BiochemFinal2011.doc

Biochemistry Final Exam 2011 name______________________________

Biochemistry – Final ExamBiology 020.305

December 15, 2011Instructions for examination:1. Print your full name clearly on each page. If there are any pages missing, obtain a new exam. No excuses will be accepted for missing pages. There are a total of 14 pages including:1 cover page 1 10 question pages 2-11 1 tally page 12 1 page with metabolic pathways page 13 1 page with codon table page 14

2. Write answers clearly and concisely in the space provided. Extra space for doodling etc. is found on last page of the exam. If you desire partial credit for anything that is NOT in the answer space provided, you must clearly indicate where we should look for that partial credit.

3. Exams not written in blue or black permanent ink will lose the privilege of regrading.

4. Calculators are not necessary and their use is not allowed. When arithmetic calculations are requested, if you are unable to do the calculations manually, set up the solution as completely as possible without actually solving for the final number.

5. This exam is closed note and closed book. Books, notes, etc., are to be placed out of sight before the beginning of the exam and are not to be referred to during the exam.

6. If you have any questions, please raise your hand and a proctor will come to you. If you prefer to address your question to a course instructor, please request that a proctor contact an instructor.

7. You will be allowed 2 hours to take this exam. Once time is called you MUST stop writing. Any writing after time is called is considered an ethics violation and we take these very seriously. Please write your name and JHED ID on the tally page. Failure to do so will cost 2 points.

8. When you are done with the exam, please show your ID to a proctor who will accept your exam. Then sign out by marking your test number on the roster. You will not be allowed to leave the room until you turn in your exam.

9. Ethics Pledge: Please make the following commitment by signing and dating underneath the following pledge:

“I agree to complete this exam without unauthorized assistance from any person, materials, or device. I will not take more than the allotted time. I will not alter my answers in any way after receipt of the graded exam.

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Signature: Date:

1. (15 points total) Circle True or False. If False, correct the statement to make it true. a) True/False: ATP is kinetically stable but thermodynamically unstable. 3

b) True/False: Protein folding is favored by the hydrophobic effect and hydrogen bonding, but disfavored by the entropy of folding.3

c) True/False: Allosteric inhibitors bind at an enzyme active site. 3

d) True/False: Covalent modification is an example of an irreversible form of protein regulation.

e) True/False: Carbohydrate modifications on proteins are important for cell-cell signaling and recognition. 3

2. 7 total. Evidence from electron microscopy shows that the complexes of the electron transport chain form supercomplexes in living systems. The supercomplex contains Complex I, Complex III, and Complex IV.

a) What is one advantage of using a supercomplex for electron transport instead of separate complexes? 3

b) Complex II (succinate coenzyme Q oxidoreductase) has never been observed in the supercomplex. How is Complex II different from the other complexes in the electron transport chain? Circle all that apply. 4

i) it participates in the TCA cycleii) it does not contain electron carriersiii) it does not transport protons across the mitochondrial inner membraneiv) it is not associated with the mitochondrial inner membrane

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3. 6 points The RNA world hypothesis states that life started with RNA molecules that could both store information (like DNA) and catalyze chemical reactions (like enzymatic proteins).

a) Which of the following are “ribozymes” that contain catalytic RNA? Circle all that apply. 2i) tRNA synthetaseii) spliceosomeiii) RNA polymeraseiv) poly A polymerasev) ribosome

b) Why are proteins better catalysts for chemical reactions than RNA? 2

Synthetic ribozymes designed to bind and cleave specific mRNAs have been tested in clinical trials as therapeutic drugs for cancer and other diseases, but none have been approved by the FDA.

c) Why do you think ribozymes might not be effective as drugs? Circle all that apply. 2i) ribozymes are single-strandedii) ribozymes are not sequence specific iii) ribozymes contain uracil instead of thymine iv) ribozymes are degraded by RNA interferencev) ribozymes are not very stable

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4. (8 total) On the molecule below, please label the following:

a) Label the polar head group by putting a box around it

b) Label the fatty acid chain(s) with a circle or circles

c) Name two properties of the fatty acid chains that will affect the melting temperature of this molecule and describe how that property will affect the melting temperature. See the example as a guide. 4

property affect

example color more blue = higher melting temperature

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2

d) What properties are shared by all lipids? Circle all that apply 2i) largely non-polarii) non-polymericiii) smalliv) highly reactivev) unstable at low temperatures

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5. 32 total. Hexokinase catalyzes the first reaction of glycolysis.

a) Circle True or False: Hexokinase increases the rate of the reaction but does not change the thermodynamics of the reaction. 2

b) Hexokinase performs a group transfer reaction to produce glucose-6-phosphate. Draw the chemical structure of glucose-6-phosphate. 4

Arora et al. (1991) studied hexokinase function by making specific mutations in one of the HK variants expressed in humans, HK-1. They created a plasmid that encoded the human HK-1 gene and moved the plasmid to an E. coli bacterial expression system. c) What must have been included in the plasmid for it to encode human HK-1 protein made in E. coli? Circle all that apply. 3

i) a version of the gene exactly as it is encoded in the human genomeii) a version of the gene that does not include intronsiii) a version of the gene that can be alternatively splicediv) a version of the gene that encodes polyA at the 3’ endv) a version of the gene that includes a Shine-Dalgarno sequence

vi) a version of the gene in which the first 3 bases transcribed make a start codonvii) a version of the gene with a bacterial promoter upstream of the coding sequenceviii) a telomere

d) Based on the sequence of HK-1, the isoelectric point is estimated to be 6.51. If the researchers used buffer of pH 8.5 during protein purification, which type of column would bind the HK-1? Circle one. 2

i) cation exchange (negatively charged column)ii) anion exchange (positively charged column)iii) a non-polar column

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5 continued. Using purified HK-1, Arora et al. found the following:

area of mutation mutation KM(app) glucose KM (app) ATP Relative vmax

glucose binding site wild-type 0.045 2.4 100

glucose binding site Ser603-->Ala 0.0084 2.0 5.9

glucose binding site Glu708-->Ala 2.28 2.7 9.6

glucose binding site Glu742-->Ala 0.62 2.6 6.4

ATP binding site wild-type 0.046 2.2 100

ATP binding site Lys558-->Arg 0.05 2.2 70

ATP binding site Lys558-->Met 0.047 2.1 29

e) The researchers chose to mutate Serine and Glutamate residues in the glucose binding site. Which of the following properties do Serine and Glutamate share? Circle all that apply. 2

i) both are small ii) both are bulky iii) both are non-polariv) both are polarv) both are mostly negatively charged at cellular pHvi) both are mostly positively charged at cellular pH

f) Why are serine and glutamate good amino acids for the binding of glucose? 2

g) Based on the data above, which mutant version of HK has the lowest affinity for glucose? Circle one. 2

i) Ser603 to Alaii) Glu708 to Alaiii) Glu742 to Alaiv) Lys558 to Metv) impossible to tell from these data

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5 continued. h) Mutations in the ATP binding site do not have a dramatic effect on ATP binding, but they have significant impact on Vmax. How can this be explained? Circle all that apply. 2

i) Lys558 is more important to catalysis than to ATP bindingii) Lys558 is important for glucose bindingiii) Vmax is estimated using high [substrate] so binding is not important in finding Vmaxiv) Vmax cannot be accurately estimated for enzymes with more than one substratev) a single amino acid change cannot affect both binding and activity

i) The KM for ATP was measured in the absence of glucose. When HK-1 binds glucose, a conformational change occurs and the affinity for ATP increases. Why is this two-step binding with a conformational change good for ensuring G6P is produced efficiently? 3

Naturally-occurring mutations in a version of HK found in red blood cells, HK-R, cause haemolytic anemia which is anemia due to the death of red blood cells.

j) HK-R is transcribed from a different promoter than HK-2 or HK-3. HK-R, HK-2, and HK-3 are: Circle one. 2

i) splice-variantsii) orthologsiii) isozymesiv) ribozymesv) non-homologous

Other symptoms of haemolytic anemia are similar to symptoms of diabetes such as high blood glucose levels. You decide to study HK-R mutations and find that HK-R mutant cells are sensitive to insulin.

k) Which of the following would be evidence that insulin signaling is functional in these cells? Circle all that apply. 2

i) dimerized insulin receptorsii) phosphorylation of insulin receptorsiii) GTP binding by Galphaiiv) accumulation of insulin inside the cellv) movement of the insulin receptor to the nucleus

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5 continued. You decide to use the presence of GLUT4 on membranes as evidence of insulin signaling in HK-mutant cells. You find that GLUT4 is induced on membranes in response to insulin, just like in wild-type cells.l) If GLUT4 is abundant on membranes, why do blood sugar levels remain high? 4

GLUT4 is a glucose transporter found only in eukaryotes. When you look at the amino acid sequence, you do not see a consistent pattern of fully-hydrophobic alpha helices. m) Why might this protein not contain fully-hydrophobic alpha helices? Circle all that apply. 2

i) residues where two helices interact may be polarii) GLUT4 is in the plasma membrane, which is a polar membraneiii) side chains of alpha helices face the inside of the helix and can therefore be polar

even within a membrane iv) alpha helices cannot pass through membranes

6. 20 total. TranslationMicroarrays are useful to determine which genes are transcribed into mRNA, but mRNA levels do not always correlate with the cellular levels of the encoded protein. a) Why is this? Circle all that apply. 2

i) more than one copy of a protein can be synthesized from a single mRNAii) mRNAs can be stabilized by short complimentary RNAs iii) mRNAs do not encode proteiniv) mRNAs can be degraded to prevent translation

One way to determine which mRNAs are translated is to isolate the mRNAs attached to functional ribosomes, determine the sequence of those mRNAs, and match the mRNA sequence to the gene sequence encoded in the DNA: a process called Ribosome Profiling. b) If you performed Ribosome Profiling on mouse cells, what type of mRNA would you expect to find associated with active ribosomes? Circle all that apply. 4

i) mRNA with a 5’ capii) mRNA with a 3’ CCA extensioniii) mRNA that contains both introns and exonsiv) mRNA base-paired to snoRNAsv) mRNA with a polyA tailvi) mRNA bound by translation initiation factors (eIFs)vii) mRNA bound by the RISC complex

c) Why are protein levels a better output for cell behavior than mRNA levels? 2

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6. continuted. Cycloheximide (shown in the figure below) can be used to inhibit translation in eukaryotic systems like the mouse cells described in part b above.

Cycloheximide has been shown to bind in the E site of the 60S ribosomal subunit by hydrogen bonding to a C in the 28S rRNA. Cycloheximide hydrogen-bonds to Cytosine in a way that mimics the C-G base pair. When Cycloheximide is bound in the E site, empty tRNAs cannot exit the ribosome.

d) On the CMP molecule below, circle the atom(s) involved in hydrogen bonding with guanine. 3

Cycloheximide doesn’t act instantly. Studies using radioactive Phenylalanine and a polyU template have shown that one peptide bond is formed after addition of Cycloheximide. e) Why does Cycloheximide block elongation only after one peptide bond is formed? 3

Another inhibitor of eukaryotic translation is Harringtonine. Harringtonine also binds the 60S subunit but only in the initiation complex.f) What is unique about the initiation complex compared to any other complex in which the ribosome is actively transcribing? Circle all that apply. 2

i) only the initiation complex includes the mRNAii) only the initiation complex includes empty E siteiii) the initiation complex includes a special initiator met-tRNAiv) only the initiation complex includes GTP

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g) On the mRNA sequence below, circle the codon in the P site when a cell has been treated with harringtonine. 2

h) Put a box around the codon that will be in the P site when a cell is treated with cyclohexamide. 2

5’ cap-ACCUGAUGGACACGUACGAUCUGGAAAUAACAGCCUAAAAAAAAAAAAAA

7. 12 points total Phosphofructose kinase (PFK) catalyzes the conversion of fructose-6-phosphate to fructose 1,6- bisphosphate during glycolysis. PFK is an important point of metabolic regulation. Similar to hemoglobin, PFK binds substrates cooperatively and has two stable conformational states, R and T.

a) Where is PFK located in cells? Circle one answer. 2i) plasma membraneii) mitochrondrial matrixiii) mitochondrial inner membraneiv) cytoplasmv) nucleus

The following graph shows the effects of ATP and AMP on PFK activity.

b) For the following statements, circle True or False. 1 point each, 6 points

i) True or False: PFK has multiple fructose-6-phosphate binding sites

ii) True or False: PFK is regulated by negative feedback inhibition

iii) True or False: ATP is an inhibitor of PFK

iv) True or False: ATP is a substrate for PFK

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v) True or False: AMP preferentially binds the T-state

vi) True or False: ATP preferentially binds the T-state

An increase in [H+] caused by lactic acid fermentation in muscles inhibits PFK activity. The graph below shows PFK activity at two different pH values.

c) Which curve represents the lower pH? Circle one answer. 2

A B

d) If a muscle creates a lot of force and uses ATP rapidly, which of the following would be true? Circle one answer. 2

i) hemoglobin is highly saturated with O2 and PFK has high activityii) hemoglobin is less saturated with O2 and PFK has high activityiii) hemoglobin is highly saturated with O2 and PFK has low activity iv) hemoglobin is less saturated with O2 and PFK has low activity

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