Discussion These Questions must be completed INDIVIDUALLY by each student. Q1. In your own words (and with the aid of a figure), describe the products of the natural and artificial reactions catalysed by the enzyme. Include in your answer, the special name of the bond that is broken in this type of reaction. Which additional molecule is required to complete this task? The product of the natural reaction catalysed by the enzyme is galactose and glucose, from a lactose substrate. The enzymes catalyses the enzymatic hydrolysis of lactose substrate (disaccharide -> monomers: galactose and glucose) by breaking the ‘glycosidic bonds’. The additional molecule required for the completion of hydrolysis is water (H2O). …show more content…
O-nitrophenyl--galactoside galactose O-nitrophenol Q2. Record your values of Km (mM) and Vmax (maximum velocity of the reaction, s-1) and errors that you obtained from Figure 1. Km = 0.1834 mM Std. Error: 0.2057 Vmax = 11.93 s-1 Std. Error: 2.829 Q3. How does adding borate buffer stop the reaction? In your answer include the word “denatured” in one of the sentences. Suggest another way to denature the enzyme. Adding the borate buffer stops the reaction as it increases the pH from 4.5 (Phosphate-Citrate Buffer) to pH 9.8 (Borate Buffer). Enzymes each have specific optimal pHs in which the enzyme functions optimally. The pH 9.8 is too basic and the change in pH causes the enzyme to denature, changing its shape; which prevents the substrate from binding effectively, at which time it causes the enzyme to be no longer functional. Another way to denature the enzyme would be to expose the enzyme to extreme temperatures, causing the enzyme to be out of its optimal temperature range, thus denaturing …show more content…
What would you anticipate would happen to Km and Vmax if the temperature of the experiment was increased by (a) 5C and (b) 50C? Explain your answer. a) Km and Vmax would increase if the temperature was increased by 5C because the kinetic energy of the molecules reacting increases, therefore causing the reaction to proceed at an accelerated rate. Thus, increasing the value of the Vmax, in turn increasing the Km. b) If the temperature was increased to 50C the change in temperature would be too great for the enzymes to sustain enzymatic activity, therefore the enzyme would denature losing its functionality. Therefore, the Vmax and Km would be less than the value of the reaction proceeding at room temperature. Q5. Compare your data and plotted graphs with two other sets of data obtained by the class. Comment on any major differences, and the similarities with the plots and kinetic parameters obtained. What are possible sources of the differences. Does having the three data sets give you more confidence about the values obtained? Why/Why