Please note – In some cases, your answers will differ from mine by a small amount. If your answer is within 0.01 or so of mine, we’re probably both right, and the difference is simply rounding error (don’t worry – rounding error won’t be an issue on the exam). Also, in writing the problems, I assumed that you didn’t know the chi-square test. If you want to use this test to check your intuition regarding some of your answers, feel free to do so.

1. We’ll start with our pink and white flower example from lecture (if you are unable to attend lecture, you can find this example on page 431, Figure 23.3 of your text). There are two alleles, on (A) is dominant and codes for pink flowers, the other (a) is recessive and codes for white flowers.

a. Suppose we have a total of 1000 flowers, 200 pink and 800 white. Of the pink flowers, 160 are

AA, and 40 are Aa. What is the frequency of each of the three possible genotypes for this locus

(AA, Aa, and aa)?

ANSWER: frequency of aa = 0.80, frequency of Aa = 0.04, frequency of AA = 0.16

b. What are the current allele frequencies (A and a) at this locus?

ANSWER: frequency of the a allele = 0.82, frequency of the A allele = 0.18

c. Assume the population is not evolving (genetic drift, migration, nonrandom mating, natural

selection and mutations are not occurring). What will the genotype and allele frequencies be in

the next generation? (big hint: this is where you use the Hardy-Weinberg equation.

ANSWER: The frequency of the alleles won’t change (A = 0.18, a = .082), but the expected genotype frequencies will be different in the next generation: frequency of AA (rounded to two decimal places) = 0.03, frequency of Aa = 0.30, frequency of aa = 0.67

d. Was this population in Hardy-Weinberg equilibrium in the first generation (where we had 200

pink and 800 white flowers)? Is it in equilibrium after one round of random mating, assuming

no evolution has occurred?

ANSWER: Technically, we should use the chi-square test to determine whether the population was in equilibrium , but we can see that the genotype frequencies in the next generation are rather different from the frequencies we started with. Thus, the population probably wasn’t in equilibrium initially. It is, of course, in equilibrium after one round of random mating.

2. Staying with the flowers…repeat a-d above, with the following starting conditions:

a. 500 flowers; 10 white, 490 pink. 450 of the pink flowers are AA, 40 are Aa.

ANSWER: The initial genotype frequencies are AA = 0.90, Aa = 0.08, and aa = 0.02. The initial allele frequencies are a = 0.06 and A = 0.94. After one round of random mating, the expected genotype frequencies are (rounding to two decimal places) AA = 0.88, Aa = 0.11, aa = 0.01 (actually its 0.0036, but we’ll round up). The allele frequencies don’t change. The population was very close to equilibrium initially; if we used the chi-square test, we wouldn’t be able to reject the null hypothesis that the population was initially in equilibrium.

b. 10000 flowers; 5000 white, 5000 pink. 2500 of the pink flowers are AA, 2500 are Aa.

ANSWER: The initial genotype frequencies are AA = 0.25, Aa = 0.25, and aa = 0.5. The initial allele frequencies are (rounding to two decimal places) a = 0.63 and A = 0.37. After one round of random mating, the expected genotype frequencies are AA = 0.14,

Aa = 0.47, aa = 0.39. The allele frequencies don’t change. The population probably was not in equilibrium initially.

c. 100 flowers; 99 white, 1 pink. The lone pink flower is Aa.

ANSWER: The initial genotype frequencies are AA = 0, Aa = 0.01, and aa = 0.99. The initial allele frequencies are a = 0.995 and A = 0.005. After one round of random mating, the expected genotype frequencies are (after rounding) AA = 0, Aa = 0.01, aa = 0.99. There’s a bit of trickery here. Only if the Aa individual self-fertilizes can we get any AA’s in the next generation (I assumed self-fertilization wasn’t happening, thus, the “0” for the frequency of AA). If you did allow self-fertilization during random mating and calculated expected frequencies as before, the frequency of AA would be tiny (0.000025). The population was effectively in equilibrium initially; the expected genotype frequencies are only marginally different from the initial frequencies.