00:01
For this question, we're looking at the codon and how this is interpreted to make different amino acids or parts of a protein.
00:09
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 matei.
00:22
So the first experience by crick and his associates were dealing with adding mutations to different dna sequences inside a virus.
00:32
So you could imagine a virus could have a dna sequence such as this of atagca per se.
00:43
What they did is they're going to add in a different number of mutations to see what effects it would have on the virus and its capabilities.
00:57
So they noticed that, of course, the unmutated virus had a normal function, two different viruses with either.
01:07
The addition of one or two nucleotides showed no correct function, and the virus that picked up three nucleotide additions also had a normal function.
01:21
And this was used to prove the importance of three nucleotides to make up one codon.
01:29
This is because in the addition of either one or two nucleotides, it's going to cause a frame shift mutation.
01:38
And this is because three nucleotides are required to make the protein.
01:45
So when one nucleotide is added, it's going to pull two nucleotides from the next three pair of nucleotides.
01:55
And that's going to alter all the next upcoming proteins, and it can severely alter the capability for the virus to function.
02:03
This is the same for the addition of two nucleotides, because it's still going to grab the first nucleotide, of every next amino acid grouping or codon coming up.
02:15
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.
02:29
And typically one mutation of a protein is not enough to completely alter and destroy the function of the cell.
02:37
So the rest of the amino acid groups, or codons located later in the dna sequence would still be normal, allowing for the virus to function normally.
02:51
As for the experiments by nirenberg and matei, they discovered that with the addition of ribosomes and, say, a certain amino acid, if they only had, say, u's or a's or g's 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.
03:26
So that just helped prove the specificity of the coding sequence where one type of codon is going to produce one type of amino acid as long as the rnas or ribosomes are present inside the solution as well.
03:54
Moving on to part b of the question, they're asking us to calculate ratios or the percent chance of different combinations of codons.
04:04
So we have three different types.
04:07
We have the first scenario where we have two guanines to one.
04:15
Cytosine.
04:17
We have one guanine to two cytosines and we have the last scenario where we have only cytosines.
04:26
And of course the ratio of g's to c's in here is five to six.
04:32
So five guanines for every one cytosine.
04:38
Now you can add these together to get the whole of the probability chance.
04:44
So this would be equal to six parts to the hole.
04:51
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.
05:03
So we'll say we're calculating the ratio of this codon, ggc, to the percent chance of randomly coding for all guanines.
05:15
So the chance in this solution of coding just for a guanine is going to be 5 out of 6.
05:24
Because pulling 6 times from the solution, one time you're likely to get a cytosine, whereas 5 times you're likely to pull the guanine.
05:35
So again, for the second guanine, you have a 5 out of 6 chance.
05:41
Whereas for the last nucleotide, the cytosine, you have a 1 out of 6 chance or probability.
05:48
And then they are just dividing this by the chance of getting all gs in this other codon.
05:56
So again, the chance of getting a guany is 5 out of 6.
06:02
So this is all they are doing in the problem.
06:06
And then they are just simplifying it and rewriting it as, say, 5 over 6 squared times, say 1 over 6 divided by 5 over 6 cubed.
06:26
So this is all they're doing in the problem.
06:28
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 1g to 1c, say you have a gcc sequence to a ggg, the chance of, or the probability of coding for this exact codon would be 5 over 6 times 1 over 6 times 1 over 6, divided by, of course, that 5 over 6 cubed.
07:02
And for the last problem, the chance of getting all these cytosines is 1 over 6 times 1 over 6 times 1 over 6 divided by 5 over 6 cubed.
07:15
You can just plug that into a calculator to see your chance for getting each of these...