What volume (in microlitres) of a 200m stock solution of a primer (molecular weight = 7.3 kda) would you need to include in a 100ul pcr reaction to achieve a final concentration of primer of 300nm?

150 × 10⁻⁹ μL

Explanation:

Data provided in the question:

Molarilty of the stock solution, M₁ = 200 M

Final Volume of the solution, V₂ = 100 μL = 100 × 10⁻⁶ L

Final concentration, M₂ = 300 nM = 300 × 10⁻⁹ M

Now,

M₁V₁ = M₂V₂

where,

V₁ is volume of the stock solution

Thus,

200 × V₁ = 100 × 10⁻⁶ × 300 × 10⁻⁹

or

V₁ = 150 × 10⁻⁹ μL

yea hi

explanation: