7.28 Spring ‘01 Problem Set #1...

7.28 Spring ‘01 Problem Set #1 Answers

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1. You have a 2000 base linear, double-stranded DNA template, a 1000 base region ofwhich you would like to amplify using polymerase chain reaction (PCR). In addition to thetemplate, you add the appropriate primers (see diagram below), a thermostable DNApolymerase (Taq), and the other necessary components for the reaction. Primer 13' ___________________________________ 5'

5' ___________________________________ 3' Primer 2

The cycle that you set the machine for is:

1) 92 oC for 30 sec. (denaturation)

2) 56 oC for 30 sec. (annealing primer to template)

3) 72 oC for 30 sec. (elongation)

To perform the full PCR reaction, this sequence of temperature variations is repeated 20times.

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 10-20 bp, willit be able to synthesize enough DNA for productive PCR to occur during the time given (step3 = 30 sec.)? What if Taq polymerase had a processivity of 100 bp?

If Taq polymerase had a processivity of 10-20 bp, Taq polymerase would fall offthe template every 10 to 20 bases and it would take about 1 second for Taq to findthe primer template junction again.Let's use 20 bp as the processivity. The template is 1000 bp long. 1000/20=50So, we would need approximately 50 binding events that should require about 50seconds plus the 1 second needed to synthesize 1000 bases. The 30 secondsgiven for elongation (step 3) are not enough to synthesize a new DNA strand ifTaq has a processivity of 10-20 bp.If Taq polymerase had a processivity of 100 bp: 1000/100=10On average the Taq enzyme would only have to find the primer template junction10 times, so 10 seconds plus the 1 second needed for adding 1000 bases, equals11 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 92oC (step 1)denatures the DNA (makes it single stranded). A topoisomerase is not neededeither since the template is linear and therefore cannot accumulate torsionalstrain.

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 thelocation of the primers. What would the majority of the products look like after 20 cycles?Draw one representative example.


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After 1 cycle:

After 2 cycles:

After 20 cycles:

d) What if you had added radiolabeled nucleotides at the beginning of the PCR reaction? Goback to your diagrams and indicate the DNA strands that would have incorporatedradiolabeled nucleotides with an asterisk *.

See diagram above.

2. You set up the in vivo assay for methyl-directed mismatch repair in E. coli that wasdiscussed 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 aheteroduplex 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 inE. coli.

__________Z_____________________+___________


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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 cellswill 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 mutationThere will be random repair, and so the mutation rate will increase, because thecell will have no way of keeping track of which strand is the newly replicatedstrand.

dam overexpressionThere will be no repair, and so the mutation rate will increase.


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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 entirechromosome 13.

a) How would you determine the fragments that contain origins of replication? Describethe attributes of the plasmid backbone that would be required for your test (a diagram isrecommended).

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 ofeach insert, you determine that the majority of the inserts can be narrowed to a region ofless than 200 bp and still retain replicator function. The exception is clone 728, whichrequires >2000 bp to function in your assay. Intrigued by this atypical yeast replicatorstructure you construct a series of substitution mutations in this region and test them inyour replicator assay. You obtain the following results.

Replication+

1 2 3 4 5 6

+

++

7

+++

++

Based on your data, which regions are necessary and which regions are sufficient forreplicator 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 onthe 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.


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c) You want to identify the origin of replication for clone 728. You restriction digest theregion and determine the map shown below.

1 2 3 4 5 6 7

Eco RI Eco RIHind III Xho I

300 bp 900 bp 750 bp

You test for origin function using the 2-dimensional gel assay. For your firstexperiment you digest the DNA of actively growing yeast cells with Eco RI. The probe forthe Southern blot is a radiolabeled HindIII to XhoI DNA fragment. You obtain the followingresult (this is an autoradiogram):

What conclusions can you make about the location of the origin of replication based on thisdata 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 center4 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 thelocation of the origin of replication. What additional 2-D gel analysis could you do thatwould 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 thatyou would radiolabel to use as probe, and the pattern that you would expect for eachpossible 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


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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 theprobe determined the pattern. The digest always determines the patternobserved. The probe need only be within the region of the digest that isappropriate.

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 ofthe pathogen to use as drug targets. You decide that the most likely candidates fordivergent proteins are those involved in initiating replication at the already identified T.assistantus origin. To do so, you take a biochemical approach, fractionating cell extractsand looking for protein fractions that can bind the origin.

4a) What assay might you use to check quickly whether your fractions contain proteins thatbind a sequence containing the origin DNA? Describe the assay and the results expected fora 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 Probewith the protein fractions containing origin binding proteins. Unlabeled DNAshould compete for the protein; this will lead to an elimination of the mobilityshift. Since you don't know the structure of the origin, you can't compete thesignal away specifically.

4b) You find multiple fractions that bind the origin DNA, and decide to use DNAseIfootprinting to check where on the origin the proteins bind. The results are shown below.


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*

4b i) Indicate on the DNA molecule where it is labeled (at the end or throughout, and onwhich strand?).

See Diagram. DNA must be end-labeled on only one strand. To line up to the gelshown, 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 thetwo assays could explain this?

Since the gel shift can detect protein binding events where only 10% of the DNAis bound, whereas footprinting only detects highly specific events where 90% ofthe DNA is bound, one possibility would be that the proteins in fraction B arebinding weakly. However, the question stated that they are binding strongly.Therefore, they must be binding nonspecifically to a different place on every pieceof 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. Notethat simply digesting your end-labeled probe with MNase instead of DNAse andlooking at the x-ray film would not work, as the information above shows thatthey are not binding specifically to one spot.

No Pr

otein

A B A+C+D


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If you thought there were other interesting proteins in this fraction, what kind of columnmight 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 outin 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 theAT-rich region of the origin DNA when they bind to it.

4ci) What assay would you use? (Note that a helicase assay would not detectlocalized unwinding.) Describe the assay and the results expected if the DNA is unwound orif 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 forfootprinting with the protein fractions, treat with P1 nuclease, run a denaturinggel and expose dried gel to film. If the DNA is unwound at all(single-stranded), itwill be cut in that region Otherwise it will stay as one big band at the top of thegel.

4cii) You perform your assay and discover that when fractions A, C, and D are alladded, 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 aprotein bound to it. (This is not surprising, since you saw a footprint over thisregion when these fractions were all subjected to DNAseI footprinting above.)

4d) In order to understand which of these purified protein fractions are necessary to formthe higher order complex at the origin that you see evidence of by DNAseI footprinting, youperform 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 arepresent, C binds if both A and D are present, and D never binds. The curve showing thefractions in which DNA alone elutes is shown first.

DNA alone:

Gel Filtration Fractions

EtB

r St

aini

ng


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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 thecomplex, you titrate the amounts of protein D, and find that protein D works even when

added at 1/1000th

the concentration of the other proteins in the reaction. What couldprotein D be doing?

One possibility is that catalytic amounts of protein D load protein C onto theorigin, perhaps by changing the conformation of protein C.

5. You are studying the E. coli dnaB DNA helicase and are interested in defining theregions 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 [3H] thymidine into


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cells containing each of the mutants. You identify two classes of mutants. The results ofyour experiment are shown below.

20 40 600

Wild Type

A

B

-20Minutes after shift to 42°C

50,000

100,000

150,000

[3 H] Th

ymidin

e Incor

porate

d (Coun

ts Per M

inute)

a) You are surprised to find that you have isolated both fast and slow stop mutations in thednaB. What aspect of dnaB function in E. coli DNA replication do you suspect is defective inthe A class of mutants. Briefly explain in molecular terms why a defect in this functionwould lead to a slow stop phenotype.

“A” class mutants in dnaB cannot interact properly with dnaC, the helicasechaperone. After the temperature shift, ongoing replication will be unaffected.New initiation of replication will not occur without recruitment and loading ofdnaB 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 purifyseveral of the mutant dnaB proteins. You assay them for DNA helicase activity and find thatmutants B1 and B2 have normal DNA helicase activity and mutant B3 has nohelicase activity. You also put these mutant proteins into a reconstituted oriC-dependent

DNA replication assay and assay for [3H] thymidine incorporation. All of the assays are

performed at 42°C (the non-permissive temperature). You find the following results:

40 600

Wild Type

Reaction Time

5,000

10,000

15,000

[3 H] Th

ymidin

e Incor

porate

d (Coun

ts Per M

inute)

20

B1

B2, B3


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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 holoenzymeand are not stimulated to a high rate of helicase activity. This dramatically slowsthe movement of the replication fork and the rate of DNA polymerization in the invitro 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 inthe in vitro replication assay. There is a fast stop phenotype in vivo because thelagging strand continually requires primase activity.

NOTE: “B1” mutants cannot have a "reduced" affinity for primase, because (aswas stated in class) these mutations have no effect on the rates of replication invivo, only increasing the size of Okazaki fragments.

c) Design an experiment to test one of your two hypotheses above. Assume that you haveaccess to all of the purified proteins involved in E. coli DNA replication, an oriC containing

plasmid, [3H] thymidine, and [

3H] 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 primaseactivity assay:

1. Mix a single-stranded template with an artificial “fork” created by an annealedoligonucleotide 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 appropriatebuffer.

3. Incubate the reaction at 42°C for several minutes.

4. Spot your reaction products on a filter and wash to remove free nucleotides. Ina 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. Withthe B2 mutant helicase, you should not observe any radioactivity retained on thefilter.

OR: For the wt helicase, you should see short readiolabeled RNA primers


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about 10 bases long on the film. For the B2 mutant helicase, you should observeno 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 asthe assay successfully tested your hypothesis.

6. To determine the mechanism behind the processivity of chromatin assembly you look forproteins associated with CAF1. One of the interacting proteins, CIP1, is of particularinterest 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 andCAF1 to either circular DNA, the same DNA linearized with a restriction enzyme (bluntcutting), or no DNA. After incubating for 15 minutes, you separate the binding reaction ona non-denaturing acrylamide gel and expose the gel to X-ray film to see where theradiolabeled CIP1 migrated on the gel.

NoDNA

Template

LinearDNA

Template

CircularDNA

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 youexplain the difference in your results with linear and circular DNA. (Hint: You get the sameresult whether the circular DNA is supercoiled or relaxed).

No, CIP1 is not recognizing the sequence of the DNA. Instead, it is recognizing thestructure of the DNA. It can interact with circular DNA, but not linear DNA of thesame sequence. One explanation for this is that the protein is topologically linkedwith the DNA, and falls off the end of a linear DNA molecule. (Note, it is also aformal possibility that the protein does recognize the specific binding site thathappens to have the restriction enzyme site you linearized with right in the middleof it. This is highly unlikely, but could be tested by cutting with a differentenzyme.) (This is also an old exam question.)


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b) You do the same experiment except that you leave out CAF1 and you get the followingresult:

NoDNA

Template

LinearDNA

Template

CircularDNA

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 aloading factor for CIP1. This is analogous to the gamma complex loading the betaclamp 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 ofchromatin 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 ineukaryotes, and thioredoxin in T7 phage increase the processivity of thepolymerase.

7. Most cells encode multiple pathways to repair DNA mismatches and DNA damage. Youhave isolated a previously uncharacterized bacterial strain and wish to determine whichrepair pathways are functional in this organism. To this end, you make a cell extract andcharacterize the repair of the DNA substrate diagrammed below.Note: the restriction enzyme HindIII cleaves the sequence:AAGCTT

TTCGAA


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To assay repair by the extract, you incubate this substrateunder 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 torestriction digestion (with enzymes marked over the gellanes), and run the products on a native agarose gel. Allthe protein factors are removed from the DNA beforeelectrophoresis. A diagram of the resulting gel is shown.

6

32

1

0.5

SizeMarkers

1 2 3 4 5 6

Enzyme: None

6

2

1

0.5

kb

Et Br sta in7 8 9

0.1

DpnI

HindIII

None

DpnI

HindIII

None

DpnI

HindIII

Assaycondit ion: 1 2 3

4

supercoiledsubst rat e

a) Based on the results of this experiment, what type of DNA repair appears to beoccurring 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 linearizedDNA. There are two explanations for this--the plasmid has been nicked orprotein factors are associated with the DNA. Since the problem tells you thatall proteins factors have been removed from the DNA, you can conclude thatthe shift is from nicked plasmid.

• In condition three (when dCTP is added), directed repair occurs by substitutinga single nucleotide, dCTP. The cleavage pattern produced by digestion with

1/ 4000bp

HindIII

1500

AAGCTTTTTGAA


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HindIII gives you 1500bp and 2500bp bands indicating that the inner strand ofthe plasmid is selectively repaired. Repair is not random since in randomrepair you would expect to see three fragments: linear DNA, 1500bp and2500bp fragments.

The last two observations should lead you to the conclusion that the DNA wasrepaired using Base Excision Repair. In base excision repair, the damagedbase is removed by a specific glycosylase. In this case, the T of the G-Tmismatch was most likely removed by the thymidine glycosylase thatspecifically removes only T in G-T mispairing generating an AP site. After thebase has been removed, the AP endonuclease comes in and cuts 5’ of the APsite thereby nicking the backbone. The single stranded DNA exonuclease canthen come in and removed the sugar moiety. When dCTP is available, the DNAPolymerase will add a single nucleotide to fill in the gap.In condition two, since there is no dCTP available, the DNA remains nickedsince the polymerase can not fill in the hole. When dCTP is provided incondition three, this gap is filled eliminating the nick and restoring sensitivityto HindIII.

b) You also wish to test if these cells have a lesion bypass repair system. Below, diagrama substrate you could use to assay for the presence of such an enzyme system. Draw (andclearly label) the products you would expect to see (for example on an autoradiogram of adenaturing gel) if the cell extract does or does not perform lesion bypass.

There were many different and creative experiments described that wouldallow you to test whether the cells have a lesion bypass repair system. The keyfactors to remember were that:

1) In order to get lesion bypass repair, you need transcription. Lesion bypassrepair occurs when the polymerase gets to a lesion that it cannot read duringtranscription. Rather than aborting transcription (which would be lethal) the celluses lesion bypass to just replicate through unreadable DNA by adding adenines.2) The substrate you choose should contain a lesion that would be need bypassrepair. The best possibility would be a pyrimidine dimer, an apurionic site, orapyrimidinic site. A mismatch mutation is not a good choice since it would notblock the polymerase during replication since both strands could be replicated. Apiece of DNA with a single strand break is also not a good choice since yourpolymerase would fall off of the strand with the break while replicating.

One example of an appropriate answer, would be to do a primer extensionassay using a template with a pyrimidine dimer.

** T T

100bp 200bp


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Your primer would be radiolabeled then you would add your cell extract, unlabeleddNTPs, and allow transcription to take place. You would then run yourtranscription products on a denaturing gel and expose the gel to film to detectradioactivity. If you had lesion bypass repair, the entire template strand would bereplicated (300bp) and without lesion bypass repair, replication would stall at thepyrimidine 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, althoughyou know that this strain encodes the enzymes necessary for bypass. Upon seeing thisresult, how might you change the way you prepared the cell extracts to increase yourchances of finding lesion bypass activity?

The most likely explanation for the fact that you did not detect lesion bypassactivity in your cell extracts is that the wild-type cells you prepared your extractsfrom were not expressing lesion bypass repair proteins. The proteins involved inlesion bypass repair are upregulated upon DNA damage, therefore you couldincrease the lesion bypass activity by treating your cells with UV irradiation priorto lysis. Another possibility would be to start with a mutant cell line such asphotolyase mutant or uvr A mutant which would cause an increase in DNA damagethus 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 inthe original extracts due to a lack of LBR protein expression it would not besuccessful.

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


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has two frameshift mutations. Both strains’ mutations are in genes required forsynthesizing histidine. Therefore, mutant strains do not grow on media lacking this aminoacid. You plate cells from each strain onto minimal media plates lacking histidine and thenexpose them to the indicated mutagens.

average colonies per plate

Mutagen Strain A Strain B

Control (no mutagen) 35 ~5

UV (1 sec) 45 30

UV (5 sec) 500 2000

UV (10 sec) 1000 2000+

MMS 350 12

9-aminoacridine 40 150

a) In this experiment, you are exposing the cells to mutagens, yet the assay countscolonies that grow, and in order to grow the cells cannot be mutant. Please explain thisapparent contradiction.

You are using the mutagens to make mutations that reverse the original mutationsin Strains A and B. Counting colonies that grow is what tells you how manymutations were made, because the mutations are RESTORING the genes to theirwild type sequences. This is why the cells can grow and be counted.

NOTE: Reversion does not have to restore the exact wild type sequence. Acompensatory mutation could also restore the histidine pathway and give you thegrowth phenotype.

b) Why do you see colonies appearing on the control plates? Propose an explanation forthe difference in the control frequencies observed between Strains A and B.

These colonies arise from spontaneous reversions. Strain A had a single basesubstitution, while Strain B had two frameshift mutations. Logically, it should beeasier to reverse the single base substitution than the two frameshifts. Thisexplains the higher incidence of spontaneous reversions in Strain A.


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c) A chemical is classified as a mutagen if it produces twice the number of mutants thatoccur spontaneously. From the data above, suggest whether UV light, MMS, and 9-aminoacridine act as effective mutagens for each strain. What do these results indicateabout 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. ForStrain B, 1+ seconds of UV light and 9-aminoacridine both are good mutagens,while the MMS is marginal. This indicates that MMS probably induces basesubstitutions, while 9-aminoacridine induces frameshifts. UV light, is apparentlycapable 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 reversionstudies instead of his suggested approach? Identify a couple of specific advantages to thereversion 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. Imaginestarting with a lawn of cells the are all wild type. When you apply mutagen, somewill die from their mutations, but how do you find those cells? The reversionsstudy allows you to identify a mutation based on the fact that cell can produce alarge, visible colony of cells which is indicative of one reversion event.

Reversion studies are often quite sensitive because they provide a simple, geneticselection for mutants in a specific target gene (for example, in the Ames test theselection 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 thatmake frameshift mutations and those that make base susbtitutions, for example).That any particular type of mutation will only be reverted by certain classes ofmutagens can also be a limitation to reversion studies. Therefore, it is necessaryto test for reversion of several different types of mutant alleles.

9. A colleague from California solicits your expertise on DNA repair. He is studying twonew strains of bacteria, P. outageous and B. outus, whose genomes are similar in size tothat of E. coli. He was able to ascertain that the mutation rate was equal to that seen in E.coli, ~1 x 10-10 mutations/ round of replication. However, he is concerned because hisstrains are now accumulating mutations at an elevated rate. Your colleague tells you thatthe problem is caused by mutations in genes encoding exonucleases. He sends you hismutant strains, as well as his purified wild-type exonucleases.

a) You decide to examine the mutation rate in the strains sent to you by monitoringspontaneous reversion of a point mutation in a gene necessary for histidine biosynthesis.You plate multiple independent cultures of each strain and estimate the number ofmutations/ round of replication as the following:

P. outageous B. outus

# of mutations/ round of replication: 1 x 10-7 3 x 10-9


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i) Given this data, do you think the defective exonuclease in each strain is involved in DNAreplication or excision repair? Why?

The data above suggests that the mutant exonuclease in P. outageous is necessaryfor mismatch repair during DNA replication, while the exonuclease in B. outus isnecessary for excision repair. DNA replication in the absence of mismatch repairwill commit approx. 1 error per 107 nucleotides. Mismatch repair increases fidelityduring DNA replication to approx. 1 error per 1010 nucleotides. Since we aremonitoring 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 provideredundancy 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 formismatch repair. We are assuming that our strain contains a single mutation. Theobserved change in mutation frequency due to this single mutation indicates thatthere are no other exonucleases that can complement the activity of the mutantexonuclease.

iii) In E. coli, what frequency of mutation (# of mutations/ round of replication) would youpredict for mutation in the following genes:

mutS 1 x 10-7 , mutS is responsible for binding mismatches that occur duringDNA replication. Thus, a mutation in this gene will cause a mutation frequency of1 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 mutationfrequency of 3 x 10-9 for reasons stated in part i).

b) Before you begin characterizing the proteins, you decide to double check that thepurified proteins sent to you are exonucleases. In order to do so, you must assay eachprotein for the proper activity. Design an experiment to test each protein for exonucleaseactivity on a double stranded DNA substrate.

1. In a buffer containing Mg2+, mix one of your purified exonucleases with auniformly labeled radioactive DNA substrate (>300bp). As a control, prepareanother 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 freeradioactive nucleotides. Measure the radioactivity in the supernatant and compareto your control experiment. If your purified protein is an exonuclease, you shoulddetect radioactivity in the supernatant.c) Given that one of your exonucleases is important during DNA replication, while the otheris important for excision repair, you suspect that the two exonucleases also differ inprocessivity. You decide to test this hypothesis by measuring the processivity of thepurified wild type exonucleases. Describe an experiment that tests the processivity of theseproteins.


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1. In a buffer without Mg2+ , mix one of your purified exonucleases with auniformly labeled radioactive DNA substrate (>300bp). The absence of Mg2+ atthis step prevents the reaction from proceeding forward.

2. Incubate for 5 minutes to allow the exonuclease to bind to its substrate.

3. Add Mg2+ plus 1000x unlabeled substrate. Incubate for an additional 5 min. at37C.

4. At this point, you can assay the processivity by observing the length of theradiolabeled substrate. Run the product of the reaction on a gel. Dry the gel ontopaper and expose to X-ray film to make an autoradiogram. Compare this band tothe observed band from the reaction containing your other exonuclease. A bandthat is close to 300bp in length will be expected from an exonuclease with lowprocessivity. However, with a highly processive exonuclease, a much smallerband 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 theresulting DNA on a non-denaturing agarose gel and stain with ethidium bromide. To yoursurprise, you see the following results:

P. outageous B. outus - + - + MboI treatment

ane: 1 2 3 4

i) What is the average size of the DNA fragments in lane 2? (MboI cuts at the sequence‘GATC’).


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The sequence GATC will appear approximately 1 time in every 44 nucleotides.Therefore, the average size of the DNA fragments in lane two is 44 , or 256nucleotides.

ii) Based on this data, do you think these strains are dependent on DNA methylation foraccurate DNA repair? If not, name two possibilities the cell might use to determine whichDNA strand is the newly synthesized strand, and therefore, the correct strand to repair.

MboI specifically digests unmethylated DNA. Genomic DNA from P. outageous wasdigested, but not the DNA from B. outus. Therefore, DNA repair is not dependenton methylation in P. outageous, but may be in B. outus. Cells whose DNA is notmethylated may use Okazaki fragments or proteins similar to PCNA to determinethe newly synthesized strand.

7.28 Spring ‘01 Problem Set #1...

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