In a population of flowers growing in a meadow, C1 and C2 are autosomal codomina

Question

In a population of flowers growing in a meadow, C1 and C2 are autosomal codominant alleles that control flower color. The alleles are polymorphic in the population, with f (C1) = 0.9 and f (C2) = 0.1. Flowers that are C1C1 are yellow, orange flowers are C1C2, and C2C2 flowers are red. A storm blows a new species of hungry insects into the meadow, and they begin to eat yellow and orange flowers but not red flowers. The predation exerts strong natural selection on the flower population, resulting in relative fitness values of C1C1 = 0.30, C1C2 = 0.60, and C2C2 = 1.0.

a.Assuming the population begins in H?W equilibrium, what is C1 allele frequency after one generation of natural selection?

b.Assuming the population begins in H?W equilibrium, what is C2 allele frequency after one generation of natural selection?

c.Assuming random mating takes place among survivors, what is the genotype C1C1 frequency in the second generation?

d.Assuming random mating takes place among survivors, what is the genotype C1C2 frequency in the second generation?

e.Assuming random mating takes place among survivors, what is the genotype C2C2 frequency in the second generation?

Explanation / Answer

Let the total no. of individuals be 100. Before selection, No. of C1C1 individuals = 0.9^2 = 0.81 or 81% = 81 individuals were yellow.

No. of C2C2 individuals = 0.1*0.1= 0.01=1% i.e. one out 100 individuals was red before selection.

Now, fitness = no. of individuals after selection/ no. of individuals before selection.

Fitness of C1C1 after selection = 0.30 = N after / 81; N after = 81*0.30 = 24.3 individuals will be yellow after selection; C1 = .492 = allele frequency of C1 after selection.

b) You can calculate it by fitness formula also, 1 = N after/ 1; So, N after = 1; C2 = 0.1 = allele frequency of C2 after selection.

c) Genotype C1C1 frequency in second generation = 0.49^2 = 0.24 (means around 24 individuals will be yellow in the second generation).

d) Genotype C1C2 frequency in second generation = 2*0.49*0.1 = 0.098 = 9.8% (around 10 individuals will be orange). The result will be same using fitness values.

e) Genotype frequency of C2C2 = 0.1*0.1 = 0.01 = 1%

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