I II III Normal Vision Normal Hearing Blindness Normal Hearing Normal Vision Deafness Blindness Deafness 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 5 Problem 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 X Bd Y II-1 must receive X Bd from her father. She must receive an X D from her mother since she is not deaf. The X from her mother must also have X b, 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.) X Bd X bD

2. Determine the Female Offspring Possible for II-1 and II-2 X Bd X bD X bD Y X II-1 can produce four different gametes. Two are parentals: X bD and X Bd. Two are recombinants: X bd and X BD. Since the genes are 26 map units apart, the frequencies of recombinants must add to 26% and the parentals must add to 74%. X bD X Bd X bd X BD Parentals add to 74% Considering only female offspring, II-2 passes his X chromosome. Recombinants add to 26% 0.37 X bD X bD 0.37 X Bd X bD 0.13 X bd X bD 0.13 X BD X bD

3. Determine the possible genotypes for III-3 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. X bD X Bd X bd X BD Parentals add to 74% Recombinants add to 26% 0.37 X bD X bD 0.37 X Bd X bD 0.13 X bd X bD 0.13 X BD X bD Probability this can happen Total possible outcomes Genotypes for III-3: 0.74 X Bd X bD or 0.26 X BD X bD

3A. An easier way to determine the possible genotypes for III-3 I II III X Bd Y X Bd X bD X bD Y 0.26 X BD 0.74 X Bd X bD III-3 is female, so she must have received the X bD chromosome from her father. She has normal vision, so she must receive an X B from her mother. This X B can be part of a recombinant chromosome, X BD, which should occur with 26% probability. Or it can be part of a parental chromosome, X Bd, 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 X Bd X bD genotype. 2.She must pass the X bd 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. Possible gametes for X Bd X bD Parental: 0.37 X Bd and 0.37 X bD Recombinant: 0.13 X bd and 0.13 X BD X bd 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. X bD X Bd X bd X BD Parentals add to 74% Recombinants add to 26% 0.37 X bD X bD 0.37 X Bd X bD 0.13 X bd X bD 0.13 X BD X bD Probability this can happen Total possible outcomes Genotypes for III-5: 0.74 X bD X bD or 0.26 X bd X bD

5A. An easier way to determine the possible genotypes for III-5 III-5 is female, so she must have received the X bD chromosome from her father. She is colorblind, so she must receive an X b from her mother. This X b can be part of a recombinant chromosome, X bd, which should occur with 26% probability. Or it can be part of a parental chromosome, X bD,which should occur with 74% probability. I II III X Bd Y X Bd X bD X bD Y X bD 0.26 X bd 0.74 X bD

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 X bd X bD genotype. 2.She must pass the X bd chromosome to her offspring. The father must pass a Y chromosome. Includes 0.37 X bd as parental and 0.13 X bd as recombinant.