Effect Of Substrate Concentration On Reaction Biology Essay

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Enzymes are biological catalysts. They speed up chemical reactions in all living things, and allow them to occur more easily. Enzymes are chemical molecules, made up of globular proteins. Each particular enzyme has a unique, 3-dimensional shape made up of a tertiary structure that is made from the twisting and folding of the polypeptide helix, which contains both the alpha helix and the beta pleated sheet. Enzymes are also biological catalysts; they differ from inorganic catalysts in that they catalyse only one reaction. A small region on the enzyme, called the active site, has a shape that fits with specific substrate molecules. The active site is where the binding occurs. The shapes of the active site, and the positions of the chemical groups and bonds within it, ensure that only those substrate molecules with a complementary structure will combine with the enzyme. (1)(6)

The enzyme alkaline phosphatise is important in the reuse of phosphate within living cells. Alkaline phosphate catalyzes the spilt of a phosphate group from varieties of compounds such as the substrate p-nitrophenylphosphate which the substrate is colourless. Conversely, one of the product, p-nitrophenol is yellow in basic solutions. The yellow colour indicates the point to which the substrate has been acted upon by the enzyme (2)

Results

Table 1: Table showing the velocities of the reaction at different p-nitrophenyl phosphate concentrations

Absorbance (405 nm)

Concentration of p-nitrophenyl phosphate

Time (sec)

0.5 mM

0.8mM

1.0mM

2.0mM

5mM

10mM

15

0.019

0.036

0.043

0.069

0.120

0.097

30

0.033

0.065

0.077

0.124

0.221

0.169

45

0.049

0.093

0.108

0.183

0.320

0.246

60

0.062

0.118

0.138

0.239

0.419

0.322

75

0.075

0.143

0.168

0.296

0.509

0.399

90

0.089

0.169

0.198

0.347

0.601

0.473

105

0.098

0.194

0.228

0.400

0.698

0.542

120

0.110

0.220

0.258

0.458

0.790

0.610

This table is showing the velocities of the reaction at the following p-nitrophenyl phosphate concentrations:- 0.5mM,0.8mM,1.0mM,2.0mM,5mM,10mM.

Table 2 A table showing the reciprocal of absorbance change per minute and substrate concentration

Concentration (mM) of p-nitrophenyl phosphate (S)

Absorbance change /min

(V)

0.5

0.044

0.8

0.072

1

0.088

2

0.148

5

0.251

10

0.316

A table showing the class data of the rates that were determined. This data will used to plot a Lineweaver-Burk graph to estimate the Km and Vmax for p-nitrophenyl phosphate.

Figure 2. Graph showing the class data of concentration against absorbance change/min

This graph is showing a curve but Km and Vmax points cannot be determined, to estimate Km and Vmax a lineweaver-burk graph needs to be plotted in the reciprocal form.

Table 3: A table showing the reciprocal of absorbance change per minute and substrate concentration

1/S (substrate concentration) units?

1/V (Velocity) units?

2

22.73

1.25

13.89

1

11.36

0.5

6.76

0.2

3.98

0.1

3.16

Results from this table will be used to plot a lineweaver-burk graph to estimate Km and Vmax for p-nitrophenyl phosphate.

Table 4: Table showing the intercepts obtained from the lineweaver-burk graph

Value at the intercept of the X-axis (1/S)

-0.2

Value at the intercept of the Y axis (1/V)

1.5

Table 5: Table showing the values of Km and Vmax

Km Value (mM)

5

Vmax µmol min-1

0.25

Km value was determined using -1/Km because both values are in mM the value remains same.

-1/Km = -1/-0.2=5mM

Vmax value was determined using 1/Vmax

To determine the value of Vmax the following steps were taken:

-converting the value of the absorbance change/min using Beer lambert law.

A=ECL

M-1cm-1A/EL=C abs/min-1/EL cm

Mol/l=M/min-1/M-1

M/min

-0.003

Mol/min -106=µmol/min

1.5/1.8-104=8.33-10-5

8.33-10-5-0.003=2.5-10-7

2.5-10-7--106=0.25µmol min-1

Discussion

Figure 3 shows that an Increase in the concentration of substrate increases the rate at which product is formed, up to a maximum value. From this point the enzyme molecule is soaked with the substrate and the rate of reaction (Vmax) depends on how fast the enzyme can process the substrate molecule. Km, the concentration of substrate allows the reaction to proceed at one-half its maximum rate(0.5Vmax). Results from figure 3 show a low Km value showing that the enzyme reaches its maximum increase in rate at a low concentration of substrate and indicates that the enzyme binds to its substrate very tightly. This is backed up with the value taken from the Brenda enzyme database. The value from Brenda enzyme shows km value to be 0.056 compared to the value from experiment 0.25 shows both are a low km value. (3)(5)

Plotting the reciprocal of absorbance change per minute and substrate concentration produces a Lineweaver Burk plot. This provides a more accurate way to determine Vmax and Km.

Vmax is determined where the line crosses the 0 axis.

Km equals Vmax times the slope of line and is determined from the intercept on the X axis.

The lineweaver Burk plot gives a more accurate estimate of Vmax and is useful in analyzing enzyme inhibition. The lineweaver plot shows as the substrate concentration decreases the rate also decreases this is because at low substrate concentrations the active sites on the enzymes are not soaked by the substrate causing the enzymes to not work at maximal capacity.

Figure 2 shows as the concentration of the substrate increases more enzymes are working. When it reaches the point when saturated, no more active sites are available for substrate binding; at this point, the enzyme reaches its maximum velocity (Vmax). As the substrate concentration increase above this point the Vmax occurs as the enzyme is soaked with the substrate. This relationship is described as a hyperbolic curve. (3)