1.
Frequency of dominant allele = p=0.5196
Frequency of recessive allele= q=0.4804
% homozygote dominant= p^2=27%
% homozygote recessive= q^2=23%
% heterozygotes = 2pq= 50%
2.
p=0.7
q= 0.3
p^2=49%
q^2=9%
2pq= 42%
3.
p=0.409
q= 0.591
p^2=16.7%
q^2= 35%
2pq= 48.3%
4.
p=0.99
q= 0.01
p^2=98.01%
q^2= 0.01%
2pq= 1.98%
Explanation:
1. The percentage of homozygous dominant organism is expressed as p^2 and its value in this question is 27%. Using the information, we can find the frequency of the dominant allele (p)
p^2= 0.27
p= √0.27= 0.5196
After we get dominant allele frequency we can find recessive allele frequency
p+q=1
q= 1- 0.5196
q= 0.4804
With recessive allele frequency and dominant allele frequency, we can find the heterozygote and homozygote recessive.
homozygote recessive= q^2= (0.4804)^2= 0.2307= 23%
heterozygote = 2pq= 2(0.4804)(0.5196)=0.499= 50%
2.
Only an organism with homozygote recessive genotype will show symptoms and signs of sickle-cell disease, so 9% is the percentage of q^2.
q^2= 0.09
q= √0.09= 0.3
p+q=1
p=1-q
p=1-0.3= 0.7
p^2= (0.7)^2= 0.49= 49%
2pq= 2(0.7)(0.3)= 0.42 = 42%
3.
The 35% population of the moth is homozygote recessive (q^2). We can use the same step as problem 2 then
q^2= 0.35
q=√0.35= 0.591
p+q=1
p=1-q
p=1-0.591= 0.409
p^2= (0.409)^2=0.1672= 16.7%
2pq= 2(0.409)(0.591)= 0.4834 = 48.3%
p=0.409
q= 0.591
p2=16.7%
q2= 35%
2pq= 48.3%
4.
Cystic fibrosis is a homozygous recessive disease with a population rate of 1/10,000. That means the percentage of the population with q^2 gene will be: 1/10000= 0.01%
Then, the rest will use same step as question 2
q^2= 0.0001
q=√0.0001= 0.01
p+q=1
p=1-q
p=1-0.01= 0.99
p^2= (0.99)^2=0.9801= 98.01%
2pq= 2(0.99)(0.01)= 0.0198 = 1.98%
