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Lab Restriction Analysis of DNA 7.0 Using the smaller fragments that resulted from the double-digestion (R.E. and R.E. combined), you will determine which of the sm...

Question

Lab Restriction Analysis of DNA 7.0 Using the smaller fragments that resulted from the double-digestion (R.E. and R.E. combined), you will determine which of the smaller fragments make up the larger fragments oblained each single-digestion (R,E. alone, R,E alone);This will likely take more than one try to figure out: Use scrap paperto help determine the small fragment combinations that equate the larger fragments.Use Tables anoorganize your data.TableFragments from RE: 1 (bp)Fragments from RE: L

Lab Restriction Analysis of DNA 7.0 Using the smaller fragments that resulted from the double-digestion (R.E. and R.E. combined), you will determine which of the smaller fragments make up the larger fragments oblained each single-digestion (R,E. alone, R,E alone); This will likely take more than one try to figure out: Use scrap paperto help determine the small fragment combinations that equate the larger fragments. Use Tables ano organize your data. Table Fragments from RE: 1 (bp) Fragments from RE: L and RE 2 Table Fragments from RE: 2 Fragments from RE: LandRE 2 Using the data in Tables and create series of DNA restriction maps (on the next page) like the ones in Figure 54 Hint: look for overlaps in the smal fragment numbers: Use scrap paper tor rough work Draw your final maps on the next page (using drawing tools on-screen, print and draw by hand) and on the assignment worksheet:



Answers

Mapping a Chromosome Segment A group of overlapping clones, designated A through $\mathrm{F}$, is isolated from one region of a chromosome. Each of the clones is separately cleaved by a restriction enzyme, and the pieces are resolved by agarose gel electrophoresis, with the results shown below. There are nine different restriction fragments in this chromosomal region, with a subset appearing in each clone. Using this information, deduce the order of the restriction fragments in the chromosome. FIGURE CANT COPY

For this question. What we have is we've taken a single DNA strand and of course, we have multiple copies of it, and we've broken it down into sort of randomized fragments between each of our different strands. So here we have six different strands each which could make up a singular section of it original DNA strand. So where I like to start from this is to start from a small strand where we can easily relate it to others. So if we look for very significant or unique feature in this first strand, we can see that there are three A's in a row. So if we find three days in a row in one of our other sequences, it's likely that would be a similar or contain this whole sequence as part of that strength. So at number two, you can see we have these three A's, and if you travel your way back, you can see that this fragment portion is similar to this strand. So here we just have this strand located right here in fragment number two. So from there we have the rest of the original DNA sequence, so we just need to look to make sure we don't have any additional DNA nucleotides on either the five prime or three prime ends of the strands. So again, if we look for the Triple A, we can see that we still have this see from the original first Strand, and we're going to have the rest of Fragment three able to be found in fragment, too. That fragment. For if we look through here, we can see that there are no Triple A's. However, you can see that there's a triple C here, which is pretty unique, which relates to the triple C found in Strand two. So here, fragment for is just this beginning portion of fragment and Strand two. So we can't grab any nucleotides from either three or four as long as we make sure to include that C from Fragment one. If we move on to five, we can see the Triple A at the very end, and it's going to continue on the five prime end until it reaches this similar strand at the beginning of Strand tube. So we don't have any new nucleotides from five either. However, if we look at six, we can see there are very different unique features here. There's a triple T, which is not found in any of the other strands. So we need to find where this strand is located in Strand number two, which contains the majority of our d N A sequence. So if we look at the very end of Strand six, we see this G c a a t. But you'll also find at the very beginning of Strand number two so you'll see this G c a a t. So here's Strand six is the nucleotides that are going to be found before strand number two. So here we just need to include this new portion of nucleotides two Strand number two, and we have our whole DNA sequence. So if we combine all those strands together and their correct sequencing of one another, we're going to get our final strand or a continuous strand of a T T. T, which you'll find in Strand number six and then you can write in strain number two a. C C T C A T a C C C T A g to t A. And then you just have to include the final see nucleotide that you can find in strand number one or some of the other strands. So here this would be the nucleotide sequence for the full strand that each of these fragments originated from, and that should complete the question.

Question here. Associate. Construct a map of the original DNA molecule indicating the relative all researching enzyme cleaner sites. So we know that you go are two here gives two fragments here which are going to be four Tila bases been six killer bases here Way. Know that hind three so high in three years. And Nico, are I together? Is three fragments which are gonna be one killer base three Killah based six killer base. This is so essentially, this would give us a total of 10 killer bases total. So there is also going to be cutting site for hind three because our to essentially cuts to there is another cut site. Actually, with this, this means that there is a single find three cleaver site and this particular DNA molecule itself is going to be a circular because of this fact

This'd is a problem. Looking at a restriction map of the second of DNA and ah, drawn it out here and it shows you a segment of DNA A. The numbers underneath represent base pairs. Eso there from this mark from position one to position to. It's 5000 base pairs from position two to Position three is 3000 from three before is 4000 from 4 to 5 is 2000 and 5 to 6 is another 5000 base pants at positions 13 and five. We see the word echo E CEO, and that is representative of the ECO R one restriction enzyme recognition site. So this means that position one Position three and in position. Five. The enzyme would be able to cut at those places because they're the regular recognition sites or presence. PST represents the recognition site for a restriction Enzyme PST. One. Those recognition sites are at position to position for position six. Finally, the other piece of information we need to know is that are pro, which is represented here in blue would overlap with this piece of DNA, so it overlaps starting here and ending over here. So now we're going to look at different patients, D N A. And we're going to cut them with certain enzymes and then try to find the pieces that will Aneel or hybridize to our pro. So in the first scenario, we have a paint patient. One patient one has all six restriction enzyme sites. And they're saying that you are digesting this DNA with the inside PST. So we're going to digest its site, too. Site for insights. Six. Now we add or pro. And the question is, what size pieces will Aneel toe are pro. So since we're going to have a DNA cut at Position six and position for this will be this problem, it will be 7000 base pairs. Since it overlaps with the probe, 7000 base pairs are going to be one of the, um, pieces of DNA we find, and we're going to cut from position for to position two, and we're going to create another 7000 base pair piece, and it too, will a meal to the pro because there is overlaps. So for this patient, we will get to pieces that are approximately 7000 base pairs that will a meal to the quote. The next question asked about a nen vivid jewel patient to impatient to has a mutation that eliminates position for so they do not have the recognition sequence for this enzyme PST. All right, what happened to me to adjust with PST now? Well, we don't have this site ending over, so we will cut at position two and we will cut a position. Six. So that means we will create one piece of DNA A this length, um, which will be 3000 plus 4000 plus 2000 plus 5000 which is 14,000 base pairs. And it will bind to this pro. And so for this patient without position of without the recognition site at position number four, we will create one piece of DNA that is 14,000 base pairs long that will bind to the pro next individual. Individual three has a mutation that eliminates positions to spy. So they do not have this, uh, echo or one recognition site. What happens if you digested with PST again? So it really doesn't matter that they're missing the site, the PST sites will be recognized. And will the DNA will be cut so we will get a piece again from this set from position sixth position for which will leave us with a 7000 based therapies. And we will have another piece from position for to position two, and that's 7000 base pairs. So we will have two pieces of DNA at 7000 base pairs that will a meal to the pro. Next question is back to, um, patient one. So they have all of the all of the enzyme sites. But in this case, now we're digesting them their DNA with both PST and eco or one what will the pieces look like? So once we digest this DNA with all with both of these enzymes, we're going to create a piece. Uh, we're going to get a cut. This piece will be 5000. These fears the next piece will be 3000 base pairs. The next piece will be 4000 base pairs and next piece will be 2000 days. Pierce and the next piece will be 5000 base pairs. Those will be produced from the inside digestion. But the next part of the question is which pieces will actually a meal to our probe and the Onley pieces that will a meal with to the probe here will be those that overlap sequence. So in this case, it we digest with both enzymes and then a meal with the probe. Who will get the 4000 base pair piece. We will get a 2000 base pair piece and we will get a 5000 base pair, please. So, finally, the last part of this question asks us about another patient. Who, um so patient five. Excuse me. Patient three. That has. Sorry. Patient three, but he has no, um site at position five. So position five site is missing. Get back to you. So, positions, position five site is missing. So the question is it we digest now with all enzymes. What will we end up with? So we will not. We will not have position five cutting, so this will end up to be one large piece. So this will be a 7000 base pair piece and we will have a 4000 base pair piece that will also a meal. So in the last scenario they give you, you will end up with a 4000 base pair peas as well as a 7000 base pair, please. I think what's really important in this problem is to realize that, um, just get out of this is to realize that the Onley pieces we care about are those that will kneel to or pro. So Onley pieces that overlap with our pro will show up on our job because the probe is going to a Neil toe on Lee, the pieces that have common on sequences.

So n is the number of past players, which means two n e will be the charge so and I equals Q. I. It was two times 1.6 times 10 to the negative 19 times, and I couldn't Now from an I find Q I and then you already have the force values You already have the force values. Fuck it. If I so we will be focusing on this part. Remember that Fourth force equals charge Times Electric failed. So find the force, plot the force in the y axis and plot que y in the y axis. And it would be something like this and find the best fit line. Yeah, on the electric field will be the slope. And from this data, the electric field will come out to be zero point triple zero to a three noted parkas. Um


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