Join thousands of students who trust us to help them ace their exams!Watch the first video
Multiple Choice
In a population in Hardy-Weinberg equilibrium, 1% of the individuals in a population show the recessive trait of a certain characteristic. In this situation, what is the value of p?
A
0.99
B
0.81
C
0.18
D
The answer cannot be determined from the information given.
E
0.9
Verified step by step guidance
1
Understand that Hardy-Weinberg equilibrium involves the equation \( p^2 + 2pq + q^2 = 1 \), where \( p \) is the frequency of the dominant allele and \( q \) is the frequency of the recessive allele.
Recognize that the percentage of individuals showing the recessive trait corresponds to \( q^2 \). In this problem, 1% of the population shows the recessive trait, so \( q^2 = 0.01 \).
Calculate \( q \) by taking the square root of \( q^2 \). Therefore, \( q = \sqrt{0.01} \).
Use the relationship \( p + q = 1 \) to find \( p \). Substitute the value of \( q \) into the equation to solve for \( p \).
Substitute the calculated value of \( q \) into \( p = 1 - q \) to find the value of \( p \).