Biomedical Homework: QUESTION 1 A normal woman mated with a man with Marfan Synd

Question

Biomedical Homework:

QUESTION 1

A normal woman mated with a man with Marfan Syndrome. The man with Marfan Syndrome had a father who was normal. What is the chance that the man and woman will produce offspring with Marfan Syndrome?

Enter the percentage without the % sign. For example, if your answer was 100%, you would just put 100. Follow this same rule for all of the following questions.

QUESTION 2

A color blind woman and a man with normal vision have a son. What is the percent probability that the son is colorblind? Enter your numberical answer without the % sign.

QUESTION 3

A normal woman whose father had Duchenne muscular dystrophy mated with a normal man. What is the percent chance that they will have a daughter with Duchenne muscular dystrophy? Enter your numerical answer without the percent sign.

QUESTION 4

An X-linked recessive gene results in hemophilia. A normal woman whose father had hemophilia marries a man with hemophilia. Use H to indicate the normal allele and h to indicate the defective hemophilia gene. Female genotype possibilities would be HH, Hh, or hh. Male genotype possibilities would be HY or hY since this gene is on the X-chromosome and males only have one copy.

(a) What are the possible genotypes for the mother of the man with hemophilia? ___________ or   ___________

(b) What are the possible genotypes for the father of the man with hemophilia? ___________ or ___________

(c) What are the chances that the first son will have hemophilia? Type in your numerical answer without a % after. ___________

(d) What are the chances that the first daughter will have hemophilia? Type in your numerical answer with a % after.  ___________

QUESTION 5

In cats, the allele B produces black fur while b produces yellow. Neither gene is dominant, and in the heterozygous state, the phenotype is a combination of yellow and black spots called tortoiseshell. The alleles B and b are X-linked. If a tortoiseshell cat has three tortoiseshell kittens and two black kittens, one of which was female:

Would the tomcat have the B or b allele?  ___________

What is the phenotype of the tomcat's fur, black, tortoiseshell, or yellow?  ___________

QUESTION 6

In cats, an X-linked pair of alleles, B and b, controls color of fur. BB (female) and BY (male) would result in a black cat. Bb would result in a female tortoiseshell cat. bb (female) or bY (male) would result in a yellow cat.

(a) A yellow cat had a litter of two tortoiseshell kittens and one yellow kitten. Is the yellow kitten male or female? ___________

(b) A yellow male cat is crossed with a tortoiseshell female cat. If the female cat has all male kittens in her litter of four, fill in the percent probability for each of the following phenotypes (remember, do not include the % after your number)

Yellow: ___________

Black: ___________

Tortoiseshell: ___________

Please write answer in specific order.

Explanation / Answer

1. Marfan Syndrome is an example of autosomal dominant disorder.

M = WT allele

Md = Dominant disease causing allele

Man's genotype = MdM

Man's father genotype = MM

Woman's genotype = MM

Parental cross: MdM X MM

Progeny: MdM MdM MM MM

The probability for progeny to develop disease = 1/2 = 0.5 = 50%

= 50

2. Color blindness is an example of X-linked recessive disorder.

XC = WT allele

Xc = Recessive disease causing allele

Man's genotype = XCY

Woman's genotype = XcXc

Parental cross: XcXc X XCY

Progeny: XCXc XcY XCXc XcY

The probability for a son to develop disease = 1 = 100%

= 100

3. Duchenne muscular dystrophy is an example of X-linked recessive disorder.

XD = WT allele

Xd = Recessive disease causing allele

Man's genotype = XDY

Woman's genotype = XDXd

Woman's father = XdY

Parental cross: XDXd X XDY

Progeny: XDXd XDY XCXd XdY

The probability for a daughter to develop disease = 0 = 0%

= 0

4.

Hemophilia is an example of X-linked recessive disorder.

XH = WT allele

Xh = Recessive disease causing allele

Man's genotype = XhY

Woman's genotype = XHXh

Woman's father = XhY

Parental cross: XHXh X XhY

Progeny: XHXh XHY XhXh XhY

Genotype of the man's mother = XhXh or XHXh

Genotype of the man's father = XhY or XHY

Probability for the first son to have the disease = 1/2 = 0.5 = 50% = 50

Probability for the first daughter to have the disease = 0%

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