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MIT 7 03 - Basic Genetic Tests

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Exam Questions from Exam 1 – Basic Genetic Tests, Setting up and Analyzing Crosses,and Genetic Mapping1. You are studying three autosomal recessive mutations in the fruit fly Drosophilamelanogaster. Flies that are homozygous for the hb– mutation are “humpbacked” (wild-type flies are straight-backed). Flies that are homozygous for the bl– mutation are“blistery-winged” (wild-type flies are smooth-winged). Flies that are homozygous for thest– mutation are “stubby-legged” (wild-type flies are long-legged).You mate flies from two true-breeding strains, and the resulting F1 flies are all arestraight-backed, smooth-winged, and long-legged. F1 females are then mated to malesthat are humpbacked, blistery-winged, and stubby-legged. In the F2 generation, among1000 progeny resulting from this cross, you observe the following phenotypes:Phenotype Numberhumpbacked, blistery-winged, and stubby-legged (26 flies)humpbacked, blistery-winged, and long-legged (455 flies)humpbacked, smooth-winged, and long-legged (24 flies)straight-backed, blistery-winged, and stubby-legged (27 flies)straight-backed, blistery-winged, and long-legged (4 flies)straight-backed, smooth-winged, and stubby-legged (442 flies)straight-backed, smooth-winged, and long-legged (22 flies)(a) The male flies that were bred to the F1 generation in order to produce the F2generation were humpbacked, blistery-winged, and stubby-legged. On each of theirchromosomes, they have the alleles hb– bl– st–. Using this notation, state thegenotype of each of the two true-breeding parental strains (i.e. the two strains in the Pgeneration).Genotype of one parental strain:Genotype of the other parental strain:?Mouse #4Mouse #1Mouse #3Mouse #2(b) How many flies are found in the class that is the reciprocal class of the humpbacked,blistery-winged, and stubby-legged flies?(c) What is the genetic distance between the hb and bl loci? (Label your answer withthe proper units.)(d) What is the genetic distance between the bl and st loci? (Label your answer withthe proper units.)(e) Draw a genetic map showing the correct order of the hb, bl, and st loci.2. The following mouse pedigree shows the segregation of two different mutanttraits. The mutant trait indicated by the dots is dominant, whereas the mutant traitindicated by the stripes is recessive. Assume 100% penetrance and no new mutations.(Squares = males, circles = females.)= expressing recessive mutant trait, caused by the “b” allele= expressing dominant mutant trait, caused by the “A*” allele= expressing both mutant traits“A*” or “a” locus“B” or “b” locusHomolog inheritedfrom mouse #1Homolog inheritedfrom mouse #2(a) Assuming that both mutant traits are due to linked autosomal genes that are 6 cMapart, fill in the following chart using the allele notation indicated by the key above.Blocks in the chart that cannot be filled in conclusively should be indicated as“inconclusive.”NOTE: One line of the chart is already filled in correctly for you.Number of“A*” allelesNumber of“a” allelesNumber of“B” allelesNumber of“b” allelesMouse #1Mouse #2Mouse #30202Mouse #4(b) Assuming that both mutant traits are due to linked autosomal genes that are 6 cMapart, fill in the boxes with the alleles possessed by mouse #4 on each of the twohomologs of this autosome that are depicted in the diagram below.(c) Assuming that both mutant traits are due to linked autosomal genes that are 6 cMapart, what is the probability that the mouse indicated by a question mark will showboth mutant traits (the trait encoded by “A*” and the trait encoded by “b”)?(d) Assuming that the recessive mutant trait is caused by a gene on an autosome andthe dominant mutant trait is caused by a gene on the X chromosome, fill in thefollowing chart using the allele notation indicated by the key above. Blocks in the chartthat cannot be filled in conclusively should be indicated as “inconclusive.”Number of“A*” allelesNumber of“a” allelesNumber of“B” allelesNumber of“b” allelesMouse #1Mouse #2Mouse #3Mouse #4(e) Assuming that the recessive mutant trait is caused by a gene on an autosome andthe dominant mutant trait is caused by a gene on the X chromosome, what is theprobability that the mouse indicated by a question mark will show only the recessivemutant trait assuming that the mouse is born female?3. You are working with a mutant strain of yeast that is dark tan (wild-type yeast arewhite). The “dark tan” phenotype of the haploid cells you are working with is caused bytwo different mutations in the same strain. The two mutations are designated drk1– anddrk2–.(a) Mating of the drk1– drk2– double mutant to wild-type yeast produces diploids thatare white. Sporulation of these diploids yields 50 tetrads. 4 of these tetrads (called“Type One”) contain four light tan spores. 37 of these tetrads (called “Type Two”)contain two dark tan spores and two white spores. 9 of these tetrads (called “TypeThree”) contain one dark tan spore, two light tan spores, and one white spore.Categorize each of the tetrad types as parental ditype (PD), tetratype (TT), ornonparental ditype (NPD).X-linkedautosomal(b) Are the drk1– and drk2– mutations linked? If so, give the distance between them.(Label your answer with the proper units.)(c) In yeast, 1 cM of genetic distance corresponds to 3,500 base pairs of physicaldistance. An average yeast gene is about 1,400 base pairs long, and the longest yeastgene is 14,700 base pairs. Keeping this information in mind, you select a “Type Three”tetrad from part (a) and mate the two light tan spores from that tetrad to each other.Can you deduce the color of the resulting diploids? If so, what color would the diploidsbe?Next you isolate a mutant strain of yeast that cannot grow on medium lacking leucine.This strain contains a single mutation you call leu1–. The leu1– mutation is near todrk1– on the same chromosome. When the leu1– mutant is mated to wild-type yeast,the resulting diploids cannot grow on medium lacking leucine.(d) You mate leu1– yeast to drk1– yeast and sporulate the resulting diploid. You growthe resulting spores on medium containing leucine. You then test for growth on mediumlacking leucine. It is apparent that you have isolated only two types of tetrads, 10tetrads of Type A and 10 tetrads of Type B. On medium lacking leucine, only twospores


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MIT 7 03 - Basic Genetic Tests

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