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15 August, 00:43

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

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  1. 15 August, 03:35
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    D) B/b; S/s (x) b/b; s/s

    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
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