HookThe grass that learned to live on poison
On the poisoned spoil heaps of old lead and zinc mines in North Wales — Trelogan in Flintshire is the classic site, studied intensively since the 1960s — a common pasture grass, Agrostis capillaris, grows where almost nothing else can. The soil is laced with heavy metals that kill ordinary plants, yet within a few metres of the mine boundary you cross from bare, toxic ground into a green sward of tolerant grass. The tolerant plants carry alleles that let them survive the metals; the pasture plants a stone's throw away do not. The grass is wind-pollinated, so pollen blows freely across that boundary in both directions — and still the two populations stay sharply different, because on the spoil only tolerant seedlings survive and off it the tolerant ones are outcompeted. Most tellingly, the mine plants have begun to flower at a slightly different time from their neighbours, which cuts down the interbreeding between them. That is natural selection, a shift in allele frequency, and the first crack of reproductive isolation — evolution caught in the act on a Welsh hillside.
That single hillside contains the whole of this section. Inheritance decides which alleles a plant can pass on; the Hardy-Weinberg principle turns those alleles into population frequencies; natural selection and reproductive isolation change the frequencies and, in time, split one species into two; and the grass itself is a population in an ecosystem, its numbers and distribution set by the harsh abiotic conditions and measurable by the sampling methods you meet in the required practical. It is also the most calculation-heavy part of the specification — genetic ratios, the chi-squared test, Hardy-Weinberg and mark-release-recapture all live here — so the marks reward clean, shown working more than almost anywhere else in Biology.
ModelMonohybrid and dihybrid crosses
Genetics begins with vocabulary the mark scheme is ruthless about. A gene is a length of DNA coding for a polypeptide; an allele is one version of that gene; the genotype is the alleles an organism carries and the phenotype is the characteristic they produce. An organism is homozygous when its two alleles match and heterozygous when they differ, and a dominant allele is expressed even when only one copy is present, whereas a recessive allele is masked unless both copies are recessive.
A monohybrid cross follows one gene. Cross two heterozygotes — say \(Tt \times Tt\) for tall and dwarf tomato plants — and the offspring appear in a 3 : 1 ratio of tall to dwarf, because three of the four genotype combinations (\(TT\), \(Tt\), \(tT\)) carry a dominant allele. A dihybrid cross follows two genes at once, and this is where students throw away marks by rushing the gametes. A parent heterozygous for both genes makes four equally likely gamete types, and combining the gametes of both parents in a Punnett square gives the famous 9 : 3 : 3 : 1 ratio — provided the two genes sit on different chromosomes and assort independently.
Cross two tomato plants each heterozygous for height (T tall, dominant to t dwarf) and fruit shape (R round, dominant to r pear-shaped): \(TtRr \times TtRr\). Each parent makes four gamete types — \(TR\), \(Tr\), \(tR\), \(tr\) — so the Punnett square has sixteen boxes. Counting the phenotypes gives 9 tall-round : 3 tall-pear : 3 dwarf-round : 1 dwarf-pear. Scale that to a real brood of 160 seedlings and you expect \(90 : 30 : 30 : 10\). Write the gametes in rings and fill every box: examiners award the ratio only when the genotypes behind it are shown, and a bare '9:3:3:1' with no working scores almost nothing.
ModelCodominance, multiple alleles and sex linkage
Not every gene obeys simple dominance. With codominance both alleles are expressed in the heterozygote — neither is masked. Human ABO blood group is the standard example, and it adds a second twist, multiple alleles: three alleles (\(I^A\), \(I^B\), \(I^O\)) exist in the population even though any one person carries only two. \(I^A\) and \(I^B\) are codominant, so genotype \(I^A I^B\) gives blood group AB, while both are dominant to \(I^O\).
Sex linkage describes genes carried on the X chromosome. Because a male is \(XY\), he has only one allele for any X-linked gene and cannot be a carrier — he expresses whatever he inherits. That is why X-linked recessive conditions such as haemophilia and red-green colour blindness are far commoner in males: a female (\(XX\)) needs two recessive alleles to be affected but only one to be an unaffected carrier. When you write a sex-linked cross, put the allele as a superscript on the X, never as a bare letter, or the diagram loses clarity and marks.
A woman who carries haemophilia, \(X^H X^h\), has children with an unaffected man, \(X^H Y\). Her gametes are \(X^H\) and \(X^h\); his are \(X^H\) and \(Y\). The four equally likely offspring are \(X^H X^H\) (unaffected daughter), \(X^H X^h\) (carrier daughter), \(X^H Y\) (unaffected son) and \(X^h Y\) (affected son). So none of the daughters is affected but half of them carry the allele, and half of the sons are affected — a \(1 : 1 : 1 : 1\) genotype ratio that means one affected child in four, always a son. Stating which sex shows the phenotype, and why, is the marking point the examiner is hunting for.
MechanismLinkage and epistasis
Two genes on the same chromosome are linked and tend to be inherited together, because they travel in the same gamete unless crossing over in meiosis separates them. Autosomal linkage shows up as a large excess of the two parental phenotypes and only a few recombinants, distorting the 1 : 1 : 1 : 1 you would expect from a test cross of unlinked genes. The rarer the recombinants, the closer together on the chromosome the two genes lie.
Epistasis is when one gene locus affects the expression of another, collapsing the dihybrid ratio into a modified form. In Labrador retrievers, the gene E must carry at least one dominant allele for any coat pigment to be deposited at all; a dog that is \(ee\) is yellow whatever its black/chocolate gene B says. So a cross of two \(BbEe\) dogs gives 9 black : 3 chocolate : 4 yellow — the '4' being every \(ee\) dog, of either B genotype. Recognising a modified ratio (9 : 3 : 4, 9 : 7, 12 : 3 : 1) as the fingerprint of epistasis, and naming which locus is doing the masking, is the higher-mark skill.
Test-cross a plant heterozygous for two linked genes, arrangement \(AB/ab\), against a double recessive \(ab/ab\). With no linkage you would expect the four offspring classes in equal numbers, a 1 : 1 : 1 : 1 ratio. Instead a brood of 100 comes out 42 \(AB\) : 41 \(ab\) : 8 \(Ab\) : 9 \(aB\). The two parental types dominate; the recombinants \(Ab\) and \(aB\) number only \(8 + 9 = 17\). The recombination frequency is \(17/100 = 17\%\), evidence the genes are linked and roughly 17 map units apart. A clean 1 : 1 : 1 : 1 would have meant no linkage — so the departure from the expected ratio is itself the data.
DataThe chi-squared test
When your observed offspring numbers do not exactly match the predicted ratio, is the difference real or just chance? The chi-squared test decides. It is used on categorical data that you can compare against a theoretical ratio — precisely what a genetic cross gives you. State a null hypothesis, that there is no significant difference between observed and expected values, then calculate
\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]
where \(O\) is each observed count and \(E\) each expected count. Compare the result against a critical value read off at probability \(p = 0.05\) and the correct degrees of freedom — the number of categories minus one. If your \(\chi^2\) is smaller than the critical value, the difference is not significant and you accept the null hypothesis; if it is larger, you reject the null and something — linkage, epistasis or selection — is skewing the ratio away from expectation.
Take the 160 tomato seedlings, expected \(90 : 30 : 30 : 10\) but observed 95 : 28 : 27 : 10. Work term by term: \((95-90)^2/90 = 25/90 = 0.278\); \((28-30)^2/30 = 4/30 = 0.133\); \((27-30)^2/30 = 9/30 = 0.300\); \((10-10)^2/10 = 0\). Summing gives \(\chi^2 = 0.711\). There are four categories, so degrees of freedom \(= 4 - 1 = 3\), and the critical value at \(p = 0.05\) is 7.815. Because 0.711 is far smaller than 7.815, the difference is not significant: accept the null hypothesis and conclude the data are consistent with a 9 : 3 : 3 : 1 ratio. That final sentence — accept or reject, stated in words — earns as much credit as the arithmetic that precedes it.
ModelPopulations and the Hardy-Weinberg principle
Individuals have genotypes; populations have allele frequencies, and the Hardy-Weinberg principle is the mathematics that links the two. It predicts that allele and genotype frequencies stay constant from one generation to the next, provided five conditions hold: a large population, no mutation, no migration, no natural selection, and random mating. If \(p\) is the frequency of the dominant allele and \(q\) the frequency of the recessive, then
\[ p + q = 1 \qquad \text{and} \qquad p^2 + 2pq + q^2 = 1 \]
Here \(p^2\) is the frequency of homozygous dominant individuals, \(q^2\) the homozygous recessive, and \(2pq\) the heterozygotes. The power of the equations is that the one phenotype you can usually count directly — the homozygous recessives, \(q^2\) — unlocks every other frequency in the population, including the carriers you cannot see. Because the principle assumes no evolution, any real deviation between generations is a signal that one of the five conditions is broken.
Cystic fibrosis is a recessive condition affecting roughly 1 in 2500 people in the UK. So \(q^2 = 1/2500 = 0.0004\), and \(q = \sqrt{0.0004} = 0.02\). Then \(p = 1 - q = 0.98\). The carrier frequency is \(2pq = 2 \times 0.98 \times 0.02 = 0.0392\) — about 3.9%, or roughly 1 in 25 people. That is the striking payoff: a disease seen in only 1 in 2500 has its allele hidden, unexpressed, in one person in twenty-five. Always start a Hardy-Weinberg calculation from \(q^2\) (the frequency you can measure), take the square root for \(q\), and only then work outwards to \(p\) and the heterozygotes.
MechanismNatural selection, drift and speciation
Natural selection acts on variation that already exists: individuals whose alleles suit the environment survive and reproduce more, so those alleles rise in frequency. It comes in three forms. Directional selection favours one extreme and shifts the mean — the engine behind antibiotic resistance, where the few resistant bacteria survive a course of treatment and repopulate. Stabilising selection favours the middle and squeezes out extremes, as with human birth mass, where historically the very small and the very large babies had lower survival. Disruptive selection favours both extremes over the middle, and is the type most likely to split a population in two.
In small populations, chance alone changes allele frequencies through genetic drift: an allele can be lost or fixed regardless of whether it is useful, which matters far more in a tiny founder population than in a large one. Speciation is the formation of a new species, and it happens when two populations become reproductively isolated so their gene pools stop mixing and their allele frequencies diverge until they can no longer interbreed to produce fertile offspring. Allopatric speciation follows a geographical barrier; sympatric speciation happens without one, through isolation such as different flowering or mating times, behavioural differences, or, in plants, polyploidy. The mine-spoil grasses of the introduction are sympatric isolation caught in its earliest stage.
CasePopulations in ecosystems and estimating their size
A population is all the individuals of one species in a habitat; the carrying capacity is the largest population the environment can sustain, set by limiting factors such as food, space, light, water and the build-up of waste. Numbers are held in check by competition — intraspecific (within a species, the main density-dependent brake on population size) and interspecific (between species, which can drive one out altogether by competitive exclusion) — and by predator-prey cycles, in which predator numbers lag behind and track those of the prey.
Succession is the directional change in a community over time. Primary succession begins on bare, lifeless ground — new rock, blown sand — colonised by pioneer species such as lichens that tolerate the harsh abiotic conditions. Each stage, or sere, changes the environment (building soil, holding water, adding shade), making it less hostile and letting later species outcompete the pioneers, until a stable climax community forms. Conservation often means deliberately halting succession — grazing or mowing a chalk grassland to stop it turning to scrub — producing a managed plagioclimax. To count a population you sample: non-motile or slow organisms with quadrats placed at random coordinates (recording frequency or percentage cover and scaling up), and motile animals by mark-release-recapture.
To estimate the woodlice under a woodland log pile, you collect 60, mark each harmlessly on its underside, and release them. Three days later you collect a second sample of 50 woodlice, of which 15 carry a mark. The population estimate is
\[ N = \frac{n_1 \times n_2}{n_3} = \frac{60 \times 50}{15} = 200 \]
where \(n_1\) is the number first marked, \(n_2\) the size of the second sample, and \(n_3\) the marked individuals recaught. The estimate is only valid if the assumptions hold: no births, deaths, immigration or emigration between samples; the marks do not rub off, harm the animal or make it easier for predators to spot; and the marked individuals mix randomly back into the population and have enough time to do so. Naming those assumptions is where the evaluation marks sit.
DataRequired practical 12 — a factor and a species' distribution
Required practical 12 asks you to investigate how a named environmental factor affects the distribution of a species — and the design choice that decides the marks is random versus systematic sampling. When you simply want the abundance of a species across a uniform area, you sample at random, generating coordinates from a random-number source so no unconscious bias creeps in. But this practical is about distribution along a gradient — light from open field into woodland, exposure up a rocky shore, trampling out from a footpath — so you use a belt transect: a line laid along the gradient with quadrats placed at fixed intervals. That is systematic, not random, and choosing it correctly is itself a mark.
At each quadrat you record the dependent variable — the species' abundance as percentage cover or frequency — and measure the independent variable, the environmental factor, with the right instrument: a light meter for light intensity, a soil-moisture probe, a pH meter. Control variables such as time of day, quadrat size and the same observer judging cover keep it fair. Reliability comes from repeating the transect in several places and taking means; a common source of error is the subjectivity of estimating percentage cover by eye, cut down by using a gridded point quadrat. A correlation between factor and abundance can then be tested with Spearman's rank correlation coefficient — but a correlation is not proof of cause, since an unmeasured factor may co-vary with the one you chose. Saying so is the classic AO3 evaluation point.
Suppose the first three quadrats along a transect running from open ground into woodland give bramble percentage covers of 5%, 20% and 45% as the canopy closes. Recording cover at ten points along the belt and pairing each with a light-meter reading turns a vague impression into data: as light intensity falls from about 1800 to 300 lux, bramble cover climbs from roughly 5% to 60%. A Spearman's rank test on the ten paired readings then tells you whether that negative correlation is statistically significant. The honest conclusion names light as correlated with — not proven to cause — the change in bramble distribution, because soil depth or moisture may change along the same transect.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
AQA rewards a full genetic diagram, never just the answer: parental phenotypes, parental genotypes, the gametes ringed, the offspring genotypes, then the phenotypic ratio. Drop the gametes and you cap your own marks. On the chi-squared test, state the null hypothesis in words, show every \((O-E)^2/E\) term, use degrees of freedom of one less than the number of categories, compare with the critical value at \(p = 0.05\), and finish with a sentence that accepts or rejects the null — that final interpretation is where candidates habitually drop the mark.
For Hardy-Weinberg, quote the assumptions when asked and always begin from \(q^2\), the only frequency you can observe directly. In ecology, the two AO3 favourites are the assumptions behind mark-release-recapture and the random-versus-systematic sampling decision — know when a belt transect is correct and why random sampling would be wrong for a gradient. Much of the maths here is Level 3 (the chi-squared, Hardy-Weinberg and mark-recapture calculations together satisfy a large slice of the 10% quantitative requirement), so show working line by line and carry units throughout. This is prime synoptic material for Paper 2 and the Paper 3 essay, which link it back to meiosis, genetic variation and biodiversity — expect to be asked to connect them rather than recall them in isolation.