Problem Set 4 - key

Problem Set 4 - key - Biology 202 Problem Set 4 2/16/07...

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Biology 202 Problem Set 4 2/16/07 From Griffiths: Chapter 4, problems 1, 2, 4, 5, 6, 7, 9, 12, 13, 14, 17, 21, 24, 25, 29, 30 Answers to the following questions are due in February 28 th 1. Two linked genes, ( A ) and ( B ), are separated by 18 cM. A man with genotype AB / ab marries a woman who is ab / ab. a. What is the probability that their first child will be Ab / ab ? The recombination frequency between these gene loci is 18%. The man is thus expected to produce a total of 18% recombinant gametes, which includes two reciprocal products (9% A / b + 9% a / B ). The mother produces only one kind of gamete ( ab ). For the first child to be Ab / ab , he/she must inherit a specific recombinant gamete ( A / b ) from the father (probability = 0.09), and any gamete ( a / b ) from the mother (probability = 1). Therefore, the final probability of the first child being Ab / ab = (0.09)(1) = 0.09 (or 9%). b. What is the probability that their first two children are ab / ab ? The father is expected to produce a total of 18% recombinant gametes (9% A / b + 9% a / B ), and 82% non-recombinant gametes (41% A / B + 41% a / b ). Use the multiplicative rule to determine the probability of the couple’s first two children being ab / ab : (0.41)(1) × (0.41)(1) = 0.1681 (or 16.8%). 2. You just bought two black guinea pigs from the pet store that are known to be heterozygous ( Bb ). You also know that black fur ( BB ) is dominant over white fur ( bb ), and that a lethal recessive allele is located only one cM away from the recessive b allele. You decide to start raising your own guinea pigs, but after mating these animals several times you discover they only produce black progeny. a. How would you explain this result? The recessive lethal allele is tightly linked to, and thus co-segregates with, the recessive b allele, which is lethal in the homozygous state ( bb ). b. If the original black guinea pigs produce an average of 10 offspring per mating, how many matings would you have to make before you’d expect to see a white guinea pig? 1% of the gametes will be recombinant (Bl or bL) 0.5% of the gametes will be Bl 0.5% of the gametes will be bL white animals can have the following genotype:
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genotype probability bL/bL (0.005)(0.005) = 0.000025 bl/bL (0.495)(0.005) = 0.002475 bL/bl (0.005)(0.495) = 0.002475 probability of white animal = 0.004975 1/0.004975 = 201 therefore on average you’d have to look at 201 progeny before seeing a white animal so on average you’d have to do 21 matings. c.
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Problem Set 4 - key - Biology 202 Problem Set 4 2/16/07...

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