Gen371 Practice 3 Answers

Gen371 Practice 3 Answers - Gen371 Practice 3 Answers 4:18...

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1/26/11 4:18 PM Gen371 Practice 3 Answers Page 1 of 7 Practice problems set 3 -- answers 1. The pedigree below shows the inheritance of red-green colorblindness, an X-linked recessive trait ( X D = normal vision, X d = colorblindness allele). For all of the questions below, assume that there is no abnormal event or new mutation; also ignore X-inactivation for this problem. (a) Below each individual (except III-1), write out their genotype for color vision. Use "?" to indicate any allele about which you are uncertain. (b) What is the probability that III-1 will be colorblind? Explain your reasoning to show how you arrived at your answer. For III-1 to be colorblind, it has to receive X d from the mom (II-2) and Y from the dad (II-3). Probability that the mom is X D X d = 1/2 Probability that she will transmit X d if she is X D X d = 1/2 Probability of Y from the dad = 1/2 Overall probability = 1/2 x 1/2 x 1/2 = 1/8 . (c) Using long and short chromosomes for the X and Y, respectively, diagram the sex chromosomes as they would appear in a proper metaphase I of meiosis in individual II-3 of the pedigree above. Be sure to mark the color-vision locus where you think it could be located. Notes: This question was on a test previously. Many people lost points because they turned II-3 into an XXYY individual. .. two X's paired with each other and two Ys separately paired with each other. Because homologous pairing and crossing over can only happen in the pseudoautosomal region, that's where the crossover has to be placed. Because colorblindness is an X-linked trait, the gene has to be present only on the X and not on the Y. So, it has to be marked on the X chromosome at a location outside the pseudoautosomal region and not marked on the Y at all. 2. As you saw in class, polydactyly is a dominant trait affecting human hand development: D = polydactyly (extra fingers) and d = its normal, recessive allele. Home Course mechanics Help hours Calendar Syllabus Lectures, Podcasts Quiz Sections Practice problems Exams GoPost Send email to class Useful links The Gradiator
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1/26/11 4:18 PM Gen371 Practice 3 Answers Page 2 of 7 The pedigree shows inheritance of polydactyly. Blood types (phenotypes), where known, are also shown. II-3 is the second wife of II-2. (a) Below each individual, write out their genotype for polydactyly and blood types, using D/d and I A /I B /i as allele designations. If more than one genotype is possible, list all possibilities. (b) Assuming that blood type and polydactyly are independently assorting traits, what is the probability that III-3 will have blood type B and have extra fingers? (Show your calculations.) Probability of dad transmitting D to offspring = 1/2 Probability of dad transmitting i = 1/2 Probability that mom has I B = 1/2 Probability that mom will transmit I B if she has it = 1/2
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This note was uploaded on 03/31/2011 for the course GENOME 371 taught by Professor Unsure during the Spring '03 term at University of Washington.

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Gen371 Practice 3 Answers - Gen371 Practice 3 Answers 4:18...

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