Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

the allele for brown eye color is dominant over the allele for blue eye color. i

ID: 217764 • Letter: T

Question

the allele for brown eye color is dominant over the allele for blue eye color. in a population of 15000 individuals 1. The allele no dimples. Determine the frequency many individuals would be expected to be ncy of the dominant and recessive allele in this population. Ho for dimples (D) is dominant. In a randomsple allele in this po many indivi 872 5 of 2000 individuals,1745 individuals had r d dominan terozygous for dimoygous dominant for dimples? How many individuals 2. 8.25 (0) +2C 02) 433 Ga : .07 he allele for brown eye color is dominant over the allele for blue eye color. In a population of 15,000 individuals, 25%were found to have blue eyes. How many individuals in this population are R :Brow homozygous and heterozygous for brown eyes? 5.000

Explanation / Answer

ANSWER 1.

As per Hardy-Weinberg Equilibrium, let
p = Frequency of Dominant Allele
q = Frequency of Recessive Allele
Also p + q = 1
Now,
p2 = Frequency of Homozygous Dominant Genotype
2pq = Frequency of Heterozygous Dominant Genotype
q2 = Frequency of Homozygous Recessive Genotype

Since allele for dimples (D) is dominant, then
q2 = Frequency of Homozygous Recessive Genotype = 1745/2000 = 349/400 = 0.8725
Therefore q = 0.934
Therefore p = 1 - q = 1 - 0.934 = 0.066
Therefore, Frequency of Homozygous Dominant Genotype = p2 = 4.345 * 10-3
Therefore, Number of Homozygous Dominant Genotype = p2 * 2000 = 4.345 * 10-3 * 2000 = 9
Therefore, Frequency of Heterozygous Dominant Genotype = 2pq = 0.123
Therefore, Number of Heterozygous Dominant Genotype = 2pq * 2000 = 0.123 * 2000 = 246

ANSWER 2.


As per Hardy-Weinberg Equilibrium, let
p = Frequency of Dominant Allele
q = Frequency of Recessive Allele
Also p + q = 1
Now,
p2 = Frequency of Homozygous Dominant Genotype
2pq = Frequency of Heterozygous Dominant Genotype
q2 = Frequency of Homozygous Recessive Genotype

Allele for Brown eye colour is dominant over blue eye colour.
q2 = Frequency of Homozygous Recessive Genotype = 25% = 0.25
Number of Homozygous Recessive Genotype = 0.25 * 15,000 = 3,750
Therefore q = (0.25)1/2 = 0.5
Therefore p = 1 - q = 1 - 0.5 = 0.5
Therefore, Frequency of Homozygous Dominant Genotype = p2 = 0.5 * 0.5 = 0.25
Therefore, Number of Homozygous Dominant Genotype = p2 * 15000 = 3,750
Therefore, Frequency of Heterozygous Dominant Genotype = 2pq = 2 * 0.5 * 0.5 = 0.5
Therefore, Number of Heterozygous Dominant Genotype = 2pq * 15000 = 7,500