Download Chemistry Problem Set 2: Amino Acids and Peptides and more Assignments Biochemistry in PDF only on Docsity!
1) Draw the complete side chain structure of each of the amino acids as they occur at pH 7.0, including the 3-letter and 1-letter codes (i.e. Ala and A for alanine). Be sure that you can categorize whether an amino acid is non-polar, polar-uncharged, acidic or basic. Next, write the proton dissociation reaction demonstrating the acid-base properties of the amino acid side chains Arg, Asp, Cys, Glu, His, Lys, Ser, Thr and Tyr, which have R-group pK (^) as of 12.5, 3.9, 8.3, 4.3, 6.0, 10.5, 13, 13, 10.1, respectively. FYI: you are going to need to learn these structures, so here is a trick that might help. After you have spent some time studying the structures, try to write all 20 amino acids down from memory. Then concentrate on the amino acids that you missed and try again.
You can refer to your notes or text to check the 20 commonly occuring amino acid structures.
Cα
HN
H 2 N NH 2 +
Arg (^) Cα
HN
H 2 N NH
Cα
HN NH+
Cα
HN (^) N
His Cα
NH 3 +
Cα
NH (^2)
Lys
Cα
HS
Cα
Cys
for pKa values above pH 7.
Cα
HO
Cα
Ser
Cα CH HO
Cα CH
Thr
Cα
OH
Cα
O-
Tyr
for pKa values below pH 7.
Cα
Cα
O O
O O- -
Cα
OH O
Asp
Cα
O OH
Glu
2) Starting from the acid form of a 0.1 M solution of the amino acid histidine (the carboxyl and imidazole R-group are protonated) draw the complete titration reaction showing the chemical structure after addition of each of 3 equivalents of OH-. What is the net charge of the molecule and the pH of a solution of the amino acid at 0, 0.5, 1, 1.5, 2, 2.5 and 3 equivalents of strong base added?
H
N
NH
H
pKa ~9.5^ pK^ a^ ~ 2.
pKa = 6.
1st eq. OH-
(^) H 3 N COO -
H
N
NH
H
2nd eq. OH -
(^) H 3 N COO -
H
N
NH
H 2 N COO -
H
N
NH
+2 charge +1 charge
isoelectric zero net charge -1 charge
Equivalents of OH-
pH 1.5 2.0 4.0 6.0 7.75 9.5 11. net charge 2.0 1.5 1.0 0.5 0.0 -0.5 -1.
Answers given assume that you have used the pKa listed above (and mentioned in class as a good average number to know for all amino acids) for the carboxy and amino (2.0 and 9.5, respectively). Many of you correctly found the exact pKa’s for Histidine (1.8 and 9.2, respectively), and that’s okay as well. However, the numbers above will be slightly different.
The 3 half equivalent points are the 3 respective pKa values. Make sure you understand why.
The 1.0 equivalent point can be estimated as the average of the pKa at 2.0 and the pKa at 6.0. The 2.0 equivalent point can be estimated as the average of the pKa at 6.0 and the pKa at 9.5.
The 0.0 equivalent point is calculated identically to how it was done for acetic acid in Problem set #1. Likewise, the 3.0 equivalent point for His is done the same as for acetate (1.0 equivalents added to acetic acid). Remember, here you need to use Kb (for the amino group of His) and the dissociation of a weak base. Please let me know if you need help with this.
4) In certain positions of the enzyme ribonuclease A the amino acid proline is involved in a cis-peptide bond. For the proper folding of this protein an additional enzyme called cis-trans proline isomerase is needed to catalyze the isomerization of the trans to cis- peptide bond of proline. Explain why this enzyme is necessary.
Cα N
Cα
O
O
Cα N
O
Cα
O
trans cis
∆ Go^ = +3.4 KJ/mol E (^) a = +88 KJ/mol
Proline isomerization
There is a relatively high kinetic barrier (or activation energy Ea ) of 88KJ/mol for the interconversion of a trans to a cis-peptide bond for each proline. Once all the cis- proline was used up from the equilibrium mixture, the protein could only be folded into the correct structure only after another trans-proline was formed by this very slow isomerization reaction into a cis-proline. This step is very slow due to a high activation energy. With a special isomerase enzyme that lowers the activation energy for this conversion, the rate of trans to cis is much faster, and no longer is the rate-limiting step for proper protein folding.
The barrier between the trans and cis-conformation is large due to the partial double bond character of the peptide bond. This double bond character also causes the planar arrangement of “the peptide plane”.
Cα N
Cα
O
H
Cα N
Cα
O-
H
5) Using the web sites introduced in class search for the protein sequence of the α- and β- subunits of hemoglobin (http://www.ncbi.nlm.nih.gov/entrez/query.fcgi). Then perform the appropriate protein sequence alignments (http://www.ncbi.nlm.nih.gov/BLAST/) in order to answer the following questions. Describe the homology between the human alpha and beta hemoglobin subunits in terms of percent identity and also in terms of percent similarity. Now describe the protein homology in the same way between a human alpha hemoglobin chain and the alpha chain of another animal of your choice.
Your answers will vary depending on which sequences you chose.
Here are some results I obtained:
Human alpha-2 (gi:22671717) aligned to human beta (gi:23268683) was 42% identical and 62% similar. The output of BLAST has a value computed called positives. This is the same as % similarity that we discussed in class.
Human alpha-2 (gi:22671717) aligned to bovine alpha (gi:13634094) was 87% identical and 91% similar. The output of BLAST has a value computed called positives.
