Question #46586
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Answer on Question #46586, Biology, Other
Question. If the homozygous recessive genotype makes up 9% of the population and the heterozygous genotype makes up 42% of the population, what is the frequency of the dominant allele in that population?
Solution. The structure of the population is: f(AA)+f(Aa)+f(aa)=1.
If the frequency of aa genotype is 0,09 and frequency of Aa genotype is 0,42 thus the frequency of AA genotype is: 1-0,09-0,42=0,49.
f(AA) =0,49.
The frequency of the dominant allele in that population we may calculate by the next way:
pA = f(AA) + 0,5 * f(Aa).
(0,5 - because the heterozygous have only one dominant allele).
Thus pA=0,49+0,5*0,42=0,7.
Answer. The frequency of the dominant allele in the population is 0,7 or 70%.
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#340153 on Dec 2023
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allele not genotype. (genotype % below) out of 100 people 9% homo reces = 9 people rr 42% hetero = 42 people w/ Rr ?% homo dom?= ? people w/ RR Genotype = 2 alleles. allele frequency = # of allele / all possble alleles at that locus 9 aa = so two aa per person. 2x 9 = 18 total a alleles 42 Aa = one a per person AND one A per person 42 a and 42 A 42% + 9% + X% = 1. (hardy w.) so AA = 49%. sample = 100 then 49 people have AA. 49 people AA = two A per person. 49 x 2 = 98 A total. + 42 = 140/200 ales