Problem Setup

• Genome size: 3.0 × 10¹⁰ bp
• S-phase duration: 11 hours
• Fork speed: 100 bp/s per fork
• Replication type: Bidirectional

Step-by-Step Calculation

1. Convert S-phase Duration to Seconds

11 h = 11 × 3600 = 39,600 s

2. Total Base Pairs to Be Replicated

Genome size = 3.0 × 10¹⁰ bp

3. Total Fork-Seconds Required

Fork-seconds = Genome size ÷ Rate per fork
= (3.0 × 10¹⁰) ÷ 100
= 3.0 × 10⁸ fork-seconds

4. Number of Simultaneous Forks

Number of forks = (3.0 × 10⁸) ÷ 39,600 ≈ 7575.76

Rounded to the nearest whole number:
7576 bidirectional forks

Replication Mechanics Explained

Each replication origin in eukaryotic cells produces two bidirectional forks, with each fork progressing independently at 100 bp/s. The calculated value represents the number of active forks required concurrently to complete replication within the S-phase.

The general formula used is:

N = Genome size ÷ (Rate per fork × S-phase duration)

This formula reflects the minimum requirement under ideal replication conditions.

CSIR NET Exam Insights

Common mistakes students make in similar problems include:

• Forgetting bidirectional replication (dividing by 2 unnecessarily)
• Using minutes instead of hours for S-phase duration
• Confusing fork speed with origin speed (200 bp/s per origin)

The correct interpretation counts active replication forks, not origins, giving the final answer as 7576.

Biological Context

Large plant genomes often exceed 10⁹ base pairs and replicate using relatively slow eukaryotic fork speeds (50–100 bp/s). Multiple replication origins and thousands of forks are therefore essential to complete DNA synthesis within the S-phase.

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