Exam #2MOLB / MICR 2021March 11th, 2008Name:______________________________TA’s name:______________________________

Answer the first 20 questions on the computer-graded answer sheet by filling in the proper bubble with a No. 2 pencil. If you change an answer, erase the undesired mark thoroughly. Mark only the best answer to each question. Each question is worth 2 points.

Be sure to fill in the boxes for your name, then fill in the corresponding bubbles beneath them correctly. Please also write your TAs name somewhere on the scantron!

Questions 21 through 29 are each worth 10 points. Please choose to answer 6 of these 9 questions. Circle the questions that you choose.This exam has 20 pages. Please check to make sure that you have them all.

*On page 19 are the skeleton outlines of glycolysis and the TCA cycle. On page 20 is a codon usage table.

The exam is worth 100 points.

Useful equations and constants:

ΔG = ΔH - TΔS

*If you feel that you really need to explain one of your answers to a multiple choice question, please do so on the lines below. I will award credit or partial credit for good justifications of missed questions.__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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1. In the following cartoon, two enzymes are shown binding to their substrates. Based on your prerequisite knowledge as well as the picture, which statement is TRUE?

a. Enzyme 2 binds to its substrate via the lock-and-key model.b. After converting the substrate to products, enzyme 1 will be unable to bind to another substrate. c. Enzymes 1 and 2 both lower reaction activation energy in part by stabilizing a high-energy complex that has a structure somewhere between that of the substrate and its products.d. all of the abovee. b and c

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2. Following is a cartoon depiction of the glycolytic enzyme phosphofructokinase (PFK). Labeled on the drawing are the enzyme active sites as well as additional sites to which the molecule phosphoenolpyruvate (PEP) binds. When PEP binds to the enzyme, the enzyme undergoes conformational changes at the active site that preclude the substrate from binding. Which statement is FALSE?

a. Phosphofructokinase is regulated by allosteric inhibition.b. Phosphoenolpyruvate is an effector molecule.c. Phosphofructokinase is regulated by feedback inhibition.d. Phosphofructokinase requires phosphoenolpyruvate as a cofactor.

3. As we remember from our study of the TCA cycle, the enzyme succinate dehydrogenase catalyzes step 6 during which succinate is converted to fumarate. Based on your knowledge of the events that occur during this step, which statement is TRUE?

a. Succinate dehydrogenase requires a cosubstrate.b. Succinate dehydrogenase requires an essential ion..c. Succinate dehydrogenase requires a prosthetic group.d. Active succinate dehydrogenase is an apoenzyme.e. all of the above

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Use the following table to help in answering questions 4 and 5:

4. The following reaction occurs during the first step of glycolysis:

Glucose + ATP Glucose 6-phosphate + ADP

Which statement is FALSE?a. This reaction is catalyzed by a kinase.b. The overall ΔGº’hydrolysis for this reaction is -16 kJ/mol.c. This reaction is overall exergonic.d. The overall ΔGº’hydrolysis for this reaction makes it more favorable (exergonic) than the following reaction: Phosphoenolpyruvate + ADP ATP + pyruvate

5. Which statement is TRUE?a. Glucose 6-phosphate could transfer a phosphoryl group to glucose to form glucose 1-phosphate.b. When phosphoenolpyruvate transfers a phosphoryl group to ADP to form ATP, this is called oxidative phosphorylation.c. The transfer of a phosphoryl group from 1,3-bisphosphoglycerate to ADP is an exergonic process.d. The hydrolysis of ATP to form ADP and inorganic phosphate requires 30 kJ/mol.

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6. During the sixth step of glycolysis, glyceraldehyde 3-phosphate is converted to 1,3 -bisphosphoglycerate. At this point in the pathway

a. two ATP molecules have been consumed and two have been generated for a net yield of zero ATP.b. a net of two NADH have been generated.c. ATP has been generated via substrate-level phosphorylation on two separate occasions. d. all reactions that have occurred havr occurred in the six-carbon stage of glycolysis.

7. In both steps 7 and 10 of glycolysisa. ATP is generated via substrate-level phosphorylation.b. ATP is consumed.c. a six-carbon sugar is cleaved to form two three-carbon sugars.d. a molecule with high phosphoryl group transfer potential is generated.

8. Based on your experience in enzyme names, as well as the type of reaction occurring, what is the name of the enzyme that catalyzes the transition step?

a. pyruvate dehydrogenaseb. pyruvate kinasec. acetyl-CoA synthetased. thiolase

9. Which reduced cosubstrate is generated in the TCA cycle?a. NADHb. FADH2

c. NADPHd. QH2

10. ATP is produced by substrate-level phosphorylation in which (one) of the following?

a. the TCA cycleb. the electron transport chainc. the transition stepd. all of the abovee. a and c

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11. Which statement/s about the TCA cycle is/are TRUE?a. The TCA cycle generates more ATP via substrate-level phosphorylation than does glycolysis (per glucose molecule).b. After step 5 of the TCA cycle, four molecules of NADH have been generated per acetyl group.c. The TCA cycle generates more reducing power per glucose molecule than glycolysis and the transition step combined.d. If 2 molecules of glucose are completely catabolized, 6 molecules of NADH are generated during the TCA cycle.e. all of the above

12. As two electrons from NADH are transferred through the mitochondrial electron transport chain, many events occur. Which one of the following correctly orders the sequence of events

1) cytochrome c is oxidized by cytochrome c oxidase2) NADH dehydrogenase transfers electrons to ubiquinone (Q) converting it to Ubiquinol (QH2).3) Complex III pumps protons from the cytosol to the mitochondrial intermembrane space.4) O2 is reduced.

a. 2, 1, 3, 4b. 2, 3, 1, 4c. 1, 3, 2, 4d. 4, 3, 2, 1e. 2, 3, 4, 1

13. The Chemiosmotic Hypothesisa. states that NADH may be oxidized only at Complex II of the mitochondrial electron transport chain.b. states that the proton motive force can serve as an energy source for the oxidative phosphorylation of ADP to form ATP.c. is valid only for strictly fermentative bacteria.d. none of the above

14. Organisms using anaerobic respiration to generate ATPa. lack the components of the electron transport chain.b. can use a molecule other than oxygen as their terminal electron acceptor.c. generate less energy than those organisms using aerobic respiration.d. all of the abovee. b and c

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15. Sulfur bacteria such as Thiobacillus produce sulfuric acid as a metabolic byproduct and are thus commonly used in biomining. These resourceful bacteria

a. are anaerobically respiring chemoheterotrophs.b. are chemolithotrophs.c. are phototrophs.d. utilize hydrogen gas as an energy source.

16. All of the following events occur when oxygenic phototrophs utilize non-cyclic photophosphorylation EXCEPT

a. NADPH is produced.b. electrons are stripped from water to replenish the electrons at the reaction center chlorophyll of photosystem II.c. ATP is synthesized.d. Electrons from the reaction center chlorophyll of Photosystem I are transferred back to this reaction center chlorophyll after traveling through the photosynthetic electron transport chain.

17. Which statement/s regarding DNA polymerase is/are TRUE?a. DNA polymerase synthesizes DNA in the 3’ to 5’ direction adding new nucleotides to the open 5’ phosphate.b. The next nucleotide that DNA polymerase would add to the elongating chain would be dTTP.

c. Unlike RNA polymerase, DNA polymerase requires a primer in order to begin synthesis of a strand of DNA. d. all of the abovee. b and c

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18. During replication of the E. coli chromosome, Okazaki fragmentsa. are formed during the synthesis of the leading strand.b. are formed during the synthesis of the lagging strand.c. are synthesized from the amino to the carboxyl terminus by a special DNA polymerase called topoisomerase.d. seal the nicks between newly transcribed RNA primers.

19. Which events occur during the initiation phase of transcriptiona. The sigma factor of RNA polymerase recognizes the promoter.b. The rho enzyme wraps ssRNA around itself.c. The 30S subunit of the ribosome binds to a sequence on the mRNA called the Shine-Dalgarno sequence.d. all of the above

20. With regards to procaryotic translation, the term wobble refers toa. the tentative association between the 30S and 50S subunit.b. the fact that tRNAs can recognize more than one codon.c. the translocation of the ribosome.d. the fact that RNA can be translated by several ribosomes at once.

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This part of the exam is worth 60 points. Every question (21-29) is worth 10 points so you must choose to answer six questions. Please circle the six questions that you choose. These are the only questions that will be graded.21. Metabolism

The TCA intermediate, oxaloacetate, is the building block for the synthesis of amino acids such as lysine, methionine and threonine (shown below):

a. (3 points) Is this pathway (by which Met and Thr are synthesized) anabolic or catabolic? Is it endergonic or exergonic? Is it oxidative or reductive? This is an anabolic, endergonic, and reductive process. It is a synthesis in which NADPH and ATP are consumed.

b. (3 points) Which enzyme on the figure above would you expect to be most likely regulated by feedback inhibition? Which molecule might you expect to act as the effector?Enzyme 1 or enzyme 2 would likely be regulated because they catalyze steps early in the pathway and, for enzyme 2, a step during which ATP is consumed. Met and Thr would likely act as effectors but other answers may suffice as long as reasoning is good.

*Be sure to answer the rest of this question on the next page9

__________________________________________________________________c. (4 points) Unlike humans, bacteria can synthesize the growth factor called folic

acid. One of the enzymes in this synthesis recognizes an intermediate in the pathway called p-aminobenzoate (PABA). This enzyme can be inhibited by the antibiotics called the sulfa drugs. Based on the following pictures of the sulfa drug, sulfanilamide and the substrate, PABA, by what mechanism would you predict that sulfanilamide inhibits the enzyme? Explain.

Competitive inhibition because the inhibitor, sulfanilamide, is structurally similar to the substrate and therefore it likely competes for binding at the active site.

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22. Metabolism(10 points) For each event, determine in which pathway/s (step/s) it occurs. The first one has been done for you.

Glycolysis and TCA

TCA

TCA

Electron transport

TCA

Glycolysis

Fermentation and electron transport

Pentose phosphate

Electron transport

α-ketoglutarate dehydrogenase oxidizes α-ketoglutarate

Ubiquinone is oxidized by the cytochrome bc1 complex

ATP is generated via substrate-level phosphorylation

O2 is reduced

FAD is reduced, forming FADH2

Citrate is rearranged to form isocitrate

The enzyme pyruvate kinase catalyzes the transfer of a phosphoryl group from a high-energy metabolite to ADP

NADH is oxidized (NAD+ is generated)

Ribose 5-phosphate is synthesized

NADH is produced Occurs in glycolysis, the transition step and the TCA cycle

Does this event occur during glycolysis, fermentation, the pentose phosphate pathway, the transition step, the TCA cycle and or electron transport?

Event

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23. MetabolismNext week in lab we will begin making wine. In order to start the process, dry yeast will first be placed into a small jar exposed to O2. This jar will be left to grow overnight.

a. (2 points) For this aerobic environment, fill out the following table indicating which processes are occurring and why or why not.

A terminal electron acceptor is present and therefore reducing power can deliver electrons to the ETC. Thus, the fate of pyruvate is further oxidation in the transition step and then TCA.

A terminal electron acceptor is present and therefore reducing power can deliver electrons to the ETC. Thus, the fate of pyruvate is further oxidation in the transition step and then TCA.

Since an externally-derived terminal electron acceptor is available, pyruvate is further oxidized and fermentation is not necessary.

A terminal electron acceptor is present and therefore reducing power can deliver electrons to the ETC. Thus, the fate of pyruvate is further oxidation in the transition step and then TCA.

Yes

Yes

No

Yes

If a process is occurring, why is it occurring? If it is not occurring, why not?

Is this process occurring or not?

Electron transport (ETC)

TCA cycle

Fermentation

Transition step

Process

b. (2 points) If, during the aerobic overnight incubation, the yeast consume 4.3 X 102 0 molecules of glucose, how many ATP are generated in total? (Assume that only 4 ATP are generated via oxidative phosphorylation from the NADH molecules reduced in glycolysis.)

4.3 X 102 0 molecules glucose * 36 ATP / glucose = 1.5 X 102 2 ATP*See table in notes.

c. (2 points) After growing the yeast starter culture overnight, we will use it to inoculate pasteurized juice. We will then stopper the bottle and starve the yeast of O2. Under these anaerobic conditions, the fate of pyrvuate will change and the yeast will live using alcoholic fermentation. Under these conditions, if the yeast catabolize an equivalent amount of glucose molecules to that used by the aerobic culture above (4.3 X 102 0), how many ATP are generated in total?4.3 X 102 0 molecules glucose * 2 ATP / glucose = 8.6 X 102 0 ATP

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d. The last step of alcoholic fermentation involves the conversion of acetaldehyde to ethanol (figure below):

Not surprisingly the enzyme that catalyzes this step is called alcohol dehydrogenase. Pyrazole is a chemical inhibitor of alcohol dehydrogenase. If pyrazole were added to the wine fermentation mixture, how would it affect each one of the following:

1) The concentration of NADH[NADH] would go up because the step which consumes it is inhibited.

2) GlycolysisGlycolysis would stop and slow because step 6 would stop and slow.

3) ATP productionBecause glycolysis is the only source of ATP in this anaerobic

environment, ATP production will slow and stop, the yeast will die.

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24. Enzymes / Pigments(10 points) Place the letter corresponding to the best description below the name.

A. A component of the electron transport chain that can be found in eucaryotes and in the bacterium Pseudomonas.

B. A replication enzyme that catalyzes the synthesis of RNA primers during formation of the lagging strand of DNA.

C. Synthesizes DNA using dNTPs as building blocks

D. An enzyme that unwinds DNA

E. A secretory enzyme that cleaves glycosidic linkages in starch.

F. An accessory pigment that increases the efficiency of light utilization in a variety of photosynthetic organisms.

G. Catalyzes the oxidation of isocitrate with the simultaneous production of NADH.

H. A secretory enzyme that hydrolyzes triacylglycerols.

I. An enzyme that wraps ssRNA around itself destabilizing the RNA-DNA hybrid

J. A glycolytic enzyme that catalyzes a phosphoryl group transfer

K. An enzyme that relieves the supercoiling ahead and behind the the replication fork.

L. An enzyme that seals the nicks between adjacent DNA fragments.

1. Oligo-1, 6-glucosidase ___E_____

2. Cytochrome c oxidase____A____

3. rho ___I_____

4. Carotenoid ___F_____

5. Lipase___H_____

6. Primase ____B____

7. DNA polymerase ___C_____

8. Isocitrate dehydrogenase ___G_____

9. Helicase ___D_____

10. Hexokinase____J____

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25. Metabolism

a. If an aerobically respiring microorganism completely catabolized a fatty acid of 22 carbons in length.

1) (4 points) How much reducing power would be generated (NADH and FADH2)?

11 ACETYL * 3 NADH / ACETYL = 33 NADH

11 ACETYL * 1 FADH2 / ACETYL = 11 FADH2

2) (4 points) How many protons could be pumped (by the proton-pumping members of the electron transport chain) from the reducing power generated?

33 NADH * 10 H+/NADH = 330 H+

11 FADH2 * 6 H+/FADH2 = 66 H+

b.(2 points) If a cell needed to synthesize a great deal of the amino acid glutamate, how would this affect the net energy yield of the TCA cycle?

Since high-energy intermediates are diverted from the TCA, the overall E yield goes down.

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26. Gene Expression(10 points) Determine whether each term is generally associated with:

A. DNA replicationB. transcriptionC. translation

___A____ dNTPs ___B____ RNA polymerase___C____ Anticodon arm ___C____ Initiation factors___B____ Hairpin loop ___A____ Terminator utilization substance___A____ 3’-5’ exonuclease ___C____ polysome___C____ Shine-Dalgarno sequence ___A____ helicase

27. TranscriptonPictured below are the promoter regions and a portion of the downstream transcribed region of two different procaryotic genes. Note that only the coding strand is pictured here!

a. (2 points) Put boxes around the -35 region and the TATA boxes within the promoters of both genes above. -35 regions = TTGCCA and TTGACC in A and B respectively. TATA boxes, TATA box = TATATT in gene A and TATAAT in gene B.b. (2 points) The promoter for which gene (A or B) is the strongest? Explain.B more strongly resembles the ideal consensus sequence of TTGACA and TATAAT.c. (2 points) What is the sequence of the RNA transcript for the portion of gene A shown here?

5’ UGGCAATTCCGATCCGGATT...3’d. (2 points) There are two ways in which procaryotic transcription is usually terminated, describe one of these ways. Hairpin loop = A region of complimentation on RNA desabilizes RNA/DNA heteroduplex.Rho enzyme = wraps RNA around itself and destabilizes heteroduplex

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28. Translationa. (2 points) Based on the sequence of the following short polypeptide chain, write

one possible sequence for the mRNA which encoded it (see attached table for assistance). Be certain to correctly label the 5’ and 3’ ends of the mRNA transcript.

Amino-f-Met-Lys-Glu-Asp-Ser-Arg-Phe-Met-Ala-carboxyl

5’ AUGAAA(AAG)GAA(GAG)GAU(GAC)UCN(AGU)(AGC)CGN(AGA)(AGG)UUU(UUC)AUGGCNUAG(or another stop codon) 3’

b. (2 points) Can you know for sure that the sequence you wrote in a. is the actual sequence of the mRNA that encoded for the peptide? Why or why not?

No, because the genetic code is degenerate.

c. (2 points) Is the entire length of the mRNA transcript represented by the sequence you wrote for a.? If not, what is missing?

No, the untranslated leader and trailer sequences are not shown.

d. (4 points) If a tRNA has the anticodon sequence 5’GAA3’, what amino acid does it carry attached to its acceptor stem? [*NOTE - show your process for partial credit]

Anticodon: 3’AAG 5’Codon: 5’ UUC 3’

Amino acid: Phe

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29. Omnivore’s Dilemma(10 points) On WebCT, we have been discussing chapter 4 of The Omnivore’s dilemma. If you have participated in the discussion (in such a way that your reading of the chapter is evident), I will award you 10 points for this question!!

EXTRA CREDIT: What is a snurp? What does it do?

Small nuclear ribonucleoprotein = splices out introns in eucaryotic precursor mRNA

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Following are some pictures and a table that may be useful:

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