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Komok [63]
3 years ago
15

Q:- In dogs, the Mexican hairless phenotype never breeds true. When Mexican hairless are interbred the offspring are 2/3 hairles

s and 1/3 normal. It is also observed that some deformed puppies are born dead. Propose a genetic explanation for these observatios.
Biology
1 answer:
drek231 [11]3 years ago
7 0

Explanation:

The problem says that the hairless phenotype never breeds true. That means that it's not the result of a homozygous genotype (H₁H₁ or H₂H₂), so it is caused by the heterozygous genotype (H₁H₂).

The <u>expected </u>offspring from the cross between two Mexican hairless would be:

<h3>P                 H₁H₂  x  H₁H₂</h3><h3>F1    1/4 H₁H₁,  2/4 H₁H₂ and 1/4 H₂H₂.</h3>

And the <u>expected</u> phenotypic ratio 3:1. However, the observed offspring shows a 2:1 ratio. What's happening?

If the observed phenotypic ratio in the offspring of a monohybrid cross (a single gene with two alleles) is 2:1, we can suspect that one of the genotypes is lethal in homozygosis and therefore does not appear in the progeny (the puppies are born dead).

If we proposed that the H₂ allele is lethal in homozygosis, then:

  • The H₁H₁ genotype would cause normal puppies --> 1
  • The H₁H₂ genotype would cause hairless puppies --> 2
  • The H₂H₂ is lethal and causes the death of puppies --> 0

The phenotypic ratios change to 2:1, as observed in the experiment.

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