4. The Rh blood type response system is controlled by the D allele. The genotypes DD and Dd are Rh (Rh positive); dd is Rh- (Rh negative). The Basque people from the border between France and Spain have the highest d allele frequency in the Rh system so far reported. In one study of 400 individuals, 230 were found to be Rh and 170 Rh-. Calculate the frequencies of the D and d alleles, the genotype frequencies, and the proportion of Rh individuals who are heterozygous. (Assume the genotypes are in Hardy-Weinberg equilibrium).

The equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1 - This is according to Hardy-Weinberg.

So p+q =1

p2+ 2pq+q2 =1

where p = frequency of dominant allele

q is the frequency of the recessive allele.

from the question, it was given that there were 170 individual out of 400, which were Rh- negative,

So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

abidhussain7972 abidhussain7972 · Answer 2 · 2020-03-06T05:45:54+01:00

Answer:

A - Frequency of D allele = 0.35

Frequency of d allele = 0.6

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425

C- Frequency of heterozygous population = 400*0.455 = 183

Explanation:

According to Hardy-Weinberg equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1.

So p+q =1

p2+ 2pq+q2 =1

here p = frequency of dominant allele

q is the frequency of the recessive allele.

Given that there we 170 individual out of 400, which were Rh- negative So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.