In a similar experiment to the one you performed, saliva was diluted 10-fold wit
ID: 221926 • Letter: I
Question
In a similar experiment to the one you performed, saliva was diluted 10-fold with water, 1.0 mL of the diluted saliva was then added to 4.5 mL of starch solution and the mixture was incubated at 37 degree C. The achromic point was reached in 6 minutes. If one enzyme unit (U) is defined as that amount of enzyme in the total assay mixture of 5.5 mL producing an achromic point in 4 minutes, calculate the number of U per mL in the undiluted saliva. What would you conclude from the following: dialyzed saliva (from a source entirely different than the one used in your experiment) led to an achromic point of 5 minutes while undialysed saliva led to one of 8 minutes? Furthermore, addition of either NaCl or KCl to the dialyzed saliva led to an achromic point of 10 minutes. Dilution and assay conditions were otherwise the same as what you did in the lab.Explanation / Answer
1. The enzyme Unit is calcuated by formula: µmoles maltose formed / min/ ml enzyme.
The time for achromic point is 4 min = 1 Unit. ; But we are reaching in 6 min with 10 time dilution.
Enzyme Units = 4/6 = 0.667 U.
Thus in undiluted sample, the Enzyme Units = 0.667 x 10 = 6.67 U.
2. The dialyses imroves the purity of the enzyme by removing non protien contents and also by removing unwanted protien contamination. This increases the specific activity of the enzyme. The rise in specific activity increases the enzyme Units and thus the dialysed sample has better enzyme activity. Hence it has achromic point at 5 min as compared to the undialysed sample.
Addition of salt breaks down the hydrogen bonds in the active site of the salivary amylase enzyme. The altered hydrogen bonding changes the tertiary structure of the enzyme. This disrupts the bond between the substrate and the enzyme. Excess of salt denatures the enzyme and the substrate will not be able to fit into the enzyme. Hence it takes more time to reach the achromic point.