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Draw DNA molecule that has five randomly spaced restriction sites far a speclfic palindrome: How many fragments would be produced ifeach site were cut by that speci...

Question

Draw DNA molecule that has five randomly spaced restriction sites far a speclfic palindrome: How many fragments would be produced ifeach site were cut by that specific restriction enzyme?Label each fragment:Rank them in order of size from largest to smallestIn this diagram, A and B are different palindrome sequences on a DNA strand Only the restriction enzyme that recognizes site B is present:Explain why only two fragments would be produced.

Draw DNA molecule that has five randomly spaced restriction sites far a speclfic palindrome: How many fragments would be produced ifeach site were cut by that specific restriction enzyme? Label each fragment: Rank them in order of size from largest to smallest In this diagram, A and B are different palindrome sequences on a DNA strand Only the restriction enzyme that recognizes site B is present: Explain why only two fragments would be produced.



Answers

A molecule of double-stranded DNA was cleaved with restriction nucleases, and the resulting products were separated by gel electrophoresis (Figure $Q 10-11$ ). DNA fragments of known sizes were electrophoresed on the same gel for use as size markers (left lane). The size of the DNA markers is given in kilobase pairs (kb), where $1 \mathrm{kb}=$ 1000 nucleotide pairs. Using the size markers as a guide, estimate the length of each restriction fragment obtained. From this information, construct a map of the original DNA molecule indicating the relative positions of all the restriction enzyme cleavage sites.

Looking at reflects a restriction fragment length polymorphisms ARB a produced by reaction with restrictions. End of nucleus is and detected by seven blocks be of the same length for mutants and normal beta globe in values C determine the sequence of bases in the DNA fragments D have in their middle short fragments of DNA. Better palindrome IQ E are used as vectors. So what are these re flips? How as we get them? Well, as the name suggests, we get them by adding restriction enzymes to our DNA. So it's based on the restriction sites. And what is special about re flips is but it distinguishes between a normal and mutant alien because the normal and mutants have different restriction sites will be given restriction enzyme. That's why we choose that particular restriction enzyme. And after we've done this we can use a seven blocks on them to separate out the DNA fragments and identify a sequence of interest. So which of these is true. That's the real A. Is true. B. So B is, for some reason the answering a text book. But this is false. If the if they were the same length, they would not be length polymorphisms. And the example even given in the text book is beta Global. Because we're looking at sickle cells. They did they determine the sequence of basis? No. To be have panoramic sequences, not necessarily other users vectors. No. So we answer here is A.

So we're giving a segment of DNA. Um, and you want to know which is part, um are we want to know if the dean a plan race will start replicating the segment from the right. Um, so the question A which will be the template for the leading shrimp so of replication is proceeding such that the DNA on the right is replicated. First, the top strand is a tip of road leading strand. And for B, we're gonna draw them all killed when the dina plan raises halfway along the segment. So here we have five prime a T t See e g t uh, a c g a tcg. And then it breaks off a c j c g t a c t Hey, see, a g t c. This is three crime. And so this is the, um, leading strand. Well, then, on the bottom here we have, um, three prime. And this is the art. A primer for the Okazaki of every fragments. Now we're gonna draw the to complete daughter molecules. Um, so we'll have five prime. And then will this repeat the opposite with a eagling t and see equaling G still continue on like that. And, um, this will be the same as the other daughter cell. And then lastly, um, is the diagram a compatible with bi directional replication from a single origin, the usual mode of replication. And, yes, it simply represents replication at one of the forks.

So in this podcast, we're continuing to like a DNA and more specifically, restriction enzymes. We're gonna be taking a look at the function of restriction enzymes with a specific example. So we have a DNA sequence, and so we know that the restriction enzymes will only cleave DNA at specific sites and so we can draw out new fragments. So the sequence of DNA fragment resulting from the cleavage with both of our restriction enzymes are as follows. So we have fascinated five prime and we have G A a g TC c g c g t t A T a g a. It's g C A t G in and that's the three prime. And so now we have the next fragments, obviously, to start with the three primary because they're anti parallel. So we have a C g t C travel T see a double G C G C a a T A t double C g t a C T T a A. And then we've got the five prime. And so these are the fragments that would get after using our restriction enzymes

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.


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