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2) A. You will need copy of the Genetic Code in order to solve this problem: Four brand new mutant strains of E: coli have been found that each have mutation in the...

Question

2) A. You will need copy of the Genetic Code in order to solve this problem: Four brand new mutant strains of E: coli have been found that each have mutation in the lac I gene Partia mRNA sequences of the mutants are shown below and the wild type mRNA sequence (WT) is shown as reference The reading frame is not indicated and must be determined by first finding the start codon Classify the type of each of the mutations below (e point; transition; transversion_ missense nonsense silent; insertion;

2) A. You will need copy of the Genetic Code in order to solve this problem: Four brand new mutant strains of E: coli have been found that each have mutation in the lac I gene Partia mRNA sequences of the mutants are shown below and the wild type mRNA sequence (WT) is shown as reference The reading frame is not indicated and must be determined by first finding the start codon Classify the type of each of the mutations below (e point; transition; transversion_ missense nonsense silent; insertion; deletion, frameshift): Each sequence can have one more than one or no mutation type Write your answer in the space to the left of each sequence. Note: indicates that the sequence continues; only the relevant portion of each sequence is shown: (4 pts] (Reference) Wild type 0 @ U A A U G C C U U @G @ @ C U U 4 C C LacI-a C G G U A 4 U G C C G A C G U U G G G C U U 4 C C_ LacI-b C G G U A 4 U G C C U U G G G 4 U U 4 C C _3' LacI-C C A U G C C U U G G G C U U A C C_ LacI-d C G G U A 4 U G C C U U G A G C U U 4 C C_ Two of the mutations above have the same phenotype as the classic lac I- mutation. One of the above mutations has the same phenotype as the classic lac I' mutation and one of the mutations has the same phenotype as the wild type allele_ lac I-. Determine which mutant strains (LacI-a_ LacI-b, Lacl or LacI-d) belong in each class. Briefly justify your answer using complete sentences pts) Lac I-phenotype lac IS phenotype lac Itphenotype



Answers

Below are several DNA sequences that are mutated compared with the wild-type sequence: $3^{\prime}-\mathrm{T} \mathrm{A} \mathrm{C} \mathrm{T} \mathrm{G} \mathrm{A}$ C T G A C G A T C-5'. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5' and 3' ends) for each mutated DNA sequence, then transcribe (indicating 5' and 3' ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the $\mathrm{N}$ and C termini). What type of mutation is each? Mutated DNA Template Strand #1: 3'-T A C T G T C T G A C G A T C-5' Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #2: 3'-T A C G G A C T G A C G A T C-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation: Mutated DNA Template Strand #3: 3'-T A C T G A C T G A C T A T C-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation: Mutated DNA Template Strand #4: 3'-T A C G A C T G A C T A T C-5' Complementary DNA sequence: mRNA sequence transcribed from template: Amino acid sequence of peptide: Type of mutation:

The strain of E. Coli contains a nonsense mutation in the gene encoding l arrival Chi nese, which allows E. Coli to metabolize revenues. So E. Coli containing this nonsense mutation are unable to metabolize the sugar. Um, a mutant of the strained possesses suppressor mutations which restore the wild type l arrive below kind. He's which allows these mutants to metabolize revenues. Um, one suppressor mutation or one was linked to the original nonsense mutation allowing the production of L A regular Chinese, albeit with reduced functionality. Another suppressor mutation S U B, which was found to not be linked to the original nonsense mutation in conjunction with our one, was found to fully restore the wild type in a type of L arrival a chi nese. However, it was also found that this suppressor mutation s u B was unable to the restore. Um, the wild type in a type of l arrive, it'll kind is by itself. Given this, the question wants us to categorize the suppressor mutations as inter or intra genic and proposed a mechanism of action for how these mutations work. Um, given that are one was linked with the original mutation. We can conclusively state that are one is intra genic Given that s u B was not linked to the original nonsense mutation. We can say that s u B is inter genic. Um, and the original mutation is nonsense, and our one partially reverses it. Um, this likely partially reverses it and partially restores functionality by changing the original mutation to a miss sense mutation. Um, and because it isn't miss since mutation, overall, it has that reduced functionality. Now, s U B is likely a mutation in a TR in a gene resulting in 80 or in a anti code on capable of binding to this miss since code on, um, generated by our one. But delivering the wild type of you know acid, um, to clarify, say, for example, the wild type amino acid is amino acid. A Now following the original nonsense mutation, we have the, um, amino acid. Now, um, being a stop code on Hello. We're following the r one suppressor mutation. Um, that particular code on now in codes amino acid B Um, however, following S u B um, the suppressor mutation which changes a tr in a, um even though the d n a is encoding specifically for amino acid B, the T or in a is recognizing amino acid B and saying that, um, it needs amino acid. A. So there is a delivery of the wrong tr in a, um, delivering the wrong, um, amino acid. However, in this case, the wrong amino acid being delivered due to the SUV mutation is the wild type of, you know, acid amino acid. A. So, with the effect of these two suppressor mutations, wildlife UNA type was able to be restored.

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.

This is a longer question. We have a wild type DNA sequence that we have four mutants. In each case we want to transcribe we want to translate and then we have to work out what mutation we've had. So we should start the wild type. So let's transcribe that. It's anti sense. So we're going to be starting anti parallel. So our sequence is going to start as followed. So it's going to be a U. G. Let's put these in threes to make our couldn't be easier. A C. You G A C. UTC and UNG. There we go. Okay so now we want to work out the amino acid sequence through remember? So starting at B. And stimulus. A Yuki is methionine. So that's quite normal for a start for them. So we have machine A C. U. Is thrilling. G A C. Is a specific assets. UTC 16 and then um GSB stock code on Swiss stopper. Okay so that's wild type. Now we can start with our first mutants. So starting with program and I'm just going to trends privates again. So we have a U G A C. A. Can already see a mutation here. T J C U T C. You're a genius. And the three permits. Okay so now I'm going to do the same thing. I'm going to go to my table of code owns and go through the so you think is still gonna be happy again. A. C. A. So that's different. What is a CIA? Well it's also dreaming Shericka. The these codes are sharing the main offices. Men T. A. C. Source path passes to see a Sistine and we still have our still colonel. Okay so be amino acid sequence has not changed. Which means this is a neutral mutation and is a base substitution. We need for our base substitution because the amino acid sequence doesn't change. So our protein will end up the same. Okay, next we're going to do sequence two. So same process again, starting five prime. We have a U G E. We have C C U. So there's a mutation again. T S c u t c u a G three prime. It's already written these out a little bit faster. Put them down so I can immediately see we have another point mutation. Have a base substitution before it was this you to an A. This time it's this A to C. Let's see if this affects feed. I mean that's a sequence. So we're starting off with caffeine in again. Now what is C c U c c U is prudent since. True. And then the others are trained. So we thought was in so we have a base substitution here that has has an effect. So the organization and we've changed three and in two program. So this is possibly going to affect function that protein. We don't know unless we know what protein is. Let's go to our first mutants. So same process E V E C. You seem to be a UK. Okay. So if I got the amino acids we have mapping in still a C. U. Is going to be three, then G A C. Is still Spartak acid. What is U K. A. But it's different. You G A is a stupid on. Oh no, we've stopped. So that's the end. We we've stopped a code on early. So this is a point mutation that has truncated our Mhm sequence from facing seconds. Finally, let's look at number four switch starts the same way you T c u t A c u g A U A G. And that's where ends. So that's strange because we don't have a full number of good answer what's gone wrong? Well this a here has been deleted so I can already see that this is a delusion mutation. But let's follow this through. So we have mckinnon still. What is C U. G. See you G is loosen. So you do see a see you A C. U. Is threatening G A. U. Is spotted acids and we can't do anything of a G. Really because it's not a full freedom. So would you stop that? So what's happened here? What? We've had a delusion mutation and this has resources in a frame shift because from the mutation onwards all of these have been shifted by one. So we're getting a very different a series of amino acids.


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