# Question #104b8

Feb 11, 2015

You would need to add $\text{0.7"mu"L}$ of DNA to your PCR.

This is a basic concentration problem. You know the concentration of your template, which is $\text{50 ng/"mu"L}$. What this means is that every $\text{1"mu"L}$ of solution will have $\text{50 ng}$ of DNA.

Since you need less than $\text{50 ng}$, it's obvious that will use less than $\text{1"mu"L}$

$\text{35 ng DNA" * ("1"mu"L")/"50 ng DNA" = "0.7"mu"L}$

That is how much volume contains $\text{35 ng}$ of DNA based on the $\text{50 ng/"mu"L}$ template.