Advertisements
Advertisements
Question
A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:
The number of homozygous dominant fruit flies.
Advertisements
Solution
In Hardy Weinberg, p2 + 2pq + q2 = 1, where 'p2' is the frequency of the homozygous dominant genotype (AA), '2pq' is the frequency of the heterozygous genotype (Aa), and 'q2' is the frequency of the homozygous recessive genotype (aa).
Given:
q = 0.4
We know p + q = 1
p = 1 − q
= 1 − 0.4
= 0.6
number of homozygous dominant fruit flies is As,
p2(AA) = (0.6)2
= 0.36
= 0.36 × 200
= 72
APPEARS IN
RELATED QUESTIONS
p2 + 2pq + q2 = 1. Explain this algebraic equation on the basis of Hardy Weinberg's principle.
Very short answer question.
State the Hardy – Weinberg equilibrium.
In a population, Hardy Weinberg equilibrium is disturbed by following factors EXCEPT ______.
Hardy Weinberg's principle explains ______
For the MN-blood group system, the frequencies of M and N alleles are 0.7 and 0.3, respectively. The expected frequency of MN-blood group bearing organisms is likely to be ______.
State and explain any three factors affecting allele frequency in populations.
Write Hardy Weinberg's equation.
Give a mathematical expression for Hardy Weinberg's principle.
A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:
The number of homozygous recessive fruit flies.
Where do a large proportion of recessive alleles exist in a population at Hardy-Weinberg equilibrium?
