MIT-SCIENCE-Lectures-fa01quizrdmans

Question 1 (20 points)

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.

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

Grading was as follows: -1 point if true/false incorrect

-1 point for failing to correct a false statement

-1 point for correcting a true statement

__false___a) In an agarose gel, plasmid DNA that is supercoiled migrates more slowly than DNA that is nicked.

Acceptable corrections (changes are underlined):

1. In an agarose gel, plasmid DNA that is supercoiled migrates more quickly

than DNA that is nicked.

2. In an agarose gel, plasmid DNA that is nicked migrates more slowly than

DNA that is supercoiled.

__false___b) Isopropanol was used to remove excess salt from precipitated miniprep DNA.

1. Isopropanol was used to precipitate miniprep DNA.

2. 80% Ethanol was used to remove excess salt from precipitated … DNA.

Note: -0.5 points for Ethanol instead of 80% Ethanol; in fact, it is the 20%

water in 80% ethanol that actually removes the excess salt.

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

___false__d) AG1111 cells contain T7 RNA polymerase and will therefore express GFP.

1. AG1111 cells do not contain T7 RNA polymerase and therefore will not

express GFP.

2. BL21 cells contain T7 RNA polymerase and will therefore express GFP

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

1. Ethidium bromide binds to DNA is a stoichiometric manner, i.e. the shorter

the DNA fragment, the fewer molecules of EtBr it binds.

2. Ethidium bromide binds to DNA is a stoichiometric manner, i.e. the longer

the DNA fragment, the more molecules of EtBr it binds.

___false__f) Calf Intestinal Phosphatase (CIP) catalyzes the removal of 5’ hydroxyls from DNA.

1. Calf Intestinal Phosphatase (CIP) catalyzes the removal of 5’ phosphates

from DNA.

___false__g) Taq polymerase is denatured each time the temperature is raised to 96?C in the PCR reaction.

1. Taq polymerase is not denatured each time the temperature is raised to 96?C

in the PCR reaction (because it is thermostable).

2. DNA is denatured each time the temperature is raised to 96?C in

the PCR reaction.

__ true___h) Type II restriction enzymes (like XbaI, SspI, and EcoRI) cut DNA at

palindromic sites.

___false__i) Two primers that have the same length (same number of base pairs) will always have the same Tm.

1. Two primers that have the same length (same number of base pairs) will

NOT always have the same Tm.

2. Two primers that have the same length (same number of base pairs) AND

the same GC content (or GC/AT ratio) 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 PCR reaction.

___false____l) EDTA is a reducing agent that sequesters divalent cations (like Mg+2) away from nucleases.

EDTA is a chelating agent that sequesters divalent cations (like Mg+2) away

from nucleases.

Question 2 (12 points)

After 7.02, you enroll in 7.17 (Project Lab), where you need to design primer pairs for a PCR experiment. The first three times you design your primers, the TA sends you back to try again because there are problems with the reverse primer. Finally, on the fourth try, you get it right. Here are the sequences you tried:

Forward primer:

5' CCGCGGCGCAGGAGAAC 3'

Reverse primer attempts (in no particular order):

CAAGAGTACAGCATACT 3'

a) 5'

CATATCGGCGTACAGCAGTTC 3’

b) 5’

CATATCGGCGTACAGCATCCT 3'

c) 5'

CATCGTGCAGTACTGCACGAT 3'

d) 5'

Which was the successful reverse primer? Explain your choice by pointing out the serious flaw in each of the “unsuccessful primers.”

The successful primer was primer c) (3 points). The other primers had the following “serious flaws” (3 points each):

Primer a): The melting temperature (Tm) of this primer is 48?C. This is too low for the PCR to be successful, as the Tm of the forward primer is 60?C (primers need similar Tms).

Primer b): This primer has a high probability of forming a “primer dimer” with the forward primer. “Primer dimers” form when the 3’ ends of the primers are complementary, forming a structure that looks like this:

5’CATATCGGCGTACAGCAGTTC3’ (reverse primer)

|||||

(forward primer) 3' CAAGAGGACGCGGCGCC 5’

The formation of this structure will reduce the efficiency of PCR, as primers can bind to each other rather than the template DNA.

Primer d) This primer is capable of forming a hairpin structure by basepairing with itself.

5' CATCGTGCAGT--|

|||||||||| |

3' TAGCACGTCA--|

Partial credit was also given for mentioning “primer dimers,” low Tm, and hairpins, but not identifying the correct primer that had the flaw.

Question 3 (18 points)

4 kb

1 kb 5 kb

You want to ligate a 1 kb insert into a 4 kb vector as shown in the diagram above. Both the vector and the insert were cut with XbaI and EcoRI, and therefore the ligase will be joining compatible cohesive ends.

You have the following reagents at the concentrations given:

Insert in TE (at 100 mg/ml)Vector in TE (at 100 mg/ml)T4 DNA ligase (4000 U/ml)

10X ligase buffer (without added ATP)5 mM ATP ddH 20

You want to set up a 20ul reaction using 0.2 mg vector DNA, 4 units of ligase, and containing ATP at a final concentration of 500 μM. You also know that the molecular weight of a 1kb fragment of DNA is 6.6x105g/mole.

a) (8 points) How much insert DNA will you add? Explain your answer/show your calculations.

In a ligation with compatible cohesive ends, you want a 2:1 molar ratio of insert to vector:0.2 mg vector X__1 mole of 4 kb vector_ = 7.6 x 1011 moles X 2 = 1.5 x 1010 moles

4 kb x __6.6 x 10

5 g__ of vector of insert

1 kb 1.5 x 1010 moles X __6.6 x 105 g__= 1 x 10-4 g X ___103 mg___ = 0.1 mg insert

of 1kb insert mole of 1kb g

insert

b) (10 points) What volume of each reaction component will you add? (Fill in the table)

Reaction component

volume added

vector DNA 2 μl insert DNA 1 μl T4 DNA ligase

1 μl 10X ligase buffer (without added ATP)

2 μl ATP

2 μl

You are studying an organism that lives in hydrothermal vents (like Thermus aquaticus !).You are working with a gene that keeps proteins folded properly at extremely high

temperatures, and you need to map a plasmid containing this gene prior to subcloning it.

After digesting the plasmid with two different restriction enzymes (XbaI and BamHI), you get the following results on an agarose gel.

Draw a restriction map consistent with these results. Be sure to indicate the location of all restriction sites, the distances between restriction sites, and the total size of the plasmid.There are actually two possible maps (diagrammed above) that fit this data, and either was acceptable.

1234M

0.51.0

1.6

2.0

3.0

4.0

5.0

6.0

7.0

8.00.5

1.0

1.6

2.0

3.0

4.0

5.0

6.0

7.0

8.0lane M: 1 kb ladder lane 1: uncut DNA lane 2: XbaI lane 3: BamHI

lane 4: XbaI + BamHI

lane M: 1 kb ladder

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

a) (2 points) Name the technique that can be used to get the sequence of the DNA flanking your Tn10 transposon.

Inverse PCR (partial credit for PCR)

After performing this technique and sequencing the DNA, you find that your Tn10 has disrupted an open reading frame of unknown function. You clone the open reading frame, express it in the mutant bacteria, and find that the bacteria now survive above 39?C!

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

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

b) (4 points) What is the advantage of using a promoter, like pGAL, that can be regulated? Using a promoter, like pGAL, that can be regulated allows the researcher to control the expression of the gene of interest (when it is “on” and “off”). This is especially important when the protein product of the gene is toxic to the cells in which it is being expressed.

c) (8 points) Attached is the GenBank entry (L28712) for the maize heat shock gene you cloned. On the GenBank entry, circle the start and stop codons of the maize heat shock gene, and underline a sequence that would make a good forward primer for your PCR reaction.

To find the start and stop codons, you need to look next to “CDS” (coding sequence) on the GenBank entry. The start codon of the maize heat shock gene begins with nucleotide #91 (i.e. the “a” of the “atg”), whereas the stop codon ends with nucleotide #813 (i.e. the “g” of the “tag” stop codon). Therefore, to receive 2 points, you needed to circle nucleotides #91-93 and #811-813.

The other six points were given for choosing an appropriate forward primer. The important factors considered were location (upstream of or including the start codon) and Tm (length and GC-rich enough for a Tm of ~50-60?C.

d) (4 points) In order to clone your PCR product into pGAL you’ll need to add the sequence for a restriction 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 should add the restriction enzyme site to the 5’ end of each primer. This is important for two reasons. First, the DNA polymerase (Taq) is going to recognize and bind to the 3’end of the primer to begin DNA synthesis. Thus, it is important that the 3’ end of the primer is complementary to the template DNA (which it is likely not to be if a restriction enzyme site is attached). Second, you want to be sure that the entire coding sequence is part of your PCR product. If you were to start your forward PCR primer at or near the start

codon, for example, and placed the RE site at the 3’ end, you would cut off part of your gene when you went to clone it into your vector.

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

e) (10 points) Another student in the lab suggests that you clone the maize gene by including an XbaI site in your forward primer and an SspI site in your reverse primer. Do you agree or disagree with this strategy? Provide two reasons for your answer.

I disagree with this strategy (2 points) for the following reasons (4 points each):

1. There is an SspI site in the middle of the maize heat shock gene ORF. If you

incorporate an SspI site in your primer, then cut with SspI to clone the insert into pGAL, you will cut your ORF in half! Thus, you will not clone the whole ORF into pGAL.2. Using XbaI in the forward primer and SspI in the reverse primer would also clone

your ORF into pGAL in the incorrect orientation. Thus, no heat shock protein would be made.

pGAL multiple cloning site

E= EcoRI S= SspI Nd= NdeI Nc= NcoI X= XbaI B= BamHI

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

5' ATG

TAA 3'