24. 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:
(1) BB cc and Bb CC, respectively
(2) Bb cc and Bb CC, respectively
(3) bb CC and BB CC, respectively
(4) bb Cc and BB CC. respectively
The correct answer is:
(1) BB cc and Bb CC, respectively.
Only this combination explains:
-
Albino male phenotype.
-
Brown female phenotype.
-
Progeny showing both black and brown (no albino).
Below is the detailed reasoning plus the SEO‑style article.
Stepwise genetics reasoning
Given:
-
Gene B controls actual coat colour:
-
BB or Bb = coloured (either black or brown, but which is which is unknown).
-
bb = albino (no pigment from B locus).
-
-
Gene C controls expression of colour:
-
cc suppresses colour → albino regardless of B genotype (epistatic suppressor).
-
C_ allows B locus to show.
-
So an animal is albino if:
-
bb (no pigment) with any C/c, or
-
cc with any B/b.
A coloured animal must be:
-
B_ C_ (because bb and/or cc would be albino).
The cross: albino male × brown female → progeny only black and brown (no albino).
Key inference:
-
Since no progeny are albino, each offspring must be B_ C_.
-
That means every gamete from the male must carry B and every gamete from the female must carry C, otherwise bb or cc (and thus albino) would appear.
Therefore:
-
Male must be BB (only B gametes) and cc (albino due to suppressor) → BB cc.
-
Female must be B_ C_ and phenotypically brown. To give C in every gamete, she must be CC. She is brown, so genotype is Bb CC (since bb is albino, brown must be heterozygous at B).
Cross: BB cc (male) × Bb CC (female):
-
Male gametes: Bc only.
-
Female gametes: BC or bC.
-
Progeny:
-
Bc × BC → BB Cc (black or brown depending on BB phenotype).
-
Bc × bC → Bb Cc (opposite colour, i.e., if BB = black and Bb = brown, or vice versa).
-
All offspring are B_ C_ (coloured) and segregate into two colour classes (black and brown) with no albino, exactly as observed.
So option (1) fits perfectly.
Why other options are wrong
-
Bb cc and Bb CC
-
Male Bb cc is albino (cc), female Bb CC is brown.
-
Cross: Bb cc × Bb CC gives some bb offspring (bbC_) and some cc offspring (B_cc) → multiple albino progeny.
-
But the question says progeny are only black and brown, no albino, so this is impossible.
-
bb CC and BB CC
-
Both parents are CC, so all progeny are C_.
-
bb CC is albino by definition; BB CC is coloured (black or brown).
-
Cross gives some bb C_ offspring → albino progeny must appear.
-
Also, parent described as brown cannot be BB CC if that genotype happened to be black, and regardless, albinos would appear, contradicting the data.
-
bb Cc and BB CC
-
bb Cc is albino (bb), BB CC is coloured.
-
Cross bb Cc × BB CC produces only Bb C_ progeny (all B_ C_), giving a single coloured phenotype (all same colour, black or brown depending on Bb).
-
There would not be two different colours (black and brown) among progeny, so this also fails.
Hence only BB cc (male) × Bb CC (female) is consistent with all observations.
Introduction
Coat colour in many mammals is controlled by interactions between a pigment gene and a separate suppressor gene, a classic example of epistasis. In this problem, gene B determines black or brown colour while bb causes albino, and gene C acts as a suppressor, where cc turns any genotype albino by blocking colour expression. Analysing an albino × brown cross that yields only black and brown offspring reveals how to deduce the exact parental genotypes.
Genetic model for coat colour
At the B locus, genotypes BB and Bb produce pigment and may appear black or brown depending on how dominance is expressed, whereas bb eliminates pigment and creates an albino coat. At the C locus, the recessive homozygous genotype cc suppresses colour from the B locus, so any B or b combination with cc becomes albino, while C_ allows the B gene’s effect to be visible. Thus, coloured animals must be B_ C_, and albinos can be either bb C_ or any B/b with cc.
Deducing parental genotypes from progeny
The key observation is that the albino male crossed with the brown female produces progeny that are all coloured (black or brown) with no albino individuals. This means every offspring is B_ C_, so every sperm from the male must carry B and every egg from the female must carry C. The albino male must therefore be BB cc (albino because of cc), and the brown female must be Bb CC (coloured, carrying C in all gametes and heterozygous at B so that both black and brown phenotypes can segregate in offspring). Crossing BB cc × Bb CC yields only B_ C_ progeny that split into black and brown, perfectly matching the described result.
Evaluation of alternative genotype options
Other proposed parental combinations inevitably generate some albino progeny or only a single coat colour class in offspring. For example, Bb cc × Bb CC would produce some bb or cc offspring that are albino, while bb CC × BB CC or bb Cc × BB CC would either yield albino progeny or only one non-albino colour. Because the observed progeny include both black and brown but no albinos, these alternatives are inconsistent with the data, confirming BB cc (male) and Bb CC (female) as the only valid genotypes.
