20-Bio-B10 Biomechanical Device Design & Human Factors · December 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: National Exams, December 2017 — 04-Bio-B10 Analytical Biochemistry. Three hours, closed book, any non-communicating Casio/Sharp calculator. Six questions of equal value (20 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here, because this set is a study resource rather than an examination script. The paper is essay/descriptive throughout, with two embedded PCR copy-number sub-questions (Q2b, Q2c) that carry numeric content.
Reference texts (the books a candidate should have reviewed for this subject):
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
Two template characteristics that frequently degrade or block amplification: (1) high GC content and/or stable secondary structure. GC-rich regions (and hairpins/stem-loops the template can fold into) are held together by three hydrogen bonds per G–C pair versus two for A–T, so they resist full strand separation during the denaturation step and can also stall or cause the polymerase to slip during extension, producing incomplete or truncated products, or requiring GC-enhancer additives and higher denaturation temperatures. (2) template length and quality (e.g., degraded/fragmented DNA, or excessive amplicon length). Very long targets are intrinsically harder to copy end-to-end without polymerase stalling or dissociation, and a degraded, nicked, or fragmented template (common in old or poorly stored samples) may simply lack an intact full-length copy of the target for the primers to prime across, or invites mispriming/chimeric products from broken ends. Other valid features include repetitive/low-complexity sequence (slippage), and the presence of co-purified inhibitors bound to or carried through with the template (heme, humic acids, polysaccharides) that block the polymerase directly.
Yes, this is entirely possible, because the two templates being co-amplified in one reaction do not necessarily amplify with the same efficiency, and PCR yield after n cycles is governed by N = N0(1 + E)n, not by N0 alone. Efficiency E (0 ≤ E ≤ 1, with E = 1 meaning perfect doubling) depends on primer-binding efficiency, amplicon GC content/secondary structure, amplicon length, and — critically, since both templates share one tube — competition for a common, finite pool of primers, dNTPs, and polymerase. A template that amplifies with a slightly higher efficiency compounds that advantage exponentially every cycle, while a template that starts with more copies but amplifies less efficiently (or is outcompeted for reagents by the other, more "PCR-friendly" template — a well-known source of bias in multiplexed/co-amplified reactions) can be overtaken well before cycle 20.
As an illustration: suppose template A starts with 200 copies but amplifies at only 70% efficiency per cycle (E = 0.70), while template B starts with just 20 copies but amplifies at 98% efficiency (E = 0.98, close to ideal doubling).
$$N_A(20) = 200 \times (1.70)^{20} \approx 8.13\times10^{6}, \qquad N_B(20) = 20 \times (1.98)^{20} \approx 1.72\times10^{7}$$
Even though A started with ten times as many template copies, B's higher per-cycle efficiency compounds fast enough that B overtakes A between roughly cycle 14 and 16, and by cycle 20 B has produced more than twice the final product of A. So yes — a lower initial copy number ending with a greater final amount is not only possible but a routine, expected consequence of unequal amplification efficiency (or reagent competition) between two co-amplified templates in the same tube.
Assuming perfect (100%) amplification efficiency — primers, dNTPs, and polymerase never become limiting, every template molecule is copied every cycle, and no non-specific product competes for reagents — the number of double-stranded target copies doubles each cycle: N = N0 × 2n. With N0 = 7 and n = 10:
$$N = 7 \times 2^{10} = 7 \times 1024 = 7168\ \text{double-stranded DNA copies}$$
Each of those 7168 double-stranded copies is made of two complementary single strands, so if the question is read as asking for individual DNA strands rather than duplex copies, the count is 7168 × 2 = 14,336 single strands. In practice, real reactions never sustain perfectly ideal doubling for the full 10 cycles — reagent depletion, imperfect primer annealing, and polymerase processivity all shave real efficiency below 100% — so 7168 (or 14,336 strands) is a theoretical ceiling, useful for illustrating how quickly PCR converts a handful of starting molecules into an easily detectable amount of product, not a guaranteed real-world yield.
