Answer:
25% of the offsprings will be BBCC
Explanation:
This is a typical dihybrid cross involving two distinct genes. One coding for fur colour and the other for claw sharpness. The allele for brown fur (B) is dominant over the allele for black fur (b) in the first gene while the allele for sharp claws (C) is dominant over the allele for dull claws (c) in the second gene.
In a cross between parents with genotypes: BbCc x BBCC , each parent will produce four possible allelic combinations of gametes as follows:
BbCc: BC, Bc, bC, bc
BBCC: BC, BC, BC, BC
Using these gametes in a punnet square (see attached image), 16 possible offsprings will be produced with four distinct genotypes:
BBCC (4)
BBCc (4)
BbCC (4)
BbCc (4)
According to the question, an offspring that is homozygous dominant for both traits will possess a genotype: BBCC
N.B: Homozygous dominant means contains same alleles for the dominant trait.
Hence, offsprings with genotype, BBCC, from this cross are expected to be 4 out of the 16 possible offsprings. Hence, the percentage is 4/16 × 100 = 25%.
