The Genetic Code - Video Tutorials & Practice Problems
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1
concept
The Genetic Code Concept 1
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4m
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Video transcript
Hey, everyone. So in this video, we're gonna talk about the genetic code. Now, the genetic code is the assignment of each code on to one of the 20 amino acids. And recall when we say codon, a codon is a nucleotide triplet. So three nucleotides that codes for an amino acid or it acts as a start or stop signal. And here we're going to say that there are 64 total coons. Now, how do we come up with the 64? Well, here, remember we have four of our nitrogenous bases and we have your cell cytosine, adenine and guanine. And we have for the triplet that's 33 of these nitrogenous bases. So we have the first letter of the triplet, the second letter of the triplet and the third letter of the triplet. So in each case, that's four, so four times four times four gives us 64 total codons. So this is how we come up with this total number of coons. And we're also gonna say here that a single amino acid can have multiple coons. If we take a look here, we have our chart which helps us to connect these codons to different amino acids. How do we use this chart? Well, if we take a look, we have our first letter here on the left side. U and let's say we're looking for the second letter. So look up here, let's say the second letter is A. So that'll put us here. All of these, start with U A and then the third letter is on this side here. Let's say the third letter was C. So we'd say it's you A C which gives us this, it would code for the amino acid Tracy. Let's say we wanted to do guu. So guu, the first letter is G, the second letter is U. The third letter is uguu codes for Valley as our amino acid. Now, here we're gonna notice that some of these coons are in different colors that are not black. So here we have a UG in green and then these three red ones uaaug A in UAG. So here we're going to say that a UG at the start of MRD acts as a start signal, a start coder. Now it codes for the amino acid methionine. So here, if a UG is found within the inside of an MRN A strand, it it's no longer the start signal. It codes for methionine amino acid. So again, it has two functions. If a UG starts off the MRN a strand, it acts as a start signal if it's found in the interior of the MRN a strand. Then it codes for methionine. Now remember this number of 64 total code outs of this 64 total of, of this 64 total co downs 61 of them are assigned to amino acids and three of them act as stop signals or stop codons. They don't code for any particular amino acid. As you can see here. It's blank here and it's blank here because those three in red are, are stop codons or stop signals. Now, how do you remember them? All you gotta remember is on the left side, you have you and then on the top and on the right, it's just a G. So this represents our first letter of our code on the second letters and the third letters. So we'd have you a A here. Actually, we will do it in red to signify the stop signal. So uaaug A and then UAG those are our three stop signals or stop codons. So remember on the left side is you and then it's just A G on the top A G on the right and then match up 1st, 2nd and 3rd letter of our code on to get the overall code on for that particular stop signal or stop code on. So just keep this in mind in terms of how do we utilize this chart? Remember A UG represents a start code on our start signal if it's at the beginning of an MRN A strand. If it's found on the interior, then it codes for Methionine instead.
2
example
The Genetic Code Example 1
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1m
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Video transcript
Here it says, determine the polypeptide sequence from the MRN A sequence given below. So step one is we're gonna break the given MRN a sequence into codons. Remember those are nucleotide triplets. So here every three, we're gonna split it. So that gives me GGCCAU AC Caug. And then UAU step two, we're gonna identify the amino acids for each coon and write the peptide sequence, right. So here we're gonna utilize our chart. So we have GG and then C which gives us glycine gly, then we have cau so C A and then U gives us this histadine. Next, we have AC and then C give us three anine A UG. Well, that gives us Met Methionine. Remember if A UG is at the start of the MRN A sequence and it starts, it's a start codon or start signal. But when it's within the MRN a chain itself in the interior, then it codes for the amino acid methionine. And then finally, we have Uau, so you a and then U gives us piracy. So this would be our peptide sequence. They were able to determine from the coons that we were given initially, right? So this is how you'd utilize this chart in order to map out the peptide sequence.
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Problem
Problem
Determine the number of bases in the information strand for the gene that codes for the peptide below:
Pro–His–Gly–Gly–Lys–Arg
A
6
B
12
C
18
D
36
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