7.03 Genetics: Answers to Problem Set 3 (2002)

2002 7.03 Problem Set 3 Answers 1. (a) During your study of the gene for an enzyme you have isolated an amber mutation in the middle of the gene which produces a truncated form of the enzyme. By placing a +1 frameshift mutation a short distance before the amber mutation and a –1 frameshift mutation a short distance after the amber mutation you create a triple mutant that restores expression of a functional, full-length enzyme. Explain how this could be. The +1 frameshift puts the amber mutation out of frame so that it is no longer read as a stop. However this throws off the frame of the rest of the protein unless a -1 frameshift is inserted after the amber to restore the frame of the end of the protein. For example: amber mutation: +1 frameshift mutation (nucleotide #4, A): 5’- …ACG GCA UAG AUG AUC… - 3’ stop 5’- …ACG AGC AUA GAU GAU C… - 3’ - 1 frameshift mutation (nucleotide #13, G): 5’ - …ACG AGC AUA GAU AUC… - 3’ (b) Working with the same gene as in part (a) you are studying intragenic suppressors of a +1 frameshift mutation near the middle of the gene. You find that a number of different –1 frameshift mutations a short distance after the +1 frameshift mutation will restore the expression of a functional enzyme. But to your surprise you find that –1 frameshift mutations placed a short distance before the +1 frameshift mutation lead to expression of a truncated enzyme. Propose a mechanism to explain these results. There must be a stop codon before the +1 frameshift that is normally not in frame and is thus not read, but is set into frame by a -1 frameshift. There is, however, no potential stop codon in the short distance after the +1 frameshift. For example: wt 5’ - …AAU AUA ACG AUC AAA TCG… - 3’ +1 frameshift (nucleotide 8, G) 5’ - …AAU AUA AGC GAU CAA ATC G… - 3’ -1 frameshift after +1 (nucleotide 11, A) 5’ - …AAU AUA AGC GUC AAA TCG… - 3’ -1 frameshift before +1 (nucleotide 2, A) 5’ - …AUA UAA GCG AUC AAA TCG… - 3’ stop (c) Organisms that live in environments exposed to light often have a relatively low proportion of A•T base pairs in their DNA (this is probably because neighboring thymine residues are particularly sensitive to reaction to UV light). The genes in such organisms tend to be biased for codons that contain G•C rather than A•T base pairs. As it turns out it is much easier to find coding sequences in organisms with a low A•T content in their DNA. To see why this is so, calculate the probability of finding an open reading frame of 50 codons by chance in DNA with equal frequencies of A•T and G•C base pairs and compare this to the frequency of finding an open reading frame of the same length in DNA that is made up of 40% A•T and 60% G•C. - In an organism that is 25% A, 25% T, 25% G, and 25% C, the probability of getting any of the three possible stop codons is: p(UAG) = 25% X 25% X 25% = 1/64 p(UAA) = 25% X 25% X 25% = 1/64 p(UGA) = 25% X 25% X 25% = 1/64 Therefore the probability of getting any non-stop codon is: 1- 3 x (1/64) = 61/64 and the probability of getting any non-stop codon 50 times in a row is: (61/64)50 = 9.1 % - In an organism that is 20% A, 20% T, 30% G, and 30% C, the probability of getting any of the three possible stop codons is: p(UAG) = 20% X 20% X 30% = 3/250 p(UAA) = 20% X 20% X 20% = 1/125 p(UGA) = 20% X 30% X 20% = 3/250 Therefore the probability of getting any non-stop codon is: 1- [ 1/125 + 2 x (3/250)] = 121/125 and the probability of getting any non-stop codon 50 times in a row is: (121/125)50 = 19.7% (c) The three stop codons are 5’UAG3’, 5’UAA3’, and 5’UGA3’. You are trying to isolate tRNA mutations that will suppress and ochre (UAA) mutation. To increase the frequency of mutations you use a mutagen that produces transition mutations (i.e. C•G to T•A and T•A to C•G base changes). Which tRNA gene can be altered by the mutagen to give the desired suppressor mutation? For your answer give the sequence of the codon normally recognized by the tRNA. In addition, write out the DNA base sequence of the segment of the tRNA gene that codes for the anticodon sequence for both the wild-type and suppressor alleles of the gene. Please show both strands of the DNA and indicate the 5’ and 3’ ends of each strand indicating which strand is used as the template for transcription of the tRNA molecule. For a mutant tRNA to recognize 5’-UAA-3', it must have a mutant anticodon 5'-UUA-3'. Therefore the mutant gene encoding the mutant tRNA must read 5'-TAA-3' on the strand that is transcribed (DNA uses Ts instead of Us). If the mutagen used to get this sequence causes transition mutations (a purine is changed for a purine or a pyrimidine is changed for a pyrimidine), then the original possibilities for the transcribed strand of the wildtype tRNA gene would be: 1. 5'-CAA-3' 2. 5'-TGA-3' 3. 5'-TAG-3' (The altered base is shown in bold.) The first possibility would encode a wildtype tRNA that recognized the codon 5'-CAA-3', which is the tRNA for glutamine. The other two possibilities would encode wildtype tRNAs that recognized 5'-UGA-3' or 5'-UAG-3', which are both stop codons. There are no wildtype tRNAs that recognize stop codons. Therefore: For the wildtype tRNA: DNA tRNA 5’ -……………T T G…………..- 3’ 3’ -……………AA C…………..- 5’ (The transcribed strand is shown in bold) 3’ Gln 5’ GUU For the suppressor tRNA: DNA 5’ -……………T T A…………..- 3’ 3’ -……………AAT…………..- 5’ tRNA 3’ Gln 5’ AUU 2. You have isolated two E. coli mutants in the PyrF gene, called PyrF-1 and PyrF-2. These mutants require a pyrimidine (such as uracil) to be added to the medium in order to grow – this phenotype is referred to as Pyr–. In order to work with these mutants, you wish to isolate a linked Tn5 insertion. To do this you start with a collection of 1000 different random Tn5 insertions in the otherwise wild type E. coli strain (these insertion strains are all Kanr and Pyr+). You grow P1 phage on the entire mixed collection of Tn5 insertion strains and then infect a PyrF-1 mutant and select for Kanr transductants. Most of the Kanr transductants remain Pyr–, but one out of 1000 is Pyr+. (a) What do you know about the position of the Tn5 insertion in this rare Pyr+ transductant? Does this frequency with which it arose (1/1000) tell you anything more specific about its position? This Tn5 insertion must be near to but not inside the PyrF gene and the distance between Tn5 and PyrF must be less than 105 bp, which is the size of the phage genome. The frequency does not tell you anything more specific than that. Next you grow P1 phage on the Pyr+ transductant isolated above and then infect the Pyr– mutant with the resulting phage. You select for Kanr transductants and then test these transductants for the ability to grow without pyrimidine. You find that 30 out of 100 Kanr transductants are Pyr+. (b) Give the distance between the Tn5 insertion and PyrF-1, expressed as a cotransduction frequency. The Tn5 insertion was cotransduced with the wildtype copy of PyrF 30% of the time. The Tn5 insertion shows about the same linkage to PyrF-2 as to PyrF-1. You construct a strain that has both the Tn5 insertion and PyrF-1 and another strain that has both the Tn5 insertion and PyrF-2. Using these strains you perform two reciprocal crosses. In the first cross, P1 is grown on the Tn5 PyrF-1 strain and the resulting phage are used to infect a PyrF-2 strain. In this transduction experiment, 7 out of 1000 Kanr transductants are Pyr+. In the reciprocal cross, P1 is grown on the Tn5 PyrF-2 strain and the resulting phage are used to infect a PyrF-1 strain. In this experiment, 50 out of 1000 Kanr transductants are Pyr+. (c) Draw a map showing the relative order of the Tn5 insertion, PyrF-1 and PyrF-2. Also include any relevant distances. There are two possible orders: Tn5, PyrF1, PyrF2 OR Tn5, PyrF2, PyrF1. If we draw out the reciprocal crosses using both orders, we can determine what the rarest outcome would be given each order. The rarest outcome will always be the quadruple crossover, and the more common outcomes will be the double crossovers. If the order is Tn5, PyrF1, PyrF2: The quadruple crossover from the first cross would give: Tn5 PyrF1 + + PyrF2 Genotype Phenotype Tn5, PyrF1 +, PyrF2 + pyr+ (in this case 7/1000) The quadruple crossover from the reciprocal cross would give: Tn5 + PyrF1 PyrF2 + Genotype Phenotype Tn5, PyrF1 -, PyrF2 - pyr- Both results are consistent with the data. If the order is Tn5, PyrF2, PyrF1, getting pyr + from the first cross would require a double crossover event and would occur (7/1000) times. Getting pyr + from the reciprocal cross would require a quadruple crossover and would occur (50/1000) times. It is not possible that a quadruple crossover event would occur more frequently than a double crossover event, and thus this order is not consistent with the data given. The order is therefore: Tn5 PyrF1 PyrF2 Once we know the order we can estimate the distance. We know that transducing Tn5 PyrF1+ PyrF2- into a host that is PyrF1 - PyrF2+ gives us (50/1000) KanR pyr+ transductants. In order to get pyr+ in this cross, we must not get a crossover between Tn5 and PyrF1, and we must get a crossover between PyrF1 and PyrF2. The probability of getting pyr+, which is (50/1000), equals the product of the probabilities of these two events, which must happen in order to get pyr+. The probability of not getting a crossover between Tn5 and PyrF1 is just the cotransduction frequency, or 30%. Therefore: p(getting pyr+) = p(not getting crossover between Tn5 & PyrF1) x p(crossover between PyrF1 & PyrF2) (50/1000) = (30/100) x p(crossover between PyrF1& PyrF2) p(crossover between PyrF1 & PyrF2) = 16.7% Therefore PyrF1 and PyrF2 would be cotransduced 83.3% of the time. Therefore: Tn5 PyrF-1 PyrF-2 30% 83.3% 3. You are studying the polysaccharides displayed on the outer membrane of a bacterial species related to E. coli. Mutants that do not produce the normal polysaccharides can be identified because they have a dry-looking colony morphology. (a) You use transposon mutagenesis to isolate two sets of mutants dry-looking colonies. Three mutants, numbered 1-3, were generated by random insertion of Tn5 r (carrying a gene for Kan ) into the bacterial genome, while three other mutants, numbered 4-6, were generated by random insertion of a second type of transposon r Tn10 (carrying a different antibiotic resistance, Tet ). You grow P1 phage on each of the three strains carrying a Tn5 insertion and then you use these phage for transduction of the Tn5 marker into each of the three Tn10 insertion mutants by selecting for Kanr after infection of the recipient strain: Donor strain 1::Tn5 1::Tn5 1::Tn5 2::Tn5 2::Tn5 2::Tn5 3::Tn5 3::Tn5 3::Tn5 Recipient strain 4::Tn10 5::Tn10 6::Tn10 4::Tn10 5::Tn10 6::Tn10 4::Tn10 5::Tn10 6::Tn10 % Kanr Tets 100% 10% 24% 2% 100% 100% 15% 100% 100% % Kanr Tetr 0% 90% 76% 98% 0% 0% 85% 0% 0% How many linked groups of mutants are there? What are the distances between the linked insertions, expressed as cotransduction frequencies? If two mutations are grouped together, then they will be cotransduced 100% of the time and the resulting phenotype of the strains will be KanR TetS. If the mutations are not grouped together, then sometimes they will be cotransduced in which case the resulting strain will be KanRTetS and sometimes they won’t in which case the resulting strain will be KanRTetR. Therefore 1::Tn5 and 4::Tn10 are in one group, and 2::Tn5, 3::Tn5, 5::Tn10, and 6::Tn10 are in another group. These insertions are grouped by position/distance only -- they cannot be considered complementation groups because we do not know the phenotypes of merodiploid bacterial strains containing pairs of insertions. The distance between 1::Tn5 and 5::Tn10 is 10%. The distance between 1::Tn5 and 6::Tn10 is 24%. The distance between 2::Tn5 and 4::Tn10 is 2%. The distance between 3::Tn5 and 4::Tn10 is 15%. (b) As it turns out, the set of Tn5 insertions (strains 1-3) were generated in an Hfr strain. In mating experiments to a Strr recipient, you find that although the 1::Tn5 Hfr r strain transfers Kan late, even after a brief mating time it is possible to isolate a few Kanr Strr exconjugants. You choose two of these exconjugate strains for further study (for reference you call these strains a and b. The set of Tn10 insertions (strains 4-6) – were generated in an F strain. You mate strains a and b with strains 4-6 selecting for Kanr and Tetr. Donor strain a a a b b b Recipient strain 4::Tn10 5::Tn10 6::Tn10 4::Tn10 5::Tn10 6::Tn10 Kanr Tetr exconjugant morphology dry dry wildtype dry wildtype wildtype Propose a mechanism to explain why the wild type colony morphology is restored in some of these matings and why the results for matings with strains a and b differ. Your answer should include diagrams of the form of F factor in strains a and b, showing any relevant genes and the position and orientation of the F factor origin of transfer. These rare exconjugates must have the F' form of the F factor, as they received the F factor from a strain that could transfer a marker early that is normally transferred very late. Donor b must have an F' that contains wildtype activity in the gene(s) in which the 5::Tn10 and 6::Tn10 insertions are inserted into, because this F' can complement the mutant phenotypes of both 5::Tn10 and 6::Tn10. The fact that the F' from donor b does not complement 4::Tn10 means that insertions 1 and 4 must be in the same gene. The F' from donor a must be wildtype for the gene in which 6::Tn10 is inserted because this F' can complement the mutant phenotype of 6::Tn10. Donor a does not carry a wildtype version of the gene in which 5::Tn10 is inserted or else it would complement. This is possible because from the cotransduction frequencies given above, we know that insertion 6 is closer to insertion 1 than insertion 5 is. This means that the F' could have recombined out of the genome and taken with it the gene in which 6 is inserted, but left the gene in which 5 is inserted behind, in the host chromosome. NOTE: The IS sequences drawn are the only relevant IS sequences for this problem, although there are more IS sequences scattered throughout the genome. (c) In studying one of the Tetr Kanr exconjugates isolated by the mating between strain b and 5::Tn10 isolated in part (b) you find a rare derivative that can transfer Tetr (on r r Tn10) early and efficiently to a Str recipient but transfers Kan (on Tn5) very inefficiently. Draw a diagram showing how this unusual strain was formed. The only way that the TetR gene could be transferred was if the F' from donor b integrated into the genome of a strain carrying 5::Tn10. The F' must recombine in a way so that the origin of transfer is in front of the TetR gene but behind the KanR gene, for only then would it transfer Tet resistance early and Kan resistance very late. NOTE: The F's will most likely integrate by homologous recombination between genes or intragenic regions, as these are large regions of homology between the F' and the host chromosome. The F' could integrate by homologous recombination between IS sequences on the F' and on the genome, but those sequences are very small. Therefore such an event is much rarer than recombination between genes or intragenic regions.