fall 10 - Who Are You CHEM 340 Biochemistry I Exam 2 ANSWER...

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Unformatted text preview: Who Are You? ______________________________________________________________________ CHEM 340: Biochemistry I, Exam 2 ANSWER KEY 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 antigen. 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 infection. 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 = 0.70 103.68µM + 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.

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