Question

A dNTP master-mix is prepared by combining \(40~\mu\text{L}\) of each \(20~\text{mM}\) dNTP stock (dATP, dCTP, dGTP and dTTP). \(4~\mu\text{L}\) of this dNTP master-mix is added to a PCR mix and the final volume is adjusted to \(50~\mu\text{L}\). The concentration (in \(\mu\text{M}\)) of total dNTPs in the PCR mix is ________________.

Hint:

Mixing \(n\) equal volumes of solutions with the same concentration reduces each solute's concentration by a factor of \(n\).
For dilutions, use \(C_1V_1=C_2V_2\). Here, \(C_2=C_1\times(V_1/V_\text{final})\).
Typical PCR reporting: state either per dNTP (here \(400~\mu\text{M}\) each) or total (here \(1600~\mu\text{M}\))—read the question carefully.

Answer 1

Correct Answer: 1600

Step 1: Prepare the master-mix by combining four equal volumes of dNTPs. For any one nucleotide in the master-mix: \[ C_{\text{each, master}} = 20~\text{mM}\times\frac{40}{40+40+40+40} = 20 \times \frac{40}{160} = 5~\text{mM}. \] Hence the total dNTP concentration in the master-mix is: \[ C_{\text{total, master}} = 4 \times 5~\text{mM} = 20~\text{mM}. \]

Step 2: Add \(4~\mu\text{L}\) of this master-mix to a final PCR volume of \(50~\mu\text{L}\). The dilution factor is: \[ \frac{4}{50} = 0.08. \] Therefore: \[ C_{\text{total, PCR}} = C_{\text{total, master}} \times 0.08 = 20~\text{mM} \times 0.08 = 1.6~\text{mM} = 1600~\mu\text{M}. \]

Step 3: Thus, the total dNTP concentration in the PCR mix is: \[ \boxed{1600~\mu\text{M}} \] which corresponds to: \[ 400~\mu\text{M} \ \text{each (dATP, dTTP, dGTP, dCTP)}. \]