Biology › Inheritance and population genetics › Hardy-Weinberg: what alleles do in a population when nothing acts on them

Hardy-Weinberg: what alleles do in a population when nothing acts on them

Scale a Punnett square up from two parents to a whole breeding population and the same probabilities become allele frequencies. The result is a null model: the state a population settles into when nothing is happening to it, and therefore the yardstick against which something happening is measured.

Before this Monohybrid inheritance · Meiosis and the origin of variation

Before you start

A dominant allele will spread through a population, because dominant alleles override recessive ones and so get passed on more. Dominance describes what happens inside one heterozygous individual, and it has nothing to do with how often an allele occurs. Polydactyly is caused by a dominant allele and is rare; the allele for blood group O is recessive and is the commonest of the three. Hardy-Weinberg makes the point exactly: with no selection, no mutation and random mating, allele frequencies do not change at all from one generation to the next, whatever the dominance relationships are.

What you should be able to do

Define a gene pool and calculate an allele frequency from genotype counts.
State the Hardy-Weinberg principle and the five conditions it assumes.
Use p + q = 1 and p² + 2pq + q² = 1 to move between allele and genotype frequencies.
Calculate carrier frequency from the frequency of a recessive phenotype, and turn frequencies into numbers of individuals.
Explain what a departure from Hardy-Weinberg proportions tells you about a population.

From individuals to a gene pool

Everything so far in this unit has followed alleles through one family. Population genetics asks a different question: not what this cross will produce, but what proportion of a whole population carries a particular allele, and whether that proportion is changing.

Population: A group of organisms of the same species living in the same place at the same time and able to interbreed.
Gene pool: All the alleles of all the genes present in a population at a given time.
Allele frequency: The proportion of all the alleles of a gene in a gene pool that are of one particular kind.

Allele frequencies are counted over alleles, not over individuals, and that is the first place answers go wrong. Every diploid individual carries two alleles of an autosomal gene, so a population of N individuals holds 2N alleles of it. A homozygote contributes two of the same allele and a heterozygote one of each.

Take a population of 100 people in which 36 are AA, 48 are Aa and 16 are aa. There are 200 alleles in total. The count of A is 2 × 36 + 48 = 120, and the count of a is 2 × 16 + 48 = 80. So the frequency of A is 120 ÷ 200 = 0.6 and the frequency of a is 80 ÷ 200 = 0.4. Notice that the recessive allele is not rare — it accounts for 40% of the gene pool — even though only 16% of people show the recessive phenotype.

Those two frequencies are conventionally called p and q. Because the gene in this example has only two alleles, every allele in the pool is one or the other, and so they must account for all of it.

p + q = 1p is the frequency of the dominant allele and q the frequency of the recessive one, for a gene with exactly two alleles. Find one and you have the other.

The second equation, and where it comes from

Now let that population breed at random. Random mating means gametes meet without reference to genotype, so a zygote is formed by drawing two alleles at random from the gene pool. The probability of drawing A is p and of drawing a is q, and the two draws are independent, which is a Punnett square with frequencies in place of alleles.

The heterozygotes appear in two cells rather than one, because Aa can be made either way round. That is the entire reason for the 2 in 2pq, and it is the term candidates most often drop.

p² + 2pq + q² = 1p² is the frequency of the homozygous dominant genotype, 2pq the heterozygous, q² the homozygous recessive. Every individual has one of the three, so the three frequencies add to 1.

Check it against the population counted above. With p = 0.6 and q = 0.4, p² = 0.36, 2pq = 2 × 0.6 × 0.4 = 0.48 and q² = 0.16 — predicting 36 AA, 48 Aa and 16 aa in a hundred people, which is exactly what was counted. That population is in Hardy-Weinberg equilibrium, and if the conditions hold it will still show those proportions in a thousand years.

The Hardy-Weinberg principle is the statement that this happens: in a population meeting certain conditions, allele frequencies and genotype frequencies remain constant from generation to generation. It is a null model, in the same spirit as the null hypothesis in a chi-squared test. It describes what happens when nothing happens, and its value lies in measuring departures from it.

Where this appears on your specification depends on the board. AQA examines it in section 3.7.2, OCR Biology A in 6.1.2 and Cambridge International in 17.2, in each case as a calculation you are expected to perform. Pearson Edexcel's Biology A (Salters-Nuffield) specification does not list it, so a SNAB candidate can read this lesson as background on how allele frequencies behave rather than as examinable content — check your own specification before assuming either way.

One limit worth noting while the equations are in front of you: they are written for an autosomal gene with two alleles. A sex-linked gene needs a different treatment, because males carry one copy and females two, and a gene with three alleles — the ABO system, for instance — needs a third term.

Working a calculation from the one genotype you can see

Here is the practical difficulty that shapes every Hardy-Weinberg question. Dominance hides genotypes. Looking at a population you can tell the homozygous recessive individuals apart, because they are the ones showing the recessive phenotype, but you cannot distinguish a homozygous dominant from a heterozygote. So the calculation always starts from the only genotype frequency you can measure directly: q².

Four steps, always in this order. The square root is the step that turns a genotype frequency into an allele frequency, and skipping it — using q² where q belongs — is the most expensive single error on this topic.

The reverse route does not exist. If 84% of a population shows the dominant phenotype, you cannot set p = 0.84, because that 84% contains both p² and 2pq and you have no way of splitting it. What you do instead is subtract: the recessive phenotype is 16%, so q² = 0.16, q = 0.4, and p = 0.6.

Carriers in a village

In a population of 4000 people, 160 have a condition caused by a recessive allele. Assuming Hardy-Weinberg equilibrium, calculate the frequency of each allele and the number of people who are carriers.

The affected people are the homozygous recessives, so q² = 160 ÷ 4000 = 0.04.

Taking the square root, q = 0.2, and since p + q = 1, p = 0.8.

Carriers are the heterozygotes: 2pq = 2 × 0.8 × 0.2 = 0.32. In a population of 4000 that is 0.32 × 4000 = 1280 people.

Worth checking: p² = 0.64, so 2560 people are homozygous dominant, and 2560 + 1280 + 160 = 4000. If your three genotype numbers do not add back to the population, something has gone wrong and you have time to find it.

Notice the result. Eight times as many people carry the allele as show the condition, and none of them knows it from their phenotype. That disproportion is why recessive conditions persist in populations at all, and it is the point most of these questions are really making.

Cystic fibrosis makes the same point with real numbers. It affects roughly 1 in 2500 births in the UK, so q² = 0.0004 and q = 0.02. Then p = 0.98 and 2pq = 2 × 0.98 × 0.02 = 0.0392 — about 3.9% of the population, or roughly 1 person in 25, carries a cystic fibrosis allele without any sign of it. Selection acts strongly against the homozygotes and barely touches the reservoir held in heterozygotes, which is why removing a recessive allele from a population is so slow.

TRY IT — Starting from the dominant phenotype

The ability to taste the compound PTC is controlled by a dominant allele T. In a sample of 500 people, 420 are tasters. Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of the recessive allele and the number of people in the sample expected to be heterozygous.

Check your answer

The non-tasters are the homozygous recessives: 500 − 420 = 80 people, so q² = 80 ÷ 500 = 0.16.

q = √0.16 = 0.4, and p = 1 − 0.4 = 0.6.

Heterozygotes: 2pq = 2 × 0.6 × 0.4 = 0.48, and 0.48 × 500 = 240 people.

So of the 420 tasters, 240 are heterozygous and 180 are homozygous dominant — which you can check as p² × 500 = 0.36 × 500 = 180, and 240 + 180 = 420.

The move that scores is turning the dominant phenotype into the recessive one by subtraction before doing anything else. Setting p = 420 ÷ 500 = 0.84 is the error the question is set to catch.

Five conditions, and what breaking them tells you

The equilibrium holds only under conditions that no real population meets exactly. Learn all five, because questions routinely ask for them and routinely receive three.

Condition	What breaks it	Effect if broken
The population is large	Small populations, bottlenecks, founder events	Genetic drift: allele frequencies change by chance alone
Mating is random	Choosing mates by phenotype, inbreeding, geographic subdivision	Genotype frequencies shift, usually towards an excess of homozygotes
There is no selection	Any genotype surviving or reproducing better than another	The favoured allele rises in frequency
There is no mutation	New alleles arising, or one allele mutating to another	Slow introduction of new variation
There is no migration	Individuals joining or leaving the breeding population	Gene flow imports or removes alleles

Read the coral curve where it peaks. Heterozygotes are commonest when the two alleles are equally frequent, and become scarce at both extremes — which is why a rare recessive allele sits almost entirely inside heterozygotes rather than in affected individuals.

So what does a departure mean? If a population's observed genotype frequencies differ significantly from the p², 2pq and q² its allele frequencies predict, at least one condition is not being met, and the pattern of the departure narrows down which. An excess of homozygotes and a shortage of heterozygotes points to non-random mating or to a sample that is really two separate populations lumped together. A shortage of one homozygote class points to selection against it.

Sickle-cell anaemia is the classic worked case. In parts of sub-Saharan Africa the Hb^S allele occurs at frequencies far higher than selection against affected homozygotes should permit. The reason is that heterozygotes are more resistant to malaria than either homozygote, so the heterozygote advantage maintains the allele at a stable intermediate frequency. Hardy-Weinberg is what makes that visible: without a model of the expected frequency there is no way to say the observed one is surprising.

The same logic underlies the connection to evolution. Hardy-Weinberg says allele frequencies do not change when nothing acts on a population; a change in allele frequencies between generations is therefore evidence that something is acting, and a change in allele frequency over time is one standard definition of evolution. The principle is not a description of any real population so much as the baseline that lets you detect one that is evolving.

Two cautions for the exam. Every calculation you do assumes equilibrium, so say so — 'assuming Hardy-Weinberg equilibrium' is often an explicit mark. And distinguish an allele frequency from a genotype frequency in your wording, because p and q count alleles while p², 2pq and q² count individuals, and a sentence that muddles them cannot be given credit even when the arithmetic under it is right.

In the exam

Allele frequencies are counted over 2N alleles, not N individuals. A homozygote contributes two copies of the same allele.
Start from q², because it is the only genotype frequency you can read off the phenotypes. Take the square root before doing anything with it.
You cannot get p directly from the frequency of the dominant phenotype. Subtract to find the recessive phenotype first.
Multiply by the population size when the question asks for a number of individuals, and state the assumption of equilibrium.
For the five conditions, give all five: large population, random mating, no selection, no mutation, no migration.

Check yourself

A recessive allele causes a metabolic disorder that affects 1 in 10 000 newborns. Assuming Hardy-Weinberg equilibrium, calculate the frequency of the recessive allele and the percentage of the population who are carriers. Then explain why a screening programme that identifies and counsels affected individuals would have almost no effect on the frequency of the allele.

Answer

The affected newborns are homozygous recessive, so q² = 1 ÷ 10 000 = 0.0001 and q = 0.01. Then p = 1 − 0.01 = 0.99.

Carriers are the heterozygotes: 2pq = 2 × 0.99 × 0.01 = 0.0198, so about 2% of the population — roughly 1 person in 50 — carries the allele.

Comparing the two figures answers the second part. For every affected individual there are around 198 carriers, so the overwhelming majority of copies of the recessive allele are in heterozygotes, who show no symptoms and would not be identified by a programme based on phenotype.

Removing the affected individuals from the breeding population therefore removes only a tiny fraction of the allele's copies, and the frequency falls very slowly. This is the general rule for rare recessive alleles: the rarer the allele, the larger the proportion of it that is hidden in heterozygotes.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 14 marks

In a population of 5000 people, 450 show a condition caused by a recessive allele. Assuming the population is in Hardy-Weinberg equilibrium, calculate the frequency of each allele and the number of people expected to be heterozygous.

Mark scheme

M1 the affected people are the homozygous recessives, so q² = 450 ÷ 5000 = 0.09
M1 take the square root to get the allele frequency: q = 0.3, and since p + q = 1, p = 0.7
A1 the heterozygotes are 2pq = 2 × 0.7 × 0.3 = 0.42
A1 0.42 × 5000 = 2100 people, and a check that 2450 homozygous dominant plus 2100 plus 450 comes back to 5000

Question 24 marks

A sample of 200 snails contains 98 of genotype CC, 84 of genotype Cc and 18 of genotype cc. Calculate the frequency of each allele in the sample, and calculate the number of heterozygotes expected under Hardy-Weinberg equilibrium in order to say whether the sample is at equilibrium.

Mark scheme

M1 each snail carries two alleles, so the sample holds 2 × 200 = 400 alleles, and the count of C is 2 × 98 + 84 = 280
A1 p = 280 ÷ 400 = 0.7 for C, and q = 120 ÷ 400 = 0.3 for c, which add to 1 as they must
M1 the expected number of heterozygotes is 2pq multiplied by the number of individuals: 2 × 0.7 × 0.3 × 200
A1 84, which is exactly the number observed, so the sample is in Hardy-Weinberg equilibrium

Question 34 marks

Explain why a dominant allele does not become more common in a population simply because it is dominant.

Mark scheme

B1 dominance describes which allele is expressed in a heterozygous individual, and says nothing about how often the allele occurs in the gene pool
B1 the two alleles of a heterozygote are equally likely to enter a gamete, because they separate at meiosis independently of which one is expressed
B1 the Hardy-Weinberg principle predicts that with a large population, random mating, no selection, no mutation and no migration, allele frequencies stay constant from one generation to the next whatever the dominance relationships
A1 the evidence agrees: polydactyly is caused by a dominant allele and is rare, while the recessive allele for blood group O is the commonest of the three

Question 44 marks

A biologist finds that the observed genotype frequencies in a population of beetles differ significantly from those predicted by its allele frequencies. Explain what this tells the biologist, and explain how the pattern of the difference helps to identify the cause.

Mark scheme

B1 at least one of the five conditions for Hardy-Weinberg equilibrium is not being met by this population
B1 an excess of homozygotes and a shortage of heterozygotes points to non-random mating, such as inbreeding, or to a sample that is really two separate populations counted together
B1 a shortage of one homozygous class points to selection acting against that genotype, so that fewer of those individuals survive to reproduce
A1 the principle works as a null model, so a change in allele frequency between generations is evidence that something is acting on the population, which is one standard definition of evolution

Question 54 marks

In parts of sub-Saharan Africa the sickle-cell allele occurs at a far higher frequency than selection against affected homozygotes should allow. Suggest an explanation for this, and suggest why the same allele is rare among populations in northern Europe.

Mark scheme

B1 heterozygotes are more resistant to malaria than either homozygote, which is a heterozygote advantage
B1 where malaria is common the heterozygotes therefore survive and reproduce better than the homozygous dominant individuals, so copies of the sickle-cell allele keep being passed on
B1 selection against the affected homozygotes pushes the frequency down at the same time, so the allele settles at a stable intermediate frequency rather than disappearing
A1 there is no malaria in northern Europe, so the heterozygote has no advantage there; selection acts only against the affected homozygotes and the allele frequency stays low

Question 62 marks

State what is meant by the gene pool of a population, and state what is meant by the frequency of an allele.

Mark scheme

B1 the gene pool is all the alleles of all the genes present in a population at a given time
B1 an allele frequency is the proportion of all the alleles of that gene in the pool that are of one particular kind

Worth remembering

A gene pool holds 2N alleles of an autosomal gene in N diploid individuals.
p + q = 1 for two alleles; p² + 2pq + q² = 1 for the three genotypes they make.
q² is the frequency of the recessive phenotype, and the square root of it is the allele frequency.
The five conditions: large population, random mating, no selection, no mutation, no migration.
A significant departure from the predicted proportions means a condition is being broken, and the pattern of the departure says which.
Examined by AQA, OCR A and Cambridge International; not listed by Edexcel's Salters-Nuffield specification.

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