Chemistry 333
Protein Structure and Function
Fall 2001

Problem 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

-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 charged side chain and so the polypeptide will be less soluble in H2O than at pH 5.5, where the histidines will have a net positive charge.

3. (d), pH = 9. To solve this problem, determine the charge of each functional group at each pH. For example, at pH=9, the charge of the terminal carboxyl is -1, the charge on the terminal amino is 0, and the charge on the side chain is +1.

4. Valine occupies more space than alanine, so the conformation of the interior of the protein probably changes significantly when Ala is changed to Val. The second mutation compensates by reducing the size of another residue in the interior, replacing the larger Ile with the smaller Gly.

5. At pH 7, Arg has a fully protonated side chain and is capable of being only a hydrogen bond donor (see Stryer, p. 33). Any amino acid side chain capable of accepting a hydrogen bond can hydrogen bond with Arg at pH = 7. This includes E, D, N, Q, S, T, Y and H (i.e., any side chain which has a lone pair of electrons on an O or an N atom).

K and R are fully protonated at pH =7, so could not accept a hydrogen bond with another R (when they are fully protonated, they do not have lone pair electrons on an N which can accept the hydrogen bond). Tryptophan also can only serve as a hydrogen bond donor, so it cannot hydrogen bond with a fully protonated arginine.

6.
A. For this problem, we want to figure out the charge of each amino acid side chain and then add them up to get the total net charge:

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 total contribution 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 be complementary; in other words, since the peptide has a net positive charge, 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).

7. We can use the Henderson-Hasselbach equation to determine the fraction of histidines which are protonated when the pKa is 7.4 and the fraction protonated when the pKa is 7.0.

when the pKa is 7.4, the pKa = pH; so the fraction of histidines protonated is 0.5.

When the pKa is 7.0, we can determine the fraction of protonated H is by:

The number of protons picked up when the pKa goes from 7.0 to 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 in alignment. I get an identity of ~40% and a score of 576 (a positive score is good). For #1 and #2, I get a 16.9% identity with a -453 score (bad). For #2 and #3, I get a 16.7% identity with a -151 score (also bad).

All three proteins are serine proteases; #1 and #3 are subtilisins taken from two different species; these two proteins are homologus. Sequence #2 is a trypsin, and is not related in sequence to the subtilisins (though, as we'll learn later, its function is similar). In your PROSITE scan of these seqeunces, you should have detected several patterns for each sequence--most importantly, #1 and #3 should contain patterns from the subtilase family, and #2 from the trypsin family.