1. (5 points totall An online survey found that of 4500 individuals of European

Question

1. (5 points totall An online survey found that of 4500 individuals of European ancestry, A said they could smell asparagus in their urine and B said they could not. a. If the S allele for odor detection is dominant over the G allele for lack of odor detection, and C individuals are heterozygotes, what are the genotype frequencies for this locus, assuming random mating? (1 point) What are the allele frequencies? (2 points) Is the population in Hardy-Weinberg equilibrium? How do you know? (2 points) b. c.

Explanation / Answer

Answer:

Based on the given information:

a) What are genotype frequencies for this locus, assuming random mating.

b) What are the allele frequencies?

c) Is the population in Hardy-Weinberg Equilibrium? How do you know?

Expected genotype frequencies if the population is in Hardy-Weinberg equilibrium can be determined as below:

We can perform a Chi Square test to determine if the population is in Hardy-Weinberg equilibrium or not.

Degree of freedom = Number of categories - 1 = 3 - 1 = 2

P-value corresponding to a Chi square value of 0.003928 is P-Value = 0.998052. The result is not significant at p < 0.05.

Since P-value is greater than the significance level 0.05, therefore, the Null Hypothesis is not rejected and the population is in Hardy-Weinberg equilibrium.

Genotype Observed Number (O) Expected Number E O-E (O-E)^2/E SS 1675 1676 -1 0.000597 GG 683 684 -1 0.001462 SG 2142 2140 2 0.001869 Total 4500 4500 Chi Square (sum) 0.003928

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