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% unbelted syndactylous 25% belted cloven 25% unbelted cloven The genotypes of the parents can best be represented as which of the followingA) B/b;8/s (x) B/B;s/s B) B/b;S/s (x) b/b;S/S C) b/b;S/s (x) B/b;s/s D) B/b;S/s (x) b/b;s/s E) B/B;S/S (x) b/b;s/s

Answer 1

D) B/b;S/s (x) b/b;s/s

Step-by-step explanation:

Parent 1 : belted syndactylous sow

Since it is showing dominant phenotype for both the traits, it can either be BBSS or BbSs

Parent 2: unbelted cloven-hoofed

Since it is showing recessive phenotype for both the traits, it can only have bbss genotype

If we assume parent 1 to be BBSS all the resulting progeny with bbss will have dominant phenotype which is not the case.

If we assume parent 1 to be BbSs:

BbSs X bbss =

BbSs : belted syndactylous

bbSs : unbelted syndactylous

Bbss : belted cloven

bbss : unbelted cloven

The progeny will be produced in 1:1:1:1 ratio which means that each of them will make 25% of the population.

Hence, parent 1 will have BbSs genotype and parent 2 will have bbss genotype