Hardy weinberg equ'm
In a population of 1000 plants, the frequency of the a allele is 5 percent. Suppose a fire causes the loss of 500 individuals who are homozygous for the A allele. In this case the fire caused the
frequency of a to change from ______ to ______.
so, the first one should be 0.05 (5%)
I am not sure how to calculate the second one.
a = 0.05
p + q = 1
p+ 0.05 = 1
p = 0.95
(0.95)^2= AA?
any hint ? thx
To calculate the change in frequency of the a allele after the fire, you can use the Hardy-Weinberg equation. The equation is p^2 + 2pq + q^2 = 1, where p represents the frequency of the dominant allele (A) and q represents the frequency of the recessive allele (a).
In this case, you are given that the initial frequency of the a allele (q) is 0.05. To find the frequency of the dominant allele (p), you can subtract q from 1: p = 1 - q = 1 - 0.05 = 0.95.
Now, you need to calculate the new frequency of the a allele after the fire. You are told that 500 individuals homozygous for the A allele were lost. Homozygous individuals are represented by p^2, so you can subtract 500 from the initial number of individuals homozygous for A (p^2 = 500) to find the new number of individuals homozygous for A: p^2 - 500.
Since p^2 + 2pq + q^2 = 1, you can express p^2 - 500 + 2pq + q^2 = 1. Since p = 0.95, you can substitute it into the equation to get (0.95)^2 - 500 + 2(0.95)(q) + q^2 = 1.
Simplifying this equation, you can solve for q to find the new frequency of the a allele.
Alternatively, you can calculate q using the formula q = √(1 - p^2). In this case, q will represent the new frequency of the a allele after the fire. Using p = 0.95, you can substitute it into the equation to find q = √(1 - 0.95^2).
Applying either method, you can find the new frequency of the a allele, which will allow you to complete the answer to the question.