Cell Biology, BIO 315

Exam II, Fall 2000

Please read each question carefully and ask the instructor for a clarification, if necessary. Multiple choice questions are worth four points and should be answered on the IF-AT form. A correct answer on the first try receives full credit. A correct answer on the second try receives two points. A correct answer on the third try receives one point.

 

  1. What is the role of chlorophyll in Photosystems I and II? (3 pts)

    2. To absorb light which excites an electron. Chlorophyll donates this excited electron to the electron transport chain.
  2. What does a kinase enzyme do? (3 pts)

    3. Adds a phosphate group.

     

    For questions 3-7. Neurons are highly polar in that one end of the cell is involved in receiving stimuli (dendritic end) and the other end is involved in sending stimuli (axonal end). Therefore, different proteins are localized in the plasma membrane at each of these ends. Researchers studying the polarity of neurons in the model organism C. elegans discovered the ACR-1 protein, an integral membrane protein that is normally only found in the plasma membrane at the dendritic end of neurons.

  3. Which of the following reasons is NOT a valid explanation for why ACR-1 is normally asymmetrically distributed in neurons?
  1. ACR-1 could be bound to the membrane skeleton.
  2. ACR-1 could be a very large protein that cannot freely diffuse.
  3. ACR-1 could be "fenced in" by another molecule.
  4. The transporter that moves ACR-1 across the membrane is only found at the dendritic end.

 

  1. Researchers later discovered that the ACR-1 protein functions as a Na+ channel that opens when a certain molecule called acetylcholine binds to its extracellular domain. When acetylcholine is not present, the channel is closed. Based on this information and your knowledge of channels, you can conclude that transport of Na+ through ACR-1 is:
  1. Not regulated, bidirectional, passive.
  2. Ligand-gated, unidirectional, active.
  3. Voltage-gated, bidirectional, passive.
  4. Ligand-gated, bidirectional, passive.

 

 

 

  1. What factor(s) will determine what direction Na+ will be transported across the membrane via ACR-1? Be specific. (6 pts)

    2. The concentration (chemical) gradient and the membrane potential (electrical gradient).

     

     

  2. The enzyme acetylcholinesterase binds its substrate, acetylcholine, and breaks it down into the products acetyl and choline. This enzymatic reaction is necessary to decrease the concentration of acetylcholine, so that the ACR-1 channels can close. Researchers identified a strain of worms that make a mutant form of acetylcholinesterase. The mutant enzyme has a lower Km than that of the normal enzyme. Which of the following could be true in these mutant worms:
  1. acetylcholinesterase has an increased affinity for acetylcholine.
  2. acetylcholinesterase has a decreased affinity for acetylcholine.
  3. The ACR-1 channel is open more often compared to wild-type worms.
  4. The ACR-1 channel is closed more often compared to wild-type worms.
  5. The mutation could be in the active site of acetylcholinesterase.
  6. The mutation could be in an allosteric site of acetylcholinesterase.
  1. i, iv, v, vi.
  2. i, iv, v.
  3. ii, iv, v.
  4. ii, iii, v.
  1. Choline is a product of the acetylcholinesterase reaction. This reaction takes place on the extracellular side of the plasma membrane, but the cells have a use for choline and they transport it inside, where it is at a high concentration, via a Na+/choline secondary active transport system.

  1. What is the specific role of Na+ in the active transport of choline into the cell? (6 pts)

    2. Na+ wants to move down its electrochemical gradient, so it will bind the transporter on the extracellular side. Na+ binding changes the conformation of the transporter so that it has an increased affinity for choline. Choline does not want to bind the transporter on its own because it is at a low concentration. Na+ will be released intracellularly causing the transporter to lose its affinity for choline.

     

  2. Would the transport of choline be different in two cells if one of them had Na+ channels in an open conformation and the other had Na+ channels in a closed conformation? (6 pts)

Circle one: YES or NO

Why or why not?

The Na+/choline secondary active transport works because a steep Na+ gradient exists. If a cell had open Na+ channels, then Na+ could freely move down its electrochemical gradient, into the cell, like it wants to. This would dissipate the Na+ gradient and therefore, the energy needed for the Na+/choline transport in that cell would be lost. Less choline would be transported.

 

 

  1. Scientists at Drugs, Inc. have been working on an antibiotic to combat a new strain of bacteria. They have been targeting one of the bacterial enzymes, BAC-1, and working to generate inhibitors to it. The latest success is a molecule called INH-1. The Lineweaver-Burk graph below shows the BAC-1 reaction in the absence (solid line) and presence (dotted line) of INH-1.

 

  1. What are the Km and Vmax values of the BAC-1 reactions in the absence and in the presence of INH-1? Be sure to show your work and include the units. (8 pts)

    2. x-intercept= -1/ Km, so Km = -1/x-intercept (+I) Km= -1/-0.2 = 5mM; (-I) Km= -1/-0.3 = 3.33mM

    y-intercept= 1/Vmax, so Vmax=1/y-intercept (+I) Vmax =1/4= 0.25 (mmoles/min) SAME FOR (-I)

     

     

     

     

     

  2. Based on the information in part A, what kind of inhibitor is INH-1? How does this kind of inhibitor work? (6 pts)

    3. Competitive inhibitor. It works by competing with substrate for binding to the enzyme active site.

     

     

     

     

     

     

  3. If you were working at Drugs, Inc. and wanted to design an inhibitor that bound tightly to BAC-1, what would you mimic it after? Please explain and be specific. (6 pts)

The transition state of a natural substrate. An enzyme binds most tightly to a substrate in its transition state.

 

 

 

(For questions 9 & 10) A recent paper in the journal Nature presented data on a natural uncoupler, mammalian uncoupling protein-3 (UCP-3). Researchers have shown by indirect immunofluorescence that UCP-3 is normally localized to the mitochondria of muscle cells. Transgenic mice were made that overexpressed UCP-3, meaning that there was more UCP-3 protein in the mitochondria of these mutant mice compared to wild-type mice. The mutant and wild-type mice were fed the exact same amount and type of food, and then researchers measured the amount of glucose in their cells at different time points after the meal. The results are shown below.

Time after meal (min)

 
  1. How do you explain the difference between the glucose concentrations in wild-type and mutant mice between 45 and 135 minutes after the feeding? Be specific about what is occuring in the mutant animal’s cells and your answer should contain the word "ATP". (6 pts)

    2. The mutant mice have lower glucose levels in their cells. This is because the mutant mice have an increased amount of the uncoupler, UCP-3. As an uncoupler, UCP-3 can "sneak" protons that have built up in the mitochondrial intermembrane space, as a result of electron transport, across the inner mitochondrial membrane. However, these protons are not being forced to do work (make ATP) as they move down their concentration gradient. Therefore, the proton gradient is being dissipated, but no ATP is being made. The mutant cells continue to oxidize glucose (do respiration) in a futile attempt to make the needed ATP. (Remember talking about DNP, an uncoupler, that was used for a while as a diet pill.)
  2. The researchers also measured muscle temperature (amount of unusable energy that is released as heat) and resting oxygen consumption in the wild-type and mutant mice. They found differences between the two types of animals. Which of the following combinations would you expect to be true?
  1. Wild-type mice had a higher muscle temperature compared to mutant mice.
  2. Mutant mice had a higher muscle temperature compared to wild-type mice.
  3. Wild-type mice had a higher resting oxygen consumption compared to mutant mice.
  4. Mutant mice had a higher resting oxygen consumption compared to wild-type mice.
  1. i and iii.
  2. ii and iv.
  3. ii and iii.
  4. i and iv.
  1. The following is a key reaction of the TCA cycle catalyzed by isocitrate dehydrogenase:

isocitrate dehydrogenase

citrate a-ketoglutarate + NADH + H+ + CO2

This reaction is subjected to both positive and negative allosteric regulation. Considering the energy status of the cell, which of the following pairs of allosteric regulators makes sense?

  1. Activated by ATP, inhibited by NAD+.
  2. Activated by NADH, inhibited by ADP.
  3. Activated by ADP, inhibited by NADH.
  4. Activated by fatty acids, inhibited by AMP.

 

  1. The following is the eighth reaction of glycolysis:

    2. Phosphoglyceromutase

    3-Phosphoglycerate 2-Phosphoglycerate

    At equilibrium, 3-Phosphoglycerate is at 20 mM while 2-Phosphoglycerate is at 18 mM. Scientists measured the prevailing concentrations of these molecules in cells and found that the concentration of 3-Phosphoglycerate was at 41 mM while 2-Phosphoglycerate was at 87 mM.

    Figure DG° ’ and DG for the reaction above. Remember your units! (R= 1.987 cal/mol K; T= 298 K) (8 pts)

    DG° ’ = -RT ln [prod]eq/[react]eq = -(1.987)(298) ln (18)/(20) = +62 cal/mol

    DG = DG° ’ + RT ln [prod]pr/[react]pr = 62 + (1.987)(298) ln (87)/(41) = +507 cal/mol

    As stated in class, the concentrations of product and reactant under prevailing conditions accidentally got switched in an edit of the exam. In reality, the DG value would have to be negative for the reaction to be catalyzed by an enzyme.

  2. The pH in lysosomes is 5 or less. The pH of lysosomes is lowered by H+ pumps in the lysosomal membrane. These H+ pumps are primary active transporters. Which of the following must be an accurate statement about the lysosomal H+ pump?
  1. When the H+ binding site faces the cytoplasm, its affinity for H+ is high.
  2. When the H+ binding site faces the cytoplasm, its affinity for H+ is low.
  3. When the H+ binding site faces the interior of the lysosome, its affinity for H+ is low.
  4. When the H+ binding site faces the interior of the lysosome, its affinity for H+ is high.
  5. The affinity of the H+ binding site is the same when facing in either direction.
    1. i. and iii.
    2. i. and iv.
    3. ii. and iii.
    4. v.
  1. Which of the following is true about the Tricarboxylic Acid Cycle?
  1. Occurs in the mitochondria.
  2. Occurs in the cytosol.
  3. Produces reduced molecules like NADH + H+ and FADH2.
  4. Oxidizes reduced molecules like NADH + H+ and FADH2.
  5. Oxidizes pyruvate to CO2.
  1. i and iii.
  2. ii and iii.
  3. i, iii, v.
  4. ii, iv, v.

 

  1. The DG for the breakdown of ATP into ADP and Pi is –7.3 kcal/mol. However, I can make up a solution of 1 mM ATP in a test tube in the morning and be fairly confident that the concentration of ATP will still be 1 mM at the end of the day. How can the breakdown of ATP be highly favorable, yet it does not happen? (4 pts)

    2. It does not happen because the energy of activation (EA) must be overcome. An enzyme lowers the EA of a reaction.

     

     

  2. The flow of H+ back into the matrix of the mitochondria causes ATP to be made because…
  1. H+ are required in the reaction ADP + Pi à ATP.
  2. H+ causes the g subunit to spin fast enough so that when it collides with ADP and Pi there is enough energy to produce ATP.
  3. H+ causes the g subunit to spin. This is necessary to change the conformation of a b subunit with bound ATP to the open binding conformation.
  4. H+ causes the g subunit to spin. This is necessary to change the conformation of a b subunit with bound ATP to the tight binding conformation.