The frequency of the caramel allele is 0.8 and the frequency of the gray allele

Question

The frequency of the caramel allele is 0.8 and the frequency of the gray allele is 0.2. Given the previous statement, which of the following statements is correct (DF 1, Chi-Square p0.05 = 3.84) a. A population with 56 caramel cats, 48 gray cats, and 95 caramel-gray cats is in Hardy Weinberg equilibrium. b. A population with 43 caramel cats, 25 gray cats, and 53 caramel-gray cats is in Hardy-Weinberg equilibrium. c. A population with 127 caramel cats, 9 gray cats, and 63 caramel-gray cats is in Hardy-Weinberg equilibrium. d. A population with 65 caramel cats, 22 gray cats, and 85 caramel-gray cats is in Hardy-Weinberg equilibrium.

Explanation / Answer

(c) A population with 127 caramel cats, 9 gray cats, and 63 caramel-gray cats is in Hardy-Weinberg equilibrium.

Frequency of Caramel(c)= 0.8

Frequency of Gray(g)=0.2

p2= f(cc)= 0.8*0.8 =0.64

q2=f(gg)=0.2*0.2=0.04

2pq=2*0.8*0.2= 0.32

Multiplying the total population with genotype above, will give Expected Population, as follows:

(c) Caramel cats (cc)=199* 0.64 =127

Gray cats (gg) = 199*0.04 =8 (Round off value)

Caramel-Gray cats(cg) = 199*0.32 = 64 (Round off value)

When we compare the above obtained Expected population of each genotype to the Observed population given in Question, we see both are same approximately.

Observed data is: cc= 127, gg= 9, and cg=63. Hence the population is in Hardy Weinberg Equilibrium.

In all the rest options, there is great difference between observed and expected values.

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