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I II III 2 3 4 5 6 Color Vision Normal Hearing Colorblind Normal Hearing Color Vision Deafness Colorblind Deafness Deafness and red-green colorblindness in humans are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below. a. What are the possible genotypes for individual III-3? What is the probability for each possible genotype? b. What is the probability that individual III-3 can have a colorblind son with deafness? c. What are the possible genotypes for individual III-5? What is the probability for each possible genotype? d. What is the probability that individual III-5 can have a colorblind son with deafness? Problem 5 Problem Set 1 Fall 2007

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Deafness and red-green colorblindness in humans are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below. What are the possible genotypes for individual III-3 ? What is the probability for each possible genotype? - PowerPoint PPT Presentation

Transcript of I II III

Page 1: I II III

I  II   III

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      Color VisionNormal Hearing

      Colorblind Normal Hearing

      Color VisionDeafness

ColorblindDeafness

Deafness and red-green colorblindness in humans are determined by recessive alleles at X-chromosome loci that are 32 map units apart. Consider the pedigree shown below.

a. What are the possible genotypes for individual III-3? What is the probability for each possible genotype?

b. What is the probability that individual III-3 can have a colorblind son with deafness?

c. What are the possible genotypes for individual III-5? What is the probability for each possible genotype?

d. What is the probability that individual III-5 can have a colorblind son with deafness?

Problem 5Problem Set 1

Fall 2007

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1. Determine the Genotype for II-1

  

    

I  II   III

Let B=color vision, b=colorblindness D= normal hearing, d=deafness

XBdY

II-1 must receive XBd from her father.She must receive an XD from her mother since she is not deaf.The X from her mother must also have Xb, since II-1 has colorblind children. (It is not possible or necessary to determine whether the chromosome from her mother is a parental or recombinant combination.)

XBdXbD

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2. Determine the Female Offspring Possible for II-1 and II-2

  

    

XBdXbD XbDYX

II-1 can produce four different gametes.Two are parentals: XbD and XBd. Two are recombinants: Xbd and XBD. Since the genes are 32 map units apart, the frequencies of recombinants must add to 32% and the parentals must add to 68%.

XbD

XbD

XBd

Xbd

XBD

Parentals add to 68%

Considering only female offspring,II-2 passes his X chromosome.

Recombinantsadd to 32%

0.34 XbDXbD

0.34 XBdXbD

0.16 XbdXbD

0.16 XBDXbD

0.34

0.34

0.16

0.16

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3. Determine the possible genotypes for III-3

  

    

III-3 is colorblind so she can have only two of the four possible genotypes shown in the Punnett square. Use the probability rule of the chance that something can happen out of the total possible outcomes to normalize these probabilities so that the sum equals 1.

XbD

XbD

XBd

Xbd

XBD

Parentals add to 68%

Recombinantsadd to 32%

0.34 XbDXbD

0.34 XBdXbD

0.16 XbdXbD

0.16 XBDXbD

0.34

0.34

0.16

0.16

0.680.160.34/0.34

0.320.160.16/0.34

Probability this can happen

Total possible outcomes

Genotypes for III-3: 0.68 XbDXbD or 0.32 XbdXbD

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3A. An easier way to determine the possible genotypes for III-3

  

    

III-3 is female, so she must have received the XbD chromosome from her father. She is colorblind, so she must receive an Xb from her mother. This Xb can be part of a recombinant chromosome, Xbd, which should occur with 32% probability. Or it can be part of a parental chromosome, XbD,which should occur with 68% probability.

I  II   III

XBdY

XBdXbD XbDY

XbD0.32 XbdXbD0.68 XbD

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4. Determine the probability that III-3 can have a colorblind son with deafness.

  

    

There are three events that must occur for III-3 to have a colorblind son with deafness:

1. She must have the XbdXbD genotype.2. She must pass the Xbd chromosome to her offspring.3. The father must pass a Y chromosome.

Y)passes P(dad x )bdX passes P(mom x )bDXbdX is P(momY)bdP(X

0.08 0.5 x 0.5 x 0.32Y)bdP(X

Includes 0.34 Xbd as parental and 0.16 Xbd as recombinant.

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5. Determine the possible genotypes for III-5

  

    

III-5 has color vision and normal hearing so she can have only two of the four possible genotypes shown in the Punnett square. Use the probability rule of the chance that something can happen out of the total possible outcomes to normalize these probabilities so that the sum equals 1.

XbD

XbD

XBd

Xbd

XBD

Parentals add to 68%

Recombinantsadd to 32%

0.34 XbDXbD

0.34 XBdXbD

0.16 XbdXbD

0.16 XBDXbD

0.34

0.34

0.16

0.16

0.680.160.34/0.34

0.320.160.16/0.34

Probability this can happen

Total possible outcomes

Genotypes for III-5: 0.68 XBdXbD or 0.32 XBDXbD

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5A. An easier way to determine the possible genotypes for III-5

  

    

III-5 is female, so she must have received the XbD chromosome from her father. She is not colorblind, so she must receive an XB from her mother. This XB can be part of a recombinant chromosome, XBD, which should occur with 32% probability. Or it can be part of a parental chromosome, XBd,which should occur with 68% probability.

I  II   III

XBdY

XBdXbD XbDY

XbD

XbD

0.32 XBD

0.68 XBd

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6. Determine the probability that III-5 can have a colorblind son with deafness.

  

    

There are three events that must occur for III-5 to have a colorblind son with deafness:

1. She must have the XBdXbD genotype.2. She must pass the Xbd chromosome to her offspring. This requires

a recombination event and is one of four types of gametes she can produce.

3. The father must pass a Y chromosome.

Y)passes P(dad x )bdX passes P(mom x )bDXBdX is P(momY)bdP(X

0.0544 0.5 x 0.16 x 0.68Y)bdP(X

Xbd is one of two recombinant gametes with a frequency of 0.16

Possible gametes for XBdXbD

Parental: 0.34 XBd and 0.34 XbD

Recombinant: 0.16 Xbd and 0.16 XBD


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