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MCB 440B

Second Examination

Thursday November 10, 2011

9:30 am – 10:50 am, 217 Noyes

Students should answer all questions. Marks are allocated to each question as indicated. Show all work and calculations clearly, and put units on answers where required. Answer each section on a separate booklet Make sure that your name is included on each set of answers handed in, and on each sheet if separate. Hand in the exam question booklet together with your answer booklets.

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1. (a) Consider the formation of a double stranded RNA duplex from two sequences that are non self-complementary (duplex concentration of 0.0001 M): Top strand + Bottom strand Duplex RNA For the melting of the duplex into the two single strands, derive an expression for the equilibrium constant at Tm using only the concentration (CT). Using this expression, calculate the Tm and ΔGo

37 if ΔHo = 30 kcal/mol and ΔSo = 100 kcal/mol K.

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Given ΔHo = 30 kcal/mol and ΔSo = 100 kcal/mol K, calculate Tm and ΔGo37.

Plug in the numbers to the equation (duplex concentration = 1.0 x 10-4 M = CT) given above to get: Tm = (30 x 103)/[1.987 ln (1.0 x 10-4/4) + 100 x 103] ΔGo

37 = ΔHo37 -T ΔSo

37 = (30 x 103 cal) – 310 K(100 x 103cal/mol*K) (b) Now consider a DNA duplex with the following chemical structure (duplex concentration of 0.00005 M): 5’- C G T T G A – 3’ | | | | | | 3’- G C A A C T – 5’ Derive an expression for Tm in terms of ΔHo, ΔSo, and CT. Using this expression, assuming standard condition of 0 M NaCl, calculate the Tm for the duplex. Show all work.

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terms ΔHo (kcal/mol) ΔSo (cal/mole)

CG/GC -10.6 -27.2 GT/CA -8.4 -22.4 TT/AA (read as AA/TT) -7.9 -22.2 TG/AC (read as CA/GT) -8.5 -22.7 GA/CT -8.2 -22.2 Symmetry 0 0 initiation with G-C 0.1 -2.8 Total -43.5 -119.5 (with symmetry correction) -43.5 -119.5 Use the expression given above:

Plug in values for ΔHo = -43.5 kcal/mol=-43,500 cal/mol, CT= 1.0 x 10-4 M and ΔSo = -119.5 cal/mol K.

oT

o

m SCRHT

Δ+

Δ=

)4/ln(

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2. For the following transition

(a) Draw a free energy diagram showing the (relative) standard state chemical potential for each of the species

(b) Using a diagram similar to that in Part (a), show how the addition of 1 M NaCl would affect the standard state chemical potential of each of these species. For which form would the standard state chemical potential be most affected? Explain.

As the DNA transitions from unstacked to stacked to dsDNA, the distance between phosphate groups in the backbone become shorter. Consequently, the addition of counterions will have the greatest stabilizing effect on dsDNA because in this form, the distance between the bridging phosphate groups is the shortest.

(c) Using a diagram similar to that in Part (a), show how increasing the G-C content would affect the standard state chemical potential of each of these

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species. For which form would the standard state chemical potential be most affected? Explain.

Increasing the GC content only has an effect on the strength of Watson-Crick pairing. Since neither unstacked ssDNA or stacked ssDNA has Watson-Crick interactions, these are not affected at all. The only form that is affected is dsDNA. (d) Because DNA is polymorphic, it can form different structures. The most drastic structural change involves a GC base pair going from B-form (right handed helical turn) to Z-form (left handed helical turn). The tendency to form Z-form DNA is enhanced if the cytosine is methylated. Using free energy diagrams, show the standard state chemical potentials for a GC base pair in B-form DNA and in Z-form DNA. Using the same free energy diagram, shows the standard state chemical potentials for a GC pair in which the C is methylated in both B-form and Z-form. Explain the diagram.

Normally B-form DNA is the more stable form as Z-form has C5 buried in a hydrophobic patch. Addition of a methyl group on cytosine would expose this to solvent in B-DNA but would increase the hydrophobic surface in Z-DNA.

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3. A closed circular plasmid of total length 8000 base pairs is isolated from a bacterium in a relaxed form (assuming that the DNA is B formed with 10.5 bp/turn, 3.5 Å per bp).

(a) What are the twist, writhe and linking numbers of the plasmid? What is the length (in Å) of the DNA? (i) Length=8,000 basepairs x 3.5 Å per basepair = 28,000 Å (ii) Twist = 8,000/10.5 = 762 (iii) Writhe = 0 (no supercoils) (iv) Linking number = 762 (b) A protein that is capable of binding single stranded DNA is added to the solution. The protein interacts with and separates 200 bases of the DNA. What are the twist, writhe and linking numbers? The ssDNA interacting protein interacts with and unwinds 200 bases of DNA (200/8000= 2.5%). Therefore, twist decreases by 2.5% or (762)*(-0.025) = 19.05 = 20 turns (not 19!)

(i) Twist = 762 – 20 = 742 (ii) Writhe = +20 (positive supercoils) (iii) Linking number = 762 (c) Adding a topoisomerase results in the plasmid DNA becoming relaxed. What are the twist, writhe and linking numbers? The topoisomerase relaxes the DNA so:

(i) Twist = 742 (ii) Writhe = 0 (relaxed) (iii) Linking number = 742 (d) The topoisomerase is removed, followed by removal of the single stranded DNA binding protein. What are the twist, writhe and linking numbers? Removing the topoisomerase locks the linking number.

(i) Twist = 762 (goes back to B-form DNA) (ii) Writhe = -20 (negative supercoil) (iii) Linking number = 742 (locked in) (e) The crystal structure of the topoisomerase bound to DNA has been solved and shows a Arg side chain interacting with an AT base pair. Draw

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the chemical structure of Arg interacting with an GC (Watson Crick) base pair. Draw the chemical structure of Arg interacting with an GC (Hoogsteen) base pair.

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4. (a) Draw the binding schemes for an enzyme (E) interacting with substrate (S), in the presence of a competitive inhibitor (I). Label the equilibrium constants KM and Ki and indicate the step that corresponds to kcat. Define the forward rate in terms of kcat.

For the above case, kcat = k2 and rate = kcat [ES] (b) Draw the binding schemes for an enzyme (E) interacting with substrate (S), in the presence of a uncompetitive inhibitor (I). Label the equilibrium constants KM and Ki and indicate the step that corresponds to kcat.

For the above case, kcat = k2. (c) For the case of uncompetitive inhibition, derive an equation for d[P]/dt

(d) For the case of uncompetitive inhibition, use the steady state approximation to derive an equation in terms of [ES], [E] and [S]. (e) For the case of uncompetitive inhibition, define the total enzyme concentration and use this to derive a relationship between the forward rate and the equilibrium constants KM and Ki. SEE NEXT PAGE.

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5. The following data were obtained for an enzyme-catalyzed reaction that follows Michaelis Menten kinetics:

k1 and k-1 are much faster than k2

k2 = 100 s-1, KM = 1.0 x 10-4 M at 280K k2 = 200 s-1, KM = 1.5 x 10-4 M at 310K (a) For a substrate concentration of 0.1 M and a total enzyme concentration of 1.5 x 10-5 M, calculate the rate of formation of product at 280K.

k2 [Eo]* [S] 100s-1 * 1.5x10-5M*0.1 M Rate = ------------------- ------------------------------- = 1.5 x10-2M s-1 KM+ [S] 1.5x10-5M+ 0.1 M

(b) Calculate the activation energy for k2. Use Arrhenius equation k = Aexp[-Eact/ RT] k’ = Aexp[-Eact/ RT’]

k/k’ = Aexp[-Eact/R(1/T-1/T’)]

ln (k/k’) = Eact/R(1/T’-1/T)] ln (100/200) = Eact/8.314 J K-1M-1(1/310K-1/298K)]

Eact= 16.67 kJ/M (c) What is (roughly) the ratio of k1/k-1 for the enzyme-substrate complex at 280K? Roughly, since k2 is slow: KM = k-1/k1 so k1/k-1 = 1/KM =1/(1x10-4 M)= 1x104 M

PEk

ESk

kSE +→+

−

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1

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(d) What is the sign and magnitude of the free energy change for the formation of ES from E and S? (e) What is the catalytic efficiency (kcat/KM) for the reaction at 310K? Since k2 = kcat because it is the slow step, k2/KM= kcat/KM =(200s-1)/(1.5x10-4M)

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Biochemistry 440b

Fall 2012

Useful Information for Examinations

Physical Constants Avogadro's number N 6.02 x 1023 mol-1

Boltzman's constant k 1.381 x 10-23 J K-1 Gas constant ( N x k ) R 8.314 J mol-1 K-1 1.987 cal mol-1K-1 Planck's constant h 6.626 x 10-34 J sec electron charge e 1.602 x 10-19 C Faraday constant ( N x e ) F 9.649 x 104 C mol-1 9.649 x 104 J mol-1 V-1 23.06 kcal mol-1 V-1 velocity of light c 2.998 x 108 m sec-1 (1 cal = 4.184 joule (J) ) Ideal Gas Law

PV nRT= General thermodynamic relationships: First Law

rev revU q wU q w

Δ = +

Δ = +

Other State Functions H U PVA U TSG U TS PV H TS

= +

= −

= − + = −

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(if only PV work)

(T constant)

(constant T and P)

rev rev

V V

rev

P P

dU w q PdV TdSdU q C dT

dA dU TdS

dH dU w dU PdVdH q C dT

dG dH TdS

= + = −

= =

= −

= − = +

= =

= −

For a general biochemical reaction: aA + bB cC + dD

, ln[ ]om A A AG RT Aµ µ= = +

Temperature-dependence van’t Hoff relationship (assumes constant 'oHΔ and 'oSΔ )

' ''ln

o oH SKRT RΔ Δ

= − + 'oHΔ

pH-dependence

'

'

, ,

' '2 1 2 1

( )2.303

( )

If is constant, then

2.303 ( )

or

r HT P

r H

o or r r H

G RT NpH

N

G G RT N pH pH

ξ

" #∂ Δ= Δ& '∂( )

Δ

Δ −Δ = Δ −

For the dissociation of a weak acid, AH A- + H+

' { } { }ln{ } { }

c do

a b

C DG G RT

A B!Δ = Δ +

' 'lnoeqG RT KΔ = −

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log[ ]eqpK K= − For a redox half-cell:

1ox redA + A eeν

− �

' ' [ ]ln[ ]

o redr A r A

ox

AG G RT

A! "

Δ = Δ +$ %& '

''

oo r

e

GE

FνΔ

= −

'' r AA

e

GEFν

Δ= −

' ' [ ]ln[ ]

o redA A

e ox

ARTE E

F Aν

" #= −% &% &' (

' ' [ ]59log at 298K

[ ]o red

A Ae ox

AE E

Aν= −

'

', ,

( ( )) 2.303 ( )( )

o

mr H

eT P

E A RT N ApH F

ξν

# $∂= − Δ( )

∂( )* +

'

', ,

( ( )) change in protons bound 59 59 at 298K

( ) number of electrons transferred

o

m r H

eT P

E A NpH

ξν

# $∂ Δ= − = − •( )

∂( )* +

For ligand binding: Independent, equivalent binding sites:

d

d

KAKAN/][1/][

+=ν or dKA /]{

1=

−θθ where

Nν

θ =

Infinite positive cooperative binding:

dN KA /][

1=

−θθ

][][

log10 AHApKpH−

+=

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Hill Plot:

Plot θ

θ−1

log vs log[A]

(the slope at 50% saturation is defined as the Hill coefficient: nH)

Cooperativity parameter for a dimer: α

α+=12

Hn