Update: Time Line of Revolutions in GeneticsCarl Correns,Hugo DeVriesRediscovery1900Gregor Mendel1866Walter Sutton,Theodor BoveriChromosome theory1902-3| | | | | | | |T. H. Morgan,A. H. Sturtevant,C. B. Bridges,H. J. Muller,Drosophila,recombination, mapping,chromosome mutations,etc.1910-1916Complementation groups, genes, and alleles.Object:Determine if two mutations are in the same or different genes.e.g. isolate three arginine auxotrophic mutations in haploid yeast strains:arg1, arg2, arg3arg1arg2arg3Cross a arg1 × α arg2, look at diploid phenotype:arg1arg2The diploid grows on minimal medium because each mutant came with the wildtype allele of the other gene, thereby complementing it. So arg1 and arg2 are indifferent genes, and they are not alleles.arg2arg3Cross a arg1 × α arg3, look at diploid phenotype:arg1The diploid does not grow on minimal medium because each mutant is in the samegene and there is no wild type copy of that gene in the diploid. arg1 and arg3 donot complement each other, so they are different mutations of the same gene, andthey are alleles.We have identified two genes, which we might call ARGA and ARGB.ARGA has three alleles: ARGA, argA1, and argA3.ARGB has two alleles: ARGB and argB2.arg3arg1Isolate two more arg auxotrophs, test growth of diploids on MM:-+++++++-+arg1arg2arg3arg4arg5arg1 arg2 arg3 arg4 arg5arg4 and arg5 are in another complemention group, so we now have three complementationgroups and have identified three genes. Each one may encode an enzyme for a different stepin the biosynthesis of arginine.Can you explain hw # 5.8 [I did this in Discussion] and 8.12 as well as another example ofphysical and genetic maps?8.12 Recessive genes a, b, c, d, e, and f are are closely linked but their order is unknown.Three deletions in the region are examined. One deletion uncovers a, d, and e; anotheruncovers c, d, and f; and the third uncovers b and c. What is the order of the genes?“Uncovers” means that in a heterozygote that has one chromosome with a deletion but isotherwise wild type and the other chromosome has recessive alleles, one sees the recessivephenotype for the gene that is “uncovered”.Deletion 1 uncovers a, d, and e, so these are in a cluster that includes none of the others:a d e or a e d or d a e.Deletion 2 uncovers c, d, and f, so they are also clustered: c d f or c f d or d c f. But d must beat the end because it is also uncovered by deletion 2, so we have a e d c f or a similar order.The third uncovers b and c, so these are adjacent to each other. We also know that b is not inany of the other clusters, so c must be at the end of the combined cluster identified bydeletions 1 and 2.The order must be as shown below, with the three deletions shown in order below it.Note that for mapping purposes, deletions are shown as lines where the deleted segmentsare, not chromosomes missing certain regions.a e d f c b (or e a d f c b)w.t.deleteddeletionw xWIntragenic recombinationThis simply means recombination between two mutations in the same gene. Itis detected by making a heterozygote for the two mutations and finding wildtype recombinant chromosomes.We would also have to do a complementation test to know that the mutationsare in the same gene. Finding a wild type recombinant merely shows thatmfg1 and mfg2 are not two cases of the same mutation (in the same base pair).mfg1mfg2mfg1 mfg2mfg+Cross with mfg/mfgmfg+mfgw.t.mfg1 and/or 2mfgmutantDuring our review tomorrow can you explain how to do problems 2 and 3 from homework 2[I reviewed these in Discussion.]You won’t be required to do a complete analysis, but might be asked to do part of it.2. In maize, a strain homozygous for two recessive mutations, liguleless (lg)and glossy (gl),was crossed to another strain homozygous for a dominant allele, Booster (B). An F1 plantwas backcrossed to the liguleless glossy parent strain. The progeny phenotypes and numbersare shown below. (For example, lg + gl 172 means that 172 plants were liguleless glossy.)Make a map with these three genes. Carry calculations to two decimal places.lg + gl 172 F1 had lg B+ gl on one chromosome+ B + 162 and lg+ B gl+ on the other. lg B gl 56 This has B and gl together, so it is a recombinant for these loci.+ + + 48lg B + 51+ + gl 43lg + + 6+ B gl 5543By comparing the double crossover genotypes to the parental genotypes, we can tell that glis in the middle.Distance lg-gl = 51 + 43 + 6 + 5 = 105 105/543 = 0.1933 = 19.33 cMDistance gl - B = 56 + 48 + 6 + 5 = 115 115/543 = 0.2118 = 21.18 cMMap lg gl B |-------------------|---------------------|19.33 21.183. In dragons, three genes (a, b, c) have been mapped on achromosome as shown below, with distances between the loci incentimorgans:a b c|----------|------| 5 3Suppose a dragon of genotype A A b b C C was mated to one ofgenotype a a B B c c and the offspring were testcrossed. What isthe expected frequency of a a b b c c dragons in the progeny fromthe test cross? Give answer to 5 decimal places.Offspring had one chromosome with A b C and the other with a Bc. So the only way to get a a b b c c is by a double crossover.A b C---------------- X X----------------a B cAnswer: (0.05)(0.03) = 0.0015 = P(double crossovers) = f(aabbcc +AABBCC)f(aabbcc) = 0.0015/2 = 0.000751.) What do we need to know about the Chi squared test for the exam?You won’t have to do one on an exam.Should we be able to recognize what is statistically acceptableaccording to this test?Yes2.) I understand what n, w, and x represent individually, but whatexactly does n!/(w!x!) without (p^w)(q^x) represent?It calculates the number of orders.3.) On slide 4 of the 15th lecture is a time line of scientificdiscoveries, are these all the names, dates, and events we need toknow for the exam?See first slide.4.) On slide 19 of the 3/2/07 discussion, in the very bottom rightcorner, you have the distance from lz to gl as 24.7. I thought thatthis distance should instead be the result of the addition betweenlz-su (10.7) and su-gl (14.7), which is 25.4. Could you please explainwhy it is actually 24.7?My error, now fixed.Could you discuss:1.problem 3 from homework sheet 2.Similar problem done earlier.2. problem 10 from Practice problems 4Similar problem done