PRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES

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1 PRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES 1. Margaret has just learned that she has adult polycystic kidney disease. Her mother also has the disease, as did her maternal grandfather and his younger brother (both of whom are now dead). As far as Margaret knows, no one else in her extended family has the disease, although she had a sister, Allison, who died in a car accident when she was 16 and might have showed symptoms if she had lived long enough. Margaret is 42 years old and has three children with her husband, Art. Anna is 20, Lydia is 18, and Tom is 15. Her husband s name is Art. a. Draw a pedigree for this family, using the proper symbols. (Hint: Do not use names.) I II III IV b. Could this disease be autosomal recessive? If so, is this mode of inheritance likely? Explain. Yes, but it's unlikely. In order for the inheritance to be autosomal recessive, I-1 and II-1 would both have mated into the family as carriers. Unless there is inbreeding, the chance of this is remote. c. Could this disease be autosomal dominant? If so, what is the chance that Lydia will get the disease when she reaches middle-age? Yes. The disease shows the classic pattern for autosomal dominant inheritance. It is seen in every generation and no one with the disease has to mate into the family. Lydia's chance of inheriting the disease allele from her mother is 0.5 because her mother had a normal father and is therefore heterozygous. If Lydia received the allele, she will eventually get the disease (if she lives long enough). d. Could this disease be X-linked recessive? Explain, using specific individuals in the pedigree to support your answer. No. III-2 is a female with the disease. If this was an X-linked recessive disease, her father (II-1) would have to be affected and he isn't. 1

2 e. Could this disease be X-linked dominant? Explain, using specific individuals in the pedigree to support your answer. Yes. Fathers pass it to all their daughters and none of their sons; mothers can pass it to either daughters or sons. This mode of inheritance is uncommon but there is nothing in the pedigree that rules it out. 2. Grant had a younger sister, Nanna, who died of cystic fibrosis when she was 22. He is 28 now, has recently gotten married to a woman named Claire, and wants to start a family. However, Claire knows she is a carrier for CF because her two older brothers, who were identical twins, died of it several years ago and she decided to get tested for the disease allele. No one else in Grant s or Claire s family has ever gotten the illness. Grant s and Caire s parents are still living and are healthy. a. Draw a pedigree for this family, using the proper symbols. I II b. Could this disease be autosomal recessive? If so, what is Grant s chance of being a carrier and what is the overall chance that Claire s and Grant s first child will have the disease? (Assume that Grant does not get genetically tested prior to getting his wife pregnant so he does not know his carrier status.) Yes - if Grant's parents are obligate carriers and Claire's parents are obligate carriers. In order for Claire and Grant's first child to be affected, both Claire and Grant must be carriers and both must pass their mutant allele to the child. These events are independent so the overall probability of an affected child is the product of the individual probabilities: (Prob. Claire is a carrier)(prob. Grant is a carrier)(prob. first child gets both mutant alleles) We know that Claire is a carrier, so the probability that she is a carrier is 1. We don't know Grant's genotype but we DO know that there are only two possibilities: AA or Aa (i.e. he is not affected). The chance of Aa is twice that of AA because Grant could get Aa in two different ways (A from Mom and a from Dad OR A from Dad and a from Mom). There is only one way he could get AA (A from Mom and A from Dad). Therefore, his chance of being a carrier is 2/3. The chance that both carriers will give mutant alleles to a child (any pregnancy) = 1/4 Putting this altogether: (1)(2/3)(1/4) = 1/6 2

3 c. Could this disease be autosomal dominant? How do you know? No. It is not seen in every generation. d. Could this disease be X-linked recessive? How do you know? No, because Nanna's father would have to be affected and he isn't. e. Could this disease be X-linked dominant? How do you know? No. It is not seen in every generation. 3. Eloise and Hank are both carriers for sickle cell anemia, an autosomal recessive disease. If the couple has three children, what's the probability that exactly two of the children will be affected and exactly one will be unaffected (healthy)? Show how you arrived at your answer. A=affected; U=unaffected. There are three possible orders: AAU, AUA, UAA Each has a (1/4)(1/4)(3/4) chance of occurring = 3/64 So the overall chance is 3 x 3/64 = 9/64 4. In maize, seed color is under the control of one gene with two alternative alleles. The dominant allele (R) codes for a yellow seed and the recessive allele codes for a dark red seed (r). A geneticist crosses a pure-breeding plant with yellow seeds to a purebreeding plant with dark red seeds. Then he crosses the offspring to each other with the following results: 522 yellow seeds 199 dark red seeds Do these data support Mendel's First Law? Explain, using a Punnett Square and a Chi-Square analysis to support your answer. The expectation from the cross is 3 yellow:1 dark red. So the expected values are yellow and dark red. Chi-Squared = ( ) 2 / ( ) 2 / = = 2.6; df = 1. p =0.107 (exact value from Excel). We fail to reject the null and the deviation is observed can be attributed to sampling error. There is nothing to suggest that Mendel's First Law is not controlling the inheritance of this gene. 5. Luisa and Mario both come from large families and want to have a large family of their own. However, Mario has a rare genetic condition called congenital hypertrichosis, a disease he shares with his mother and his younger sister. Hypertrichosis is dominant and is caused by a mutation in a gene on the X chromosome. If Mario and Luisa have eight children, 3 boys and 5 girls, which of the children will have the disease? Explain. In X-linked dominant inheritance, the father passes the disease allele to his daughters but not to his sons. (He passes the Y chromosome to his sons.) Therefore, 5 children (all of the girls) will be affected. The boys will not be affected. 6. The following cross data was obtained from Drosophila. Your experimental hypothesis is that there should be equal numbers of offspring in each of the four classes. 3

4 # observed phenotype 502 wild wing, wild eye 417 scalloped wing, wild eye 519 scalloped wing, lobed eye 406 wild wing, lobed eye a. What is your null hypothesis? The deviation from a 1:1:1:1 ratio that is observed in the data can be attributed to sampling error. b. What is the Chi-Square value associated with the data? (HINT: Make sure you can calculate Chi-Square values without Excel because you will probably be asked to do one by hand on the lab exam. A Chi Square table will be provided.) Observed Expected Chi-Squared = ( ) 2 /461 + ( ) 2 /461 + ( ) 2 /461 + ( ) 2 /461 = = c. What is the approximate p value associated with the data? (See the 2 table on the below.) Df = 3; The p value associated with this Chi-squared value is < (Actual P Value via Excel = 7.5 X 10-5 ) d. Do you reject or fail to reject your null? Briefly explain. The p value is well below This indicates that the deviation we observed is very unlikely to be due to sampling error. Therefore, our experimental hypothesis is not supported. Perhaps the genes are linked... (Note: The data suggest that the genes are linked at a distance of 44.6 cm.) 7. Garrison has Becker muscular dystrophy. He is married to Joanne, and they have one daughter, Lisa, who also has the disease. Garrison's parents are still alive and are both normal. His younger brother, Rob, is also normal. Joanne has two older siblings. The oldest, Mark, is married and has four children, born in the following order: Elizabeth, Peter, Jane, and Andrew. Joanne's sister, Marrisa, does not have any children. Joanne's parents are both dead. As far as Joanne knows, no one in her pedigree (other than her daughter) have ever been diagnosed with Becker muscular dystrophy. a. Draw a pedigree for this family, including all the persons mentioned above and using the proper symbols. Wherever possible, include the names of the individuals below their symbols. 4

5 I II Mark Marrisa Joanne Mark Rob III Elizabeth Peter Jane Andrew Lisa b. Is it possible that this disease is X-linked dominant? Clearly explain your reasoning. No. Dominant diseases are present in every generation. Only two generations in the pedigree have affected members. In particular, Garrison's mother would have to be affected for this disease to be X-linked dominant. c. Is it possible that this disease is X-linked recessive? Clearly explain your reasoning. Yes. Garrison is alive, so the disease is not fatal and can be passed on by affected males. Garrision got the disease allele from his mother, who is a carrier. Rob inherited his mother's normal allele so is not affected. 8. Gina's brother Joe died of a rare X-linked recessive disease that causes severe birth defects. Gina has just married Albert, who lost a brother to the same disease. a. What is the probability that Gina is a carrier? Briefly explain your answer. 1/2. Her father must have been normal and her mother is a carrier. Gina has a 1/2 chance of getting her mother's mutant X chromosome. b. What is the probability that Albert is a carrier? Briefly explain your answer. Zero. Males cannot "carry" X-linked recessive alleles. If they have the allele, they will exhibit the phenotype. c. What is the probability that Gina and Albert's first child will be affected? Draw a Punnett Square to support your answer. For the couple to have an affected child, all of the following must be true: Gina must be a carrier: 1/2 Gina must pass the allele to her child: 1/2 The child must be a boy: 1/2 These are independent events, so the overall probability is (1/2) 3 = 1/8 If Gina is a carrier, the Punnett Square looks like this: X H X h X H X H X H X h X H Y X H Y X h Y 5

6 d. If Gina and Albert have 5 boys, what is the probability that none will be affected? Show your calculation. The chance of Gina being a carrier is 1/2. The chance of an affected child from a carrier mother is 1/2. Therefore, the overall probably of each boy being affected is 1/4. Each pregnancy is an independent event, so the overall probability is (1/4) 5 = 1/1024 e. If Gina and Albert have 3 boys, what's the probability that exactly one will be affected (any order)? Show your calculation. Given that all the children are boys, the chance that any one of them will be affected is 1/4 (as explained above). There are three mutually exclusive orders (AUU, UAU, UUA), each with a probability of (1/4)(3/4)(3/4) = 9/64. So the overall probability is 27/ Mouser, a female cat, has just given birth to 8 kittens. a. Assuming that the kittens have just been born and you have no information about the litter other than the number of kittens, what is the probability that all the kittens are male? (1/2) 8 = 1/256 b. What's the probability that there is at least one male in the litter? (Hint: There's a fast way to do this calculation.) 1 minus the probability that all the kittens are female = 1 - (1/256) = 255/256 = 99.6% c. You pick up a kitten from the litter and find that it is male. Does this change the probability that all the kittens are male? Explain. Yes, because we now know that at least one of the kittens is a male. The equation changes as follows: (1)(1/2) 7 = 1/128 d. What's the probability that the litter contains a mix of sexes (one or more males and one or more females)? 1 minus the combined probabilities that the kittens are all males OR all females. Probability all males = 1/256; Probability all females = 1/256 Combined probabilities = 2/256 = 1/128 Chance litter contains a mix of sexes = 1 - (1/128) = 127/128 = 99.2% e. What's the probability that there are 4 males and 4 females (an even mix)? P = {8!/[4!(8-4)!} x {(.5 4 )(.5 4 )} = = 70/256 (This equation is in your textbook. If you didn't use it, I can only marvel if you got the correct answer using the ordering method!) 6

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