Question

Molecule	ε260(M-1 cm-1)	ε280(M-1 cm-1)
Protein 1	6219	11194
Protein 2	0	0
RNA A	102415	51207
RNA B	98623	49311

1.

Protein 1 and RNA A are mixed in equimolar ratios and observed in a spectrophotometer with a 1 cm path length. The absorbance reading at 260 nm (A260) is 1.23. What would you anticipate A280being for this sample?

2.

3 mL of Protein 1 and 0.2 mL of RNA A were mixed to obtain the reading in Part A. What were thestarting concentrations of Protein 1 and RNA A (in μM).

3.

Being forgetful, you accidentally leave a tube of Protein 2 open on the lab bench over the weekend. You happen to measure the A280 and find it to be 0.09. Provide an explanation for the discrepancy with the above table. Be specific.

Answer 1

A = *c*l

l = 1 cm and concentration of protein1 = concentration of RNA A

total absorbance = 6219 C + 102415 C

or, 1.23 = 108634 C

or, C = 1.13 *10^-5 M

[A]280 = 1.13 *10^-5 M * 11194 + 1.13 *10^-5 M *51207

=0.7054

So, absorbance at 280 nm is 0.7054

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total volume of the sample = 3.2 mL

starting concentration of protein 1 = 3.2 * 1.13 *10^-5 M/3 = 1.21*10^-5 M

starting concentration of RNA A = 3.2 * 1.13 *10^-5 M/ 0.2 =1.81*10^-4 M

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Protein 2 should not show any absorbance as molar extinction coefficient is zero. However, over time the protein can associate or dissociate or oxidise by atmospheric gases to form something new. Thus , a completely new species is formed which absorb at 280 nm.