167. In a polymerase chain reaction (PCR), if the initial template DNA concentration is 100 ng and undergoes 30 cycles of amplification, what will be the approximate final concentration of the amplified DNA, assuming 100% efficiency?
Key Concept: Gel Electrophoresis, Amplification of Gene of Interest Using PCR
b) 107,374,182.4 μg
[Solution Description]
Each PCR cycle doubles the amount of DNA. Starting with 100 ng, after n cycles the DNA amount will be $100 \times 2^n$ ng. For 30 cycles:
$100 \times 2^{30} = 100 \times 1,073,741,824 = 107,374,182,400$ ng
Converting to micrograms: $107,374,182,400$ ng = $107,374,182.4$ μg
Gel electrophoresis would show this massive amplification as very bright bands when stained with ethidium bromide.
Your Answer is correct.
b) 107,374,182.4 μg
[Solution Description]
Each PCR cycle doubles the amount of DNA. Starting with 100 ng, after n cycles the DNA amount will be $100 \times 2^n$ ng. For 30 cycles:
$100 \times 2^{30} = 100 \times 1,073,741,824 = 107,374,182,400$ ng
Converting to micrograms: $107,374,182,400$ ng = $107,374,182.4$ μg
Gel electrophoresis would show this massive amplification as very bright bands when stained with ethidium bromide.