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06. (a) Template-:ssisted biosynthesis of nucleic acids are critical for the cell" $ ability to replicate its DNA and t0 decode the information stored in DNA f...

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06. (a) Template-:ssisted biosynthesis of nucleic acids are critical for the cell" $ ability to replicate its DNA and t0 decode the information stored in DNA for prolein synthesis. Briefly describe the propenies of nucleic acid strands that facilitate high fidelity *copying' of the infonation stored within themWith the aid of diagrams. describe the essential features of the polymerase chain reaction (PCR)(c) The DNA sequence of gene that encodes baclenal prolein of interest given below

06. (a) Template-:ssisted biosynthesis of nucleic acids are critical for the cell" $ ability to replicate its DNA and t0 decode the information stored in DNA for prolein synthesis. Briefly describe the propenies of nucleic acid strands that facilitate high fidelity *copying' of the infonation stored within them With the aid of diagrams. describe the essential features of the polymerase chain reaction (PCR) (c) The DNA sequence of gene that encodes baclenal prolein of interest given below You have been supplied Wuth sll quantity o chromosomal DNA from the organism from which the protein is derived and you decide [0 use PCR t0 clone the gene. Design suitable primers for Ihe amplification, incomorating into them Ndel and BamHI restriction sites So that these are located upstream and downstream of the gene, respectively Write both primets in the 5-3" direction: Note the following: only the 5'-3 strand of the gene shown the recognition sequences for Ndel BamHI are CATATG and GGATCC. respectively: and the translational start and stop codons ure underlined: CATGCAGGATTTTTTTTGCTCAGTATCGAATACAAACATGGGACAGTAAGGACG GTTTTAGAAAGCGTTTTTATGCTGGCTGGAAAGGATGGATGTC ATG AAG GTT ATC AAG GGG TTA ACG GCT GGG CTG ATT TTT CTG TTT TTG TGT GCA TGC GGA GGA CAG CAG ATTAAA GAT CCG CTC AATTAC GAG GTG GAG CCT TTT ACA TTT CAA AAC CAA GAC GGC AAG AAC GTT TCT TTA GAG AGT TTA AAA GGA GAA GTA TGG CTG GCG GAT TTT ATT TTT ACC AAT TGT GAA ACT ATA TGT CCG CCA ATG ACC GCT CATATG ACC GAT CTGCAA AAA AAA CTG AAA GCC GGA AAT ATA GAT GTC CGC ATC ATA TCA TT AGT GHT GAT CCA GAA AAC GAT AAG CCG AAA CAG CTG AAG AAA TTT GCC GCA AAT TAT CCA TTA TCT TTT GAT AAC TGG GATTTT CTC ACG GGA TAC AGC CAGAGT GAG ATT GAG AAG TAAGCTAAGAGACTAGGCTTCGCAAATCTTCTAAAACATGATACACTTTCCACTA GTAGAATGGGAAGGAGCAACAAGATTGAAGCTGCGCATTTTT



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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.

Let's talk about Port A Her a Onley clone a not BC or Dean is activated by F gs. So which indicates to us that the DNA binding site, which is Onley located in Clone A, is located somewhere each Wayne, the three prime end of Exxon, one of E one s and the five prime end of Exxon three e three. That's for be part. We can see that cortisol is responsible for repressing transcription and clones A and B transcription Olbricht. Repression is not occurring because of cortisol. Includes scene. And so the DNA binding site must be located in the three prime region of E three s or in the flanks of the three prime int. And finally see part clones C and D or activated by E. T and not Clem being so. What this tells us is that the DNA side involved must be located to the three prime side uh, e three Jeanne. Oops, let's talk parentheses around the three. You need to be around the G. What's just trying? Get in here. There we go. Or again in the three prime sequences in the flights of the global gene

For this question, we are looking at DNA transcription in the presence of different proteins, including the opt Immers of the nuclear zones. We also have the H1 protein, and we're looking at this special protein in this experiment called gal, for G p 16. So first we can take a role. Look at the role of D. N. A. And the optimizers and H. One that combined to it. So DNA, of course, is going to be that single string.

For this question. We're looking at DNA transcription and the role of certain proteins involved with it, including the nuclear. So um optimizers, We also have the protein H1. And during this experiment they're looking at a special protein called Gal for VP 16. So here we can start by taking a look at DNA transcription and the role of our customers and this H1 his stone protein. So of course DNA is going to be in that double stranded structure and you're going to have special regions associated with it. You're going to have a promoter region and you can also have certain repressor and other regions involved with this. The role of October's and the H1. Histone protein is to rabble up this double stranded DNA in order to condense it so it can fit in the nucleus. And you're going to have to kind of confirmations of the histone proteins. You have a methylated form where you have methyl groups attached to the DNA and the histone proteins and you have an assimilated form. And there you have a Seattle groups attached to your histone proteins. And those are going to affect the way that your DNA reps around those his stones. So in your methylated form, DNA is going to be closely wound to each of the histone proteins. And you're going to have very little space in between. In the assimilated form, it's going to be in a looser confirmation and you'll have much more space in between each of your his stones. And this is going to affect the way your genes can be read. So of course inside of that DNA we have those genes of interest. But in order to transcribe these genes we need preliminary races and other proteins able to activate and attach to those specific regions. So in the methylated form, you can imagine that the headstones that are close together. leave very little space for your preliminary races and other proteins to buy. So there is going to be very difficult to transcribe those jeans. Whereas in the assimilated form of your headstones there's plenty of space for your preliminary races to access as well as your other transcription factors and proteins. So you're set elated form is going to increase the rate of transcription. Or as your methylated form can decrease or even deactivate gene transcription. So here are the roles of the October's and H1 is to bind that protein and to down regulate or decrease the amount of proteins created from the genes. So knowing that we can take a guess at the role and function of this special protein, gal for BP 16. So since this protein is going to function and prevent the function and decreased transcription caused by the nucleus, oh mok tumors and your histone proteins. You can imagine that gal for VP 16 is going to probably bind to your his stones or to the DNA. Around where those headstones. Woodbine. And this is going to prevent your assimilated form from becoming methylated and closing that gene area and from down regulating gene transcription. So if it's going to bind to either the his stones or the DNA in either of these areas, it's going to increase the transcription and it's going to prevent it from being inhibited since it's also going to increase the rate of transcription. Since in the graph were given that the naked DNA, meaning just this DNA alone without any additional proteins has a rate of 100 and our DNA Plus this gal for VP 16 has a rate of 1000. This protein has a good chance of binding to this promoter region because the promoter region has a tendency to increase the rate at which the genes downstream are transcribed. So if this protein binds to this region here, it's going to increase the times the problem arises. Transcribe the genes and it's going to probably increase the affinity for the area for your different transcription factors and that's going to increase the overall rate that we see in the data provided. So overall the important pieces of information to get from this is that your october's and your histone proteins decrease the rate of transcription by binding it into these confirmations that are closed where the genes can be hidden and inaccessible to your preliminary races. The gal for VP 16 likely prevents that binding by those ah customers or H. One and leaves it in this open confirmation where your prelim erases and transcription factors can access it. It also likely has an increasing or stimulating effect, possibly by binding to the promoter or some other stimulating region that's going to increase the rate at which that gene area or region of the D. N. A. Is transcribed.

This question is making sure you understand restriction enzymes, how they cut and how they're sticky or blunt ends will react with different portions of DNA. Create a new hours double strand set of DNA. So part A is asking us. But a Nico are one enzyme cut would look like so in one of the tables in the textbook. It gives us the sequence for ICO, our wine. It's going to look for a G, A, A, T T C and its complementary strand of C T T A A. G. And here the ICO are one enzyme is going to cut along this pathway. It's going to separate the two ends and create two fragments of DNA along this point. So for part A, we're going to have a long strand of DNA mixed with It's a and its complementary par shin of the strand with a T. T. A sticky end, and this is going to be separated from the rest of the fragment of the single strand of DNA that we just cut you have. It's three prime and five prime, so these are the kind of things you'd see. Just remember, we have a longer strand of DNA continuing out along both sides. But this is the site we're looking at. That was just cut for B. It's just asking us to attach a nucleotide triphosphate to the sticky ends. So we're just going to attach the complementary bases for each of these fragments. So here on the three prime Strand, we have our t t a. So we're going to attach our complimentary A T T. And then on the second fragment we're going to attach are complementary pairs like ways. So we have a okay to to. So these are new fragments we've created with blunt ends. Now, since we've added our nuclear tied try phosphates to each of those for part C, it's just asking us to attach kind of these blunt ends together to create a single strand of DNA has just been reassembled. So here we'll have our G A T T fragment, and it's going to attach and, like, eight to our A a two TC and Mike wise on the bottom. That's May. So here we have our new DNA strand when are previously made and nucleotide triphosphate added strands have been connected. Just remember, we have our five, prime to three Prime and the complementary strand as well. For Part D, it's asking us to take our original sticky and fragments and to digest any single strand of DNA that's present. So here we're going tohave are sticky ends made with just an original eco r one cut and we're going to digest and remove our UNP aired nucleotides. So in the end, we're going to get very short fragments of just a G here, iguana zine and a cytosine and the other fragment with the site is seen on top and a quantity nonviolent. So here's our new strand we've created. If we're going to digest those UNP aired nucleotides from the sticky ends for party, it's going to ask what happens if we stick our original eco r one cut to what we've created here with just our single nuclear tides. So if you imagine we're going to take our sticky and pears Ah, and we're going to attach this three prime to five prime to our single nucleotide fragment of just the guanine from part D. So here we're just taking this single nucleotide fragment of the five prime and attaching it to the ico are one cut here. Likewise on top. It's just gonna be the complementary, but with this cytosine attached or with a guanine, because we're attaching our a a TTC to our single nuclear Tiggy from before. So for this instance, we are attaching to this five prime to three prime end here, and we are attaching our A, t, t c and G fragment. So this is what they're looking for, which is, ironically, just the ICO are one restriction site as well. Basically, what we've done is we've cut our original strand of DNA removed, are single nucleotides and added them back again for part F. There's actually a misprint in the book. Um, we're looking for the restriction kind of product if we're using a P V U two enzyme. Yeah, So if you notice in the book, the ICO are five. Restriction site is listed as the same as the PV You one PV YouTube PTU to actually has a different restriction site of C A, G, C T G, and its complementary G T c g A C. And it's going to create blunt ends by cutting straight through the center. Try that and read So for going for a PV you to cut, we're going to make the fragments C a g by prime three prime GTC and the fragments ctg g a c So this is what they're looking for for part G. We're going to attach this fragment we've made and add it to the blunt ends that we made in part B. So we're going to add our TV You to to this section here. So, Fergie, we take our five prime three prime initial strands C G G c, and we're going to add it to those blunt ends of our five prime to three Prime and our three prime five Prime. We're going to make this new strand here. So we've just attached are blunt ends together along this point, and this is our new strand with me for part h. They're asking us to apply this knowledge to kind of make us try and get a different plasmid structure where it would no longer be digested by an eco r one site, but it will have a bam h one site. So from what we've learned in the previous questions, what we could dio is we could take our initial Iko are one G A, T T C and its complementary strand Cut it with the ICO are one. And we can either remove this single strand DNA using a different enzyme, or we can fill in here using our nuclear tied complementary pairs. And then we can just add additional nuclear tights on here That would give us a bam H one site as long as we don't begin with C and G along this next nuclear tied we add on. If we add in this CG pair equal are one would still be able to cut it. But if we add, say A and A and A T here equal are one will no longer recognize this sequence and it won't be able to cut now it takes a lot of time and scientific error. Um, restriction enzyme aren't perfect, and they don't give you 100% outcome and yield. So a better way to do this is to create our own DNA strand. But they have listed is we're going to make this artificial strand here and its complement so feel notice. What we're now including in here is our a t T See are these editions to this nucleotide sequence. So basically, this sequence is made by cutting it away with a Nico are one enzyme on both ends and you'll notice the Bam H one restriction enzyme of G A, T, C C, and its complement is located right here in the center. So by artificially creating this sequence of DNA, we have an opportunity to attach it to a Nico are one enzyme site, and it automatically has a bam H one enzyme site included. So, after we've introduced equal art one toe wherever this is attached, we can now make a new cut down the center of this sequence with a bam H one cut. So if you look in this previous example, if we're going to add DNT piece here, we create a blunt end, and blunt ends aren't very good at connecting up ends of DNA. Um, it's easier with sticky ends because complementary pairs like to stick and fix along to make double stranded DNA. Rather than bear up to blunt ends, there's just more free energy available and sticky ends make the reaction more spontaneous and more likely to occur and give a stronger hold rather than just attaching across the DNA rather than across the compliments. So for H, this is our better option. Two years, due to its advantages of sticky ends and its utility of both ends for a Nico are one enzyme and the inclusion of the Bam H one site straight through the center finally per part II. They're asking us to create different strands that are compatible with Iko are one and or PST one. So just as a reminder, all right, the equal are one cut site and the PST one cut site. So for equal are one It's the G A, T T c. And for PST wine, we have CTG, C, A G, and our method of cutting creating sticky ends. So what we want to do is create a sequence where you have hey is both being c N t. Do you is neither for being able to do an ICO are one site and for C, just a pst one site. So, for a what we can do is we can make an eco r one site along the far ends of the strand of DNA, and we can make the PST one sites closer into the center. So if we start with this sequence of DNA a long way, we're cutting. So here we'll have a a a T T. C bound to its G here would be the ICO, our wine cut site in the center. We could use whatever nucleotide sequence we want. It doesn't matter, because we're not going to cut along these lines. And along the right side, we'll have our PST one cut site. So here is our c g C. A sequence. If we wanna have unequal are one on Lee, What we're going to do is we can just change this sequence. This base pair here MPs Taiwan will no longer recognize our strand so we can have a a t t c whatever we want in the center. And then we could just add in a guanine here instead on create are complementary sequence in the middle. So here we stole of our eco r one cut site. However, on this and this sequence is not recognized by any of the enzymes. So for part c, we do just the opposite and change this base pair here and leave this original water alone. So here we'll put in a guanine instead of that side is seen. Keep our cytosine on this strand and drawn are complementary. So here we have our PST one cut site and no cuts site over here. Lastly, for our neither we can change both of these sites so that neither iko are one or P s t one will recognize them. We have a a t t. G. So there's no cut and G T g c A. So there's no cut, so you'll notice neither this strand or this strand is recognized by either equal are one or P s t. One. So there will be no cut along these points. So this sums up how we've located where to cut along Our enzyme restriction site had to attach blunt and sticky ends and how to form different pairs so that were either removing the restriction cut site or adding it into it. And this is what this question is looking to address


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