Exam 1-Key. Biology II Winter 2013

Transcription

1 Exam 1-Key Biology II Winter 2013 Multiple Choice Questions. Circle the one best answer for each question. (2 points each) 1. Which of the following is part of the Cell Theory? A. All organisms are composed of organs B. All cells contain a nucleus C. Cells can exist in many shapes or sizes. D. All cells come from cells E. None of the above 2. Enzymes speed up a reaction by: A. changing the k eq so that it is >1 B. stabilizing the transition state of the substrate C. lowering the activation energy of the reaction D. changing the G of the reaction E. both B and C are correct 3. Triglycerides are formed by linking glycerol to fatty acids by the process of : A. dehydration B. hydrolization C. peptide bond formation D. amphipathic bond formation E. both B and E. 4. Alpha helices are an example of protein. A. Primary structure B. Tertiary Structure C. catalysis D. Secondary Structure E. hydrophobicity 5. Which of the following does NOT support the idea of Mitochondria arising through the endosymbiosis theory? A. they divide independently B. they are similar in size to prokaryotic cells C. they have cell walls similar to prokaryotic cells. D. they have DNA sequence that is similar to prokaryotic genes E. they have two membranes. 6. Gibbs free energy (G) is defined as: A. Energy available within a system to do work B. Total amount of Energy in a system C. Amount of energy released into a system D. Energy that has no cost to the system. E. The energy used by John Travolta in Saturday Night Fever Version A p. 1 of 7 January 19, 2012

2 7. is when a molecule binds to a site other than the active site and decreases the ability of an enzyme to bind substrate. A. competitive inhibition B. allosteric activation C. Lowering the activation energy D. Decreasing the K m E. allosteric inhibition 8. The describes how cell size is limited to the size we see associated with eukaryotic cells. A. function of the cell B. cell wall C. surface area to volume ratio D. cytoskeleton E. genome size 9. Head group and fatty acid chains make phospholipids. A. amphipathic B. hydrocarbons C. allosteric D. oligomeric E. polypathic 10. Which of the following is describes K m : A. k 1 /k cat B. substrate concentration equal to V max /2 C. k eq >1 D. [Product]/[substrate] E. Both A and B 11. Matching. The statements below describe cellular structures involved in the HIV life cycle. From the list on the right, choose the viral life cycle stage that occurs at or uses the cellular sites listed on the left. There can be more than one letter per answer, and a letter may be used more than once, or not at all. (2 points each) A,F Lipid Rafts. A. Attachment B,C Microtubules B. Uncoating C. Reverse Transcription E Rough Endoplasmic reticulum D. Integration A Endosome E. Assembly E Cytoplasmic ribosomes F. Budding Version A p. 2 of 7 January 19, 2012

3 12. After learning about the Cell Theory in class, you had a discussion with your roommate. He said the Cell Theory is all very nice, but his theory is that some living things (yet to be discovered) are not made up of cells. Because both of these ideas are just theories, he believes that his is just as valid as yours. Would you agree? (2 points) Hopefully, as a scientist, you would not agree. While the general public thinks of a theory as just an idea that somebody had that isn t known to be a fact, scientists have a different idea. For a scientist, a theory is an idea that was once just a hypothesis but now is supported by a considerable weight of data and is therefore broadly accepted in the scientific community. A physicist might call this a law, while a member of the general public would typically just call it a fact, such as the theory that the earth orbits the sun (not vice-versa) or the theory all matter is made up of atoms. 13. As we discussed in class, drugs that inhibit microtubules can block the ability of HIV to infect a cell by preventing the virus from uncoating. a. What is the normal function of microtubules in a cell? (2 points) Part of the cytoskeleton, involved in structure and transport, such as transport of materials to/from the nucleus. b. Would a drug that inhibits microtubules be a good choice to test as a possible anti-hiv drug for use in humans? Defend your answer. (3 points) No! Certainly, inhibiting microtubules would stop the virus, but it would also inhibit the important normal activities of microtubules in the cell, probably killing the cell. 14. There are 4 things an enzyme does NOT change, what are they? What does an enzyme change in regards to a chemical reaction? (5 points) Does not change: -Equilibrium -Keq -amount of products made - G it does change the rate of the reaction. Version A p. 3 of 7 January 19, 2012

4 15. Proteins are made up of amino acid polymers. Draw the reaction that links (a) two amino acids, and label the (b) peptide bond (c) N-terminus, and (d) C-terminus of the growing peptide chain (2 points each). R1 O R2 O H3N C C--OH + H3N C C OH H H H2O R 1 O H R 2 O N-terminus H3N C C N C C OH C terminus Peptide Bond 16. Below are the data some BIO 102 students collected in their sporulation independent experiment: Percent sporulation for yeast grown in buffer with the addition of: 1% glucose 1% sucrose 1% lactose Group Group Group Group Group a. What hypothesis were these students testing? (1 point) Since our lab data supported the hypothesis that yeast sporulate in response to starvation (glucosefed yeast did not sporulate), these students appear to have been testing the hypothesis that yeast will sporulate when given sugars they can t metabolize. b. What two statistical calculations should the students now do? (2 points) Average each set of data and calculate the standard deviation. c. Which set of data represents these students control? (1 point) 1% glucose. We have used this before and know what to expect in regards to sporulation, so this is an ideal control. We KNOW what will happen in the experiment! Version A p. 4 of 7 January 19, 2012

5 17. You have isolated an enzyme that has potentially flavor saving activity for brussel sprouts. The bitter taste of brussel sprouts is associated with a chemical called glucosinolate. The enzyme Glucosinolate Convertase (GC) catalyzes a reaction that converts glucosinolate to yummy glucose. The reaction glucosinolate glucose has a G of a. Based on the information above, what type of reaction is the conversion of glucosinolate to glucose, and how does it relate to k eq? (2 points) Based on the G being positive, the reaction is endergonic. Since it is endergonic, then k eq <1 b. You notice that your father really enjoys brussel sprouts, however you can t stand them (which is completely normal!). You successfully isolate the GC enzyme from your father and yourself. You determine that the K m for glucosinolate is higher in your GC enzyme compared to your father s. What does this mean in terms of how his GC appears to catalyze this reaction quicker? Does this change the V max of the enzyme? Explain your answer. (2 points) Since your father s GC has a lower Km, that must mean his enzyme has a higher affinity for glucosinolate. Therefore, it will find the substrate faster than your enzyme. This has no effect on Vmax. Vmax does not change if Km changes, just the rate at which you reach Vmax will change. c. You have also noted that your enjoyment of brussel sprouts increases when they are wrapped in bacon. You determine that high sodium content is responsible for increasing the taste of brussel sprouts. Adding high amounts to your enzyme reaction appears to lower the K m of your GC enzyme to the same level as your father s enzyme. What is happening here? (2 points) It appears that sodium is acting as an activator for GC. It is likely helping GC to stabilize the transition state of glucosinolate to lower the activation energy. 18. The cell theory states that the cell is the smallest living unit of a living thing; if a virus is not a cell, then it cannot be living by that definition. But, is a virus a cell? Certainly, it has genes and reproduces, two key characteristics of a cell. List two specific ways in which viruses do not meet the definition of a cell. (2 points) 1) Viruses do not have a cell membrane (envelope of an enveloped virus does not function as a membrane) 2) Viruses do not have both DNA and RNA 3) Virus genome can be DNA or RNA 4) Viruses have no ribosomes or other cellular structures 5) Viruses have no metabolism outside a host cell 19. Describe how cholesterol is important for lipid rafts, and how lipid rafts can help the budding step of the HIV life cycle. (2 points). Cholesterol decreases the fluidity of the cellular membrane. Therefore in lipid rafts, there are discrete regions in the membrane that have a high concentration of cholesterol and therefore very little movement in the membrane. This is useful in organizing HIV glycoproteins to discrete locations in the cellular membrane. This will help localize capsid proteins and HIV genomes to where the virus will eventually bud from the cell. Version A p. 5 of 7 January 19, 2012

6 20. HIV protease has been shown to break the peptide bond between the Capsid and Reverse transcriptase units of the HIV GAG polypeptide. In the presence of a drug that blocks HIV protease activity, HIV replication is blocked. a. Why is it important for HIV protease to break the peptide bond between Capsid and Reverse transcriptase on the GAG polypeptide? (2 points) Capsid and Reverse transcriptase do not function properly when attached to each other. Capsid protein can t interact and form the capsid, and Reverse transcriptase won t make a DNA copy of the RNA genome when capsid is still attached. b. Draw an Michaelis-Menton graph expected in this reaction with (1) with enzyme with no drug, and (2) with enzyme and drug. (3 points) Enzyme w/o drug Rate of Product formation Enzyme w/ drug [Substrate] 21. Draw the structure of a triglyceride, which is a building block for phospholipid synthesis. In this molecule, include at least one unsaturated and one saturated hydrocarbon tail (3 points). 22. There are very few viruses that can infect both your dog and you. Which step of the virus life cycle is responsible for this host specificity? (2 points) The attachment step: a virus can only infect a cell if it can bind to an appropriate receptor on the outside of the cell. Most viruses that infect humans bind receptors that are not present on the cells of other mammals and vice-versa. Version A p. 6 of 7 January 19, 2012

7 BONUS: What do we call regions in the cell membrane that are dense with cholesterol? (1 point) Lipid rafts. Version A p. 7 of 7 January 19, 2012