Practical Report on Beta-Galactosidase & Enzymes Kinetics





Dr Ian Ashton




The most important reason for studying enzymes is of course to understand their catalytic activity and at a practical level how to make best use of them to catalyse reactions and make products. To understand the activity one needs a knowledge of the enzyme which in turn provides information on its mechanism, it’s role in the cell it originates from, and most importantly if you are going to use it as a catalyst in a process, how it’s activity can be controlled. Two of the key parameters when describing the properties of an enzyme are Km and V max. The maximum possible rate of reaction is V max and will occur when all of the enzyme present is in the form of an enzyme –substrate complex, i.e. when all the enzyme is fully engaged in the reaction process. In a typical enzyme reaction :-


E + S « ES « E + P


A plot of reaction velocity, v, versus [S] is a hyperbolic curve that approaches V max but never quite reaches it, so it cannot be measured directly ( it would need a substrate concentration of ¥ ). The substrate concentration that gives half the maximum velocity corresponds to a value for the equilibrium constant K for the overall reaction known as Km, the Michaelis constant. Km is the dissociation constant of the ES complex and in effect is a measure of how strongly the enzyme binds to the substrate. Both Km and Vmax also give a measure of the specificity of the enzyme as the values for different substrates, or potential substrates, can be compared. The units of Km are concentration units (molarities). The lower the value of Km the better because this means that only a low substrate concentration is needed for all the enzyme to be mostly in the form of the ES complex, i.e. working most efficiently. The units of Vmax are velocity units, such as moles of substrate consumed per unit time, per unit of enzyme.


Inhibitors reduce the catalytic power of an enzyme and in order to do this they must bind to it in some way, often to the same active site as the substrate but sometimes or to more distant parts of the molecule, but still being able to influence the activity by causing a conformational change. The different types of inhibition (competitive, non-competitive or uncompetitive) can be distinguished from their effects on Vmax and Km while the strength of inhibitor binding is indicated by Ki.


In this practical simple measurements will be used to monitor the progress of an enzyme reaction (the hydrolysis of ortho-nitro-phenyl-galactoside, ONPG, a colorimetric substrate) by the enzyme β-galactosidase (also known as lactase) , in the presence of different concentrations of substrate (ONPG) and inhibitor (galactose ) to obtain values of Km and Vmax for β-galactosidase with ONPG substrate and Ki for galactose.






  • 0.1 M potassium phosphate buffer, pH 7.3
  • 2M Na2CO3
  • ONPG stock solutions, contains 9mg/ml of ONPG in 0.1 M potassium phosphate, pH 7.3.
  • Galactose solution (20%w/v), in 0.1 M potassium phosphate, pH 7.3.
  • β-galactosidase (from E. coli) solution contains 30units/ml of enzyme made up in 0.1M potassium phosphate buffer pH 7.3
  • It is important to know and control the pH because of it’s effects on enzyme kinetics, and hence the values of Km and Vmax.


In order to measure Km and Vmax need to measure the consumption of substrate by a fixed amount of enzyme at different substrate concentrations. Repeating the same experiment in the presence of a minimum of two different levels of an inhibitor enables a value of Ki for that inhibitor to be obtained. The consumption of the substrate ONPG by β-galactosidase (a bacterial enzyme) will be measured. The enzyme hydrolyses ONPG to galactose and the yellow O-nitrophenol, which can be quantified by measuring absorbance in a spectrophotometer at 420nm. For accuracy, experimental readings should always be carried out in duplicate or triplicate if possible. To this end, the practical class will be divided into at least three groups, with each group performing a set of readings as described below and then pooling the data obtained in order to have replicate values.


Set up the following three sets of reaction mixtures in the plastic cuvettes provided.


Before you start:

  • Dilute the stock ONPG solution x20 with buffer (pH 7.3) to obtain working stock.
  • Dilute the 20% galactose solution with buffer to obtain a 1.08% working stock solution.


Set 1


Cuvette no. ONPG working stock solution (ml) buffer pH 7.3


Total volume


1 0 3 3
2 5 2.995 3
3 10 2.99 3
4 25 2.975 3
5 50 2.95 3
6 100 2.90 3
7 150 2.85 3
8 250 2.75 3
9 500 2.50 3













Calculate the volume of the 1.08% galactose working stock solution that has to be added to the cuvettes in order to have 1mM and 2 mM galactose in each of the assay mixtures for sets 2 and 3 respectively.


The MW of galactose is 180.




Set 2


Cuvette no. 1mM galactose

Vol of 1.08% working stock galactose


ONPG working stock solution (ml) buffer pH 7.3


Total volume


1   0   3
2   5   3
3   10   3
4   25   3
5   50   3
6   100   3
7   150   3
8   250   3
9   500   3






Set 3


Cuvette no. 2mM galactose

Vol of 1.08% working stock galactose


ONPG working stock solution (ml) buffer pH 7.3


Total volume


1   0   3
2   5   3
3   10   3
4   25   3
5   50   3
6   100   3
7   150   3
8   250   3
9   500   3


Add the enzyme (50ml) to each cuvette at 1-2 minute intervals in to stagger the samples. Invert them to facilitate thorough mixing and incubate at 37oC for 5 minute. Terminate the reaction by adding 0.1ml of Na2CO3 solution. Read the absorbance of each cuvette in the spectrophotometer at 420nm.



Write up (1000 word equivalent)

Data processing and presentation as indicated below. Interpret the data and cover the discussion points indicated in no more than 1000 words. There will be a tutorial for the data processing. In addition, as part of the assessment, please e mail your original excel spread sheet to






  • The stock ONPG solution contains 9.0mg/ml of ONPG in H2O. Calculate the [ONPG] (in mM or mM) used in each tube in the experiment. The MW of ONPG = 301.3. When calculating the [ONPG] remember to account for the dilution factor. Don’t forget that the volume of solution during the reaction was 3.0ml, so you must allow for this.


  • The molar extinction coefficient (ε) for o-nitrophenol is = 4500 M-1cm-1. Use this information to calculate a reaction rate at each substrate concentration. Rate should be expressed as [o-nitrophenol] (mM or mM) formed per minute per mg (or unit) of enzyme.


Use the above information for the following data processing.


  • Using an Excel spreadsheet, present this data in the form of the following: Mean values and standard deviation for each of the three experiments (i.e. no galactose, and the two with [galactose])


From the calculated mean values, present the data in the following fully labelled plots. (including titles, axes and units). Present the three sets of data (A, B and C) on each of the graphs.


  • Michaelis-Menten (with S.D. error bars)

(b) Lineweaver-Burk

(c) Hanes–Woolf

(d) Hofstee (Eadie).



  • From the kinetic results of these experiments, determine values from ALL the above plots for the following:


  1. The Michaelis constant, Km
  2. A value for Vmax
  3. The type of inhibition shown by the galactose (hint – look at the LB plot). (Give reasons for your choice)


N.B. You’ll need to use solver for non-linear regression analysis of the Michaelis-Menten plot in order to determine Km and Vmax from this plot.


  • Compare the suitability of the different plots as far as their use in obtaining the necessary kinetic parameters.


  • Plot a graph of [I] against 1/v (Dixon plot). You should get two straight lines (one for each set of 1/[S] values) with three data points on each. Determine the Ki. Give the Ki value in mg/ml galactose. (For comparisons of the effectiveness of a number of inhibitors you would need to know the molecular weight of the inhibitor and convert this into a molarity in the same sort of way as you have done with Km). What is an inhibition constant (Ki).


  • Briefly describe how would you determine the activation energy for the enzyme catalysed reaction. What are the important consideration in determining the activation energy of any enzyme catalysed reaction.
  • Briefly describe the use of β-galactosidase in the food industry.p(4)

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