GENE3200 Practice 10 102819 Pedigrees KEY PDF

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Discussion key, Dr. Bedell...

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GENE 3200 – Bedell - Fall 2019 Practice Problem Set 10: Oct 28 and Oct 29

KEY

1. Alkaptonuria is a rare metabolic disorder in which affected persons produce black urine. This disorder has autosomal recessive inheritance. Bob and Maggie are about to have their first child and they know they have close relatives with this disorder (see pedigree below). (cont) What is the probability that the child of Bob and Maggie (IV-1) will have the disorder? Don't forget that the question mark in the symbol for IV-1 indicates you should consider all possible genotypes. For your answer, you should use the table and write genotypes and probabilities for individuals needed to solve this problem, and write all necessary equations. The numbers of rows and columns in each table are not necessarily the ones needed to work the problem. Use A for the dominant allele and a for the recessive allele. Use three decimal points for your calculations. Maggie's father (II-3) must be Aa because Maggie's grandmother (I-3) is aa Since Maggie's mother (II-2) has a sibling who is aa (II-1), both of Maggie's maternal grandparents (I-1 and I-2) are Aa. So, Maggie's mother (II-2) has 1/3 chance of being AA and 2/3 chance of being Aa. Bob has a sibling with the disorder (III-3), so both parents (II-4 and II-5) must be Aa. So, Bob has 2/3 chance of being Aa. If Maggie and Bob are both Aa, then the chance of their child being aa is 1/4 There are two ways that both Maggie and Bob can be Aa

II-2

II-3

III-1

III-2

IV-1

1/3 AA

Aa

1/2 Aa

2/3 Aa

1/4 aa

1/3 x 1/2 x 2/3 x 1/4 = 2/72

2/3 Aa

Aa

2/3 Aa

2/3 Aa

1/4 aa

2/3 x 2/3 x 2/3 x 1/4 = 8/108

2/72 = 0.028

8/108 = 0.074

Total

Total = 0.028 + 0.074 = 0.102

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GENE 3200 – Bedell - Fall 2019 Practice Problem Set 10: Oct 28 and Oct 29

KEY

2. The pedigree below is for a very rare, autosomal recessive disorder. What is the probability that the child of III-2 and III-3 is a heterozygous carrier? Note that the symbol for IV-1 is not filled and it does not have a question mark. This means that IV1 is unaffected. For your answer, you should use the table and write genotypes and probabilities for individuals needed to solve this problem, and write all necessary equations. The numbers of rows and columns in each table are not necessarily the ones needed to work the problem. Use A for the dominant allele and a for the recessive allele. Use three decimal points for your calculations. Since this disorder is very rare, we can assume that II-3 is AA. IV-1 can be Aa if either or both of his parents are Aa. III-2 has to be Aa because his father (II-1) is aa. III-3 can be either AA or Aa and the probability of her genotypes depends on whether II-4 is AA or Aa. Since II-4 has a sibling that has the disorder, both I-3 and I-4 must be Aa. So, II-4 has 1/3 chance of being AA and 2/3 chance of being Aa. If II-4 is AA, then III-3 has to be AA. If II-4 is Aa, then III-3 has 1/2 chance of being Aa and 1/2 chance of being AA. Note that because IV-1 doesn't have the disorder, he cannot be aa. There are three way that IV-1 could be Aa.

II-3

II-4

III-3

III-2

IV-1

Total

AA

1/3 AA

AA

Aa

1/2 Aa

1/3 x 1/2 = 1/6

AA

2/3 Aa

1/2 AA

Aa

1/2 Aa

2/3 x 1/2 x 1/2 = 2/12

AA

2/3 Aa

1/2 Aa

Aa

2/3 Aa

2/3 x 1/2 x 2/3 = 4/18

Total = 1/6 + 2/12 + 4/18 = 0.167 + 0.167 + 0.222 = 0.556

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