Translation: Protein Synthesis - Video Tutorials & Practice Problems
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Translation: Protein Synthesis Concept 1
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Hey, everyone. So we're gonna say that similar to transcription. The process of translation consists of three steps. We have our initiation step, our elongation step and our termination step. First, let's take a look at our initiation step here within it. We're gonna say we have ribosomes that consist of a small and large ribosome subunit. We're gonna say each subunit is made up of proteins and our RN A. So remember our RN A is ribosome RN A. Now we're gonna say that the MRN A is going to bind to the small ribosome subunit. So here we have our MRN A and it's going to bind to this small ribosome subunit. So here this would be a next, we're going to say for B, we're gonna say that uh Matine carrying TRN A which has the anti codon of U AC binds to the start codon, which is a UG. So here goes our MRN A that is bonded to the small ribosome subunit. Remember these are the coons, the three nucleotide sets that exist on the MRNATRN A which is this large structure here comes in in docks and pairs up complementary wise with the codon of MRN A. So here's them pairing up here. We see the hydrogen bonds that are forming between the nucleotide and nitrogen spaces involved. So this would be B and remember when we're talking about our nitrogen in spaces because we're dealing with Mrnarn A in general, we're gonna use ell instead of thyme. Now, c which is the initiation step is completed when the large subunit joins the small subunit complex. So we're gonna see here, we have our TRN A which is has its anti codons paired up with the coons of MRN A. And now we have this large subunit basically coming in and docking and pairing up with these small um ribosome subunit together. This represents our ribosome complex. So why those are more complex is the small and large subunits together? So this would complete our initiation step of translation.
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example
Translation: Protein Synthesis Example 1
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Which of the following is not a part of the initiation step of translation activation of TRN A through amino A cell. Trn A synthetic nowhere do we talk about needing to activate our TRN A when it comes to the initiation step? So this is our answer. If we look at our other choices, we have combination of the large rivals O subunit with a small subunit complex. This is actually the last step that concludes the initiation step of translation. So this is a part of it binding of MRN A to the small ribosome subunit. This is one of the initial steps. When it comes to initiation of uh step of translation, we need the MRN A to initially bind to the small subunit of the ribosome complex binding of met, which is a finding trn A with the start code on a UG through complementary base pairing. Remember MRN A comes in with its code on its code on its dark code on is a UG A meth finding carrying TRN A comes in with its anticodon to match up with those nucleotides. So yes, this is a part of the process. So here the only thing that's not a part of the initiation step of translation would have to be option A.
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Translation: Protein Synthesis Concept 2
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Everyone. So now we're gonna take a look at the second step of translation, which is elongation. Now, here in an elongation, we're gonna see a second TN A approaches and binds to the next coon in the complex. So remember we had our large subunit and our small subunit joined together to complete the initiation step. We had our mething carrying TRN A come and pair up with the coons of MRN A. Now a second trn A with another type of amino acid comes and docks in this second slot here, we're going to say here that a peptide bond forms between methionine M and the second amino acid. In this case, s sine and D passes. Now we're gonna have E here. So our second TN A has docked in with its anti codon to pair up with the codons of MRN A. What happens now is this me finding is gonna break off this Tr Nate and it's gonna come over here to us to help make this bond, this peptide bond. So now this TRN A no longer has its amino acid attached, it no longer has methionine. So what's going to happen is that we have translocation where the whole ribosome moves to the next codon after the first trn eight leaves the complex. So basically what happens is this whole structure, the large and the small they're gonna shift over to the right because here this TN A is empty now, we don't need it. So we shift the whole thing over. So now here's this coon that was within the complex. It's now out here exposed and we're gonna jettison this empty TRN A. So this would be step f so this is translocation, the whole complex shifts over to this slot here, which has this growing peptide sequence going on. And what's gonna happen is over time, we're gonna make that peptide chain longer and longer, adding more and more to it. Now, here we're gonna say steps DNF keep repeating. So we have another one come in, the amino acid is gonna break off and help to make the chain longer and longer. So it keeps on repeating. Now, if we take a look here, this is just happening on one location. But in reality, a single MRN A can be translated by multiple ribosomes simultaneously. So here is our MRN A here and here's a ribosome complex, a ribosome complex. Here's another one. They're making these peptide chain, these sequences at the same time. So we can make multiple peptide sequences at the same time off of the same MRN a strand, right. So just remember when we're talking about elongation. We're basically starting to add and chain amino acids together through these peptide bonds. And this is the way that the ribosome complex works where a new TRN A with a new amino acid attached comes and docks in next to the older one. In this case, methionine containing TRN A. And we're just gonna try to make the peptide bonds to make our peptide sequence longer and longer.
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example
Translation: Protein Synthesis Example 2
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Here, we're told a section of a protein has a sequence of phenyl alanine, valen and Arginine. These here represent the three letter codes. What is a possible MRN a codon sequence to do this? We're gonna have to utilize our chart here which gives us the 1st, 2nd and 3rd letter codes for our codon. Remember a coon is just a triplet of nucleotides, three of them together. Now, if we take a look here, we need to find pheno al uh phenyl alanine here. But remember that some amino acids will have multiple codons. If we look at this chart, we have it here and we have it here. And those are the only places. So we have to see where do we see these codes in our options here. So Fenny Alanine could be all Uuucuguugc if we look at our options here, OK. The only one that matches up with this is D where we have UUU. But let's verify the other amino acids we have Valley here. Valley. If we look, where do we see it? Well, we see it down here and GU A is this Gu A and then we have Arjun nine here, where do we see it? We see it in more than one place, we see it here and we see it here. We need a G A which matches up with this A G A. So we utilize this chart in order to find the amino acid and the three letter code associated with it in terms of the cod on doing that gave us D as our only answer.
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Translation: Protein Synthesis Concept 3
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Hey, everyone. So in this video, we're gonna take a look at the last step of translation, which is our termination step. We're gonna say that translation terminates when the Russom encounters a stop codon. Remember your three stop codons are uaauag or UG A in phase G, we're gonna say the release factor protein binds to the stop codon. So if we take a look here, we have our stop code on here and we're gonna see our release factor comes in and attaches this will initiate phase G. We can see that we have a growing chain of amino acids connected together by peptide bonds. We have our release factor protein that's attached to the stock code on. What's gonna happen. Next is the peptide chain. The growing one is gonna be hydrolyzed and released from the last trn A. So we're going to sever this link here. And as a result of this hydrolysis here is our released peptide chain here. Next I we've released the peptide chain that we wanted to grow. So we no longer need our ribosome complex. So the whole complex just disassembles. So we can see that our TRN A which no longer has any amino acids attached to it has broken off. We have our release factor pro protein that's broken off. We have our large ribosome sub unit as well as our small ribosome subunit. Here, they've broken apart and then we have our MRN A chain right there. That's just free floating right there. Now, this is important to know that methionine is generally removed from the protein backbone after translation. Remember Mathia is attached to the start code on a UG. So here we no longer need any of this. So everything is just broken apart. And this is the end of translation. We've created the peptide chain that we were looking to make, right. So this concludes translation as a whole the third and final step.
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example
Translation: Protein Synthesis Example 3
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In this example, question it states write the sequence of the peptide translated from the following DNA template sequence, right? So here they're giving us this DNA template sequence. What we need to do first is we need to basically transcribe it into our MRN A. So here this would be our five prime end and this would be our three prime end. So remember since we're going from DNA to MRN A, remember RN A doesn't use thymine, it doesn't use T, it uses U uracil. So our pairings would be A with U and still G with C. So since this is a A, this would be U, this T would become an A, this would become C and there goes our first coon. Next we have CC A, then we have you, you, you, then we have CC A. So now we just have to find each one of these coons. So U A and then C give us Tyr. So that means the answer is gonna be either A or B CC and then a give this Prolene here pro so, so far they're the same. Next, we have you, you, you, so you, you and then you give us phenylalanine. So phe so it looks like B is the answer. But let's just make sure that Prolene is the last amino acid here. So we need CC A. So CC A again. So again, it's Prolene. So we had to transcribe our DNA template into MRN A because we needed your cell because your cell is the nitrogenous space used within this chart. Once we decoded the MRN A codons, we can determine the amino acid attached to it. Right. So again, that gives us option B as our final answer.
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Problem
Problem
Tuftsin is an immunostimulator tetrapeptide having the following sequence:
Thr–Lys–Pro–Arg
Write a possible sequence for the gene (Informational & template strand) that codes for this tetrapeptide.
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