Question

In a certain group of African people, 4% are born with sickle-cell disease (homozygous recessive). In this case, only homozygous recessive individuals have sickle-cell disease. Heterozygous individuals not only don't have sickle-cell disease, they are resistant to malaria. If this group is in Hardy-Weinberg equilibrium, what percentage of the population is heterozygous and resistant to malaria?

Answer 1

32% of the population is heterozygous and resistant to malaria.

Step-by-step explanation:

The population is in Hardy-Weinberg equilibrium, and according to the H-W principle:

a) Allele frequencies in a population are stable and is constant from generation to generation.

b) The gene pool remains constant.

c) The sum total of all the allelic frequencies is 1

So mathematically this statement can be written as:

p2 + 2pq + q2 = 1

or p+q = 1

Where frequency of occurrence of allele "A" is p and of "a" is q (fothe frequr a gene of two alleles A and a)

p2 is the frequency of AA individuals

q2 is the frequency of aa individuals

2pq is the frequency of Aa individuals

Now in the question it is clearly given that 4% of the population has homozygous recessive disease condition, so q2 = 0.04; it implies q = 0.2

As we know p+q =1; so p = 0.8

The percentage of the population who are heterozygous can be given by 2pq = 2 * 0.2 * 0.8 = 0.32

=> 32% of the population is carrier for sickle cell anemia and at the same time resistant to malaria also.