7.28 Spring 01 Problem Set #1 Answers

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1 1. You have a 2000 base linear, double-stranded DNA template, a 1000 base region of which you would like to amplify using polymerase chain reaction (PCR). In addition to the template, you add the appropriate primers (see diagram below), a thermostable DNA polymerase (Taq), and the other necessary components for the reaction. Primer 1 3' 5' 5' 3' Primer 2 The cycle that you set the machine for is: 1) 92 o C for 30 sec. (denaturation) 2) 56 o C for 30 sec. (annealing primer to template) 3) 72 o C for 30 sec. (elongation) To perform the full PCR reaction, this sequence of temperature variations is repeated 20 times. a) It takes Taq polymerase 1 second to find a primer bound to a single stranded template, but only 1 millisecond to add a base. If Taq polymerase has a processivity of bp, will it be able to synthesize enough DNA for productive PCR to occur during the time given (step 3 = 30 sec.)? What if Taq polymerase had a processivity of 100 bp? If Taq polymerase had a processivity of bp, Taq polymerase would fall off the template every 10 to 20 bases and it would take about 1 second for Taq to find the primer template junction again. Let's use 20 bp as the processivity. The template is 1000 bp long. 1000/20=50 So, we would need approximately 50 binding events that should require about 50 seconds plus the 1 second needed to synthesize 1000 bases. The 30 seconds given for elongation (step 3) are not enough to synthesize a new DNA strand if Taq has a processivity of bp. If Taq polymerase had a processivity of 100 bp: 1000/100=10 On average the Taq enzyme would only have to find the primer template junction 10 times, so 10 seconds plus the 1 second needed for adding 1000 bases, equals 11 seconds. The 30 seconds given are enough time for productive PCR to occur. b) Do you need to add a helicase to the reaction mixture? Do you need a topoisomerase? Why or why not? You do not need to add a helicase since heating the DNA to 92 o C (step 1) denatures the DNA (makes it single stranded). A topoisomerase is not needed either since the template is linear and therefore cannot accumulate torsional strain. c) After one cycle (steps 1-3), draw the DNA strands that are present in the reaction tube. Indicate where primer 1 and 2 are. Draw the products after 2 cycles, again indicate the location of the primers. What would the majority of the products look like after 20 cycles? Draw one representative example. 1

2 After 1 cycle: After 2 cycles: After 20 cycles: d) What if you had added radiolabeled nucleotides at the beginning of the PCR reaction? Go back to your diagrams and indicate the DNA strands that would have incorporated radiolabeled nucleotides with an asterisk *. See diagram above. 2. You set up the in vivo assay for methyl-directed mismatch repair in E. coli that was discussed in lecture using a heteroduplex of lambda phage DNA. a) Draw the products expected from each strain after a single round of DNA replication of a heteroduplex of lambda DNA if the mutant (Z) strand is methylated and the wildtype (+) strand is not. Note that dam encodes the enzyme that methylates the A of a GATC tract in E. coli. Z + 2

3 Wildtype E. coli (repair based on methylated strand) Z Z muts - E. coli (no repair, just replication. Both kinds of phage released form a single E. coli cell) Z Z and + + dam - E. coli (just like wildtype, since adding methylated DNA) Z Z b) Would your answer for a) change if neither strand is methylated? If yes, how? If neither strand is methylated, there will be random repair. Single E. coli cells will release EITHER mutant or wildtype phage, but not both.) Z Z OR + + c) How would the following mutations affect the mutation rate of E. coli? Explain. dam null mutation There will be random repair, and so the mutation rate will increase, because the cell will have no way of keeping track of which strand is the newly replicated strand. dam overexpression There will be no repair, and so the mutation rate will increase. 3

4 3. You are in the process of mapping the replicators on yeast chromosome 13. To this end, you have obtained a set of 220 yeast plasmids with inserts that cover the entire chromosome 13. a) How would you determine the fragments that contain origins of replication? Describe the attributes of the plasmid backbone that would be required for your test (a diagram is recommended). This was an exam question, and the points were awarded as follows: 3 pts: Diagram of Vector and mention of 3 out of 4 of: Centromere, Insert, No Replication Origin, Selectable Marker. 1 pt: Transform into plasmids into yeast. 1 pt: Select for marker with appropriate media. b) You identify 12 plasmids that you think contain replicators. Using deletion analysis of each insert, you determine that the majority of the inserts can be narrowed to a region of less than 200 bp and still retain replicator function. The exception is clone 728, which requires >2000 bp to function in your assay. Intrigued by this atypical yeast replicator structure you construct a series of substitution mutations in this region and test them in your replicator assay. You obtain the following results. Replication Based on your data, which regions are necessary and which regions are sufficient for replicator function. 3 pts: Regions 2, 4, and 6 are each necessary for replicator function. 2 pts: Regions 1-7 are the only regions that are demonstrated to be sufficient on the basis of this experiment. -1 pt: Regions 2, 4, and 6 are sufficient. This is not demonstrated here. You would need to show that you can change all the intervening sequences and leave 2, 4, and 6 to demonstrate this

5 c) You want to identify the origin of replication for clone 728. You restriction digest the region and determine the map shown below. Eco RI Hind III Xho I Eco RI 300 bp 900 bp 750 bp You test for origin function using the 2-dimensional gel assay. For your first experiment you digest the DNA of actively growing yeast cells with Eco RI. The probe for the Southern blot is a radiolabeled HindIII to XhoI DNA fragment. You obtain the following result (this is an autoradiogram): What conclusions can you make about the location of the origin of replication based on this data and the data in 3a and 3b. 3 pts: The RI to RI fragment has an origin. 3 pts: The origin of replication is located in a position that is off center 4 pts: The origin is located in either regions 2 or 6. The data is ambiguous on this point. d) A colleague points out that your experiment does not definitively demonstrate the location of the origin of replication. What additional 2-D gel analysis could you do that would show without a doubt where the origin of replication is? For each experiment, describe the restriction digest of the yeast DNA that you would use, the DNA fragment that you would radiolabel to use as probe, and the pattern that you would expect for each possible origin site. Assay for 2 being the origin: (2 pts) Cut DNA with XhoI and RI. (1 pt) Probe with HIII and Xho I. (2 pt) Explanation of answer/drawing of autoradiogram 5

6 Assay for 6 being the origin: (2 pts) Cut DNA with Xho I and RI. (1 pt) Probe with HIII and Xho I. (2 pt) Explanation of answer/drawing of autoradiogram Other digests and probes are possible and were awarded points similarly. In both parts C and D many people were confused about whether the digest or the probe determined the pattern. The digest always determines the pattern observed. The probe need only be within the region of the digest that is appropriate. 4) You are studying the mechanism of DNA replication in the newly identified pathogen T. assistantus. You are hoping to find proteins involved specifically in the DNA replication of the pathogen to use as drug targets. You decide that the most likely candidates for divergent proteins are those involved in initiating replication at the already identified T. assistantus origin. To do so, you take a biochemical approach, fractionating cell extracts and looking for protein fractions that can bind the origin. 4a) What assay might you use to check quickly whether your fractions contain proteins that bind a sequence containing the origin DNA? Describe the assay and the results expected for a protein fraction that binds the DNA and one that doesn t. Gel shift A complete answer required the following information: Labeled probe corresponding to the origin Protein fractions Run products on a nondenaturing gel and expose to film Explanation/description of results (words or figure): The mobility of the labeled probe should be shifted upon incubation of the Probe with the protein fractions containing origin binding proteins. Unlabeled DNA should compete for the protein; this will lead to an elimination of the mobility shift. Since you don't know the structure of the origin, you can't compete the signal away specifically. 4b) You find multiple fractions that bind the origin DNA, and decide to use DNAseI footprinting to check where on the origin the proteins bind. The results are shown below. 6

7 A B A+C+D No Prot ein * 4b i) Indicate on the DNA molecule where it is labeled (at the end or throughout, and on which strand?). See Diagram. DNA must be end-labeled on only one strand. To line up to the gel shown, it must be labeled at the 3 end of the left strand, as shown, or at the 5 end of the right strand. 4bii) Fraction B bound the DNA in your first assay, but is not footprinting in this assay. Assuming that the proteins in this fraction can bind the DNA strongly, what difference in the two assays could explain this? Since the gel shift can detect protein binding events where only 10% of the DNA is bound, whereas footprinting only detects highly specific events where 90% of the DNA is bound, one possibility would be that the proteins in fraction B are binding weakly. However, the question stated that they are binding strongly. Therefore, they must be binding nonspecifically to a different place on every piece of DNA. You decide that fraction B might contain histones. What assay would you use to check this? Digest unlabeled DNA with MNase, look at EtBr staining for 160 bp ladder. Note that simply digesting your end-labeled probe with MNase instead of DNAse and looking at the x-ray film would not work, as the information above shows that they are not binding specifically to one spot. 7

8 If you thought there were other interesting proteins in this fraction, what kind of column might you run to separate away the histones from the other proteins? Since histones are extremely positively charged, you might run a charged column. If you run a positively charged column, for instance, the histones should come out in the flow through. (not sticking to the column at all). 4c) You then want to know if any of the proteins that you ve fractionated are unwinding the AT-rich region of the origin DNA when they bind to it. 4ci) What assay would you use? (Note that a helicase assay would not detect localized unwinding.) Describe the assay and the results expected if the DNA is unwound or if it is not. DNA unwinding assay. Just like footprinting, but use an enzyme that cuts ssdna, like P1 nuclease. So, incubate the same probe end-labeled on one strand used for footprinting with the protein fractions, treat with P1 nuclease, run a denaturing gel and expose dried gel to film. If the DNA is unwound at all(single-stranded), it will be cut in that region Otherwise it will stay as one big band at the top of the gel. 4cii) You perform your assay and discover that when fractions A, C, and D are all added, the DNA unwinds on one strand, but not the other. How is this possible? Perhaps one strand of the unwound DNA is protected from the nuclease by a protein bound to it. (This is not surprising, since you saw a footprint over this region when these fractions were all subjected to DNAseI footprinting above.) 4d) In order to understand which of these purified protein fractions are necessary to form the higher order complex at the origin that you see evidence of by DNAseI footprinting, you perform gel filtration assays with a plasmid containing the origin, and protein fractions A, C, and D. Draw the following results: A binds the origin no matter what other proteins are present, C binds if both A and D are present, and D never binds. The curve showing the fractions in which DNA alone elutes is shown first. DNA alone: EtBr St aining Gel Filtration Fractions 8

9 Protein A is labeled: Protein C is labeled: Protein D is labeled: e) Intrigued by the fact that protein D seems important and yet never associates with the complex, you titrate the amounts of protein D, and find that protein D works even when added at 1/1000 th the concentration of the other proteins in the reaction. What could protein D be doing? One possibility is that catalytic amounts of protein D load protein C onto the origin, perhaps by changing the conformation of protein C. 5. You are studying the E. coli dnab DNA helicase and are interested in defining the regions of the protein involved in its interactions with other replication factors. To this end, you isolate multiple temperature sensitive mutations in the dnab protein. To begin to classify the different dnab mutants, you determine the incorporation of [ 3 H] thymidine into 9

10 cells containing each of the mutants. You identify two classes of mutants. The results of your experiment are shown below. [ 3 H] Thymidine Incorporated (Counts Per Minute) 150, ,000 50, Wild Type A B Minutes after shift to 42 C a) You are surprised to find that you have isolated both fast and slow stop mutations in the dnab. What aspect of dnab function in E. coli DNA replication do you suspect is defective in the A class of mutants. Briefly explain in molecular terms why a defect in this function would lead to a slow stop phenotype. A class mutants in dnab cannot interact properly with dnac, the helicase chaperone. After the temperature shift, ongoing replication will be unaffected. New initiation of replication will not occur without recruitment and loading of dnab at the origin by dnac. This gives a slow stop phenotype. b) To look more carefully at the function of the B class of mutant proteins you purify several of the mutant dnab proteins. You assay them for DNA helicase activity and find that mutants B1 and B2 have normal DNA helicase activity and mutant B3 has no helicase activity. You also put these mutant proteins into a reconstituted oric-dependent DNA replication assay and assay for [ 3 H] thymidine incorporation. All of the assays are performed at 42 C (the non-permissive temperature). You find the following results: [ 3 H] Thymidine Incorporated (Counts Per Minute) 15,000 10,000 5, Wild Type B1 B2, B Reaction Time 10

11 What activities of dnab do you suspect will be defective in the B1 and B2 activities. Explain your reasoning in each case. B1 mutants in dnab cannot interact with the τ subunit of the Pol III holoenzyme and are not stimulated to a high rate of helicase activity. This dramatically slows the movement of the replication fork and the rate of DNA polymerization in the in vitro reconstituted system, and it leads to a fast stop phenotype in vivo. B2 mutants in dnab have completely lost the interaction with dnag (primase). Without recruitment of primase to the origin, there can be no polymerization in the in vitro replication assay. There is a fast stop phenotype in vivo because the lagging strand continually requires primase activity. NOTE: B1 mutants cannot have a "reduced" affinity for primase, because (as was stated in class) these mutations have no effect on the rates of replication in vivo, only increasing the size of Okazaki fragments. c) Design an experiment to test one of your two hypotheses above. Assume that you have access to all of the purified proteins involved in E. coli DNA replication, an oric containing plasmid, [ 3 H] thymidine, and [ 3 H] uridine. Draw any substrates needed, how you would detect the products, and the results you would expect for wild type and mutant proteins. To assay B2 mutants for a loss of the primase interaction, perform a primase activity assay: 1. Mix a single-stranded template with an artificial fork created by an annealed oligonucleotide with primase and B2 mutant helicase, or wt helicase as a control. If you used a double-stranded oric-containing plasmid as a template, you needed to add other proteins like dnaa and dnac or use extracts from B2 mutant or wt cells in order to create a replication fork. 2. Include [ 3 H] UTP and other rntps (in order to make RNA) in the appropriate buffer. 3. Incubate the reaction at 42 C for several minutes. 4. Spot your reaction products on a filter and wash to remove free nucleotides. In a scintillation counter, measure the radioactivity that remains. OR: Separate your reaction products on a denaturing acrylamide gel and visualize by autoradiography. 5. With a wt helicase, you should observe retention of radiolabeled primers. With the B2 mutant helicase, you should not observe any radioactivity retained on the filter. OR: For the wt helicase, you should see short readiolabeled RNA primers 11

12 about 10 bases long on the film. For the B2 mutant helicase, you should observe no primer formation. NOTE: Other assays to test other hypotheses were graded on a similar scale (i.e. describing substrates, product detection, and both wt and mut results) as long as the assay successfully tested your hypothesis. 6. To determine the mechanism behind the processivity of chromatin assembly you look for proteins associated with CAF1. One of the interacting proteins, CIP1, is of particular interest to you. To study it further you make a radiolabeled form of the protein. a) Suspecting that CIP1 might be a DNA binding protein you add radiolabeled CIP1 and CAF1 to either circular DNA, the same DNA linearized with a restriction enzyme (blunt cutting), or no DNA. After incubating for 15 minutes, you separate the binding reaction on a non-denaturing acrylamide gel and expose the gel to X-ray film to see where the radiolabeled CIP1 migrated on the gel. No DNA Template Linear DNA Template Circular DNA Template + Given that CIP1 forms a stable complex with circular DNA but not with linear DNA of the same DNA sequence, is CIP1 a sequence-specific DNA binding protein? How could you explain the difference in your results with linear and circular DNA. (Hint: You get the same result whether the circular DNA is supercoiled or relaxed). No, CIP1 is not recognizing the sequence of the DNA. Instead, it is recognizing the structure of the DNA. It can interact with circular DNA, but not linear DNA of the same sequence. One explanation for this is that the protein is topologically linked with the DNA, and falls off the end of a linear DNA molecule. (Note, it is also a formal possibility that the protein does recognize the specific binding site that happens to have the restriction enzyme site you linearized with right in the middle of it. This is highly unlikely, but could be tested by cutting with a different enzyme.) (This is also an old exam question.) 12

13 b) You do the same experiment except that you leave out CAF1 and you get the following result: No DNA Template Linear DNA Template Circular DNA Template + What do you think that the role of CAF1 is? CIP1 does not associate with the DNA without CAF1 around, so CAF1 must be a loading factor for CIP1. This is analogous to the gamma complex loading the beta clamp in E. coli, or RFC loading PCNA in eukaryotes. c) How do the DNA binding properties of CIP1 help to explain its role in the processivity of chromatin assembly. By topologically linking CAF1 to the DNA, CIP1 can prevent CAF1 from falling off, thereby increasing its processivity. This is how the beta-clamp in E. coli, PCNA in eukaryotes, and thioredoxin in T7 phage increase the processivity of the polymerase. 7. Most cells encode multiple pathways to repair DNA mismatches and DNA damage. You have isolated a previously uncharacterized bacterial strain and wish to determine which repair pathways are functional in this organism. To this end, you make a cell extract and characterize the repair of the DNA substrate diagrammed below. Note: the restriction enzyme HindIII cleaves the sequence:aagctt TTCGAA 13

14 AAGCTT TTTGAA 1/ 4000bp 1500 To assay repair by the extract, you incubate this substrate under three conditions: (1) DNA, no cell extract, +ATP, + dctp. (2) DNA, +cell extract, +ATP (3) DNA, +cell extract, +ATP, + dctp. After these incubations, you recover the DNA, subject it to restriction digestion (with enzymes marked over the gel lanes), and run the products on a native agarose gel. All the protein factors are removed from the DNA before electrophoresis. A diagram of the resulting gel is shown. HindIII Assay condit ion: Enzyme: None DpnI HindIII None DpnI HindIII None DpnI HindIII kb supercoiled 1 subst rate Size 1 2 Markers Et Br sta in a) Based on the results of this experiment, what type of DNA repair appears to be occurring in the cell extract? Give two specific observations that lead you to this conclusion. The data provided indicate that: The plasmid substrate that you are using is fully methylated. (DpnI is able to cleave the plasmid.) In condition two (when there are no dntps available), the plasmid DNA is nicked. The plasmid DNA runs much higher than the supercoiled and linearized DNA. There are two explanations for this--the plasmid has been nicked or protein factors are associated with the DNA. Since the problem tells you that all proteins factors have been removed from the DNA, you can conclude that the shift is from nicked plasmid. In condition three (when dctp is added), directed repair occurs by substituting a single nucleotide, dctp. The cleavage pattern produced by digestion with 14

15 HindIII gives you 1500bp and 2500bp bands indicating that the inner strand of the plasmid is selectively repaired. Repair is not random since in random repair you would expect to see three fragments: linear DNA, 1500bp and 2500bp fragments. The last two observations should lead you to the conclusion that the DNA was repaired using Base Excision Repair. In base excision repair, the damaged base is removed by a specific glycosylase. In this case, the T of the G-T mismatch was most likely removed by the thymidine glycosylase that specifically removes only T in G-T mispairing generating an AP site. After the base has been removed, the AP endonuclease comes in and cuts 5 of the AP site thereby nicking the backbone. The single stranded DNA exonuclease can then come in and removed the sugar moiety. When dctp is available, the DNA Polymerase will add a single nucleotide to fill in the gap. In condition two, since there is no dctp available, the DNA remains nicked since the polymerase can not fill in the hole. When dctp is provided in condition three, this gap is filled eliminating the nick and restoring sensitivity to HindIII. b) You also wish to test if these cells have a lesion bypass repair system. Below, diagram a substrate you could use to assay for the presence of such an enzyme system. Draw (and clearly label) the products you would expect to see (for example on an autoradiogram of a denaturing gel) if the cell extract does or does not perform lesion bypass. There were many different and creative experiments described that would allow you to test whether the cells have a lesion bypass repair system. The key factors to remember were that: 1) In order to get lesion bypass repair, you need transcription. Lesion bypass repair occurs when the polymerase gets to a lesion that it cannot read during transcription. Rather than aborting transcription (which would be lethal) the cell uses lesion bypass to just replicate through unreadable DNA by adding adenines. 2) The substrate you choose should contain a lesion that would be need bypass repair. The best possibility would be a pyrimidine dimer, an apurionic site, or apyrimidinic site. A mismatch mutation is not a good choice since it would not block the polymerase during replication since both strands could be replicated. A piece of DNA with a single strand break is also not a good choice since your polymerase would fall off of the strand with the break while replicating. One example of an appropriate answer, would be to do a primer extension assay using a template with a pyrimidine dimer. ** T T 100bp 200bp 15

16 Your primer would be radiolabeled then you would add your cell extract, unlabeled dntps, and allow transcription to take place. You would then run your transcription products on a denaturing gel and expose the gel to film to detect radioactivity. If you had lesion bypass repair, the entire template strand would be replicated (300bp) and without lesion bypass repair, replication would stall at the pyrimidine dimer. (100bp) -LBR +LBR 300bp 100bp c) Using the assay in 5b you find that your cell extract does not support lesion bypass. However, an extract you prepared from wild-type E. coli cells also had no activity, although you know that this strain encodes the enzymes necessary for bypass. Upon seeing this result, how might you change the way you prepared the cell extracts to increase your chances of finding lesion bypass activity? The most likely explanation for the fact that you did not detect lesion bypass activity in your cell extracts is that the wild-type cells you prepared your extracts from were not expressing lesion bypass repair proteins. The proteins involved in lesion bypass repair are upregulated upon DNA damage, therefore you could increase the lesion bypass activity by treating your cells with UV irradiation prior to lysis. Another possibility would be to start with a mutant cell line such as photolyase mutant or uvr A mutant which would cause an increase in DNA damage thus stimulating the LBR system. Many people suggestion fractionation and purification of the LBR activity. This is a possibility and a reasonable approach, however if you had no activity in the original extracts due to a lack of LBR protein expression it would not be successful. 8. Your introductory lab course has set up an Ames Test experiment for you to conduct. You are given two strains to study. Strain A has a single base substitution, while Strain B 16

17 has two frameshift mutations. Both strains mutations are in genes required for synthesizing histidine. Therefore, mutant strains do not grow on media lacking this amino acid. You plate cells from each strain onto minimal media plates lacking histidine and then expose them to the indicated mutagens. average colonies per plate Mutagen Strain A Strain B Control (no mutagen) 35 ~5 UV (1 sec) UV (5 sec) UV (10 sec) MMS aminoacridine a) In this experiment, you are exposing the cells to mutagens, yet the assay counts colonies that grow, and in order to grow the cells cannot be mutant. Please explain this apparent contradiction. You are using the mutagens to make mutations that reverse the original mutations in Strains A and B. Counting colonies that grow is what tells you how many mutations were made, because the mutations are RESTORING the genes to their wild type sequences. This is why the cells can grow and be counted. NOTE: Reversion does not have to restore the exact wild type sequence. A compensatory mutation could also restore the histidine pathway and give you the growth phenotype. b) Why do you see colonies appearing on the control plates? Propose an explanation for the difference in the control frequencies observed between Strains A and B. These colonies arise from spontaneous reversions. Strain A had a single base substitution, while Strain B had two frameshift mutations. Logically, it should be easier to reverse the single base substitution than the two frameshifts. This explains the higher incidence of spontaneous reversions in Strain A. 17

18 c) A chemical is classified as a mutagen if it produces twice the number of mutants that occur spontaneously. From the data above, suggest whether UV light, MMS, and 9- aminoacridine act as effective mutagens for each strain. What do these results indicate about they types of mutations induced by each of these treatments? For Strain A, 5+ seconds of UV light and MMS both acts as effective mutagens. For Strain B, 1+ seconds of UV light and 9-aminoacridine both are good mutagens, while the MMS is marginal. This indicates that MMS probably induces base substitutions, while 9-aminoacridine induces frameshifts. UV light, is apparently capable of inducing both types of mutations. d) Your lab partner asks you why you didn t just start with wild type (His+) cells, mutagenize them, and look for His- cells. How do you justify using the mutant reversion studies instead of his suggested approach? Identify a couple of specific advantages to the reversion analysis approach. What are the limitations of determining mutagenic specificities (i.e. of deciding whether something is or is not a mutagen) by reversion studies? The lab partner s suggestion is much harder and less quantitative. Imagine starting with a lawn of cells the are all wild type. When you apply mutagen, some will die from their mutations, but how do you find those cells? The reversions study allows you to identify a mutation based on the fact that cell can produce a large, visible colony of cells which is indicative of one reversion event. Reversion studies are often quite sensitive because they provide a simple, genetic selection for mutants in a specific target gene (for example, in the Ames test the selection is for His+). They also allow you to test for specific types of mutations (the earlier parts of the question had you distinguish between mutagens that make frameshift mutations and those that make base susbtitutions, for example). That any particular type of mutation will only be reverted by certain classes of mutagens can also be a limitation to reversion studies. Therefore, it is necessary to test for reversion of several different types of mutant alleles. 9. A colleague from California solicits your expertise on DNA repair. He is studying two new strains of bacteria, P. outageous and B. outus, whose genomes are similar in size to that of E. coli. He was able to ascertain that the mutation rate was equal to that seen in E. coli, ~1 x mutations/ round of replication. However, he is concerned because his strains are now accumulating mutations at an elevated rate. Your colleague tells you that the problem is caused by mutations in genes encoding exonucleases. He sends you his mutant strains, as well as his purified wild-type exonucleases. a) You decide to examine the mutation rate in the strains sent to you by monitoring spontaneous reversion of a point mutation in a gene necessary for histidine biosynthesis. You plate multiple independent cultures of each strain and estimate the number of mutations/ round of replication as the following: P. outageous B. outus # of mutations/ round of replication: 1 x x

19 i) Given this data, do you think the defective exonuclease in each strain is involved in DNA replication or excision repair? Why? The data above suggests that the mutant exonuclease in P. outageous is necessary for mismatch repair during DNA replication, while the exonuclease in B. outus is necessary for excision repair. DNA replication in the absence of mismatch repair will commit approx. 1 error per 10 7 nucleotides. Mismatch repair increases fidelity during DNA replication to approx. 1 error per nucleotides. Since we are monitoring spontaneous mutations, the frequency of observed mutations in B. outus is low, but still higher than in wild type strains. ii) E. coli possesses three exonucleases (ExoI, ExoVII, and RecJ) that can provide redundancy when one of the others is mutated. Based on the data above, is it likely that P. outageous possesses a similarly redundant system? It is unlikely that P. outageous possesses more than a single exonuclease for mismatch repair. We are assuming that our strain contains a single mutation. The observed change in mutation frequency due to this single mutation indicates that there are no other exonucleases that can complement the activity of the mutant exonuclease. iii) In E. coli, what frequency of mutation (# of mutations/ round of replication) would you predict for mutation in the following genes: muts 1 x 10-7, muts is responsible for binding mismatches that occur during DNA replication. Thus, a mutation in this gene will cause a mutation frequency of 1 x 10-7 mutations/ round of replication for the reasons stated above. uvra 3 x 10-9, uvra is involved in excision repair and will cause a mutation frequency of 3 x 10-9 for reasons stated in part i). b) Before you begin characterizing the proteins, you decide to double check that the purified proteins sent to you are exonucleases. In order to do so, you must assay each protein for the proper activity. Design an experiment to test each protein for exonuclease activity on a double stranded DNA substrate. 1. In a buffer containing Mg 2+, mix one of your purified exonucleases with a uniformly labeled radioactive DNA substrate (>300bp). As a control, prepare another reaction in the same manner, without the addition of the exonuclease. 2. Incubate for 5 minutes at 37C. 3. Pellet the undigested DNA and remove the supernatant containing the free radioactive nucleotides. Measure the radioactivity in the supernatant and compare to your control experiment. If your purified protein is an exonuclease, you should detect radioactivity in the supernatant. c) Given that one of your exonucleases is important during DNA replication, while the other is important for excision repair, you suspect that the two exonucleases also differ in processivity. You decide to test this hypothesis by measuring the processivity of the purified wild type exonucleases. Describe an experiment that tests the processivity of these proteins. 19

20 1. In a buffer without Mg 2+, mix one of your purified exonucleases with a uniformly labeled radioactive DNA substrate (>300bp). The absence of Mg 2+ this step prevents the reaction from proceeding forward. at 2. Incubate for 5 minutes to allow the exonuclease to bind to its substrate. 3. Add Mg 2+ plus 1000x unlabeled substrate. Incubate for an additional 5 min. at 37C. 4. At this point, you can assay the processivity by observing the length of the radiolabeled substrate. Run the product of the reaction on a gel. Dry the gel onto paper and expose to X-ray film to make an autoradiogram. Compare this band to the observed band from the reaction containing your other exonuclease. A band that is close to 300bp in length will be expected from an exonuclease with low processivity. However, with a highly processive exonuclease, a much smaller band will be observed. d) Finally, you are interested in studying the methylation state of the DNA in your strains. After purifying genomic DNA from each strain, you digest the DNA with MboI. You run the resulting DNA on a non-denaturing agarose gel and stain with ethidium bromide. To your surprise, you see the following results: P. outageous B. outus MboI treatment ane: i) What is the average size of the DNA fragments in lane 2? (MboI cuts at the sequence GATC ). 20

21 The sequence GATC will appear approximately 1 time in every 4 4 nucleotides. Therefore, the average size of the DNA fragments in lane two is 4 4, or 256 nucleotides. ii) Based on this data, do you think these strains are dependent on DNA methylation for accurate DNA repair? If not, name two possibilities the cell might use to determine which DNA strand is the newly synthesized strand, and therefore, the correct strand to repair. MboI specifically digests unmethylated DNA. Genomic DNA from P. outageous was digested, but not the DNA from B. outus. Therefore, DNA repair is not dependent on methylation in P. outageous, but may be in B. outus. Cells whose DNA is not methylated may use Okazaki fragments or proteins similar to PCNA to determine the newly synthesized strand. 21