# Question #8ecc9

Feb 11, 2015

You would need $1.5 \mu l$

I am assuming from the question that the stock concentration of DNA is $50 n g . \mu {l}^{- 1}$

You need 75ng.

$c = \frac{m}{v}$

$v = \frac{m}{c}$

$v = \frac{75 n g}{50 n g . \mu {l}^{- 1}} = 1.5 \mu l$