12.37. The pedigrees indicated here were obtained with three unrelated families whose members express the same completely penetrant disease caused by a dominant mutation that is linked at a distance of 10 cM from an SSR marker locus with three alleles numbered 1, 2, and 3. The SSR alleles present within each live genotype are indicated below the pedigree symbol. The phenotypes of the newly born labeled individuals-A, B, C, and D-are unknown. a. What is the probability of disease expression in each of these newborn babies? b. Why would a human geneticist be unlikely to use this SSR marker for diagnosis of the genetic disease? 13 22 13 22 12 32 12 32 12 32 12 32 12 32 12 32

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please explain why child D is 18% I don't understand the given explanation. explain thoroughly and in simple terms

12.37. The pedigrees indicated here were obtained with three unrelated families
whose members express the same completely penetrant disease caused by a
dominant mutation that is linked at a distance of 10 cM from an SSR marker
locus with three alleles numbered 1, 2, and 3. The SSR alleles present within
each live genotype are indicated below the pedigree symbol. The phenotypes of
the newly born labeled individuals-A, B, C, and D-are unknown.
a. What is the probability of disease expression in each of these newborn
babies?
b. Why would a human geneticist be unlikely to use this SSR marker for
diagnosis of the genetic disease?
13
22
13
22
12
32
12 32
12
32
12
32
12
32
12
32
A
D
Transcribed Image Text:12.37. The pedigrees indicated here were obtained with three unrelated families whose members express the same completely penetrant disease caused by a dominant mutation that is linked at a distance of 10 cM from an SSR marker locus with three alleles numbered 1, 2, and 3. The SSR alleles present within each live genotype are indicated below the pedigree symbol. The phenotypes of the newly born labeled individuals-A, B, C, and D-are unknown. a. What is the probability of disease expression in each of these newborn babies? b. Why would a human geneticist be unlikely to use this SSR marker for diagnosis of the genetic disease? 13 22 13 22 12 32 12 32 12 32 12 32 12 32 12 32 A D
In the pedigree at the right, the phase of the alleles in the male parent is not known.
However, we can use the information that the sibling of Child D is affected to
12-24
ight © 2021 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior
written consent of McGraw-Hill Education.
chapter 12
determine the likelihood of one phase or the other. The brother of Child D inherited
Dml from his father. This means that the chance is 90% that the genotype of the father
is D ml / d m2, and the chance is 10% that his genotype is d m1 / D m2. Child D
inherited m2 from his father. The probability that Child D inherited m2 D from his
father is the weighted sum of the chances of inheriting this chromosome given each of
the father's possible genotypes above. If the father's genotype is D m1/ d m2, then
using the same logic as we did above for Children A-C, there are 9 d m2 gametes
produced for
the father's genotype is d m1/D m2, the relative chance of D m2 is 90%. Therefore,
the chance that Child D has the disease (inherited D m2) is: (0.9)(0.1) + (0.1)(0.9)
= 0.09 + 0.09 = 0.18, or 18%.
every
1 D m2 gamete made, and so the relative chance of D m2 is 10%. If
%3D
Transcribed Image Text:In the pedigree at the right, the phase of the alleles in the male parent is not known. However, we can use the information that the sibling of Child D is affected to 12-24 ight © 2021 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. chapter 12 determine the likelihood of one phase or the other. The brother of Child D inherited Dml from his father. This means that the chance is 90% that the genotype of the father is D ml / d m2, and the chance is 10% that his genotype is d m1 / D m2. Child D inherited m2 from his father. The probability that Child D inherited m2 D from his father is the weighted sum of the chances of inheriting this chromosome given each of the father's possible genotypes above. If the father's genotype is D m1/ d m2, then using the same logic as we did above for Children A-C, there are 9 d m2 gametes produced for the father's genotype is d m1/D m2, the relative chance of D m2 is 90%. Therefore, the chance that Child D has the disease (inherited D m2) is: (0.9)(0.1) + (0.1)(0.9) = 0.09 + 0.09 = 0.18, or 18%. every 1 D m2 gamete made, and so the relative chance of D m2 is 10%. If %3D
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