two 1]

water: activates the enzymes (and is needed to break down and transport the stored food) [1]

oxygen: for aerobic respiration, releasing the energy needed for growth [1]

⚠ If you missed marks here: Name the tubes — “seeds need water, oxygen and warmth” with no evidence is recall, not use of the table, and the question says ‘using the tubes’. For (ii), “oxygen to breathe” is too loose at this level: the mark needs aerobic respiration releasing energy. Note light is deliberately absent from the table — seeds do not need it.

1 mark — all three conditions each tied to its pair of tubes (2 vs 1 water, 3 vs 1 oxygen, 4 vs 1 temperature) 1 mark — each comparison changes only one condition — tube 1 acts as the control 1 mark — water activates enzymes / mobilises the stored food 1 mark — oxygen for aerobic respiration, releasing energy for growth

A germinating bean is pinned on its side inside a dark box. Two days later its shoot has bent and is growing upwards.
(i) Explain, in terms of auxin, how the bend is produced. [2]
(ii) Explain why doing this experiment in darkness makes the conclusion trustworthy. [1]

Model Answer — 2(c)

auxin accumulates on the lower side of the horizontal shoot [1]

cells on the lower side elongate more, so the shoot bends upwards, away from gravity [1]

darkness removes light as a stimulus, so the bending cannot be phototropism — the only stimulus left is gravity [1]

⚠ If you missed marks here: Auxin gathers on the LOWER side — and in a shoot it speeds elongation, so the lower side outgrows the upper and the tip turns up. Writing “auxin flows to the top so the top grows” reverses the mechanism. The darkness mark is about controlling a variable: with light present you could not tell gravitropism from phototropism.

1 mark — auxin accumulates on the lower side of the shoot 1 mark — lower-side cells elongate more, bending the shoot upwards 1 mark — darkness rules out phototropism — gravity is the only stimulus acting

Question 3 — A Condition That Will Not Skip

Total: 10 marks — a dominant-allele pedigree, probability, and what a ratio really promises

(a) [4]

Fig. 3.1 shows a family with a rare condition caused by a dominant allele D. People with genotype dd are unaffected. Shaded symbols show people who have the condition.
Deduce the genotypes of I-1, II-2 and III-1. For I-1 you must show why his genotype is forced by the pedigree — “could be DD or Dd” is not a deduction. [4]

Model Answer — 3(a)

I-1 is affected, so he has at least one D. His son II-3 is unaffected (dd) and received one allele from each parent — so I-1 must carry a d to give. I-1 is exactly Dd [2: genotype 1, forcing argument via II-3 1]

II-2 is affected, so she has a D; her mother I-2 is unaffected (dd) and could only give her a d. II-2 is Dd [1]

III-1 is affected, so she has a D; her father II-1 is unaffected (dd) and gave her a d. III-1 is Dd [1]

⚠ If you missed marks here: With a DOMINANT condition the deduction runs through the unaffected relatives: every unaffected person is dd, full stop, and every d they hand to an affected child pins that child as Dd. If you treated shaded symbols as recessive (ff-style) out of habit, every deduction inverts — check which allele the question says is dominant before touching a pedigree.

1 mark — I-1 shown to be exactly Dd, forced via his unaffected son II-3 (dd) 1 mark — the forcing logic stated: an unaffected child must receive d from BOTH parents 1 mark — II-2 deduced Dd via her dd mother I-2 1 mark — III-1 deduced Dd via her dd father II-1

(b) [3]

II-1 and II-2 plan another child. Draw a genetic diagram (Punnett square) for this cross and use it to state:
(i) the probability that the child will have the condition [working + answer, 2]
(ii) the probability that the child will be an affected daughter. [1]

Model Answer — 3(b)

gametes: II-2 (Dd) gives D or d; II-1 (dd) gives only d [1]

offspring 1 Dd : 1 dd, so the probability of an affected child is 1/2 [1]

P(affected daughter) = 1/2 × 1/2 = 1/4 [1]

⚠ If you missed marks here: The affected parent here is Dd, not DD — that came from part (a), and using DD gives every child the condition and no marks. For (ii), sex and genotype are independent events, so the two halves multiply; adding them (to get 1) or forgetting the sex factor (leaving 1/2) are the standard slips.

1 mark — correct gametes: D or d from II-2, d only from II-1, in a Punnett square 1 mark — offspring 1 Dd : 1 dd read off — probability affected = 1/2 1 mark — probability of an affected daughter = 1/2 × 1/2 = 1/4

In a different family, two affected people who are both Dd have children together.
(i) Give the expected ratio of affected to unaffected children, showing the genotypes behind it. [2]
(ii) The couple in fact have four children, and all four are affected. Does this prove the expected ratio wrong? Explain. [1]

Model Answer — 3(c)

Dd × Dd gives genotypes 1 DD : 2 Dd : 1 dd [1]

DD and Dd both show the condition, so the phenotype ratio is 3 affected : 1 unaffected [1]

no — each fertilisation is an independent chance event; a ratio is a probability, and small numbers of children often differ from it [1]

⚠ If you missed marks here: Keep the two ratios apart: 1 : 2 : 1 is the GENOTYPE ratio, 3 : 1 is the PHENOTYPE ratio, and quoting 1 : 2 : 1 as “the ratio of affected children” muddles them. Part (ii) is testing whether you know a ratio is a probability, not a promise — four affected children in a row has probability (3/4)⁴ ≈ 0.32, which is hardly surprising.

1 mark — genotype ratio 1 DD : 2 Dd : 1 dd from the Punnett square 1 mark — phenotype ratio 3 affected : 1 unaffected (DD and Dd look the same) 1 mark — ratio is a probability — each fertilisation is independent, so four affected children does not disprove it

Question 4 — Written on the X

Total: 10 marks — sex linkage in X notation, and where new alleles come from

(a) [2]

Haemophilia is a condition in which blood fails to clot properly. It is caused by a recessive allele h carried on the X chromosome; the normal allele is H. Explain why haemophilia is far more common in males than in females.

Model Answer — 4(a)

a male has only one X chromosome, so a single copy of the recessive h allele is always expressed — there is no second allele to mask it [1]

a female is only affected if she is X^hX^h — she must inherit an h from both parents, which is far rarer [1]

⚠ If you missed marks here: The answer must be about the NUMBER of X chromosomes, not vague “boys are more susceptible”. And be exact about the female case: one h makes her a carrier, not affected — saying “girls cannot get it” overshoots, as part (b) is about to show.

1 mark — males have one X, so a single h allele is always expressed 1 mark — females need h from both parents (X^hX^h) — much rarer; one copy only makes a carrier

(b) [5]

A carrier woman (X^HX^h) has children with a man who has haemophilia (X^hY).
Draw a full genetic diagram in X notation. Give the genotype and phenotype of every possible child, and state the probability that a child of this couple has haemophilia. [5]

Model Answer — 4(b)

parents X^HX^h × X^hY; gametes X^H or X^h from the mother, X^h or Y from the father [1]

four offspring: X^HX^h, X^hX^h, X^HY, X^hY [1]

phenotypes: carrier daughter, daughter with haemophilia, unaffected son, son with haemophilia [1]

probability a child has haemophilia = 2/4 = 1/2 [1]

an affected daughter is possible with THIS couple because the father supplies X^h — the rule of thumb “only boys are affected” fails whenever the father is affected and the mother is a carrier [1]

⚠ If you missed marks here: The trap in this cross is X^hX^h: if you wrote that every daughter is merely a carrier, you copied the standard textbook cross (carrier × normal man) instead of THIS one. Genotype and phenotype are both asked for — a Punnett square with no phenotype labels loses that mark. And the probability asked for is over ALL children (1/2), not among sons only.

1 mark — parents and all four gamete types shown in X notation 1 mark — all four offspring genotypes correct 1 mark — all four phenotypes correct, including the affected daughter X^hX^h 1 mark — probability of haemophilia = 1/2 across all children 1 mark — stated why an affected daughter is possible here (father gives X^h)

(i) The h allele first arose by mutation. State what a mutation is, and name one factor that increases the rate of mutation. [2]
(ii) Explain how the h allele can persist for many generations in a family in which nobody has haemophilia. [1]

Model Answer — 4(c)

a mutation is a random change in the base sequence of DNA / of a gene [1]

ionising radiation (X-rays, gamma rays, ultraviolet) or certain chemicals increase the rate [1]

the allele can hide in carrier females (X^HX^h): the dominant H masks it, so it is passed silently from mother to daughter until an X^h finally meets a Y [1]

⚠ If you missed marks here: “Random” is doing real work in the definition — a mutation is not a response to need, and radiation raises only the RATE, it does not aim the change. For (ii), the recessive allele is invisible in every carrier, which is exactly why X-linked conditions seem to appear from nowhere after generations of unaffected family.

1 mark — mutation defined as a random change in the DNA/base sequence of a gene 1 mark — ionising radiation or a named chemical mutagen increases the rate 1 mark — allele persists hidden in carrier females, masked by the dominant H

Question 5 — The Rats That Would Not Die

Total: 10 marks — natural selection with data, a Lamarckian trap, and selective breeding

(a) [5]

The poison warfarin was first used against rats in the 1950s. Table 5.1 shows the percentage of rats in one city that were resistant to warfarin.
Explain the change from 0.5% to 45% in terms of natural selection. [5]

year	1960	1975	1990	2005
rats resistant to warfarin (%)	0.5	9	28	45

Model Answer — 5(a)

variation already existed: the resistance allele arose by random mutation, before (and not because of) the poisoning [1]

warfarin is the selection pressure [1]

susceptible rats are killed; resistant rats survive [1]

the surviving resistant rats reproduce, passing the resistance allele to their offspring [1]

over many generations the proportion of rats carrying the allele rises — matching the climb from 0.5% to 45% in the table [1]

⚠ If you missed marks here: Five marks means five separate steps — variation first, pressure named, differential survival, reproduction and inheritance, and the population-level change tied back to the figures. The two killers are “the rats became resistant” (individuals never change — the population’s make-up changes) and leaving the table unused: quote 0.5% and 45%.

1 mark — variation pre-existed — resistance allele from random mutation, not caused by warfarin 1 mark — warfarin named as the selection pressure 1 mark — differential survival: susceptible killed, resistant survive 1 mark — survivors reproduce, passing the allele on 1 mark — proportion carrying the allele rises over generations, matching 0.5% → 45%

(b) [2]

A student writes: “The rats got used to the poison by eating small amounts, so their bodies became resistant, and they passed this resistance on to their young.”
Give two different reasons why this explanation is wrong.

Model Answer — 5(b)

individual rats do not become resistant by exposure — a rat is resistant (or not) from birth, according to the alleles it inherited [1]

characteristics acquired during an organism’s lifetime are not inherited — only alleles pass to offspring, so ‘getting used to it’ could never be passed on [1]

⚠ If you missed marks here: This is the Lamarckian account, and the mark scheme wants it dismantled at both ends: no individual acquires resistance by practice, and nothing acquired is heritable. Restating natural selection without saying what is WRONG in the student’s version answers a different question.

1 mark — individuals do not acquire resistance by exposure — resistance is set by inherited alleles 1 mark — acquired characteristics are not inherited — only alleles pass to the young

A wheat breeder wants a variety that is both high-yielding and resistant to a fungal disease.
(i) Describe how selective breeding could produce this variety. [2]
(ii) State one way this process differs from the natural selection described in part (a). [1]

Model Answer — 5(c)

cross plants showing the highest yield with plants showing the best fungal resistance, then select the offspring that show both features [1]

breed from those selected offspring and repeat over many generations [1]

in selective breeding humans choose which plants breed, for usefulness to us; in part (a) the environment (warfarin) determined which rats survived to breed [1]

⚠ If you missed marks here: The two process marks are choose-the-parents and repeat-for-generations — a single cross does not fix two features in a variety. For the difference mark, name the selecting agent (humans vs the environment); “one is faster” or “one is artificial” without saying who selects is too thin at Challenge level.

1 mark — cross high-yield with resistant plants, select offspring showing both features 1 mark — breed from the selected offspring, repeating over many generations 1 mark — difference: humans choose the parents for usefulness; in (a) the environment selected

Self-Assessment

Section A is scored automatically after you submit. Tick your Section B marks, then click Calculate Grade.

Your Score

Out Of

Percentage