M. Match the scenarios with the concept that complete… (2023)

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..

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