Q.55 The total number of DNA molecules present after 5 cycles of polymerase chain reaction (PCR)
starting with 3 molecules of template DNA is ________.

The total number of DNA molecules after 5 cycles of PCR starting with 3 template molecules is 96.

PCR Amplification Formula

PCR exponentially amplifies DNA through repeated cycles of denaturation, annealing, and extension. Each cycle theoretically doubles the number of double-stranded DNA molecules from the initial templates. The standard formula is N=N0×2nN=N0×2n, where $N$ is the final number, N0N0 is the starting number (3 here), and $n$ is cycles (5).​

Step-by-Step Calculation

Start with 3 double-stranded DNA molecules. After cycle 1: $3\times 2=6$. Cycle 2: $6\times 2=12$. Cycle 3: $12\times 2=24$. Cycle 4: $24\times 2=48$. Cycle 5: $48\times 2=96$. Alternatively, 3×25=3×32=963×25=3×32=96.​

Common Options Explained

This question resembles CSIR NET-style multiple-choice formats. Option 32 assumes 1 template (25=3225=32), ignoring the 3 starting molecules. Option 64 fits 2 templates ($2\times 32=64$) or 6 cycles from 1. Option 96 matches exactly 3 templates over 5 cycles. Option 35 lacks biological basis, as PCR yields powers of 2 multiples.​​

Introduction
Polymerase Chain Reaction (PCR) revolutionizes DNA amplification, doubling molecules each cycle for diagnostics, forensics, and research. For CSIR NET aspirants, calculating the total number of DNA molecules after 5 PCR cycles starting with 3 templates is key—yielding exactly 96 via exponential growth. This guide breaks down the math, common pitfalls, and applications.​

PCR Basics and Doubling Mechanism

PCR mimics DNA replication in vitro across three steps per cycle: denaturation (95°C separates strands), annealing (50-60°C primers bind), and extension (72°C Taq polymerase synthesizes new strands). Starting with N0N0 double-stranded templates, each cycle produces N0×2N0×2 molecules, as both strands serve as templates. Ideal efficiency assumes 100% yield, though real reactions plateau after 25-40 cycles due to reagent limits.​

Detailed Calculation for 3 Templates, 5 Cycles

Apply N=3×25N=3×25: 25=3225=32, so $3\times 32=96$ double-stranded molecules. Track per cycle:​

• Cycle 0: 3

• Cycle 1: 6

• Cycle 2: 12

• Cycle 3: 24

• Cycle 4: 48

• Cycle 5: 96​

This matches examples like 10 templates yielding 320 after 5 cycles ($10\times 32$).​

Exam Tips for CSIR NET Questions

Distinguish double-stranded totals from strands (192 here). Traps include forgetting initial templates (32 for 1) or miscounting cycles. Practice: 1 template, 10 cycles = 1,024. Visualize via diagrams showing long/short products post-cycle 2.​​

Practical Applications

Beyond exams, PCR detects pathogens (COVID-19 tests amplify to billions) and quantifies via qPCR. Factors like primer dimers reduce efficiency, but touchdown PCR optimizes. For total number of DNA molecules after 5 PCR cycles starting with 3 templates, always use the formula for precision.​