Question

Suppose that individuals in a population of annual plants have leaves that are divided into leaflets, toothed along the edges, or smooth along the edges. The degree to which the leaves are divided is determined by a single gene with two alleles. The genotype DD results in leaves with smooth edges, the genotype Dd results in leaves with toothed edges, and the genotype dd results in divided leaves. Calculate the genotype frequencies of a population that contains 128 individuals with smooth leaves, 46 individuals with toothed leaves, and 4 individuals with divided leaves. Express the frequencies to three decimal places. Assume that this generation of individuals is in Hardy–Weinberg equilibrium

DD genotype frequency:

Dd genotype frequency:

dd genotype frequency:

Using the genotype frequencies, calculate the allele frequencies for D and d. Express each frequency to two decimal places.

D allele frequency:

d allele frequency:

Answer 1

Smooth leaves (DD)-128
Toothed leaves (Dd)- 46
Divided leaves(dd)- 4

Total number of individual plants = 128+46+4= 178
Genotype frequency of DD = 128/178= 0.719
Genotype frequency of Dd = 46/178=0.258
Genotype frequency of dd=4/178=0.022

According to Hardy Weinberg Equation = p+q=1

Where p is the allele frequency of the dominant allele ( here it is D)

and q is the frequency of the recessive allele ( here it is d)

(p+q)²=1 = p²+2pq+q²=1

Here p² is the genotype frequency of D = 0.719

so allele frequency of D =p = √ p²=√ 0.719 = 0.85

Similarly q² is the genotype frequency of d = 0.022

so allele frequency of d =q = √ q²=√ 0.022 =0.15