Question

The following genotypes of two independently assorting autosomal genes determine coat color in rats: \(A-B-(\text { gray }) ; A-b b\) (yellow) \(; a a B-\) (black); \(a a b b\) (cream) A third gene pair on a separate autosome determines whether any color will be produced. The \(C C\) and \(C c\) genotypes allow color according to the expression of the \(A\) and \(B\) alleles. However, the ce genotype results in allbino rats regardless of the \(A\) and \(B\) alleles present. Determine the \(F_{1}\) phenotypic ratio of the following crosses: (a)AAbbCC \(\times\) aaBBcc; (b) \(A a B B c c \times A A B b c c\) (c) \(A a B b C c \times A a B b c c\)

Short answer

What is the F1 phenotypic ratio for the following crosses for a three-gene system determining coat color in rats: (a) AAbbCC x aaBBcc: 100% gray (b) AaBBcc x AABBcc: 100% albino (c) AaBbCc x AaBbcc: 3:3:1:1:4 (gray:black:yellow:cream:albino)

Step-by-step solution

Analyze the given genotypes

Start by writing down the parent genotypes: AAbbCC and aaBBcc. Note that gene A is heterozygous in one parent and homozygous recessive in the other parent, gene B is homozygous in both parents but with different alleles, and gene C is homozygous in both parents but with different alleles.

Determine the genotype probabilities for the offspring

Using the rules of probability, determine the offspring's genotype probabilities for each gene: - For gene A: (Aa) - For gene B: (AB or Ba) - For gene C: (Cc)

Combine the probability to determine the F1 phenotypes

The combined genotype of the offspring will be AaBbCc. According to the provided information on the genotype and phenotype relationship, this genotype will result in gray coat color because the dominant alleles A and B are both present and the genotype Cc allows color according to the gene A and B alleles being present. Therefore, the \(F_{1}\) phenotypic ratio for this cross is 100% gray. ##Cross (b): AaBBcc \(\times\) AABBcc##

Analyze the given genotypes

Start by writing down the parent genotypes: AaBBcc and AABBcc. Note that gene A is heterozygous in one parent and homozygous dominant in the other parent, gene B is homozygous dominant for both parents, and gene C is homozygous recessive for both parents.

Determine the genotype probabilities for the offspring

Using the rules of probability, determine the offspring's genotype probabilities for each gene: - For gene A: (AA or Aa) - For gene B: (BB) - For gene C: (cc)

Combine the probability to determine the F1 phenotypes

The possible genotypes of the offspring are AABBcc and AaBBcc. However, since the genotype cc is homozygous recessive for gene C, it results in albino rats regardless of the A and B alleles present. Therefore, the \(F_{1}\) phenotypic ratio for this cross is 100% albino. ##Cross (c): AaBbCc \(\times\) AaBbcc##

Analyze the given genotypes

Start by writing down the parent genotypes: AaBbCc and AaBbcc. Note that gene A is heterozygous in both parents, gene B is heterozygous in one parent and homozygous recessive in the other parent, and gene C is heterozygous in one parent and homozygous recessive in the other parent.

Determine the genotype probabilities for the offspring

Using the rules of probability, determine the offspring's genotype probabilities for each gene: - For gene A: (AA, Aa, aA, or aa) - For gene B: (BB, Bb, bB, or bb) - For gene C: (CC, Cc, or cc)

Combine the probability to determine the F1 phenotypes

There are multiple possible genotypes for the offspring, so the phenotypic outcome will vary based on the final combination. However, knowing the relationship between the genotypes and phenotypes, we can do a weighted calculation by the number of each phenotype: - 3 gray:[\((AA,B_,C_)\), \((Aa,B_,C_)\) or \((Aa,B_,Cc)\)] - 3 black:[\((aa,B_,C_)\), \((aa,B_,C_)\) or \((aA,B_,Cc)\)] - 1 yellow:[\((A_,bb,C_)\)] - 1 cream:[\((a a b b C_)\)] - 4 albino:[\((_,_,cc)\)] Therefore, the \(F_{1}\) phenotypic ratio for this cross is \(3:3:1:1:4\) (gray:black:yellow:cream:albino).