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Hallowing saemcnt pupponad gene scquences? Which oithe nasimany Mote motits culimonthal dols Ncanderthal and Denlsovan Neancerthal and humans TNn all motikincommor ...

Question

Hallowing saemcnt pupponad gene scquences? Which oithe nasimany Mote motits culimonthal dols Ncanderthal and Denlsovan Neancerthal and humans TNn all motikincommor dces Krindertha moneril Neinderthal Denisnvanss navt nurnt peclctall had around the sanc nurber ofmonfs ICAOn Thethrec archaic and Icdcrn humlan renes for neurologicl developmentthan Ior thrce archaic and modemn human spcces shared odontology norz of the above What is cxasperating about Lh CCRS deta 32 allele? Thcre has been cure foun

Hallowing saemcnt pupponad gene scquences? Which oithe nasimany Mote motits culimonthal dols Ncanderthal and Denlsovan Neancerthal and humans TNn all motikincommor dces Krindertha moneril Neinderthal Denisnvanss navt nurnt peclctall had around the sanc nurber ofmonfs ICAOn Thethrec archaic and Icdcrn humlan renes for neurologicl developmentthan Ior thrce archaic and modemn human spcces shared odontology norz of the above What is cxasperating about Lh CCRS deta 32 allele? Thcre has been cure found for this detective allele Jule to tino consistent drug cure for It Thls alic @ Nnluly children-~adults are not protected Thls allele only confers imtunity pujulnon: { that have the highest amount of HIV: Itis (oundat the highest frequency amcineofhv it [s found at thehighest fre queno populations that have the lowst effective at treating discases than sene therapy berause: Today, enzyac therapy has becn more tkebedyoiten relects both therapies therapyactually does not work 1; well Fencinetapy EIme the body rejerts gene therapy more olten than enzyme therJpy pricucal for eflectivl MJyCke onehas to Ininoducelle ene Ia tno Inanyceusmnin Q Eta bceneliectite multedceeellfon Cene ttnpy hemowhill the curinion cold mlecal chucken PJx Alonthranu Which ufthe ollowIngktNue? Aelcn Jtnstrn Meaelrituur inscrt needed Fene: Jn-Firo therupy USES Insett necdedEcne: -lla uerpy Tuns Fk [0 [Lene relecuon cr-Mivo uenpynunS -neertul supermutated gencs small sample of Ilvcr cells Inlroducing Bood Fene Into In-Vna Lefupy conducted Feny thase; Tcalnzertn Wealeeredcclks hackinto Lhe patcnt Which afthcfollowingite? cingcxnerimenea wncn gene therapy vector oncolyc unustsAn" being used jor ecnctnerni Adeno-ueociatcd viruscsaneteingused Ecne therapy and C none ofthe above



Answers

Gamow (1954) proposed that the structure of DNA deduced by Watson and Crick (1953) could be interpreted as a way of forming roughly 20 "words" of the common amino acids from the four "letters" A, T, C, and G that represent DNA nucleotides.
Crick and coworkers (1961) used a method developed by Benzer to induce mutations in the DNA of a virus by the insertion of a single nucleotide. The mutant could not infect the bacterium Escherichia coli and neither could viruses with a second insertion of a second DNA nucleotide. However, a third nucleotide insertion restored the ability of the virus to infect the bacterium.
In 1961, Nirenberg and Matthaei conducted a series of experiments to better understand the flow of genetic information from gene to protein. They discovered that in solutions containing the contents of ruptured E. coli bacterial cells from which DNA had been removed, polymers containing only one repeating amino acid, phenylalanine, would be synthesized if synthetic mRNA composed of only the single nucleotide, uracil (U), was added to the solution in which phenylalanine was also present. In solutions containing mRNA with only adenine (A) or cytosine (C) and the amino acids lysine or proline, polymers containing only these amino acids would be synthesized. The researchers found that when ribosomes were removed by filtration, these polymers did not form. Nirenberg and Leder (1964) extended this work to include other nucleotides.
A. Summarize the conclusions regarding the encoding and decoding of heritable information supported by these studies. Explain how these studies provided evidence to support the Triplet Code. Khorana (1960) developed a technique for synthesizing RNA composed of predictable distributions of repeated pairs or triplets of nucleotides. He found, for example, that RNA synthesized when A and U were present in relative concentrations of 4:1, respectively, will produce RNA sequences with these distributions determined by their relative probabilities: AAU:AAA, AUA:AAA, and UAA:AAA; $0.8^{2} \times 0.2 / 0.8^{3}=1 / 4$ [calculated as follows: i) 4/5 of the bases are A, so the likelihood of selecting A is 0.8; ii) the selection is repeated to determine the second letter of the three-letter codon; iii) the likelihood of selecting a U is 1 in 5; iv) the probability of selecting the set AUU is the product; v) similarly, the probability of AAA is $(4 / 5)^{3}$ ;
and vi) the ratio of these probabilities is their relative likelihood]: AUU:AAA, UUA:AAA, and UAU:AAA; $0.8 \times 0.2^{2} / 0.8^{3}=1 / 16$ and UUU:AAA; $0.2^{3} / 0.8^{3}=1 / 64$
B. Based on Khorana’s findings, calculate the relative distributions of the following ratios of concentrations of RNA triplet sequences from mixtures in which the relative concentrations of guanine and cytosine, G:C, are 5:1.
C. Based on the work of Nirenberg, Matthaei, Leder, and Khorana, the following table was constructed (taken from Khorana's Nobel Prize address):
D. Describe the effects of the codons UAA, UAG, and UGA on protein synthesis.

So for this problem, we are to use Thebe 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 secrets we know from the blurb before this problem that the canonical Pam site is five crime and G g and being any nuclear tied. 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 eso at the end of the Exxon, we have tea. Oops, T g and then in the first intron, we have another t g. This first part of the problem also asks us which site we would use to produce, Annul, 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 framed shift somewhere within this coding sequence and leave us with a no Leo. The next part of this problem part be asked us to determine the percentage of imprecisely repaired genes that we could say with confidence would be no Leal'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're 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. Um, however, if it adds three or if it had six thes 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 thesis obstruction 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. Eso the addition of six or the subtraction of six nucleotides and then eight, uh, 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 seventies. Next, we are to diagram the cut pc FD three vector, um, and where to ignore the blue segment that would be removed. Now this would be cut using the BBS won recognition site. So we have are five prime and and three prime ends. Eso this would be the left site or the orange side eso we've got t t Uh huh, A c and then we've got our matches and then we've got an overhang. So see a g c. And so that is where it would cut on the left side. On the right side, we would have our overhang G t t t. And then we have our pairs. So t a g a chief. And here is our overhang and that would be the factor. So the next part of this problem Part D assess designed to 24 nucleotide pieces of DNA that could a Neil together and fit inside the cut plasma. Um, that would be useful for expressing single guide Arne. So to design a single guide RNA, we have to go back to Thebes, blurb before this problem and look at how cast nine bubbles. Um, the genomic target and now 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 compliment. 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 than our 24 um, blip piece of DNA essentially just needs to be the genomic target. Um, what? We have to include four overhanging pieces. So if you look here, I have already typed it out. But these first and last four are compliments to the cut portion of the plasma, which is C A, g C and G T t t eso. We know that that would Aneel and then this 24 20 base pair section of DNA. On the top side is simply the last, uh, 20 nuclear tides of the coding Exon. And this is because we want the single guide RNA to bubble the genomic tart, 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 cast nine would cut in the N I PP one gene. Now, if we've done our design correctly, then according to the figure before this problem, our cast 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? Nuclear tied and then we see after three cast nine cuts. So if we look back to our ah 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 a G. And then we've got the intro on with another G eso if we go three nucleotides back from the T, we've got another T A g and and A and it would cut between the A and the C before it, um, specifically between the history and and the Syrian coat on. So part f of this problem asked us to outline how we would go about making this nor mutation with a competent plasma. Um, one common way is to inject a newly fertilized EG, and you would inject it with the plasma expressing the single guide RNA as well as another plasma expressing cast nine. And once they're injected, um, the placements are expressed. Andi, the cast 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. 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. Eso They ask how you would modify the technique to create a knockin, um, to change the, uh um, the three men right after the initiative, initiating met to an al Ani um now to do this essentially, you would just have to create a complementary piece of DNA. So, um, some sort of piece of DNA that could be expressed on a plasma that iss complimentary to the gene eso the NYPD 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 a t g representing the Met. Um, and then the A in the three ning would be changed to a G Wow. And then you could have c t completing the sequence and a pair to make this a complementary piece of DNA, and then you would need to get this expressed within the cell, likely via plasma. But now, if you have this chunk of DNA expressed, hopefully you would get a, um hm ology directed repair instead of no, no homologous 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 3. 19 to ah, excuse me to Anel Anin, and that is the end of the problem.

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.

How old their students today for this question, I understand. It's very long and extensive talking about um the uh the uh 10 dr nucleus, I mean nucleotide poly a tail and the terminal trans transfer is um the ATP and hello with the strands of it because they were bringing up, they want us to draw the project chin to predict the computer print out for reaction to which contains the creek strand as a template in the following box. And they also want, just wants to remember that all DNA sequences occur in the 5-3 direction. And that the sequence of the innocent strengths read from from the five to the three and so from left to right in the print out, so ringing it. So basically that's like the whole synopsis. So based off a ring that sorry, if I hit the mic based off a rain dad, what would have set up be So based off if this set up, we see that it would be a two little D's and a big G. Which stands for blue and black. Thank God T. T. C. Which is like who? And finally you have D. D. A. In D. D. A. Basically is green and DDT, his stands for red. And yep, that is it. Hope your father is very helpful and please take care and stay safe. And they and if you didn't get the answer, we'll try to just practice and work on green strands from right to left to Mhm. I mean from left to right. Excuse me. From left to right to fully grasp on what you read in, yep, that is it. I hope if I was very helpful and please take care. Okay, bye.


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