Introduction to Bioorganic Chemistry and Chemical Biology: ANSweRS To CHAPTeR 6
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Course
CHEM180 (CHEM180)
Institution
British Columbia Institute Of Technology (BCIT
)
Answer 6.1
Solve the equation.
Convert temperature in °C to K by adding 273.
Plug the numbers into ΔG = ΔH – TΔS.
Kassociation = e–ΔG/RT.
This is the equilibrium constant for association, so take the inverse: Kd = 1/Kassociation.
Calculated using Kd = koff/kon:
Answer 6.3
when the ...
solve the equation convert temperature in °c to k by adding 273 plug the numbers into Δg Δh – tΔs kassociation e–Δgrt this is the equilibrium constant for association
so take the inverse kd
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British Columbia Institute of Technology (BCIT
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Chemistry
CHEM180 (CHEM180)
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Introduction to Bioorganic Chemistry and Chemical Biology 1
Answers to Chapter 6
(in-text & asterisked problems)
Answer 6.1
Solve the equation.
Convert temperature in °C to K by adding 273.
Plug the numbers into ΔG = ΔH – TΔS.
Kassociation = e–ΔG/RT.
This is the equilibrium constant for association, so take the inverse:
m
er as
Kd = 1/Kassociation.
co
Protein Protein ΔH ΔS Kd
eH w
(kcal mol–1) (cal K–1 mol–1)
o.
mutant TCR β S. aureus –15.8 –21 at 25 °C 9.6 × 10–8 M
chain 8.2
rs e
enterotoxin C3
ou urc
p67phox Rac•GTP –7.3 52 at 18 °C 1.7 × 10–6 M
complex
iso-1- iso-1-cc –2.6 18.5 at 25 °C 1.1 × 10–6 M
o
cytochrome c peroxidase
aC s
vi y re
Answer 6.2
Calculated using Kd = koff/kon:
Protein Small ligands kon (M–1 s–1) koff (s–1) Kd
ed d
chymotrypsin proflavin 1 × 108 8300 8.3 × 10–5
ar stu
, 2 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to Chapter 6
Answer 6.3
When the concentration of NADH is 3 × 10–7 M, the ratio of bound to unbound alcohol
dehydrogenase is 1:1. From there, the ratio of bound to unbound enzyme can easily be
estimated at other concentrations of NADH.
KD = 3 × 10–7 M
bound unbound
+ NADH
enzyme•NADH enzyme
quantities of curcumin might be effective for colorectal cancer in the GI tract, but not for
aC s
systemic cancers like leukemias.
vi y re
Answer 6.5
A Galactose binds most tightly because it has the lowest Km; however, the affinities of
all three substrates are within a factor of two.
ed d
B Galactose is isomerized more than 100 times faster than the other two substrates
ar stu
on the basis of the kcat/Km values: galactose (3700 mM–1 s–1), glucose (13 mM–1 s–1),
xylose (20 mM–1 s–1).
C If the system is at equilibrium, when the concentration of glucose is 10 Km (340 mM),
the ratio of glucose–enzyme complex to free enzyme will be 10:1. In a typical mam-
is
malian cell, the intracellular glucose concentration is less than 1 mM. Of course the
amount of free enzyme is likely to be small because galactose and other sugars can
Th
occupy the enzyme active site.
Answer 6.6
A The substrate with the lowest Km binds most tightly: LRRASLG.
sh
B The substrate with the highest kcat is phosphorylated fastest (once it binds): LR-
RASLG.
C The relative rates of phosphorylation will be proportional to kcat/Km. LRRASLG is bet-
ter than LRAASLG by a factor of 1507.
Substrate Km (μM) kcat (s–1) kcat / Km (M–1 s–1)
LRAASLG 12200 8.7 0.00071
LHRASLG 804 19.8 0.0246
LRRASLG 31 33.1 1.07
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