1. Consider a gene with two alleles. What equation describes the frequencies of the alleles in a population? p + q = 1 (p = the frequency of the first allele, q the frequency of the 2nd allele, and since there are no other alleles, the frequencies of these two must sum to 1. 2. Consider a gene with two alleles. What equation describes the frequencies of the genotypes in a population if we assume the population is in genetic equilibrium? p2 + 2pq + q2 = 1 In this case, the frequencies of the homozygotes (AA or aa, for example) are p2 and q2, respectively. The frequency of the heterozygote (Aa) is pq + qp, which we normally write as 2pq. 3. Do we have an equation that that allows us to describe genotype frequencies if we cannot assume genetic equilibrium? If so, what is it? No, we don’t. 4. Imagine a population of Daphnia (a microscopic crustacean, similar to a crayfish) in Lake Superior. There are billions and billions of Daphnia in Lake Superior. There is no migration into or out of Lake Superior. A single gene controls whether the animals are dark or or clear-colored. Mutations in this gene are so rare, that we can assume they don’t matter. Daphnia do not pick their mates based on color, so mating is random in terms of color. However, dark Daphnia are more likely to be seen by fish, and therefore much more likely to be eaten than the clear Daphnia. Is it reasonable to assume the gene for color is in equilibrium? Why or why not? No. There is selection on color, due to the fish predation (i.e., color affects survival). 5. Consider a population of dragonflies, some with long wings others with short wings. Imagine that winglength is governed by a gene with two alleles, S (for long wings) and s (for short wings). The allele frequency of s is 0.8. What is the frequency of S? 0.2 (since p + q = 1, it follows that 1 – p = q, and 1 – 0.8 = 0.2) 6. In the previous question, do we have to assume genetic equilibrium to answer it? No. Allele frequencies always sum to 1, regardless of equilibrium. (It is also true that genotype frequencies also always sum to 1. However, if the population is NOT in equilibrium, then we can’t say that p2 and q2 are the frequencies of the homozygotes or that 2pq is the frequency of the heterozygote).7. In our dragonfly example, the individuals with the genotypes SS or Ss have long wings; only the individuals with ss have short wings. You go out and count long and short winged dragonflies, and when you’re done, you’ve counted 250 short-winged dragonflies and 750 long-winged dragonflies. Assume Hardy-Weinberg equilibrium, and calculate: a. the allele frequencies for S and s. (note: they won’t be the same as in Question 5) 250 individuals have short-wings and are therefore ss. 750 individuals have long-wings and could be SS or Ss. Since we don’t know which, we cannot directly calculate the allele frequencies. However, by assuming HWE, we can calculate it indirectly. First, focus on the 250 individuals that are short-winged. We know their genotype is ss, and we therefore can calculate that what the genotype frequency of this homozygote is. Since there are 1000 individuals total, the frequency of the ss genotpe is 250/1000, which equals 0.25. Since we usually call the recessive allele q, this genotype frequency is q2. q2 = 0.25, therefore q = 0.5. (the square root of 0.25; this step is often forgotten) Now, since p + q = 1, we can also see that p = 0.5. (1 – 0.5 = 0.5) Thus, the allele frequencies for S and s are both 0.5. b. the frequency (proportion) of the total population that are heterozygotes (Ss). This is simply the genotype frequency of Ss, which is equal to 2pq, assuming HWE. 2pq = 2 x 0.5 x 0.5 = 0.5 Thus, the frequency of the total population that are heterozygous is 0.5 c. The total number of dragonflies that are long-winged, but do NOT carry the s allele. We just figured out that the frequency of heterozygotes is 0.5. This is all of the long-winged dragonflies that DO carry the s allele. We start by converting this to real numbers of individuals. total number of individuals = 1000. frequency of heterozygotes = 0.5 number of heterozygotes = 1000 x 0.5 = 500 total number of long-winged dragonflies = 750 # of non-carrying long-winged dragonflies = # of long-wings – the # of heterozygotes = 750 – 500 = 2508. In Whistling-Ducks, individuals have primary wing feathers that are black, white or gray. At the autosomal gene that governs wing feather color, imagine there are two co-dominant alleles, B and W. BB individuals are black, WW individuals are white and BW individuals are gray. You are studying a population where there are 45 Black individuals, 45 White individuals and 10 gray individuals. a. calculate the frequencies of each allele 45 Black Ducks are BB, so they have a total of 90 B alleles 45 White Ducks are WW, so they have a total of 90 W alleles 10 Gray ducks are BW, so they have 10 B alleles and 10 W alleles. The total number of B alleles = 90 + 10 = 100 The total number of W alleles = 90 + 10 = 100 The total number of all alleles = 100 + 100 = 200 (or you could say 2 x (45 + 45 + 10) = 200) Frequency of B allele = 100 B alleles / 200 total alleles = 0.5 Frequency of W allele = 100 W alleles/ 200 total alleles = 0.5 b. Based on these allele frequences, calculate the genotype frequencies you would expect to see if the population were in genetic equilibrium. If the population were in HWE, we would expect the genotype frequencies to be: freq(BB) = p2 = (0.5)2 = 0.25 freq (BW) = 2pq = 2 x 0.5 x 0.5 = 0.5 freq (WW) = q2 = (0.5)2 = 0.25 c. Based on your calculations, do you think the population is in equilibrium? Why or why not? No. The actual genotype frequencies are f(BB) = 45/100 = 0.45, f(BW) = 10/100 = 0.1 and f(WW) = 45/100 = 0.45. These are substantially different from what is expected at equilibrium. d. Being as specific as you can, what evolutionary forces would be a good hypothesis for why the population is/is not in equilibrium? Disruptive selection. Why? There are more individuals than we expect of each extreme (45 individuals each of black and white, rather than the expected 25) and fewer of the middle (10 gray individual rather than the predicted 50). An argument could also be made for non-random mating (geese may choose mates of the same color as they are), or the beginning of speciation.9. Molecular