laczlacylaci

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The answer is C.

The key says that they knew it was BamHI because lane 2 is absent in the blot, so we know it cut gene X.

From my understanding, after the blot, probes are introduced to the solution.
Anything that the probe does not bind does not show up on the blot. Anything that the probe does bind to, does show up on the blot. (ie. lane 1 and 3)

I think I am just not understanding how they came to the conclusion that since lane 2 did not show up on the blot, we know that they cut through gene x.

Is it because the probe needs to complementarily bind to the whole gene x and not a chopped up one, and that's why we can deduce the gene was cut in some way?

Please correct me if I am wrong, as I am pretty weak in microbio techniques.
 

lpp06

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You hit the head on the nail!

The Blot represents the movement of a gene with a bound probe through a medium. Since we are concerned with Gene X - the probe introduced will have a complimentary sequence to Gene X. If gene X has been partially digested by one of the endonucleases - in this case, BamHI - the probe will not bind and the gene will not appear on the blot. Lane 2 lacks Gene X, meaning BamHI was used to fragment the plasmid - this leaves us with options A and C.

Now you must use the gel diagram to determine the length of fragments produced by the combination of endonucleases used. Lane 2 shows 2 fragments - one greater than 0.5kb (but less than 0.8kb) and one slightly larger than 1.0kb. Since we know from the blot that BamHI was used, we now must select another endonuclease that produces fragments of the above size. This leaves C as the only correct answer since HindIII will cut the plasmid into fragments of these sizes.
 
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theonlytycrane

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Solid answer ^ :)

For any technique pros: shouldn't a control lane be shown under "S" on the blot which is ran for gene X? This would make it more clear that gene X is not shown up in lanes 2 / 4 or possibly showing up at different spots indicating that something went wrong (it got cut).
 
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lpp06

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Although no pro here, I'd assume the control lane in the digestion, showing the fragments produced by the actions of all enzymes, would have cut Gene X. When you transfer the gel to a blot and wash that lane with the probe, it won't find its complement and won't produce a positive result.

The blot is used to determine the unknown location or presence of a gene on the map of the plasmid, so it'd be tough to set up a control against the blot itself.

The wiki on Southern Blotting is pretty informative and simple - my browser won't let me post the link :/
 
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Rozayy

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You hit the head on the nail!

The Blot represents the movement of a gene with a bound probe through a medium. Since we are concerned with Gene X - the probe introduced will have a complimentary sequence to Gene X. If gene X has been partially digested by one of the endonucleases - in this case, BamHI - the probe will not bind and the gene will not appear on the blot. Lane 2 lacks Gene X, meaning BamHI was used to fragment the plasmid - this leaves us with options A and C.

Now you must use the gel diagram to determine the length of fragments produced by the combination of endonucleases used. Lane 2 shows 2 fragments - one greater than 0.5kb (but less than 0.8kb) and one slightly larger than 1.0kb. Since we know from the blot that BamHI was used, we now must select another endonuclease that produces fragments of the above size. This leaves C as the only correct answer since HindIII will cut the plasmid into fragments of these sizes.
Hey I'm sill confused on the second part where we decide between A and C. Lane 2 shows two fragments of sizes around-- 1kb and 0.6kb. are we suppose to add these up and infer that it is closest to 1.7kb that HindIII would cut? or would the 1kb be the size of BamHI and the 0.6 is the hindIII but how do we know the sizes of each? Lane 4 is also missing a band so that would also be an indication of BamHI used?
 

lpp06

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Hey I'm sill confused on the second part where we decide between A and C. Lane 2 shows two fragments of sizes around-- 1kb and 0.6kb. are we suppose to add these up and infer that it is closest to 1.7kb that HindIII would cut? or would the 1kb be the size of BamHI and the 0.6 is the hindIII but how do we know the sizes of each? Lane 4 is also missing a band so that would also be an indication of BamHI used?
To reiterate how we came to the conclusion that BamHI was used:

Because the Blot for Gene X returned a negative result (no band show; i.e. Gene X did not bind to the compliment) - BamHI must have been used to cut GeneX

This leaves us with options A or C

An important note here is that the diagram does not show the length between the BamHI and HindIII site - but we can assume it is 0.65kb since we know the total length of the plasmid and the length of all the other fragments.

To respond to your first point: the entire plasmid is 1.7kb long, no matter which endonuclease is used. Look at each lane - they all add up to 1.7kb. Even Lane 4 (one band at .8kb and the other at .9kb).

Let's focus on lane 2: You are correct with judging the lengths of the bands (one is more than 0.5kb and less than 0.8kb, and the other is slightly more that 1kb). Now visualize the action as each endonuclease cuts the plasmid. First, we know from above that BamHI must cut it, so now visualize that the plasmid is cut and becomes a straight line (I know this is biologically inaccurate but it helps to explain). Next, we must determine the next endonuclease that cuts the plasmid. We know the resultant pieces must be one of 0.5 - 0.8kb and the other must be slightly more than 1kb. Pretend that A occurs - BamHI cuts the plasmid followed by EcoRI. The result would be a fragment with a length of 0.8kb and 0.9kb - this does not match lane 2 and leaves us with C as are only remaining, and correct answer. For the sake of exercise - simulate the action of BamHI followed by HindIII: BamHI cuts the plasmid and creates one long fragment, then HindIII cuts and produces 1 fragment of 0.65 kb and another of 1.05 kb.
 

Rozayy

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thanks, excellent explanation.
To reiterate how we came to the conclusion that BamHI was used:

Because the Blot for Gene X returned a negative result (no band show; i.e. Gene X did not bind to the compliment) - BamHI must have been used to cut GeneX

This leaves us with options A or C

An important note here is that the diagram does not show the length between the BamHI and HindIII site - but we can assume it is 0.65kb since we know the total length of the plasmid and the length of all the other fragments.

To respond to your first point: the entire plasmid is 1.7kb long, no matter which endonuclease is used. Look at each lane - they all add up to 1.7kb. Even Lane 4 (one band at .8kb and the other at .9kb).

Let's focus on lane 2: You are correct with judging the lengths of the bands (one is more than 0.5kb and less than 0.8kb, and the other is slightly more that 1kb). Now visualize the action as each endonuclease cuts the plasmid. First, we know from above that BamHI must cut it, so now visualize that the plasmid is cut and becomes a straight line (I know this is biologically inaccurate but it helps to explain). Next, we must determine the next endonuclease that cuts the plasmid. We know the resultant pieces must be one of 0.5 - 0.8kb and the other must be slightly more than 1kb. Pretend that A occurs - BamHI cuts the plasmid followed by EcoRI. The result would be a fragment with a length of 0.8kb and 0.9kb - this does not match lane 2 and leaves us with C as are only remaining, and correct answer. For the sake of exercise - simulate the action of BamHI followed by HindIII: BamHI cuts the plasmid and creates one long fragment, then HindIII cuts and produces 1 fragment of 0.65 kb and another of 1.05 kb.