
If the frequency of a recessive phenotype in a stable population is 25%, the frequency of the dominant allele in that population is
A. 50%
B. 75%
C. 25%
D. 10%
Answer
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Hint: As we all know the monohybrid cross is a cross between parents who are heterozygous for a single trait. Each trait can express themselves in two forms i.e. dominant and recessive in which the one that expresses itself the most is dominant while the latter is the one that fails to express itself comparatively. Generally, the first generation of progeny shows a ratio of 1:1 in both genotypes as well as phenotype. While the second generation of progeny shows a genotype ratio of 1:2:1 and phenotype ratio of 3:1.
Complete answer:Punnett square
As given, the frequency of a recessive phenotype is 25% (tt).
So 25% is the phenotypic ratio of 3:1 of the F2 generation of a monohybrid cross.
Therefore, the frequency of the dominant phenotype will be 75%.
The dominant phenotype can be heterozygous or homozygous - TT, Tt, and Tt as shown above in the table.
Calculating the frequency of alleles by using Hardy-Weinberg law,
\[p^2\] + 2pq + \[q^2\] = 1
In this \[q^2\] = 25% while \[p^2\] + 2pq = 75%
Therefore, \[q^2\] GT ye = 0.25, so q will be 0.5
Value obtained of q is 0.5, putting the value in the equation of p + q = 1
p = 1 - 0.5 = 0.5 = 50%
Option (A)- 50%
So the frequency obtained of the dominant allele is 50%.
Option (B)- 75%
75% is the frequency of the dominant phenotype as mentioned above but not of the dominant allele.
Option (C)- 25%
25% is the mentioned value in the question.
Option (D)- 10%
Incorrect value.
Thus, the correct option is Option (A)- 50%.
Note: According to the law, p denotes the frequency of the dominant allele while q denotes the frequency of the recessive allele in the population. \[p^2\] denotes the percentage of homozygous dominant individuals while \[q^2\] denotes the percentage of homozygous recessive individuals and 2pq denotes the percentage of heterozygous individuals.
Complete answer:Punnett square
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
As given, the frequency of a recessive phenotype is 25% (tt).
So 25% is the phenotypic ratio of 3:1 of the F2 generation of a monohybrid cross.
Therefore, the frequency of the dominant phenotype will be 75%.
The dominant phenotype can be heterozygous or homozygous - TT, Tt, and Tt as shown above in the table.
Calculating the frequency of alleles by using Hardy-Weinberg law,
\[p^2\] + 2pq + \[q^2\] = 1
In this \[q^2\] = 25% while \[p^2\] + 2pq = 75%
Therefore, \[q^2\] GT ye = 0.25, so q will be 0.5
Value obtained of q is 0.5, putting the value in the equation of p + q = 1
p = 1 - 0.5 = 0.5 = 50%
Option (A)- 50%
So the frequency obtained of the dominant allele is 50%.
Option (B)- 75%
75% is the frequency of the dominant phenotype as mentioned above but not of the dominant allele.
Option (C)- 25%
25% is the mentioned value in the question.
Option (D)- 10%
Incorrect value.
Thus, the correct option is Option (A)- 50%.
Note: According to the law, p denotes the frequency of the dominant allele while q denotes the frequency of the recessive allele in the population. \[p^2\] denotes the percentage of homozygous dominant individuals while \[q^2\] denotes the percentage of homozygous recessive individuals and 2pq denotes the percentage of heterozygous individuals.
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