In a mammal, coat colour is governed by gene B, The coat colour is either black or brown, depending on whether the genotype is BB or Bb. It is not known which of these genotypes lead to the black and brown colours. The genotype bb results in albino coat colour. Further, the genotype cc suppresses the expression of coat colour resulting in albino coat colour. An albino male was crossed with a brown female and the resulting progeny had individuals with either black or brown coats. From this observation it can be inferred that the genotype of the male and female that were crossed are:

Question

Prepp Team · Accepted Answer

Coat Colour Genetics Analysis

This problem involves the inheritance of coat colour in a mammal, controlled by two genes: gene B and gene C. Let's break down the information given:

Gene B determines if the colour is black or brown based on genotypes \(BB\) or \(Bb\). The specific phenotype for each is unknown initially.
Genotype \(bb\) results in albino coat colour.
Gene C affects the expression of the colour determined by gene B.
Genotype \(cc\) suppresses coat colour expression, leading to an albino phenotype. This is an example of recessive epistasis, where the homozygous recessive genotype at the C locus (\(cc\)) masks the expression of the B locus alleles.

Analyzing the Parental Cross

We are told an albino male is crossed with a brown female, and their progeny consist of individuals with either black or brown coats, but no albino individuals.

Male Parent: Albino

The male is albino. This can be due to either of the albino conditions:

Genotype \(bb\) at the B locus, regardless of the C locus genotype (as \(bb\) is albino).
Genotype \(cc\) at the C locus, provided there is at least one dominant allele at the B locus (\(BB\) or \(Bb\)) that could potentially express colour.

Female Parent: Brown

The female is brown. This means:

She must have at least one dominant allele at the B locus (\(BB\) or \(Bb\)) to have colour.
She must have at least one dominant allele at the C locus (\(CC\) or \(Cc\)) to allow colour expression (since \(cc\) suppresses colour).
Given she is brown, she cannot be genotype \(BB\) if \(BB\) results in black, or she cannot be genotype \(Bb\) if \(Bb\) results in black. However, this distinction isn't immediately necessary to find the genotype pair. She definitely cannot be \(bb\) or \(cc\).

Analyzing the Progeny

The progeny are exclusively black or brown. There are no albino individuals. This is a crucial observation.

No albino progeny means that no offspring have the genotype \(bb\) and no offspring have the genotype \(cc\).
This implies that every offspring received at least one dominant B allele (\(B\)) or at least one dominant C allele (\(C\)) such that the combination \(bb\) or \(cc\) is avoided in the progeny. Specifically, if a \(bb\) individual would be albino regardless of C, then no \(bb\) individuals appeared. If a \(cc\) individual would be albino, then no \(cc\) individuals appeared. Since *both* \(bb\) and \(cc\) cause albinism, the absence of albino progeny strongly suggests that neither the \(bb\) nor the \(cc\) genotypes occurred in the offspring generation.
To avoid \(cc\) in the progeny, at least one parent must contribute a \(C\) allele to every offspring. This means at least one parent must be homozygous dominant \(CC\).
To avoid \(bb\) in the progeny, the parents must not both carry the \(b\) allele in a way that allows two \(b\) alleles to combine in any offspring.

Evaluating the Option and Determining Parent Genotypes

Let's examine the option identified as correct and see if it fits all the conditions:

Male genotype: \(BB\)\(cc\)

Female genotype: \(Bb\)\(CC\)

Check male: \(BB\)\(cc\). Has \(BB\) for colour potential, but \(cc\) suppresses colour. This genotype results in albino. Fits the male phenotype.

Check female: \(Bb\)\(CC\). Has \(Bb\) for colour potential, and \(CC\) allows colour expression. This genotype results in a coloured phenotype (specifically brown, as stated). Fits the female phenotype.

Now, let's determine the possible gametes from each parent:

Male (\(BB\)\(cc\)): Can only produce gametes with alleles from BB and cc, so only \(Bc\).
Female (\(Bb\)\(CC\)): Can produce gametes with alleles from Bb and CC, so \(BC\) and \(bC\).

Let's perform the cross using a Punnett square:

Gametes >	\(BC\)	\(bC\)
\(Bc\)	\(BBCc\)	\(BbCc\)

The resulting progeny genotypes are \(BBCc\) and \(BbCc\).

Let's analyze the phenotypes of the progeny:

\(BBCc\): Has genotype \(BB\) at the B locus. Has genotype \(Cc\) at the C locus. Since \(cc\) suppresses, \(Cc\) allows colour expression. So, this individual will have colour based on the \(BB\) genotype.
\(BbCc\): Has genotype \(Bb\) at the B locus. Has genotype \(Cc\) at the C locus. Since \(Cc\) allows colour expression, this individual will have colour based on the \(Bb\) genotype.

The progeny are a mix of \(BBCc\) and \(BbCc\). Both genotypes express colour because they have at least one dominant C allele. Importantly, neither progeny genotype is \(bb\) (avoiding albino due to gene B) nor \(cc\) (avoiding albino due to gene C suppression).

The problem states the progeny are either black or brown. This is consistent with having two different B locus genotypes (\(BB\) and \(Bb\)), where one results in black and the other in brown. For instance, if \(BB\) results in black and \(Bb\) results in brown (or vice versa), then the progeny will indeed be a mix of black and brown individuals.

Since this pair of parental genotypes (\(BB\)\(cc\) and \(Bb\)\(CC\)) perfectly explains the albino male, the brown female, and the resulting progeny of only black and brown individuals, it is the correct inference.