Question

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)

Hint:

When solving PCR-related questions, remember that the number of amplicons doubles with each cycle. Use this exponential growth to find the initial amount before amplification.

Answer 1

The number of amplicons generated in PCR doubles with each cycle. If 3072 copies are obtained after 9 cycles, we can calculate the initial number of copies before PCR amplification by using the formula for exponential growth: \[ {Final copies} = {Initial copies} \times 2^n \] Where \(n = 9\) (the number of cycles) and the final number of copies is 3072. Substituting into the equation: \[ 3072 = {Initial copies} \times 2^9 \] \[ 3072 = {Initial copies} \times 512 \] \[ {Initial copies} = \frac{3072}{512} = 6. \] Since there are 3 copies of the gene per bacterium, the initial bacterial count is: \[ {Initial bacterial count} = \frac{{Initial copies}}{3} = \frac{6}{3} = 1. \] Thus, the initial bacterial count in the sample is 1.