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How many different kinds of polypeptides, each composed of 12 amino acids, could be synthesized using the 20 common amino acids?

The book's answer is $20^{12}$.

However, I disagree. This result distinguishes the following two polypeptides even though they are exactly the same polypeptide:

(Met)x11-Glu

and

Glu-(Met)x11

Am I right? How would I otherwise answer this question? My background in probabilities is very weak.

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3 Answers 3

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Judging from what you have said, I assume that combinatorics is not a problem to you.

I believe your problem is that you think Glu-(Met)x11 is equivalent to (Met)x11-Glu, just turned around. However, that is not a correct mindset. Amino acids are not symmetrical molecules, therefore reversed linear combination does not create a turned-around (be it chiral or not) form of a similar amino acid. My guess is that you write amino acid names in consecution when representing peptides. This method has a small problem that it does not incorporate directionality of molecules. Drawing the whole molecule may be a solution, but that is too much chore.

My solution is to add a small mark (perhaps an arrow, or a > mark) to represent a specific side of the amino acid molecule (amine group or carboxyl group), so that mirroring the sequence (so that the mark would be at the "other" side of each amino acid) would actually create a peptide molecule different from reversing the sequence (so that the mark is still at the "normal" side of each amino acid). If you think that doesn't look good, you can draw a frame around each amino acid, where one side of the amino acid has a dent and the other side has a bump (to represent the amine and carboxyl groups) so that only a bumped side can connect to a dented side. Mirroring the sequence would also reverse the positions of the bumps and dents, thereby creating a different peptide molecule than merely reversing the sequence while keeping the positions of the bumps and dents in place.

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  • $\begingroup$ Tricky indeed. Very detailed and visual explanation. I forgot about the difference between the C- and N- terminus. Thanks. $\endgroup$
    – yolo123
    Commented Jan 18, 2015 at 4:57
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Think of the amino acid choices as 12 seats. In the first seat, we have 20 choices. In the next seat, we have 20 choices, and this continues. Therefore, we have that $$ \underbrace{20\cdots 20}_{12\text{ times}} = 20^{12} $$ For your question about the the polypeptides, (Met)x11-Glu is not the same as Glu-(Met)x11, order matters.

Up to this point, we have considered the sequence of amino acids that make up a protein. This is the first of several levels of protein structure, illustrated in Fig. 4.4. The sequence of amino acids in a protein is its primary structure. The sequence of amino acids ultimately determines how a protein folds [1].

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The individual amino acids are not symmetrical. Thus, your two example peptides are not chemically equivalent, and the $20^{12}$ figure is correct.

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