Enzyme Lab Write-up PDF

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Enzyme Lab Write-up...

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Enzyme reaction kinetics: lactate dehydrogenase

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Lab 3: Enzymes Write-up

Name: ID#: TA: Section: 1. Draw the progress curve (plot of A340 vs. time) for each substrate concentration. (3 marks)

Figure 1: Progress Curves for the absorbance of sodium pyruvate over a time period of 40 seconds

2. Calculate V o for each of the curves and tabulate them in the space below. Disregard the negative sign. Show only one sample calculation here. (2 mark) 1 A } x t l 1 A } x Vo = { 3 −1 t (6.23 x 10 M )( 0.6 cm) 1 −0.29 Vo = x 3 −1 10 s (6.23 x 10 M )( 0.6 cm ) 1 −0.29 Vo = x 3738 10 s Vo = −0.000007758 M x 1 000 000 Vo = −7.758 µ M BIOC*2580 – Introductory Biochemistry

Vo = {

Enzyme reaction kinetics: lactate dehydrogenase

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Table 1: Calculated Vo values for the progress curves of the five samples graphed in Figure 1 Sample Vo number Sample 1 7.758 µM Sample 2 6.688 µM Sample 3 4.013 µM Sample 4 4.280 µM Sample 5 1.873 µM

3. Calculate the substrate concentrations [S] and tabulate them in the space below. Show only one sample calculation here. (2 marks) C1V1 = C2V2 C 1V 1 C2 = V2 1 mM x 100 µ L C2 = 250 µ L C2 = 0.4 mM Table 2: Substrate concentration values of the five samples calculated using Table 3.1 in the lab manual Sample [S] number Sample 1 0.4 m M Sample 2 0.32 m M Sample 3 0.24 m M Sample 4 0.16 m 4. Plot the Michaelis-Menten graph (Vo vs. [S]). Do the points fall on M a hyperbolic curve? Estimate V max and find the value of KM. (3 Sample 5 0.08 m marks) M

BIOC*2580 – Introductory Biochemistry

Enzyme reaction kinetics: lactate dehydrogenase

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9 Vmax 8

Initial Velocity, Vo (µM/s)

7 6 5 ½ Vmax

4 3 2 1 0 0.05

KM

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Substrate Concentration, [S] (mM)

Figure 2: The Michaelis-Menton graph comparing the initial velocities (V o) in µM/s presented in Table 1 and substrate concentrations [S] in mM presented in Table 2 of the five samples involved in this lab experiment Vmax = highest V o point + 10% Vmax = 7.758 x 1.1 Vmax = 8.534 KM = substrate concentration at ½ Vmax ½ Vmax = 8.534/2 ½ Vmax = 4.267 KM = 0.16 mM/1000 KM = 0.00016 M As seen in Figure 2, the points on this graph do fall on a hyperbolic curve.

BIOC*2580 – Introductory Biochemistry

Enzyme reaction kinetics: lactate dehydrogenase

BIOC*2580 – Introductory Biochemistry

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