7.03 Genetics: Answers to Problem Set #4 (1993)

7.03 Problem Problem Set 4 1 a) The complementation groups FA, FB, FD and FF have been identified groups. and C3/FD FA and C2/FB and C1/and affect the same gene ( they do not complement by both FF and C4, are alleles that each other). The complementation groups FC and FE have onlybeen identified inBloomingtonand the complementationgroup C.5has onlybeen identified inCologne. b) C2 C3 C1 / .................... / / / / Del 4 ............. t Del 5 / / FA FB -_.................................. /-................... /-................................... / (FD // .................... /- ........................................ .............. / _ Del 2 Del 3 C4 / Del I -4 --/ C5 / / FE FC) /. FF This is just one possible way to write the map. There is ambiguity with respect to the order of :FI), FE and FC. These three genes have to be containbA between C3/FB and C5 but the order is arbitrary (8 possibilities) [Remember that FE and FC are In different genes since the original crosses were made with a representative allele of each complemenmdon group isolated by each research group] c) A three factor cross can be done in order to determine the order of the genes. In this case the generation of the transheterozygotes and double mutants is more difficult because this organism is diploid and these mutations can not be bred into homozygozity because they are Iethal. The first step would be to generate a transheterozygote ml +])* + x + + m2 * the wild type chromosomecarries +3* + a markerthat gives a dominant CI_) non-lethal ml + + m2 by crossing two single mutants. phenotype The progeny that do not carry * dominant marker are the transhetemzygotes mutations. for the two ,rib. Once you have a female transheterozygote you cross it to a m/ale fly. What you want to isolate is the double mutant which is a product of a recombination ml + + x event. + + m2 + + " ml +/+ + These are the type of progeny that you would get out of + m2/+ ÷ such a cross. ml m2/+ + Since you can't distinguish between these possibilities by phenotype you would have to cross each of them to the original single mutants and isolate the progeny that do not complement with either of the single mutants, that is the double mutant. Once you have each of the double mutants FC FE, FC FD and FD FE you would now be able to do the experiment to determine order. To do this you would fixst have to generate a transheterozygote between a double mutant chromosome and a single mutant chromosome(to obtain this think'about the transheterozygote generated above) For example: FD + FC + + FE X + Del D* + The majority of the viable progeny will be of the genotype: FD FC +/D* and + + FE/D* Lethal progeny that will not be recovered: FD FC +/Del and + + FE/Del Only progeny that is viable in trans to the deletion will be the products of rare "interesting" recombinants of the genotype: + + +/Del Transheterozygous females are crossed to males which carry one of the deletions used in (b) (Dell, Del2, Del4 or Del5) which are deleted for all three genes. To distinguish the chromosome that carries the deletion from its homologue which carries the wild-type alleles for all seven genes studied this chromosome is marked with a dominant marker D* as described above. Most progeny of this cross carry the chromosome marked with D* in trans to the mutants and you are not interested in those. You search the viable progeny for the products of recombination. Only recombinants that are wild-type for all three genes will be viable in trans to the deletion chromosome. If FE mapped on either side of FC or FD, wild-type recombinants can be produced by a single cross-over event. If however, FE mapped between FC and FD then the frequency of viable progeny would be very low since it would require two-cross over events to generate the + + + chromosome. You would perform crosses with the three possible combinations (FD FC/FE; FD FENC; FC FE/FD). By comparing the frequency of viable progeny in trans to the deficiency you can determine the relative order of the three genes as we did in phage genetics. Any answer that suggests to resolve the order of the three genes by three factor crosses or by recombination distances will get partial credit. d) This stock fails to complement the mutant alleles C5, C4/FF, FE and FC. Thus this stock most likely carries a deletion for this region. Since this deletion faiis to complement the FC mutation but does complement the lethality caused by the FE and FD/C1 mutations you can refine the order of genes: FA/C2--FB/C3-- (FE--FC)--FD/C 1--C5--FF/C4 The relative order of FE and FC is still not determined. _j Firstdeterminethe ge__s of_-_z_ee in_als: _ \ _j. Let (Xc)representsthe X chromosome with the recessive allele forcolorblindness. The son isC_c)(Xc)Y The mother is(Xc)(X+) The fatheris(X+)Y Clearlythe son must have receivedthe Olc)chromosomes from hismother. The nondisjunction eventthatoccurredin meiosistwo ofthe mother isdiagrammed below: