Chem/Biochem 471
Half Exam 3
10/22/10
Page 1 of 2
Name:_________ KEY ___________________
1. The following reaction is the fourth step in glycolysis:
fructose-1,6-diphosphate (FDP)
→ dihydroxyacetone phosphate (DHAP) + glyderaldehyde-3-phosphate (G3P)
0'
ΔG298
= 23.8kJ / mol
a) What is the equilibrium constant for this reaction and is it spontaneous under standard
conditions (25˚C)?
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0'
⎡ −ΔG ⎤
⎡
⎤
−23.8kJ / mol
K = exp⎢
⎥ = exp⎢
⎥
−3
⎣ (8.3145 ×10 kJ / mol)(298K ) ⎦
⎣ RT ⎦
K = 6.74 ×10 −5
Not spontaneous because ∆G0’ > 0 or K < 1.
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b) If the entire reaction were diluted 10-fold, would this shift the equilibrium toward reactants or
products? Explain.
Shift is towards products because diluting 2 products vs 1 reactant will favor products (Le
Chatelier) OR more quantitatively: dilution will reduce the numerator of Q more than the
denominator, making Q < 1, so decreasing ∆G which will shift the equilibrium towards products.
ΔG = ΔG 0' + RT lnQ
Standard conditions: Q=(1)(1)/(1); 10-fold dilution makes Q=(0.1)(0.1)/(0.1) = 0.1 so RTlnQ is
negative which lowers ∆G.
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c) Suppose the enzyme aldolase is added to an aqueous solution of these species to catalyze the
reaction and establish equilibrium. What will be the final value of ∆G?
∆G = 0 at equilibrium
d) Determine whether the reaction will be spontaneous in red blood cells, which contain the
following concentrations of these species (assume 25˚C):
[FDP] = 35 µM
[DHAP] = 130 µM
[G3P] = 15 µM
(130 ×10 −6 )(15 ×10 −6 )
ΔG = ΔG + RT lnQ = 23.8kJ / mol + (8.3145 ×10 kJ / molK )(298K )ln
(35 ×10 −6 )
ΔG = −0.5kJ / mol
Yes it will be spontaneous.
0'
€
−3
Chem/Biochem 471
Half Exam 3
10/22/10
Page 2 of 2
Name:_________KEY___________________
2. In most cases native proteins are in equilibrium with their denatured form:
protein(native) ↔ protein(denatured)
For ribonuclease (a protein), the following equilibrium concentration data for these two forms
were experimentally determined for a total protein concentration of 1 × 10-3 M:
Temperature (°C)
Native
Denatured
-4
100
8.6 × 10 M
1.4 × 10-4 M
25
1 × 10-3 M
1.4 × 10-7 M
a) What is Keq for the denaturation reaction at 100 ºC?
K eq (100˚C) =
1.4 ×10 −4
= 0.16
8.6 ×10 −4
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b) If Keq at 50 ºC is 2.58 x 10-3 and the total protein concentration is 1 × 10-3 M, what is the
concentration of the denatured protein at 50˚C?
Let x = [denatured], which makes [native] = 1 x 10-3 – x
[denatured]
x
K eq (50˚C) = 2.58 ×10 −3 =
=
[native]
1×10 −3 − x
(both in units of M).
2.58 ×10 −6 − 2.58 ×10 −3 x = x
x=
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2.58 ×10 −6
−6
−3 = 2.57 ×10 M
1+ 2.58 ×10
c) Briefly describe how you would calculate ΔH0 for the denaturation reaction: what formula
would you use and what values would you plug in? Assume ΔH0 is independent of temperature.
, Plug in K values at 2 temps and calculate ∆H0:
could use 0.16 at 100˚C=373K, 2.57 × 10-3 at 50˚C=323K, or 1.4 × 10-4 at 25˚C=298K.
d) Is denaturation spontaneous under standard conditions (25˚C)? Briefly explain your answer.
Denaturation is not spontaneous under standard conditions (25˚C) because K < 1 (reactants are
favored over products). OR K < 1 makes ∆G0 = -RTlnK > 1 which means it’s not spontaneous.