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Chemistry 333Protein Structure and Function
Fall 2001Problem set #1 answer key
1.
| Amino Acid | charge at pH 2 | charge at pH 7 | charge at pH 12 |
| Glutamtic acid | 0 | -1 | -1 |
| Aspartic acid | 0 | -1 | -1 |
| Lysine | +1 | +1 | 0 |
| Arginine | +1 | +1 | +1 (50%) |
| Histidine | +1 | +1 (25%) | 0 |
| Tyrosine | 0 | 0 | -1 |
| Cysteine | 0 | 0 | -1 |
2. At pH = 7.8, the histidines will have a neutrally chargedside chain and so the polypeptide will be less soluble in H2O than at pH 5.5, where the histidines willhave a net positive charge.
3. (d), pH = 9. To solve this problem, determine the chargeof each functional group at each pH. For example, at pH=9, thecharge of the terminal carboxyl is -1, the charge on the terminalamino is 0, and the charge on the side chain is +1.
4. Valine occupies more space than alanine, so the conformationof the interior of the protein probably changes significantlywhen Ala is changed to Val. The second mutation compensates byreducing the size of another residue in the interior, replacingthe larger Ile with the smaller Gly.
5. At pH 7, Arg has a fully protonated side chain and is capableof being only a hydrogen bond donor (see Stryer, p. 33). Any aminoacid side chain capable of accepting a hydrogen bond can hydrogenbond with Arg at pH = 7. This includes E, D, N, Q, S, T, Y andH (i.e., any side chain which has a lone pair of electrons onan O or an N atom).
K and R are fully protonated at pH =7, so could not accepta hydrogen bond with another R (when they are fully protonated,they do not have lone pair electrons on an N which can acceptthe hydrogen bond). Tryptophan also can only serve as a hydrogenbond donor, so it cannot hydrogen bond with a fully protonatedarginine.
6.
A. For this problem, we want to figure out the charge of eachamino acid side chain and then add them up to get the total netcharge:
All K's are +1; there are 4 K's, and so they contribute +4 to the total peptide charge.
All R's are +1; there are 3 R's, and so they contribute +3 to the total peptide charge.
All E's are -1; there are 4 E's, and so they contribute -4 to the total peptide charge.
All D's are -1; there are 2 D's, and so they contribute -2 to the total peptide charge.
So, after totaling the K's, R's, D's, and E's, we have a net of +1 so far.
Histidine has a pKa close to our pH (pH = 7, pKa = 6.0 from the table in Stryer). Using the Henderson Hasslebach equation, we can determine that about 9% of the histidines have a positive charge; to the total contribution form histidine is 9% of +1 = +0.09
All other side chains will be neutral at pH 7; so, the totalcontribution of the side chains to the total charge is +1.09
Now let's consider the contributions from the termini.
At pH = 7, the carboxyl terminal will have a full -1 charge.
At pH = 7, we need to use Henderson-Hasselbach to determine the proportion of N termini which are charged; since the pKa of the N terminus is 7.8, using Henderson - Hasselbach we get a net charge of ~+0.86.
For overall net charge:
Side chains +1.09
Termini: -0.14
Total: +0.95
B. For a receptor to bind this peptide, its charge must becomplementary; in other words, since the peptide has a net positivecharge, you might expect the receptor to have a net negative charge.So, the receptor should contain acidic residues (acidic residues,like Asp and Glu, have a net negative charge at pH 7).
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7. We can use the Henderson-Hasselbach equation to determinethe fraction of histidines which are protonated when the pKa is7.4 and the fraction protonated when the pKa is 7.0.
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when the pKa is 7.4, the pKa = pH; so the fraction of histidinesprotonated is 0.5.
When the pKa is 7.0, we can determine the fraction of protonatedH is by:
The number of protons picked up when the pKa goes from 7.0to 7.4 is given by the difference 0.5 - 0.28 = 0.21.
So, 0.21 protons/hemoglobin molecule are picked up.
8. ELVIS is alive and well (found in ~21 sequences).
9. Answers will vary.
10. Sequences #1 and #3 should show the most similarity inalignment. I get an identity of ~40% and a score of 576 (a positivescore is good). For #1 and #2, I get a 16.9% identity with a -453score (bad). For #2 and #3, I get a 16.7% identity with a -151score (also bad).
All three proteins are serine proteases; #1 and #3 are subtilisinstaken from two different species; these two proteins are homologus.Sequence #2 is a trypsin, and is not related in sequence to thesubtilisins (though, as we'll learn later, its function is similar).In your PROSITE scan of these seqeunces, you should have detectedseveral patterns for each sequence--most importantly, #1 and #3should contain patterns from the subtilase family, and #2 fromthe trypsin family.
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