Answer:
1. 2.8%
2. 97.1%
Step-by-step explanation:
- If the birth rate of this disorder is 1 in every 5000 births it means that the frequency of the recessive homozygous genotype is 1/5000=0.0002
If we name the gene for the disorder with A, than the genotypes are aa (homozugous recessive), Aa (heterozygous, carrier), AA (dominant homozygous).
According to the Hardy-Weinberg equation:
frequency of AA is p2 (P)
frequency of Aa is 2pq
frequency of aa is q2 (Q) and
p2+2pq+q2=1 (P+Q=1)
Q=0.0002 P=1-0.0002=0.9998
q=
0.0002=0.014
p=
0.9998=0.999
2pq=0.028=2.8%
2. According to the Hardy-Weinberg equation:
frequency of AA is p2 (P)
frequency of Aa is 2pq
frequency of aa is q2 (Q) and
p2+2pq+q2=1 (P+Q=1)
The frequency of the recessive homozygous genotype (Q) is 1/5000=0.0002
Q=0.0002
q2=0.0002 2pq=0.028 p2=?
p2= 1-0.0002-0.028 =0.9718=97.1%