23) To determine the restriction map of a circular plasmid,...

23) To determine the restriction map of a circular plasmid, you perform the follow complete digests with Asc1, BamH1, and a third enzyme, EcoRI. You run the digested DNA on agarose gel, and obtain the following fragments (all sizes are in kb): EcoRI 10 90 Asc1 100 BamH1 100 EcoRI + BamH1 3 7 90 EcoRI + Asc1 45 55 10 BamH1+ Asc1 52 48 A) Draw a restriction map of the this plasmid that is consistent with all the data provided. Be sure your map includes: 1) the location of all restriction sites; 2) the distances between restriction sites; and 3) the total size of the plasmid written in the center. (10 points) B) If you run a triple digest, using EcoRI, Asc1, and BamH1, how many fragments do you expect to get? What size will they be? (5 points)

Transcript

00:01 So to determine the restriction map of a circular plasmid, you perform the following digest with acsc1, femh1, and third enzyme, ecor1.

00:12 Now you run the digest dna on agarose gel and obtain the following fragment.

00:17 Now there is one mistake, and i want to point out later.

00:23 So when you digest with ecor1 only, you're going to have two fragment 10 plus 90.

00:30 And they are kb, obviously.

00:34 And if you digest with asc only, then you produce a linearized 100 kb band.

00:44 And if you digest with femh1, then you end up having a 100 kb band as well.

00:51 So from there, you can see that for a single digestion, ecor1 cut the plasmid twice, produced two fragments.

01:01 And both asc1 and femh1 cut the plasmid only once, linearized the plasmid.

01:09 Now then you have double digestion.

01:11 The first double digestion is e plus b.

01:16 And you have 3, 7, and 90 kb.

01:22 So you can see that 3 plus 7 plus 90 equals 100.

01:26 And if you look at e and b, you can see that most likely that the 10 kb band is being broken into 7 and 3 by femh1.

01:40 So that you can see that it split into two smaller fragments.

01:45 So this is what our expectation is.

01:48 The next one is the e plus a double digestion.

01:53 Now in this case, you can see there is a calculation mistake.

01:57 45 plus 55 already equals 100.

02:02 And then you have a 10.

02:03 So this is practically impossible because the total size of the plasmid is 100.

02:09 So you cannot have 45 plus 55 plus 10, which is 110 kb.

02:17 So the actual size from what i see from the femh1 and asc1 double digestion, most likely you will have 245 base pair band, which add up to 90.

02:34 And you have additional 10 kb.

02:36 So you can see 45 plus 45 plus 10 equals 10.

02:40 So this is the mistake that i was just talking about.

02:42 That it's impossible to produce a 45, 55, 5 plus 10 kb, which is going to be longer than 100 in total.

02:51 Now, so from there, we know that we actually produced 245 kb band with a 110 kb band by equar1 and asc1 digestion.

03:00 Lastly, when you have b plus a, femh1 plus asc1, you have 52 kb and 48 kb.

03:15 So from there, we can actually already somehow draw a map and then try to put them all together.

03:23 Let's say this is first plasmid that we digested with e.

03:30 Equar1 produced 10 and 90.

03:33 So you can tell that most likely you have an e here and e here.

03:39 So the shorter fragment is 10.

03:41 The longer fragment is 90...

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