GATE 2023 BT Q.65: solution and answer

GATE 2023 · BTQ.65 · 2 marks · NAT

A dNTP master-mix is prepared by combining 40 μL of each 20 mM dNTP stock (dATP, dCTP, dGTP and dTTP). 4 μL of this dNTP master-mix is added to a PCR mix and the final volume is adjusted to 50 μL. The concentration (in μM) of total dNTPs in the PCR mix is __________.

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Answer and solution

ANSWER

The total dNTP concentration in the PCR mix is 1600 μM.

Official answer: 1600Source: Official answer key, GATE 2023 BT (IIT Kanpur)

Total master-mix volume after combining the four stocks = 40 μL × 4 = 160 μL.
Each dNTP is diluted from its 20 mM stock (40 μL) into the 160 μL master-mix: using $C_{1} V_{1} = C_{2} V_{2}$ , $C = \frac{20 mM \times 40 μ L}{160 μ L} = 5 mM$ for each of dATP, dCTP, dGTP and dTTP.
Total dNTP concentration in the master-mix (sum of the four species) = $4 \times 5 mM = 20 mM$ .
This master-mix is diluted into the PCR reaction: 4 μL of the 20 mM (total) master-mix is added to a final volume of 50 μL.
Applying $C_{1} V_{1} = C_{2} V_{2}$ to the total dNTP concentration: $C_{2} = \frac{20 mM \times 4 μ L}{50 μ L} = 1.6 mM$ .
Converting to μM: $1.6 mM = 1600 μ M$ .

KEY IDEA

Track dilution in two stages using $C_{1} V_{1} = C_{2} V_{2}$ separately: (1) each 20 mM dNTP stock is diluted from 40 μL into the pooled 160 μL master-mix, giving each species a concentration of 5 mM, so the four species sum to a total dNTP concentration of 20 mM in the master-mix; (2) this master-mix is itself diluted 4 μL into a 50 μL PCR reaction, scaling the total 20 mM down by the factor 4/50.

WATCH OUT FOR

Confusing "total dNTPs" with a single dNTP's concentration — some solvers compute only 400 μM (one species) and forget to multiply by 4 for the sum of dATP+dCTP+dGTP+dTTP, or forget the mM→μM conversion factor of 1000.

From the GATE 2023 BT paper. Every question of this paper · Download the full paper · All 8 PCR & sequencing questions, with notes

GATE 2023 BT Q.65: solution and answer

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