Quiz 4, BICH 410 (TuTh 12:45-2), Thursday, Dec. 5, 1996
Clearly circle the letter corresponding to your answer for each problem. There are seven problems, six of which count 3 points, and one is worth 2 points, and one bonus question for 2 points. No partial credit given for multiple-choice questions.
gas constant R 8.315 J/mol-K; Faraday constant F 96.5 kJ/mol-volt
1. In the biological degradation pathway of glucose, the enzyme pyruvate kinase catalyzes this reaction:
phosphoenolpyruvate + ADP <-> pyruvate + ATP
Given the information below, calculate the standard free energy change for the above reaction.
ATP <-> ADP + Pi; G°' = -30.5 kJ/mol
phosphoenolpyruvate <-> pyruvate + Pi; G°' = -61.9 kJ/mol
glucose + Pi <-> glucose-6-phosphate; G°' = +13.8 kJ/mol
- A. -78.6 kJ/mol
- B. -92.4 kJ/mol
- C. -31.4 kJ/mol
- D. +31.4 kJ/mol
- E. +92.4 kJ/mol
2. Which of the following statements is NOT true about the sodium-potassium ATPase?
- A. This transport system is an example of primary active transport.
- B. This transporter protein is reversibly phosphorylated.
- C. This transport system creates a membrane potential with a net positive charge inside the cell.
- D. This transporter is inhibited by vanadate.
- E. This is an example of an antiport transport system.
3. An integral membrane protein will commonly be solubilized by extraction with:
- A. a solution of high ionic strength.
- B. a solution containing detergent.
- C. hot water.
- D. a buffer of alkaline or acid pH.
- E. valinomycin.
4. Which of the following types of membrane transport systems is NOT saturable with the molecule or ion that is being transported?
- A. ion channels
- B. passive transport
- C. primary active transport
- D. facilitated diffusion
- E. secondary active transport
5. Consider the transport of K+ from the blood, where its concentration is about 4 mM, into an erythrocyte that contains 150 mM K+. The transmembrane potential is about 60 millivolts, inside negative relative to outside. The free energy change for this transport process is (assume 37°C):
- A. about 10 J/mol.
- B. about 4 J/mol.
- C. about 10 kJ/mol.
- D. about 15 kJ/mol.
- E. impossible to calculate with the information given.
Answer: Actually, none of these answers is correct, because I blew it. If the question stated "inside positive relative to the outside", as I thought, then:
Gt = RT ln(C2/C1) + ZF(membrane potential)
Gt = (8.315 J/mol-K)(310 K)/(1000 J/kJ) ln 150/4 + (1)(96.5 kJ/mol-V)(.06 V)
Gt = 9.34 + 5.79 = 15.1 kJ/mol
6. (2 pts) A common secondary structure for a transmembrane protein is:
- A. a single beta strand containing hydrophobic sidechains.
- B. an alpha helix containing hydrophobic sidechains.
- C. an alpha helix containing hydrophilic sidechains.
- D. two antiparallel beta strands containing hydrophobic sidechains.
- E. a single beta strand containing hydrophilic sidechains.
7. When a bacterium such as E. coli is shifted from a warmer growth temperature to a cooler growth temperature, it compensates by:
- A. synthesizing thicker membranes to insulate the cell.
- B. putting longer-chain fatty acids into its membranes.
- C. putting more saturated fatty acids into its membranes.
- D. putting more unsaturated fatty acids into its membranes.
- E. increasing its metabolic rate to generate more heat.