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Chemical Engineering 170 October 11, 2004 Midterm Exam Closed Book and Closed Notes One 8.5 x 11 in. page of notes allowed
Section 1. Short Answers
- (6 pts.) Recently site-directed mutagenesis was used to modify the cofactor specificity of the enzyme phosphite dehydrogenase (Woodyer et al., Biochemistry 2003 , 42, 11604). Shown in Figure 1 on the last page are interactions between the wildtype (WT), the E175A mutant, and the E175A, A176R double mutant enzymes and the nicotinamide cofactor NADP (the 2’- phosphate of NADP is shown in red and yellow). The numbers shown are distances of the dotted lines in angstroms.
a) Which of the above three enzymes has the highest Km toward NADP?
WT (+1)
b) Briefly explain (1-2 sentences) your answer above.
Electrostatic repulsion between the two negative charges (+2)
c) Which of the above three enzymes has the lowest Km toward NADP?
E175A, A176R double mutant (+1)
d) Briefly explain (1-2 sentences) your answer above.
Electrostatic attraction between NADP and R176 (+2)
- (12 pts.) As shown in Figure 2 on the last page, the enzyme tRNA guanine transglycosylase (TGT) catalyzes replacement of guanine (G) by 7-aminomethyl-7-deazaguanine (PreQ1) at a specific position of four tRNAs. A proposed mechanism of the TGT-catalyzed reaction involves the following steps: (i) nucleophilic attack of the side chain of Asp102 on the C1' of the targeted nucleotide (tRNA-G) to assist the departure of the guanine base and form a TGT–tRNA covalent intermediate; (ii) replacement or exchange of guanine by PreQ1 in the guanine/PreQ1-binding pocket in TGT; and (iii) covalent linkage of PreQ1 to the tRNA through a nucleophilic attack of the N9 of PreQ1 (shown in blue on the last page) on the C1' of the targeted ribose to form the product.
a) Write a mechanistic equation (e.g., E + tRNA-G? E⋅tRNA-G → E + tRNA-PreQ1)
consistent with the mechanism described above.
E + tRNA-G? (E•tRNA-G)? E-tRNA? (E-tRNA•PreQ1)? E + tRNA-PreQ1? E + tRNA-PreQ
or
E + tRNA-G? (E•tRNA-G)? E-tRNA? (E-tRNA•PreQ1)? E + tRNA-PreQ1 (+6)
b) Using site-directed mutagenesis, you propose to generate two different mutant enzymes, with each mutant containing a replacement for Asp in position 102. Assuming each mutant is to retain at least some catalytic activity, which two amino acids would you use to replace Asp102?
Glu and Ser
c) The above reaction is inhibited by 9-deazaguanine, in which the N9 of guanine is replaced by carbon. What type of inhibitor is 9-deazaguanine?
Competitive
d) Assume that the Ks value of the enzyme for tRNA-G is 10 μM, and that the enzyme-catalyzed reaction proceeds 10^7 -times faster than the uncatalyzed reaction. What, then, is the theoretical lowest value of Ki, the inhibition constant for 9-deazaguanine?
T T
S n
e K K
K
k
k (^) 10 M 107 μ = = →
(+1) (+1)
K (^) T ≈ Ki = 10 × 10 −^7 μ M= 1 pM(+1)
- (6 pts.) Give brief definitions for the following:
a) Operon a group of prokaryotic genes under the control of a single operator (i.e. expressed under the control of one promoter and regulatory mechanism) (+2)
b) Codon a three-base sequence in mRNA that codes for a specific amino acid in a protein or causes the termination of translation (+2)
c) Diauxic growth growth occurring in two phases between which a temporary lag occurs; due to the switching in metabolism from one limiting growth substrate to another; generally carbon (+2)
- (3 pts.) Name three components usually found on an E. coli expression vector (plasmid).
promoter, terminator, multiple cloning site (restriction enzyme sites), antibiotic resistance
marker, origin of replication, gene of interest (each +1)
- (6 pts.) Name three advantages and three disadvantages of immobilized enzymes.
Advantages: can be reused continuously, easier product recovery, suitable for many reactor types, require less space (lower capital costs), often exhibit increased stability, introduce potential for manipulation of catalytic properties (each +1)
Disadvantages: loss in activity, mass transfer limitations, additional cost (reagents, carrier), impractical for solid substrates (each +1)
- (7 pts.) Write a complete expression for the rate of substrate consumption by a growing cell population. Label all terms discussed in class.
=
n
j (^) P s
P i xs
x si j i
j i Y
r mx Y
r r /^1 /
,
Y j i
r j
x
m
Y i
r x
r q x i
j i
j
i
P s
P
i
xs
x
si si
yieldcoefficientofproduct on substrate
rateofformationofproduct
cellbiomass
maintenancecoefficient
yieldcoefficientonsubstrate
growthrateofcells
rateofsubstrate consumed
/
/
, ,
μ
Is this an unstructured or structured model? Segregated or nonsegregated?
Unstructured (+1), Nonsegregated (+1)
An organism’s doubling time is 40 minutes. Starting with a single cell, how many cells will be
present after 6 hours?
9 doubling times 40 minutes
1 doublingtime 1 hour
60 minutes 6 hours× × = 29 = 512 cells (+1)
- (14 pts.) A peptide has four independently ionizable groups with these pK’s:
pK 1 (COOH) = 2.
pK 2 (ImidazoleNH+) = 6.
pK 3 (Phenolic OH) = 10.
pK 4 (NH 3 +) = 10.
a) What is the approximate (or exact) net + or – charge on each group at these pH’s (complete
the following table)?
Net + or - Charge COOH of pK 1 Imid. of pK 2 OH of pK 3 NH 3 +^ of pK 4
pH 3 -0.90 +1.0 0 +1.
pH 7 -1.0 +0.10 0 +1.
pH 10 -1.0 0 -0.5 +0.
b) What is the net + or – charge on the peptide at
pH 3: +1.1 pH 7: +0.10 pH 10: -1.
[ ]
[ ]
[ ]
[ ][ ]
[ ]
[ 0 ] [ ] [ ] [ ] ECS
S
K ECS
S C
K K ECS
E = E + EC + ECS = S^ C + S +
[ ][ ] [ ]
[ ]
[ ]^0
S
K
S C
KK
E
ECS
S C S
[ ][ ] [ ]
[ ]
[ ]
[ ] 0
2 2
S
K
S C
K K
E
k ECS k dt
dP S C S
( [ ]) [ ][ ]
[ ][ ][ ]
[ ] [ ][ ]
[ ] 2 [ 0 ][ ][ ] 2 0
K K C S C
k E S C K K K C S C
k E S C dt
dP S C S S C + +
1 [ ]
[ ]
[ ] 2 [ 0 ][ ]
S
C
K
K
k E S dt
dP C S (^) +
= (this form is also acceptable)
(c) Under what condition would the reaction velocity to be first order in C? [ C ]<< K C (+2) (d) Under what condition would the rate be independent of cofactor concentration? [ C ]>> K C (+2)
- (19 pts.) As we discussed in class, Eadie-Hofstee plots are useful for determining kinetic
parameters of enzyme-catalyzed reactions. Shown below are two such plots (note the axes have
been switched from what we discussed in class): one for the enzyme chymotrypsin immobilized
to small (~ 10 μm) catalyst particles (O), and one for chymotrypsin immobilized to large
spherical (~ 120 μm) porous particles (*). The reaction velocity, v, is defined per unit catalyst
volume; s 0 is the bulk substrate concentration, ea is the active enzyme loading per weight of
catalyst, and ρp is the particle density.
a) Why does the plot for the large catalyst particles curve downward near the y-intercept,
and approach the straight line for the small catalyst particles at large values of v? Please limit
your answer to a few sentences.
b) Now we will derive an expression for the y-intercept of the immobilized enzyme data in terms of the Thiele modulus of the catalyst. If φ is sufficiently large, as we will assume it is for the large particles, the effect of curvature can be neglected and the particle can be treated as a slab.
i) Write a steady state material balance on substrate through the particle in terms of d^2 S/dr^2.
m
c (^) Sr K
v Sr vr dr
d S D
2 (^ ) max
2 (+3)
ii) Use the following identity
2 2 2
1 2
d S d dS dr dS dr
= ^
and the material balance from (i) to show that, if S(r) << Km,
1 / 2 max r R m eff
dS S R dr K D
ν
≅
where S(R) is the substrate concentration at the particle surface.
2 2 2
2 v r dr
dS dS
D d dr
dS dS
d dr
d S c =
=
=
0 2 (^0 ) ( ) 2
sr
ds (^) dr S SR c R
D
vrds dr
dS d (+6)
(^1) max S R DK
v dr
dS R c m
^ =
Figure 1 (for Problem 1)
Figure 2 (for Problem 2)
