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A midterm exam for a chemical engineering course, covering topics such as protein structure, damkohler number, enzyme inactivation, and enzyme kinetics. It includes short answer and short problem questions.
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Chemical Engineering 170 October 16, 2003 Midterm Exam Closed Book and Closed Notes One 8.5 x 11 in. page of notes allowed
Section 1. Short Answers
Hydrogen bond
Hydrophobic Interactions
Ionic Interactions (electrostatic)
Van der Waals
Disulfide bonds
meaning.
KS S o
V max Da = Maximum TransportRate
Maximum Reaction Rate Da = ;
Km D eff
R V max 3
φ =
If φ >>1 then Vmax>> KmDeff, hence the system must be transport- limited. Reactant molecules are consumed before they diffuse very far, and the reaction is limited to a thin region near the periphery of the particle. Thus, the observed reaction rate is much smaller than the rate in the
absence of an internal concentration gradient (ηI<<1)
i. glutamate changed to aspartate
Neutral – similar size and pKa’s.
ii. leucine changed to valine
Neutral – both branched-chain hydrophibics of similar size.
iii. valine changed to glycine
Not neutral – G is very small, allows significant free rotation around peptide bonds.
iv. cysteine changed to alanine
Not neutral – A is hydrophobic, C is polar and can even dissociate.
v. serine changed to threonine
Neutral – both contain –OH in side chains and similar size.
vi. tryptophan changed to phenylalanine
Neutral – both are hydrophobic aromatics of similar size.
vii. arginine changed to lysine
Neutral – both are large and basic.
viii. histidine changed to alanine
Not neutral – H is polar and ionizes with a pKa around neutrality, A is non-polar.
The two-stage model of enzyme inactivation is: N ←→ K^ D → k I
The corresponding expression for kobs is:
k kobs
, where
Section 2. Short Problems
E + S ES E + 2 P
You are also given the following information:
kp = 5 x 10-3^ cm/sec Km = 0.5 mM v 'max= 100 μM cm-2^ min- So = 1 mM ηE = 0.
a) Write an expression that relates the flux of product from the surface to the rate of substrate consumption at the surface. Please define any parameters in your expression that are not given above.
m
' * max o
p K S
2 v S k P P
m o
o
' E max o
p (^) K S
2? v S k P P
P* = concentration of product at surface S* = concentration of substrate at surface
b) Calculate the measured rate of appearance of product in the bulk per unit area.
0.5mM 1 mM
1 mM cm min
2 0.40 100 μM K S
2? v S k P P 2 m o
o
' E max o
p 53.3^ μM cm
-2 (^) min-
However, the mutant enzyme still follows general enzyme kinetics:
Use these definitions in answering the following questions:
3
3 (^3) k
k K = − ; 1
1 2 m k
k k K
a) Assuming the reaction occurs in a closed system, begin by writing mass balances for the enzyme complexes (ES and D):
3
2 k 1 E S k 1 ES k 2 ES k 3 ES k D dt
d ES = − − − − + −
3
2 k 3 ES k D dt
d D = − −
b) Using the pseudo-steady-state-hypothesis for each enzyme complex, derive an expression for [ES] in terms of Eo (and other parameters, including constants, but no other enzyme complexes):
3
2 k 3 ES k D dt
d D = = − − => 3
2 2 3
k
k D = = − also, since k (^) 3 [ ES ]^2 = k − 3 [ D ]:
3
2 k 1 E S k 1 ES k 2 ES k 3 ES k D dt
d ES = = − − − − + −
1
1 2 S
k
k k E = m
k 1
k (^) -
k 2
k 3
k (^) -
C 6 H 12 O 6 + a O 2 + b NH 3 → c (C4.4H7.3N0.86O1.2) + d H 2 O + e CO 2
Assume that the cells can convert 2/3 (wt/wt) of the substrate carbon to biomass.
a) Calculate each of the stoichiometric coefficients, a, b, c, d , and e.
Amount of C per mol of glucose: 72g; 2/3*(72g) = 48g converted to biomass
Biomass balance: 48 g/mol = (4.4)(c)12g/mol c = 0.
N balance: 14b = (0.86)(0.909)14 b = 0.
H balance: 12 + 3(0.782) = (0.909)(7.3) + 2d d = 3.
C balance: 6(12) = 0.909(12)(4.4) + 12e e = 2
O balance: 6(16) + 2(16)a = (0.909)(1.2)(16) + (3.86)(16) + 2.16e a = 1.
b) Calculate the yield coefficient YX/S (g dw cell/g substrate).
YX/S = (0.909)(MW biomass)/(MW glucose) = (0.909)(91.34)/
YX/S = 0.461 g biomass/ g glucose
c) How would you expect this coefficient to compare to YX/S for hexadecane (i.e., larger or smaller)? Why?
It should be lower because hexadecane is more reduced than glucose (i.e. YX/SHex^ > YX/SGlu^ ). In other words, hexadecane has a greater degree of reductance, so it has more electrons/energy to form biomass.