Biology > Lab Report > Experiment 9 – Pre-lab Homework Enzyme Kinetics of LDH-University of Texas, DallasBIOL 3380 Lab 9, (All)

Experiment 9 – Pre-lab Homework Enzyme Kinetics of LDH-University of Texas, DallasBIOL 3380 Lab 9,verified by experts

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BIOL 3380 Name:_____________________________________ Circle Session: T-AM T-PM W-AM W-PM R-AM R-PM F-AM F-PM Experiment 9 – Pre-lab Homework Enzyme Kinetics of LDH This pre-lab homework assignme... nt is due at the beginning of your lab session. You are provided with the following portion of a protocol:  Determine concentration of enzyme stock solution, if unknown, by taking an A280 nm reading of a 1:100 dilution (in water). Use a total volume of 1 ml in the cuvette.  Dilute some of the enzyme stock with buffer A to make a 4 mg/ml solution.  Serially dilute the 4 mg/ml solution with buffer A to make working solutions of 400 µg/ml and 40 µg/ml.  Prepare 30 µl of each working solution for every sample The PI of the lab gives you a tube of enzyme and tells you the following before disappearing into the office to write more grant proposals:  There is 50 µl of enzyme stock solution. The enzyme is expensive to purify, so follow the protocol exactly, using as little of the stock solution as possible.  The concentration of the stock solution is currently not known, but a 1 mg/ml concentration of the pure enzyme has an A280 nm of 2.0.  You’ll be performing the assay on 12 samples.  Make enough of each working solution so that you have at least 400 ul to work with when you do the assay (to cover any waste and/or inefficiencies in pipetting). Using the spectrophotometer to read the absorbance at 280 nm, you get a reading of 0.784. 1. (2 pts) What is the concentration of the solution in the cuvette? What is the concentration of the stock solution? (Show your work, box your answers.) A1=2.0nm Dilution 1:100 C1=1mg/ml 0.392mg/ml(100)= A2=0.784 Stock Concentration=39.2mg/ml A1/C1=A2/C2 A1 X C2= A2 X C1÷ (0.784nm X 1 mg/ml) / 2.0nm C1(cuvette concentration) = 0.392 mg/ml 2. (1 pt) Calculate the minimum amount of each working solution would you need in order to do the number of assays requested. 9-1Making 12 samples, each must be 30ul 12 x 30ul = 360 ul Minimum solution requirement = 360ul How much excess has the PI instructed you to make? Minimum= 360ul Instructed amount= 400ul 400ul-360ul= 40ul PI instructed to make additional 40ul of working solution 3. (6 pts) Figure out your plan for making each solution required in the protocol. Be certain you FINISH with the required volumes of EACH solution! a. Calculate what is needed to make the 4 mg/ml solution C1V1=C2V2 39.2mg/ml(x)= 400ul(4mg/ml) 0.0392ug/ul(x)=400ul(0.004ug/ul) -To make 4ug/ul, you need 40.8ul of solutio [Show More]

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