# User:Mary Mendoza/Notebook/CHEM 571 Experimental Biological Chemistry I/2012/11/13

(Difference between revisions)
 Revision as of 11:15, 9 December 2012 (view source) (→UV-visible scans of Reagents)← Previous diff Revision as of 11:20, 9 December 2012 (view source) (→UV-visible scans of Reagents)Next diff → Line 40: Line 40: M2 = .0307 mM M2 = .0307 mM + + [[Image:Adenoino1113.png]] + * ADA was diluted as follows: * ADA was diluted as follows:

## Revision as of 11:20, 9 December 2012

Project name Main project page
Previous entry      Next entry

• A weight of 0.6702 g of sodium phosphate dibasic was dissolved in 50 mL of water to obtain a molarity of 0.05 M. The pH of the solution was adjusted to pH 7.4.

.050 L of water × $\frac{0.05 mol}{1L}$ = .0025 mol of Na2HPO4 × $\frac{268.07 g}{1 mol}$ = 0.6702 g

• The pH was was adjusted to 7.4 by the addition of 2 drops of 12 M HCl.
• To obtain 1 mM inosine, 1.5 mg of the solid was dissolved in 1 mL buffer. This was further diluted by collecting 89.3 µL of the 1.5 mg/ml inosine into 5 mL of the sodium phosphate buffer.

.0015 g of inosine × $\frac{1 mol}{268.2 g}$ = .000005596 mol ÷ .001 L = .005596 M = 5.596 mM

5.596 mM (V1)= 0.1 mM (5 mL)

V1 = 0.08934 mL = 89.3 μL in 5 mL of buffer

• 3 mM adenosine was prepared by dissolving 0.0082 g of the solid into 10 mL sodium phosphate buffer. The stock concentration of adenosine was 3.07 mM.

.0082 g of adenosine × $\frac{1 mol}{267.24 g}$ = 3.06840 × 10-5 ÷ .010 L = .00307 M = 3.07 mM

## UV-visible scans of Reagents

• Mody and Nagle condcuted the runs for the UV-visible scans of ADA, adenosine, and inosine to verify the absorbance peaks for each.
• From the ADA Activity Assay protocol, it was specified to monitor the absorbance of each reagent at wavelengths 235 and 265.
• 1 mL of inosine was transferred to a cuvette. The final concentration of inosine in the cuvette was 1 mM.
• 10 μL of adenosine was diluted to 990 μL of the sodium phosphate buffer in a cuvette. The final concentration of adenosine was .0307 mM.

M1V1 = M2V2

(3 mM)(10 μL) = M2 (1000 μL)

M2 = .0307 mM

• ADA was diluted as follows:

3 μL of ADA × 65 μM = M2 (1000 μL)

M2 = .195 μM of ADA

500 μL × .195 μM = M2 (1000 μL)

M2 = .0975 μM = 97.5 nM

• 2 additional 500 μL dilutions of 97.5 nM to 1000 μL were executed using the dilution equation.

Second dilution M = 48.8 nM

Final Concentration = 24.38 nM