possible peptide formation and ratio of acid to base

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agw1200
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possible peptide formation and ratio of acid to base

Post by agw1200 » Fri Oct 03, 2008 7:58 pm

I have two homework problems that are stumping me....

How many different peptides of 15 residues can be made from the 20 common amino acids?

-is this a simple 15^20 caluclation or am I overlooking something. 15^20 = 3.3*10^23. This number seems quiet large.

Calculate the ratio of conjugate base to acid form for each of the ionizable groups of glutamic acid in a medium of pH 4.0

-when calculating the ionizable groups of a amino acid, do you condisder the R-group or just the carboxyl and ammine groups?

pH = pka + log [base]/[acid] since pH < pka (4 < 4.5) does this mean both of the COO- groups (r and carboxyl groups) are in the form COOH and the ammine group has a (+) charge?

therefore, 4 = 4.5 + log [NH3+] \ 2 [COOH] ? or is this problem more simple than I am making it.... and the ratio is 1:2 (+):neutral ionizable groups?

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mith
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Post by mith » Fri Oct 03, 2008 8:18 pm

by your logic if there was one residue, it would be 1^20? Meaning there's only one outcome?
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blcr11
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Re: possible peptide formation and ratio of acid to base

Post by blcr11 » Sat Oct 04, 2008 12:26 am

1 residue: 20 = 20^1 possible sequences
2 residues: 20x20 = 20^2 possible sequences
3 residues: 20x20x20 = 20^3 possible sequences
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15 residues: ?

It is a large number, just not 15^20. I think you had the right idea, but you didn’t think it through from the beginning.

You consider the ionizable groups separately. There are three pKa’s, one for the alpha-carboxylate (pKa1 = 2.2), one for the alpha-amino group (pKa2 = 9.7) and one for the gamma-carboxylate (pKa3 = 4.3). The Henderson-Hasselbalch relation for each group is:

pH – pKa1 = log[alpha-COO-]/[alpha-COOH]

pH – pKa2 = log[alpha-NH2]/[NH3+]

pH – pKa3 = log[gamma-COO-]/[gamma-COOH]

What you should find is that two out of three of these equilibria heavily favor one of the ionization states, while the other has a significant proportion of both ionization states.

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