In guinea pigs, smooth coat (S) is dominant over rough coat ( s ) and black coat (B) is dominant over white coat (b). In the cross SsBb x SsBb, how many of the offspring will have a smooth black coat on average

9 out of 16 offsprings

Explanation:

This question involves two genes in guinea pig; one coding for coat texture and the other for coat color. The alleles for smooth coat (S) and black coat (B) are dominant over rough coat (s) and white coat (b).

In this question, two heterozygous guinea pigs were crossed i.e. SsBb × SsBb. The gametes produced by each parent is as follows: SB, Sb, sB, sb. Using these gametes in a punnet square (see attached image), the proportion of offsprings will be:

Smooth black coat (S_B_) = 9

Smooth white coat (S_bb) = 3

Rough black coat (ssB_) = 3

Rough white coat (ssbb) = 1

Hence, according to this question, 9 out of 16 offsprings will have a smooth black coat on average.

after making a punnet square using bv bv bv bv you will find that the ratio will be 9: 7 or a 128.6% probability.

hope this

the missing space is phylum