Answer:
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.
EXPLANATION:
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.
