Answer key to Prof Urnov s section 1 The following is Fig 5 13 from your textbook It shows one of the most majestic achievements of the scientific method a map of the Drosophila genome obtained without any knowledge of the chemical basis of heredity i e solely via genetic analysis a 5 pts We talked about the fact that unlinked genes recombine at a frequency of 50 That said for some reason the genetic distance on Chr 1 between two linked genes aristaless antenna and blistered wings is a whopping 107 map units How could this impossible value have been experimentally measured Did they observe recombination 107 of the time Answer the value was measured indirectly by adding map distances that span intervening genes b 20 pts A white eyed male with normal veins on its wings and a body of wild type color is mated to a red eyed female with thick veined wings and a black body All the progeny of this cross are wild type Female progeny of this cross are mated to white eyed males with thick veins on their wings and a black body One thousand progeny from this cross are phenotyped Write out the phenotypic classes observed in this progeny and the number of flies in each class assuming perfect compliance with the genetic distances shown above and perfect compliance with Mendel s laws Preliminary considerations 1 The white locus is unlinked so can be considered separately as per Mendel s second law 2 The thick veins and black body loci are separated by 32 map units so 32 of chromatids that emerge from meiosis in the female are recombinant and the remaining are parental Answer note for each class half will be w and half will be w Normal veins wild type body color Thick veined wings black body Normal veins black body Thick beined wings normal body 340 340 160 160 Questions on Prof Cline s section For each question below 2 6 BRIEFLY EXPLAIN your answer even if your answer is there is no logical basis from Genetic principles and or vocabulary for an answer other than can t say based on the information provided Remember not to overthink the questions A 2 Consider two different dominant gain of function alleles of the same gene in the German Shepherd dog Q H is an antimorph while Q is a hypermorph The phenotype of the null Q allele is no tail and the mutant allele is fully recessive a 2 pts What can you say about the likely phenotype of a QA German Shepherd Since QA is an antimorph we expect it to antagonize the wildtype gene function Since we are told it is dominant it must have a mutant phenotype From the phenotype of the null mutant dogs we know this gene is involved in tail length hence we can anticipate that the tail will be shortened although how much we cannot say since we don t know how powerful the antimorphic effect is b 2 pts Would you expect the mutant phenotype of QA dogs to be qualitatively similar to that of QH dogs or instead be qualitatively different Qualitatively different whatever it might be A priori there is no basis for predicting what effect increasing the activity of a gene above the wildtype level to a pathological level will likely have Information on the null phenotype or the qualitatively related QA phenotype is no help But since we know that QH does have a mutant phenotype and that the universe of possible phenotypes that would be qualitatively different from QA is huge while only one would be qualitatively similar the most we can say is that the phenotype is more likely to be qualitatively different c 2 pts QA QH dogs are wildtype Would it be correct therefore to say that the two gain of function alleles complement No we only use the term complementation or the complementation test to apply to the behaviour of loss of function alleles It doesn t really make sense to say one of these alleles provides the function that the other lacks which is what complementation means since neither of them really lack anything except proper control The wildtype phenotype that we see with this heteroallelic combination is not the result of restored proper control as complementation would demand but of a balance between the effects of two improperly controlled alleles d 3 pts If we reverted the dominance of QA and QH generating QA and QH respectively what would the likely phenotype be of a QA QH dog The most likely event to revert the dominance of any gain of function mutant allele is the induction of a loss of function mutation is cis to the gain offunction Hence the QA QH dog would be likely to have no tail or at least a very short tail the phenotype of an animal homozygous for the null allele or a hypomorph e 2 pts In light of the information given you in part c above what if anything could you predict A H A H about the phenotype of a Q Q dog if you were told that Q and Q satisfy expectations for a A H cis trans test for allelism Assume that Q is a doubly mutant allele created by recombination A H between Q and Q not very likely of course It will be abnormal in some way The only expectation from a cis trans test is that mutations that are allelic will have a different phenotype in cis vs trans Since you were told that the trans arrangement is wildtype and that these mutations satisfy the demands of the cis trans test we can infer that QA H Q arrangement must have a mutant phenotype of some kind 3 2 pts The mutagen ethylmethane sulfonate will generate null alleles but it is particularly effective at generating a wide spectrum of hypomorphic point mutants as well Brenner and crowd used proflavin as the mutagen for the rII gene of phage T4 in their famous experiments exploring the nature of the genetic code They demonstrated that the genetic code was commaless in part by showing that certain of their point mutant alleles could be suppressed by certain other of their point mutant alleles Do you think that some of the mutations they induced in rII by proflavin would likely be suppressed by some of the mutations that one might typically induce in rII by ethylmethane sulfonate No Proflavin induces frameshift mutation and Brenner s suppressors had to be frameshift mutations but of the opposite sign Ethylmethane sulfonate is not effective at making frameshift mutations but instead generates base substitutions of the transition category 4 2 pts To move DNA around fly workers generally use an engineered immobile genetic source of P element transposase called delta 2 3 derived from DNA that was originally introduced into an M strain animal After many generations will female flies from the delta 2 3
View Full Document
Unlocking...