7.5 Hardy–Weinberg Equilibrium

1. AIt describes the genotype frequencies found in most natural populations, so they can be predicted.
   A student who thinks real populations usually meet the conditions picks this. The conditions are never all met, so the model gives a baseline for comparison, not a description of real populations.
2. BIt gives the allele frequencies that a population returns to after it has been disturbed.
   A student who thinks equilibrium is a restoring force picks this. The model predicts no change only when no evolutionary force acts; it does not return a population to earlier allele frequencies.
3. CIt predicts genotype frequencies for a nonevolving population, which real data can be compared with. Correct
   The model gives the frequencies expected if no evolution were happening. Comparing real data with these expected values, as a null hypothesis, shows whether and how a population is departing from equilibrium.
4. DIt shows that dominant alleles increase in frequency until the recessive alleles are lost.
   A student who thinks dominant alleles spread because they are dominant picks this. The model shows the opposite: without evolutionary forces, dominant and recessive alleles keep the same frequencies.

CED 7.5.A.1 · Read this in Fix

1. AThe fraction of individuals in the population that show the b phenotype
   A student who takes a phenotype frequency to be an allele frequency picks this. Because b is recessive, only bb individuals show its phenotype, and in equilibrium their frequency is q², not q; Bb individuals also carry b.
2. BThe fraction of the copies of the gene in the population that are b Correct
   Allele frequency is measured over copies of the gene: in a diploid population of N individuals there are 2N copies, and q is the fraction of them that are b.
3. CThe fraction of individuals that carry at least one copy of allele b
   A student who counts individuals instead of allele copies picks this. Bb individuals carry one copy of b and bb individuals carry two, so the fraction of carriers is not q.
4. DThe frequency of the recessive allele b, which is lower than that of B
   A student who thinks dominant alleles are the more common ones picks this. Recessive alleles can be more common than dominant ones; q can take any value from 0 to 1.

CED 7.5.A.2 · Read this in Fix

1. A27%
   A student who takes the percentage of black beetles as the frequency of the recessive allele picks this. Using q = 0.16 and p = 0.84 gives 2pq = 0.27, but 16% is the frequency of the aa genotype, q², so q = 0.4.
2. B48% Correct
   Black beetles are homozygous recessive, so q² = 0.16 and q = √0.16 = 0.4. Then p = 1 − 0.4 = 0.6, and the heterozygote frequency is 2pq = 2 × 0.6 × 0.4 = 0.48, or 48%.
3. C24%
   A student who counts heterozygotes only once picks this: pq = 0.6 × 0.4 = 0.24. A heterozygote can form from an A egg and an a sperm or from an a egg and an A sperm, so the frequency is 2pq = 0.48.
4. D15%
   A student who takes the brown beetles (84%) to be the homozygous dominant genotype, p², picks this: p = √0.84 = 0.92, q = 0.08 and 2pq = 0.15. Brown beetles include heterozygotes, so 84% equals p² + 2pq; only the black beetles give a single genotype, q² = 0.16.

Working Black beetles are aa, so q² = 0.16 and q = 0.4; p = 1 − q = 0.6. Heterozygotes: 2pq = 2 × 0.6 × 0.4 = 0.48 = 48%. Distractors: 0.16 used as q gives 2 × 0.84 × 0.16 = 0.27 (27%); pq without the 2 gives 0.24 (24%); 0.84 taken as p² gives p = 0.917, q = 0.083, 2pq = 0.15 (15%).

CED 7.5.A.2 · Read this in Fix

1. A0.53
   A student who uses the Hardy–Weinberg shortcut picks this: q = √(110/500) = 0.47, so p = 0.53. The shortcut works only if the population is in equilibrium; here the genotypes were counted directly, so allele copies can be counted, giving 0.60.
2. B0.42
   A student who treats a genotype frequency as an allele frequency picks this: 210/500 = 0.42 is the frequency of AA plants. Copies of A are also carried by the Aa plants.
3. C0.60 Correct
   Each plant has two copies of the gene, so there are 1,000 copies. AA plants carry 2 × 210 = 420 copies of A and Aa plants carry 180, so the frequency of A is (420 + 180)/1,000 = 0.60.
4. D0.78
   A student who counts the plants that carry A picks this: (210 + 180)/500 = 0.78. Allele frequency counts copies of the allele: AA plants carry two copies and Aa plants one, out of 1,000 copies in all.

Working Total copies = 2 × 500 = 1,000. Copies of A = 2 × 210 + 180 = 600. Frequency of A = 600/1,000 = 0.60 (and a = (2 × 110 + 180)/1,000 = 0.40). Distractors: 1 − √(110/500) = 0.53; 210/500 = 0.42; (210 + 180)/500 = 0.78.

CED 7.5.A.2 · Read this in Fix

1. AThe frequency of b will be 0.30, and the frequency of Bb snails will be 0.42. Correct
   The null hypothesis is that the population is not evolving at this gene and is in Hardy–Weinberg equilibrium. The allele frequencies therefore stay at q = 0.30 and p = 0.70, and the heterozygote frequency is 2pq = 2 × 0.70 × 0.30 = 0.42.
2. BThe frequency of b will be 0.30, and the frequency of Bb snails will be 0.21.
   A student who counts heterozygotes only once picks this: pq = 0.70 × 0.30 = 0.21. Bb snails form from a B egg with a b sperm or a b egg with a B sperm, so their expected frequency is 2pq = 0.42.
3. CThe frequency of b will be 0.50, and the frequency of Bb snails will be 0.50.
   A student who thinks equilibrium means that the two alleles have equal frequencies picks this. A population in equilibrium keeps the allele frequencies it has, here q = 0.30; it does not move toward 0.50.
4. DThe frequency of b will be 0.20, and the frequency of Bb snails will be 0.32.
   A student who expects a dominant allele to spread picks this, predicting that B rises to 0.80. Dominance does not change allele frequencies; in a nonevolving population q stays at 0.30.

Working Null hypothesis: no evolution at this gene, so allele frequencies are unchanged (q = 0.30, p = 0.70) and genotypes are in Hardy–Weinberg proportions: BB = p² = 0.49, Bb = 2pq = 0.42, bb = q² = 0.09. Distractors: pq = 0.21; q = 0.50 with 2pq = 0.50; q = 0.20 with 2pq = 2 × 0.80 × 0.20 = 0.32.

CED 7.5.A.1 · Read this in Fix

1. AIn each population of 10, the frequency of A moved steadily in one direction over time.
   A student who pictures genetic drift as a directional trend picks this. Each small-population line zigzags up and down from one generation to the next, and the three lines go in different directions.
2. BThe populations of 10 changed their allele frequencies to adapt to being so small.
   A student who thinks populations change because they need to picks this. The changes are random: the three small populations went in different directions, which is not what adaptation to a shared condition would produce.
3. CAfter each change, the frequency of A returned toward 0.5 in all six populations.
   A student who thinks equilibrium restores allele frequencies picks this. Two small populations reached 1.0 and 0 and stayed there, and the third ended at 0.35; nothing pushed the frequencies back to 0.5.
4. DChanges in A were far larger in the populations of 10, rising in one and falling in the other two. Correct
   The three small populations moved far from 0.5: one rose to 1.0, one fell to 0 and one fell to 0.35. The three large populations stayed between 0.46 and 0.54. Chance changes in allele frequency (genetic drift) are much larger in small populations.

CED 7.5.A.1.i · Read this in Fix

1. AThe frequency of d on the island will stay at 0.1, as the migrants bring no allele the island lacks.
   A student who thinks migration matters only when it introduces a new allele picks this. Both alleles are already on the island, but the migrants carry them in different proportions, so adding migrants changes the island's allele frequencies.
2. BThe frequency of d on the island will fall, as the dominant allele D will spread among the lizards.
   A student who thinks dominant alleles spread because they are dominant picks this. Dominance does not change allele frequencies; the migrants, with more d, push the frequency of d up, not down.
3. CThe frequency of d on the island will rise toward 0.6, as migrants carry d at a higher frequency. Correct
   Each generation the migrants add alleles in which d is more common (0.6) than among the island lizards (0.1), so the island frequency of d rises over the generations toward the mainland value.
4. DThe frequency of d on the island will settle at 0.5, the equilibrium value for a gene with two alleles.
   A student who thinks Hardy–Weinberg equilibrium means equal allele frequencies picks this. There is no equilibrium value of 0.5; with continued migration the island frequency moves toward the frequency in the migrants, 0.6.

CED 7.5.A.1.ii · Read this in Fix

1. AMutation produces the alleles the moths need for their surroundings, so the frequencies change.
   A student who thinks mutations arise to meet a need picks this. Mutations are random with respect to what the moths need; the chemical increases the number of mutations, not the number of useful ones.
2. BMutation converts copies of one allele into another allele, so the allele frequencies change. Correct
   A mutation in the DNA of a germ-line cell turns a copy of one allele into a different allele, which offspring can inherit. This changes allele frequencies, so the condition of no new mutations is not met.
3. CMutation kills the moths that carry the new alleles, so the population becomes too small.
   A student who thinks all mutations are harmful picks this. Many mutations have little or no effect on survival; the reason mutation violates the condition is that it changes allele frequencies directly.
4. DMutation has no lasting effect, as equilibrium restores the original allele frequencies.
   A student who thinks Hardy–Weinberg equilibrium is a restoring force picks this. The model does not return allele frequencies to earlier values; once mutation changes them, the new frequencies are passed on.

CED 7.5.A.1.iii · Read this in Fix

1. ASelf-fertilization changed both the genotype frequencies and the allele frequencies.
   A student who thinks nonrandom mating changes allele frequencies picks this. Calculating p from each generation gives 0.50 every time: AA and aa rose by equal amounts, so the alleles kept the same frequencies.
2. BThe population stayed in Hardy–Weinberg equilibrium, as p and q each remained 0.5.
   A student who thinks constant allele frequencies are enough for equilibrium picks this. With p = q = 0.5, equilibrium predicts genotype frequencies of 0.25, 0.50 and 0.25; generation 2 has only 0.10 heterozygotes, so the population is not in equilibrium and the random mating condition is not met.
3. CIndividual Aa plants turned into AA or aa plants as the whole population evolved.
   A student who thinks individuals change their genotypes as a population evolves picks this. Each plant's genotype is fixed; the heterozygote frequency fell because Aa parents produced AA and aa offspring by self-fertilization.
4. DSelf-fertilization changed the genotype frequencies but not the allele frequencies. Correct
   The frequency of A is AA + ½Aa: 0.30 + 0.20 = 0.50 in generation 0, 0.40 + 0.10 = 0.50 in generation 1 and 0.45 + 0.05 = 0.50 in generation 2. Heterozygotes fell from 0.40 to 0.10 while AA and aa rose equally, so genotype frequencies changed but allele frequencies did not.

Working p = f(AA) + ½f(Aa): generation 0, 0.30 + 0.20 = 0.50; generation 1, 0.40 + 0.10 = 0.50; generation 2, 0.45 + 0.05 = 0.50. Heterozygote frequency halves each generation: 0.40 → 0.20 → 0.10. Under random mating with p = 0.5 the expected frequencies are 0.25, 0.50, 0.25.

CED 7.5.A.1.iv · Read this in Fix

1. AIt will fall each generation but persist, since Rr carriers survive and pass r on. Correct
   Selection removes only the r copies in rr individuals. Most r copies are in Rr carriers, who survive and pass r to about half their offspring, so r declines each generation and declines more slowly as it becomes rare.
2. BIt will fall to zero in the next generation, since every rr individual dies before breeding.
   A student who thinks selection against a recessive phenotype removes the allele at once picks this. Before the disease, 2pq = 0.32 of the population were Rr carriers, so many r copies survive in carriers.
3. CIt will fall at first, then return to 0.2 as the population restores its equilibrium.
   A student who thinks Hardy–Weinberg equilibrium is a restoring force picks this. No process pushes r back to 0.2; while the disease continues, r keeps declining.
4. DIt will stay at 0.2, since selection acts on phenotypes rather than on alleles.
   A student who thinks selection changes phenotypes but not allele frequencies picks this. The rr phenotype leaves no offspring, so its r copies are not passed on, and the frequency of r falls.

CED 7.5.A.1.v · Read this in Fix

1. AGraph 1
   A student who thinks Hardy–Weinberg equilibrium means equal allele frequencies picks Graph 1, with 0.25, 0.50 and 0.25. That ratio applies only when p = q = 0.5; here p = 0.8.
2. BGraph 2 Correct
   Graph 2 shows AA = p² = 0.64, Aa = 2pq = 0.32 and aa = q² = 0.04, which sum to 1 and match p = 0.8, q = 0.2.
3. CGraph 3
   A student who uses allele frequencies as genotype frequencies picks Graph 3, with AA = 0.80 and aa = 0.20. Allele frequencies are p and q; genotype frequencies are p², 2pq and q², and heterozygotes are 0.32 of the population.
4. DGraph 4
   A student who counts heterozygotes only once picks Graph 4, with Aa = pq = 0.16. Heterozygotes form in two ways, so Aa = 2pq = 0.32; the bars in Graph 4 also add to only 0.84.

Working p = 0.8, q = 0.2. AA = p² = 0.64, Aa = 2pq = 2 × 0.8 × 0.2 = 0.32, aa = q² = 0.04 (total 1.00). Distractor graphs: 0.25/0.50/0.25 (equal allele frequencies assumed); 0.80/0/0.20 (allele frequencies used as genotype frequencies); 0.64/0.16/0.04 (heterozygotes as pq).

CED 7.5.A.2 · Read this in Fix

1. ASet up 6 populations of 10 flies and 6 of 1,000 flies, all with A at 0.5; track A for 20 generations. Correct
   Population size is the only difference between the two groups, every population starts at the same allele frequency, and six replicates per size show whether drift is consistently larger in small populations rather than by chance in one population.
2. BSet up 1 population of 10 and 1 of 1,000 flies, both with A at 0.5; track A for 20 generations.
   A student who thinks one population per condition is enough picks this. Drift is random, so a single small population could change a lot or a little by chance; replicates are needed to compare sizes.
3. CSet up 8 populations of 10 flies with A at 0.2 and 8 of 1,000 flies with A at 0.8; track A for 20 generations.
   A student who thinks only the tested factor needs to be set picks this. The two sizes start at different allele frequencies, so any difference in the change could come from the starting frequency rather than from population size.
4. DSet up 5 populations of each size, with A at 0.5 and a predator that eats aa flies; track A for 20 generations.
   A student who thinks allele frequencies change only under natural selection picks this. A predator that removes aa flies adds selection, which would change allele frequencies in every population and obscure the effect of drift that the experiment is meant to measure.

Working Independent variable: population size (10 vs 1,000). Controlled: starting frequency of A (0.5), strain, gene with no effect on fitness, number of generations (20). Replication: 6 populations per size. Dependent variable: change in the frequency of A.

CED 7.5.A.1.i · Read this in Fix

1. ARegion 1 alone, totaling 0.49
   A student who thinks the dominant phenotype frequency is p² picks this. Region 1 is the AA genotype; the Aa offspring in regions 2 and 3 also show the dominant phenotype.
2. BRegions 1 and 2, summing to 0.70
   A student who takes a phenotype frequency to be an allele frequency picks this. Regions 1 and 2 together are the A eggs, whose frequency is p = 0.7; region 3 (an a egg with an A sperm) also gives an Aa offspring with the dominant phenotype.
3. CRegions 1, 2 and 3, totaling 0.75
   A student who treats the model like a Punnett square for Aa × Aa, with four equally likely boxes, picks this: three of four regions give 3/4 = 0.75. The regions differ in area because p = 0.7 and q = 0.3, so regions 1, 2 and 3 total 0.49 + 0.21 + 0.21 = 0.91.
4. DRegions 1, 2 and 3, summing to 0.91 Correct
   Offspring with the dominant phenotype are AA (region 1, 0.7 × 0.7 = 0.49) and Aa (regions 2 and 3, each 0.7 × 0.3 = 0.21). Together they are 0.49 + 0.21 + 0.21 = 0.91, which is also 1 − q² = 1 − 0.09.

Working Region areas: 1 = 0.7 × 0.7 = 0.49 (AA); 2 = 0.7 × 0.3 = 0.21 (Aa); 3 = 0.3 × 0.7 = 0.21 (Aa); 4 = 0.3 × 0.3 = 0.09 (aa). Dominant phenotype = 0.49 + 0.21 + 0.21 = 0.91 = 1 − q².

CED 7.5.A.2 · Read this in Fix