7 03 Fall 2003 1 of 6 Answer Key Problem Set 5 1 a Genetic properties of gln2 and gln 3 Both are uninducible as they give decreased glutamine synthetase GS activity Both are recessive as mating them with wildtype produces normal GS activity Both are trans acting as when either mutation is crossed to gln1 complementation takes place if either gln2 or gln3 were a cis acting element regulator of gln1 we would not see complementation as the two mutations would be in the same gene i e the promoter is considered to be part of the gene it acts on Since we see complementation in both cases we know that both mutations must be trans acting Therefore both gln2 and gln3 are mutations in genes for positive regulators b Linear models gln2 gln3 gln1 gln3 gln2 gln1 Parallel model gln2 gln1 gln3 c The parallel model from part b best fits the experimental results as we are told that the addition of glutamate glu and glutamine gln independently regulate the GS activity i e the results of the addition of each is different from the addition of both glu gln2 gln1 gln gln3 d We would expect the gln2 gln3 double mutant to be uninducible e The cis acting elements in the promoter are Deletion1 300 to 250 GLN2 function activity pattern looks like gln2 when deleted UAS Upstream Activation Sequence involved in activation Deletions 3 4 200 to 100 GLN3 function activity pattern looks like gln3 when deleted UAS Upstream Activation Sequence involved in activation Deletion 6 50 to 0 TATA sequence 1 7 03 Fall 2003 2 of 6 f We would expect to see no change in beta galactosidase expression in Deletion 1 in a gln2 mutant as in e we determined that the region deleted in 1 is necessary for GLN2 function i e deleting both the cis acting element and gln2will look the same as deleting either one We would expect 0 units of betagalactose from Deletion 4 or low baseline level as deletion 4 is necessary for GLN3 function and we know that when both the gln2 and gln3 pathways for activating GLN1 are mutated neither pathway can activate GLN1 2 a We can use the information from the order of the BACs to figure out where each STS is For instance BAC A is positive for STS 3 which is not present in any other BAC therefore it must be in the region of A that does not overlap with B or C There are three ambiguities in the data resulting in three possible maps Most of the ambiguity lies in whether BAC D contains both STS 51 and STS 52 as PCR only provides the qualitative answer that STS 5 is present we don t get quantitative data as to how many versions of STS5 are present on the BAC Case1 displays the possibility that BAC D contains both STS5s Given this case we cannot determine the order of STS51 and STS 2 thus the parenthesis Case 2 displays the possibility that STS51 is not on BAC D If this assumption is true then we can determine the order of STS51 and STS2 Case 3 displays the possibility that BAC D contains only STS51 If this is the case we cannot determine the relative order of STS52 and the second STS1 CASE1 3 7 1 51 2 4 52 7 1 5 2 4 5 1 5 2 4 5 1 D A E B 1 6 C F CASE 2 3 7 1 2 4 52 7 1 5 2 4 5 1 5 4 5 1 D A B 1 C 2 E 6 F 2 7 03 Fall 2003 3 of 6 Case Three 3 7 1 51 2 4 7 1 5 2 4 52 1 5 2 4 5 1 D A B 1 E 6 F Therefore there are quite a few ambiguities in our map The first is that we do not know the order of STS 51 relative to STS2 It is possible that STS 2 comes before STS 51 or STS 51 comes before STS 2 Also we do not know if D is carrying STS 51 and 52 or if it is carrying just 52 Consider the possibility that D contains only STS 52 we would have to assume that STS 2 came after STS51 and the STS 5 we are detecting cames from STS 52 In addition there is a third possibility which is that the STS 5 contained on BACD is only STS51 If that is the case then we don t know the order of STS5 with respect to the later STS1 We need an assay to distinguish between the possibilities b PCR just amplifies DNA to which primers bind One could imagine that the PCR primers could anneal to both sites and amplify both sites This would result in a heterogeneous mixture of STS51 and STS52 D 51 52 However PCR alone does not help us distinguish between the three possibilities This sequencing assay is our assay to distinguish between the three cases Sequencing can tell us whether BACD contains both STS 51 and STS52 just STS51 or just STS52 Given these data it looks as if BAC D is carrying both STS5s which allows us to throw out cases two and three There is a remaining ambiguity we still do not know the order of STS51 relative to STS2 c depending on your answer it could confirm refute Most answers should be refined as we can now distinguish between the three cases There are some remaining ambiguities but the map is significantly more refined after the sequencing assay of STS5 d You would expect both sequences to be present in the genome BACs are derived from the mouse genome and are therefore a reflection of what is present in the genome If we find to STS 5 in twice our BACS there are probably at least two STS 5 on the mouse chromosome Only BAC D contains both STS5s e STS1 is present more than once BACs A B C would have the same sequence E F would have the same sequence It is possible that the STS1 from BACs A B C have a different 3 7 03 Fall 2003 4 of 6 sequence than those from STS1 from BACs E F Primers that distinguish between these two STS1 would be convenient as we would not have to be concerned about an ambiguity in our data in the future like there was with STS5 and BAC D f Design primers that include the non homologous sequence in STS51 and STS52 These primers should specifically amplify either STS51 or STS52 g Yes STS51 and STS52 should both be present in BAC D and the mouse genome refer to d 3 a The integration of Pamylase LacZ into the amylase gene is unlikely to occur when the construct is microinjected into the male pronucleus of a fertilized egg Integration of the construct into the endogenous amylase gene locus could occur via homologous recombination or by chance non homologous insertion into the endogenous amylase gene Homologous recombination …
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