Life Science 1A Final Exam January 19, 2006sites.fas.harvard.edu/~lsci1a/F06final02key.pdf · Life Science 1A Final Exam January 19, 2006 ... Explain briefly. Cen trosome Choice A:

Name: TF: Section Time

Life Science 1A Final Exam

January 19, 2006 Please write legibly in the space provided below each question. You may not use calculators on this exam. We prefer that you use non-erasable pen when writing your answers. A significant number of completed exams will be photocopied by the teaching staff. Please write your name on each page of the exam. There are TEN multi-part questions in this exam. We recommend that you first read through all questions and begin with the questions that are easiest for you. Be sure to take a look at all questions before the end of the exam! Please write your TF’s name and section time on the front page only. Question 1 / 20 Question 6 / 38 Question 2 / 20 Question 7 / 44 Question 3 / 36 Question 8 / 28 Question 4 / 26 Question 9 / 36 Question 5 / 36 Question 10 / 16 Total /300


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1. You decide to use FRAP (Fluorescent Recovery After Photobleaching) to study microtubule dynamics. Your photobleaching set-up is as follows:

Photobleachone section

Microtubule growth w/fluorescent tubulin

Further growth w/ fluorescent tubulin ?

Centrosome Centrosome Centrosome

a. (6 points) Of the two choices below, which does the microtubule resemble after continued growth (assume catastrophe does not occur)? Explain briefly.

Centrosome

Choice A:

Centrosome

Choice B:

[Choice A: Microtubule growth occurs by the addition of tubulin dimers to the end of the microtubule (the “+” end), not to the origin at the centrosome. Thus, through the properties of FRAP, the photobleached section should appear stationary, as fluorescently-active tubulin dimers are added sequentially to the end]


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Shown below is the drug colchicine, which affects microtubule dynamics:

OCH3

OH3C

OH3C

OOH3C

NH

O

CH3

Colchicine

b. (4 points) Circle any chiral centers present on colchicine [Caveat: colcichine actually has a second source of stereochemistry, called atropisomerism, as a result of the rings clashing against each other… of course, the students won’t know this… just FYI]

c. (4 points)Draw a box around any amide bonds in colchicine.

d. (6 points) Colchicine inhibits the formation of microtubules by binding to tubulin dimers. With which phase of the cell cycle would you expect colchicine to interfere the most?

[Since microtubules are key to spindle formation and chromosome segregation, the inhibition of microtubule formation would probably lead to cell death (apoptosis) during M phase [should induce apoptosis in mitosis], specifically by initiating the spindle checkpoint. As a point of interest, since some plant cells can tolerate the missegregation of chromosomes (I guess there’s no spindle checkpoint), botanists use colchicine to generate polyploidy plant lines, such as seedless watermelons. There is a checkpoint, but cells eventually pass it and reenter interphase without either segregating their chromosomes or dividing and thus double their ploidy, frustrating prospective seed-spitters the globe over]


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2. Scientists can use analogs of ATP and GTP to analyze protein function.

Shown below are GTP, and two closely related analogs of GTP. The first analog, GMppCp, has a terminal (gamma) phosphorous-carbon bond that cannot be hydrolyzed. The second analog, 3’ deoxy GTP, is missing the 3’ hydroxyl group. Assume that GTP, GMppCp, and 3’ deoxyGTP bind to the proteins below with the same affinity (same Kd).

O N

N

N

NH

O

OHHO

O

P

O

O

O

PO

O

O

P

O

O

O

GTP

N

H

H

GMppCpO N

N

N

NH

O

OHHO

OP

O

OO

PCH2

O

O

P

O

OO N

H

H

O N

N

N

NH

O

OH

OP

O

OO

PO

O

O

P

O

OO

3' deoxyGTPN

H

H

a. (4 points) What effect should GMppCp have upon the activity of Ras?

Explain.

[Ras will always be in the ‘on’ state. GMppCp mimics GTP, however it cannot be hydrolyzed to GDP. Therefore Ras-GMppCp cannot be inactivated by hydrolysis.]

b. (4 points) What effect should 3’ deoxyGTP have upon the activity of Ras? Explain. [Ras should function normally. 3’ deoxyGTP will still activate Ras, and can still by hydrolyzed to 3’ deoxyGDP leading to Ras inactivation.]


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c. (4 points) Could RNA polymerase use GMppCp in the transcription of mRNA? Explain. [Yes. Both the bond between the alpha and beta phosphate is cleaved in RNA synthesis, and that bond is still cleavable in GMppCp. Alternatively give points for a No – if they say that without the hydrolysis of pyrophosphate into phosphate, the free energy of synthesis would be positive.]

d. (4 points) Could RNA polymerase use 3’ deoxyGTP in the transcription of mRNA? Explain.

[No. RNA synthesis occurs in the 5’ to 3’ direction, where each new nucleotide is added to the 3’ hydroxyl of the growing chain. Without the 3’ OH group, no synthesis could occur.]

e. (4 points) GTP modulates the propensity of microtubules to grow (by

adding subunits) or shrink (by losing them). Explain how GMppCp would affect microtubule elongation and contraction.

[Since GMppCp is non-hydrolyzable, tubulin dimers present in a microtubule will almost always be in the ‘GTP’ bound state. This state is more apt to grow rather than shrink. Therefore microtubules will elongate more frequently to longer lengths, and contraction will rarely occur if ever.]


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3. Adenosine triphosphate (ATP) is used by the cell to fuel many biochemical reactions.

a. (4 points) Describe the chemical features of phosphate esters that allow

ATP hydrolysis to be regulated in time and space in the cell. [Because ATP is kinetically stable, its hydrolysis is extremely slow without an enzyme. Therefore, the cell can “choose” when and where to hydrolyze it using enzymes. The source of this kinetic stability is the high concentration of negative charges on phosphate groups, which repel any nucleophile, including water, which needs to enter to attack the electrophilic phosphate atom.]

b. (6 points) Describe three methods that protein kinases use to catalyze the

transfer of a phosphate group from ATP to an amino acid sidechain.

[-maximize proximity of reactants -facilitate ideal reaction orientation -neutralization of negative charge on phosphate by metal ion coordination (Mg +2) and basic amino acids (e.g. lysine) thereby lowering the transition state energy.

-activation of Tyrosine nucleophile by base catalysis (from an aspartate)]

c. (8 points) Draw an energy diagram for the uncatalyzed and enzyme-catalyzed hydrolysis of ATP. Explain how the enzyme affects the kinetics and/or thermodynamics of this reaction.

Reaction Coordinate

Fre

e E

nerg

y


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The enzyme affects the kinetics by binding strongly the transition state and stabilizing it, thereby lowering its energy. So the reaction goes faster because this energy barrier is lower. The enzyme does not affect the thermodynamics of the reaction. No enzyme does this. d. (8 points) XP is a high-energy, phosphate-containing compound.

Phosphate hydrolysis of XP to X + Pi releases 30 kcal/mol of energy (compared to 7 kcal/mol released in the hydrolysis of ATP to ADP +Pi). The reaction diagrams for the uncatalyzed hydrolysis of ATP and XP are shown below. Based on your understanding of kinetics and thermodynamics together with your answer to part (a), would XP be a better or worse energy currency for the cell? Explain your answer.

[It would be a worse energy currency. Even though thermodynamically it would release more free energy, the kinetic barrier is so low that you could not store energy. [you could give extra points to those who point out that this is an inefficient way to run the cell since burning 30 kcal for each energetically unfaovrable reaction you couple ot XP hydrolysis is far more than you need for most reactions.]


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e. (4 points) In the amino acid sequence shown below, circle and name any amino acid residues that could serve as substrates for phosphorylation by protein kinases.

HN

HC C

HN

O

CH

CH

CHN

HC C

HN

CH2

HCO

O

CH

CHN

HC C

CH2

O

O

CH2

CH2

CH3

H3C

OH

CH3

OH

CH2

CH2

CH2

NH3+

SH

f. (6 points) Draw the phosphorylated versions of the amino acids that you indicated in part (e) above. You only need to draw those residues that are modified by phosphorylation, but be sure to include the entire side chain.

NH3+

O

O

-OP

O

O--O

phosphothreonine

NH3+

O

O

-O

P

O

O--O

phosphotyrosine


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4. The Her2 receptor is a receptor tyrosine kinase whose activation leads to cell growth and proliferation. In many breast cancers, Her2 receptor mutations have been isolated.

a. (4 points) The ligand for the Her2 receptor causes receptor dimerization.

What is the effect of receptor dimerization for many receptor tyrosine kinases?

[Receptor dimerization activates the receptor. Additional – brings the receptors close enough to transactivate one another through their kinase activity.]

b. (4 points) A point mutation in the transmembrane domain of the Her2 Receptor converting a Val to a Gln causes the receptor to dimerize in the absence of ligand. What effect would this mutation have upon the activation state of the receptor? [Always on, even in the absence of the ligand.]

c. (4 points) In order to study receptor tyrosine kinases, scientists can

genetically modify the mutant Her2 receptor described in part b and eliminate the intracellular domain. If this modified receptor is the only one expressed in a cell, will it respond to the Her2 ligand? Explain. [No. The intracellular domain has the kinase domain which is required for intracellular signaling. If it is absent, the kianse activity would be absent as well.]

d. (8 points) If a heterodimer were to form containing one subunit of the receptor described in part b and the other subunit of the receptor in part c, would this complex be able to activate downstream targets? Explain.

[No. Receptor activation occurs via trans-activation (one receptor phosphorylates the other one. If this heterodimer forms, the monomer with the functional kinase domain will have nothing to phosphorylate, while the monomer missing the kinase domain will be unable to phosphorylate the other monomer.]

e. (6 points) Activation of Her2 leads to the phosphorylation of MAP kinase

(MAPK). Why do normal cells require either a phosphatase that removes the phosphate group(s) from MAP kinase or a MAP kinase that is rapidly degraded by proteolysis?


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[Once the ligand for Her2 is gone, you want signaling to stop, meaning that you need to get rid of the phosphorylated MAP kinase that is carrying the signal. You can do this in two ways: removing the phosphate with a protein phosphatase OR degrading the phosphorylated MAP kinase, since any new MAP kinase you make will not be phosphorylated.]

5. Chronic Myelogenous Leukemia (CML) is a cancer caused by a specific

chromosomal translocation between chromosomes 9 and 22. The result of this translocation is the rapid proliferation of immature lymphoblasts (white blood cell precursors).

a. (6 points) In healthy individuals, immature lymphoblasts represent on

average less than 5% of the cells made in the bone marrow. The percentage of immature lymphoblasts is relatively constant in the bone, and individual lymphoblasts have a lifetime of about a week.

In untreated CML patients, the percentage of immature lymphoblasts increases to greater than 30% of the bone marrow in approximately 4-5 years. If the lymphoblast population is maintained at 30% for one year, have the lymphoblast populations of these untreated CML patients now reached a new steady state or a new equilibrium? Briefly explain. [New steady state. The inwards and outwards reactions are not the same reaction in reverse for steady state, while they are the reverse in an equilbrium. For the size of the lymphoblast population, the ‘reactions’ are the rate of birth and rate of death, which are not reverse reactions of one another.]

b. (6 points) Upon treatment with Gleevec, how do you predict the

lymphoblast populations will change? In your answer mention if after successful treatment with Gleevec for one year the populations will have reached a new steady state, a new equilibrium, or neither. [Population shrinks. New steady state has been achieved as Gleevec has decreased the rate of birth, but not the rate of death.]

c. (6 points) Based on its mechanism of action why does Gleevec cause fewer side effects than traditional chemotherapeutic agents like methotrexate? Explain in three sentences or less.

[More specific towards only cancerous cells. Traditional chemotherapeutics affect all cells, but affect cancerous cells slightly more than non-cancerous cells.]


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d. (6 points) Is Gleevec a cure for CML? Explain your answer. In order to gain a deeper insight into CML, scientists studied, c-Src a kinase related to the Abl kinase affected in CML. Below is the scheme showing the normal regulation of c-Src.

Scientists have found a mutant of c-Src. When the DNA from the mutant Src gene was sequenced, the reading frame was determined and it read as follows (in both wild type and mutant Src, TAC is the codon at position 499, and CAG and TAG are the codons at position 500, respectively):

499 500

WILD TYPE Src: 5’ TAC CAG 3’ MUTANT Src: 5’ TAC TAG 3’

e. (4 points) How does this particular mutation affect the translation of src?

[Results in a truncation of all the amino acids after residue 499. You lose the inactivating c-terminal tail.]

f. (8 points) This mutation affects the amino acid at position 500 (refer to

schematic for location of Tyr 527 in protein). Why did this mutation cause the mutant Src to be constitutively active?

[C-terminal tail stabilizes a closed-inactive form. Without it, that form is unstable and the equilibrium shifts to favor the open states. If those states are favored, then the number of open-active molecules increases, which results in increased signaling and growth.]


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6. Growing and dividing human cells take 24 hours to proceed through the cell cycle and complete division into two identical daughter cells.

a. (8 points) Label the four phases of the cell cycle on the diagram below and

indicate which of the phases are part of interphase.

Daughter cell

M

G1

S

G2

G0

G1, S, and G2 are part of interphase b. (4 points) Cells can enter and exit the cell cycle. Indicate with arrows from

which phase this entry and exit occurs, and name this resting state.

c. (8 points) Growing human cells take 24 hours to double their mass. What

will happen to the average mass of each cell in a population of proliferating cells if the duration of the cell division cycle is reduced to 18 hours, without affecting the time it takes for them to double each cell’s mass?

[the cells are now dividing faster than they are doubling in mass, and as a result they will get smaller in each successive cell cycle. People who assume that mass increases linearly over cell cycle time will conclude that their mass will fall by 75% in every cell cycle. Anyone who is smart enough to imagine they grow exponentially (dm/dt = km) will get a more complicated answer and should be given a small gold star.]


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By measuring the amount of DNA inside each cell in a population of cells, it is possible to construct a graph that show how many cells in the population contain various amounts of DNA. Such a graph is shown below for a population of cells that are growing and proliferating. The population is asynchronous, meaning that different individual cells are in different phases of the cell cycle. The graph has been divided into three sections: A, B, and C.

A B C

Relative Amount of DNA

Num

be

r o

f C

ells

1 20

d. (6 points) Which phase/phases of the cell cycle does section C of the

graph correspond to? Explain in two or fewer concise sentences. [Both the G2 and M phases come after the S phase, and thus, cells in these phases will contain ~ twice as much DNA as their G0 / G1 counterparts].

e. (6 points) Which phase/phases of the cell cycle does section A of the graph correspond to? Explain in two or fewer concise sentences.

[Both the G0 and G1 phases come before the S phase, and thus, cells in these phases will contain half as much DNA as their G2 / M counterparts].

f. (6 points) Which phase/phases of the cell cycle does section B of the graph most likely correspond to? Explain in two or fewer concise sentences.

[In the S phase, cells are in the process of replicating their DNA, and thus the cells will have somewhere between 1 and 2 “equivalents” of DNA, relative to the G1 / G0 cells].


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7. Ras is a member of a large family of small GTPases and plays a key role in

many intracellular signaling pathways. When Ras is bound to GTP, it is capable of activating many protein kinases, while when Ras is bound to GDP, Ras is inactive. For this reason, the binding of GTP is a molecular “switch” for many biochemical reactions inside the cell. The figure below shows GTP in the active site of Ras, where the numbers represent the amino acid residues of Ras.

NH

N

N

O

NH2N

O

OH

HH

HHOH

OPO

O

O-

POP-O

OO

O-O-

HN

O

HN

O

NH

O

HN

O

NH3+

117

146

119

Mg2+

OH HO

35

17

HN

O

+H3N

16

NH

OO

OH


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You make mutations in the GTP-binding pocket of Ras and examine the effects on the binding of GTP. For parts a-c, first draw the new amino acid. Then considering the size and the of the substitutions give the most likely reason why each mutation has the stated effect.

a. (4 points) Residue 16 is mutated to an Arginine, resulting in Ras that still

binds GTP. NH2

HN

NH2+

H2N

O

OH

Arginine charged or uncharged okay [Arg and Lys are both positively charged, thus the ionic interaction with the phosphate group is preserved. The side chains of both amino acids are also of similar size.]

b. (4 points) Residue 119 is mutated to a Leucine, resulting in Ras that

cannot bind GTP.

NH2

O

HO

Leucine [Polar Aspartic Acid is replaced with nonpolar Leucine. The hydrogen bond between the Aspartic Acid and the guanine ammonia group is lost, however it is more likely that steric hindrance of the slightly larger leucine group or even misfolding of the protein prevents binding of GTP.] c. (4 points) Residue 146 is mutated to a Tyrosine. No effect on GTP

binding is seen. H2N

OHOHO

Tyrosine [The sidechain is not directly involved in GTP binding, instead it’s the backbone of the protein that interacts with GTP. Changing the side chain in this case does not perturb the backbone, and therefore GTP binding is not disrupted.]


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ATP and GTP are structurally very similar however Ras is extremely selective in only being activated by GTP and not ATP. Biochemical analyses have determined that the Kd for GTP binding to Ras is 5 x 10-12 M and the kon is 2 x 106 M-1 s-1. The binding of ATP to Ras is much weaker, however the kon for ATP is identical at 2 x 106 M-1 s-1.

d. (8 points) Write the chemical equilibrium for the dissociation of Ras-GTP

to Ras and GTP. Label kon next to the appropriate arrow in your reaction. [Ras-GTP Ras + GTP]

e. (8 points) Write the rate equations below for both the forward and reverse reactions of Ras-GTP dissociating to Ras and GTP. Forward rate = koff [Ras-GTP] Reverse rate == kon [Ras][GTP]

f. (8 points) At equilibrium, how do these rate equations relate to the equilibrium constant for the dissociation of Ras-GTP to Ras and GTP (Kd). At equilibrium the rate of the forward and reverse reaction is equal:

koff [Ras-GTP] = kon [Ras][GTP] and Kd= [Ras][GTP] so Kd= koff

[Ras-GTP] kon

g. (8 points) Given your answers above, explain why Ras preferentially binds GTP over ATP in terms of specific kinetic or thermodynamic parameters.

[The preferential binding of Ras to GTP versus ATP means that the Kd value for Ras binding to GTP must be lower. Since Kd= koff /kon and the kon values are the same for both GTP and ATP associating with Ras the difference in the binding must result from ATP having a higher koff value than GTP.]

koff

kon


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8. Recall the Dupont-Merck HIV protease inhibitors and examine the analogs below:

N N

HO OH

O

O

NH

O

HN

N

S S

N

Dupont-Merck Inhibitor + HIV protease Ile 50 and Ile 50' partially shown

H

N

H

NIle 50 Ile 50'

N N

HO OH

O

NH

O

HN

N

S S

N

Analog #1

N N

HO OH

O

O

NH

O

HN

N

S S

N

Analog #2

a. (8 points) How would you expect the Kd value for the complex between Analog #1 and HIV protease to compare with that of the Dupont-Merck inhibitor? Explain your answer in terms of specific enthalpic and/or entropic differences.

[[[no water molecule released upon binding into bulk solvent for analog #1, therefore less positive delta S(water). Also loses the possibility of making H-bonds between analog #1 and the Ile residues in HIV protease (they would have to remember this, but could get nearly full credit without this additional answer), which


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means less negative delta H(protease-inhibitor) upon binding than in the Dupont inhibitor.

b. (8 points) How would you expect the Kd value for the complex between Analog #2 and HIV protease to compare with that of the Dupont-Merck inhibitor? Explain your answer in terms of specific enthalpic and/or entropic differences.

[[[with analog 2 the ring is broken and the analog is more floppy than the Dupont inhibitor, therefore more ordering of the inhibitor upon binding is necessary. As a result, delta S (inhibitor) is more negative upon binding.

c. (8 points) Inhibition of the HIV protease prevents the formation of mature

viral proteins. One of the proteins affected is integrase. How would an inability to make integrase would this affect future viral lifecycles?

[If virus infected a new cell it would not be able to insert its viral DNA into the host cell’s genome (DNA)]

d. (4 points) HIV protease itself must be cleaved from a gag/pol precursor polyprotein substrate by another HIV protease molecule. What type of feedback loop does this represent for untreated wildtype virus?

[positive feedback]


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9. The enzyme triose phosphate isomerase, or “TIM,” catalyzes the interconversion between two metabolic intermediates, as shown below:

H

OPO32-O

OH Keq = .042OPO3

2-HO

O

Compound BCompound A

Energy viametabolism

TIM(enzyme)

a. (4 points) If compound A and B come to equilibrium in a flask, which

compound is present in higher concentrations?

A

For each circumstance below, choose the set of arrows (A, B, or C) that best describes the relative rates of the forward and reverse reactions.

A B A BA B

Choice A Choice B Choice C

b. (4 points) Enzyme has JUST been added to a solution containing equal amounts of compound A and compound B. Choice: ___C (not A)___

c. (4 points) A solution containing compound A, compound B, and the enzyme has reached equilibrium. Choice: __ B____

d. (4 points) The enzyme is present within a cell, where compound B is constantly being consumed to generate metabolic energy. Choice: __ A (not B)____


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The first step of the reaction catalyzed by TIM is shown below:

OPO32-

OHH

OHO

OEnyzmeglutamate

NNH

Enyzmehistidine OPO3

2-

OHH

OHO

OEnyzmeglutamate

NNH

Enyzmehistidine

OPO32-

OH

OHO

OEnyzmeglutamate

NN

Enyzmehistidine

H

H

‡

e. (8 points) How does the enzyme glutamate assist in catalyzing the

reaction? Name the specific type of catalysis in your answer and explain how it accelerates the reaction. [base catalysis-glutamate draws a hydrogen away from Compound A helping the carbon to form a double bond]

f. (8 points) How does the enzyme’s histidine assist in catalyzing the reaction? Name the specific type of catalysis in your answer and explain how it accelerates the reaction.

[acid catalysis-histidine acts as a proton donor, it protonates carbonyl group in Compound A]


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10. Arachidonic acid is one of the essential fatty acids required by most mammals.

HO

OArachidonic acid

a. (4 points) Label each double bond as cis or trans

all cis

b. (4 points) How many total geometric isomers of arachidonic acid are possible?

24=16

c. (4 points) Give one physical property you could use to distinguish the geometric isomers of arachidonic acid from one another. Physical property such as physical length, boiling point, melting temperature, etc.

d. (4 points) Using the following graph determine the pKa for the carboxylic acid group of arachidonic acid.

4.7, anything between 4.5 and 5 okay


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The Genetic Code

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Life Science 1A Final Exam January 19, 2006sites.fas.harvard.edu/~lsci1a/F06final02key.pdf · Life Science 1A Final Exam January 19, 2006 ... Explain briefly. Cen trosome Choice A: