I  II   III

1 2 3 4 5 6 7

      Normal VisionNormal Hearing

      Blindness Normal Hearing

      Normal VisionDeafness

BlindnessDeafness

In humans, deafness (d) and blindness (b, due to the disease retinitis pigmentosum) are determined by recessive alleles at X-chromosome loci that are 26 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 son who is both blind and deaf? 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 son who is both blind and deaf?

Problem 5Problem Set 1

Fall 2010

1. Determine the Genotype for II-1

      

I  II   III

Let B= normal vision, b= blindness 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 blind children. (It is neither possible nor necessary to determine whether the chromosome from her mother is a parental or recombinant combination.)

XBdXbD

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 26 map units apart, the frequencies of recombinants must add to 26% and the parentals must add to 74%.

XbD

XbD

XBd

Xbd

XBD

Parentals add to 74%

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

Recombinantsadd to 26%

0.37 XbDXbD

0.37 XBdXbD

0.13 XbdXbD

0.13 XBDXbD

0.37

0.37

0.13

0.13

3. Determine the possible genotypes for III-3

0.740.130.37/0.37

      

III-3 has normal 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 74%

Recombinantsadd to 26%

0.37 XbDXbD

0.37 XBdXbD

0.13 XbdXbD

0.13 XBDXbD

0.37

0.37

0.13

0.13

Probability this can happen

Total possible outcomes

Genotypes for III-3: 0.74 XBdXbD or 0.26 XBDXbD

0.260.130.13/0.37

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

      

I  II   III

XBdY

XBdXbD XbDY

0.26 XBD0.74 XBd XbDXbD

III-3 is female, so she must have received the XbD chromosome from her father. She has normal vision, so she must receive an XB from her mother. This XB can be part of a recombinant chromosome, XBD, which should occur with 26% probability. Or it can be part of a parental chromosome, XBd, which should occur with 74% probability.

4. Determine the probability that III-3 can have a blind and deaf son.

      

There are three events that must occur for III-3 to have a blind and deaf son:

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.

0.0481 0.5 x 0.13 x 0.74Y)bdP(X

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

Possible gametes for XBdXbD

Parental: 0.37 XBd and 0.37 XbD

Recombinant: 0.13 Xbd and 0.13 XBD

Xbd is one of two recombinant gametes with a frequency of 0.13

5. Determine the possible genotypes for III-5

      

III-5 is blind 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 74%

Recombinantsadd to 26%

0.37 XbDXbD

0.37 XBdXbD

0.13 XbdXbD

0.13 XBDXbD

0.37

0.37

0.13

0.13

0.740.130.37/0.37

0.260.130.13/0.37

Probability this can happen

Total possible outcomes

Genotypes for III-5: 0.74 XbDXbD or 0.26 XbdXbD

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 colorblind, so she must receive an Xb from her mother. This Xb can be part of a recombinant chromosome, Xbd, which should occur with 26% probability. Or it can be part of a parental chromosome, XbD,which should occur with 74% probability.

I  II   III

XBdY

XBdXbD XbDY

XbD

XbD

0.26 Xbd

0.74 XbD

6. Determine the probability that III-5 can have a blind and deaf son.

      

There are three events that must occur for III-5 to have a blind and deaf son:

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

must pass a Y chromosome.

0.065 0.5 x 0.5 x 0.26Y)bdP(X

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

Includes 0.37 Xbd as parental and 0.13 Xbd as recombinant.