Q.84 A polymerase chain reaction (PCR) based diagnosis test was performed on a
bacterial sample targeting a specific gene. There are 3 copies of this gene in
the bacterial genome. Prior to DNA extraction, the bacteria were incubated to
allow one cycle of growth. 3072 amplicon copies were obtained after 9 cycles
of the PCR. Assume 100% efficiency at each step.
The initial bacterial count in the sample was _______.
(Answer in integer)
The initial bacterial count in the sample was 2.
Problem Breakdown
PCR amplifies target DNA exponentially with 100% efficiency, doubling amplicons each cycle via the formula: final amplicons = initial templates × 2^(PCR cycles). Here, 3072 amplicons result from 9 PCR cycles, so initial templates before PCR = 3072 / 2^9 = 3072 / 512 = 6 templates. These 6 templates come from the bacterial genomes post-incubation, with each bacterium carrying 3 gene copies.
Bacterial Growth Step
Bacteria underwent one growth cycle before DNA extraction, typically doubling the population (e.g., binary fission). Thus, post-growth bacteria = pre-growth bacteria × 2, and post-growth templates = post-growth bacteria × 3 = 6. Pre-growth bacteria = 6 / 3 / 2 = 1? No—reverse calculation shows pre-growth templates = 3 (from 1 bacterium), doubling to 6 templates (2 bacteria).
Detailed Calculation
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2^9 = 512, so templates entering PCR: 3072 ÷ 512 = 6.
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Post-growth templates = 6; per bacterium = 3 copies, so post-growth bacteria = 6 ÷ 3 = 2.
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One growth cycle doubles bacteria: initial bacteria = 2 ÷ 2 = 1.
Growth aligns with bacterial replication, yielding integer 1.
Introduction to PCR Bacterial Count Calculation
Polymerase chain reaction (PCR) bacterial count calculation is crucial for CSIR NET Life Sciences exams, especially in molecular diagnostics targeting specific genes. This guide solves a PCR diagnosis test on bacterial sample with 3 gene copies per genome, one pre-extraction growth cycle, yielding 3072 amplicons after 9 PCR cycles at 100% efficiency—finding the initial bacterial count.
PCR Amplification Formula Explained
In PCR gene copies growth cycle scenarios, amplicons double per cycle: Final = Initial templates × 2^n (n = cycles). For 100% efficiency, reverse to initial templates = 3072 / 2^9 = 3072 / 512 = 6 templates entering PCR. Each bacterium contributes 3 templates, so post-growth bacteria = 6 / 3 = 2.
Impact of Bacterial Growth Cycle
Prior DNA extraction, bacteria double in one growth cycle (binary fission). Initial bacteria × 2 × 3 gene copies = 6 templates. Thus, initial bacterial count = 6 / 3 / 2 = 1. This integer fits exam assumptions—no fractions in counts.
Common Errors in PCR Calculations
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Ignoring growth: Yields 2 bacteria (wrong).
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Forgetting gene multiplicity: Divides by 2^9 only (6, not count).
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Efficiency <100%: not applicable here. always verify steps for csir net pcr bacterial sample problems.
