In a population of wild hogs, fur color is determined by alleles at one locus; a

Question

In a population of wild hogs, fur color is determined by alleles at one locus; allele A for black, and allele a for white. The alleles are co-dominant; so, the heterozygotes have a mixed (ie.,roan) phenotype. A sample 500 hogs is evaluated for fur color; 33 white, 172 black, and 295 roan.

a. What are the frequencies of the A and a alleles?

b. Assuming that the population is in Hardy-Weinberg Equilibrium, what is the expected frequency of roan hogs?

c. Assuming that the population is in H-W E, what is the expected number of white hogs?

d. How many degrees of freedom would you use in a Chi-Square test to assess if that population was in Hardy-Weinberg Equilibrium?

e. (2pts) what is the Chi-square value for testing if this population is in Hardy-Weinberg equilibrium?

f. What is the approximate probability (ie., the “P value”) that those observed data do not deviate from the frequencies expected of a population that is in Hardy-Weinberg equilibrium?

g. Is that population in Hardy-Weinberg equilibrium? Choose either “yes” or “no” and briefly explain your choice:

Explanation / Answer

AS PER CHEGG POLICY, PROCEEDING TO ANSWER A MINIMUM OF FIRST QUESTION, OR, FIRST 4 SUB-QUESTIONS / PARTS / SUBPARTS, THAT IS, IN THIS CASE ONLY QUESTION 1.a to 1.d, UNLESS SPECIFIED/INDICATED TO DO SPECIFIC QUESTIONS BY THE QUESTION UPLOADER:

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

Answer a.
Frequencies of the A and a alleles
p2 = Frequency of Homozygous Black Genotype (AA) = 172/500 = 0.344
2pq = Frequency of Heterozygous Roan Genotype (Aa) = 295/500 = 0.590
q2 = Frequency of Homozygous White Genotype (aa) = 33/500 = 0.066
As per Hardy-Weinberg Equilibrium, let
p = Frequency of Black Allele (A) = (172*2 + 295) / 1000 = 0.639
q = Frequency of White Allele (a) = (295 + 33*2) / 1000 = 0.361

Answer b.
2pq = Expected Frequency of Heterozygous Roan Genotype (Aa) = 2 * 0.639 * 0.361 = 0.461
Number of Heterozygous Roan Genotype (Aa) = 0.461 * 500 = 230.5

Answer c.
q2 = Expected Frequency of Homozygous White Genotype (aa) = 0.361 * 0.361 = 0.130
Number of Homozygous White Genotype (aa) = 0.130 * 500 = 65

Answer d.
We can use 2 degrees of freedom in a Chi-Square test to assess if that population was in Hardy-Weinberg Equilibrium because it is (number of possibilities - 1).

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