Answer:
25%
Explanation:
This cross is a dihybrid cross involving two genes in the guinea pig. One gene is coding for hair color while the other is coding for hair length.
The allele for black hair (B) is dominant over that of brown hair (b) while the allele for short hair (H) is dominant over that of long hair (h).
In a cross involving the parents with genotypes Bbhh (black long hair) and bbHh (brown short hair), both parents will undergo meiosis to produce gametes as follows:
Bbhh: Bh, Bh, bh, bh
bbHh: bH, bH, bh, bh
Using punnet square (attached picture) for this dihybrid cross, 16 offsprings will be produced. According to the question, an offspring with black and long hair phenotype will either have BBhh or Bbhh genotype.
Note that, in the heterozygous state of a gene, the dominant allele will express itself.
Checking the drawn punnet square attached, one will notice that the four circled offsprings have the above genotypes, hence they are phenotypically black and long haired.
Since there are 16offsprings and only four are black & long haired, we then calculate the percentage by saying:
4/16 × 100 = 25%
Therefore, 25% of the offsprings will be black and long haired.
