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100 points possible. Mean score: 79.6. Range: 43.5 - 94.5.

1. (9 pts) List the three essential features of a plasmid vector used in cloning, and describe the function of each.

[full-credit student answer]

a. Origin of replication: this feature is essential for the DNA polymerase to recognize the foreign plasmid and begin replication. This orgin of replication must be specific for the organism into which the vector will be transformed.

b. Marker: The selectable marker allows us to separate cells that have taken up the plasmid from those that have not. One type of marker could be a marker that codes for ampicillin resistance. If this marker was used, we could treat all the cells with ampicillin and those that have the marker live, those that don't die.

c. Polylinker: This is a region that allows us to add inserts of DNA into our vector using restriction enzymes. The sequence of the polylinker is known to us and chosen with specific restriction enzymes in mind. With the restriction enzymes we cut the vector open and leave sticky ends. We know the sequence of the sticky ends and this allows us to make complementary sticky ends on our insert so we can incorporate the insert into the vector.

2. For each of the following types of cloning vectors, provide the following information: the approximate size of DNA insert that the vector can carry, a primary advantage of this type of vector, and an example of a situation in which this vector type would be a good choice.

[full-credit student answer]

a. plasmids:

(1 pt) approximate insert size: 10 - 20 kb (actually the lower limit is as small as you like...)

(2 pts) advantage: creating a plasmid is a rather simple procedure in comparison to other methods

(2 pts) example of situation in which plasmids would be the way to go: when working with simple organisms whose genome can be cut into 10-20 kb pieces without there being too many pieces

 

b. lambda phage

(1 pt) approximate insert size: 25 kb (20 - 25 kb)

(2 pts) advantage: a little larger than plasmids and the transformation rate is extremely high. It is possible to transform more bacteria per plate than with a plasmid.

(2 pts) example: Use this when working with more complex organisms whose genome would take up too many agar plates to use a plasmid.

 

c. COS vectors

(1 pt) approximate insert size: 45 - 50 kb

(2 pts) advantage: larger fragments allow one to clone larger genes. Still have the extremely high transformation rate of lambda phage.

(2 pts) example: when making genomic libraries of more complex organisms that include introns and thus have bigger genes.

 

d. artificial chromosomes

(1 pt) approximate insert size: 1 mb (1000 kb)

(2 pts) advantage: can place entire chromosomes into yeast cells. Allows you to ask questions outside of coding DNA and study whole chromosomes. (You would not be able to study entire human chromosomes this way, but you can study many genes at once.)

(2 pts) example: to study chromosomal abnormalities caused by problems in non-coding DNA.

 

3. Expression vectors that carry the T7 promoter can be used either for expression of cloned genes in bacteria or for in vitro transcription/translation systems.

[full-credit student answers, with additions from me]

a. (5 pts) In bacterial systems, what is the potential advantage of using a T7 promoter rather than the lac promoter? Explain.

The potential advantage of using a T7 promoter vs. the lac promoter is that T7 RNA polymerase specifically polymerizes RNA associated with the T7 promoter [i.e. transcribes genes that are immediately downstream of the T7 promoter]. When using a T7 promoter on a plasmid vector, we use genetically engineered bacteria that have the T7 polymerase gene integrated into their genome [under control of the lac promoter, so that expression of T7 RNA polymerase is inducible with IPTG.] So all of the T7 RNA polymerase molecules only recognize the T7 promoter, [and hence only transcribe the gene of interest.] Whereas when using a lac promoter, it relies on the bacteria's own RNA polymerase which will also be busy polymerizing RNA elsewhere in the genome. So more protein is produced with using a T7 polymerase system due to the high specificity [and efficiency] of T7 RNA polymerase.

b. (6 pts) What are the steps involved in in vitro transcription/translation, once an expression construct has been created? (An expression construct, in this case, is a plasmid carrying the gene of interest under the control of a T7 promoter.)

To this plasmid DNA, one wants to add T7 polymerase, NTP's, and buffer. [This mixture is incubated to allow T7 polymerase to make RNA copies of the gene-of-interest, using the NTP's as building blocks. The reaction is then stopped chemically, and then...]

To this mixture eukaryotic cell extract, with all mRNA removed, is then added. At this step labeled amino acids can also be added, to yield a labeled protein. [The only protein that should be synthezied, and hence the only protein in this mixture that will be labeled, is the protein of interest.]

 

c. (2 pts) What is a primary advantage of using a bacterial expression system rather than an in vitro approach?

The bacterial system can have much higher yield than the in vitro approach. [MUCH!]

d. (2 pts) What is a primary advantage of using an in vitro approach rather than a bacterial expression system?

The in vitro approach yields a product that is pretty pure. [and is specifically labeled]

4. As a graduate student, you are studying a protein that is important for a normal rate of cell growth in yeast. You've obtained a strain carrying a mutation in the gene for this protein. As you predicted, your mutant strain grows more slowly than wild-type strains--it takes longer for cells of your strain to establish full-size colonies on an agar plate. When you grow your cells in this way, then, you can see their phenotype because they have a small colony size.

One day, you go to the incubator to take out an agar plate on which you have placed some of your mutant cells. You notice that one of the colonies on your plate is quite large relative to the others, even though all of the cells were derived from a single culture of genetically identical cells. "Aha!" you say to yourself. "I remember this from Bio311 at Lewis & Clark! I've isolated a suppressor!"

To make sure, you make an agar plate with cells from your new "suppressor" colony on one side and cells from your original strain on the other side. Sure enough, your "suppressor" colonies grow up nice and big, demonstrating a more rapid rate of growth.

[again, full-credit student answers with comments from me...]

a. (2 pts) Is this a genomic or a high-copy suppressor? Explain

Genomic. A random mutation seems to have reversed the effects of the mutant gene.

[No high-copy plasmids were added to this strain, so the change in phenotype must be due to a spontaneous genomic mutation. Even if this mutation altered expression of a gene, it would be considered a genomic suppressor.]

[Even though you did not mutagenize this strain, spontaneous mutations can (and did) arise!]

b. (6 pts) When you go into your advisor's office to share the good news, your advisor asks you how you know that you have isolated a suppressor and not a revertant. Mumbling to yourself, you go back out into the lab to do the experiment. You cross your "suppressor" haploid strain to a wild-type haploid strain to get a diploid, and then you sporulate that diploid and test its haploid progeny for colony size.

What pattern will you see if your "suppressor" is really a revertant? What will you see if you're right, and it's a suppressor? Explain.

Revertant: all progeny would have a wild-type phenotype because the [original mutation] would no longer exist.

Suppressor: Some of the progeny will retain mutant phenotype because during synapsis, it is likely the suppressor gene was swapped off the mutant gene's chromosome in some cases because they were not the same gene. It is also just as likely [actually more likely, since there are 16 chromosomes in yeast] that the suppressor gene wasn't even on the same chromosome to begin with and because chromosomes independently segregate in meiosis, some progeny are bound to receive a copy of the mutation but not the suppressor gene.

[If the original mutation and the suppressor are unlinked, about 1/4 of the progeny will show the original mutant phenotype, because they're inherited the original mutation and not the suppressor.]

[Be careful of the word "mutation," which means a change in DNA sequence. Yeast colonies will not "show the mutation" unless you isolate their DNA and sequence it. They may, however, show the mutant phenotype. Also, a suppressor will not "return a strain to WT", just to a WT phenotype.]

c. (4 pts) If you've isolated a suppressor, how could you determine whether you've got an interaction suppressor or a bypass suppressor? Explain.

[my answer]

Test the suppressor for allele specificity. If you have (or can make) yeast strains that carry other mutations in the same gene as your original mutation, you can do this test. Cross each of these strains with a haploid strain carrying your suppressor mutation, and sporulate your diploids to obtain haploid progeny that carry each of these mutations in combination with the suppressor. Test their phenotype. If the suppressor is able to allow all of them to show normal growth, it is NOT allele specific and is therefore probably a bypass suppressor. If the suppressor only suppresses the phenotype of some of the mutations, it IS allele specific and is therefore probably an interaction suppressor.

5. a. (1 pt) What do the initials "VNTR" stand for?

Variable number of tandem [or trinucleotide, or tetranucleotide] repeats.

b. (3 pts) How can the length of a particular VNTR region in an individual's genome be determined?

 [full-credit student answer]

By flanking each end of a known VNTR site with primers and producing many copies through PCR reactions you can obtain enough DNA of that region to be run on an agarose gel [and visualized with ethidium staining]. The size is determined by comparing it to a standard ladder of known DNA fragment lengths on the electrophoresis gel.

c. (10 pts) Explain how VNTR regions can be used as genetic markers to identify the chromosomal region involved in a particular human genetic disease.

VNTR regions are subject to frequent mutations, namely insertions or deletions of an entire repeated sequence of DNA, due to slippage when [replicating] such repeated sequences. The mutations are genetically passed on through generations. They are of use to scientists in search of a disease causing gene, when the scientist has a few generations of families in which the gene runs who are willing to give blood samples to be analyzed. Using the technique in b) or by sending the samples to a company willing to determine the lengths of 400 VNTR regions for both alleles of all the family members, the scientist has at his/her disposal the information to make some inferences based on the data. The scientist's task is to look carefully through all VNTR site lengths and try to find one that was always present in about the same length in both alleles of an individual with the disease (assuming recessive and genetic ) and in one allele of each of the parents. The conclusions drawn would not be that the mutated gene was within the VNTR region (though it could be ) but that the gene was somewhere nearby. This can be inferred because a nearby gene would be inherited along with the VNTR site neighboring it, since genetic recombination is low between close genes.

6. In your job in the R&D department of WhizBang Biotech, incorporated, you have been given the task of isolating a new source of thermostable DNA polymerase for PCR reactions. Your company hopes to market this new enzyme as a cheaper alternative to the ever-popular Taq and Pfu polymerases already available.

You have on hand the cloned gene for Taq polymerase and a collection of thermostable bacteria of various species, from which can readily isolate genomic DNA. You design an approach based on the assumption that a DNA polymerase that will be useful in PCR will be similar in sequence to Taq polymerase.

a. (6 pts) Describe a simple method for screening these organisms for the presence of a gene homologous to the gene for Taq polymerase. (Assume that the genomes of these organisms have NOT been sequenced.)

[full-credit student answer, with my comments]

A Southern blot (or zoo blot) could be used for this task. DNA would be isolated from the genomes of the thermal bacteria that were to be used, and a couple of organisms that are known not to have the protein, as a control [this is an extra touch, not required.] The DNA would be [digested with restriction enzymes,] run in an agarose gel, transferred to a membrane, probed with a radioactively-labeled probe designed from the Taq polymerase gene, and then washed. A picture would be taken [i.e. a piece of Xray film would be exposed to the membrane and then developed]. DNA to which the probe bound should have a similar polymerase gene.

b. (12 pts) Your procedure in step a identified a likely organism. Describe how you could clone the gene for its polymerase, using the Taq polymerase gene as a probe. (Please describe these techniques step by step--impress me with how clearly you understand them!)

 [my answer]

I would construct a genomic library from the organism of choice (see notes below for an explanation of this approach instead of a cDNA library.) To do this, I'd grow a culture of the organism, isolate and purify its genomic DNA, and then digest this DNA with a restriction enzyme. I'd use a frequent-cutting restriction enzyme for this purpose, and perform a partial digest. This should give me a collection of overlapping fragments that span the genome, increasing my chances of isolating my gene intact.

Next, I'd digest a plasmid vector with the same restriction enzyme (or one that generated compatible "sticky ends" to the first enzyme) to open its polylinker. I'd mix these linear vector fragments with my genomic DNA digest and set up an overnight ligation. This should allow each vector fragment to take up a genomic DNA insert.

Next, I'd transform this ligation mixture into E. coli and select transformants. If I were using a plasmid vector that encodes ampicillin resistance, I'd plate my cells on plates containing ampicillin. I would attempt to do this in such a way as to get about 200 - 300 E. coli colonies per plate, and a total of many thousands of E. coli colonies. Each colony should carry a different plasmid, containing a different inserted piece of genomic DNA. Within each colony, all the cells should have the identical plasmid. In this way, the bacterial transformants amplify each plasmid for me, making many many copies of each within each colony.

Next, I'd make "filter lifts" of each bacterial plate, by pressing a circular membrane to the surface of each and lifting it off with some bacterial cells "attached." I would lyse these cells on the membrane, denature their DNA, and "fix" the DNA to the membrane with a chemical treatment (or by UV irradiation).

Next, I'd make a solution containing my radiolabeled probe from step a), made from the Taq polymerase gene. I'd wash this solution over my membranes and let hybridization occur overnight. Then I'd wash off all non-specifically bound probe, dry the membranes, and expose them to film.

When I develop the film, I should see circular spots on some of the membranes, corresponding to colonies whose DNA was able to hybridize with the probe. I'd go back to my original plates, isolate these colonies, and grow liquid cultures of each.

Next, I'd prepare DNA from each of my "positive" cultures. I'd purify the plasmid DNA from each culture and sequence it, using vector sequences to design my primer. I'd compare this sequence to the known Taq polymerase gene sequence and look for similarities between the two. If I find significant homology, I have isolated and cloned a new polymerase gene! (If not, I've isolated a false positive, and I should keep looking.)

NOTE 1: Why not a cDNA library? Many of you suggested this. There are several reasons why it would not be a good idea in this case...

1. bacterial mRNA is VERY unstable, with a half-life of seconds or minutes. It is therefore very hard to isolate and handle in the lab.

2. Bacteria don't poly-adenylate their mRNA, so you couldn't isolate it with a poly-T column. You could get around this another way, probably, but...

3. There's no compelling reason to use a cDNA library when working with an organism with a small genome, especially with an organism that lacks introns (as bacteria do).

NOTE 2: Why not just lift the positive band from the Southern blot and clone it into a vector? This would be really nice if you could get it to work, because it would save you virtually all of the steps described above. Alas, it is not to be. When the DNA is transferred from the agarose gel to a membrane for probing, it is denatured (so the probe can anneal to it) and then attached to that membrane chemically, or by UV cross-linking. There's no going back from that point to intact DNA, at least not by any method of which I'm aware. Also, if this gel contained a digest of genomic DNA, there are likely to be many bands of the same size at each location, only one set of which correspond to the gene you want.

7. You've successfully obtained a plasmid carrying the DNA polymerase gene from your thermostable organism, as described in problem #6. You would now like to obtain the DNA sequence of this gene. The DNA sequence of the polylinker section of your plasmid vector is as follows:

5' GGTAGGTACCTAGCTGCTAAGTCGTGAATTCTGCGTAGCTGCGATCGTACGTGGTGTC 3'
3' CCATCCATGGATCGACGATTCAGCACTTAAGACGCATCGACGCTAGCATGCACCACAG 5'

Your insert has been cloned into the EcoRI site of this region, which is indicated by bold type above. (EcoRI cuts this site between the G and the first A.)

a. (2 pts) Give the nucleotide sequence of a 15-base oligonucleotide primer that you could use for your sequencing reactions (there is more than one correct answer to this question--pick one)

There are too many correct answers to list. You needed to give a 15-base sequence that...

• matches the sequence above
• does not completely span the EcoRI site and go beyond it
• reads 5' to 3' *toward* the EcoRI site, not away from it

This last point was the most common error. DNA polymerase will add new nucleotides (and ddNucleotides) on to the 3' end of your primer. If this end is oriented away from the EcoRI site, you will get sequence of your vector (which you presumably already know) instead of your insert.

b. (10 pts) Describe how you would set up these reactions. What would be included in each? What is the purpose of each ingredient?

The most common error here was not reading the question, and giving me an explanation of how sequencing works and a picture of a sequencing gel. Instead, I asked for a list of ingredients in each reaction and an explanation of what each is for.

The second most common error was to imply that PCR is somehow involved in this process, perhaps preceding the sequencing reactions themselves. PCR is not required here. You have a plasmid clone of this gene, which means that you can grow a bacterial culture carrying your plasmid and get as much DNA as you want or need.

[full-credit student answer, with comments]

I would set up 4 test tubes. Each test tube would have template DNA, primer, dNTP's, polymerase, buffer, and a ddNTP of A, T, C, or G, respectively. [The ddNTP will be present at about 1/100th the concentration as each of the dNTP's.] The DNA serves as the basis [template] of the reaction. The primer serves as a chain which the polymerase can use as a starting point to begin the reaction of copying the complementary sequence. Buffer (together with temperature) provides the suitable conditions for the reaction. dNTP is neccessary for the reaction, in order for the polymerase to have something to add to the chain. [The dNTP's also provide the energy for the synthesis reaction, since two phosphates are released each time a nucleotide is incorporated. This is the same energy source as is present in vivo.] The ddNTP's are special because when they are selected to be put on the end of the chain the reaction stops. The 3' "OH" group [which is missing in the ddNTP's] is needed to continue the series of condensation reactions. Thus, what will happen is that there will be a stop of the sequence at variable sites that correspond to the respective ddNTP added, either A,T, C, or G.

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