Question

Coat colors of dogs depend upon the action of at least two genes. At one locus a dominant epistatic inhibitor of coat color pigment (A) prevents the expression of color alleles at another independently assorting locus, producing white coat color. When the recessive condition exists at the inhibitor locus (aa), the alleles of the hypostatic locus may be expressed, aaB_ producing black and aabb producing brown. When dihybrid white dogs are mated together, determine the phenotypic proportions expected in the progeny.

• A. 9:3:3:1
• B. 1:2:1
• C. 12:3:1
• D. 9:7

Answer 1

Answer: (C)

12:3:1

Answer 2

The problem describes the inheritance of coat color in dogs controlled by two independently assorting genes. Let the two loci be the inhibitor locus and the color locus.

Let the alleles at the inhibitor locus be A (dominant, epistatic inhibitor) and a (recessive, allows color expression).
Let the alleles at the color locus be B (dominant, black) and b (recessive, brown).

The phenotypic expression is as follows:

• Genotypes with at least one dominant allele 'A' at the inhibitor locus (A_ _ _) result in white coat color because the inhibitor prevents pigment expression. This includes genotypes AABB, AABb, AAbb, AaBB, AaBb, Aabb.
• Genotypes with the recessive condition 'aa' at the inhibitor locus and at least one dominant allele 'B' at the color locus (aaB_) result in black coat color. This includes genotypes aaBB, aaBb.
• Genotypes with the recessive condition 'aa' at the inhibitor locus and the recessive genotype 'bb' at the color locus (aabb) result in brown coat color.

The parents are dihybrid white dogs. A dihybrid individual is heterozygous for both genes. Since the dogs are white, they must have at least one 'A' allele. Therefore, the genotype of the dihybrid white dogs is AaBb.

The cross is AaBb x AaBb. This is a standard dihybrid cross. The possible gametes for each parent are AB, Ab, aB, and ab, each produced in equal proportion (1/4).

The genotypes of the progeny and their expected proportions can be determined using a Punnett square or by considering the segregation of each gene independently. The genotypic ratio for a dihybrid cross (AaBb x AaBb) is:

9/16 A_B_ (AABB, AABb, AaBB, AaBb)
3/16 A_bb (AAbb, Aabb)
3/16 aaB_ (aaBB, aaBb)
1/16 aabb (aabb)

Now, let's determine the phenotype for each genotypic class based on the given rules:

• A_B_ (9/16): These genotypes have at least one 'A' allele. According to the rule, A_ _ _ results in white coat color. So, 9/16 of the progeny are white.
• A_bb (3/16): These genotypes also have at least one 'A' allele. According to the rule, A_ _ _ results in white coat color. So, 3/16 of the progeny are white.
• aaB_ (3/16): These genotypes have 'aa' and at least one 'B' allele. According to the rule, aaB_ results in black coat color. So, 3/16 of the progeny are black.
• aabb (1/16): This genotype has 'aa' and 'bb'. According to the rule, aabb results in brown coat color. So, 1/16 of the progeny are brown.

To find the total proportion of each phenotype, we sum the proportions of the corresponding genotypic classes:

• White: Proportion from A_B_ + Proportion from A_bb = 9/16 + 3/16 = 12/16
• Black: Proportion from aaB_ = 3/16
• Brown: Proportion from aabb = 1/16

The phenotypic proportions expected in the progeny are White : Black : Brown = 12/16 : 3/16 : 1/16. This ratio simplifies to 12 : 3 : 1.

This is a classic example of dominant epistasis, where the ratio is 12:3:1.