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Answers to lesson exercises
1 a) kcat/knooncat = 4.6 x 103/0.5x10-6 s-1/s-1M-1 = 9.2x109 M = Ceff (Effective
concentration)
b) Enzyme catalyzed reactions are monomolecular, while this uncatalyzed
reaction is bimolecular. The reacting groups are close to one another with correct
orientation in the active site, which is entropically favorable.
c) kcat/KM = 4.6x103/1,2x10-4 = 3.8x107 M-1s-1< kdiff = 108-109 M-1s-1
The rate is almost as high as the diffusion limited rate
(kcat/KM is the rate constant by which E and S combine to form products at low
concentrations of S.)
d)
2) a) The reason to excite at 295 nm is to obtain a specific contribution from the
tryptophans. At 280 nm also Tyr and Phe contribute to the fluorescence.
b) The Trp mutant that gives the most blue-shifted spectrum is the Trp that is the most
exposed. The reason to this is that the removed Trp must have been red-shifted, i.e.
exposed and absence of one exposed Trp will give rise to a blue-shifted spectrum.
Thus, Trp94 is the most exposed Trp in barnase.
c) The reason to the increased fluorescence intensity when Trp 94 is mutated must
depend on energy-transfer to this Trp, which in turn is quenched.
d) The emission is increased when His18 mutated to a Gly despite the fact that none of
these are fluourescent amino acids. The reason must be that His18 is a quencher of one of
the three Trp:s, likely Trp94 (see c))
3) See the Scatchard plot below:
Scatchard formula:
r/[A] = nKa – Ka r,
where r = molar bound ligand/ [protein]total
And [protein]total = 1.2/60 000 = 2.0x10-5 M
[A] = free ligand concentration, i.e. the concentration of cAMP on the outside of the
dialysis tube.
Molar bound ligand is calculated as the difference in concentration between cAMP on the
inside and the outside of the dialysis tube.
Outside (M) Inside (M)
Bound ligand (M)
r
r/[A]
-6
-6
-6
0.7x10
2.7x10
2.0x10
0.1
1.4x105
-6
-6
-6
2.3x10
8.3x10
6.0x10
0.3
1.3 x105
4.5x10-6
1.5x10-5
1.1x10-5
0.5
1.2 x105
-5
-5
-5
1.0x10
2.6x10
1.6x10
0.8
0.8 x105
-5
-5
-5
3.0x10
5.4x10
2,4x10
1.2
0.4 x105
7.0x10-5
9.8x10-5
2.8x10-5
1.4
0.8 x105
-4
-4
-5
1.5x10
1.8x10
3.0x10
1.5
0.1 x105
Kd = 1/Ka
Ka = the slope of the curve = 1x105
Thus, Kd = 1x10-5 M
n = 1.6
The reason to a non-integer number of binding sites may be
a) that the molecular weight determined by ultracentrifugation may not be quite exact
b) a bad preparation of protein with a certain amount of denatured protein which does not
bind cAMP.
4)
E TyrAMP
E TyrAMP#
E TyrATP
E Tyr
C35G
ΔΔG (kcal/mol)
β (=)
1.65
1.24
1.24/1.65 = 0.75
0
0
0
0
H48G
ΔΔG (kcal/mol)
β (=)
1.83
1.61
1.61/1.83=0.9
0.81
0.81/1.83=0.4
0.37
0.37/1.83=0.2
Solid line: the wild-type
Dashed line: the C35G mutant
C35 does not affect the binding of the substrate, but affects the transistion state almost as
much (75 %) as the final product. Thus, C35 is a catalytic group.
H48 has a binding function to a certain degree, but above all H48 has a catalytic function.