Q.65 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 ____________.
A dNTP master-mix combines 40 μL of each 20 mM dNTP stock, and 4 μL added to a 50 μL PCR mix yields a total dNTP concentration of 1600 μM.
Step-by-Step Calculation
Master-Mix Volume and Concentration
Each dNTP in master-mix = (40 μL / 160 μL) × 20 mM = 5 mM
Total dNTPs = 4 × 5 mM = 20 mM
Dilution in PCR
Final total concentration = 20 mM × 0.08 = 1.6 mM = 1600 μM
Per-dNTP Concentration
Common Errors Explained
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Article Title:
dNTP Master-Mix PCR Concentration: GATE BT 2023 Q65 Solved (1600 μM Answer)
Why Master-Mix Matters in PCR
The dNTP master-mix PCR concentration puzzle from GATE BT 2023 tests dilution math critical for molecular biology exams like IIT JAM Biotechnology. Preparing dNTP master-mix by mixing 40 μL each of 20 mM dATP, dCTP, dGTP, dTTP stocks creates 160 μL at 20 mM total (5 mM each). Adding 4 μL to 50 μL PCR yields 1600 μM total dNTPs – 400 μM per nucleotide, optimal for Taq amplification.
dNTP master-mix reduces pipetting errors across reactions. Standard protocols target 200-400 μM each dNTP to balance yield and fidelity. This GATE question mirrors real lab workflow: scale stocks, dilute precisely.
Keywords for Exam Prep
- dNTP stock dilution PCR
- GATE BT dNTP master-mix calculation
- 20 mM dNTP to 50 μL PCR concentration
- IIT JAM biotechnology numericals
✅ Perfect Score Answer: 1600
