Homework SourseIf the eggtoadult survival rates of genotypes are 90 85 and
If the egg-to-adult survival rates of genotypes are 90, 85, and 75 percent, respectively, and their fecundity values are 50, 55, and 70 eggs per female, what are the approximate absolute fitnesses (R) and relative fitnesses of these genotypes?
What are the allele frequencies at equilibrium?
Suppose the species has two generations per year, that the genotypes do not differ in survival, and the fecundity values are 50, 55, and 70 in the spring generation and 70,65, and 55 in the fall generation.
Will polymorphism persist, or will one allele become fixed?
What if the fecundity values are 55,65, 75 in the spring and 75, 65, 55 in the fall?
Solution
If the egg-to-adult survival rates of genotypes are 90, 85, and 75 percent, respectively, and their fecundity values are 50, 55, and 70 eggs per female, what are the approximate absolute fitnesses (R) and relative fitnesses of these genotypes?
What are the allele frequencies at equilibrium?
Suppose the species has two generations per year, that the genotypes do not differ in survival, and the fecundity values are 50, 55, and 70 in the spring generation and 70,65, and 55 in the fall generation.
Will polymorphism persist, or will one allele become fixed?
What if the fecundity values are 55,65, 75 in the spring and 75, 65, 55 in the fall?
Suppose the genotypes are A1A1, A1A2 and A2A2.
a) Approximate absolute fitnesses (R) and relative fitnesses of these genotypes are:
Abslute fitness of A1A1 = 0.9 x 0.5 = 0.45
Abslute fitness of A1A2 = 0.85 x 0.55 = 0.4675
Abslute fitness of A2A2 = 0.75 x 0.70 = 0.525
Relative fitness of A1A1 = 0.45 / 0.525 = 0.857
Relative fitness of A1A2 = 0.467 / 0.525 = 0.889
Relative fitness of A2A2 = 0.525 / 0.525 = 1.0
b) What are the allele frequencies at equilibrium?
Selection coefficient = 1 – relative fitness
Selection coefficient of A1A1 = 0.143
Selection coefficient of A1A2 = 0.111
Selection coefficient of A2A2 = 0
Frequency of A2 = p = t / (s+t) ; t = generation; s =selection coefficient.
P = 1/ (0.143+1)= 0.87
Frequency of A1= q = 1 - 0.87 = 0.13
c) Since the two generations a year and the fecundity just reverses, so the overall relative fitness remain the same, slowly the allele A2 will persist.
