Question

In a large random mating population of plants, 16,800 individuals express the phenotype of the dominant allele (genotypes AA and Aa) and 3200 express the phenotype of the recessive allele (genotype aa).

a. What are the frequencies of the two alleles in this population?

b. How many plants would be expected to heterozygous?

c. You decide to select for the recessive phenotype and pull up 50% of the plants expressing the dominant phenotype. What will be the gene frequencies of A and a in the next generation?

d. If you had pulled up all of the plants expressing the dominant phenotype, what would be the proportions (frequencies) of AA, Aa, and aa individuals in the next generation?

Answer 1

a. According to Hardy-Weinberg equilibrium, p²+q²+2pq = 1 where

p= frequency of dominant allele f(A)

and q is the ferquency of recessive allele f(a)

number of dominant alleles = 16,800 = AA and Aa = p²+2pq

recessive allele = 3200 =aa = q²

The frequency of recessive alleles, q² = (number of recessive alleles / Total number of alleles) = 3200 / (3200+16800) = 0.16; Hence q= 0.4

p+q= 1 ; p= 1- 0.4 = 0.6

So p= frequency of dominant allele, A= 0.6 and q= frequency of recessive allele, a = 0.4

b. number of heterozygous plants = ?

The allele frequency of heterozygous individuals, 2pq = 2x0.4x0.6 = 0.48

48% of the total population is heterozygous.

0.48 x 20,000 = 9600 is the number of heterozygous where 20,000 is the total population.

c. [(0.5)(0.16)(0.6)]/(1-0.5)(0.5*0.16)]= -0.083

that means,

(50% x number of recessive plants x allele frequency of recessive gene) / (1- 50%) (50% of number of recessive plants)

a=0.4-0.083=0.317 and A= 1-0.317= 0.683  

[1-50%, because from the total 50% were pulled up or avoided]

d. if all the dominant plants were pulled up, p² = 0 and p=0

In p+q=1 is q=1 and q² =1, the number of recessive plant

Hence AA = 0, Aa=0 and aa=1