Color in McIver beetles is determined by two alleles at one gene locus. McIver beetles must have a heterzygous genotype (genotyp
1 answer:
Answer: The frequency of brown beetles is 0.32.
Explanation: The frequency of A1 allele is 0.8. As p+q=1, or the sum of dominant and recessive frequencies equals 1 or 100%:
1 - 0.8 = 0.2
In Hardy-Weinberg principle,
2pq represents the frequency of heterozygote individuals, so:
genotype A1A2 = 2*0.8*0.2 = 0.32.
Thus, the frequency of brown beetles (A1A2) in the population is 0.32.
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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!