Unit 7 · Topic 7.5 end-of-topic test
Suggested time: about 47 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.
Biologists genotype the 400 ladybugs of one hedge at a spot-pattern gene with alleles R and r. The table shows the census.
Which of the following is the genotype frequency of Rr?
A census of one population of ground beetles gives the genotype frequencies in the table at a wing-vein gene with alleles T and t.
Which of the following is the frequency of allele t?
In a population of clover, a leaf-mark gene has two alleles. The marked-leaf allele is dominant to the plain-leaf allele. A biologist calls the marked-leaf allele allele 1 and writes its frequency as p. In the gene pool, p = 0.35.
Which of the following is q?
A population of guppies mates at random at a tail-color gene with alleles T and t. In the gene pool, q = 0.20 for allele t.
Which of the following gives the expected share of tt offspring, and why?
A field holds 300 sunflowers that mate at random. At a petal-color gene, p = 0.60 for allele Y and q = 0.40 for allele y.
Which of the following is the number of Yy sunflowers the Hardy–Weinberg model expects?
Of the 250 salamanders in a pond, 36% are yellow. Yellow (y) is recessive to black (Y), the salamanders mate at random, and no force acts on this gene.
Which of the following is the expected number of carriers, the Yy salamanders?
Biologists record the head color of both birds in 100 mating pairs of buntings. Head color is set by one gene. The table shows the pairs.
According to the table, which of the following Hardy–Weinberg conditions is broken?
Biologists census a mountain meadow’s plants at a petal-color gene with alleles R and W in generation 1 and again in generation 50. The table shows the genotype frequencies. From the generation-1 census, a biologist calculates the genotype frequencies the Hardy–Weinberg model predicts for generation 50: 0.36, 0.48 and 0.16.
Which of the following best explains the purpose of the predicted frequencies?
A biologist graphs the frequency of allele W in a population of pigeons over eight generations.
Which of the following describes the population at this gene over the eight generations?
The table gives the frequency of allele K in a population of dormice for generations 0 to 4. A student constructs a line graph from the table. Four graphs are shown.
Which graph is the correct and complete construction of the table?
Researchers sample 100 plants from each of two populations of one plant species in generation 1 and again in generation 6, and genotype them at one gene with alleles A and a. The table shows the counts.
Which statement about the two populations over the five generations is correct?
The frequency of allele S in a population of limpets was 0.44 twelve years ago and is 0.55 today.
Which of the following is the percent change in the frequency of S?
Biologists genotype 400 minnows from one stream at a fin-stripe gene with alleles S and s. The table shows the census.
Which of the following is the number of Ss minnows the Hardy–Weinberg model expects for this census?
Biologists genotype 300 dragonflies from one marsh at a wing-spot gene with alleles G and g. The table shows the observed counts above the counts the Hardy–Weinberg model expects from the census’s own p.
Which of the following is chi-square for the three genotype classes?
For a census of 250 tortoises, chi-square for the three genotype classes against the Hardy–Weinberg expectation is 7.20. Use the chi-square table above.
Which of the following is the verdict at p = 0.05, and why?
Owls hunt the voles of a meadow and take every white (ww) vole before it breeds. Gray (W) is dominant to white (w). The graph shows q, the frequency of w, over five generations.
Which of the following explains why the fall in the frequency of w slows?
A biologist has genotyped a census of wasps at one gene and has calculated p, q and the three expected genotype shares.
Which of the following is the next step toward a verdict on Hardy–Weinberg equilibrium?
In a garden’s snails, shell shape is set by one gene.
Which of the following describes a population that meets the random-mating condition?
(a) Describe the genotype make-up expected before the beetles arrived, giving the three genotype frequencies. (1 point)
A full-credit answer: p = 0.80 and q = 0.20.
YY: p² = 0.64.
Yy: 2pq = 2 × 0.80 × 0.20 = 0.32.
yy: q² = 0.04.
Check the box for each point your answer earns
Common slip: Writing 0.80 and 0.20 as genotype frequencies. p and q are shares of copies; the genotype shares are p², 2pq and q².
(b) Explain why the arrival of the beetles takes the population out of Hardy–Weinberg equilibrium at this gene. (1 point)
A full-credit answer: The beetles kill yy plants before they set seed, so plants with the yy genotype leave no offspring.
Survival to breeding now depends on genotype: natural selection acts on this gene, so the no-natural-selection condition is broken.
Each generation copies of y are removed from the gene pool, so q falls and the frequencies no longer stay the same.
Check the box for each point your answer earns
Common slip: Saying the beetles break random mating. The beetles change who survives to breed, not who mates with whom.
(c) Predict how the frequency of y changes over the next twenty generations. (1 point)
A full-credit answer: The frequency of y falls.
It falls quickly in the first generations and ever more slowly after that, and it is still above zero after twenty generations.
Check the box for each point your answer earns
Common slip: Predicting that y is gone within a few generations. Only yy plants are killed; the copies of y in Yy plants are untouched.
(d) Justify your prediction in part (c). (1 point)
A full-credit answer: Only yy plants are killed.
A Yy plant has yellow petals, so the beetles leave it, and its copy of y stays in the gene pool.
As y becomes rare, the yy share, q², shrinks faster than the Yy share, 2pq.
So more of the remaining copies of y sit in Yy plants, and the beetles reach fewer copies each generation.
Check the box for each point your answer earns
Common slip: Justifying the slowing by the beetles becoming fewer. The slowing comes from where the copies of y sit, not from the beetles’ numbers.
(a) Calculate p, the frequency of allele A, from the census. (1 point)
A full-credit answer: Copies of A: 2 × 312 + 216 = 840, of 1,200 copies in all.
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Check the box for each point your answer earns
Common slip: Dividing 312 by 600. That is the genotype frequency of AA; the Aa fish each carry one copy of A too.
(b) Calculate the number of Aa cichlids the Hardy–Weinberg model expects among the 600. (1 point)
A full-credit answer: q = 1 − 0.70 = 0.30.
Expected Aa share: 2pq = 2 × 0.70 × 0.30 = 0.42.
Expected Aa count: 0.42 × 600 = 252.
Check the box for each point your answer earns
Common slip: Forgetting the 2: 0.21 × 600 = 126. Aa forms two ways, so its share is 2pq.
(c) Determine, using a chi-square test at p = 0.05, whether the census is consistent with Hardy–Weinberg equilibrium at this gene. Use the chi-square table below. (1 point)
A full-credit answer: Expected counts: AA 0.49 × 600 = 294, Aa 252, aa 0.09 × 600 = 54.
AA: (312 − 294)² / 294 = 1.10.
Aa: (216 − 252)² / 252 = 5.14.
aa: (72 − 54)² / 54 = 6.00.
Chi-square = 12.2; three classes give two degrees of freedom, critical value 5.99.
12.2 is greater than 5.99, so reject the null hypothesis: the census is not consistent with Hardy–Weinberg equilibrium.
Check the box for each point your answer earns
Common slip: Comparing chi-square with 7.81. Three genotype classes give 3 − 1 = 2 degrees of freedom, not 3.
(d) Explain how the pattern in the counts suggests which Hardy–Weinberg condition is broken. (1 point)
A full-credit answer: The census has fewer Aa cichlids than expected, 216 against 252, and more of both homozygotes.
When like mates with like, heterozygotes become rarer and homozygotes commoner while p stays the same.
So the pattern suggests that random mating is broken.
The verdict alone says only that some force acts; a shortage of heterozygotes could also come from selection against Aa fish.
Check the box for each point your answer earns
Common slip: Naming natural selection because the chi-square test rejected the null. Rejecting says a force acts; the pattern in the counts is what suggests one.