A human population of 1000 people includes 200 people who have albino skin (rece
ID: 87304 • Letter: A
Question
A human population of 1000 people includes 200 people who have albino skin (recessive homozygotes, aa).. It also has 200 carriers (heterozygotes, Aa) of albinism (tested by DNA analysis), who are normal-skinned. What is the frequency of heterozygous genotype in this population?
a) 0.8
b) 0.6
c) 0.4
d) 0.2
In the same population, what is the frequency of recessive allele (a)?
a) 0.8
b) 0.7
c) 0.3
d) 0.2
In the same population, what is the frequency of dominant allele (A) ?
a) 0.8
b) 0.7
c) 0.3
d) 0.2
What would be the expected frequency of recessive homozygous genotype if this population was in Hardy-Weinberg equilibrium?
0.70
0.42
0.30
0.09
What would be the expected frequency of heterozygous genotype if this population was in Hardy-Weinberg equilibrium?
0.70
0.42
0.30
0.09
a.0.70
b.0.42
c.0.30
d.0.09
Explanation / Answer
Ans. Given,
aa = 200
Aa = 200
So, AA = 1000 – (aa + Aa) = 1000 – (200 + 200) = 600
#1. Frequency of heterozygote genotype = No. of heterozygote (Aa) / Population size
= 200 / 1000
= 0.2
Correct option. D
#2. Allelic frequency = (No. of given allele / Total number of alleles in population)
Being diploid each individual has 2 alleles of the gene. The homozygotes have two copies of the respective allele, whereas the heterozygotes have 1 copy of both the alleles. That is, AA individuals have 2 copies of allele A, whereas Aa individuals have one copy each of allele A and a.
Allelic frequency of allele a, f(a) =
(No. of Allele a in population / Total no. of alleles for the trait)
= (2 x no. of aa + 1 x no. of Aa) / (2 x 1000)
= (2 x 200 + 1 x 200) / 2000
= 600/ 2000
= 0.3
Correct option. C.
#3. Allelic frequency of allele A, f(A) =
(No. of Allele A in population / Total no. of alleles for the trait)
= (2 x no. of AA + 1 x no. of Aa) / (2 x 1000)
= (2 x 600 + 1 x 200) / 2000
= 1400/ 2000
= 0.7
Correct option. B.
#4. Hardy- Weinberg Equation for 2 allele system (for a population at HW equilibrium) is given by-
p + q = 1 - equation 1
(p + q)2 = p2 + q2 + 2pq = 1 - equation 2
Where,
p = allelic frequency of allele dominant allele
q = allelic frequency of allele recessive allele
p2 = genotypic frequency of homozygous dominant
q2 = genotypic frequency of homozygous recessive
2pq = genotypic frequency of heterozygote
Now,
Expected frequency of recessive homozygous genotype (aa) = q2 = [f(a)]2
= (0.3)2
= 0.09
Correct option. D
#5. Expected frequency of heterozygous genotype (Aa) = 2pq
= 2 x f(A) x f(a)
= 2 x 0.7 x 0.3
= 0.42
Correct option. B
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