Respuesta :
Answer:
The correct answer is:
0.25
Explanation:
Since, the frequency of dominant yellow-flowered plants is 50% (i.e. 50/100 = 0.5) therefore, the frequency of dominant allele is 0.5 and recessive allele is 0.5 .
Lets say dominant allele is P and recessive allele is q.
According to the Hardy-Weinberg equilibrium, the frequency of dominant yellow-flowered plants is determined by the following formula:
2Pq
Substituting the frequency of dominant and recessive allele in the above formula: 2(0.5)(0.5) = 0.5
Similarly, according to the Hardy-Weinberg equilibrium, the frequency of homozygous recessive genotype in the population is determined by the following formula:
q2
Substituting the frequency of dominant and recessive allele in the above formula: (0.5)2 = (0.5)(0.5) = 0.25
In this way, the frequency of homozygous recessive allele is calculated with the help of Hardy-Weinberg equilibrium.
A population in the Hardy-Weinberg equilibrium, will have a constant
genetic variation from one generation to the next.
- The frequency of the h o m o z y g o u s recessive genotype in the population is; 0.25.
Reasons:
The given frequency of the dominant h o m o z y g o u s flowered plant = 50% = 0.5
The Hardy-Weinberg equilibrium model is presented as follows;
p² + 2·p·q + q² = 1
p + q = 1
Where;
p = The frequency of the dominant allele
q = The frequency of the recessive allele
p² = The dominant h o m o z y g o u s frequency
q² = The recessive h o m o z y g o u s frequency
Required:
The frequency of the h o m o z y g o u s recessive genotype.
Solution:
Frequency of the dominant h o m o z y g o u s flowered plant, p = 0.5
Therefore;
q = 1 - p = 1 - 0.5 = 0.5
Which gives;
q² = 0.5² = 0.25
The frequency of the h o m o z y g o u s recessive genotype, q² = 0.25
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