Exam (elaborations)

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 ...

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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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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

creatine kinase ADP 0.2 × 108 18,000 9.0 × 10–4

G-3-P dehydrog. NAD+ 0.2 × 108 1000 5.0 × 10–5

lactate dehydrog. NADH 10 × 108 10,000 1.0 × 10–5
is

alcohol dehydrog. NADH 0.3 × 108 9 3.0 × 10–7
Th

lysozyme (N-Ac-Glu)2 0.4 × 108 100,000 2.5 × 10–3

ribonuclease 3ʹ-UMP 0.8 × 108 11,000 1.4 × 10–5
sh

Protein Large ligands kon (M–1 s–1) koff (s–1) Kd

tRNASer tRNASer 2 × 108 11 5.5 × 10–8
synthetase

trypsin protein inhibitor 0.07 × 106 0.0002 2.9 × 10–9

insulin insulin 1 × 108 20,000 2.0 × 10–4

β-lactoglobulin β-lactoglobulin 0.00005 × 108 2 4.0 × 10–4

α-chymotrypsin α-chymotrypsin 0.000004 × 108 0.7 1.8 × 10–3

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, 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

[NADH] [enzyme•NADH] : [enzyme]
3 × 10–6 M 10 : 1
Introduction to Bioorganic Chemistry and Chemical Biology | A6019
10–7
3 ×Van M & Weiss | 978-0-8153-4214-4
Vranken 1 : 1
www.garlandscience.com
3 ×©10–8 M design
1 by www.blink.biz
: 10
3 × 10–9 M 1 : 100

A When [NADH] = 3 μM, the ratio of bound to unbound enzyme is 10/11 ≈ 91%.
B When [NADH] = 3 nM, the ratio of bound to unbound enzyme is 1/101 ≈ 1%.

m
Answer 6.4

er as
cancer cells

co
[Curcumin] Dead / Live Percentage

eH w
(µM) viable
5.5 1 : 1 50

o.
O OH
11 2 : 1 33

rs e MeO OMe
55
550
10
100
:
:
1
1
9.1
1.0
ou urc
HO curcumin OH 5500 1000 : 1 0.10
Introduction to Bioorganic Chemistry and Chemical Biology | A6117
Van Vranken & Weiss | 978-0-8153-4214-4
Curcumin has poor bioavailability. At an oral dose of 8 g of curcumin per day, the peak
© www.garlandscience.com design by www.blink.biz
serum concentrations of curcumin reach only 1.8 μM. Hypothetically, eating large
o

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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