Question: What if a new breeding occurred between two different cats. Both cats were heterozygous hairy cats, but one parent loved sinks and the other parent disliked sinks.

a) List the genotype of each parent based on the information above.

b) Determine how many of the potential offspring would be hairless and love sinks

Question

Jazlyn Conway

Biology

30 August, 16:09

Question: What if a new breeding occurred between two different cats. Both cats were heterozygous hairy cats, but one parent loved sinks and the other parent disliked sinks.

a) List the genotype of each parent based on the information above.

b) Determine how many of the potential offspring would be hairless and love sinks

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Answers (1)

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Trevin Bennett · Answer 1

Answer and Explanation:

Available dа ta:

New breeding occurred between two different cats. Both cats were heterozygous hairy cats, One parent loved sinks and the other parent disliked sinks.

Let us say that:

the gene for the cat hair trait is H, being "H" the dominant allele expressing hair, and "h" the recessive allele expressing hairless. the gene for the sinks-liking trait is S, being "S" the dominant allele expressing sinks-liking, and "s" the recessive allele expressing sinks-desliking

If both cats are heterozygous for the hair trait, then their genotypes would be Hh in both animals.

One of the cats likes sinks, so its genotype would be S-. The "-" symbol means that we do not know if this is a homozygous or heterozygous animal. The other cat disliked sinks, so its genotype for the trait would be ss.

Cross: HhS - x Hhss

Gametes) HS hS H - h-

Hs hs Hs hs

Punnet square: HS hS H - h-

Hs HHSs HhSs HHs - Hhs-

Hs HHSs HhSs HHs - Hhs-

hs HhSs hhSs Hhs - hhs-

hs HhSs hhSs Hhs - hhs-

a) List the genotype of each parent based on the information above.

Parent 1: HhS - - --> there are two possibilities: HhSs or HhSS

Parent 2: Hhss

b) Determine how many of the potential offspring would be hairless and love sinks

If parent 1 is HhSs, the potential offspring that would be hairless and sinks-loving would be 2/16=1/8. Its genotype would be hhss.

HS hS Hs hs

Hs HHSs HhSs HHss Hhss

Hs HHSs HhSs HHss Hhss

hs HhSs hhSs Hhss hhss

hs HhSs hhSs Hhss hhss

But if parent 1 is HhSS, there would not be any individual hairless and sinks-loving. All of them would be heterozygous for sinks-liking.

HS hS HS hS

Hs HHSs HhSs HHSs HhSs

Hs HHSs HhSs HHSs HhSs

hs HhSs hhSs HhSs hhSs

hs HhSs hhSs HhSs hhSs