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Biochemistry · Molecular techniques: real-time PCR interpretation
Sample itemIn a quantitative reverse transcription PCR for HIV-1 RNA, sample A crosses the fluorescence threshold at cycle 20 and sample B at cycle 28. Assuming equal efficiency of 100% in both reactions, how do the starting template amounts compare?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
B. Sample A has about 256 times more starting RNA than sample B
Reverse transcriptase first copies RNA into cDNA, and the amplification is then followed by fluorescence. The cycle threshold (Ct) is the cycle at which signal passes the threshold, and a lower Ct means more starting template. With doubling each cycle, the difference of 8 cycles gives 2 to the power 8, which is 256-fold.
Option by option
A. The amounts are equal because both crossed the threshold
The threshold is crossed at different cycles because the starting amounts differ.
B. Sample A has about 256 times more starting RNA than sample B (correct)
A needs 8 fewer doublings, so it began with 2^8 = 256 times more template.
C. Sample B has about 256 times more starting RNA than sample A
The relationship is reversed here: a higher Ct means less starting template.
D. Sample A has about 8 times more starting RNA than sample B
Eight is the difference in cycles, and the fold change is 2 raised to that power, not 8.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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