Ask a New Question

Question

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
4 years ago

Answers

Related Questions

for the hardy-weinberg equation, is p always for the dominant allele and q recessive? is the value... Question 13 Hardy-Weinberg Equilibrium - ScienceAid a q=0.2; p=0.8 b q=0.04; p=0.16 c... Hardy-Weinberg Equilibrium - ScienceAid a q=0.04; p=0.16 b q=0.8; p=0.2 c q=0.2; p=0.8... Hardy-Weinberg Equilibrium - ScienceAid a q=0.2; p=0.8 b q=0.1; p=1.0 c q=0.04; p=0.16... Hardy-Weinberg Equilibrium - ScienceAid Assume that a population is in Hardy-Weinburg equilibrium... The Hardy-Weinberg principle is written as the equation p 2 + 2 pq + q 2 = 1. What does p represent?... 23. The Hardy-Weinberg principle is written as the equation p 2 + 2 pq + q 2 = 1. What does p repr... 25. The Hardy-Weinberg principle is written as an equation: p 2 + 2 pq + q 2 = 1. What does the q... The Hardy-Weinberg equilibrium states that allele and genotype frequencies in a population will ____... Is the hardy Weinberg equilibrium always equal to 1? True or false
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use