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Answer:
The correct answer would be:
- Frequency of recessive allele (q) = 0.2 or 20 percent
- Frequency of dominant allele (p) = 0.8 or 80 percent
- The frequency of homozygous dominant individuals (p²) = 0.64 or 64 percent.
- The frequency of homozygous recessive individuals (q²) = 0.04 or 4 percent.
- Frequency of heterozygous dominant indivduals (2pq) = 0.32 or 32 percent.
It can be explained with the help of the Hardy-Weinberg equation:
p + q = 1
p² + q² + 2pq = 1
q² represent the frequency of individuals with homozygous recessive genotype. It can be calculated by dividing the number of individuals with homozygous recessive genotype with a total number of individuals in a population.
Thus, q² = [tex]\frac{64}{1600}[/tex]
⇒[tex]\frac{4}{100}[/tex] ⇒ 0.04
Hence, q = [tex]\sqrt{0.04}[/tex] = 0.2
As we know p + q = 1
thus, p = 1 - 0.2 = 0.8
Hardy-Weinberg equilibrium states the constant relation of the genetic variation in population in absence of disturbing factors. The frequencies are p = 0.8, q = 0.2, [tex]\rm p^{2}[/tex] = 0.64, 2pq = 0.32 and [tex]\rm q^{2}[/tex] = 0.04.
What is Hardy-Weinberg equilibrium?
It states the frequencies of the recessive, dominant, heterozygous and purebred individuals in a population.
The Hardy-Weinberg equation can be shown as,
[tex]\begin{aligned} \rm p + q &= 1\\\\\rm p^{2} + q^{2} + 2pq &= 1\end{aligned}[/tex]
Here [tex]\rm q^{2}[/tex] represents the frequency of purebred recessive allele and can be estimated as:
[tex]\begin{aligned} \rm q^{2} &= \dfrac{64}{1600}\\\\&= 0.2\end{aligned}[/tex]
As from the above equation we know, p + q = 1
So, substituting the value of q in the equation, p = 0.8
Therefore, the frequencies are p = 0.8, q = 0.2, [tex]\rm p^{2}[/tex] = 0.64, 2pq = 0.32 and [tex]\rm q^{2}[/tex] = 0.04.
Learn more about Hardy-Weinberg equilibrium here:
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