In a population that is in Hardy-Weinberg equilibrium, 16% of the individuals show recessive trait. The frequency of dominant allele in the population is:
Correct answer: D. 0.6
- A. 0.36
- B. 0.4
- C. 0.48
- D. 0.6
Explanation
In Hardy-Weinberg equilibrium, the frequency of the recessive allele (q2) can be calculated as 0.16 (since 16% show the recessive trait). To find the frequency of the dominant allele (p), you can use the equation p + q = 1, where q represents the frequency of the recessive allele. So, q2= 0.16 implies q = √0.16 = 0.4. Then, p = 1 - q = 1 - 0.4 = 0.6. Therefore, the frequency of the dominant allele in the population is 0.6.
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