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BIOS E-1a f11 exam 01 KEY

BIOS E-1a f11 exam 01 KEY - Name_TF_Credit BIOS E1A Midterm...

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Unformatted text preview: Name_________________________________TF__________________Credit_____________ BIOS E1A Midterm Exam, Fall 2011 A. Be sure to write your full name on the top of each of page of the examination. B. IF YOUR ANSWER CANNOT BE READ IT CANNOT BE GRADED. C. Remember, if you do not show your reasoning, you cannot get partial credit. D. The point value for each question is noted. ALLOCATE YOUR TIME WISELY. E. Write each answer only on the same page as the pertinent question. THE SPACE PROVIDED IS MEANT TO BE SUFFICIENT - BE CAREFUL & BE CONCISE. F. Read carefully, think before you write, good luck, and do well! page 2: __________ (50 pts) page 3: __________ (15 pts) page 4: __________ (12 pts) page 5: __________ (12 pts) page 6: __________ (11 pts) TOTAL: _______ (100 pts) Page 1 of 6 Name_________________________________TF__________________Credit_____________ Multiple Choice (each question 2 pts) 1. D 11. ABCD 2. A 12. C 3. ABCD 13. B 4. C 14. C 5. D 15. C 6. AB 16. C 7. A 17. A D (ACCEPT BOTH OR EITHER) 8. C 18. D 9. C 19. B AC 20. D 10. Matching (1pt each) 21. Electron Transport Chain 22. Citric Acid Cycle 23. Glycolysis 24. ATP synthase 25. Anion 26. Starch 27. Glycogen 28. Cellulose 29. Protons 30. Acids Page 2 of 6 Name_________________________________TF__________________Credit_____________ True or False – 1 pt each 31. TRUE FALSE 32. TRUE FALSE 33. TRUE FALSE 34. TRUE FALSE 35. TRUE FALSE 36. TRUE FALSE 37. TRUE FALSE 38. TRUE FALSE 39. TRUE FALSE 40. TRUE FALSE 41. TRUE FALSE 42. TRUE FALSE 43. TRUE FALSE 44. TRUE FALSE 45. TRUE FALSE 46. TRUE FALSE Page 3 of 6 Name_________________________________TF__________________Credit_____________ 47. (12 pts) The citric acid cycle is an oxygen ­dependent process. However, none of the reactions of the cycle directly involves oxygen as a reactant. A) (2pts) Why is the pathway oxygen ­dependent? NAD+ must be regenerated. B) (2pts) One step of the citric acid cycle generates an ATP equivalent via substrate level phosphorylation. How does substrate level phosphorylation differ from oxidative phosphorylation? Substrate level involves direct transfer of a phosphate group from a reactant while oxidative phosphorylation utilizes the energy harnessed by the ETC to generate a proton gradient that drives ATP production (it is an indirect process). C) (2 pts) In this metabolic pathway, conversion of malate to oxaloacetate is an endergonic reaction. How, then, do cells accomplish the reaction and complete the cycle? Couple it to an exergonic reaction D) (3 pts) The conversation of malate to oxaloacetate is catalyzed by the enzyme malate dehydrogenase. Indicate (with an "x") if the observed kinetic parameters at the left would be altered by the following factors. Give only one answer for each. Condition Km Vmax Neither Both Competitive Inhibitor X 6M Urea (denaturing agent) X Doubling substrate concentration X You have identified the active site – the region where substrates bind and undergo chemical transformation ­ of malate dehydrogenase. To understand the importance of residues within this region, mutant forms of the enzyme were generated that differ from the wild ­type by a single amino acid. In the Lineweaver ­Burke graph below, the solid line represents the wild ­type enzyme and dashed line represents a mutant of malate dehydrogenase with a single amino acid change in the active site. E) (3 pts) What is a simple interpretation of this data? Does the mutation likely affected the catalytic mechanism, substrate binding or both. This is a qualitative answer – no number values are given or needed. It appears that the mutation has decrease Km (1 pt) without changing Vmax (1pt). This suggests that the mutation affects the binding site but not the catalytic mechanism (1 pt). Page 4 of 6 Name_________________________________TF__________________Credit_____________ 48. (12 pts) Shown in the schematic on the left is the orientation of the transmembrane protein, Protein A, within the ER membrane. A) (2 pt) The N ­terminus of protein A is indicated in the schematic. What is the designation for the other end of the protein? What is the basis for these designations? Free carboxyl end= C terminus, free amino end = N terminus B) (1 pts) Would this protein be considered an integral membrane protein or a peripheral membrane protein? Integral membrane B) (2 pt) Do you expect any differences in the chemical properties of the amino acids found in the region shaded black versus those in the regions shaded gray? Why or why not? Those in black would be hydrophobic and those in gray hydrophilic. C) (2 pt) Circle your answer from the choices below. When this protein completes trafficking to the plasma membrane, the N ­terminus will be: A. found on the extracellular face of the cell B. in the cytosol C. found in the orientations described in A and B D. There is not enough information to make a prediction. D) (2 pt) Where (smooth ER, rough ER, Golgi or cytsol) might the sugar residues have been added to the protein? Rough ER and Golgi You suspect that the transmembrane protein, Protein A, is held within certain regions of the membrane via attachment to cytosketal components. You generate a form of Protein A that is tagged with green fluorescent protein. You then perform a FRAP experiment to test your hypothesis. As a negative control, you use a GFP ­tagged form of protein A. Protein A is known to move freely within the membrane. E) (3 pts) Which data set would you as the results of an experiment comparing the FRAP profiles of Protein A ­CFP with that of a negative control sample Protein X ­CFP? Clearly indicate your choice of data sets (X, Y or Z) and explain your choice. Data Set X Data Set Y Data Set Z Dat Data Set Y (1 pt)– in the control you would expect rapid recovery of fluorescence Control – indicating free mobility (1 pt). In the Protein Protein X A experiment, you would expect to see limited mobility and a slower and incomplete recovery of fluorescence. (1pt) Protein A Page 5 of 6 Prot e Name_________________________________TF__________________Credit_____________ 49. (4 pts) A plant breeder has developed a new frost ­resistant variety of tomato that contains higher levels of unsaturated fatty acids in membrane lipids than those found in standard tomato varieties. However, when temperatures climb above 95 °F, this frost ­resistant variety dies, whereas the standard variety continues to grow. Provide a likely explanation of the biochemical basis of increased tolerance to cold and increased susceptibility to heat of this new tomato variety. More unsaturated fatty acids will cause an increase in membrane fluidity because unsaturated fatty acids contain “kinks” and cannot pack as tightly as saturated fatty acids (2 pt). At cold temperatures, the fluidity increase from the extra unsaturated fatty acids counterbalances the tendency of lipids to solidify at low temperature (1 pt). At high temperatures, the fluidizing effects of the extra unsaturated fatty acids add to the fluidizing effect of higher temperature, and the membrane of the new plant loses its integrity (1 pt). 50. (7 pts) Brush border cells are involved in the uptake of dietary glucose from the intestine and the transfer of this glucose into the bloodstream. The series of transporters important in this process are shown in the figure below. A) (2 pts) What are two sources of energy utilized in the transport of glucose into the blood stream? ATP and concentration gradient B) (3 pts) Identify the uniporter, symporter and antiporter present in this system. Please use the number designation for each transporter in your answer. 1 – symporter 2  ­ uniporter 3  ­ antiporter C) (2 pts) What would happen to the glucose concentration (increase, decrease, or remain the same) in the cell if ATP were no longer available? Explain your answer. The glucose concentration in the cell would decrease. It would equilibrate with that found in the intestine. ATP is used to maintain the sodium gradient used by Na ­Glucose symporter to move glucose into the cell against its gradient. If the sodium gradient is not maintained, glucose is not transported into the cell and the levels would equilibrate with the capillary as glucose moved down its gradient through the glucose uniporter. Page 6 of 6 ...
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