Assume that in guinea pigs, the brown coat color allele (B) is dominant over red (b), and the solid color allele (S) is dominant over spotted (s). The F₁ offspring of a cross between true-breeding brown, solid-colored guinea pigs and red, spotted guinea pigs are crossed. What proportion of their offspring (F₂ generation) would be expected to be red and solid-colored?a. 1/9
b. 1/16
c. 3/16
d. 9/16
e. 3/4

C (3/16)

Explanation:

This is a typical dihybrid cross involving two genes; one coding for coat color and the other coding for coat appearance in guinea pigs. The brown color (B) is dominant over the red color (b) in the first gene while the solid color (S) is dominant over the spotted color (s) in the second gene. Two guinea pigs that are truebreeding for both genes i.e homozygous were crossed i.e. BBSS and bbss. A heterozygous F1 generation of BbSs guinea pigs will be produced from this cross.

If the heterozygous F1 offsprings were self-crossed I.e. BbSs v BbSs, each parent will undergo meiosis to produce gametes. According to Mendel's law of independent assortment, each allele of each gene will segregate into gametes independently of one another. The following possible gametes combination will be produced by the BbSs guinea pig:

BS, Bs, bS and bs.

If these gametes are crossed using a punnet square (see attachment), 16 possible offsprings will be produced. According to the punnet square, only 3 of the 16 F2 offsprings will be expected to be red and solid-colored i.e. bbSS or bbSs genotypes. Hence, the probability of producing a red, solid-colored

F2 guinea pig is 3/16.