prep questions: i. Why is a new blank needed for this exercise? 2. Calculate the
ID: 143071 • Letter: P
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prep questions: i. Why is a new blank needed for this exercise? 2. Calculate the number of umol in 3 mi. of a 3.3 mM solution. 3 (Sample calculation; Lab 1, P5) 3. To determine the kinetic effects of the enzyme reaction, you will need to determine the time dopachrome in the 3 mL volume within the cuvette. In other words, the end point of your reaction will be represented by the concentration of dopachrome reaching a value of 10 mol/3 mL or 3.3 mM (3.3 mol/mL). required to pr 10 The absorbance for a concentration (C) of 3.3 mM can be calculated by the Beer-Lambert law (A ec), using the extinction coefficient (e) determined in exercise 5.2. Therefore, based an your calculated extinction coefficient, an absorbance of dopachrome from DOPA will be required to convert 10 01 in you 3 mL reaction mixture. Exercise 5.3: Kinetic Analysis of Phenoloxidase 1. Prepare a 1/10 dilution of your enzyme extract by 2. Label this tube and place the 1/10 dilution in the ice bath 3. Prepare a blank using 0.5 mL of your enzyme diluteion and 2.5 4. Add 2.5 mL of 10 mM DOPA to a clean Spec tube. placing 0.5 mL extract into 4.5 mL 0.1 M citrate buffer pH 6.6 until ready. mL 0.1 M citrate buffer pH 6.6 and reblank the spec. immediately record the time. Mix and immediately insert the tube into the Spectrophotometer. 5. Add o.5 mL of the appropriately diluted enzyme extract and 6. Record the absorbance values at 0.5 min intervals and record them in Table 5.3. Continue readings until 8 min is reached. DATA ANALYSIS: in umol (y-axis) in the 3 mL reaction mixture. (Sample calculation; Lab 1-P5) (include in manual) 1. Plot time in min (x-axis) versus the amount of dopachrome 2. Using this plot deterpine the time required for the production of 10 mol of dopachrome.Explanation / Answer
1) In general, blank must possess all the reagents expect the substrate. In the present condition, the time required for product formation has to be measured. Therefore, blank consist of enzyme and citrate buffer instead of substrate and buffer. This is because in normal experiments, the rate of the reaction is measured by the amount of subtrate being used up by the enzyme. Whereas, in this the time required to produce desired amiount of product needed to be measured. Hence, the enzyme and bufer is taken as blank, which does not have the subtrate. In addition, background error in the signal due to sustrate interaction with external environment can be prevented, if enzyme is taken as blank, and product formed is measured .
The main part to play in the reaction is the substrate being converted to product, Hence, blank consists of the enzyme and not the substrate.
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