### - BUGS McGill

```Lecture 2: Spectroscopy and Enzyme Kinetics
By: Urshila
Overview
 Spectroscopy
 Absorbance
 Beer-Lambert Law
 Sample Questions
 Enzyme Kinetics
 Rates and Michaelis-Menten
 Enzyme Inhibition


Competitive vs. Non-competitive
Inhibition constant
 Sample Questions
Absorption
 Transmittance is the fraction of
light passed through an object

•Absorbance is measured
using a spectrophotometer

•It consists of 4 main parts:
•Light Source
•Monochromator (slit)
•Sample (in a cuvette)
•Light detector and recording
device (usually a computer)
Transmittance values range from
0 (all of the light is absorbed) to 1
(all of the light is transmitted)
T = [I]/[I0]
 Absorbance is the amount of
light absorbed by an object


Absorbance is measured as the
negative log of transmittance (T)
Abs (or O.D) = -log10 (T)
Beer- Lambert Law
 Absorbance is proportional to the
A=ε*c*d
concentration of a substance (c) and
the distance that light travels through
the substance (d = path length)
 This is denoted by the Beer-Lambert
Law where:
 A= absorbance
 ε = molar extinction coefficient
(mol-1 * L * cm-1 )
 c= molar concentration (mol/L)
 d= path length (cm)
Exceptions to the Beer-Lambert
 Complex Solutions
 Like Iodine, equilibrium
between colored and
non-colored species
 Nonhomogenous
Solutions
 Scattering Solutions
 Particles shield each
other from light at high
concentration
 Particles can absorb,
transmit, or scatter light
Things to remember!
 Rules of thumb
 Protein: 1 O.D. @ 280 nm = 1 mg/ml
 DNA (ss) 1 O.D. @ 260 nm = 30 ug/ml
 DNA (ds) 1 O.D. @ 260 nm = 50 ug/ml
 Bacteria : 1 O.D. @ 600 nm = 5* 108 bacteria/ml
 SPF
 SPF = 1/T
 Doubling the thickness of sunscreen, doubles the
absorbance and decreases transmittance by a factor of
100. (Recall: A = -log10 (T) )
Thinking Time
 1. A researcher is studying the absorbance level of a
bacterium at 600 nm and discovers that the absorbance
spectrum does not follow the Beer-Lambert Law. What
is a plausible explanation for this situation?
 2. In what other situations would the Beer-Lambert
Law not apply?
 3. If you are measuring the spectra of a mixture, what
modifications should be made to the Beer-Lambert
Law?
 1. The large size of bacteria (relative to molecules in a solution)
cause light to be refracted in a phenomenon known as
scattering. This results in a deviation from Beer’s Law (unless
the bacterial suspension is dilute)
 In bacterial suspensions less light is transmitted to the detector and
the increased absorbance value is proportional to the cell density.
(This is only valid in dilute suspensions where o.d. < 1)
 2. Other deviations in Beer’s Law are observed in complex
solutions and suspensions.
 3. In mixtures, the absorption spectrum is a linear sum of the
component spectra. Therefore:
A = (ε*c 1 + ε*c 2 ) d
Problem Solving
 As the head of a sunscreen company you wish to test the
effectiveness of a new sunscreen (product 2) versus a
competitive brand(product 1) which has an SPF of 20. The
absorbance of Product 1 is double that of Product 2 when
the thickness of Product 2 is 4 times that of Product 1.
 What is the SPF of your new sunscreen?
 Sunscreen 2 has:
 0.5 Absorbance, 4 times thickness
 Since A=εcd, the normal Abs of (sunscreen 2) is 0.5/4 =
0.125 of (sunscreen 1)
 T1 = 1/SPF = 1/20 = 0.05
 A1 = -log(T1) = 1.3
A2=(0.125) (1.3) = 0.1625
 T2 = 10-A2 = 10-0.1625 = 0.688
 SPF2 = 1/T2 = 1/0.688 = 1.45
Enzyme Kinetics
 Enzymes are catalysts that
accelerate the rate of chemical
reactions
 They function by lowering the
activation energy, but do not alter
the overall energy of the reaction
(∆ G) or the position of
equilibrium
 Examples: pepsin, urease
Reaction Rates
 The rate of an enzyme catalyzed
reaction can be measured as a
function of product formation
(∆P/∆t)
 Alternatively, enzymatic rate can
be measured as a function of
substrate loss (∆S/∆t)
 The rate of product formation is
usually preferred (much easier
to quantify)
 Enzyme kinetic experiments
measure the initial velocities or
rates of reaction (linear portion
of the curve)
Graph of V vs. [S]
 The plot to the left shows the Vo



Vmax= maximum rate of activity
Km= concentration of substrate at half
maximal activity
Vmax/Km = physiological efficiency

(initial rate) values at varying
substrate concentrations for a given
enzymatic reaction.
Vmax represents the maximum
rate of activity and occurs when the
enzyme is saturated with substrate.
Km is the [S] at half maximal
velocity and represents the binding
affinity of the enzyme
At low [S], V0 is nearly
proportional to [S]
At high [S], V0 approaches Vmax
and becomes independent of
substrate concentration
Reaction Rate Equations
 Reaction scheme for an enzyme-catalyzed reaction:
 The rate of ES formation can be represented as follows:
 Two assumptions can be made to further simplify this equation:
The initial concentration of product ([P]) is ≈ 0. Therefore the term k-2
[E][P] can be eliminated.
2. The rate of change of [ES] is ≈ 0 (pseudo steady state)
1.
 The Steady State assumption states
that the concentration of ES complex
remains constant over time
 Thus, the rate of formation of ES is
equal to the rate of degradation and
∆[ES]/∆t = 0
Simplified Reaction Mechanism
Simplified Rate Equation
Michaelis Constant
 From steady state, we have:
 K1[E][S] = K-1[ES] + Kcat[ES] (Formation = Destruction)
 Isolate for [ES]: K1[E][S] = (K-1 + Kcat)[ES]
 Rearrange: [E][S]/[ES] = (K-1 + Kcat)/K1 = KM
Deriving the Michaelis-Menten Equation
 Recall :
 Substitute [E] = [Et]- [ES] and rearrange the equation as follows:
 Recall:
Final Form!
Starting Equations
 [ET] = [ES] + [E]
 K1[E][S] = K-1[ES] + Kcat[ES]
 KM = (K-1 + Kcat)/K1
 Vo = kcat[ES], Vmax = kcat[ET]
Total Enzyme
Michaelis Constant
Reaction Rate
 Get rid of all the E’s and ES’s!
 Start by solving Steady State for [E] and plugging into
the Total Enzyme equation.
Derive it Yourself!
Michaelis-Menten Equation
*Learn the derivation!
 Km: an "aggregate" constant (sum of rate constants for breakdown of ES
divided by rate constant for formation of ES):
 The Michaelis-Menten equation explains the Vo vs. [S] curve:
1.
At very low [S] ([S] << Km), Vo approaches (Vmax/Km)[S]. Vmax and Km are
constants, so linear relationship exits between Vo and [S]at low [S].
2.
3.
When [S] = Km, Vo = 1/2 Vmax
At very high [S], ([S] >> Km), Vo approaches Vmax (velocity independent of
[S])
Plotting the Data
Lineweaver-Burke Plot
Slope = Km/Vmax
Yint = 1/Vmax
Xint = -1/Km
Slope = -Km
Yint = Vmax
Xint = Vmax/Km
** The problem with the Lineweave- Burke plot is that the data points are not well
distributed. The Eadie-Hofstee plot provides a more even weighting of the data points.
Turnover Rate and Efficiency
 The turnover rate of an enzyme, k2, is defined as the maximum
number of molecules of substrate that an enzyme can convert to
product per unit of time.
k2= Vmax/[E]T
 The turnover rate is measured in units of sec-1 or min-1
 The physiological efficiency of a reaction is defined as Vmax/ Km
Sample Calculation
Practice Questions 
 1. If you double enzyme concentration, how is the Vmax value
affected? How is the Km value affected?
 2. The Vmax of an enzymatic reaction is 20mM per min and you
have a total enzyme concentration of 0.1 M. What is the
turnover rate of the enzyme?
 3. The activity of 6 ug of enzyme is measured by absorbance in a
1ml, 1cm cuvette. The change in absorbance is measured as 0.18
O.D./min and the molar extinction coefficient is measured at
220 M-1 cm-1 . Calculate the activity in umol/min/ug.
 1. Vmax = kcat[ET], Doubles. Km = (k-1 + kcat)/(k1), no
change
 2. Vmax = kcat[ET]
 Kcat = Vmax/[ET] = (20 mM/min)/0.1 M = 0.2 min-1 = 12 s-1
 3. OD/min  µmol/min
 A=εcd
c = A/(εd) = 0.18/(222M-1cm-1 x 1 cm) = 810.8µM
 (810.8µM/min)/(6µg) = 135µM/min/µg
Enzyme Inhibition
 Inhibitors are compounds that block enzyme activity
 Two main types of inhibitors are competitive and non-competitive.
 Competitive Inhibitors:
 Compete with the substrate for the enzyme active site
 These inhibitors function by decreasing the binding affinity of an
enzyme for its substrate
 Km increases; Vmax remains unchanged
 Non-competitive inhibitors :
 Inhibitor binds to a site other than the active site and does not
compete with the substrate for binding
 These inhibitors function by decreasing the turnover rate of an
enzyme
 Vmax decreases; Km remains unchanged
Competitive vs. Non-Competitive Inhibition
Competitive Inhibition:
Non-competitive Inhibition:
Reaction Scheme
Reaction Scheme
**Enzyme binds either I or S
Lineweaver-Burke Plot
**Enzyme can bind both I and S
Lineweaver-Burke Plot
Inhibition Table
Type of
Inhibition
Km
V max
Physiological
efficiency
None
Km
Vmax
Vmax/Km
Competitive
Kmapp = Km (1 + [I]/Ki )
Vmaxapp = Vmax
(Vmax is unchanged)
Vmax/ Km*(1 + [I]/Ki )
Non-competitive
Kmapp = Km
(Km is unchanged)
Vmaxapp = Vmax/(1 + [I]/Ki )
Vmax/ Km *(1 + [I]/Ki )
*Note: Ki is the concentration at which the inhibitor is 50% effective
(1 + [I]/Ki ) = α (inhibition factor)
Tips for enzyme problems!
 Use your units to help avoid errors

Moles are not Molar (mol/L)
 Be explicit with conversion factors

E.g. 1 = 1 ml/ 10-3 L


Do they make sense? Are the values too high or too low?
Use reasonable precision

1. Why would lowering enzyme concentration be
equivalent to

2. The Vmax of a reaction is 33.2 mol/L/year. What is the Vmax in
umol/ml/sec?

3. In the presence of an inhibitor ([I] = 1 mM), the Km is increased
by a factor of 3. Calculate the Ki value. What type of inhibition is
this?
Challenge Question
The kinetics of an enzyme are measured as a function of substrate
concentration in the presence and in the absence of 2 mM inhibitor (I). Plot
the data as an Eadie-Hofstee plot
(a) What are the values of Vmax and KM in the absence of inhibitor? In its
presence?
(b) What type of inhibition is it?
(c) What is the binding constant of this inhibitor?
(d) If [S] = 10 μM and [I] = 2 mM, what fraction of the enzyme molecules have a
bound substrate? A bound inhibitor?
Hint: fractional [ES] = ratio of occupied sites to active sites= V0/ Vmax
Solution
(a) In the absence of inhibitor, Vmax is 47.6 μmol minute-1, and KM is 1.1 ×
10-5 M. In the presence of inhibitor, Vmax is the same, and the apparent
KM is 3.1 × 10-5 M.
(b) Competitive.
(c) 1.1 × 10-3 M.
(d) fES is 0.243, and fEI is 0.488.
```