Problem 13-8 from 19th ed (2009) of Problem Book

The entire question and the entire answer are included here. Parts A and D are the same as A and B respectively in the 2008 edition.

Question:

13-8.  Consider the plasmid described in problem 13-4. Assume it is 3150 BP long, and contains no repeated sequences. The order of restriction sites (going in the clockwise direction) is Eat1, EatII, EatIII, EatI, EatIV, EatIV. Assume the first 2 sites are 150 base pairs apart, and all other restriction sites are 600 BP apart.

A. Suppose you digest the plasmid with one or more of the restriction enzymes listed.  (See the table below for the enzyme(s) used in each case, A-1 to A-4.) After digestion, you separate the resulting pieces on a gel, and stain the gels for DNA. (No probe – just a stain containing a dye that binds to DNA.)

For each of the cases, fill in the following table. See notes below table for explanation of (i) & (iii).

 Case: A-1 A-2 A-3 A-4 Enzyme(s) Used: EatI EatII EatIII + EatIV All 4 enzymes (i) How many pieces?* (ii) How many bands will you see on the gel? (iii) How long is shortest**?

Notes:

*(i) = How many different pieces of DNA will you get? (Two pieces are different if they have different sequences.)

**(iii) = How long is the DNA in the band that migrates the fastest on the gel?

B. Suppose you have a labeled probe that hybridizes to the drug resistance gene. The probe is 100 bases long. You cut up your DNA and run gels as an A. Then you do a blot of your gel, hybridize to the labeled probe, and examine the labeled bands on the blots. When you cut up the DNA with all 4 restriction enzymes (case A-4), you get one labeled band corresponding to 150 BP.

B-1. If the probe is 100 bases long, why does the position of the labeled band correspond to 150 BP?

B-2. How many labeled bands will you see if you cut up the DNA with just EatI, just EatII, or EatII + Eat IV (cases A-1 to A-3)?  Draw the blots, indicating the approximate positions of the labeled bands.

C. You have a second plasmid that is almost the same as the one described above. The only difference is it has one RFLP – it lacks the EatI site in the drug resistance gene.

C-1.  Plasmid #2 probably has a (deletion) (insertion) (insertion or deletion) (substitution) (any of these).

C-2. You cut up plasmids #1 & #2 with all 4 enzymes and compare the gels. (You use staining, no probes.) Will you be able to tell the difference?  Explain why or why not. .

C-3. You cut up plasmids #1 & #2 with all 4 enzymes, and run gels, but this time you do blots and hybridize to the labeled probe. Will you be able to tell the difference?  Explain why or why not. .

D. Suppose you insert a fragment of human gene X into the plasmid in the EatII restriction site (whether this is the best way to do it or not). Assume the human DNA fragment has one EatI restriction sites (& no EatII sites). You then cut up the chimeric plasmid with EatII, and separate the resulting pieces on a gel.  You have a labeled probe for gene X. Suppose the probe is 1 KB long and the human DNA is 2 KB long. You make a blot of your gel and hybridize to the probe.

D-1. How many labeled bands will you find on the blot?

D-2. The fastest migrating labeled band should correspond to a DNA fragment that is (<1 KB) (2 KB) (>1 but <2KB) (2KB) (>2KB). Explain briefly.

D-3. Suppose you repeat your experiment. You make your chimeric plasmid the same way, but this time you cut it up with EatI & EatII. Then you run a gel, make a blot and hybridize to the same probe as before.

(i). Suppose you find one labeled band on your blot. This band will correspond to a DNA fragment that is (the same size as in B-2) (shorter) (longer) (can’t predict).

(ii). Suppose you find two labeled bands on your blot. How will the intensity of the two bands compare? You expect that (one band will be darker than the other) (both bands will have the same intensity of label) (you can’t predict).

For both (i) and (ii) explain how you could get the results given. What is different about the two cases?

(iii) Is it possible to get a blot with more than 2 bands? Explain why or why not.

13-8.    A.

 Case: A-1 A-2 A-3 A-4 Enzyme(s) Used: EatI EatII EatIII + EatIV All 4 enzymes (i) How many pieces? 2 1 3 6 (ii) How many bands? 2 1 3 2 (iii) How long is shortest? 1350 KB 3150 KB 600 KB 150 KB # times cut? 2 1 3 6 All the pieces: 1350 & 1800 KB Only 1; whole length 600, 1200, & 1350 KB 150 & 5 dif. Pieces all 600 KB long

B-1. The fragments are separated by size first, and then probe is added. It’s the size of the fragment that matters, not the size of the probe that hybridizes to it. In this case the fragment is 150 BP.

B-2.  You get one labeled band each time. You cut the plasmid into the same pieces as above, but only one fragment in each case hybridizes to the probe and is labeled.

 Case: A-1 A-2 A-3 A-4 Enzyme(s) Used: EatI EatII EatIII + EatIV All 4 enzymes Labeled band 1350 KB* 3150 KB 1350 KB* 150 KB

* These two DNA fragments are the same length, and both include the section that hybridizes to the probe, but the two 1350 KB fragments correspond to different sections of the plasmid.

C-1. Any of these. Any of these could have change the base sequence at the restriction site so it is no longer recognized by EatI.

C-2. Yes. The original plasmid gives the 6 pieces listed above (2 dif. Sizes). Plasmid #2 will give 5 pieces. Four will be 600 long as above, but the remaining one will be 750 KB (600 + 150) instead of 150 KB.

C-3. Yes. You will get one labeled band each time, but they will be different sizes. The fragment that hybridizes to the probe will be 750 KB (plasmid #2) instead of 150 (plasmid #1)

D.-1. One. You’ll cut out the inserted gene fragment, and it’s the only piece that hybridizes to the probe.

D-2. 2 KB. See answer to B-1.

D-3.     (i). Smaller.

(ii). Both bands will be equal in intensity.

(iii). No.

Explanations for (i) & (ii):
In both cases, the fragment of gene X was cut out of the plasmid by EatII and cut into two ‘halves’ by EatI, each ‘half’ smaller than the original. In case (ii), one ‘half’ must be smaller than the other, but in case (i) they could be the same size.

In case (i), one half was at least 1 KB long, and contained the entire sequence complementary to the probe. The molecules in the gel containing that sequence hybridized to the probe and formed a labeled band. The molecules containing the other ‘half’ wouldn’t hybridize. (It doesn’t matter if they were smaller, in a different band, or the same size, in the same band.) In this case, the sequence that hybridized to the probe was entirely on one side of the EatI site

In (ii), the original fragment was cut into two unequal sized ‘halves’, and each part contained some of the sequence complementary to the probe. In this case, the probe overlapped the spot where the EatI restriction site cut. Therefore the probe overlapped with sequences in each ‘half’ and could hybridize to either one. Some molecules of probe hybridized to one ‘half’ (forming one labeled band) and some to the other ‘half’ (forming a second labeled band). Each molecule in the gel could trap one molecule of probe, so the two bands should be equally labeled.

Explanation for (iii): There is only one site for EatI in the insert, so the maximum # of pieces that can be generated is 2. One or both of them may hybridize to the probe, as explained above.