Questions on population genetics and Hardy-Weinberg equilibrium

Question 1

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.

Accepted Answer

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 2

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.

Accepted Answer

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 3

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

Accepted Answer

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 4

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.

Accepted Answer

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 an excess of heterozygotes over the 2pq prediction points instead to selection favouring heterozygotes, as with the sickle-cell allele where malaria is common

Question 5

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.

Accepted Answer

B1 heterozygotes are more resistant to malaria than Hb^AHb^A homozygotes and do not have sickle-cell anaemia, so they are fitter 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 6

In a population of 300 rabbits, the frequency of the allele for black fur, B, is 0.3 and the frequency of the allele for brown fur, b, is 0.7. Assuming Hardy-Weinberg equilibrium, calculate the number of rabbits expected to be heterozygous, and calculate the number expected to show the recessive brown phenotype.

Accepted Answer

M1 heterozygote frequency = 2pq = 2 × 0.3 × 0.7 = 0.42 A1 number of heterozygous rabbits = 0.42 × 300 = 126 M1 recessive phenotype frequency = q² = 0.7² = 0.49 A1 number of brown rabbits = 0.49 × 300 = 147

Question 7

Two neighbouring populations of the same beetle species have different allele frequencies for a colour gene: population 1 has p = 0.8 and population 2 has p = 0.2. A new road is built directly connecting the two populations' habitats, greatly increasing the movement of beetles between them. Suggest what would happen to the allele frequencies of the two populations over the following generations, and suggest what this shows about one of the Hardy-Weinberg conditions.

Accepted Answer

B1 with much more movement of beetles, and therefore alleles, between the two populations, allele frequencies would be expected to move towards each other over successive generations as migration mixes the two gene pools B1 population 1's frequency of p would be expected to fall from 0.8 and population 2's frequency of p would be expected to rise from 0.2, tending towards some intermediate, shared value if migration continues B1 this shows that 'no migration' is one of the conditions required for Hardy-Weinberg equilibrium, and breaking it by itself is sufficient to change allele frequencies between generations, without any selection, mutation or drift being involved B1 the change would be fastest at first, while the two populations differ most, and would slow as their frequencies converge

Question 8

State the five conditions that must hold for a population to be in Hardy-Weinberg equilibrium.

Accepted Answer

B1 the population is very large, so that chance sampling has a negligible effect on allele frequencies B1 mating is random with respect to the genotype concerned, and there is no selection acting on that genotype B1 there is no mutation and no migration into or out of the population, so no alleles are added to or removed from the gene pool

Question 9

A biologist samples a population of snails and finds significantly more homozygotes and significantly fewer heterozygotes than the Hardy-Weinberg equations predict from the measured allele frequencies, although the allele frequencies themselves have not changed. Explain what this pattern suggests about the population.

Accepted Answer

B1 the excess of homozygotes and shortage of heterozygotes shows the population's genotype frequencies do not match Hardy-Weinberg predictions, so at least one of the five conditions is not being met even though the allele frequencies are unchanged B1 this pattern, more homozygotes and fewer heterozygotes without a change in allele frequency, points to non-random mating, such as inbreeding or assortative mating, rather than to selection, mutation or migration, since those would typically also shift the allele frequencies themselves B1 it could also indicate that the sample is really two separate, genetically somewhat different subpopulations counted together, since combining two differently shifted populations increases apparent homozygosity in the same way

Question 10

Compare a population that is in Hardy-Weinberg equilibrium with a population that is evolving, referring to what happens to allele frequencies between generations in each.

Accepted Answer

B1 in a population at Hardy-Weinberg equilibrium, allele frequencies remain constant from one generation to the next, because all five conditions are being met, whereas in an evolving population allele frequencies change from one generation to the next, because at least one of those conditions is being broken B1 the change in an evolving population is most often caused by selection, but can also be caused by drift, mutation or migration, whereas a population at equilibrium has none of these acting to change its frequencies B1 the Hardy-Weinberg principle acts as a null model for evolution: a population's frequencies are compared against the constant values the equations predict, and a measured change is exactly what evolution means in this context

Question 11

Describe the sequence of steps needed to calculate the frequency of heterozygous carriers of a recessive condition in a population, starting from the number of people affected by the condition and the size of the population.

Accepted Answer

B1 first, divide the number of affected people by the population size to obtain q², the frequency of the homozygous recessive genotype B1 second, take the square root of q² to obtain q, the frequency of the recessive allele, and then calculate p as 1 − q B1 third, substitute p and q into 2pq to obtain the frequency of heterozygotes, and multiply this frequency by the population size to convert it into a number of people

Question 12

A recessive genetic disease affects 1 in 90 000 people. A newspaper article states that 'about 1 in 300 people carry the allele, since the square root of 1/90000 is about 1/300'. Suggest what error the article has made in describing 1 in 300 as the carrier frequency, and suggest the calculation that should have been done instead.

Accepted Answer

B1 the article's arithmetic finds q, the frequency of the recessive allele in the gene pool, not the carrier frequency; carriers are heterozygotes, and heterozygote frequency is given by 2pq, not by q alone B1 with q ≈ 1/300, p is very close to 1, so the carrier frequency 2pq is roughly twice the reported figure, closer to 1 in 150 than to 1 in 300 B1 the correct calculation is to find q as before, then calculate p = 1 − q, and substitute both into 2pq

Question 13

Red-green colour blindness is caused by a recessive allele of a gene on the X chromosome, and the Y chromosome carries no copy of the gene. In one population, 8% of males are colour-blind. Assuming the population is in Hardy-Weinberg equilibrium, calculate the percentage of females expected to be colour-blind.

Accepted Answer

M1 a male has one X chromosome and is colour-blind whenever his single allele is the recessive one, so the frequency of that allele is q = 0.08 M1 a female is colour-blind only if homozygous recessive, with frequency q² = 0.08² A1 0.0064, so 0.64% of females, about one in 160

Question 14

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

Accepted Answer

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

Question 15

Give the two equations of the Hardy-Weinberg principle, stating what each symbol represents.

Accepted Answer

B1 p + q = 1, where p and q are the frequencies of the two alleles of the gene B1 p² + 2pq + q² = 1, where p² is the frequency of the homozygous dominant genotype, 2pq the heterozygous genotype and q² the homozygous recessive genotype