Unformatted text preview: Who Are You? ______________________________________________________________________ CHEM 340: Biochemistry I, Exam 2
October 13, 2010
Part I: Multiple Choice.
Question Answer Question Answer 1 B 11 B 2 C 12 C 3 D 13 D 4 C 14 C 5 B 15 A 6 B 16 D 7 A 17 D 8 B 18 B 9 B 19 C 10 C 20 B Part II: True or False.(10 pts.-5 pts. each)
21) __F___ An epitope can have several antigens, but a particular antigen is found on only one epitope.
An epitope is a speciﬁc portion of a larger molecule that is a ligand for an antibody or a cell. The
larger molecule is the antigen. An antigen can have several epitopes, but a given epitope is found on one
22) __F___ If one were to isolate IgG antibodies from two different people that were infected with the
same cold virus, one should expect that the amino acid sequences of the variable domains of their IgG
antibodies would be the same.
Not necessarily so. The virus is the antigen, but any part of the virus could be an epitope which is
what would be recognized by the antibody. If a different part of the virus is recognized in the two
different people, then their variable domains would also be different because sequence of the variable
domains would be complementary to the different epitopes.
23) __T___ The very low Kd values observed with antibodies and their ligands is an important component
of the immune response.
When the antibody is bound to the ligand, it targets it for destruction to prevent infection. The
antibody-ligand complex is the signal for destruction. If the antibody did not bind tightly, then the
complex would not have a long lifetime and the ligand would not be destroyed and could cause
1 24) ___F___ The fact that an antigen can have several different IgG antibodies that will bind to it is
inconsistent with antibodies being very speciﬁc for their ligands.
If different parts of the antigen are recognized by different IgG antibodies, but that speciﬁc part
is recognized by only that antibody. So an antigen can have different antibodies that recognize it, but
only a speciﬁc part is recognized by a given antibody. Part III: Other Stuff
25) (30 pts.)
____M____ It is found primarily in muscle.
____H____ It is found primarily in the blood.
____ M___ It is always greater than 90% saturated under all physiological pO2 conditions.
____ M___ It functions as a monomeric protein.
____ H___ It is involved in oxygen transport.
___ M H__ It increases the solubility of oxygen in the aqueous environment of the cell.
_____ H___ It undergoes a conformational change upon oxygen binding and oxygen release.
_____ H___ A single molecule contains 4 iron atoms.
____ M____ It has an experimental Hill constant of 1at all physiological pO2 values
_____X___ It has an experimental Hill constant of 4 at some physiological pO2 values.
____ M____ It has a theoretical Hill constant of 1.
_____ H___ It has a theoretical Hill constant of 4.
___ M H___ There is no beta structure in the protein.
___ M H___ It is capable of binding O2 tightly in vivo.
____ H____ It is capable of binding O2 relatively weakly in vivo.
___ M H___ It binds CO more tightly than it binds O2.
____ H____ It is allosteric.
____ H____ It forms new salt bridges when it releases oxygen.
___ M H___ It gives fresh raw meat its red color.
____ X____ It catalyzes a reaction between O2 and Fe2+
____ H____ In a living organism, its O2 afﬁnity is pH dependent
_____ H___ It binds both CO and CO2.
____ X____ It transports protons to muscle tissue.
____ H____ The pCO2 affects O2 afﬁnity.
_____ X___ It has an approximate molecular weight of 150,000 daltons. 2 Who Are You? ______________________________________________________________________
26) (15 pts.) The enzyme Gitoffmacaze has one binding site and is known to bind only to the protonated
form of the drug Ahwannakry with a Kd of 45µM. The drug only has one ionizable group and it has a
pKa of 8.2. a. (10/15) ⎡ A− ⎤
pH = pK a + log ⎣ ⎦
[ AH ]
⎡ A− ⎤
7.4 = 8.2 + log ⎣ ⎦
[ AH ]
⎡ A− ⎤
−0.8 = log ⎣ ⎦
[ AH ]
⎡ A− ⎤
0.158 = ⎣ ⎦
[ AH ]
The ratio of A- to AH is 0.158 to 1, so 1/1.158 (86.4%) of the molecules is in the AH
(protonated) form. b. (5/15)
The ligand is the AH form so we just need to know how much of that form is present. From Part a,
we know that at this pH 86.4% will be in the AH form, so: 0.864 × 120µM = 103.68µM
and θ= 103.68µM
103.68µM + 45µM Therefore, 70% of Enzyme X would have the drug bound at pH 7.4 and since the total enzyme
concentration is 70 nM, 70% of that would be: 0.70 × 70 nM = 49 nM 3 BONUS SECTION (Optional)
For questions 27-30, CHOOSE ONLY ONE to answer for extra credit. If you answer more than
one, only the ﬁrst one will be graded. Write your answer on the back of this page.
27) (10 pts. max.)
The shape of the curve indicates that the release of oxygen occurs cooperatively as the pH
increases. Myoglobin only has one binding site for oxygen so its binding mode can’t be cooperative.
Therefore the cooperativity that is being observed has to be due to another process. We know that
extremes in pH can lead to protein denaturation and that protein denaturation is a cooperative process
as well. Furthermore, since protein structure is required for function, if Mb loses structure, it will also
lose its function. So even though it appears that oxygen releases is occurring in a cooperative way, what
really is going on is the protein is unfolding and when it unfolds, it can’t bind oxygen. 28) (10 pts. max.)
Helix 1 and curve B, helix 2 and curve A, helix 3 and curve C. The three helices are the same
length, so that can’t factor into the differences. At pH 5, all three have positive interactions at their
termini with the dipoles of their helices having negatively charged residues at the N-terminal end and
positively charged residues at the C-terminal end (His will carry a positive charge at pH 5).The only
signiﬁcant difference is the number of glycine residues. Glycines introduce ﬂexibility into helices and the
more glycines present the less stable the helix would be. As for the melting curves, the less stable the
helix, the lower its Tm because it takes less thermal energy to cause structural disruption and the more
the curve will be shifted to the left. Based on this logic, Helix 3 would be the most stable because it has
no glycines and its melting curve would be farthest to the right (curve C), Helix 2 would be the least
stable because it has the most glycines of the three and its melting curve would be farthest to the left
(curve A). Helix 1 would be the middle curve because its glycine count is in between that of helices 1
and 3. 29) (10 pts. max.)
a. To make a protein more stable, one would try to increase the number of productive or positive
interactions compared to its native structure. At pH 3-5, Asp would be negatively charged because of
its low pKa and Lys would be positively charged because of its high pKa. In the native structure, the
Tyr-224 and Asp-188 interaction is through a hydrogen bond. An ionic interaction between two
residues would be stronger than a hydrogen bond interaction.
b. The size difference in Lys and Tyr may be the reason. Lys has a much longer side chain than Tyr and it
may simply be too large to be accommodated at position 224. 4 Who Are You? ______________________________________________________________________ 30) (5 pts. max.)
Under nitrogen, Hb is probably in the T-state. However, when the crystal was moved, it may have
been exposed to oxygen. Since oxygen is a gas, it could have diffused into the crystal and bound to the
deoxy Hb. This would have induced the T to R conformational change, which involves a signiﬁcant
movement in the protein. This movement probably caused the crystal to crack. 5 ...
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This note was uploaded on 12/04/2011 for the course CHEM 340 taught by Professor T.baird during the Fall '11 term at S.F. State.
- Fall '11