Pigment in the mouse is produced only when the \(C\) allele is present. Individuals of the \(c c\) 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 \(F_{1}\) and \(F_{2}\) 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

To determine the F1 genotypes, we will do a cross between AACC and aace genotypes. -For the color gene, the cross is between CC and cc, resulting in all offspring being Cc. -For the color type gene, the cross is between AA and aa, resulting in all offspring being Aa. So, the F1 offspring will have the genotype AaCc.

As previously stated, a color is produced only if at least one allele C is present. In our case, F1 offspring are AaCc. Therefore, since the genotype Aa results in agouti color, all F1 offspring will have an agouti phenotype. The F1 phenotypic ratio is 100% agouti.

To determine the F2 genotypes, we will cross two F1 individuals, both with the genotype AaCc. Both A/a and C/c will have a 1:1 ratio in the gametes. So we can organize this cross in a two-factor Punnett square to find the F2 genotypes, which are: - 1 AACC - 2 AaCC - 1 aaCC - 2 AACc - 4 AaCc - 2 aaCc - 1 ACcc - 2 Aacc - 1 aacc >Total: 16 individuals

Now, based on the genotypes obtained, let's determine the F2 offspring phenotype: - 1 AACC, 2 AaCC, 2 AACc, and 4 AaCc: All C allele present, these individuals have an agouti phenotype. Total: 9 Agouti - 1 aaCC, 2 aaCc: All C allele present, these individuals have a black phenotype. Total: 3 Black - 1 ACcc, 2 Aacc, 1 aacc: No C allele present, these individuals have no color. Total: 4 Colorless Therefore, the F2 phenotypic ratio is 9 agouti : 3 black : 4 colorless. (b)

We have to determine the genotypes of the female parents, so we'll check the given three phenotypic ratios, one by one: (1) 8 agouti (2) 9 agouti (3) 4 agouti 8 colorless 10 black 5 black 10 colorless

(1) 8 agouti: Since all offspring are agouti, the female parent should have an agouti phenotype and hence have the genotypes AA or Aa. Also, since no colorless and black mice are produced, the female must have at least a CC genotype. The female genotype likely is AAC-. (2) 9 agouti: A similar situation to the previous case, as no black or colorless offspring are produced, the female parent should have an agouti phenotype with at least a CC genotype. The female genotype likely is AAC-. (3) 4 agouti, 8 colorless, 10 black, 5 black, 10 colorless: This phenotypic ratio suggests that the female parent produces colorless, black, and agouti offspring. Therefore, a comprehensive genotype for the female parent in this case is A-Cc or Aacc. Since the female parent must be agouti, only AaCc remains valid. The female genotype is AaCc.