In mice, the waltzing allele (w) that causes the mouse to run in circles due to an inner ear defect is recessive to the non-waltzing allele (W). In an ideal mouse population exhibiting Hardy–Weinberg equilibrium of 150 mice, 15% of the alleles are non-waltzing. What is the frequency of the recessive allele? Record your answer as a whole number percentage.
5 years ago
Answered By Chelsea W
85%
5 years ago
Answered By Maikol S
97.75
5 years ago
Answered By Maikol S
From the first question of Hardy Weinberg equilibrium one may be able to find the percentages for dominant and recessive individuals:
p+q=1
However, from this equation, we can’t determine the percentage of dominant and recessive alleles.
Therefore, we must take into consideration the second equation of Hardy Weinberg equilibrium:
p2 + 2pq + q2 = 1
where:
p2= homozygous dominant
2pq= heterozygous dominant
q2= homozygous recessive
The allele “q” shows up in the places in this equation, and this two must be take into consideration.
Knowing that “p = 15% or 0.15 (from the information in the original question), and that “q = 85% or 0.85” (100% - 15 % = 85%), all we have to do now is substitute the decimal value for “q” (0.85) into the second equation of Hardy Weinberg equilibrium:
5 years ago
Answered By Chelsea W
85%
5 years ago
Answered By Maikol S
97.75
5 years ago
Answered By Maikol S
From the first question of Hardy Weinberg equilibrium one may be able to find the percentages for dominant and recessive individuals:
p+q=1
However, from this equation, we can’t determine the percentage of dominant and recessive alleles.
Therefore, we must take into consideration the second equation of Hardy Weinberg equilibrium:
p2 + 2pq + q2 = 1
where:
p2= homozygous dominant
2pq= heterozygous dominant
q2= homozygous recessive
The allele “q” shows up in the places in this equation, and this two must be take into consideration.
Knowing that “p = 15% or 0.15 (from the information in the original question), and that “q = 85% or 0.85” (100% - 15 % = 85%), all we have to do now is substitute the decimal value for “q” (0.85) into the second equation of Hardy Weinberg equilibrium:
2pq + q2
2(0.15)(0.85) + (0.852) = 0.9775
Or
97.75%
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