Protein synthesis.

[TieuBachHo]

Suppose that you have a protein synthesis system that is actively synthesizing a protein designated A. Furthermore, you know that protein A has four trypsinsensitive sites, equally spaced in the protein, that, on digestion with trypsin, yield the peptides A1, A2, A3, A4, and A5. Peptide A1 is the amino-terminal peptide, and A5 is the carboxyl peptide. Finally, you know that your system requires 4 minutes to synthesize a complete protein A. At t = 0, you add all 20 amino acids, each carrying a 14C label.

Q1) At t = 1 minute, you isolate intact protein A from the system, cleave it with trypsin, and isolate the five peptides. Which peptide is most heavily labeled?
a)A1
b)A5
c)A2
d)A4

Q2) At t = 3 minutes, what will be the order of labeling of peptides from greatest to least?
a)A1 > A2 > A3 > A4
b)A5 > A4 > A3 > A2
c)A1 > A2 > A3 > A4 > A1
d)A1 = A2 = A3 = A4

Answer for 1 is B and 2 is B. The thing I don't understand is if protein systesis is from amino-terminal to carboxyl-terminal. Then A1 should be synthesized first right? Therefore, it must be heavily labeled. How could A5 be at t = 1 minute😕
And for question 2 ... at t = 3 minutes, at least 3 peptides are completed, then they must be equal in term of labelling. B is just doesn't make any sense to me ... Thank you for all your helps.

My reasoning must be off somewhere🙁

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[capn jazz]

Do you think it's possible that once the trypsin has cleaved the protein into fragments, they are open to degradation and so the most RECENTLY synthesized peptides are most abundant because they have had less time to be degraded?

 

[TieuBachHo]

Thank you for your reply,
The problem states that protein A has equally spaced 4 trypsinsensitive sites. I am not sure that is for a complete protein A or for an incomplete protein A at any given time.

I approached this problem thinking that it is for a complete protein A. Here is my reasoning.

To the answer of question #1. At t = 1 minute, you have an incomplete polypeptide A right? If the protein is synthesized from A1 to A5 then the polypeptide A is basically A1 and a portion of A2. <=== Obviously, this must not be right 🙂

To the anser of question #2 makes more sense with my approach when they only have 4 peptides but going the A5 to A2 instead of A1 to A4. 1 peptide is short due to its incompleteness. <=== This would contradict with question #1 though. 🙁

I am just dumbfounded 😕😕😕

PS: It doesn't say anything about degradation but that would make more sense I think

 

[capn jazz]

You're missing this part: "At t = 1 minute, you isolate intact protein A from the system" meaning you get the entire peptide A, containing all 5 peptides that make it up.

 

[Freakyp]

You're missing this part: "At t = 1 minute, you isolate intact protein A from the system" meaning you get the entire peptide A, containing all 5 peptides that make it up.

Yes, they screwed up. First off with the intact protein which implies a complete one. However, it says it takes 4 minutes.
Amino to Carboxyl. I reread the question, I think they assume that the system was producing protein BEFORE. If this is like bacteria, then that means it naturally has its OWN non-radiactive Amino acids. As time proceeds, the downstream ones will get incorporated at more than the early ones. So that's why they get their A5 to A1 range. Yea, as neal said, it's too specific and illogical to actually be on the mcat. I took Biochem 3 years ago and can't believe I completely forgot a basic tenet, good catch neal.

HTH

 

[TieuBachHo]

Freakyp,

The problem has another question (a 3rd question) that asks, "what does this experiment tell you about the direction of protein synthesis?" And the answer is, "Synthesis is from the amino terminus to the carboxyl terminus." This just throws me completely off 😡

By the way, thank you all!

Edited:
Ahhhhh .... maybe you are right! my interpretation was wrong. I keep picturing the cellular protein synthesis that goes from the amino terminus of the AA to the caboxyl end of the polypeptide. They probably mean the amino terminus of the polypeptide to the carboxyl terminus of the AA. That would make more sense. 🙂)

 

[TieuBachHo]

Now that would leave me another question though,

Why the order for question #2 is A5 > A4 > A3 > A2. Wouldn't it be A5 = A4 = A3 > A2 since all 3 peptides A5, A4, A3 are completely synthesized under 3 minutes?😕😕😕

 

[0Complications]

Now that would leave me another question though,

Why the order for question #2 is A5 > A4 > A3 > A2. Wouldn't it be A5 = A4 = A3 > A2 since all 3 peptides A5, A4, A3 are completely synthesized under 3 minutes?😕😕😕

No, because the protein is formed from the C-terminus to the N-terminus and you were given " Finally, you know that your system requires 4 minutes to synthesize a complete protein A", so it has not yet synthesized the complete protein at 3 minutes and the protein has to be synthesized in order from the C-terminus (A5), so you can't get an A3 without an A4 and you can't get an A2 without an A4 and an A3, so the order has to follow A5 > A4 > A3 > A2 and if it were to go for 4 minutes it would be ...A5 > A4 > A3 > A2 > A1.

It seems like the gap is that you weren't assuming it has to be made in order. Imagine that you are constructing a house, and the foundation is the C-terminus, well you aren't going to build the foundation, then put a roof on it, you need to build it in steps, just like a protein needs to be built in steps, based on the mRNA sequence, which would be equivalent to the housing blueprints. Order matters.

 

[TieuBachHo]

OComplications,

I thought the question was asking Labeling order from greatest to least. Doesn't this mean when you say A5>A4>A3>A2 then A5 must have the most 14C labeling and so on so forth? Not the order of synthesis?

 

[neil100]

Suppose that you have a protein synthesis system that is actively synthesizing a protein designated A. Furthermore, you know that protein A has four trypsinsensitive sites, equally spaced in the protein, that, on digestion with trypsin, yield the peptides A1, A2, A3, A4, and A5. Peptide A1 is the amino-terminal peptide, and A5 is the carboxyl peptide. Finally, you know that your system requires 4 minutes to synthesize a complete protein A. At t = 0, you add all 20 amino acids, each carrying a 14C label.

Q1) At t = 1 minute, you isolate intact protein A from the system, cleave it with trypsin, and isolate the five peptides. Which peptide is most heavily labeled?
a)A1
b)A5
c)A2
d)A4

Q2) At t = 3 minutes, what will be the order of labeling of peptides from greatest to least?
a)A1 > A2 > A3 > A4
b)A5 > A4 > A3 > A2
c)A1 > A2 > A3 > A4 > A1
d)A1 = A2 = A3 = A4

Answer for 1 is B and 2 is B. The thing I don't understand is if protein systesis is from amino-terminal to carboxyl-terminal. Then A1 should be synthesized first right? Therefore, it must be heavily labeled. How could A5 be at t = 1 minute😕
And for question 2 ... at t = 3 minutes, at least 3 peptides are completed, then they must be equal in term of labelling. B is just doesn't make any sense to me ... Thank you for all your helps.

My reasoning must be off somewhere🙁

This is to the producer of this question. Logically speaking, there is a glaring contradiction in this question.

In the background, it says that 4 minutes are required to make a complete protein A. Then 1st question says you have intact protein A at one minute. You can't have an intact protein A because it's not protein A. The protein would not have all of the polypeptides needed to be called or reffered to as protein A. So, it won't have all of the four sensitive trypsin digestion sites. So, by letting the synthesis go on for a minute isn't going to give you nice five cleaved polypeptides since there won't be those 4 trypsin sensitive sites.

Btw, Proteins ARE MADE and NAMED from N to C terminal.
Analogous to DNA or RNA synthesis occuring in 5' to 3' AND READ in that order.

Also, the real MCAT exam will not have questions that do not make sense logically. Don't worry about questions like these. Ill give you a question on protein synthesis if you want that would be more like the MCAT material. It will make you use the concepts of protein synthesis and will make logical sense.

 

[csau]

Hello,

Despite what many of you think, this question is very logical. However, whether or not a question like this would actually be in the MCAT is debatable. I am very confident in my interpretation of the question so please hear me out.



First off, the passage mentions that the system is actively producing proteins. Furthermore, it mentions that at t = 1, an intact protein can be obtained. This means that protein synthesis has been occurring even before 14C was added. The other questions of how heavily labeled a peptide is also indicates that there were unlabeled amino acids previously in the solution.

Now that we've established the situation, we can approach the questions.



Q1) At t = 1 minute, you isolate intact protein A from the system, cleave it with trypsin, and isolate the five peptides. Which peptide is most heavily labeled?

They keyword here is intact. Any new peptides being synthesized will not be completed (since it requires 4 minutes to complete). Assuming that translation was occurring at a constant rate, peptides that were 3/4 completed at t = 0 would be complete and intact by t = 1. Therefore, the only possible radioactive labeling could occur on the whole of A5 and only part of A4 (because the last 1/4 of the peptide contains the whole of A5 and only part of A4).



Q2) At t = 3 minutes, what will be the order of labeling of peptides from greatest to least?

Following the same thought process as before, only peptides that were translated 1/4 of the way would be complete and intact after 4 minutes. This is not the only case, though. Any peptide that was past 1/4 completed at t = 0 would be intact by t = 3, e.g. a peptide that was 2/4 completed at t = 0, or a peptide that was 3/4 completed at t = 0. Considering this, at t = 3 minutes we would have a mix of proteins that were labeled a little bit (labeled on A5), labeled moderately (labeled on A4 and on A5), and so on. Now, we can use the information to find the right answer.

A5 > A4 > A3 > A2

A different, quicker approach would be to consider the peptide that would be labeled the least. The value of 1/4 is 0.25. Following the solution above would reveal that the first 25% of the protein sequence would be unlabeled, which includes the whole of A1 and part of A2. The answer that puts > A2 > A1 at the end would be the correct answer.



One last thing to remember is that translation is indeed from N-terminus to C-terminus.

csau

 

[TieuBachHo]

CSAU,
👍 ... Brilliant analysis which I also think you are correct. Now that makes sense when the problem states " ... you have a protein synthesis system that is actively synthesizing a protein designated A ..." Thank you very much!!! 😀

On a side note: "The answer that puts > A2 > A1 at the end would be the correct answer."
... > A1 wouldn't be correct because if any synthesizing polypeotide has yet completed A1 wouldn't yield an intact protein A. On the other hand, if protein A is intact then peptide A1 must have been completed before adding 14C under 3 minutes. The question actually was asking for the largest to least, so A1 wouldn't be included because it has no labeling.

Thank you again and I actually like this problem! LOL 😀

 

[csau]

You're welcome.


If the A1 peptide has no labeling, wouldn't consider it to have the least amount of label?

 

[Fifty 3rds]

Csau, I have to double the props to you. That was a great answer. I was struggling with this. Even if a question like this doesn't show up on the MCAT, it is important to examine why most of us were getting this problem wrong. Artifacts in our thinking. We have to examine everything in the question and not presuppose anything!

Seriously, great catch csau!

 

[neil100]

Yeah good explanation csau.

When I first read the problems and answered them I got them right. But, I would have chosen the right answers simply because of educated guessing and the nature of the question/choices.

I think the wording of this question can be a little better, no doubt. I mean there isn't anything intrinsically difficult about the question but it's just that the wording is a bit confusing. I think they only should change that word "intact". This is the word that also confused me a bit and other people. Otherwise, everything else makes sense and is not dubious. They could say something like "at t=1, a complete or fully synthesized (or something which clearly suggests that the protein has undergone complete synthesis at t=1) protein A is extracted and then....".

It was because of this word, me and others were saying the question doesn't seem to be logically sound.
Or even say that at t=3 minutes of protein synthesis, 20 14c labeled amino acids are added to the system. And then ask at t=4 blah blah blah... you know". This conveys the same message but in a heck of a lot clearer way. People who don't know that protein is synthesized from N to C terminus will still get it wrong regardless. But, the crux of this question is like csau said is knowing the direction of protein synthesis.


And yes, like csau and I have said, protein synthesis IS from N to C terminus and READ in that fashion. Analogus to DNA/RNA synthesis from 5' to 3'.

Also, I don't understand the whole tricking part (like playing the game of semantics, giving tougher calculations, etc.)that these private companies try to do. I mean the real exam that I took and the AAMC practice test questions are sometimes tricky (due to knowledge, or making connections, etc.) but they word questions very nicely. You know exactly what they are asking 99.9% of the time. Anyways, good catch csau.