Question 1

You have been asked to grow large batches of E. coli cells from which 7.02/10.702students can purify β-galactosidase during the Biochemistry module. These "fermentations"are conducted at the 10-liter scale inside bioreactors, which provide better control of the cellenvironment than the shake flasks you used to grow E. coli in the Genetics module.

You decide to study β-galactosidase transcription in E. coli during the fermentationprocess. To do this, you first need to isolate total RNA from cells at different time pointsof the fermentation.

During your RNA isolation, you take precautions to ensure that the RNA in yoursamples does not get degraded. In addition, you want to be sure that your RNA samplesare properly denatured before running them on an agarose gel.

a) Explain the difference between RNA degradation and RNA denaturation.

RNA degradation alters the primary structure of RNA (i.e. breaking theRNA backbone such that larger pieces of RNA are broken into smaller ones).

RNA denaturation eliminates the secondary structure of the RNA (usually bybreaking hydrogen bonds). This results in the conversion of a foldedpolymer into a linear one, with no change in the length of the RNA moleculeitself.

Lane 1 of the EtBr-stained agarose gel below shows the "expected" pattern of bands from abacterial RNA preparation. (Note: bacterial ribosomes contain 23S and 16S rRNAsubunits, which migrate differently on a gel than eukaryotic 28S and 18S rRNAsubunits.)

Lane 2 should show an RNA"smear," with any discretebands migrating as lengthssmaller than the originalrRNAmolecules.

b) In lane 2 of the gel, draw what you'd expect to see if the RNA in your sample waspartially degraded, and explain your answer in one or two sentences.

You would see a smear of bands because degradation will cut the mRNAs into piecesof different sizes. The gel will be "brighter" at lower sizes because the majority ofthe mRNA pieces are now smaller than they were previously.

7.59.5

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2.4

1.4

MW 1 2

Question 1 (continued)

You successfully isolate RNA from samples harvested at three time points duringthe E. coli fermentation, and perform a Northern blot using a protocol similar to that usedin 7.02/10.702 lab. Alas, when you develop the film at the end, you see NO BANDS, noteven in your positive control lane. You decide to look back at your notes to figure outwhat mistake(s) you might have made.

For each of the following four scenarios:1. IDENTIFY the mistake;2. Explain WHY the mistake resulted in a lack of bands on your film;3. Propose how you would FIX the mistake if you were to redo the entire

procedure.

Note: each scenario is independent of the others, and you should assume that all othersteps in the protocol were performed "as in lab."

c) Probe labeling reaction containing zcyt1 cDNA template, Klenow fragment, 1 mM each ofdATP, dGTP, dCTP, 0.65 mM dTTP, 0.35 mM DIG-dUTP, random hexamer primers, andbuffer (Tris-HCl, MgCl2, DTT, BSA).

1. Used zcyt1 cDNA as your template2. zcyt1 is not expressed in E. coli, so probe has nothing to anneal to3. use the lacZ gene (or gene encoding β-gal) as your template for probe

labeling

d) Second wash conditions: 0.5x SSC, 95˚C.

1. Temperature of the wash is too high2. This temperature leads to hydrolysis of RNA (or at least loss of probe

binding)3. lower temperature of wash (~60˚C as in 7.02)

e) Antibody step: 30 minute incubation with rabbit antibody specific for β-galactosidase, conjugated to alkaline phosphatase.

1. Used rabbit anti β-galactosidase antibody, which is specific for a protein2. you are trying to detect a DIG-labeled RNA probe3. use anti-DIG antibody (conjugated to AP)

f) Probe detection solution: NBT/BCIP in detection buffer.

1. Used the wrong substrate2. Cleavage of NBT/BCIP leads to a colored product, not emission of light—no

darkening of film3. Use substrate that emits light (CSPD)

Question 1 (continued)

g) Your coworker says that you can obtain data from the Northern blot developed withNBT/BCIP without redoing any steps of the protocol. Explain how this is possible.

Since cleavage of NBT/BCIP leads to the formation of a colored (blue) product,you can determine the presence/absence of your mRNA band, its intensity, andits molecular weight simply by looking for colored bands on the nylonmembrane itself!

Your labmate has performed the Northern blot procedure correctly using RNAisolated from cells harvested at 1 hour, 4 hours, and 6 hours into the fermentation. Sheobserved the following results:

EtBr-stained gel Nylon membrane(arrows indicate position of MW

bands)

Lane 1: MW standards (all sizes inkb)

Lane 2: RNA from "1 hour" timepoint

Lane 3: RNA from "4 hour" timepoint

Lane 4: RNA from "6 hour" timepoint

Lane 5: βgal cDNA fragment

Northern Blot Film(arrows indicate position of MW bands)

h) What three conclusions can you draw from this Northern blot result? Explain howyou arrived at these conclusions by referring to specific lanes of the gel/membrane/film.

1X 2X 4X 1 2 3 4 5

7.59.5

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2.41.4

7.59.5

4.42.41.4

1 2 3 4 5

7.59.5

4.42.41.4

1 2 3 4 5

Question 1 (continued)

1. The lacZ mRNA is 4.4 kb long; this was determined by comparing themobility of the bands on the Northern blot film to the molecular weightstandards.2. The lacZ gene is expressed at all time points tested during the fermentation, asyou see a band on the blot in all the lanes, and each of these lanes contains totalRNA (from the gel).3. Twice as much lacZ is expressed at 1 hour as is expressed at 4 hours, andthere is 1/4 the amount of lacZ mRNA at 6 hours as at 4 hours. This isdetermined by comparing the amount of signal on the blot and correcting for theamount of total mRNA loaded or transferred (1X, 2X, 4X) (see nylonmembrane).

Question 2

You have identified an interesting zebrafish mutant strain. When compared to wildtype embryos at 24 hours post fertilization, these mutant fish have abnormally smallbrains. You have cloned the gene whose function is lost in your mutant; you name thegene lilbrain.

You decide to study the expression pattern of lilbrain during the development of wildtype embryos by Northern blot. To do so, you need to create a lilbrain-specific-probe.Unfortunately, you have run out of the random-hexamer labeling kit used to make yourprobe in 7.02/10.702 lab.

a) Describe how you could use PCR to make a lilbrain-specific probe by completing thefollowing table:

Template used zebrafish genomic DNA

Amount of template required very low (ng)

Primer length 18-25 ntsDNA polymerase used Taq polymerase

Denaturing temperature 94-95˚C (above 90 is fine)

Annealing temperature ~50-65˚CExtension temperature 72˚C

b) PCR performed with the dNTPs used in lab will generate a labeled DNA probe, not anRNA probe. Why might an RNA probe be a better choice than a DNA probe?

RNA-RNA duplexes are more stable than DNA-RNA duplexes.

Question 2 (continued)

After successfully preparing your probe, you isolate total RNA from wild typezebrafish embryos at the blastula, gastrula, segmentation, pharyngula, and hatching stagesof development. After separating the RNA on a denaturing agarose gel, you perform aNorthern blot using your labeled, lilbrain specific probe.

Your Northern results are consistent with a role for lilbrain in brain development, andyou next want to see where lilbrain is expressed in the developing zebrafish embryo.

d) What experiment discussed in lecture is the best way to visualize if the lilbrain gene isexpressed specifically in the brain? __in situ hybridization___

e) Name one thing that this experiment has in common with a Northern blot, and onedifference between this experiment and a Northern blot.

Common: both use hybridization of a (complementary) probe to detect aspecific mRNA

Difference: in a Northern, the RNA is probed on a membrane; in the in situ, theRNA is probed in a tissue, cell, or animal

You perform the experiment named in part d), and you observe lilbrain mRNA notonly in the brain, as you expected, but throughout the entire embryo. You are not sure ifthis is due to non-specific binding of the probe, or if the lilbrain gene is really expressedall over the embryo. You decide to alter the stringency conditions to distinguish betweenthese two possibilities.

f) In order to reduce non-specific probe binding, one could do any of the following(circle the correct choice from each pair):

i. increase / decrease salt concentration of washesii. increase / decrease temperature of washes

iii. increase / decrease formamide concentration during hybridization

iv. make a shorter /longer probev. use a probe with more / fewer mismatches to the lilbrain mRNA

g) Provide molecular explanations as to how altering the salt and formamideconcentrations affects the ability of a probe to bind to a target mRNA.

Increasing salt concentrations means that there are more Na+ ions to coatthe negatively charged backbones of RNA/DNA. This leads to less repulsionbetween probe and target. Decreasing salt concentrations--> less Na+ ions-->more repulsion.

Question 2 (continued)

Formamide is able to form hydrogen bonds with both the probe and thetarget mRNA. Thus, in order for the probe and target mRNA to hybridize inthe presence of formamide, the amount of hydrogen bonds that they form toeach other (i.e. their complementarity) must be very high. At low formamideconcentrations, the probe can bind to less complementary targets (becauseformamide isn't "competing" enough).

After altering the stringency conditions appropriately, you find that lilbrain mRNA isindeed found throughout the embryo (and not just in the brain). As mutations in thelilbrain gene only affect brain morphology, you suspect that translation of the lilbrainmRNA is somehow regulated.

You hypothesize that another protein, encoded by the brainiac gene, normally blockstranslation of lilbrain mRNA in the parts of the embryo where the Lilbrain protein is notneeded.

h) How could you adapt the experiment named in part d) to detect the Brainiac proteininstead of the lilbrain mRNA? Be brief but specific.

Use an anti-Brainiac primary antibody, detect the 1˚ antibody with a 2˚Abconjugated to AP, and use a AP substrate to visualize the location of theBrainiac protein in the embryo.

i) If your hypothesis is correct, where in the embryo would you expect to find theBrainiac protein expressed? Explain your reasoning briefly.

Answer: You would expect to find Brainiac protein expressed everywhereEXCEPT for the brain.Reasoning: Brainiac expressed-->no lilbrain translation-->no brain

developmentIf Brainiac is expressed everywhere EXCEPT the brain, then lilbrain will ONLY betranslated in the brain, leading to appropriate brain development.

Question 3

Please mark whether each of the following statements is true or false.If a statement is false, correct it by crossing out and/or substituting words or

phrases. gray

(For example: __ False__ The winter sky over Boston is usually blue).

NOTE: corrections that make the statement true are noted in BOLD

Question 3 (continued)

___false__ a) In an agarose gel, plasmid DNA that is supercoiled migrates FASTERthan DNA that is nicked.

___false__ b) Isopropanol was used to PRECIPITATE miniprep DNA.

___true__ c) Ligation of two fragments of DNA by T4 DNA ligase requires ATP.

___false__ d) BL21 cells contain T7 RNA polymerase and will therefore expressGFP.

___false__ e) Ethidium bromide binds to DNA is a stoichiometric manner, i.e. theLONGER the DNA fragment, the more molecules of EtBr it binds.

__false___ f) Calf Intestinal Phosphatase (CIP) catalyzes the removal of 5’PHOSPHATES from DNA.

__false__ g) Taq polymerase is NOT denatured each time the temperature is raisedto 96˚C in the PCR reaction.

___true_ h) Type II restriction enzymes (like XbaI, SspI, and EcoRI) cut DNA atpalindromic sites.

___false__ i) Two primers that have the same length (same number of base pairs)AND SEQUENCE will always have the same Tm.

___true__ j) Taq DNA polymerase requires a primer to begin DNA synthesis.

___true___ k) The extension time (time that Taq polymerase spends at 72˚C)determines the maximal size of the product that can be produced in a PCRreaction.

___false__ l) EDTA is a CHELATING agent that sequesters divalent cations (like Mg+2) away from nucleases.

Question 4

The fictitious T. unafish species normally has two pectoral fins, one on each side of itsbody. Development of these fins requires the proper expression of the pec gene.Researchers have identified "finless" T. unafish variants which lack pectoral fins. These"finless" fish can arise in two different ways:

1) transposon mutagenesis of both copies of the pec gene;2) treating the T. unafish with caffeine during development, which drastically

reduces pec transcription

These effects can be observed by performing a Northern blot and probing for pec mRNA,as shown below:

lane 1: molecular weight markerslane 2: untreated fishlane 3: pec::transposon fishlane 4: caffeine treated fish

Note: all lanes contain equal amounts oftotal RNA

(Note: the mRNA from pec::transposon fish is larger because of the transposon insertion)

Despite their lack of pectoral fins, these "finless" fish are capable of mating andproducing offspring. You isolate total RNA from the offspring of two "finless" fish, andperform Northern blots with a pec specific probe.

a) What do you expect to see on the Northern blot if the offspring are the result of amating of two "pec::transposon" fish, and why? (Be sure to indicate both the size andintensity of the pec mRNA band as compared to that seen in "untreated" fish.)

You would expect to see a band that is the same intensity as the "untreated"fish, but this band will be larger than that seen in the untreated fish. This isbecause both parents will give the offspring one copy of the pec::transposongene.

b) What do you expect to see on the Northern blot if the offspring are the result of amating of two "caffeine treated" fish, and why? (Be sure to indicate both the size andintensity of the pec mRNA band as compared to that seen in "untreated" fish.)

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1 2 3 4

Question 4 (continued)

You would expect to see a band that is the same size and same intensity as the"untreated" fish. This is because both parents will contribute a "wild type"copy of the gene to the offspring.

c) What will be the phenotype(s) of these two sets of offspring (wild type or "finless"),and why? Write your answers in the table below:

Parental fish Offspringphenotype

Explanation

pec::transposon finlessThe offspring will inherit two pec::transposon copies;these fish cannot form pectoral fins because thefunction of the pec gene is disrupted by thetransposon insertion.

caffeine-treated wild typeThe offspring will inherit two "wild type" copies ofthe pec gene from their parents. Since the effects of ateratogen are not heritable, the transcription levelswill be unaffected in the offspring. Thus, these fishwill have normal pectoral fins.

Question 5

You have identified an enzyme (SweT) required to metabolize the fictional sugarsweetose in E. coli. You wonder if this enzyme has been conserved during evolution, soyou compare the protein sequence of SweT from E. coli with the SweT proteins from twoother bacterial species, B. subtilis and S. aureus.

While the three SweT proteins are remarkably similar, you do notice a fewdifferences. The E. coli SweT protein (E) is 140 amino acids in length. The B. subtilisSweT protein (B) has a deletion of 30 amino acids at the C-terminus, whereas the S.aureus SweT protein (S) has two point mutations in the sequence from amino acids 76-80. These differences are diagrammed below:

Question 5

a) Draw the structure of the following pentapeptide (aa 76-80) at pH= 7.0:

N terminus---Met-Tyr-Lys-Trp-Glu—C terminus

b) You run the three SweT proteins (E, B, and S) together on a gel filtration column.Which protein will elute from the column last? Explain your answer.

SweT from B. subtilis will elute last from the column because it is smaller.The matrix in a gel filtration column is made of small spheres with holes inthem. The largest proteins will pass through the column by moving aroundthe spheres and elute quickly. Smaller proteins, however, will be able toenter the holes in the spheres and this will impede their progress. Thus,smaller proteins elute later.

Met-Tyr-Lys-Trp-Glu CN

aminoacids1-75

aminoacids76-80

aminoacids

80-140E. coli SweT (E)

CLys-Tyr-Lys-Lys-GluN

B. subtilis SweT (B)CMet-Tyr-Lys-Trp-GluN

S. aureus SweT (S)

Question 5 (continued)

c) You load samples of all three purified SweT proteins on both an SDS-PAGE and a nativegel. Draw the relative migration position of each SweT protein on each gel. (Note: On eachgel, lane E contains E. coli SweT; B = B. subtilis SweT; S = S. aureus SweT).

d) You accidentally pool the tubes containing purified E. coli SweT and S. aureus SweT.What kind of column would you use to separate them, and why?

You would use an anion exchange column. Due to the point mutations in S.aureus SweT, there are an additional two lysines, which will increase theoverall positive charge of the protein. E.coli SweT and S. aureus SweT nowhave different charges and should have different retention properties on anappropriately equilibrated anion exchange column.

e) You also test the ability of all three species to grow on plates with sweetose as the solecarbon source (M9 Swe plates). You find that B. subtilis and E. coli can grow on theseplates, but S. aureus cannot. Based on what you’ve learned about the SweT proteins,how do you interpret this result?

The point mutations in S. aureus SweT may be in the active site of theprotein or may affect the folding of the protein. This change may

SDS-PAGE GEL NATIVE GEL

-

+

-

+

E B S E B S

Met-Tyr-Lys-Trp-Glu CN

aminoacids1-75

aminoacids76-80

aminoacids

80-140E. coli SweT (E)

CLys-Tyr-Lys-Lys-GluN

B. subtilis SweT (B)CMet-Tyr-Lys-Trp-GluN

S. aureus SweT (S)

prevent the protein from acting on sweetose, which would prevent S.aureus from using sweetose as an energy source.

Question 6

You are interested in studying the regulation of genes in the E. coli his operon(histidine biosynthesis). You do transposon mutagenesis using“miniTn10,kan,’lacZ” like in the GEN module, and identify a his::lacZtranslational fusion. You think your insert is in one of three genes--hisD, hisB, orhisA—and decide to use PCR to pinpoint your transposon’s location. Here is amap of the his operon (the his promoter is indicated with an arrow, and eachgene’s size is noted beneath it):

a) You first construct three sets of Forward and Reverse primers (HisD-For,HisD-For, HisD-Rev, HisB-For, HisB-Rev, HisA-For, and HisA-Rev). Indicate onthe map above approximately where these primers will hybridize. (Assume thatthese primers correspond to the Ara primers you used in the 7.02 lab)

Question 6 (continued)

hisB5'3' 5'

3'

1.6 kb 0.8 kb 1.1 kb

hisAhisD

hisB5'3' 5'

3'

1.6 kb 0.8 kb 1.1 kb

hisAhisDD-For A-For B-For

D-Rev A-Rev B-Rev

You run the following PCR reactions, and load a sample of each PCR reactiononto an agarose gel. Note that the LacZ primer is the same one you used in theRDM module, and that you allowed a 2.5 minute extension time for your PCR.You observed the following:

b) Based on this data, where do you think the transposon has inserted? Howconfident are you about this conclusion, and why?

Based on the data in lanes 5-7, it appears that the transposon is insertedin HisA. However, since this is a negative result, it is hard to have a lotof confidence in it. You can say that it is NOT is HisB/D. If it wasreally in HisA, you would have expected a band in lane 9 with the LacZprimer

You notice that none of the PCR reactions using the LacZ primer gaveproducts. You decide to design a new primer specific for the transposon, anduse this in a new set of PCR reactions. Here is a diagram of the

miniTn10,lacZ,kan transposon:

c) You decide to design a primer that hybridizes to the inverted repeat sequenceof the transposon. Your undergrad TA strongly discourages you from doingthis. Why?

Question 6 (continued)

There are two inverted repeat sequences in the miniTn10transposon—one on each end. Thus, your primer could bind toeither/both sites (rather than a unique site) making your analysis

lacZ kanr

4.9 kb

KanR (for part d)

1.0

1.5

0.5

1 2 3 4 5 6 7 8 9 10 Lane DNA used Primers used1 DNA size standard none2 Wild type HisD-For, HisD-Rev3 Wild type HisA-For, HisA-Rev4 Wild type HisB-For, HisB-Rev5 his::lacZ mutant HisD-For, HisD-Rev6 his::lacZ mutant HisA-For, HisA-Rev7 his::lacZ mutant HisB-For, HisB-Rev8 his::lacZ mutant HisD-For, LacZ9 his::lacZ mutant HisA-For, LacZ10 his::lacZ mutant HisB-For, LacZ

complicated. (Also, you may get primer dimers (primers annealing toeach other) formed, depending on how the primer was designed. Thiswould lower your PCR efficiency.)

d) Instead, you design the primer KanR-For, which hybridizes as shown on thediagram above. What primers (of HisD-For, HisD-Rev, HisB-For, HisB-Rev,HisA-For, and HisA-Rev) could you use in PCR with KanR-For to help youdetermine the location of your his::lacZ transcriptional fusion? (Assume that youuse a 2.5 minute extension time for PCR). What would the size of the PCRproduct you obtain tell you about where your transposon inserted?

You could use the three Rev primers (HisD-Rev, HisA-Rev, and HisB-Rev) with the KanR-For primer in PCR. (Based on results in part b, youcould also choose to use HisA3’ to confirm your initial suspicion!). Thesize of the PCR product would tell you how far from the 3’ end of thegene that your transposon inserted.

Question 7

You have two E. coli strains that are identical in every way except that one has anintact copy of the lacZ gene (LacZ+) and one has a deleted lacZ gene (LacZ-). Thestrains are unlabeled, and you need to figure out which strain is which!

In the chart below, describe five different types of experiments/techniques thatyou could use to help you distinguish the two strains. Tell me in what moduleyou learned about the experiment/technique, and what you’d expect to see ifyour strain is LacZ+, and if it is LacZ-!!

(Remember: Many experiments learned in 7.02 can be adapted to study any geneor protein of interest!)

Here are a number of experiments learned about in 7.02 that would work. Forexperiments other than PCR, cells would need to be grown in lactose or IPTG-containing media to induce the production of lacZ mRNA/protein in order to“see” a signal.

Growth on platescontaining IPTG (or

lactose) and Xgal

GEN blue colonies white colonies

β-galactosidase assaywith ONPG on crude

lysate from each strainPBC

yellow colorproduced

no yellow colorproduced

Western blot with anti-β-galactosidaseantibody, AP

conjugated secondaryantibody, and

developing reagent(NBT/BCIP)

PBCband on Western blotat the expected size of

β-galactosidase

no band on Westernblot at the expected

size ofβ-galactosidase

PCR with primersspecific to the

lacZ geneRDM

PCR product onagarose gel that is thesize of the lacZ gene

no PCR product of theappropriate size

Northern blot withlabeled probe specific

to the lacZ geneDEV

band on film of thesize of the lacZ

mRNA

NO band on film ofthe size of the lacZ

mRNA

In situ hybridizationwith lacZ-specific

probeDEV

signal in cells wherelacZ mRNA is present

NO signal in cellswhere lacZ mRNA is

NOT present

Question 8

You perform a transposon mutagenesis using λ1205 and pNK/BW140 E. coli cells. Youselect for mutagenized colonies by plating the cells on LB Kan plates. You then wish toscreen the KanR colonies for the following: 1) insertions in any gene required formaltose utilization (malT, malP, or malQ); and 2) insertions in any amino acidbiosynthetic gene.

a) Describe how you would screen the KanR colonies for insertions in a mal gene.

Presumably, an insertion of the transposon into a gene required for maltosemetabolism would result in the bacteria losing the ability to use maltose as asugar source. You could test for this in two ways. One would be to matchyour mutants onto MacMalKan plates. Cells that can not process maltosewill be white. Alternatively, you can patch your mutants on to M9GluKan

plates and M9MalKan plates. All cells should grow on M9GluKan, butmutants in the mal genes will not grow on M9MalKan.

Question 8 (continued)

b) Describe how you would screen the KanR colonies for those with insertions in anyamino acid biosynthetic gene.

Since amino acid synthesis is required to live, you could patch cells onLBKan and M9GluKan. If there is a disruption in an amino acidbiosynthesis gene, the cell will grow on LBKan but not on M9GluKan. Notgrowing on M9GluKan does not prove that there is a mutation in an aminoacid biosynthesis gene. To show that you do have a mutant in an amino acidbiosynthesis gene and to determine which biosynthesis pathway is disrupted,you can patch your mutants on M9GluKanAA where AA represents one ofthe twenty essential amino acids. Whichever amino acid-supplemented plateyour mutant grows on, that is the amino acid your mutant can not produceand therefore has a mutation in a gene required for synthesis of that aminoacid.

c) What is the expected frequency of mal mutants and amino acid biosynthetic mutants(approximately) among the KanR colonies?

(# of mal genes + # of amino acid biosynthetic genes)

(~4300 genes in the E. coli genome - # of essential genes)

d) What do you predict is the frequency of Tn10 insertions into essential genes (genesrequired for growth) among the KanR colonies? Briefly explain your answer.

The expected frequency is 0. If the transposon disrupts an essentialgene, then cell will not grow on any medium.

Question 9

You are studying an organism that lives in hydrothermal vents (like Thermusaquaticus!). You are working with a gene that keeps proteins folded properly atextremely high temperatures, and you need to map a plasmid containing this gene prior tosubcloning it. After digesting the plasmid with two different restriction enzymes (XbaIand BamHI), you get the following results on an agarose gel.

Question 9 (continued)

Draw a restriction map consistent with these results. Be sure to indicate the location ofall restriction sites, the distances between restriction sites, and the total size of theplasmid.

M 1 2 3 4 M

0.5

1.0

1.62.0

3.04.05.06.07.08.0

0.5

1.0

1.62.0

3.04.05.06.07.08.0

lane M: 1 kb ladderlane 1: uncut DNAlane 2: XbaIlane 3: BamHIlane 4: XbaI + BamHIlane M: 1 kb ladder

~1.85 kb

~0.75 kb

BamHI

~1 kbXbaI

~2.5 kb

~1.2 kb

~7.3 kb

XbaI

XbaI

BamHI

LigateTransformMini-prep

Question 10

After 7.02, you decide to redo the miniTn10 transposon mutagenesis from theGenetics module and see what other types of mutants you can isolate. One of the mutantsyou isolate has a remarkable phenotype—it turns blue and dies rapidly when thetemperature of the incubator reaches 39˚C. To begin to understand this phenotype, youdecide to obtain the sequence of the DNA where the Tn10 inserted.

a) Name the technique that can be used to get the sequence of the DNA flanking yourTn10 transposon.

You can use techniques learned during the RDM module: Digest the genomicDNA that contains the transposon with Restriction Enzyme, e.g. EcoRI.Digest a plasmid with EcoRI. Ligate the genomic DNA fragments into theplasmid. Transform the ligation mixture (note that your ligation is not ahomegenous population where all plasmids have the same insert) intobacterial cells to amplify the plasmid DNA. You now have a library ofgenomic DNA fragments. Retrieve the DNA by doing a mini-prep. PerformPCR on the resulting DNA using a primer specific to the plasmid and aprimer specific to the transposon. Sequence your PCR product to determinethe sequence of the genomic DNA. BLAST this sequence against the E.coligenome to determine the gene into which the transposon inserted.

After performing this technique and sequencing the DNA, you find that your Tn10has disrupted an open reading frame of unknown function. You clone the open readingframe, express it in the mutant bacteria, and find that the bacteria now survive above39˚C!

You BLAST the sequence of the E. coli protein through the database and discover aremarkable similarity to a heat shock protein in maize (corn). Heat shock proteins protectorganisms from high temperatures, and you wonder if the maize protein is capable ofrescuing your E. coli mutation. (In other words, can the maize protein, if expressed in themutant E. coli, prevent the cells from dying at high temperatures?)

To test this, you decide to clone the maize protein into an expression vector usingPCR. The expression vector you choose allows you to clone proteins under the control ofa galactose-inducible promoter (pGAL); these proteins will be produced when thebacteria containing the plasmid are grown in media containing galactose.

b) What is the advantage of using a promoter, like pGAL, that can be regulated?

+ EcoRI E

EE

EE

EE

E

E

E

E

E

E +EE

Question 10 (continued)

Using an inducible promoter allows you to express your gene of interest andtherefore generate a protein product only when you want that protein product to bepresent. In this case, you may only want to express you’re the gene for the maizeprotein when the cells are at 39 ˚C.

c) In order to clone your PCR product into pGAL you’ll need to add the sequence for arestriction enzyme to each of your primers. To which end of the each of the primers (5’or 3’) should the restriction enzyme site be added? Why?

You want to add restriction enzyme sites onto the 5’ end of the primer. Since thepolymerase moves 5’ 3’, you need to have good complementarity between theprimer and the target sequence at the 3’ end. Putting a restriction site on the 3’ endwould disrupt this complementarity.

Finally, you need to choose the restriction enzymes that you will incorporate intoyour PCR primers and that you will use to cut the vector (pGAL) DNA. The followingare maps of the multiple cloning site of pGAL and the DNA sequence that encodes themaize heat shock protein.

d) Another student in the lab suggests that you clone the maize gene by including anXbaI site in your forward primer and an SspI site in your reverse primer. Do you agreeor disagree with this strategy? Provide two reasons for your answer.

NdE S NcX B

pGAL multiple cloning site

E= EcoRIS= SspINd= NdeINc= NcoIX= XbaIB= BamHI

E S BB

restriction map of the open reading frameencoding the maize (corn) heat shock protein

5' ATG TAA 3'

pGAL promoter

5'3'

3'5'

Question 10 (continued)

The forward primer anneals to the 5’ end of the coding sequence of agene and the reverse primer anneals to the 3’ end. By following thesuggested cloning scheme, the XbaI recognition site would be at the 5’end of the coding sequence and SspI recognition site at the 3’ end. Ifyou were to insert the PCR product into pGAL like using theseenzymes, the coding sequence would be in the wrong orientation withrespect to the Gal promoter. The second problem with the suggestedcloning scheme is the SspI recognition site is also present in the codingsequence of the gene, thus digesting your PCR product with SspI wouldtruncate your gene.

Question 11

You have two E. coli strains, JAF1 and DAK2. JAF1 is phenotypically AmpR, andDAK2 is phenotypically AmpS. You suspect that the AmpR gene is carried on aplasmid in JAF1, rather than on JAF1's chromosome, and want to design an experimentto test this hypothesis.

In designing your experiment, you may assume that you have access to the followingmaterials:

1. all reagents, plates, plasmids, and equipment used in the 7.02 RDMmodule

2. Overnight cultures of JAF1 and DAK23. Competent cells of JAF1 and DAK2

a) Describe a simple experiment you could perform to test the hypothesis that the AmpRgene is carried on a plasmid in JAF1 and not on JAF1's chromosome.

(Note: there is no need to describe the details of a particular procedure you might use,simply stating what you would do is sufficient—e.g. ligate A + B together).

Here are the steps you'd want to carry out:

Miniprep plasmid DNA from JAF1 overnight culture Transform this DNA into competent DAK2 cells Plate transformed cells on LB Amp plates

b) What specific experimental result will tell you that your hypothesis is correct?

colonies on LB Amp plates

Question 11 (continued)

c) Identify one positive and one negative control you would include in your experimentabove, and how you will use these controls to help you interpret your "experimental"results.Here are the best controls, and how you'd use them to interpret the data:

Positive control:• transform pET into the competent DAK2 cells; plate on LB Amp

plates• colonies on LB Amp plates tell you that your DAK2 cells are

competent

Negative control:• transform ligation buffer into competent DAK2 cells; plate on LB

Amp platesif you don't see any colonies, confirms that your plates had Amp, that yourcells were really AmpS to begin with, you had no contamination, etc.

(d) If the AmpR gene was carried on JAF1’s chromosome, name a technique that wouldallow you to transfer this gene to DAK2? _____________________

P1 transduction

Question 12

The fictional bacteria B. experimentus contains multiple DNA polymerases thatsynthesize DNA using dATP, dCTP, dGTP, and dTTP as substrates. Ininvestigating what chromatography steps to use to purify these proteins, youfind that polymerase A elutes first while polymerase B elutes second on each ofthe columns listed below (a-c).

Describe the properties of polymerase A and polymerase B that are indicatedby their behavior on each of these columns:

a) The column is an anion exchange column equilibrated with a buffercontaining Tris at pH 7.5 and NaCl at 100mM. The column was eluted with thesame buffer containing increasing concentrations of NaCl up to 1M.

An anion exchange column has positive charge on the resin and thereforebinds negatively charged patches on proteins. The more a protein isnegatively charged, the stronger it interacts with the column and a highersalt concentration is required to elute the protein. Since polymerase A elutesfirst, it elutes at a lower salt concentration than polymerase B and is,therefore, less negatively charged than polymerase B.

Question 12 (continued)

b) The column is a gel filtration column equilibrated and eluted with a buffercontaining Tris at pH 7.5 and NaCl at 100mM.

A gel filtration column separates proteins based on their size using a resinthat has holes in it. Big proteins can not fit in the holes and thus flow rightthrough the column and elute first. Small proteins fit into the holes and thisimpedes their flow through the column, causing them to elute last. Sincepolymerase A elutes first, it must be larger than polymerase B.

c) The column is a dCTP-agarose column equilibrated with a buffer containingTris at pH 7.5 and NaCl at 100mM. The column was eluted with the same buffercontaining 100mM dCTP.

A dCTP-agarose column is an affinity column and specifically binds proteinsthat interact with dCTP. Since polymerase A elutes first, this suggests that itdoes not interact with dCTP as strongly as polymerase B does.

d) Starting with 30 ml of a cleared lysate, you wish to purify polymerase B awayfrom polymerase A using the three columns mentioned above, plus ammoniumsulfate precipitation. Which of the following would be a logical order of thesteps of protein purification? Circle one, and explain your answer briefly.

gel filtration --> ammonium sulfate ppt. --> anion exchange --> dTTPagarose

ammonium sulfate ppt. -->gel filtration --> dTTP agarose --> anionexchange

dTTP agarose --> gel filtration --> anion exchange --> ammonium sulfateppt.

The best choice is the middle one. The third choice is poor because affinitycolumns do not work well with crude lysate. The first choice is okay, but anammonium sulfate cut is usually performed first as an easy, general methodof separation. The affinity and ion exchange columns are much more specificmethods of separation and are almost always performed at the end of aprotein purification.

Question 13

Listed below are seven potential strains (A-G) that could result from the transposonmutagenesis performed in 7.02 lab (using strain pNK/KBS1 and lambda1205). On thechart below, CLEARLY indicate the growth (G or NG) and/or color phenotypes that youwould expect on each plate for each strain.

A. The strain was never infected by lambda1205 (and thus did not receive miniTn10).

B. MiniTn10 inserted into the araC gene in the same orientation and reading frame asthe araC gene is transcribed.

C. MiniTn10 inserted into the promoter of the araC gene, blocking araC transcription.

D. MiniTn10 inserted into the araA gene in the same orientation, but different readingframe, as the araA gene is transcribed.

E. MiniTn10 inserted into the araB gene in the same orientation and reading frame as thegene is transcribed.

F. MiniTn10 inserts into the gene encoding succinate dehydrogenase (constitutivelyactive promoter, not essential for growth) in the same orientation and reading frame asthe gene is transcribed.

G. MiniTn10 inserts into the thrC gene (required for threonine biosynthesis) in the sameorientation and reading frame as the gene is transcribed. (Note: thrC transcription isrepressed in the presence of threonine.)

A B C D E F GM9 Ara Leu Kan NG NG NG NG NG G NG

M9 Glu Leu Kan NG G G G G G NG

Mac Ara Kan NG white white white white red red

Mac Lac Kan NG white white white white white white

LB Xgal Kan NG blue white white white blue white

LB Ara Xgal Kan NG blue white white blue blue white

Note: NG= no growth; G = growth

Question 14

You perform transposon mutagenesis with a modified miniTn10 transposon containingthe AmpR gene (instead of KanR) and screen for cysteine mutants. You make P1 lysatesfrom your mutant to stabilize and map its location relative to the mal (maltose utilization)and arg (arginine biosynthesis) genes. In addition to the cys::miniTn10 mutation, yourmutant is also Mal- and Arg+.

a) What characteristics must the strain that you are using to map have?

AmpSMal+Arg—Cys+

b) You have decided to select for AmpR colonies. What plates will you use for mappingyour mutation?

LB Amp (to select for recipients that took up fragments of DNA with thetransposon)M9 Glu Arg Amp (growth = amibiguous, no growth = Arg-)M9 Mal Arg Amp (growth = Mal+, no growth = Mal-)or Mac Mal Amp (red = Mal+, white = Mal-)

c) You collect the following classes and data:

AmpR Mal+ Arg- 420AmpR Mal- Arg+ 190AmpR Mal+ Arg+ 10AmpR Mal- Arg- 380

What is the order of the cys, mal, and arg genes? SHOW YOUR WORK!

Question 15

You've come a long way since the beginning of the semester! Demonstrate yourunderstanding of the major goals of each of the previous course modules (and thetechniques used to reach these goals) by completing the statements given below.

Microbial Genetics (GEN) module:

cys mal arg

Question 15 (continued)

1. In this module, your goal was to create a mutation in a gene involved the process

of ________arabinose___________.2. These mutations were made using the miniTn10 transposon, which created an

____insertion_____ (a type of mutation) in the gene of interest.3. The kan gene in the miniTn10 transposon made the cells resistant to kanamycin,

which allowed you to ___select for cells that integrated the transposon_.

4. The lacZ gene in the miniTn10 transposon served as a __reporter____ gene;observing ____blue color____ told you that your gene of interest was being

___expressed___ under those particular growth conditions.5. P1 transduction was used to stabilize these mutations by moving them away from the

enzyme ____transposase___, which is required for ___moving the transposon intothe genome (transposition)__.

Question 15

Protein Biochemistry (PBC) module:1. In this module, you learned that ammonium sulfate separates proteins based ondifferences in __solubility__, whereas gel filtration, ion exchange, and APTG-

agarose columns take advantage of the fact that proteins differ in their ___size__,___charge___, and _______affinity_______, respectively.

2. You used ONPG as a substrate for your activity assay (instead of lactose) because

_one of the products (ONP) from cleaving ONPG produces a yellow color_.3. You used SDS-PAGE to separate denatured proteins based on their __size__, and

the Bradford assay was used to determine ______concentration_____.

Recombinant DNA Methods (RDM) module:1. In this module, you used PCR to ___determine which ara gene contained the

transposon.

2. You transferred gfp from the pUGFP vector to the pET vector because the pUGFP

vector lacked a _promoter__, which is required for __expression______ of gfp DNA.3. To accomplish this, restriction enzymes were used to ___cut the DNA at specifc

sites____, and ligase was used to ____join the insert and vector by connectingthe 3’-OH to the 5’-phosphate____.

4. You learned that "transformation" is a method for ____moving DNA intobacteria__, and that the "miniprep" protocol allows you to ___purify the amplifiedplasmid from bacteria___.

4. BL21 cells containing the pET-GFP plasmid glowed green under UV light becausethese cells contain the enzyme __T7 polymerase___, which is required for

__transcribing GFP which is under the T7_____.