Topic 18 has almost no facts to forget. It has one sentence structure to get right, and nearly every mark in the topic is won or lost on it. The moment you write that an organism changed itself — that the bacteria toughened up, that the cactus grew spines because the desert is dry, that a species improved — the mark scheme stops giving you anything, however fluent the rest of the paragraph is. The cure is always the same two moves: make the population the subject of the sentence instead of the individual, and put the variation before the selection. Antibiotic resistance is the centrepiece here, and it is the case you already half know — Topic 15 told you that the antibiotic does not create resistance; this is the topic where you finally get to say why. Twelve traps, six data-led walkthroughs, six lookalike pairs, a concept map built on three frameworks, six badly-scoring answers and ten full practice questions below.
Twelve traps, spread across 18.1, 18.2, 18.3 and 18.4. Six of them are the same fault wearing six different costumes — writing as though the organism did the changing. Learn to hear it.
Six challenge-level questions with real data, worked in the order you should actually think about them. Try each part yourself before you reveal the next step — the reveals get harder, not easier.
(a) Describe the change shown in Fig. 3.1 between 1998 and 2026. [2] (b) Calculate the mean increase in the percentage of resistant samples per year between 2010 and 2018. Show your working. [2] (c) The value for 1998 is 2 % and not 0 %. Explain why this is important. [2] (d) Explain, in terms of natural selection, why the percentage of resistant samples increased. [4] (e) The hospital reduced its use of this antibiotic in 2023. Suggest what the graph might show for the years after 2026, and explain your suggestion. [2]
A “describe” mark is almost never given for “it goes up”. Read the two endpoints off the axis: 2 % in 1998 rising to 63 % in 2026, an increase of 61 percentage points. Then say what the shape does: the curve is not a straight line — it is shallow at first, steepest in the middle (roughly 2010–2018) and it begins to level off after 2022. Two marks, and the second one lives entirely in that sentence about the shape.
In 2010 the value is 21 %; in 2018 it is 47 %. The change is 47 − 21 = 26 percentage points, over 2018 − 2010 = 8 years. So the mean increase is 26 ÷ 8 = 3.25 % per year. Write the subtraction and the division down: on a Paper 4, a correct answer with no working gets the marks, but a wrong final number with correct working usually still collects the method mark, and a naked wrong number gets nothing. Give the unit: % per year.
The starting value is deliberate. It tells you that resistant bacteria were already present in 1998, before the heavy use of this antibiotic. So the resistance was not produced by the antibiotic; it had already arisen by random mutation, producing a new allele. The antibiotic did not cause the change — it acted as the selection pressure that determined which bacteria survived to reproduce. If the graph had started at zero, you could not have said any of that, which is exactly why it does not.
1. There was variation in the bacterial population: a few individuals carried a resistance allele produced by a random mutation, often on a plasmid. 2. Bacteria reproduce very rapidly, producing far more offspring than can survive, and they compete for nutrients and space. 3. When the antibiotic is used, the bacteria without the allele are killed; those with it are better adapted to this environment and are more likely to survive and reproduce. 4. The survivors pass the allele to their offspring, so the proportion of the population that is resistant increases with every generation. Nothing in those four sentences has a bacterium changing itself.
If the antibiotic is used far less, the resistant bacteria lose their advantage. Producing the extra protein that gives resistance costs resources, so where there is no antibiotic the non-resistant bacteria may grow and reproduce faster and out-compete them. The sensible prediction is therefore that the curve levels off and then falls slowly — slowly, because the resistance allele does not disappear; it is simply no longer favoured. Saying “it will drop back to 2 % immediately” misses that. This part is really a test of Trap 10: being better adapted is always relative to this environment.
Fig. 2.1 shows one population of bacteria in a patient at three times. Between A and B an antibiotic was given. Between B and C several hours passed and no further antibiotic was given.
(a) State the number of type Q cells in A and in C. [1] (b) Suggest what type P and type Q represent. [2] (c) Explain what happened between A and B. [2] (d) Explain what happened between B and C, and why the population in C is different from the population in A. [3] (e) A student says the antibiotic turned type P bacteria into type Q bacteria. Explain why this is wrong. [2]
Count them. Panel A has 30 cells: 28 type P and 2 type Q. Panel B has the same 2 type Q, plus 28 that are drawn crossed out — dead. Panel C has 30 cells and all of them are type Q. That counting is the answer to almost every part below, and it is worth doing before you write a single word of explanation, because the numbers tell you that the two type Q cells in A are the ancestors of everything in C.
Type P cells are killed when the antibiotic arrives, so type P are the bacteria that are not resistant (sensitive to this antibiotic). Type Q cells survive, so type Q are the bacteria that carry an allele giving resistance to it — an allele produced by a random mutation, and in bacteria often carried on a plasmid. Notice you were not told this. You worked it out from which cells were still alive in B, which is exactly the kind of reading a challenge paper is testing.
A → B: the antibiotic kills the 28 sensitive bacteria, leaving only the 2 that already carried the resistance allele. That is the selecting step, and it is the only thing the antibiotic does. B → C: the 2 survivors reproduce. Bacteria divide roughly every 20 minutes, so 2 cells become 4, 8, 16, 30 within a couple of hours — and because they divide by mitosis-like division they produce genetically identical offspring, all carrying the resistance allele. The population in C is therefore the same size as in A but is now entirely resistant: the individuals were replaced, and the proportion carrying the allele went from 2 in 30 to 30 in 30.
Follow any individual cell through the three panels and it never changes. The type Q cells in A are already type Q, before the antibiotic arrives — that is the variation, and it was produced by a random mutation, not by the drug. The antibiotic cannot alter the DNA of a living bacterium into a resistant form; all it can do is kill the ones that lack the allele. What changed is the composition of the population, not any individual in it. If you can say that sentence about this diagram, you can say it about every natural selection question on the paper.
(a) State which chart shows continuous variation and which shows discontinuous variation, and give one piece of evidence from the charts for each. [3] (b) Using Chart A, calculate the percentage of the 64 students who were 165 cm or taller. [2] (c) Using Chart B, calculate how many of the 1000 people surveyed were blood group B. [1] (d) Explain why height varies in the way shown in Chart A but blood group does not. [3] (e) Both charts show variation within a single species. Explain why variation of either kind matters to a population living in a changing environment. [2]
Chart A shows continuous variation: the bars touch and cover a continuous range of heights from 145 to 180 cm, with every value in between possible — a range of phenotypes between two extremes. Chart B shows discontinuous variation: there are only four categories with no intermediates — there is no blood group between A and B, and the bars are separated for exactly that reason. Say “the bars touch” or “there are no values in between”; that is the evidence mark, and it is the part most answers leave out.
For (b), the classes at 165 cm and above are 165–170 (14 students), 170–175 (7) and 175–180 (3). That is 14 + 7 + 3 = 24 students. As a percentage of 64: 24 ÷ 64 × 100 = 37.5 %. For (c), Chart B is already in percentages, so 15 % of 1000 = 150 people. Two easy marks, and the only way to lose them is to mis-read which bars the boundary includes — check the axis labels before you add.
Height is continuous because it is caused by genes and the environment together: many genes contribute, and on top of that diet, health and other environmental factors move an individual within the range their alleles allow. That combination of many small influences is what produces a smooth spread with most people near the middle. ABO blood group is discontinuous because it is determined by genes only — and by a single gene, so the phenotype falls into a limited number of categories with no intermediates. No amount of diet, exercise or illness moves anybody from group O to group A.
Natural selection cannot act on a population in which every individual is the same — there would be nothing to select between. Because there is variation, some individuals will by chance be better adapted when the environment changes; those are more likely to survive and reproduce and to pass their alleles on, so the population can become better suited to the new conditions over many generations. A population with very little variation has no such individuals available, and a single new disease or a change in climate can remove all of it. That is one sentence of 18.1 and one of 18.3, and challenge papers join them constantly.
(a) Describe the relationship between the two sets of results in Table 7.1. [2] (b) Calculate the percentage decrease in the proportion of mosquitoes killed between year 1 and year 9. [2] (c) Explain, in terms of natural selection, why the percentage killed fell. [4] (d) The village council proposes to double the dose of the same insecticide. Suggest why this may not work in the long term. [2] (e) Suggest one other method of reducing the number of cases of malaria in the village. [1]
The command word is “relationship”, so one column must be described in terms of the other. As the percentage killed falls from 99 % to 26 %, the number of malaria cases rises — but not smoothly. It falls at first (240 to 205 by year 3, while the spray is still killing 95 %) and then rises steeply to 470 by year 9, nearly double where it started. That non-monotonic shape is deliberate: the spraying did work at the beginning, and the marks are for noticing both halves.
The trap in this calculation is answering “73 %” because 99 − 26 = 73. That is the decrease in percentage points, not the percentage decrease. A percentage change is always change ÷ original value × 100: 73 ÷ 99 × 100 = 73.7 % (to 1 d.p.). Read the question wording every time — “percentage decrease” and “decrease in percentage” are two different sums and Cambridge uses both.
Variation: in the mosquito population there was already genetic variation — a few individuals carried an allele giving resistance to the insecticide, produced by a random mutation before spraying began. Many offspring and a struggle to survive: mosquitoes produce very large numbers of offspring, most of which do not survive. Selection: each spraying kills the mosquitoes without the allele; the resistant ones are better adapted to this environment and are more likely to survive and reproduce. Inheritance: they pass the allele to their offspring, so the proportion of the population that is resistant increases every generation — which is exactly what a falling percentage killed measures. Notice this is word for word Walkthrough 1 with “bacteria” and “antibiotic” swapped out. That is the point of the topic.
Doubling the dose will kill more mosquitoes at first, including some of the less resistant ones. But it does not remove the resistance allele from the population — it applies an even stronger selection pressure, so the only survivors are the most resistant individuals of all, and they are the ones that reproduce. The population becomes more resistant, faster. There are also costs the question will accept: harm to other species including the mosquitoes’ predators and useful insects, and the expense. The examiner-friendly alternative is to use a different insecticide, because a mosquito resistant to one is not automatically resistant to another, or to reduce reliance on insecticide altogether.
Fig. 4.1 shows a cross section through the leaf of a plant that grows on a sand dune, where water drains away rapidly and the wind is almost constant.
(a) Name the features labelled A, B and D. [3] (b) Explain how B and D each reduce the loss of water from this leaf. [4] (c) Suggest the function of the features labelled C. [2] (d) Define the term adaptive feature. [2] (e) Explain how a population of this species came to have these features. [4]
Start with the shape. The leaf is rolled into a cylinder, so one surface is now on the outside, exposed to the air, and the other is enclosed. A is the thick band along the outer surface: a thick waxy cuticle. B is a stoma sitting at the bottom of a pit rather than flush with the surface: a sunken stoma. D marks the rolled edge itself: the rolled leaf. C are the fine projections growing into the enclosed space: hairs. E points at the thickness of the leaf tissue. A plant like this — adapted to survive in very dry conditions — is a xerophyte.
B, the sunken stoma: it lies in a pit, so water vapour collects in the pit just outside the stoma. That makes the air there more humid, which reduces the concentration gradient of water vapour between the inside of the leaf and the air outside, so less water vapour diffuses out. D, the rolled leaf: rolling encloses the stomata inside a chamber, away from the moving air, so the humid air is not blown away and, again, the gradient stays small. This is Topic 8.3 in disguise: the factors that increase transpiration are wind, low humidity and heat, and every feature here attacks one of them.
The hairs trap a layer of still, humid air next to the stomata, which does the same job as the rolling by a different route: less air movement over the stomata means the water vapour is not carried away, so the gradient is smaller and less water is lost. Now the definition, which is worth having word-perfect: an adaptive feature is an inherited feature that helps an organism to survive and reproduce in its environment. Two marking points: inherited, and survive and reproduce. Answers that say “a feature that helps it live in its habitat” get one at best, because they leave out both.
Most of the marks in (e) are lost by candidates re-describing the features. The question asks about origin, so it wants the chain. There was variation in the ancestral population — caused by mutation, some plants had thicker cuticles or more sunken stomata than others. More offspring were produced than could survive, and there was competition for water. The plants with those features lost less water, so they were better adapted to the dry conditions and were more likely to survive and reproduce. They passed those alleles to their offspring, so over many generations the proportion of the population with the features increased. That is adaptation: a process, acting on a population, over generations.
(a) Calculate the percentage decrease in the number of birds between year 1 and year 3. [2] (b) Calculate the percentage increase in the mean beak depth of the birds present. [2] (c) Explain, using the data, why the mean beak depth of the surviving birds was greater than that of the original population. [4] (d) The chicks hatched in year 3 had a mean beak depth of 10.0 mm rather than 9.4 mm. Explain what this shows. [2] (e) A student writes that “the finches adapted their beaks to the harder seeds”. Explain why this is not an acceptable answer. [2]
(a) The number falls from 751 to 90, a change of 661. 661 ÷ 751 × 100 = 88.0 % — nearly nine birds in ten died. (b) Mean beak depth rises from 9.4 to 10.1 mm, a change of 0.7 mm. 0.7 ÷ 9.4 × 100 = 7.4 %. Hold on to the contrast between those two numbers: an enormous death rate produced only a small shift in the mean. That is what real selection looks like, and it is why it takes many generations to produce anything dramatic.
The third row is the row most candidates skip. The mean hardness of the seeds available rose from 4.5 to 6.8, because the drought killed the plants that produced the small soft seeds and left the tough ones. The footnote tells you a deeper beak can crack a harder seed. So the environment changed in a specific, measurable way, and that change is what determined which birds could feed. Always look for the row that describes the environment — that is the selection pressure, and quoting its figures is usually a mark.
Variation: beak depth shows continuous variation, so before the drought the population already contained birds with beaks deeper and shallower than 9.4 mm. Nothing new was created. Struggle for survival: the drought reduced the food supply and the remaining seeds were harder (4.5 → 6.8), so there was intense competition for food. Selection: birds with deeper beaks could crack the harder seeds, so they were better adapted to these conditions and were more likely to survive; birds with shallower beaks were more likely to starve, which is why 88 % of the population died. Result: the survivors were not a random sample — they were the deeper-beaked ones, so the mean of the survivors is higher at 10.1 mm.
The chicks hatched after the drought. They never had to crack a hard seed, and yet their mean beak depth is 10.0 mm, close to their parents’ 10.1 and well above the original 9.4. That is the evidence that beak depth is inherited — the surviving birds passed on the alleles for deeper beaks, so the change is carried into the next generation. Without this row, all you could show is that deep-beaked birds survived; with it, you can show that the population itself has changed. And it kills part (e) stone dead: no bird altered its own beak. The birds that could already feed survived and reproduced, and their offspring inherited the alleles. An individual cannot adapt; a population becomes adapted, over generations.
Six pairs that look like the same question and are not. In Topic 18 the pair is nearly always a thing against a process, or an individual against a population.
Click each node. The whole topic is three frameworks: where the variation comes from, what the environment then does to it, and what changes when a human does the selecting instead.
Six answers that read fluently and score badly. Four of them fail on the same fault. See whether you can hear it before you reveal it.
Ten Cambridge-style challenge questions. Every one draws on more than one sub-topic, because that is how this topic is examined. Write your answer first, then reveal the model answer and the examiner’s notes.
(a) Suggest what W, X, Y and Z represent. [4] (b) Explain how X and Y are adaptive features of a plant whose leaves float on the surface of a pond. [3] (c) State one feature of this leaf that would be a disadvantage to a plant growing on a dry sand dune, and explain why. [1]