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A gene that is normally expressed only in pancreatic cells is shown below_ The upstrcam region of this gene including the core promoter containing the TATA and CAAT...

Question

A gene that is normally expressed only in pancreatic cells is shown below_ The upstrcam region of this gene including the core promoter containing the TATA and CAAT box (core promoter designated as D) and upstream regulatory region (A, B and C) are shown You fused the entire region (A to D) or part of it where cach of these regions were individually deleted to B-Galactosidase (LacZ) gene and then the DNA was introduced into pancreatic cells or into kidney cells_IAI B | Upstream region Core pro

A gene that is normally expressed only in pancreatic cells is shown below_ The upstrcam region of this gene including the core promoter containing the TATA and CAAT box (core promoter designated as D) and upstream regulatory region (A, B and C) are shown You fused the entire region (A to D) or part of it where cach of these regions were individually deleted to B-Galactosidase (LacZ) gene and then the DNA was introduced into pancreatic cells or into kidney cells_ IAI B | Upstream region Core promoter IacZ gene coding sequence The data in the following table are the results of 8-Galactosidase activity from this experiment. Region Deleted Cell Type Transformed Percentage of 8-Galactosidase Activity None Pancreatic 100 Pancreatic Pancreatic 100 Pancreatic 100 Pancrcatic None Kidney <l Kidney Kidney 100 Kidney <l Kidney If we assume that the upstream region has one silencer and one enhancer, answer the following questions: i) Where are the silencer and enhancer located? ii) Why don we detect the presence of the silencer in the pancreatic cells? iii) Why isn't this gene normally expressed in kidney cells?



Answers

are about CRISPR/Cas9 and relate to the following diagram, which shows the complex between an sgRNA and the genomic site it targets so that Cas9 can cut the genomic DNA at the positions shown. You should note the so-called protospacer adjacent motif or $P A M$ site located adjacent to the target site. The PAM site $\left(5^{\prime}-\mathrm{NGG}-3^{\prime}\right.$ on one strand, $\mathrm{N} \text { is any base })$ must be present in the position indicated for cleavage to occur. One way to think about this situation is that the sgRNA brings the Cas9 enzyme to the adjacent PAM site to initiate cleavage. F. Port and S. Bullock at the University of Cambridge (UK) designed the elegant plasmid vector $p C F D 3$ for the expression of sgRNAs in Drosophila. The following figure shows a part of this vector. The orange sequences are part of a strong promoter (transcription from this promoter starts at the $\mathbf{G}$ in bold-which must be present- and goes from left to right). The purple sequences are a portion of the tracrRNA component of the sgRNA. After cutting the $p C F D 3$ plasmid with the restriction enzyme $B b s I$ (whose recognition site is also shown in the following figure), you will replace the blue sequences in the figure with sequences that will allow the expression of an sgRNA that targets a Drosophila gene called NiPp 1 The last part of the jigsaw puzzle you will need is the following sequence, which shows part of the $N i P p l$ gene including the triplet corresponding to the start codon. Capital letters are in the gene's first exon with the coding region in blue; lowercase letters are in the first intron. The NiPp1 protein is 383 amino acids long. Your assignment is to generate a knockout allele of this gene by inducing Cas9 to produce a doublestrand break into the gene that will be repaired imprecisely by nonhomologous end-joining (NHEJ). a. Identify the two PAM sites in this sequence. Which of these PAM sites would you want to use in order to produce a null allele of the NiPp1 gene? Why would you prefer this site? b. If you targeted Cas9 to the proper location in the NiPp1 gene, and the resultant double-strand break was repaired imprecisely by NHEJ (so that a few usually $\leq 6$ bp are deleted or added at that location), about what percentage of the imprecisely repaired genes could you say with confidence would be null alleles? Explain. c. Diagram the $p C F D 3$ vector after it has been cut with the $B b s$ I enzyme. Don't worry about the small blue fragment that will be removed; the emphasis here is to show the $5^{\prime}$ -overhangs that will be made. d. Design two 24 -nt DNA oligonucleotides that you could anneal together and clone into $B b s I$ -cut $p C F D 3$ vector so that the recombinant plasmid could express an sgRNA useful for making null mutations in the NiPp1 gene. e. Show exactly where Cas9 would cut the NiPp 1 gene. f. Briefly outline what you would do with your recombinant plasmid to make a null mutation in the fly NiPp 1 gene. g. Briefly outline how you would modify this technique to generate a knockin allele in which the first amino acid in the NiPp1 protein after the initiating Met (that is, Thr) would be changed to Ala.

So for this problem, we are to use the given pc FD three vector to express single guide RNA in Drosophila to create a knock out of this N I PP one gene given in the problem statement. So the first part of this problem part a asked us to find the two Pam sites within the sequence. We know from the blurb before this problem that the canonical Pam site is five prime and G g and being any nucleotide. So all we need to do is scan through the sequence to find any NGS. The two that are present in this sequence are near the end. So at the end of the Exxon, we have t oops t g and then in the first entry on, we have another t g. This the first part of the problem also asks us which site we would use to produce a null illegal and why we would prefer that site. So this first Pam site would be the ideal one for us to designer cas nine system around because it would cause a bubble within this Exxon for it to cut. And hopefully we would get a break, causing a frame shift somewhere within this coding sequence and leave us with Leo. The next part of this problem part be asked us to determine the percentage of imprecisely repaired jeans that we could say with confidence would be no Khalil's. So the first possibility is that it prepares exactly where it broke so right at zero and it repairs without any addition. However, this problem statement said that we are only to consider the percentage of the imprecisely repaired genes, so that's eliminated as a possibility. Um, the next option is that it can add up to six nucleotides, however, if it adds three or if it had six. And these could possibly still maintain function because this would cause a frame shift of a full three nucleotides, which would be a single amino acid. Now the same goes for the removal of three or the removal of six um, now that could still maintain some function. Um, it's unlikely, but it could be possible. So now we know that with the addition of 1 to 4 or five and the subtraction of 1 to 4 or five nucleotides, we would have a frame shift that would most certainly disrupt function. So that means we had 12 possibilities. Um, so the addition of six or the subtraction of six nucleotides and then eight, um, known Knowles. So now it is just simple math. 8/12 is equal to 66.6 percent of the imprecisely repair genes would be null alleles. Okay, Next, we are to diagram the cut pc FD three vector, um, and where to ignore the blue segment that would be removed. Um, Now, this would be cut using the BBS one recognition site. So we have our five prime end and three prime ends, so this would be the left side or the orange side. Um, so we've got t t uh, a C and then we've got our matches, and then we've got an overhang, So c a gc. And so that is where it would cut on the left side. On the right side, we would have overhang G two t t. And then we have our pairs, so t a g a key. And here is our overhang. And that would be the vector. So the next part of this problem part D access to design to 24 nuclear tired pieces of D N A that couldn't kneel together and fit inside the cut plasma had, um that would be useful for expressing single guide, aren't it? So, to design a single guide, RNA, um, we have to go back to the blurb before this problem and look at how cas nine bubbles, Um, the genomic target and how single guide RNA fits. So we've got our five prime and three prime here, and the figure looks something like this with our genomic target site, our Pam. And then we've got our complement. And then we've got our single guide RNA that fits here. And so this is our genomic target. So if our single guide RNA is a compliment to the complement of the genomic target, then our 24 um, blip piece of DNA essentially just needs to be the genomic target. Um, what? But we have to include four overhanging pieces. So if you look here, I have already typed it out. But these first and last for our compliments to the cut portion of the plasma, which is C A, g c and G T t t. So we know that that wouldn't heal and then this 24 20 base pair section of D N A. On the top side is simply the last, uh, 20 nucleotides of the coding Exon. And this is because we want the single guide RNA to bubble the genomic, um, genomic target just upstream of the Pam site. And so we have the single guide RNA paired to that region. So next part E asks us to show exactly where cas nine would cut in the n i pp one gene. Um, Now, if we've done our design correctly, then according to the figure before this problem, our CAS nine should cut three base pairs or nucleotides upstream of the Pam. And this is simply shown in the figure before the problem where we have our n g g. And then we've got our bubble and our genomic target site, and they show a base pair another base pair. Excuse me, nucleotide. And then we see after three cast nine cuts. So if we look back to our, uh, and a p p one Jean, um, and we find our and nucleotides towards the end of the Exxon that account for our Pam, we've got a t that you and then we've got the interim with another G. So if we go three nucleotides back from the T, we've got another T A, G and N A. And it would cut between the A and the sea before it, um, specifically between the history and and the Syrian code on So part f of this problem asked us to outline how we would go about making this nor mutation with a component plasma. One common way is to inject a newly fertilized egg. And, um, you would inject it with the plasma expressing the single guide RNA as well as another plasma expressing cas nine. And once they're injected, um, the plasmids are expressed, um, and the cas nine protein and the single guide RNA are expressed. And then if the design process is gone, um, as desired, the genomic target is altered into a no oops. Um, and then you have a fertilized egg with the no alil that will grow into a fully functioning organism Now apart G is very similar. So they asked how you would modify the technique to create a knockin, um, to change the, uh um, the 39 right after the initial initiating met to an Al Ani. Um, now, to do this, essentially, you would just have to create a complimentary piece of D N A. So, um, some sort of piece of DNA that could be expressed on a plasma that is complimentary to the gene. So the a P B one gene, except that in this piece of DNA, you have a few base pairs of overhang on each direction. So essentially you would have something like 80 g representing the Met. Um, and then the A in the 39 would be changed to a. G. Uh huh. And then you could have see t completing the sequence and a pair to make this a complimentary piece of DNA, and then you would need to get this expressed within the cell, like, leave you a plasma. But now, if you have this chunk of DNA expressed, hopefully you would get a, um um ology directed repair instead of No, no, I'm all yes and joining. And if you've done the design correctly, the HDR would lead to uh huh 80 g g c t. In the beginning of the Exxon, which would change the 39 two, uh, excuse me to an l A nine. And that is the end of the problem.

Question consists of three parts. Port A. On the basis of the results of these experiments is the C e g, an operator or a regulator G and explain your reason. So the CGE is going to be a regulator gene, and this is because the sea gene is transacting. It acts on not just the present molecule of DNA, but on different DNA molecules as well, specifically affecting the expression off the aging on a different DNA molecule. So that is how we know that it is transacting and therefore regulator dream regulator genes are transacting. So this is why we believe that CG is trans acting now in terms of part B Do these experiments suggest that a rabbit knows operandi is negatively or positively control? Well, from the experimental data, we find that the rabbit knows opera is going to be positively controlled. Why? Well, because if we look at the data, the CG appears to be the regulator gene that is necessary for the transcription of the aging. So for the aging to be expressed, we need to have a functional CGE present within this up when the sea gene is absent and a rabbit knows is also absent. The aging is not expressed, which means that the C G must encode when regulator protein that is required in order to allow for the transcription of the A G. So that means that we must have a specific chemical present for transcription toe occur. And in the context of this question, that chemical is going to be both the see Gene that is functional and a rabbit knows. So if both the rabbitohs and the siege in our present then transcription occurs, which is the characteristic of positively controlled gene expression. And finally, for part C, what type of mutation is C superscript? See? So the C superscript C is going to be a continuous activation of transcription off the A gene. So this means that this kind of mutation leads to constitutive expression of the aging so that aging is always going to be continuously transcribed or is always on. So that is the kind of mutation that we find with the sea superscript see

The question here asked us, um, to firstly summarize how missiles and and stalls experiment, um illustrates the manner in which DNA is copied for transmission between generations. So we know that missile, um, sin and stalls experiment revolved around two different isotopes where we have a light isotopes and a heavy isotope whereby they wanted to see how, um, the particular isotopes of DNA here is going to basically get transmitted. So there are three main theories that they have. They had to conservatives semi conservative and disperse it. Theories from what they saw was that the particular set of DNA here was going to essentially replicate in a set by can serve a tive manner indicating that essentially DNA here is going to be wrapped in a double um, Helix. So that is going to be part a here for part B. It states that when we have DNA polymerase, it can only add on to the three primes strand. So, essentially, if we have a double stranded sequence of DNA, this can be three times five prime five prime, three prime and essentially on. What is going to do is that we're gonna have replication on two different sides, so it's gonna go in opposite directions and it's gonna expand like such here. Um, and lastly, for part C here, it wants us to explain the function of Tele memories so tall Marie's here is essentially he enzyme that works to repair the telomeres off particular chromosomes here. So the telomeres here, if we recall here, is going to function in order to make sure that the cell is able to, um, stay alive. So it basically tells us whether or not the cell should undergo apoptosis or should it rather just, um, still be viable? So if telomerase does not occur, but we're going to get is a lot of, um, a popped ASUs or programs all death, and this could be particularly an official in some circumstances. However, in other circumstances it could make it such that you'll have premature death of cells and therefore it could lead thio development of various diseases. For example, if telomerase also miss functions on particular cells and it allows it to replicate fully without end, it's gonna be because of cancer. Here

Look, buddy, this is Ricky. And today we're gonna work. Walkthrough problem 31 from chapter 33. All right, so say we have a little guy home. Steve and Steve needs a new torch. The only problem is that cardiac cells on the new heart have thes, um, agency one complex. Are these m A C one proteins that Steve does not have? So why is this a problem? Well, this is a problem, because natural killer cells see foreign M A C one proteins, and I think that a virus is invaded. And so they worked on the tackle. So eight garnets. This video is helpful.


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