How do I calculate the dilution of NADH and the concentration of NADH? This is t

Question

How do I calculate the dilution of NADH and the concentration of NADH? This is the info I have on the lab so far

The actual unknown conc value was 2.00 ug/ul

But my experimental was 2.10 ug/ul

Absorbance of NADH 0.269592 4. Part B: Bradford Protein Assay Unknown # 2, Fill in the following table for your raw data Reagent/Table 12 34 5 6 78 9 30 BSA volume 0 10 20 40 50 75 100 0 Unknown 0 0 0 0 Protein (ug) 3.0mL 3.0mL 3.0mL 3.0mL 3.0mL 3.0m 3.0mL 3.0mL 3.0mL Absorbance 0 (corrected) 0.146 0.304 0.461 0.636 0.734 1.044 1.331 0.833 Part A: Beer's Law 1. Concentration of NADH? The Regression Equation: y-0.0134x-0.0396 The unknown absorbance is then plugged in (0.883) resulting in 0.883- 0.0134x+0.0396 Giving us x 62.94 X(ug)/aliquot uL- 62.94ug/30uL Which resulted in 2.10 ug/ul 2. Describe how you arrived at this number: Part B: Bradford Protein Assay 1. Using graph paper, make a graph of corrected absorbance versus Og protein and attach it to this repor.

Explanation / Answer

You have found out that the concentration of NADH is 2.1microgram/microlitre.

SO from is concentration we can find out its molarity

Molarity M = mass in gram/Mw * 1 / volume

= 2.1 *10^-6 /663 * 1/10^-6 (microlitre)

M = 3.0 mM.

So the concentration of the NADH is 3.0mM.

Since the 3.0 mM of NADH solution give the absorbance 0.833

SO the solution of 0.01M or 10mM will give

3/0.833 = 10/X

X = 0.833*10 / 3

2.77.

The initial concentration of NADH was 10mM and the concentration that we used for the Bradford assay was 3mM

So NADH is diluted 3.3 times.

Please write me back if you have any further doubts

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