C2005/F2401 '06 -- Recitation Problems #9

(Modified from Exam #3 of '01) You have bacteria that are completely normal. They have a normal lactose operon (inducible by lactose) and a normal tryptophan operon (repressible by tryptophan). They have no plasmids. Under optimal conditions, these cells can make 10⁷ molecules of β-galactosidase per cell from the lac operon or 10⁵ molecules of trp synthetase per cell from the trp operon.

A-1. What are optimal conditions for production of the two enzymes (at the same time)? What do you have to add to the medium? If you want the cell to produce both enzymes, you need to (add lactose) (add trptophan) (add both) (add neither).
A-2. Which operon has a stronger promotor? (lac operon) (trp operon) (neither -- both promotors are the same strength) (can't predict).

In the remaining questions, you transform these bacteria with recombinant plasmids that have the gene for enzyme X (not normally made by these bacteria) next to a promotor and operator. All the plasmids are made correctly so that the enzyme X gene is transcribed under optimal conditions. The promotor and operator on the plasmids come from either the lactose or tryptophan operons. (See table below for properties of plasmids 1-4).

B. You transform your bacterial cells with plasmid 1 and the cells start to make enzyme X. To stop the cells from making more enzyme X, you would (add lactose) (add tryptophan) (remove lactose) (remove tryptophan) (add both) (take away both) (doesn’t matter – none of these changes will have any effect). Explain briefly.

C. You repeat your experiment, but this time you transform your normal bacterial cells with plasmid 2.
C-1. If you want to get maximal production of enzyme X from cells transformed with this plasmid, what must be present in the medium? (both tryptophan & lactose) (only lactose) (only tryptophan) (neither), AND
C-2. How should maximum production of enzyme X compare in cells with plasmid 1 vs plasmid 2? Maximum enzyme X production should be (higher with plasmid 1) (higher with plasmid 2) (same with either plasmid) (beats me -- can’t tell without more info.) Explain both answers briefly.

D. Suppose that enzyme X has quaternary structure – it consists of two separate subunits. Both subunits are needed to form active enzyme X. All plasmids involved here carry the structural genes for both subunits. Both genes are transcribed as part of a single polycistronic mRNA. Assume that enzyme X is needed for growth in the presence of sillimycin (an antibiotic). You transform cells with two plasmids at the same time -- you use plasmids 3 & 4. Cells transformed with either plasmid alone do not produce active enzyme X. (See table.)
D-1. You mix your normal bacteria with a solution containing both plasmids and place the bacteria on medium containing sillimycin. All of the bacteria which get both plasmids produce colonies. To get this result, it is necessary to include (lactose) (tryptophan) (both of these) (neither of these) in the growth medium AND
D-2. The structural mutations in plasmids 3 & 4 are (identical) (in different base pairs but in the same gene) (in different genes) (different, but can't tell if they're in the same gene or not).
D-3. Suppose you could isolate one molecule of polycistronic mRNA (for enzyme X) from a bacterium that is growing on sillimycin. (This is one of the bacteria transformed with both plasmids.) You translate the mRNA in a test tube using added tRNA, ribosomes, etc. Do you expect the protein you make to be enzymatically active? (yes) (no) (beats me). Explain your choices briefly.

Plasmid	Promotor	Operator	Enzyme X gene(s)
Plasmid 1	lac	lac	normal
Plasmid 2	trp	lac	normal
Plasmid 3	lac	constitutive lac operator	Substitution mutation in structural gene
Plasmid 4	trp	trp	Substitution mutation in structural gene

All promotors and operators are normal except in plasmid 3. "lac" means from the lactose operon; "trp" means from the tryptophan operon.

C2005/F2401 '06 -- Answers to Recitation Problems #10 updated

1. See below. The * indicates the points of cutting. You assume the original sequence is a palindrome, so there must be a G on the 5' end in the original sequence.

5' G*T C G A C	-->	G	+	TCGAC
3' C A G C T* G		CAGCT		G

2. A. (i) Digest both DNAs with the same enzyme, either Cut 1 or Cut 2. To make a recombinant DNA, you must cut the two DNA's so that the pieces will have overlapping complementary ends. In other words, you need overhanging ('sticky") ends that will pair up. In this case, the DNA cut by one enzyme will not pair up to DNA cut by the other. The sticky ends of the DNA cut by the two enzymes look the same at first glance, but they aren't equivalent. If you cut two DNA's with the two enzymes, the ends will look like this:

	Cut 2			Cut 1
5'...C	3'	5' TTAA	G... 3'	5'...G	AATT 3'	5'	C...3'
3'...G	AATT 5'	3'	C... 5'	3'...C		3' TTAA	G...5'

It looks like you can just flip one fragment over, say the right-most fragment from Cut 2, and ligate it to the left-most fragment from Cut 1. But that won't work because of problems with the 3' and 5' ends. You have to connect a 3' to a 5' -- you can't connect two 3' ends or two 5' ends. Try it and see.

(ii). You need ligase. First you mix the pieces and let them hybridize. This is the equivalent of pairing up say, the left fragment from Cut 1 (or Cut 2) with the right fragment from a different DNA cut with the same enzyme. The hydrogen bonds between the "sticky" overlapping ends will hold the two fragments together temporarily, but you need ligase to join the fragments permanently (covalently).

B. (i). a. No, because the plasmid will not have a functional origin of DNA replication. In this case, the insert goes in the middle of the ori, not in the middle of a Drug resistance gene or other selectable marker.
b. Drug sensitive. Since the plasmid can't replicate, only one of the descendants of a single transformed cell will have a plasmid. All the rest will have no plasmid and no drug resistance gene.

(ii) (a). Plasmids must be integrated into the chromosome. (b) Integrated plasmids must contain inserts of human DNA. (c) Both.
Explanation: You selected colonies that contain the human DNA and are drug resistant. Each original cell that got a plasmid (with a human DNA insert) must have divided many times to produce a colony -- without losing the plasmid (& insert). So we know that the plasmid DNA and the human DNA have been replicated and passed on at each cell division. However insertion of the human DNA into the plasmid destroys the origin of replication of the plasmid. So the simplest solution is that you have selected cells in which the plasmid (plus insert) has become integrated into the chromosome by crossing over. Therefore the plasmid doesn't need an origin -- it (along with the human DNA insert) is replicated using the chromosomal origin. In the presence or absence of the drug, the integrated plasmid will continue to be replicated and passed on, just like the rest of the bacterial chromosome. If the plasmid were not integrated, it would probably be lost in the absence of the drug, since there would be no selection against cells that had lost their plasmids (& lost their drug resistance). But since the plasmid is integrated, it and its insert, will both be faithfully replicated and passed on, and so both should be detectable by hybridization.

C. (i). The 4 bands are of equal intensity, but that doesn't mean the DNA (or length of exon) in each of them is the same size. It means the amount of probe sticking to the material in each band is the same. (The DNA fragments in different bands must be different lengths to end up in different places on the gel/blot.) The equal intensity means the DNA in each band "captures" or hybridizes to the same amount of labeled probe. In this experiment, each fragment of genomic DNA hybridizes to one cDNA molecule. The length of the hybridized (hydrogen bonded double stranded) region varies, depending on the length of the exon(s) in the fragment, but there is always one probe molecule "stuck to" each genomic fragment. If the genomic DNA were cut up to give several different pieces, and some of the pieces were the same length (ended up in the same band), then the bands would not be of equal intensity. Also if the probe were cut up into many pieces, then different exons might hybridize to (trap) different amounts of probe.

(ii). 3 introns and 4 exons. The DNA must be cut in 5 places, to produce 4 different pieces of DNA. In won't be cut in any of the exons, because the cDNA (= exons only) has no sites for the restriction enzyme. The two "outside" cuts are not in introns -- they are in spacers outside the gene. The other three cuts must be in introns. (Only spacers between exons, inside genes, qualify as introns.) Each of the 4 pieces contains at least one exon, and so hybridizes to the cDNA probe. All the pieces hybridize to the same probe, but each piece hybridizes to a different stretch of bases in the probe.

D. (i). One radioactive band. (ii). Band will be longer than 200 bp.
Why one band? The DNA fragment that contains the exon corresponding to the probe will hybridize to the radioactive probe. None of the other fragments will hybridize. You will get the same number of DNA fragments on the gel or blot as before, but only one will trap probe and form a radioactive band.
Why longer than 200 bp? The piece of genomic DNA that hybridizes to the probe will be longer than the probe. The probe is a single exon (of 200nt), but the genomic DNA fragment includes some intronic sequences on each end, because all the restriction sites are in introns.
The length of the DNA fragments is given in bp (base pairs), because the DNA fragments are double stranded when they are separated on the gel. The length of the probe is given in nt (nucleotides), because the probe is single stranded. The double stranded DNA fragments are separated first, and then denatured and hybridized to the single stranded probe.

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