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Question

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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:

For this question we're looking at the code on and how this is interpreted to make different amino acids or parts of a protein. So the first part of this question, we're looking at different experiments by Crick and his associates, as well as experiments by Nirenberg and Matty. So the first experience by Crick and his associates were dealing with adding mutations, two different DNA sequences inside a virus. So you can imagine a virus could have a a DNA sequence such as this, of a t a g c a per se. What they did is they are going to add in a different number of mutations to see what affects it would have on the virus and its capabilities. So they noticed that of course the um mutated virus had a normal function. Two different viruses with either the addition of one or 2 nucleotides showed no correct function. And the virus that picked up three nucleotide editions also had a normal function. And this was used to prove the importance of three nucleotides to make up. One code on this is because in the addition of either one or two nucleotides, it's going to cause a frame shift mutation. And this is because three nucleotides are required to make the protein. So when one nucleotide is added, it's going to pull two nucleotides from the next three pair of nucleotides and that's going to alter all the next upcoming proteins and it can severely alter the capability for the virus to function. This is the same for the edition of two nucleotides because it's still going to grab the first nucleotide of every next amino acid grouping or code on coming up. However, the last option appeared to be normal because with the addition of three nucleotides, this would only correspond to one mutated protein inside the virus. And typically one mutation of a protein is not enough to completely alter and destroy the function of the cell so the rest of the amino acid groups, or Cody johNS located later in the DNA sequence, would still be normal, allowing for the virus to function normally. As for the experiments by Nirenberg and maturity, they discovered that with the addition of ribosomes and say a certain amino acid, if they only had say use or Azour gs inside of a solution, it would only create one type of amino acid. Whereas the addition of different nucleotides and amino acids would correspond to a completely different type of amino acid. So that just helped prove the specificity of the coding sequence where one type of code on Is going to produce one type of amino acid as long as the are in a. Zor ribosomes are present inside the solution as well. Moving on to part B of the question, they are asking us to calculate ratios are the percent chance of different combinations of code ons. So we have three different types. We have the first scenario where we have two guan means To one side a zine. We have one wanting to to cida zines. And we have the last scenario where we have only cida zines And of course the ratio of GS two season here is 5- six. So five wanting things for everyone cytosine. Now you can add these together to get the whole of the probability chance. So this would be equal to six parts to the whole when calculating this inside the problem. They make it a little complicated but it's just calculating simple chances and probabilities for each of the amino acid sequences. So we'll say we're calculating the ratio of this code on G. C. To the percent chance of randomly coding for all guan means. So the chance in this solution of coding just for a guangming is going to be five out of six because Pulling six times from the solution one time you're likely to get a cytosine, whereas five times you're likely to pull the guarani. So again for the second quantity you have a five out of six chance, whereas for the last nucleotide, the Cytisine, you have a one out of six chance or probability. And then they are just dividing this by the chance of getting all Gs in this other code on. So again the chance of getting a quantity is five out of six. So this is all they are doing in the problem and then they are just simplifying it and rewriting it As say 5/6 Squared times, say 1/6 divided by 5/6 Cube. So this is all they're doing in the problem. So you should be able to calculate the probabilities of this out between all of these where the ratio or the percent chance of getting one G 21 C. So you have a G c C sequence to a g g G. The chance of Or the probability of coding for this exact code on would be 5/6 Times 1/6 times 1/6 Divided by of course that 5/6 Cube. And for the last problem, the chance of getting all these side Busines is 1/6 times 1/6 times 1/6 divided by 5/6. Cute! You just plug that into a calculator to see your chance for getting each of these. And of course the other possible code ons such as G c G will have this same ratio because instead of this 5/6 here you have a 1/6. And instead of the 1/6 you would have the 5/6 and you should still get the same probability or the same chance of this occurring for part C. They give us an amino acid table for this and they are saying we have an equal proportion of cida zines and guan means and they are asking us the percentages for getting each of the amino acids in these groups. So if you've never read from one of these tables before you start on the left to check for the first amino acid. So say we have a C C c sequence that we are looking to code and amino acid for. So you'll start with C. You'll move on to the second position up on top. So you move over from the table until you get to the calm with the next see. And from there you move on to the 3rd position of the table where you pull from the left. So if you do that for any amino acid you should get the amino acid that is coded for by the code on. So for example C C C would code for the amino acid of pro lean. So keeping that in mind And the 1-1 ratio of wanting to cytosine. We should consider the different possibilities we have. We have a code on where we'll have three CS. 20 Gs. We'll have code ons with two Cida Zines, 21 guanine will have a one cytosine to guanine means and will have zero citizens to three Guangdong's and you can just right in the different combinations of these. So these should be some of the code ons and DNA sequences that you get. And then from there you can just use the table to figure out what amino acids these would code for. And you should get prowling for the C. C. C. Which I demonstrated above. You should also get pro lean here, should get arginine, Ellen, E, argentine, Alan E clay scene and placing and then you can just total these up to get a ratio. So we'll have to pro lean to to Argentine 2 to Alan E. Too too classy. That can be coded for in this scenario. Of course the percent chance of getting each of these is going to be equal because the chance of getting either G or C in each of these positions is one out of two. So no matter what happens, you have a one out of eight chance of getting any of these nucleotide or code on sequences. So you can just total these up And you should find that there are eight parts and because these are all equal and they have to, each of these amino acids has a one out of four or 25% chance of occurring or being coded for when the G to the C ratio is equivalent for the last part were being asked about our different stop sequences of say you A G. We also have you A and you G. A. When you look on the table and you code for each of these, you should see the stop inside the table rather than a three letter word for an amino acid or form a thin. The role of these is they do not code for any amino acid. So when we are creating our polymers of proteins and amino acids instead of adding an amino acid to the chain, it's going to add nothing. So it's basically going to break off and end the protein sequence. So these are responsible for terminating the elongation of the protein and for finalizing it. So these are going to allow the amino acid chain and protein to disassociate from the ribosomes and to enter the cytoplasm in their complete form. Now, with those for it should help you understand the purpose of the code on With its reliability and it's three nucleotide sequence.

But this question. We have a really long list of be important sums in the straps of as and descriptions to match. So I'm gonna get through descriptions and match them to be. It's, um barracks, describing a intervening sequences found in many eukaryotic treats that these are the entrance is what we want to spice up before. Translation over. Then we have be conserved. Nucleotide sequence in oh, character promoters involves initiation conscription. It's going to one of our sequences on this one is Vita sequence here must be have C Small army molecule was in meat nuclear about Karadzic's owes its components of spices own, displacing the excision of insurance. Of course, this is Snr have d A sequence in the non templates around matters of ICO. Life facilitates from applies, so this is never sequences, sweetheart. Top secret E. Coli is a very important model organism worth remembering. Lots about E, but only memories. But cancel eyes is the synthesis of all our days. Except for the five s subpoena. We've got three irony primaries. Is it one butts? Does most of the represent Laurene? It's only been raised. Then we have if he subunits of pro Carol tick arm in eight primaries that is responsible for the initiation of transcription. That's promotions. This is be Sigma factor. Yes, and E. Coli promoter sequence. Okay, go on the sequence. 35 nucleotides upstream initiation sites. A recognition sites for significant. So this recognition for Sima fax it iss TJ this the lots of these, but it covers all of the important sums. Make the irony play Marie's to left that capitalizes for synthesis of tea on a on a small nuclear. So for this one, h he's going to be are in a play Marie's three. No. Yes, we have I a poly Adina seen fracked, but it's added to be three prime end of Mostar Aquatic. So this is the pulley A to is well described by his name. Okay, we are in a primaries but transcribes nuclear genes. So this is the only one left arm in a primaries to So this makes the Aymara nays and we have K a conserved sequence and the sequence in the non template strand of oh, character promoters. 80 nucleotides upstream from the transcription start sites. Schools be this one last one hit. Indeed. This is a sequence phone them. We have L segments of their character, Teen corresponding to sequences and we finally processed are a transcript. So after spicing, this is what is left. And that's the Exxon's. Yeah, and finally we have and a population approaching transcripts, too many clips of their parents. So and this is it's true genius are in a that Sullivan.

Here, we have a multipart question. So for part A, we have the prompt that reads a mutation that prevents the binding of the drivers up to the five time end of the M R and a five time untranslated region. What effect will this half? Well, the answer is that no gene expression will occur for part B. We're looking at a mutation that changed the TRP code ons in region one of the M R five prime untranslated region into co dance for Allan E. So what occurs in this situation? Well, if those Allen and Co. Downs were replaced by trip to fan coat on, then under conditions of Low Allen in the stalling of the Riva Zone will not occur. So the attenuated will form stopping transcription derives, almost stalled when all in the alunan is low. So transcription structural genes will occur only when Allan E is look right. Um, so that would be our answer right here. That the fact that in attenuate reforms that starts transcription all right? No, the RAV ism will stall when all Alan in his left. So the transcription of the structural genes so transcription of those structural genes in the opera will only occur when Alan in is low. For part, C mutation creates a stop code on early in region one of the M R and a five prime untranslated region. So in this case, transcription will not occur because the regions three and four are now free to form the attenuated. So Regions three and four can form the attenuated purport D deletions in region two of the m R five prime untranslated region will cause an attenuate er to form and transcription will not look her part e we have that our delish ins in the region three of the m r five prime untranslated region will cause that TRP operate on to be unable to form the attenuate er in region three due to the fact that Region three contains that delicious So attenuation does not occur due to the fact that we do not have, um we have that delish in in region three. Since we do not have a good copy of Region three, which results in continuous transcription of the TR key structural cheats moving on to f, we are giving the problem deletions in region four of the m r five prime untranslated region. Well, these will prevent the formation of the attenuate er by the five prime untranslated region Marnie causing transcription to occur. And finally, with G, there is a deletion of the string of Adami nucleotides that followed region for in the fire prime you tr well, for the attenuate er hairpin to function as a terminator. That's obvious presence of a string of your soul nucleotides following following region for in in the m r and a five prime you tr. So the delicious of this string of adenine nucleotides in DNA will result in no string of your so nuclear times following region for so no termination occurs because we do not form that attenuate er and thus transcription proceeds.


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