Question

Pigment of chicken feathers is regulated by two genes, the gene for feather pigment C, and a gene that inhibits pigment production, I. A mother chicken, who is homozygous recessive for gene c, and heterozygous for gene I, is crossed with a father chicken, who is heterozygous for the pigment gene, and heterozygous for gene I. • Inhibition is dominant to uninhibited, pigmented feathers are dominant to unpigmented feathers. • The inhibitor gene (I) will prevent expression of pigmented chicken feathers. Here is a DihybridCrossReview.pdf reviewing the FOIL method for determining gametes and performing a dihybrid cross if you need a refresher. A) What are the potential genotypes of the gametes produced by the mother for these two genes? • Use the letters C, c, I, and i for the genes. C is for color and I is for inhibitor • Enter your answer as a list of haplotypes separated by spaces such as AB ab Ab aB. Do not write a sentence. • Do not list duplicate haplotypes. • Alphabatize your haplotypes, e.g. Ab, not bA. B) What are the potential genotypes of the gametes produced by the father for these two gametes? • Use the letters C, c, I, and i for the genes. C is for color and I is for inhibitor • Enter your answer as a list of haplotypes separated by spaces such as AB ab Ab aB. • Do not write a sentence. • Do not list duplicate haplotypes. • Alphabetize your haplotypes, e.g. Ab, not bA. C) What is the probability of one of the offspring of these two chickens having pigmented feathers? % • Enter your answer as a percent, but do not include the percent symbol, e.g. 75, not 75%, 0.75, or 3/4 • Report your answer to one decimal place, e.g. 66.7, not 66.66 or 66.60. D) In a population of 150 chickens, produced from the same parents, how many chickens would you expect to have white feathers? • Round your answer to the lowest whole number since you cannot have a partial chicken, e.g. 65.8 would round to 65, as would 65.4, 65.5.

Answer 1

Mother = ccIi

Father = CcIi

A. cI, ci

B. CI, Ci, cI, ci

Gametes	cI	ci
CI	CcII	CcIi
Ci	CcIi	Ccii
cI	ccII	ccIi
ci	ccIi	ccii

C. 2/8 (Ccii, ccii)

D. 6/8 (white feathers)

# = 6/8 × 150 = 112.5

Around 112 - 113

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