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• 1. A nonhemophiliac man marries a nonhemophiliac woman whose father was a hemophiliac. What kinds of children can they have and in what percentages?
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Sex –Linked Genetics #1

Jan 12, 2016

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1. A nonhemophiliac man marries a nonhemophiliac woman whose father was a hemophiliac. What kinds of children can they have and in what percentages?. Sex –Linked Genetics #1. - PowerPoint PPT Presentation
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Page 1: Sex –Linked Genetics #1

• 1. A nonhemophiliac man marries a nonhemophiliac woman whose father was a hemophiliac. What kinds of children can they have and in what percentages?

Page 2: Sex –Linked Genetics #1

Sex –Linked Genetics #1 Step 1: Determine the letters you will use to represent the alleles. For this one I chose “B” for normal blood and “b” for the hemophiliac gene.

Step 2: Determine the alleles of the parents. The non-hemophiliac man must be XBY. The non-hemophiliac woman with a hemophiliac father must be XBXb, because she had to get Xb from her afflicted father.

Step 3: Set up your Punnett square and do the cross. Determine the genotypes and expected percent of each type of child from the cross.

Page 3: Sex –Linked Genetics #1

XB Y

XB

Xb

XB XB XB Y

XBXb Xb Y

25% Normal girl25% Normal boy25% Carrier girl25% Hemophiliac son

Page 4: Sex –Linked Genetics #1

• 2. Two normal-visioned parents have a color-blind son. What are the genes of the parents? What are the chances of their having a color blind girl? Explain.

Page 5: Sex –Linked Genetics #1

Sex-Linked Genetics #2Step 1 : Choose your alleles. I will use “C” for normal color vision and “c” for the red/green color blind condition.

Step 2 : Determine the genotypes of the parents. Both parents have normal vision, yet they have a color-blind son. Because, the dad can not have the “c” allele (it’s sex-linked remember and he has normal vision) the wife must be a carrier. Dad = XCY and Wife= XCXc

Step 3 : Set up your Punnett square and do the cross!

Page 6: Sex –Linked Genetics #1

XC Y

XC

Xc

XCXC XCY

XCXc XcY

There are no chances of this couple having a color blind girl. The father would have to be color blind to contribute the second defective Xc allele needed to produce a color-blind daughter.

There are no chances of this couple having a color blind girl. The father would have to be color blind to contribute the second defective Xc allele needed to produce a color-blind daughter.

Page 7: Sex –Linked Genetics #1

• 3. A normal-visioned man marries a normal-visioned woman whose father was color-blind. They have two daughters who grow up and marry. The first daughter has five sons, all normal-visioned. The second daughter has two normal-visioned daughters and a color-blind son. Diagram the family tree, including the genes of all the people mentioned.

Page 8: Sex –Linked Genetics #1

Sex-Linked Genetics #3Step 1 : As this is another problem about color blindness, we already have our alleles ready from the previous problem. “C” for normal vision, “c” for the red/green color blind condition. Use what you know from the problem to assign genotypes and do Punnett square crosses to help answer all the questions. A family pedigree might be a great way to show clearly the genes of all the people mentioned. Step 2 : Figure out the genotypes of the parents. Both have normal vision, so the man must be XCY . The woman’s father was color blind she has at least one daughter who has a color blind son. That son had to get his Xc from his mom who had to get it from her mom. So, the woman must be a carrier. (XCXc)

Step 3 : Do the crosses. Explain the expected genotypes of all the people involved.

Page 9: Sex –Linked Genetics #1

XC Y

XC

Xc

XCXC XCY

XCXc XcY

Dad

MomThese are their possible offspring. Mom’s dad was color blind, so she got her Xc from him. (see cross above) The problem says they had two daughters, so we will focus on them. One daughter had 5 sons, all with normal vision. She must not be the carrier daughter, otherwise odds are she would have had a color-blind son. Do we know if her husband is color blind or not? The other daughter had 2 normal vision daughters and 1 color-blind son, so she must be a carrier of the Xc allele.

Xc Y

XC

XC

XCXc XCY

XCXc XCY

Grandpa

Page 10: Sex –Linked Genetics #1

The first daughter would have the genotype, XCXC and all her sons would be XCY and they would have normal vision. If her husband was color- blind it would only effect their daughters, (50% would be expected to be color-blind) but, because they had no daughters we have no way of knowing from this information if he is color-blind or not. (see grandpa to see a Punnett square for this situation.)The second daughter would have to be the carrier. Her daughters have normal vision. However, one or both of them could be carriers of the trait. The son, however, was not so lucky and got the defective X. With no second X to mask the effects, he will be color-blind.

XC Y

XC

Xc

Normal vision

Color-blind sonCarrier

Normal vision

Page 11: Sex –Linked Genetics #1

Pedigree for this family

?