1. Problem 5 Problem Set 1 Winter 2011 I II III 1 2 3 4 5 6 7
In humans, deafness (d) and blindness (b, due to the disease retinitis pigmentosum) are determined by recessive alleles at X-chromosome loci that are 18 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?
Normal Vision
Normal Hearing
Blindness Normal Hearing
Deafness
Blindness Deafness I II
III
Problem 5
Problem Set 1 Winter 2011

2. 1. Determine the Genotype for II-1
Let B= normal vision, b= blindness D= normal hearing, d= deafness I II
III
XBdY
XbD
XBd
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.)

3. 2. Determine the Female Offspring Possible for II-1 and II-2
XbD
XBd X
XbDY
Considering only female offspring,
II-2 passes his X chromosome.
XbD
Parentals add to 82%
0.41
XbD
0.41 XbDXbD
0.41
XBd
0.41 XBdXbD
0.09 XbdXbD
Recombinants add to 18%
0.09
Xbd
0.09 XBDXbD
0.09
XBD
II-1 can produce four different gametes.
Two are parentals: XbD and XBd. Two are recombinants: Xbd and XBD. Since the genes are 18 map units apart, the frequencies of recombinants must add to 18% and the parentals must add to 82%.

4. 3. Determine the possible genotypes for III-3
XbD
Probability this can happen
Parentals add to 82%
0.41
XbD
0.41 XbDXbD
Total possible outcomes
0.41
XBd
0.41 XBdXbD
0.09 XbdXbD
Recombinants add to 18%
0.09
Xbd
0.09 XBDXbD
0.09
XBD
III-3 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.
Genotypes for III-3: 0.82 XbDXbD or 0.18 XbdXbD

5. 3A. An easier way to determine the possible genotypes for III-3II
III
XBdY
XbD
XBd
XbDY
0.18 Xbd
XbD
0.82 XbD
XbD
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 18% probability. Or it can be part of a parental chromosome, XbD,which should occur with 82% probability.

6. 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:
She must have the XbdXbD genotype.
She must pass the Xbd chromosome to her offspring.
The father must pass a Y chromosome.
Includes 0.41 Xbd as parental and 0.09 Xbd as recombinant.

7. 5. Determine the possible genotypes for III-5
XbD
Parentals add to 82%
0.41
XbD
0.41 XbDXbD
Probability this can happen
0.41
XBd
0.41 XBdXbD
Total possible outcomes
0.09 XbdXbD
Recombinants add to 18%
0.09
Xbd
0.09 XBDXbD
0.09
XBD
III-5 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.
Genotypes for III-5: 0.82 XBdXbD or 0.18 XBDXbD

8. 5A. An easier way to determine the possible genotypes for III-5II   
III
XBdY
XbD
XBd
XbDY
XbD
0.82 XBd
0.18 XBD
III-5 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 18% probability. Or it can be part of a parental chromosome, XBd, which should occur with 82% probability.

9. 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:
She must have the XBdXbD genotype.
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.
The father must pass a Y chromosome.
Possible gametes for XBdXbD Parental: 0.41 XBd and 0.41 XbD Recombinant: 0.09 Xbd and 0.09 XBD
Xbd is one of two recombinant gametes with a frequency of 0.09