Question

Pigment in the mouse is produced only when the \(C\) allele is pres- ent. Individuals of the ce genotype have no color, If color is present, it may be determined by the \(A\) and \(a\) alleles. AA or Aa results in agouti color, whereas aa results in black coats. (a) What \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) genotypic and phenotypic ratios are obtained from a cross between \(A A C C\) and aace mice? (b) In the three crosses shown here between agouti females whose genotypes were unknown and males of the aacc genotype, what are the genotypes of the female parents for each of the following phenotypic ratios? (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 colorless 10 black \(\quad 5\) black 10 colorless

Short answer

(a) F1 genotype is AaCe, and F1 phenotype is agouti. (b) F2 genotypic ratio is 1AACC:4AACe:6AaCC:4AaCe:2aaCC:4aaCe:1aaee, and the phenotypic ratio is 15 agouti:6 black:1 colorless. (c) The genotypes of the female parents for different phenotypic ratios are: (1) AaCc for 8 agouti (2) AaCe for 9 agouti (3) Aace for 4 agouti: 8 colorless: 10 black.

Step-by-step solution

Cross the parent generation to obtain F1

Cross the parent genotypes \(A A C C\) and aace to obtain the F1 generation. It's easier to deal with the two alleles separately: \(A A \times aa\) results in all offspring carrying the Aa genotype for the first gene. \(C C \times ce\) results in all offspring carrying the Ce genotype for the second gene. Thus, the F1 genotype for their offspring will be AaCe.

Cross the F1 generation to obtain F2

Cross AaCe x AaCe to obtain F2 genotypic ratios: For the first gene (A, a): Aa x Aa produces: 1 AA, 2 Aa, 1 aa (1:2:1 ratio) For the second gene (C, e): Ce x Ce produces: 1 CC, 2 Ce, 1 ee (1:2:1 ratio) Now, we will combine the results from these crosses: 1:2:1 × 1:2:1 = 1:4:6:4:2:4:1:2:1 The genotypic ratio for the F2 offspring is 1AACC:4AACe:6AaCC:4AaCe:2aaCC:4aaCe:1aaee.

Determine the F2 phenotypes based on the genotypic ratios

According to the problem: 1. Color is determined by A and a alleles only if the C allele is present. 2. AA or Aa results in agouti color, whereas aa results in black coats. Therefore, we can list the phenotypes using the given information: 1 AACC: Agouti 4 AACe: Agouti 6 AaCC: Agouti 4 AaCe: Agouti 2 aaCC: Black 4 aaCe: Black 1 aaee (Colorless): Colorless The phenotypic ratio for the F2 offspring is 15 agouti:6 black:1 colorless. (a) F1 genotype: AaCe F1 phenotype: Agouti (b) In all the crosses, the male parent has aacc genotype, and we know that the female parents are agouti. Thus, the female parent must have one A allele and at least one C allele.

Determine Female Parent Genotypes for the Phenotypic Ratios

(1) 8 agouti: Male genotype: aacc Possible female genotype: AaCc (2) 9 agouti: Male genotype: aacc Possible female genotype: AaCe (Self-cross AaCe x AaCe to verify that this combination can produce the 9 phenotypic ratios) (3) 4 agouti: 8 colorless 10 black: Male genotype: aacc Possible female genotype: Aace (Self-cross AaCe x AaCe to verify that this genotype combination can produce the 4 agouti: 8 colorless: 10 black phenotypic ratio) Therefore, the genotypes of the female parents are: (1) AaCc (2) AaCe (3) Aace