Consider two genes in a large, randomly mating population of turtles with no mov
ID: 94282 • Letter: C
Question
Consider two genes in a large, randomly mating population of turtles with no movement of individuals in and out of the population and no mutation. The two alleles at one gene, L and M, do not affect fitness. The two alleles at the other gene, T and t, do affect fitness -- they affect the thickness of turtle shells, and the degree to which they are protected from predation. TT individuals have thick shells, which repel predators, and survive best. Tt individuals have medium shell thickness and survive 88% as well as do TT individuals. tt individuals have thin shells and are easy for predators to eat; they survive only 14% as well as do TT individuals.
a) In a population of ADULT turtles, the frequency of individuals with the MM genotype is 0.06. Calculate the frequency of individuals with the LM genotype.
b) In a population of gametes at the start of a generation, the frequency of the t allele is 0.23. Calculate the frequency of zygotes with the Tt genotype.
c) Continuing from part (b), calculate the frequency of the adults that survive from the zygotes that have the Tt genotype.
d) Continuing from part (c), calculate the frequency of the t allele in the gametes that will produce the next generation.
Explanation / Answer
A.Since there are no fitness differences, this gene is in Hardy-Weinberg Equilibrium.
Freq. MM = 0.06 = p2
Freq. M = p = the square root of p2 = the square root of 0.06 = 0.24
Freq. L = q = 1-p = 1-0.24 = 0.76
Freq. LM = 2pq = 2(0.24)(0.76) = 0.36
B.In a population of gametes at the start of a generation, the frequency of the t allele is 0.23. Calculate the frequency of zygotes with the Tt genotype.
Freq. t = 0.23 = q. Freq. T = p = 1-0.23 = 0.77
Zygote genotype frequencies are in Hardy-Weinberg proportions, so:
Freq. Tt = 2pq = 2(0.77)(0.23) = 0.35
C.To calculate adult genotype frequencies, we need to know the relative fitnesses of the genotypes. These are:
TT survives the best, and the highest relative fitness is always 1, so wTT=1
Tt survives 88% as well as TT, so wTt = 0.88
tt survives 14% as well as TT, so wtt = 0.14
Now, to calculate adult genotype frequencies, the first step is to calculate the average population fitness, wbar:
wbar = wTTp2 + wTt2pq + wttq2
wbar = (1)(0.77)2 + (0.88)(2)(0.77)(0.23) + (0.14)(0.23)2 = 0.91
Freq. Tt = wTt2pq / wbar = (0.88)(2)(0.77)(0.23) / 0.91 = 0.34
D.To calculate the frequency of t, we need to use the formula Freq. t = Freq. tt + (1/2) Freq. Tt
To use this, we first calculate Freq. Tt = wttq2 / wbar = (0.14)(0.23)2 / 0.91 = 0.0081
Freq. t = Freq. tt + (1/2) Freq. Tt = (0.0081) + (1/2)(0.34) = 0.18