
There is an irregular mating population. If the frequency of an autosomal recessive lethal gene is 0.4, then the frequency of the carriers in a population of 200 individuals is
A. 30
B. 80
C. 96
D. 104
Answer
571.8k+ views
Hint: In population genetics, the Hardy–Weinberg principle, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
Complete answer:
According to the Hardy-Weinberg principle,
In a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, respectively, the expected genotype frequencies under random mating are f(AA) = p2^ {} for the AA homozygotes, f(aa) = q2^ {} for the aa homozygotes.
And f(Aa) = 2pq for the heterozygotes.
In the absence of selection, mutation, genetic drift, or other forces, allele frequencies p and q are constant between generations, so equilibrium is reached.
The sum of the entries is p2^ {} + 2pq + q2^ {} = 1, as the genotype frequencies must sum to one.
Again p + q = 1, the binomial expansion of (p + q)2^ {} = p2^ {} + 2pq + q2^ {} = 1 gives the same relationships.
Now,
According to Q,
Frequency of autosomal recessive (q) lethal gene is 0.4
∴ Frequency of dominant gene (p) is p + q = 1
⟹ p = 1-q
⟹p = 1- 0.4
⇒p = 0.6
Now, the carrier individual will be 2pq = 2 ×0.4×0.6
= 0. 48
Now, the frequency of carriers in total population = 0.48 X 200
= 96.
So, the correct answer is “Option C. 96”.
Note:The factors that affect Hardy – Weinberg principle are:
a. genetic drift,
b. mate choice,
c. assortative mating,
d. natural selection,
e. sexual selection,
f. mutation,
g. gene flow,
h. meiotic drive,
i. genetic hitchhiking,
j. population bottleneck,
k. founder effect and
L. inbreeding.
Complete answer:
According to the Hardy-Weinberg principle,
In a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, respectively, the expected genotype frequencies under random mating are f(AA) = p2^ {} for the AA homozygotes, f(aa) = q2^ {} for the aa homozygotes.
And f(Aa) = 2pq for the heterozygotes.
In the absence of selection, mutation, genetic drift, or other forces, allele frequencies p and q are constant between generations, so equilibrium is reached.
The sum of the entries is p2^ {} + 2pq + q2^ {} = 1, as the genotype frequencies must sum to one.
Again p + q = 1, the binomial expansion of (p + q)2^ {} = p2^ {} + 2pq + q2^ {} = 1 gives the same relationships.
Now,
According to Q,
Frequency of autosomal recessive (q) lethal gene is 0.4
∴ Frequency of dominant gene (p) is p + q = 1
⟹ p = 1-q
⟹p = 1- 0.4
⇒p = 0.6
Now, the carrier individual will be 2pq = 2 ×0.4×0.6
= 0. 48
Now, the frequency of carriers in total population = 0.48 X 200
= 96.
So, the correct answer is “Option C. 96”.
Note:The factors that affect Hardy – Weinberg principle are:
a. genetic drift,
b. mate choice,
c. assortative mating,
d. natural selection,
e. sexual selection,
f. mutation,
g. gene flow,
h. meiotic drive,
i. genetic hitchhiking,
j. population bottleneck,
k. founder effect and
L. inbreeding.
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