Question

2. Assume a population is in Hardy-Weinberg equilibrium for a given genetic autosomal trait. What proportion of individuals in the population are heterozygous for the gene if the frequency of the recessive allele is 1%.

Answer 1

Question

2. The population is in Hardy-Weinberg equilibrium

Let's A is the dominant allele

a is the recessive allele

Frequency of the recessive allele (a) =1 percent

=1/100= q

Therefore frequency of dominant allele is (A) =99 percent

=99/100=p

Please note that according to Hardy Weinberg equilibrium

p +q= 1

Where p is frequency of dominant allele and q is the frequency of recessive allele

The proportion of individual that are heterozugous are 2pq

=2* 1/100*99/100= 1.98