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Unformatted text preview: Genetics Homework #4 key 1. The recessive mutations a, b, and c identify three autosomal genes in D. melanogaster. The following progeny were obtained from a testcross of females heterozygous for all three genes. abC 243 aBC 241 aBc 15 abc 10 ABc 235 Abc 226 AbC 12 ABC 18 What conclusions are possible concerning the linkage relationships of these three genes? Calculate any appropriate map distances. If you rewrite the 8 different classes so that they are paired with the opposite phenotypes, you’ll notice a slight problem. There are two pairs (4 classes) with about equal amounts of progeny (somewhere around 240), and then the other two pairs also have equal amounts (~15). It’s impossible to distinguish which are the nonrecombinants, and which are the double crossovers, so we can’t establish gene order. However, you need to realize that there is the possibility that all three genes are not linked. In order to confirm this, you must look at each gene pair. Ab 238 Ac 461 bc 236 aB 256 aC 484 BC 259 ab 253 ac 25 bC 255 AB 253 AC 30 Bc 250 The only gene pair that shows linkage when examined this way is AC. B is clearly not linked to these other two. To calculate the map distance between A and C, divide the number of recombinants (55) by the total number of progeny (1000) for a map distance of 5.5 mu. 2. A genetic map of a region with three linked genes, T, R, and J, is shown below. T 30 cMR 10 cMJ Suppose that you cross a pure‐breeding strain TRJ with a pure‐breeding trj strain, and then testcross the F1 progeny. a. What percentage of the testcross progeny would you expect to be phenotypically Tr? b. What percentage of the testcross progeny do you expect to be phenotypically TRJ? Assume no interference in this region. c. Same question as part (b), but now assume that interference = 0.2. To solve parts a and b, you first need to determine what the 8 expected phenotypic classes will be in the F2, and then use the information in the given map to determine how many offspring you would predict for each class. I’ll use region I to describe the region in between T and R, and region II for RJ. To do the calculations, you need to realize that the expected % of DCO will be 30% * 10% = 3%. If 3% of the progeny are from DCO, and we need to have 30% RF in region I, then the expected % of SCO (I) is 30%3% = 27%. Likewise, for region II, we expect 103=7%. This leaves 1002773 = 63% as NCO. Each of these will be split between the two reciprocal classes. Phenotype Class % TRJ NCO 31.5 trj NCO 31.5 Trj SCO (I) 13.5 tRJ SCO (I) 13.5 TRj SCO (II) 3.5 trJ SCO (II) 3.5 TrJ DCO 1.5 tRj DCO 1.5 (a) Phenotypic class Tr would be 13.5+1.5 = 15% (b) Phenotypic class TRJ would be 31.5% Part c adds a level of complication that will force you to recalculate all of the above percentages. If interference is 0.2, then we don’t see 20% of the expected DCO, or we could say that we only see 80% of them. If we expect 3% DCO, then with interference, we would only see 0.03 * 0.8 = 2.4%. We then need to recalculate both of the SCO percentages so that they still add up to 30 and 10, which then means that we take the remainder out of 100% to get the NCO classes. Each DCO class is 1.2%, each SCO (I) is 13.8%, each SCO (II) is 3.8%, and each NCO is then 31.2%, which is the answer to the question. ...
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This note was uploaded on 09/01/2011 for the course BICD 100 taught by Professor Nehring during the Winter '08 term at UCSD.
 Winter '08
 Nehring
 Genetics, Mutation

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