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.