What happens if rna polymerase malfunctions

Mismatch Repair : In mismatch repair, the incorrectly-added base is detected after replication. The mismatch-repair proteins detect this base and remove it from the newly-synthesized strand by nuclease action. The gap is now filled with the correctly-paired base. Once the bases are filled in, the remaining gap is sealed with a phosphodiester linkage catalyzed by DNA ligase.

This repair mechanism is often employed when UV exposure causes the formation of pyrimidine dimers. DNA Ligase I Repairing Chromosomal Damage : DNA damage, due to environmental factors and normal metabolic processes inside the cell, occurs at a rate of 1, to 1,, molecular lesions per cell per day. Without molecules that can mend such breaks, cells can malfunction, die, or become cancerous.

Nucleotide Excision Repairs : Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lying adjacent to each other can form thymine dimers. In normal cells, they are excised and replaced. Mutations, variations in the nucleotide sequence of a genome, can also occur because of damage to DNA. Such mutations may be of two types: induced or spontaneous. Induced mutations are those that result from an exposure to chemicals, UV rays, X-rays, or some other environmental agent.

Spontaneous mutations occur without any exposure to any environmental agent; they are a result of natural reactions taking place within the body. Mutations may have a wide range of effects. Some mutations are not expressed; these are known as silent mutations. Point mutations are those mutations that affect a single base pair. The most common nucleotide mutations are substitutions, in which one base is replaced by another.

This means that they insert the wrong base. Insertion of the wrong base leads to a mutation — a change in the sequence of the DNA. The proofreader is an enzyme called exonuclease , which recognizes the mismatched A-C base pair, and removes the offending A. DNA polymerase then tries again, and this time inserts the correct G:. Even though DNA polymerases have proofreading abilities, they still make mistakes — on the order of about one misincorporation per 10 7 to 10 9 nucleotides polymerized.

But the RNA polymerases of RNA viruses are the kings of errors — these enzymes screw up as often as one time for every 1, — , nucleotides polymerized. This high rate of mutation comes from the lack of proofreading ability in RNA polymerases. Therefore the mutations remain in the newly synthesized RNA. Given a typical RNA viral genome of 10, bases, a mutation frequency of 1 in 10, corresponds to an average of 1 mutation in every replicated genome.

If a single cell infected with poliovirus produces 10, new virus particles, this error rate means that in theory, about 10, new viral mutants have been produced. This enormous mutation rate explains why RNA viruses evolve so readily. For example, it is the driving force behind influenza viral antigenic drift. Here is a stunning example of the consequences of RNA polymerase error rates.

Tens of millions of humans are infected with HIV-1, and every infected person produces billions of viral genomes per day, each with one mutation. Over 10 16 genomes are produced daily on the entire planet. As a consequence, thousands of mutants arise by chance every day that are resistant to every combination of antiviral compounds in use or in development.

I cannot overemphasize the importance of error-prone nucleic acid synthesis in RNA viral evolution and disease production. Please bear with me as we diverge slightly from influenza virus; these concepts will be an important and enduring component of your toolbox of virology knowledge. This isn't my field I'm a software guy , but I have a certain fascination with your field. I say this, because I have a speculation I'd like to throw out there, but it's likely kinda ill informed.

Nevertheless, I'm curious as to what you might think. You refer to this viral RNA synthesis as error-prone, which I understand what you mean when you say this.

If so, might there be a drug strategy which instead of targeting the replication cycle of the virus, instead attacks its error rate. The idea being that by lowering the error rate of the viral RNA synthesis one reduces the rate at which the virus produces immune-evading, or anti-viral drug evading mutations.

You are not at all full of crap; in fact those are excellent ideas. Often amateurs or should we say, those not in the field have excellent ideas because they are not saddled with the baggage of familiarity. Nirenberg and Leder extended this work to include other nucleotides. 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 developed a technique for synthesizing RNA composed of predictable distributions of repeated pairs or triplets of nucleotides. 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.

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'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. You just plug that into a calculator to see your chance for getting each of these. 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.