The human genome is 3×10⁹ bp in length.

How many fragments would...

Question

The human genome is 3×10⁹ bp in length.

How many fragments would be predicted to result from the complete digestion of the human genome with the following enzymes: Sau3A (˘GATC), BamHI (G˘GATCC), EcoRI (G˘AATTC), and NotI (GC˘GGCCGC)?

Accepted Answer

Hi, everybody. Welcome back. Let's look at our next problem. I'd like to start with a quick note for those listening with audio description. Normally on audio description, uh the time sign or the multiplied sign is read aloud as multiplied by. In this particular problem, we have a lot of numbers written in scientific notation. So for the sake of avoiding a really lengthy uh problem description, I'm going to be reading those numbers out loud uh as just times rather than multiplied by. So let's move on to our question. How many fragments would be predicted to be generated by the complete digestion of the human genome with S OS A U three A bam H one eco R one and not one enzymes combined choice. A 1.75 times 10 to the seventh fragments. B 2.821 times 10 to the seventh fragments, C 2.75 times 10 to the eighth fragments or D 3.75 times 10 to the eighth fragments. So we're given the information that the piece of DNA we're cutting up is the entire human genome. So let's start by thinking about how long is that piece of DNA that's going to be important when calculating how many fragments. So just above my problem, I'll jot down the length of the entire human genome, which is roughly 6.4. So that doesn't fit above I right below 6.4 times 10 to the ninth base pairs. So that's the piece of DNA we'll be working with and we'll be cutting it up with a mixture of four different restriction enzymes. So we want to look at how often do each of those enzymes cut? On average, we do have a formula to calculate that. Once we know how often we can expect them to cut, we can look at how many fragments we'd expect each individual enzyme to generate from our segment of HU of DNA. So we want to start by calculating how often does each enzyme cut on average. And we have a formula for that. So how often does a restriction enzyme cut on average? And you would calculate that by taking and I have an open parentheses, the number of base pairs in its cleavage site or its recognition site closed parentheses. And that number raised to the fourth power. And again, that's going to tell us how many base pairs. So every however many base pairs, we can expect our restriction enzyme to cut a piece of DNA. Well, if we've got the length of our DNA segment and how often our enzyme cuts, we can then get to how many fragments will be generated when you digest a particular piece of DNA by a particular restriction enzyme. So we would get to that by dividing our length of DNA segment divided by how often the enzyme cuts, that division will give us the number of fragments that will be generated. So if we can calculate the number of fragments generated by any individual restriction enzyme, we calculate this for each of our four enzymes and then just add together all of those fragments to calculate the expected total number of fragments. So let's get started. Our first enzyme is A or S A U three A. So first, we need to know how many base pairs are there in the cleavage site. Well, S3 A has four base pairs in its cleavage site. So the calculating how often does it cut? We would say four to the fourth power which equals 256 base pairs. So every 256 base pairs, we would expect that to cut. So let's do our division. We know our DNA segment is the entire human genome, 6.4 times 10 to the ninth base pairs divided by 256 space pairs. And that will give us 25 times 10 to the sixth fragments. So fairly straightforward, that's our calculation. Now we get to bam H one. Well, Ban H one and ECO R one both have six base pairs. In their recognition sites. So we can just do one calculation for both of them. We just have to remember to add that number of fragments in twice since we have two enzymes cutting at that frequency. So for B mh one and ECO R one, they both have six space pairs in that recognition site. So we'll use sixth to the fourth, which equals 4096 base pairs. So you'd expect it to cut every 4096 base pairs. So let's take our length of our genome 6.4 times 10 to the ninth base pairs divided by 4096. And that will equal one 0.56 times 10 to the sixth fragments. Finally, we just have one enzyme left to calculate. Not one and not one has eight base pairs in its restriction site or its recognition site. Excuse me. So eight to the fourth power just going to give us a really big number 65,000 base pairs for the frequency of its cutting. It's a very rare cutter with an eight base pair cleavage site. So we do our last division problem 6.4 times 10 to the ninth base pairs divided by 65,536 space pairs. And that will give us two, excuse me, not two but 9.77 times 10 to the fourth fragments. So now I'm just going to highlight for the purpose of collecting all my numbers. Three A generated 25 times 10 to the six fragments. B mh one and ECO R one each generated 1.56 times 10 of the six fragments. And not one generated 9.77 times 10 of the four fragments. So we're almost there for each enzyme. We've calculated how many fragments we'd expect to generate. If we cut just with that enzyme, last step, just add all these numbers together to give us the total number of fragments if we combined all of these enzymes, so total fragments will equal 25 times 10 to the sixth. And I'll put that in parentheses plus sign. And now remember to add in twice our number from B mh one and equal R one since it's two enzymes. So in parentheses, 1.56 times 10 to the sixth plus sign. And then in parentheses, 1.56 times 10 to the six that sign. And in parentheses, 9.77 times 10 to the fourth, add all those together. And we get our final number of 2. times 10 to the seventh fragments. And that is going to be our final answer. Which if we look over at our answer, choices, we see that corresponds to choice B which is 2.821 times 10 to the seventh fragments. See you in the next video.

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The human genome is 3×10⁹ bp in length. How many fragments would be predicted to result from the complete digestion of the human genome with the following enzymes: Sau3A (˘GATC), BamHI (G˘GATCC), EcoRI (G˘AATTC), and NotI (GC˘GGCCGC)?