Question

Equations:

p + q=1

p² + 2(p)(q) + q²

1) In a population with Hardy-Weinberg equilibrium, the frequency of a dominant allele for a gene is 85% (0.85) while the frequency of the recessive allele for the same gene is 15% (0.15).

0.0005

0.013

0.675

0.723

0.255

0.50

0.023

2) proportion of individuals will be homozygous recessive?

0.0005

0.013

0.675

0.723

0.255

0.50

0.023

3) proportion of individuals will be homozygous dominant?

0.0005

0.723

0.013

0.675

0.255

0.50

0.023

4) proportion of individuals will be heterozygous?

this is all the information I have

Answer 1

P=0. 85, q=0.15

Ans2) homozygous recessive=q2 =. 15*.15=0.023

Ans3)homozygous dominant =p2=. 85*.85=.0723

Ans4)heterozygous=2pq=2*.85*.15=0.255