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I was asked the following question in a myPAT take-home test,

Find the maximum number of tripeptides can be synthesized by using three different amino acids.

Let those amino-acids (Z) be A, B, C.

So, taking three Z's at a time, we've only one set, ZZZ. So, total 3 tripeptides.

Next, we can take two same and one distinct (eg. AAZ),

For AAZ type, we've four sets, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

Finally, take 3 distinct Z, for that we get total of 6 (i.e., 3!) tripeptides.


I am unsure if I've done it in a valid way, as answer given is 18. Where am I wrong?

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    $\begingroup$ 1. This question is phrased very poorly. Without reading your working, I would have answered $20 \times 19 \times 18 = 6840$. The question wording isn't your fault, but it's very hard to answer a question that doesn't have a clear meaning. Even the given answer, 18, isn't helpful unless there's some explanation of how they arrived at this magic number. 2. This is more of a maths question, not chemistry. Maybe post it there with some extra clarification. $\endgroup$ – orthocresol Apr 11 at 15:58
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    $\begingroup$ @porphyrin - I'd agree that is how the book answer is found. However that seems wrong to me. For a three chain peptide, let's say that the carboxylic acid is on the third group as written. Then bcc isn't the same as ccb. In bcc the c has a unbonded carboxylic acid, but in ccb the b has a unbonded carboxylic acid. So they are two different peptides. The correct answer should be 27. $\endgroup$ – MaxW Apr 11 at 17:42
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    $\begingroup$ (1) because polypeptides have C- and N-termini, AAX and XAA are not the same thing. (2) I can't answer the question about optical activity without knowing the context in which the question is asked. I would suggest that you can't count different enantiomers (they would be different amino acids), but I am not the author of this question. (3) if you guys can't reverse engineer the book solution, then that really suggests that there's no sensible answer. I could accept $3^3 = 27$, but I would still find it extremely arbitrary because it's based on a specific interpretation of the question. $\endgroup$ – orthocresol Apr 11 at 17:56
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    $\begingroup$ I should also be clear that it is not your Stack Exchange question that is poorly phrased, but rather the question given to you that was poorly phrased. Sorry if that was not clear the first time. This whole debate in the comment section (which I am also part of) is not great, though. @Maurice your approach double-counts AAA, BBB, and CCC. 30 - 3 = 27 bringing you in line with everybody else. $\endgroup$ – orthocresol Apr 11 at 18:01
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    $\begingroup$ @orthocresol: I understood what you've said, by the first comment itself. You're always vivid in your words :). It seems that, with the help of everyone, I've got the point which I was missing. By the next time, I'll try to avoid asking these kind of questions. Thanks @all! $\endgroup$ – Rahul Verma Apr 11 at 18:18
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There is a misconception that runs through the questions, the answer key and the comments: For tripeptides, the peptide linkage is directional (i.e. the tripeptide has an amino terminal and a carboxy terminal), so ABC is different from CBA.

To get the number of tripeptides combined from three types of amino acids, you have three choices for the first, three independent choices for the second and another three independent choices for the third position, for a total of 3 x 3 x 3 = 27 possibilities.

[OP] I am unsure if I've done it in a valid way, as answer given is 18.

The answer key is incorrect. I have no way of reading the question that 18 would be the answer, even considering the poor wording of the question.

[OP] For AAZ type, we've three set, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

There are three choices for Z, two remaining choices for A, and three positions for Z (AAZ, AZA, ZAA). So, total of 3 x 2 x 3 = 18 tripeptides containing two different amino acid types of the initial set of three types.

[Orthocresol] This question is phrased very poorly. Without reading your working, I would have answered 20×19×18=6840.

This answers how many different tripeptides made from (20) standard amino acids contain no amino acid twice. The question, certainly poorly stated, says "synthesized by using three different amino acids". So your reagents are a set of three amino acids types, and you are making tripeptides from them.

[Maurice] With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.

That is indeed a math problem, not a biochemistry problem. You can make AAA from A and B, but also from A and C, so you counted AAA twice (and BBB and CCC). So your answer overshot by 3.

[Rahul Verma] Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides?

The linkage does not introduce new chiral centers (different from e.g. combining sugars), so each sequence XYZ designates a single molecule (given that you start with enatiomerically pure amino acids).

[porphyrin] The list is aaa , aab , aac , aba , abb , abc , aca , acb , acc , baa , bab , bac , bba , bbb , bbc , bca , bcb , bcc , caa , cab , cac , cba , cbb , cbc , cca , ccb , ccc which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.

The list is correct, but there is no reason to remove palindromes (9 in the list) or divide non-palindromes by two (18 in the list).

Update

I missed all the comments that came in while I was writing this. They now reflect the misconceptions and how these can be resolved; what is not resolved is the poorly worded question and its incorrect answer key - maybe someone who has the original question can take on the task of sending an erratum.

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  • $\begingroup$ very nice summary. $\endgroup$ – MaxW Apr 11 at 19:15
  • $\begingroup$ @MaxW Sorry I did not mention your comment from an hour ago. I started writing the answer, then had to step out, and did not check for more comments when I came back to post it. $\endgroup$ – Karsten Theis Apr 11 at 19:17
  • $\begingroup$ Doesn't bother me at all. I really like the fact that you gave the answer, and then pointed out mistakes in some of the comments. The mathematical use of permutations and combinations can get tricky. Knowing why some of the other methods didn't work helps in understanding. // LOL -- "Results! Why, man, I have gotten a lot of results. I know several thousand things that won't work." Thomas A. Edison $\endgroup$ – MaxW Apr 11 at 19:34
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    $\begingroup$ Ugh. I missed it. Re:"9 x 9 x 9 = 27 possibilities." It should be $ 3 \times 3 \times 3 = \pu{27 possibilities}$. $\endgroup$ – MaxW Apr 11 at 21:19

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