Question

In hogs, a dominant allele B results in a white belt around the body. At a separate locus the dominant allele S causes fusion of the two parts of the normally cloven hoof resulting in a condition known as syndactyly. A belted syndactylous sow was crossed to an unbelted cloven-hoofed boar, and in the litter there were: 25% belted syndactylous, 25% belted cloven, 25% unbelted syndactylous, 25% unbelted cloven. The genotypes of the parents can best be represented as:

a. b/b;S/s × B/b;s/s
b. B/b;S/s × b/b;S/S
c. B/B;S/S × b/b;s/s
d. B/b;S/s × B/B;s/s
e. B/b;S/s × b/b;s/s

Answer 1

e. B/b,S/s x b/b;s/s

Step-by-step explanation:

Here, there is a dihybrid cross between two loci: B (for a white belt around the body) and S (for syndactyly). For each locus, genotypes and phenotypes should be:

- B/B or B/b: Belted hog

- b/b: wild type or unbelted.

- S/S or S/s: syndactylous hog.

- s/s: wild type or cloven-hoofed hog.

So, if a belted syndactylous sow was crossed with an unbelted cloven-hoofed boar, there are different possible gentotypes for the first animal:

Belted syndactylous sow: B/b;S/s, B/B;S/S, B/b;S/S or B/B;S/s.

Unbelted hoofed boar: b/b;s/s (because the wild type phenotype is recessive).

Progeny has the following phenotypes frequencies:

25% belted syndactylous

25% belted cloven

25% unbelted syndactylous

25% unbelted cloven

"Unbelted" and "cloven" are recessive conditions, therefore, both parents must have recessive alleles to inherit them to their progeny. For this reason, the most probable answer is "e. B/b;S/s x b/b;s/s". However, in the image, there is the Punnett Square that confirms this answer.