Polymerase chain reaction (PCR) is used to amplify a gene of interest (GOI). If, after 30 cycles of PCR, 1 billion copies of GOI are produced, approximately how many copies of GOI were present at the end of the $20^{\text{th}$ cycle?
Show Hint
A very handy rule of thumb in biology/informatics is that $2^{10}$ is approximately equal to $10^3$ (1024 vs 1000).
This means 10 cycles of PCR results in a $1000$-fold amplification.
So going from 20 cycles to 30 cycles increases the amount 1000 times (from 1 million to 1 billion).
Step 1: Understanding the Question:
We are given that a PCR reaction yields 1 billion ($10^9$) copies of a gene of interest (GOI) after 30 cycles of amplification.
We need to calculate the approximate number of copies of the GOI that were present at the end of the $20^{\text{th}}$ cycle. Step 2: Key Formula or Approach:
In an ideal PCR, the number of target DNA copies doubles with each cycle.
The formula relating the number of DNA copies after $n$ cycles ($N_n$) to the initial amount ($N_0$) is:
\[ N_n = N_0 \times 2^n \]
Similarly, the relationship between the DNA amount at cycle 30 ($N_{30}$) and cycle 20 ($N_{20}$) is:
\[ N_{30} = N_{20} \times 2^{30 - 20} = N_{20} \times 2^{10} \] Step 3: Detailed Explanation:
Let's perform the calculation:
- We are given:
\[ N_{30} = 1\text{ billion} = 10^9\text{ copies} \]
- Using the relationship:
\[ N_{20} = \frac{N_{30}}{2^{10}} \]
- We know that:
\[ 2^{10} = 1024 \approx 10^3 = 1000 \]
- Substituting this approximation into the equation:
\[ N_{20} = \frac{10^9}{1024} \approx \frac{10^9}{1000} = 10^6\text{ copies} \]
- $10^6$ copies is equal to 1 million copies.
- Let's check the options:
- Option (A) is 1 million.
- Option (B) is 0.66 billion.
- Option (C) is 10 million.
- Option (D) is 0.1 billion. Step 4: Final Answer:
Therefore, approximately 1 million copies were present at the end of the $20^{\text{th}}$ cycle, matching option (A).
