The Hardy-Weinberg equation is as follows:
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!
Where:
(convert all % to decimals)
p= homozygous dominant
q= homozygous recessive
pq= heterozygous
While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.
Using the 1st equation, we can solve for the other dominant/recessive frequency:
1-0.2=0.8
Meaning that:
p= 0.8 & q=0.2
If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.
2(0.8)(0.2)= 2(0.16)= 0.32
So, the heterozygous frequency would be:
0.32
Hope this helps!
