IGCSE Biology Paper 4 (Theory / Extended)

Topic 16: Reproduction -- Challenge Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0610

Instructions

Answer all questions in the spaces provided.
Show all working for calculations.
Use appropriate scientific terminology.
Your answers will be automatically graded when you submit.

Question Navigation

This paper covers the whole of Topic 16. Like a real Cambridge paper it ranges across every sub-topic — 16.1 asexual reproduction, 16.2 sexual reproduction, gametes and fertilisation, 16.3 sexual reproduction in plants, 16.4 sexual reproduction in humans, 16.5 sexual hormones and the menstrual cycle, and 16.6 sexually transmitted infections — and it mixes them inside single questions. All three Topic 16 papers do; they differ in the angle they come at it from, not in what they cover.

Question 1 — Two Bloodstreams That Must Never Meet

Total: 12 marks

Fig. 1.1 shows a fetus inside the uterus, drawn in section. Fig. 1.2 is an enlarged view of a small part of structure P, where a finger of fetal tissue is pushed into a pool of the mother’s blood. Nothing has been named for you. Work from position, shape and the direction of the arrows.

(a) [3]

Name the structures labelled P and S on Fig. 1.1, and name the ball of cells that implanted in the lining of the uterus at the start of this pregnancy.

Model Answer — 1(a)

P — the placenta [1]

S — the amniotic fluid (R is the amniotic sac, the membrane holding it in; S is the liquid itself) [1]

the ball of cells that implanted is the embryo [1]

⚠ If you missed marks here: The one that catches almost everybody is writing zygote for the third mark. A zygote is a single cell — the cell made the instant the two nuclei fuse. By the time it reaches the uterus it has divided into a ball of cells, and a ball of cells is an embryo. The other trap is swapping R and S: the sac is the membrane, the fluid is what the membrane contains, and the fluid supports the fetus and cushions it against bumps — it does not feed it.

1 mark — P: placenta 1 mark — S: amniotic fluid 1 mark — embryo

(b) [3]

In Fig. 1.2, V is a space filled with the mother’s blood and W is a fetal capillary. Name one substance that travels in the direction of the green arrows and is needed by fetal cells for aerobic respiration; name the excretory product of the breakdown of excess amino acids that travels in the direction of the orange arrows; and name the process by which both of them cross.

Model Answer — 1(b)

oxygen — the green arrows run from V into the fetal tissue, so they carry substances from the mother’s blood to the fetus (accept glucose, amino acids, water, mineral ions, some of the mother’s antibodies) [1]

urea — the orange arrows run outwards, from the fetal tissue into the mother’s blood; urea is the waste made when excess amino acids are broken down (carbon dioxide also travels this way) [1]

both cross by diffusion, down a concentration gradient, through the thin barrier between the two bloodstreams [1]

⚠ If you missed marks here: You did not need to be told which way the arrows point — V is labelled as the mother’s blood, so arrows heading into the fetal finger of tissue must be carrying things to the fetus. The vocabulary trap is writing that the placenta “gives” or “supplies” oxygen, or that the umbilical cord “feeds the baby”. Nothing is handing anything over: there is more oxygen in the mother’s blood than in the fetal blood, so oxygen diffuses down its concentration gradient, and that phrase is the mark.

1 mark — oxygen (or glucose), moving from the mother to the fetus 1 mark — urea, moving from the fetus to the mother 1 mark — diffusion, down a concentration gradient

Calcium ions are at a higher concentration in the fetal blood than in the mother’s blood, yet calcium ions keep moving from the mother’s blood into the fetal blood throughout the pregnancy. Name the process responsible and explain how it is possible.

Model Answer — 1(c)

active transport [1]

the ions are moved against the concentration gradient, from a lower to a higher concentration, by protein carriers in the cell membranes of the placenta, using energy released by respiration in those cells [1]

(this is why the placenta contains cells packed with mitochondria — a purely diffusive organ would not need them)

⚠ If you missed marks here: If you wrote diffusion, re-read the direction: diffusion can only ever move a substance down a gradient, so it physically cannot produce this result. And “because the fetus needs calcium” is not a mechanism — needing something does not make it move. The two words that earn the second mark are protein carriers and energy from respiration.

1 mark — active transport 1 mark — against the concentration gradient, by protein carriers, using energy from respiration

(d) [2]

Explain why it is important that the blood of the mother and the blood of the fetus never mix.

Model Answer — 1(d)

the mother’s blood is at a much higher pressure, and that pressure would burst the very thin fetal capillaries [1]

the fetus is genetically different from the mother, so her immune system could attack the fetal cells — and if their blood groups differ the red blood cells would clump together [1]

(accept also: a pathogen in the mother’s blood would then reach the fetus directly and in quantity)

⚠ If you missed marks here: A great many people picture the placenta as the place where the two bloodstreams join, and then cannot say why the question is even being asked. They never join. They are brought within a fraction of a millimetre of each other across an enormous surface area, and everything crosses that thin barrier. If all you wrote was “the blood types might be different”, that is one idea out of two — the pressure point is the one people forget.

1 mark — the mother’s higher blood pressure would burst the fetal capillaries 1 mark — the mother’s immune system could attack the genetically different fetal cells / blood groups may differ

(e) [2]

Even though the two bloodstreams stay separate, some pathogens and some toxins still reach the fetus and can damage it. Suggest why.

Model Answer — 1(e)

the barrier between the two bloodstreams is an extremely thin, partially permeable membrane, not a filter — it lets small molecules through and cannot tell a useful one from a harmful one [1]

so small molecules such as nicotine or alcohol, and small pathogens such as viruses, cross by exactly the same route as oxygen and glucose and enter the fetal blood, where they can interfere with the development of the fetus [1]

⚠ If you missed marks here: The commonest wrong answer treats the placenta as a sieve that sorts good things from bad. It has no such ability. A membrane can only sort by size and solubility, so anything small enough to be a useful nutrient is small enough to be a poison. That is the whole answer, and it is also why this fact is on the syllabus at all.

1 mark — the barrier is a thin partially permeable membrane, not a filter 1 mark — small molecules and small pathogens such as viruses cross it by the same route as nutrients

Question 2 — A Giant and a Swimmer

Total: 12 marks

Fig. 2.1 shows two human cells drawn to the same scale. Fig. 2.2 shows one of those cells at a much higher magnification. The scale bar applies to Fig. 2.1 only.

A student printed Fig. 2.1 and measured it with a ruler. On her printed copy the 50 µm scale bar measured 20 mm, and the large cell measured 48 mm across, not counting structure A.

(a)(i) [2]

Calculate the actual diameter of the large cell in Fig. 2.1, in µm. Show your working.

Model Answer — 2(a)(i)

20 mm on the print represents 50 µm, so 1 mm represents 50 ÷ 20 = 2.5 µm [1]

actual diameter = 48 × 2.5 = 120 µm

actual diameter = 120 µm [1]

⚠ If you missed marks here: Do not divide 48 by 20 and stop — that gives you 2.4, which is how many scale bars wide the cell is, not a size. You then have to multiply by what one bar is worth. And an answer with no unit, or with mm instead of µm, scores nothing: an egg cell 120 mm across would be the size of a grapefruit.

1 mark — 1 mm on the print represents 2.5 µm (or equivalent working) 1 mark — actual diameter = 120 µm

(a)(ii) [1]

Calculate the magnification of the large cell on her printed copy of Fig. 2.1.

Model Answer — 2(a)(ii)

magnification = image size ÷ actual size = 48 000 µm ÷ 120 µm = ×400

image size 48 mm converted to 48 000 µm, giving a magnification of ×400 [1]

⚠ If you missed marks here: If you got 0.4 you divided 48 by 120 without converting the units — the two numbers must be in the same unit before you divide, and 1 mm = 1000 µm. If you wrote ×400 µm, take the unit off: a magnification is a ratio of two lengths, so the units cancel and it has none.

1 mark — magnification = ×400 (48 000 µm ÷ 120 µm)

(b) [3]

Structures E, G and H in Fig. 2.2 are adaptive features of the cell shown. Name each one and explain how it helps that cell to carry out its function.

Model Answer — 2(b)

E — the acrosome: it contains enzymes that digest a way through the jelly coat, so the sperm can reach the membrane of the egg cell [1]

G — mitochondria (many of them): aerobic respiration in them releases the energy needed for the flagellum to beat [1]

H — the flagellum: it beats from side to side so the cell can swim along the oviduct to the egg cell [1]

⚠ If you missed marks here: Three specific words. It is a flagellum, never a tail — Cambridge writes flagellum in the syllabus and marks accordingly. Mitochondria do not “give” or “make” energy; they are where aerobic respiration releases it. And the acrosome does not drill, pierce or burst its way in — it holds enzymes, and enzymes digest.

1 mark — E: acrosome, enzymes digest a way through the jelly coat 1 mark — G: mitochondria, aerobic respiration releases energy for movement 1 mark — H: flagellum, beats so the cell can swim

Compare the male gamete and the female gamete. Give one difference in size, one difference in motility and one difference in numbers. Use Fig. 2.1 to support the difference in size.

Model Answer — 2(c)

size — the egg cell is far larger, about 120 µm across against a sperm head of about 5 µm, because the egg cell carries a large volume of cytoplasm containing energy stores; the sperm carries almost no cytoplasm [1]

motility — the sperm cell is motile, swimming by beating its flagellum; the egg cell cannot move itself and is carried along the oviduct [1]

numbers — many millions of sperm cells are released at one time, whereas usually only one egg cell is released, about once a month [1]

⚠ If you missed marks here: “The egg is bigger” on its own is half an answer — a comparison question wants both sides and, where a figure gives you numbers, the numbers. The idea people get backwards is motility: the egg cell does not swim towards the sperm. It is moved along the oviduct by the muscular wall and by cilia, which is why the sperm has to do all the travelling.

1 mark — size: egg cell about 120 µm, far larger than the sperm, with much more cytoplasm 1 mark — motility: sperm is motile using its flagellum, the egg cell is not 1 mark — numbers: millions of sperm, usually one egg cell

(d) [3]

The nuclei of both gametes are haploid. State what happens to structure A in Fig. 2.1 the instant one sperm nucleus fuses with the egg nucleus, explain why that change matters, and state the term used to describe the nucleus of the zygote that is formed.

Model Answer — 2(d)

A is the jelly coat, and it changes so that no further sperm can get through it [1]

this matters because only one sperm nucleus then fuses with the egg nucleus, so the zygote receives exactly one set of chromosomes from each gamete; if a second sperm nucleus entered, the zygote would have the wrong number of chromosomes and would not develop [1]

the nucleus of the zygote is diploid [1]

⚠ If you missed marks here: Haploid and diploid describe nuclei, not whole cells and certainly not half a cell — a sperm is a complete, living cell that happens to carry a haploid nucleus. And be precise about fertilisation: it is the fusion of the nuclei of two gametes. Writing that “the sperm joins the egg” describes the arrival, not the event Cambridge is naming.

1 mark — the jelly coat changes so no other sperm can enter 1 mark — so only one sperm nucleus fuses, giving the correct number of chromosomes 1 mark — the zygote nucleus is diploid

Question 3 — Two Crops, Two Ways of Moving Pollen

Total: 12 marks

Two crop plants, species J and species K, are grown side by side at a field station. You have not met either species. Everything you need is in Table 3.1 — work only from the data.

Table 3.1 — mean measurements taken from 200 flowers of each species
Feature measured	Species J	Species K
mean petal length / mm	14.0	1.5
petal colour	bright yellow	pale green
mean mass of nectar per flower / mg	3.8	0.0
position of the anthers	held inside the flower	hanging outside the flower on long filaments
shape and position of the stigma	small and sticky, inside the flower	large and feathery, hanging outside
mean number of pollen grains per flower	9 400	1 269 000
mean diameter of one pollen grain / µm	42	21
surface of the pollen grain	spiky and sticky	smooth and dry

(a) [4]

State which of the two species is wind-pollinated. Give three pieces of evidence from Table 3.1 and explain, in each case, how that feature suits wind pollination.

Model Answer — 3(a)

species K [1]

the anthers hang outside the flower on long filaments, so the pollen is fully exposed to moving air and is easily shaken or blown away [1]

the stigma is large and feathery and hangs outside, giving a very large surface area to catch pollen grains drifting past [1]

the pollen grains are small (21 µm), smooth and dry, so they are light, do not stick to each other and are carried a long way by the wind [1]

(any three of the above, or: small dull green petals and no nectar, because there is no insect that has to be attracted, so producing petals and nectar would be a waste of the plant’s resources)

⚠ If you missed marks here: Quoting the table is only half of what is asked. “The stigma is feathery” states a fact; the mark is for saying what being feathery achieves — a large surface area to catch airborne grains. Watch the pollen size too: the grain of the wind-pollinated species is the smaller one. Large, spiky, sticky grains are built to catch on the hairs of an insect, and that extra bulk would only make them fall out of the air faster.

1 mark — species K 1 mark — anthers hang outside on long filaments, exposed to moving air 1 mark — large feathery stigma outside the flower, large surface area to catch pollen 1 mark — small, smooth, dry pollen grains carried easily by the wind (or: dull small petals and no nectar)

(b) [2]

Explain why one flower of species K produces so many more pollen grains than one flower of species J.

Model Answer — 3(b)

wind pollination is random and undirected — the overwhelming majority of grains land on soil, water or the wrong species and are wasted, so an enormous number must be released for a few to reach a stigma of the same species [1]

an insect visiting species J carries pollen directly from one flower to another flower of the same species, so a far smaller number of grains achieves the same result [1]

⚠ If you missed marks here: “Because the wind blows it away” earns nothing on its own — that is what wind pollination is, not a reason. The reason is wastage: the wind has no aim. The second mark is the comparison, and comparison questions want both halves written down, so say what the insect does differently.

1 mark — wind transfer is random, so most grains are wasted and never reach a stigma 1 mark — an insect carries pollen directly to another flower of the same species, so fewer grains are needed

Calculate how many times more pollen grains one flower of species K produces than one flower of species J. Show your working.

Model Answer — 3(c)

1 269 000 ÷ 9 400

correct method: the larger number divided by the smaller number [1]

answer = 135 times more [1]

⚠ If you missed marks here: “How many times more” means divide, not subtract. If you wrote 1 259 600 you found the difference, which is a different quantity and answers a different question. Also keep the ratio the right way up: dividing the small by the large gives 0.0074, which would mean species K made fewer.

1 mark — correct method: 1 269 000 divided by 9 400 1 mark — answer = 135

(d) [4]

In one year the number of bees at the field station fell sharply. The yield of species J fell by 62% that year, while the yield of species K did not change.

(i) Explain these results. [2]
(ii) Species K can also transfer pollen from an anther to a stigma of the same flower. State the term for this, and give one disadvantage of it to a wild population of species K. [2]

Model Answer — 3(d)

(i) species J is insect-pollinated and so relies on pollinators; with fewer bees, far less pollen is transferred from anther to stigma, so fewer ovules are fertilised and fewer seeds and fruits form [1]

(i) species K is wind-pollinated and does not rely on pollinators at all, so the fall in bee numbers made no difference to it [1]

(ii) self-pollination [1]

(ii) the offspring show less variation, so if the environment changes — a new pathogen arrives, or a drought — the whole population is likely to be affected in the same way and few or none survive, meaning the population has a poor capacity to respond to change [1]

⚠ If you missed marks here: Two things to keep straight. Self-pollination is still sexual reproduction — gametes are made and nuclei fuse — so it does not produce clones, only offspring with less variation than cross-pollination gives. And keep the two words apart: fewer bees means less pollination, and that is why less fertilisation follows. Pollination is the transfer of pollen grains from anther to stigma; fertilisation is the fusion of nuclei that happens afterwards.

1 mark — species J relies on pollinators, so less pollen transferred and fewer ovules fertilised 1 mark — species K is wind-pollinated and does not rely on pollinators 1 mark — self-pollination 1 mark — less variation, so poor capacity to respond to a change in the environment

Question 4 — Ten Thousand Plants, One Parent

Total: 12 marks

Cassava is a tropical root crop. A new cassava plant can be grown by cutting a stem into short pieces and planting them: each piece grows into a complete plant. Cassava can also be grown from seed. A research station planted two plots of equal area on the same day.

Plot 1 — every plant grown from a stem cutting taken from a single high-yielding parent plant.
Plot 2 — every plant grown from seed collected from many different plants.

In year 3 a leaf-spot fungus that had never been recorded in the region before reached the field station.

Table 4.1 — mean yield of cassava roots / tonnes per hectare
Year	Plot 1 (grown from stem cuttings)	Plot 2 (grown from seed)
1	27.4	19.5
2	28.0	20.1
3	11.2	17.8
4	3.5	16.9

(a) [2]

Growing cassava from stem cuttings is asexual reproduction. State two features of asexual reproduction that show it is not sexual reproduction.

Model Answer — 4(a)

no gametes are involved and there is no fusion of nuclei [1]

there is only one parent, and the offspring are genetically identical to that parent and to one another [1]

⚠ If you missed marks here: The definition Cambridge marks contains the words no gametes and no fusion of nuclei. A surprising number of people write that asexual reproduction uses “only one gamete” — there are none at all. And write “genetically identical”, not “the same”: two cassava plants from the same parent may end up different heights because one got more light, but they are still genetically identical.

1 mark — no gametes and no fusion of nuclei 1 mark — one parent only, offspring genetically identical

(b) [2]

Use Table 4.1 to give two advantages to the grower of raising cassava from stem cuttings, as seen in years 1 and 2.

Model Answer — 4(b)

the yield is higher — 28.0 tonnes per hectare from cuttings against 20.1 from seed in year 2, about 8 tonnes per hectare more [1]

every plant is genetically identical to the chosen high-yielding parent, so the desirable characteristic is kept exactly, the crop is uniform in size and ripens together and can be harvested in one operation (accept: no flowering, no pollinator and no seed are needed, so a large crop can be raised quickly and predictably) [1]

⚠ If you missed marks here: When a question says “use Table 4.1”, quote a figure. “The yield is higher” may be given the mark; “28.0 against 20.1 tonnes per hectare” certainly is, and it costs you four seconds. Uniformity is the advantage growers care about most and the one students most often leave out.

1 mark — higher yield, quoting figures from Table 4.1 1 mark — genetically identical to the chosen parent, so a uniform crop keeping the desirable characteristic

Explain the results for years 3 and 4 in both plots.

Model Answer — 4(c)

every plant in plot 1 is genetically identical, so if one plant has no resistance to the new fungus then no plant in the plot has any [1]

the whole plot is therefore affected at once and the yield collapses, from 28.0 to 11.2 and then to 3.5 tonnes per hectare [1]

the plants in plot 2 grew from seed, so they are genetically different from one another; some are less badly affected than others, so the plot as a whole keeps producing and the yield falls only slightly, from 20.1 to 16.9 [1]

⚠ If you missed marks here: The answer to avoid is “the fungus attacked plot 1 harder”. The fungus did nothing different in the two plots; the plots were different. This is the single most important disadvantage of asexual propagation and it is worth being able to state in one sentence: identical plants share identical weaknesses, so one new disease can take the entire crop.

1 mark — plot 1 plants are genetically identical, so none has resistance if one has none 1 mark — the whole plot is affected and the yield collapses to 3.5 t/ha 1 mark — plot 2 plants are genetically different, some less affected, so the yield falls only slightly

(d) [2]

Calculate the percentage decrease in the yield of plot 1 between year 2 and year 4. Show your working.

Model Answer — 4(d)

decrease = 28.0 − 3.5 = 24.5 ; percentage decrease = (24.5 ÷ 28.0) × 100

correct method: divides the decrease by the year 2 value and multiplies by 100 [1]

answer = 87.5% [1]

⚠ If you missed marks here: A percentage change is always divided by the starting value, not the finishing one and not the bigger one out of habit. Dividing by 3.5 gives 700%, which should have set an alarm off — a decrease cannot exceed 100%.

1 mark — correct method: decrease divided by the year 2 value, times 100 1 mark — answer = 87.5%

(e) [3]

The nucleus of a cell in the parent cassava stem is diploid.

(i) State what is meant by a diploid nucleus.
(ii) State whether the nucleus of a cell in a plant grown from a cutting of that stem is haploid or diploid.
(iii) Explain why the plants in plot 1 are genetically identical to the parent plant.

Model Answer — 4(e)

(i) a nucleus containing two sets of chromosomes [1]

(ii) diploid [1]

(iii) the new plant is built by the division of existing cells of the one parent plant, with no gametes and no fusion of nuclei, so every new cell receives a copy of exactly the same genetic material [1]

⚠ If you missed marks here: Part (ii) is a trap for people answering on autopilot: nothing in asexual reproduction halves anything, so the cutting stays diploid from start to finish. In (i), “a full set of chromosomes” is too vague to mark — say two sets. And remember these words describe the nucleus, not the cell.

1 mark — diploid = a nucleus containing two sets of chromosomes 1 mark — the cutting-grown plant has diploid nuclei 1 mark — produced by division of existing cells of one parent, no gametes, no fusion of nuclei

Question 5 — Four Slides and a Sunny Windowsill

Total: 10 marks

When a pollen grain lands on a stigma it grows a pollen tube down through the style. Pollen will also grow a tube in a drop of sugar solution on a microscope slide, so the process can be watched in a school laboratory. A student investigated the effect of sucrose concentration on the growth of pollen tubes. His method is set out below, exactly as he wrote it.

The student’s method

I brushed some pollen off one anther of a lily flower from the school garden.
I put one large drop of sucrose solution on each of four slides: 0% on slide A, 5% on B, 10% on C and 20% on D.
Using the same paintbrush each time, without cleaning it, I added pollen to each drop in turn.
I left slide A on a sunny windowsill, slides B and C on the laboratory bench, and slide D on a shelf next to an open window.
After about an hour I looked at each slide under the microscope and measured the longest pollen tube I could find.
I did not have time to repeat the investigation.

Table 5.1 — the student’s results
Slide	Sucrose concentration / %	Length of the longest pollen tube found / µm
A	0	0
B	5	110
C	10	260
D	20	90

(a) [4]

Identify four faults in the student’s method. In each case state why the fault matters.

Model Answer — 5(a)

temperature was not controlled — a sunny windowsill, a bench and a shelf by an open window are at three different temperatures, and temperature affects the rate of growth, so a difference between slides may have been caused by temperature rather than by sucrose [1]

time was not controlled — “about an hour” is not a measured time, so the tubes on different slides may have had different lengths of time in which to grow [1]

the same paintbrush was reused without cleaning — sucrose solution and pollen were carried from one slide to the next, so the concentrations tested were not the ones he thought he was testing [1]

only the single longest tube was measured, and there were no repeats — one unusual tube decides the whole result, no mean can be calculated, and there is no way of knowing whether the result would happen again [1]

(accept also: the number of pollen grains added to each drop was not controlled; the drop sizes were not measured; pollen from only one flower was used)

⚠ If you missed marks here: Naming the fault is only half a mark’s worth of thinking. “He did not control temperature” needs the consequence attached: so the difference might be caused by temperature and not by sucrose at all. That sentence is the whole point of controlling a variable. The fault most people walk straight past is measuring only the longest tube — picking the biggest result on each slide is not sampling, it is choosing the answer you want.

1 mark — temperature not controlled, so differences may be caused by temperature not sucrose 1 mark — time not controlled, so slides had different times to grow 1 mark — same uncleaned brush, so solutions and pollen were carried between slides 1 mark — only the longest tube measured and no repeats, so no mean and no check on reliability

(b) [2]

The student concluded: “Pollen tubes grow best in a 10% sucrose solution.” State one thing his results do support, and one reason why this conclusion is not valid.

Model Answer — 5(b)

supported: pollen tubes did grow in sucrose solution but not in water alone, and of the four concentrations he tested the longest tube was found at 10% [1]

not valid: only four concentrations were tested and nothing between 10% and 20%, so the best concentration could lie anywhere in that gap; and because temperature and time were not controlled the differences between slides cannot safely be blamed on the sucrose at all [1]

⚠ If you missed marks here: There is a real difference between “the best of the four I tried” and “the best there is”, and examiners love it. He tested four points on a scale that runs from 0 to 20 and then announced the peak. Notice also that he has no evidence at all that 0% kills the pollen — slide A was the one on the sunny windowsill, so it may simply have dried out.

1 mark — supported: tubes grew in sucrose but not in water, longest at 10% of those tested 1 mark — not valid: too few concentrations / uncontrolled variables mean the cause is not established

Describe two changes that would make his results more reliable.

Model Answer — 5(c)

keep every slide at the same stated temperature, for example in a water bath at 25 °C, and leave every slide for exactly the same measured time, for example 60 minutes timed with a stopclock [1]

measure a fixed number of tubes on each slide, for example 20, using an eyepiece graticule, calculate a mean, and repeat the whole investigation at least three times (accept: use a clean brush for each solution; test more concentrations between 5% and 20%) [1]

⚠ If you missed marks here: “Be more accurate” and “do it properly” earn nothing. An improvement has to be a thing you could hand to another student as an instruction: a stated temperature, a stated time, a stated number of tubes, a stated number of repeats. Note too that repeating improves reliability while controlling variables improves validity — they are different problems and this method has both.

1 mark — control temperature at a stated value and use an exactly measured time 1 mark — measure many tubes per slide, calculate a mean, and repeat the whole investigation

(d) [2]

In a flower, the pollen tube grows down through the style. Describe what happens once it reaches the ovule.

Model Answer — 5(d)

the tube enters the ovule through a small opening, and the pollen nucleus travels down the tube into the ovule [1]

the pollen nucleus then fuses with a nucleus in the ovule — this is fertilisation — forming a zygote, which develops into the embryo of the seed [1]

⚠ If you missed marks here: The word that must appear is nucleus. Fertilisation is not “the pollen reaching the ovule” and it is certainly not “the pollen landing on the stigma” — that earlier event is pollination, and confusing the two is the most expensive mistake in this whole topic. Fertilisation is the moment two nuclei fuse.

1 mark — the tube enters the ovule and the pollen nucleus travels down it 1 mark — the pollen nucleus fuses with a nucleus in the ovule (fertilisation), forming a zygote

Question 6 — The Hormones That Hold a Pregnancy Together

Total: 12 marks

Fig. 6.1 shows the concentrations of three hormones measured in the blood of one woman from the start of a pregnancy until the birth at week 40.

(a) [4]

(i) Name the structure in the ovary that produces progesterone after ovulation.
(ii) Name the hormone that causes ovulation and the formation of that structure.
(iii) Name the part of the ovary that produces oestrogen during the first half of the menstrual cycle.
(iv) Name the organ that produces most of the oestrogen and progesterone from about week 12 of a pregnancy onwards.

Model Answer — 6(a)

(i) the corpus luteum (the structure left in the ovary after the egg cell has been released) [1]

(ii) LH (luteinising hormone) [1]

(iii) the developing follicle in the ovary [1]

(iv) the placenta [1]

⚠ If you missed marks here: Part (iv) is the one people never see coming. The placenta is usually filed away as an exchange organ, but it is also a hormone-producing organ, and that hand-over from ovary to placenta is exactly why a pregnancy can continue for months after the corpus luteum has gone. In (ii), FSH and LH are easy to swap: FSH makes a follicle grow, LH causes ovulation.

1 mark — corpus luteum 1 mark — LH 1 mark — the developing follicle in the ovary 1 mark — the placenta

(b) [2]

(i) Use Fig. 6.1 to state the concentration of progesterone in the blood at week 20.
(ii) FSH stays close to zero throughout the pregnancy. Suggest why.

Model Answer — 6(b)

(i) 75 arbitrary units [1]

(ii) the high concentrations of oestrogen and progesterone inhibit the release of FSH from the pituitary gland, so no further follicles develop, no further egg cells mature and no second pregnancy can begin [1]

⚠ If you missed marks here: For (i), read across from the curve to the axis rather than estimating — the point at week 20 sits exactly on a gridline. For (ii), “because she is already pregnant” is a description, not an explanation. The mark is for naming the inhibition: two hormones already in the blood are switching a third one off, which is what a control system looks like.

1 mark — 75 arbitrary units 1 mark — high oestrogen and progesterone inhibit the release of FSH, so no further follicles develop

State two roles of progesterone during pregnancy, then describe what Fig. 6.1 shows happening to the concentration of progesterone at the very end of the pregnancy and explain why that change is important.

Model Answer — 6(c)

progesterone maintains the thick lining of the uterus, so the embryo stays implanted and the pregnancy continues [1]

progesterone (with oestrogen) inhibits FSH, so no further egg cells mature and the lining is not shed — menstruation does not happen during pregnancy [1]

the concentration rises to a peak at about week 38 (152 units) and then falls, to about 138 units by week 40; this fall allows the muscular wall of the uterus to contract, so labour and birth can begin [1]

⚠ If you missed marks here: “Progesterone helps the baby grow” is vague and earns nothing. Be specific: it maintains the uterus lining. And when a question says “describe what Fig. 6.1 shows”, quote the figures and the weeks — a description of a graph that contains no numbers has thrown away the only thing the graph was giving you.

1 mark — maintains the thick lining of the uterus so the pregnancy continues 1 mark — inhibits FSH, so no further egg cells mature and menstruation does not occur 1 mark — the concentration falls near the end (peak about 152 at week 38 down to about 138), allowing the uterus wall to contract

(d) [3]

In some mammals the corpus luteum can be removed surgically. In one such species, removing the corpus luteum at week 6 of pregnancy always ended the pregnancy, but removing it at week 14 had no effect at all. Explain these results.

Model Answer — 6(d)

at week 6 the corpus luteum is the main source of progesterone, so removing it makes the concentration of progesterone in the blood fall sharply [1]

without progesterone the lining of the uterus is not maintained and breaks down, so the embryo is lost and the pregnancy ends [1]

by week 14 the placenta has developed and is producing enough progesterone on its own, so removing the corpus luteum no longer changes the concentration and the pregnancy is unaffected [1]

⚠ If you missed marks here: You had never met this experiment, and you were not expected to have. Everything you needed was in part (a): the corpus luteum makes progesterone early on, the placenta takes over from about week 12. The question is simply asking you to notice that week 6 is before the hand-over and week 14 is after it. When an unfamiliar result comes with two times in it, the times are almost always the clue.

1 mark — at week 6 the corpus luteum is the main source of progesterone, so removing it makes progesterone fall 1 mark — the uterus lining is not maintained, breaks down, and the embryo is lost 1 mark — by week 14 the placenta produces enough progesterone, so removal has no effect

Question 7 — Bringing the Numbers Down

Total: 10 marks

HIV is a pathogen that causes a sexually transmitted infection. Table 7.1 shows the number of people newly infected with HIV in one country in 2010 and in 2022, sorted by the route through which the virus reached them. A national public-health programme ran throughout that period.

Table 7.1 — number of people newly infected with HIV in one year
Route of transmission	2010	2022
sexual contact	14 000	6 850
mother to child, during pregnancy, at birth or through breast milk	3 300	210
sharing needles used to inject drugs	2 100	640
transfusion of donated blood	600	0
total	20 000	7 700

(a) [2]

Calculate the percentage decrease in the total number of new infections between 2010 and 2022. Give your answer to one decimal place and show your working.

Model Answer — 7(a)

decrease = 20 000 − 7 700 = 12 300 ; percentage decrease = (12 300 ÷ 20 000) × 100

correct method: the decrease divided by the 2010 figure, multiplied by 100 [1]

answer = 61.5% [1]

⚠ If you missed marks here: If you calculated 38.5% you worked out what fraction of the 2010 figure remains, not how much it fell by. Read the word in the question: decrease. And a percentage change always divides by the earlier value.

1 mark — correct method: 12 300 divided by 20 000, times 100 1 mark — answer = 61.5%

(b) [2]

Apart from transfusion of donated blood, which route of transmission fell by the greatest percentage between 2010 and 2022? Support your answer with a calculation.

Model Answer — 7(b)

mother to child [1]

(3 300 − 210) ÷ 3 300 × 100 = 93.6%

a fall of 93.6%, compared with 51.1% for sexual contact and 69.5% for shared needles [1]

⚠ If you missed marks here: If you answered “sexual contact” you compared the raw numbers: that route did fall by 7 150 people, easily the largest drop, but as a percentage of where it started it is the smallest of the three. Whenever a question puts the word percentage in bold, the biggest number in the table is usually the wrong answer.

1 mark — mother to child 1 mark — supported by a calculation giving about 93.6%

HIV is the pathogen; AIDS is the condition it may lead to. Explain how an HIV infection can eventually leave a person unable to fight off other pathogens.

Model Answer — 7(c)

HIV infects and destroys lymphocytes — a type of white blood cell — so their number in the blood falls over a period of years [1]

lymphocytes are the cells that produce antibodies, so the person makes far fewer antibodies and the immune system is progressively weakened [1]

the person can then no longer destroy pathogens that a healthy person would deal with easily, and suffers a series of infections; that collection of illnesses is AIDS [1]

(note: HIV is a virus, so antibiotics have no effect on it whatsoever)

⚠ If you missed marks here: HIV and AIDS are not two names for the same thing. HIV is the virus; AIDS is the condition that develops years later once enough lymphocytes have been destroyed. Saying “HIV weakens the immune system” is true but is not an explanation — the mark is for saying which cells it destroys and what those cells do. Watch out too for the belief that antibiotics can treat HIV; antibiotics kill bacteria and do nothing at all to a virus.

1 mark — HIV infects and destroys lymphocytes, so their number falls 1 mark — lymphocytes produce antibodies, so fewer antibodies are made and immunity is weakened 1 mark — the person suffers infections a healthy person would resist; this condition is AIDS

(d) [3]

Suggest three measures the national programme could have used to control the spread of HIV, and link each measure to the figures in Table 7.1.

Model Answer — 7(d)

screening every unit of donated blood for HIV before it is used, so infected blood is never transfused — which is why that route fell from 600 to 0 [1]

testing pregnant women and treating those who are infected with antiretroviral drugs, and advising on how the baby is fed, so far fewer babies are infected across the placenta, during birth or through breast milk — that route fell by more than 90%, from 3 300 to 210 [1]

education, widespread testing, contact tracing and treatment of infected people, and supplying clean needles so that needles are not shared — which is why the sexual-contact route roughly halved and the shared-needle route fell from 2 100 to 640 [1]

⚠ If you missed marks here: The question asked you to link each measure to the figures, so a list of measures with no reference to the table cannot score full marks. The one row that gives itself away is the transfusion route falling to exactly zero — a number that clean only happens when every unit is checked before use. And note that HIV is not spread by sharing a cup, a toilet seat or an insect bite; the routes in the table are the routes.

1 mark — screening donated blood, linked to the transfusion route falling to 0 1 mark — testing and treating pregnant women, linked to mother-to-child cases falling from 3 300 to 210 1 mark — education, testing, contact tracing, treatment and not sharing needles, linked to the other two routes falling

Self-Assessment

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