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Maslowich
3 years ago
7

A woman with normal blood clotting ability whose father was a hemophiliac and whose mother was normal with no family history of

hemophilia marries a man whose blood has normal clotting ability. Recall that hemophilia is an X-linked trait. What is the probability that their first child will be a son with hemophilia
Biology
1 answer:
Alborosie3 years ago
5 0

Answer:

The answer is 25%.

Explanation:

Hemophilia is a recessive illness that is x-linked.

If the husband has normal clotting ability, that means he has the dominant gene since he has only one X chromosome.

Since there is a probability that their child will have hemophilia, this means that the woman, who has 2 X chromosomes but has normal blood clotting abilities, has one dominant and one recessive hemophilia gene.

The probability that their first child will be a male is 50% and the probability that he will have hemophilia is also 50% since he will only get the Y chromosome from the father and the mother has one dominant and one recessive.

So the probability that their first child will be a son with hemophilia is 25%.

I hope this answer helps.

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[Chapter 3-4] A cross was carried out between the following two pureline parental strains:Parent 1: A/A; b/b; D/D; e/e; F/FParen
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Following the selfing of the F1 from the cross between the two parental strains, what proportion of the F2 individuals will phenotypically resemble either one of the two parental lines?

= 7.03% of F2 progeny will resemble parent 1

= 2.34% of F2 progeny will resemble parent 2.

Explanation:

Parent 1: AA; b/b; D/D; e/e; F/F

Parent 2: a/a; B/B; d/d; E/E; f/f

F1 Progeny: A/a; B/b; D/d; E/e; F/f

knowing that the F1 progeny is selfed, the possible gamete outcomes are:

ABDeF , ABDEf, ABdeF, ABdEf

AbDeF , AbDEf , AbdeF , AbdEf

aBDeF , aBDEf , aBdeF , aBdEf

abDeF , abDEf , abdeF , abdEf

All gametes are either eF or Ef (the two parental genotypes) and never EF or ef.

At the A, B and D loci, the F1 are heterozygotes. Therefore, in the F2 progeny, for each locus, 3/4 of the progeny will have the dominant trait while 1/4 will have the recessive phenotype. Further, 1/2 of all F2 progeny will have the haplotype eF and the other half will be Ef.

<u>Proportion of progeny resembling Parent 1: </u>

Parent 1 has the Phenoytpe AbDeF. As previously discussed, 3/4th of all F2 progeny has the phenotype A, 1/4th of F2 progeny has the phenotype b, 3/4th of F2 progeny has the phenotype D and 1/2 of F progeny has the phenotype eF.

Thus, the proportion of progeny resembling Parent 1

= 3/4 x 1/4 x 3/4 x 1/2 = 0.0703125

= 7.03% of F2 progeny will resemble parent 1.

<u>Proportion of progeny resembling Parent 2</u>:

Parent 2 has the Phenoytpe aBdEf. As previously discussed, 1/4th of all F2 progeny has the phenotype a, 3/4th of F2 progeny has the phenotype B, 1/4th of F2 progeny has the phenotype d and 1/2 of F progeny has the phenotype Ef.

Thus, Proportion of progeny resembling parent 2

= 1/4 x 3/4 x 1/4 x 1/2 = 0.0234375

= 2.34% of F2 progeny will resemble parent 2.

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