Answer:
The correct answer is "40.6%". A further explanation is provided below.
Explanation:
The given values are:
The proportion of people with resistance is,
aa = 8%
or,
= 0.08 (q²)
q² = 0.08
q = √0.08
= 0.28284
Now,
⇒ [tex]p+q=1[/tex]
⇒ [tex]p=1-q[/tex]
[tex]=1-0.28284[/tex]
[tex]=0.717157[/tex]
hence,
For the HIV resistant allele, the proportion of people will be:
= [tex]Aa[/tex]
= [tex]2pq[/tex]
On substituting the values, we get
= [tex]2\times 0.71715\times 0.28284[/tex]
= [tex]0.4056[/tex]
= [tex]40.6[/tex] (%)
The proportion of the population with the carrier of HIV-resistant allele is 40.56%.
Hardy Weinberg equilibrium is the relationship between the variation from one generation to the next. The equation for the relationship can be given as:
[tex]p^2+2pq+q^2=1[/tex]
Where the proportion of dominant alleles is p².
The proportion of recessive alleles is q².
The proportion with the caries are 2pq.
The given recessive allele proportion in the population is 0.08. The proportion of dominant alleles can be given as:
[tex]q^2=0.08\\q=0.2828[/tex]
The dominant alleles can be given as:
[tex]p+q=1\\p+0.28284=1\\p=0.717157[/tex]
The carriers in the population can be given as:
[tex]\text{Carriers}=2pq\\\text{Carriers}=2\;\times\;0.717157\;\times\;0.2828\\\text{Carriers}=0.4056\\\text{Carriers}=40.56\%[/tex]
Thus, the carriers for the HIV gene in the population are 40.56%.
Learn more about Hardy Weinberg equilibrium, here:
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