Translation: Protein Synthesis - Online Tutor, Practice Problems & Exam Prep

Translation is a three-step process: initiation, elongation, and termination. During initiation, the small ribosomal subunit binds to mRNA, and tRNA carrying methionine pairs with the start codon AUG. In elongation, additional tRNAs bring amino acids, forming peptide bonds and lengthening the chain through translocation. Termination occurs when a stop codon is reached, prompting a release factor to hydrolyze the peptide chain from tRNA, leading to the disassembly of the ribosomal complex. This process is crucial for protein synthesis, involving amino acids and peptide bonds.

Translation: Protein Synthesis Concept 1

Hey, everyone. So we're going to 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 going to say we have ribosomes that consist of a small and large ribosomal subunit. We're going to say each subunit is made up of proteins and rRNA. So remember, rRNA is ribosomal RNA. Now we're going to say that the mRNA is going to bind to the small ribosomal subunit. So here we have our mRNA, and it's going to bind to this small ribosomal subunit. So here this would be a. Next, we're going to say for b, we're going to say that, methionine carrying tRNA, which has the anticodon of UAC, binds to the start codon which is AUG. So, here goes our mRNA that is bonded to the small ribosomal subunit. Remember, these are the codons, the three nucleotide sets that exist on the mRNA. tRNA, which is this large structure here, comes in and docks and pairs up complementarily with the codon of mRNA. So here's them pairing up. Here we see the hydrogen bonds that are forming between the nucleotides and nitrogenous bases involved. So this would be b. And remember, when we're talking about our nitrogenous bases, because we're dealing with mRNA, RNA in general, we're going to use uracil instead of thymine. Now, c, which is the initiation step, is completed when the large subunit joins the small subunit complex. So we're going to see here, we have our tRNA which has its anticodons paired up with the codons of mRNA, and now, we have this large subunit basically coming in and docking and pairing up with the small ribosomal subunit. Together, this represents our ribosomal complex. So ribosomal complex is the small and large subunits together. So this would complete our initiation step of translation.

Translation: Protein Synthesis Example 1

Which of the following is not a part of the initiation step of translation? Activation of tRNA through aminoacyl tRNA synthetase. Nowhere do we talk about needing to activate our tRNA when it comes to the initiation step, so this is our answer. If we look at our other choices, we have the combination of the large ribosomal subunit with the small subunit complex. This is actually the last step that concludes the initiation step of translation. So, this is a part of it. The binding of mRNA to the small ribosomal subunit is one of the initial steps in the initiation of translation. We need the mRNA to initially bind to the small subunit of the ribosome complex. The binding of Methionine tRNA with the start codon AUG through complementary base pairing is also a part of the process. Remember, mRNA comes in with its codon. Its start codon is AUG. A Methionine carrying tRNA comes in with its anticodon to match up with those nucleotides. So, yes, this is part of the process. Here, the only thing that is not a part of the initiation step of translation would have to be option a.

Translation: Protein Synthesis Concept 2

Everyone, now we're going to take a look at the second step of translation, which is elongation. Here in elongation, a second tRNA approaches and binds to the next codon in the complex. Remember, we had our large subunit and our small subunit joined together to complete the initiation step. We had our methionine-carrying tRNA come and pair up with the codons of mRNA. Now, a second tRNA 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, serine, and it passes.

Now, we have another event here. The second tRNA has docked in with its anticodon to pair up with the codons of mRNA. What happens now is this methionine is going to break off from this tRNA and come over here to help make this bond, this peptide bond. Now, this tRNA no longer has its amino acid attached. It no longer has methionine. What's going to happen is that we have translocation where the whole ribosome moves to the next codon after the first tRNA leaves the complex.

Basically, what happens is this whole structure, the large and the small subunits, are going to shift over to the right. Because here, this tRNA is empty now. We don't need it, so we shift the whole thing over. Now, here's this codon that was within the complex, it's now out here exposed. And we're going to jettison this empty tRNA. So this would be step f. This is translocation. The whole complex shifts over to this slot here, which has this growing peptide sequence going on. And what's going to happen is, over time, we're going to make that peptide chain longer and longer, adding more and more to it.

Now, here we're going to say steps d and f keep repeating. So we have another tRNA come in, the amino acid's going to break off and help to make the chain longer and longer. It keeps on repeating. Now, if we take a look here, this is just happening at one location, but in reality, a single mRNA can be translated by multiple ribosomes simultaneously. So here is our mRNA here and here's a ribosomal complex, here's another one. They're making these peptide chains, these sequences at the same time. So, we can make multiple peptide sequences at the same time off of the same mRNA strand.

Just remember, when we're talking about elongation, we're basically starting to add and chain amino acids together through these peptide bonds. This is the way the ribosomal complex works, where a new tRNA with a new amino acid attached comes and docks in next to the older one, in this case, a methionine-containing tRNA. And we're just going to try to make the peptide bonds to make our peptide sequence longer and longer.

Translation: Protein Synthesis Example 2

Here, we're told a section of a protein has a sequence of Phenylalanine, Valine, and Arginine. These here represent our 3-letter codes. What is a possible mRNA codon sequence? To do this, we're going to have to utilize our chart here which gives us the first, second, and third letter codes for our codon. Remember, a codon is just a triplet of nucleotides, three of them together. Now if we take a look here, we need to find phenylalanine first. 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. Phenylalanine could be UUU, UUC, UGU, UGC. If we look at our options here, the only one that matches up with this chart is D, where we have UUU. But, let's verify the other amino acids. We have Valine here. Valine, if we look, where do we see it? Well, we see it down here (GUA). And then we have Arginine 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 AGA which matches up with this AGA. So, utilizing this chart, we find the amino acid and the three-letter code associated with it in terms of the codon. Doing that gave us D as our only answer.

Translation: Protein Synthesis Concept 3

Hey, everyone. So in this video, we're going to take a look at the last step of translation, which is our termination step. We're going to say that translation terminates when the ribosome encounters a stop codon. Remember, your 3 stop codons are UAA, UAG, or UGA. In phase G, we're going to say the release factor protein binds to the stop codon. So, if we take a look here, we have our stop codon here, and we're going to 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 stop codon. What's going to happen next is the peptide chain, the growing one, is going to be hydrolyzed and released from the last tRNA. So, we are going to sever this link here. And as a result of this hydrolysis, here is our released peptide chain here.

Next, we've released the peptide chain that we wanted to grow, so we no longer need our ribosomal complex. So, the whole complex just disassembles. So, we can see that our tRNA, which no longer has any amino acids attached to it, has broken off. We have our release factor protein that's broken off. We have our large ribosome subunit, as well as our small ribosomal subunit, they've broken apart, and then we have our mRNA 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, methionine is attached to the start codon AUG. 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.

Translation: Protein Synthesis Example 3

In this example question, it states, "Write the sequence of the peptide translated from the following DNA template sequence." Alright. So here, they're giving us this DNA template sequence. What we need to do first is we need to transcribe it into our mRNA. So here, this would be our 5' end and this would be our 3' end. Remember, since we're going from DNA to mRNA, RNA doesn't use Thymine; it uses Uracil. So our pairings would be A with U and still G with C. So since this is an A, this would be a U, this T would become an A, this C would become a G, and there goes our first codon. Next, we have CCA. Then we have UUU. Then we have CCA. So now we just have to find each one of these codons. So UA, and then C gives us tyrosine (TYR). So that means the answer is going to be either A or B. CC and then A gives us proline (PRO). So, so far they're the same. Next, we have UUU. So, UU and then U gives us phenylalanine (PHE). So, it looks like B is the answer, but let's just make sure that proline is the last amino acid here. So we need CCA, so CCA again, so again, it's proline. We had to transcribe our DNA template into mRNA because we needed Uracil, as Uracil is the nitrogenous base used within this chart. Once we decoded the mRNA codons, we can determine the amino acid attached to it. Alright. So again, that gives us option B as our final answer.

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.

Here in this practice question it says, Tufson is an amino stimulator tetrapeptide having the following sequence. So these are the three-letter codes of our amino acids. So we have threonine, lysine, proline, and arginine. Here it says, write a possible sequence for the gene when we're talking about the informational and the template strand that codes for this tetrapeptide. Alright. So, what we have to do is look to see where we find each one of these amino acids based on our chart here of codons. So threonine. If we look for threonine, it's here. Lysine would be here. Proline let's see. We see proline here. Arginine. Arginine we find in 2 places, here and here. Now here we're going to talk about the number of combinations that are possible. So for Lysine oh, no. Threonine first, there are 4 codons attached to it. These 4 here. So it'd be 4. Lysine has 2 codons attached to it, 2. Proline has 4 also. And then arginine has 6. So that'd be 4 times 2, which is going to give us 8. 8 times 4 which gives us 32. And then 32 times 6 here would give us a 192 possible combinations. So again, that's that many combinations that can happen. Here we're just going to choose our own set of codons for each one of these amino acids. Again, the combination, there are 192. I'm just choosing 1 out of those 192 possibilities. So, here we're going to say our mRNA, We're going to say again our mRNA is 5 prime to our 3 prime. So we say for threonine here, let's use we'll use a ACA as one codon for threonine. For lysine, there are only 2 possibilities. Let's use the AAG one. Proline, there are 4 possibilities. Again, we could choose any of them. I'm going to use the CCU one. Let's use that one. And then, Arginine, there are 6 choices. I'm just going to choose one of them. So let's see. We'll use the CGU one. Alright. So we have our peptide, which was this, and I've just chosen one of the codons for each one of these. Next, we're going to do the informational strand. And remember, the informational strand and the mRNA strand, they're copies of each other. The only difference is since the informational strand is DNA-based, all the uracils would just be thymine. So, it's going to be the same exact thing. So we're going to have 5 prime to 3 prime. So here this would be ACA. This would still be AAG. And now it's CCT. Remember, wherever we see uracil it gets changed to thymine since we're dealing with the DNA informational strand. Then we're going to see, CGT. Again, we have a uracil, so we have to change it to Thymine. Then we have finally our template strand. Now, remember our template strand here is going to be anti-parallel to our mRNA. So it's going to be 3 prime on this end and then 5 prime on this end here. Alright. So then here, based on that, we're just going to have to write this. And remember, it's these 2 connected together because they're part of DNA. DNA is a double helix. So this would be since it's DNA, we'd say A is connected to T and G is connected to C. So this is A, so this would be T, this would be GC, TTAGC, and then since this is CGT, this would be GCA. So this is how we break down all of the different strands attached to our peptide sequence. So again, from this peptide sequence, we can figure out the number of combinations possible based on the number of codons associated with a given amino acid. Choose whatever you want to make the mRNA strand. The informational strand is the DNA copy of that. Right? So or the original strand. MRNA is just an RNA copy of it. So wherever you see a uracil, for the DNA parts of thymine. And then the informational strand and the template strands, they're complementary strands to each other, so they have to be anti-parallel. So, if this is 5 prime to 3 prime, then this has to be 3 prime to 5 prime. Since it's DNA, these are the base pairings, A with T, G with C. Do that, you'll be able to write the information for all the strands involved within this given practice question, anything that's given to you like it.

Here’s what students ask on this topic:

What are the three main steps of translation in protein synthesis?

Translation in protein synthesis consists of three main steps: initiation, elongation, and termination. During initiation, the small ribosomal subunit binds to mRNA, and tRNA carrying methionine pairs with the start codon AUG. In elongation, additional tRNAs bring amino acids, forming peptide bonds and lengthening the chain through translocation. Termination occurs when a stop codon is reached, prompting a release factor to hydrolyze the peptide chain from tRNA, leading to the disassembly of the ribosomal complex. This process is crucial for protein synthesis, involving amino acids and peptide bonds.

How does the initiation step of translation occur?

During the initiation step of translation, the small ribosomal subunit binds to the mRNA. A tRNA carrying methionine, which has the anticodon UAC, pairs with the start codon AUG on the mRNA. This forms a complex with the small ribosomal subunit. The initiation step is completed when the large ribosomal subunit joins this complex, forming the complete ribosomal assembly. This setup is essential for the subsequent steps of translation, where amino acids are added to form a polypeptide chain.

What happens during the elongation step of translation?

In the elongation step of translation, a second tRNA binds to the next codon on the mRNA within the ribosomal complex. A peptide bond forms between the amino acid carried by the first tRNA (methionine) and the amino acid carried by the second tRNA. The ribosome then translocates, moving the mRNA and tRNAs to the next codon. This process repeats, with new tRNAs bringing amino acids, forming peptide bonds, and elongating the polypeptide chain. Multiple ribosomes can translate a single mRNA simultaneously, producing multiple polypeptide chains.

What triggers the termination of translation?

Translation terminates when the ribosome encounters a stop codon (UAA, UAG, or UGA) on the mRNA. A release factor protein binds to the stop codon, prompting the hydrolysis and release of the polypeptide chain from the last tRNA. This leads to the disassembly of the ribosomal complex, including the separation of the large and small ribosomal subunits, the release factor, and the mRNA. The newly synthesized polypeptide chain is then free to undergo further folding and modifications to become a functional protein.

What role do tRNAs play in the translation process?

tRNAs (transfer RNAs) play a crucial role in translation by bringing specific amino acids to the ribosome, where they are added to the growing polypeptide chain. Each tRNA has an anticodon that pairs with a complementary codon on the mRNA. During elongation, tRNAs sequentially bind to the mRNA codons, and their attached amino acids form peptide bonds, extending the polypeptide chain. The accuracy of tRNA-mRNA pairing ensures that the correct amino acids are incorporated according to the genetic code.