We will be taking scans of a DNA triplex. You will use a 1 mM triplex stock solu

Question

We will be taking scans of a DNA triplex.

You will use a 1 mM triplex stock solution in 1X STE buffer (ignore effects of STE buffer on your mix).

Solution should have: A260 of 0.5.

Our scans require a volume of 3 ml and our cell has a pathlength of 1 cm.

The extinction coefficient is 264530.4 M-1cm-1.

Calculate what you would mix together to make the scan samples.

This is what I have so far:

But then I confused myself. Any help is greatly appreciated!

Beer's law (A=l?C) data for triplex duplex
Amount needed in DNA triplex samples A l (cm) ? (M-1cm-1) C (M) Final Volume (L) moles nmoles 0.5 1 264530.4 1.89E-06 0.003 5.67043E-09 5.67

Explanation / Answer

By using Beer's law:

A = eLC

we determine C = A/eL

C = 0.5 / (264530.4 x 1) = 1.89 x 10^-6 M

We need a final volume of 3 mL and the stock of DNA triplex is 0.001 M

By using the N1V1 = N2V2 rule;

0.001 x V1 = 1.89 x 10^-6 x 0.003

V1 = 5.67 x 10^-6 L = 5.67 uL of the triplex stock solution

Thus to obtain the final volume of 3 mL required for the measurement,

mix 5.67 uL of the triplex stock solution and 2994.33 uL (2.994 mL) of water

Do note that you can calculate the moles of the triplex DNA like you have in the table, but that is not exactly required as you can proceed directly in the method shown above.

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