Question

You have recovered a cloned DNA segment from a vector and determine that the insert is 1300 bp in length. To characterize this cloned segment, you isolate the insert and decide to construct a restriction map. Using enzyme I and enzyme II, followed by gel electrophoresis, you determine the number and size of the fragments produced by enzymes I and II alone and in combination, as recorded in the following table. Construct a restriction map from these data, showing the positions of the restriction-enzyme cutting sites relative to one another and the distance between them in units of base pairs.

Enzyme Restriction Fragment Sizes (bp)
I 350, 950
II 200, 1100
I and II 150, 200, 950

Accepted Answer

Welcome back. Here's our next problem. It says a 1500 base base pair, clone DNA segment was recovered from a vector and its restriction map must be constructed in order to characterize this clone segment. The clone segment was digested with two restriction enzymes. And then electrophoresis on Aguero's gel. The band size, observed band sizes observed in the gel are as follows, enzyme A 450 base pairs and 10 50 base pairs, Enzyme B, 200 base pairs and 1300 base pairs, enzymes A. And B. 200 base pairs, 250 base pairs, 10 50 base pairs. Using this information, create a restriction map and then calculate the distance between the cutting sites of enzymes A and B. In base pairs on the cologne DNA segment. Well, I've put up the start of a little diagram to help us think through this and calculate these things. So we know that our entire segment is 1500 base pairs long. So I've drawn out that segment three different times. And we'll think about when we cut with enzyme A. Enzyme B. And then A. And B. Together, remembering that we want to look for the distance between the cutting sites and make a little map of where they are. So we know that enzyme a. We get uh two bands of 450 base pairs and 1050 base pairs. So we know that this um cutting site. If we start at zero will be 450 base pairs From the end, giving us 450 base pairs and 1050 base pairs. So let's go down to answer or I'm sorry, enzyme b. Enzyme B. We have are fragments of 200 base pairs and 1300 base pairs. So we know our cutting site here. It's going to be about 200 base pairs from the end. And that would give us those two fragment links. So we cut with enzymes A. And B. We know that we cut in this site here at 200 base pairs. Enzyme B. is going to cut there and then Enzyme A. is going to cut at its 450 base parasite. And let's make sure that gives us the fragments we observe. So we would get Our 200 base pair fragment here at the beginning. So that will give us our 200 base pair. And then this distance here between the cutting site at 200 from enzyme B. And the cutting site at 450 from enzyme A. Will be 250 base pairs apart. And then finally this last part here we know will be 1050 base pairs. And these are indeed the three band length that we expect when we cut. So here's our restriction map here with A. And B. Cutting and it says, calculate the distance between The cutting sites of enzymes a and b. In base pairs. So that would be this distance here between these two cutting sites, which is 250 base pairs. So come over here and we see the choice B is 250 base pairs. And again, that is the distance between the cutting sites of enzymes a and b. Given the different lengths were given. When the two enzymes cut this 1500 base pair fragment. See you in the next video.

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