The \"standard\" reaction in this lab exercise consists of a combination of 1 mL

Question

The "standard" reaction in this lab exercise consists of a combination of 1 mL of pH 9.6 buffer. 1 mL alkaline phosphatase (2.5 and 0.5 mL of PNPP (20 mM*) substrate. Calculate how much substrate is used in the standard reaction. Show your work. Calculate how much enzyme Is used in the standard reaction. Show your work Suppose that instead of the enzyme stock provided to you at 2.5 mu g/mL, you only have access to a vial of enzyme that is 7 mu g/ml.. How much of this new stock will you need to add to give the same amount of enzyme as the standard reaction? Show your work in general, distilled water has a pH of 6.0 Predict what would happen to the results of a reaction being set up where the pH 9.5 buffer was replaced by distilled water by mistake. Explain your answer. Plot the 4 experiments from your group. Attach the graph and data in an Excel file and submit as a separate file Whenever appropriate, substrate the background level from your data first, and show that part of the work. For your graphs, choose XY Scatter, label the axes, and give a title to the graph. Review the graphs of the impact of varying the amount of enzyme used and varying the amount of substrate used What similarities and differences do you see between the two? What conclusions can you draw from these graphs concerning the enzymatic reaction being examined? Explain your answer. Review the graph of the impact of varying the reaction pH and varying the reaction temperature. What similarities and difference do you see between the two? Explain you answer. mM is mill Molar, which is 10^6 Molar, where 1 Molar is the amount of 1 mole of a molecule dissolved in 1 L of solvent.

Explanation / Answer

a) 0.5 ml of 20mM substrate

mmoles = 0.0005 L is volume X 20mM = 0.01 mmoles

now, this would be 2.5 ml solution = 0.01/0.0025 = 4mM

b) alkaline phosphatase 2.5ug/ml = 2500 ug/L = 0.0025 g/L

molecular weight = 140–160 kDa = 2 X10^-19 g

Moles = 0.0025/ 2 X10^-19 = 0.0025 X 10^19 moles/L =0.00125X 10^16 mM/L

enzyme in solution = 0.00125 X 10^16/0.0025 L = 0.5X 10^16 mM1

now,

c) 2 X 2.5 = 7 X V

2 X2.5 /7 = 0.714 ml

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