LSM 1101- Lab Report

(Practical 3: Enzyme Kinetics)

Group 1- Bench 6

Date of Experiment: 11th Sept 2009

Members name:

Karthig S/O R.Kunasakaran (U0804997)

Kang Bo Han Abraham (U0900023)

Practical 3

Q3.1.3

Part 1

Wavelength(nm)Absorption Values

NAD+ NADH400 0.026 0.036380 0.027 0.127360 0.039 0.396340 0.035 0.548330 0.040 0.449320 0.030 0.214300 0.343 0.377280 1.506 1.383260 0.906 0.762

Graph of Absorbance against Wavelength(nm)

250 300 350 4000

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Absorbance of NAD+Absorbance of NADH

Part 2

NADH has much higher absorbance than NAD+ over the 300-400nm range. NAD+ only has a higher absorbance when both have peaks in the graph at 280nm.

Part 3

The wavelength of choice for measuring NADH in a mixture containing both NAD+ and NADH is 340nm. Both NAD+ and NADH peak at 280nm but only NADH peaks at 340nm. Thus in order to differentiate between the 2 solutions in a mixture 340nm will be the wavelength of choice.

Q3.2.3

Part 1

Final concentration of NADH(mM) A340

Volume of 1mM NADH used(ml)

Volume of Buffer used(ml)

0.00 0.000 0.00 3.000.01 0.044 0.03 2.970.05 0.215 0.15 2.850.10 0.502 0.30 2.700.15 0.878 0.45 2.550.20 1.185 0.60 2.40

Part 2

0 0.05 0.1 0.15 0.2 0.250

0.2

0.4

0.6

0.8

1

1.2

1.4

f(x) = 5.72689747003995 x

Absorbance Against Concentration for NADH

Part 3

Beer-Lambert Law: A = Ɛcl. Where A = Absorbance, Ɛ = Molar extinction coefficient, c = concentration of solution and l = light path length.

Curvette length is 1cm therefore light path length = 1cm

Concentration of solution used in calculation = 0.15mM = 1.5x10-7

A = Ɛcl

0.878 = Ɛ(1.5x10-7)(1)

Ɛ = 5.85x10-6

3.3 Determining Km and Vmax of lactate dehydrogenase

Results:

Tube

No.10mM

NAD+ (ml)200mM

lactate (ml)0.05M buffer pH 9.0

(ml) LDH (ml)mM lactate (final

[])1 0.1 0.03 2.67 0.2 22 0.1 0.06 2.94 0.2 43 0.1 0.12 2.58 0.2 64 0.1 0.15 2.55 0.2 85 0.1 0.3 2.4 0.2 10

Table 3: Volumes used for dilution of 200mM lactate (ml) for Tubes 1-5

Time (sec)

A340

Tube 1 Tube 2 Tube 3 Tube 4 Tube 50 0 0 0 0 030 0.031 0.052 0.074 0.086 0.12360 0.049 0.081 0.111 0.132 0.18290 0.064 0.103 0.141 0.165 0.232120 0.078 0.122 0.166 0.195 0.271150 0.09 0.139 0.188 0.22 0.304180 0.102 0.154 0.206 0.242 0.339210 0.111 0.168 0.224 0.263 0.365240 0.119 0.181 0.24 0.28 0.396

Table 4: Absorbance of Tubes 1-5 at wavelength 340nm over 240 seconds at 30 second intervals

Graph of A340 against time

Discussion of results

Initial velocity = Gradient of the tangent of the curve = yx

Hence, a tangent has to be drawn at t=0 for each curve.

Tube Sample

Initial Velocity, Vo (Absorbance units/min)

10.061.50

=¿0.040

20.081.00

=¿0.080

30.131.40

=¿0.093

40.171.60

=¿0.106

50.100.60

=¿0.167

Table 5: Initial velocity, Vo (Absorbance units/min) of Tubes 1-5

In order to get the initial velocity in the form of moles/min, we must apply the Beer-Lambert equation, which states that:

Abs = ε c l where A = absorbance

ε = molar absorption coefficientc = molar concentrationl = path length

c (mol.) = Abs/ ε lV0 (mol./min) = Abs/min / ε l

From 3.2.3, ε = 5.85 mM-1cm-1

The initial velocity from the graph of absorbance against reaction time is similar to the value of y in the equation of the linear graph of absorbance against concentration for NADH, which is y=5.7269.

Hence:

Initial velocity in,mMmin−1=V

5.7269× 10-3 =V 1M min−1

Initial velocity in, mM min−1= MV1000

(where M=V 1 and V=3ml)

= 3V '1000

= V 11moles min−1

Tube No.

Initial velocity, Vo

(Absorbance/min)Vl (M/min) Vll (moles/min) 1/[S] (mM-

1)1/Vll

(moles/min)-

1

1 0.040 6.99× 10-6 2.097× 10-8 0.500 4.77× 107

2 0.080 1.40× 10-5 4.20× 10-8 0.250 2.38× 107

3 0.093 1.62× 10-5 4.86× 10-8 0.175 2.06× 107

4 0.106 1.85× 10-5 5.55× 10-8 0.100 1.80× 107

5 0.167 2.92× 10-5 8.76× 10-8 0.050 1.14× 107

Table 6: Calculation of Initial velocity to moles per min.

0 5 10 15 20 250

1

2

3

4

5

6

7

8

9

Michaelis - Menten Curve

[S] mM

Vll (

mol

es/m

in) (

10-

8 )

Vmax

0.5Vmax

From Michaelis – Menten Curve,

Vmax = 8.00 × 10-7

0.5 Vmax = 4.00 × 10-7

KM = 7.80 mM

0.000 0.100 0.200 0.300 0.400 0.500 0.6000.00

1.00

2.00

3.00

4.00

5.00

6.00

f(x) = 7.82795624560554 x + 1.16035282455594

Lineweaver - Burk Plot

1/[S] (mM-1)

1/Vl

l (m

oles

/min

)-1 (1

07)

From Lineweaver – Burk Plot,

1V o

=( K mVmax ) 1[S ]

+ 1Vmax

- 1Km

=¿-0.15 mM

1Vmax

=¿1.20 × 107

Thus, KM = 1

0.15 = 6.67 mM

Vmax = 1 / 1.20 × 107

= 8.33 ×10-8 molesmin-1

Michaelis-Menton plot Lineweaver-Burk plot

Vmax 8.00 × 10-7 8.33 ×10-8 molesmin-1

KM 7.80 mM 6.67 mM

The KM and Vmax values obtained from Michaelis – Menten curve and the Lineweaver – Burk plot are different.

As the concentrations of substrate used in the experiment are not high enough, the Michaelis-Menten curve obtained from the experimental results does not level off. Hence, the values of Km and Vmax are estimated by extrapolation. Since Km is determined at half Vmax, an inaccurate Vmax would cause an inaccurate Km obtained.

In the Lineweaver – Burk plot, the Michaelis – Menten curve is transformed to a straight line from a hyperbolic curve and their inverses are plotted. Hence, the absolute value of the x-intercept of this straight line is the affinity, 1/KM, of the enzyme for the substrate. The y-intercept is 1/Vmax. It is from these intercepts that we are then able to calculate the Vmax and Km which are independent of each other.

Hence, the Lineweaver-Burk plot is the more accurate measure of both Vmax and Km.

However, the disadvantage of the Lineweaver- Burk plot is that it places undue weight on the points obtained at low concentrations of substrate (the highest values of 1/[S] and 1/Vo). These are the points at which the precision of determining the rate of reaction is lowest, because the smallest amount of product has been formed.