2020 Enzyme kinetics Practical worksheet 2021 PDF

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BIOS5003 Biochemistry of Cell Function

Enzyme Kinetics Practical Worksheet 1. Enzyme activity assay  Plot absorbance against time, and determine the initial rates of the increase in absorbance for samples A and B. This is the slope of your graphs at time zero (i.e. tangent to the curve if the time course is not linear) in absorbance units per minute (ΔA 340.min-1). CHECK YOUR UNITS!!!! (Remember to convert seconds to minutes if necessary, and define volume used). Specific activity  From your measurements of enzyme activity, calculate the specific activity of each of the G6PDH samples in absorbance units per minute per mg of protein. Enzymatic activity of G6PDH using sample A 1

Absorbance at 340 nm

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

10

20

30

40

50

60

70

80

90

70

80

90

Time (second)

Enzymatic activity of G6PDH using sample B 0.8

Absorbance at 340 nm

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

10

20

30

40

50

Time (seconds)

1

60

BIOS5003 Biochemistry of Cell Function

(Show your working below, or on a separate sheet and DOUBLE-CHECK YOUR UNITS!!!) The concentration of Sample A is: 0.100 mg/ml The concentration of Sample B is: 0.200 mg/ml Sample A:

Sample B:

Enzyme activity …0.43….ΔA340.min-1.(0.1ml-1) Enzyme activity …0.5…. ΔA340.min-1.(0.1ml)

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Protein concentration ……0.1...... mg.(0.1ml -1) Protein concentration….0.2...mg.(0.1ml -1) 0.100 mgml-1 x 0.1 ml = 0.01 mg/mL

0.200 mgml-1 x 0.1 ml = 0.02 mg/mL

Specific activity

Specific activity

0.43(min-1/mL-1)/0.01 (mg/mL) = 43 ΔA340.min-1.mg-1

0.5 (min-1/mL-1)/ 0.02 (mg/mL) = = 25 ΔA340.min-1.mg-1

 Which sample is the purest? Sample …A…. is the purest because It is the highest ‘specific activity’ value of the two samples. Therefore, a high activity value a high purity level, as the sample here is completely pure and not contaminated with other proteins present in the sample. 2. Enzyme kinetics Kinetic analysis  Please complete the following table with your data and calculations Volume of G6P

Concentration of G6P

Tris buffer

ΔA340.min-1 without DHEA

[s]/v without DHEA

ΔA340.min-1 with DHEA

[s]/v with DHEA

200 l

1 mM

0 l

0.436

2.3

0.123

8.1

2

BIOS5003 Biochemistry of Cell Function

100 l

0.5 mM

100l

0.455

1.1

0.087

5.7

50 l

0.25 mM

150 l

0.338

0.7

0.077

3.2

25 l

0.125 mM

175 l

0.256

0.5

0.076

1.6

10 l

0.05 mM

190 l

0.108

0.5

0.068

0.7

Hanes Woolf plot-G6PDH enzyme activity in the presence of DHEA vs absence of DHEA 9 8

[S]/v mM/△A340.min-1

7 6 5 4 3 2 1 0 5 .1 .3 .7 .1 .5 .9 .3 .7 .1 .5 .9 .3 .7 .1 .5 .9 .3 .7 .1 .5 .9 .3 .7 .1 .5 .9 .3 .7 .1 .5 .9 .3 .7 .1 . -0 -0 0 0 1 1 1 2 2 3 3 3 4 4 5 5 5 6 6 7 7 7 8 8 9 9 9 10 10 11 11 11 12 12 13 [S] mM

 Plot rates versus concentration as a Hanes Woolf plot (remember to include axis labels and units).

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BIOS5003 Biochemistry of Cell Function

 How can you determine the Km and Vmax values using this plot? (include a diagram in your answer) The Hanes-Woolf plot is a simple rearrangement of the Michaelis-Menten equation. By applying the equation graphically, the kinetic parameters of Km and Vmax can be determined from the straight line. The Km value is obtained when the straight plotted line is extrapolated backwards until it intercepts the X-axis. The value at the X-axis is the Km; therefore, it is the positive value version of this. The Vmax is obtained from the Y-axis of the graph, where Y= Km/Vmax. Thus, Vmax is obtained by a simple rearrangement of the equation, i.e. Vmax= Km/Y-axis.  What are your Km and Vmax values? Km

Vmax

Without DHEA

0.1

0.45

With DHEA

0.04

0.10

 What kind of inhibitor of G6PDH is DHEA? -Uncompetitive inhibitor  Briefly, explain your reasoning DHEA is an uncompetitive reversible inhibitor. A reversible inhibitor which is uncompetitive binds to the enzyme (G6PDH) after the substrate (G6P) has bound to its active site, creating and enzyme-substrate-inhibitor complex (ESI).This enzyme inhibitor has the effect of reducing the concentration of the ‘free’ ES complexes (enzyme-substrate complexes), relative to total enzyme concentration. Therefore, both Km and Vmax values are reduced proportionally. When the values of the enzyme reactions in the presence and absence of the DHEA inhibitor were plotted on the Hanes-Woolf plot, the solution containing the uncompetitive inhibitor has a lower total enzyme concentration. Thus, it gives both a lower Km and Vmax values, and this was shown by the two constructed Hanes-Woolf plots. A lower resulting Km and Vmax value when using DHEA is indicative of a uncompetitive inhibitor.  What were potential sources of inaccuracies and uncertainty during these experiments?  Human error; inaccurate pipetting of the reagents, i.e. incorrect pipetted volumes.  The spectrophotometer was calibrated incorrectly (i.e. not set at 340 nm).  The time-intervals between absorption readings were either too short or too long.  Mechanical error of the spectrophotometer, i.e. incorrect absorbance readings.  Incorrect determination and calculation of the Km and Vmax values from the graphs.  If the enzyme does not follow the Michaelis-Menten kinetics.  If the enzyme, substrate or inhibitor is ineffective (i.e. denatured or degraded).  If a Lineweaver-Burke was used instead of the Hanes-Woolf plot, it would be less reliable.  Calculated Km and Vmax are just approximations, and therefore, might not reflect the true enzyme-inhibitor enzyme kinetics

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BIOS5003 Biochemistry of Cell Function

Don’t forget to attach your graphs to your report and bring them to the online feedback workshop, 9:00 am Friday week 10!

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