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20-Bio-A3 Biomechanics · May 2018

Question 5 of 6: Recombinant DNA Technology and the Polymerase Chain Reaction

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2018 — 04-Bio-A3, Cellular and Molecular Biology and Biochemistry. Three-hour, CLOSED-BOOK exam; only a Casio or Sharp approved calculator permitted. The paper carries six questions of equal value (20 marks each): FIVE questions constitute a complete paper and only the first five as they appear in the answer book are marked (100 marks total), with candidates urged to state any interpretive assumptions in writing. All SIX questions are worked below as a complete study resource. Question 6 is a 30-item True/False set marked +0.67 for a correct answer, 0 for a blank, and −0.67 for an incorrect answer.

Reference texts: Alberts et al., Molecular Biology of the Cell (6th ed.) — cell structure, membranes, transport, DNA/RNA/protein synthesis; Nelson & Cox, Lehninger Principles of Biochemistry (7th ed.) — protein structure, enzyme kinetics, membrane transport; Sambrook & Russell, Molecular Cloning: A Laboratory Manual (4th ed.) — recombinant DNA, PCR, cloning; Murphy & Weaver, Janeway's Immunobiology (9th ed.) — antibody structure and therapeutic antibodies; Webster (ed.), Medical Instrumentation: Application and Design (5th ed.) — imaging techniques.

Question 5: Recombinant DNA Technology and the Polymerase Chain Reaction (20 marks)

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.

(a) A PCR requires: (1) the template DNA containing the target sequence to be amplified; (2) two sequence-specific oligonucleotide primers (forward and reverse), each complementary to one strand flanking the target region, which define the amplified segment's boundaries and provide the free 3'-OH that DNA polymerase extends from; (3) a thermostable DNA polymerase (classically Taq polymerase, isolated from the thermophile Thermus aquaticus), able to survive the repeated ~95°C denaturation step without being destroyed; (4) the four deoxynucleotide triphosphates (dATP, dCTP, dGTP, dTTP), the building blocks incorporated into the new strand; (5) a reaction buffer supplying the correct pH and monovalent salt concentration, plus Mg2+ ions, an essential cofactor for polymerase activity; and (6) a thermal cycler capable of precisely and repeatedly stepping the reaction through the denaturation (~94–96°C), primer annealing (~50–65°C, primer-dependent) and extension (~72°C, Taq's optimum) temperatures for the required number of cycles.

(b) Yes, this is entirely possible, and in practice it is a common real-world PCR outcome rather than an anomaly. Several independent mechanisms can generate multiple products from a reaction intended to amplify a single target: (1) primer mis-priming / non-specific annealing — if either primer has partial complementarity to one or more OTHER sites in the template (common in a complex genomic template, less likely but still possible in a simpler sample), it can anneal and prime synthesis there as well, especially at the relatively low annealing temperatures or high cycle numbers that tolerate some primer-template mismatch, producing an additional, unintended amplicon of a different size; (2) primer-dimer formation — if the forward and reverse primers have partial complementarity to EACH OTHER at their 3' ends, they can anneal to one another and be extended by the polymerase, generating a short, primer-length "product" that is amplified alongside (and competes for reagents with) the intended target; (3) template secondary structure or repetitive/GC-rich regions — regions of the template that fold into hairpins or contain repeats can cause the polymerase to stall, slip, or terminate early, producing truncated products of varying length, or to jump and re-anneal to a different position (template switching), producing recombinant/chimeric products; (4) heteroduplex formation late in the reaction — as reagents (primers, dNTPs, polymerase) become limiting after many cycles (24 cycles is well into the regime where amplification is slowing/plateauing), single strands from slightly different amplicons (or from incompletely extended earlier products) can anneal to each other rather than to fresh primer, forming heteroduplex or partial-length species that run as additional bands; and (5) if the "single strand" being amplified is itself part of a larger genomic or plasmid template with more than one region that the primer pair can bind (e.g. a repeated or paralogous sequence elsewhere), the primer pair will legitimately amplify more than one locus. In summary, multiple bands after 24 cycles are a diagnostic sign to check primer specificity and annealing stringency (e.g. by raising the annealing temperature, redesigning primers, or using a hot-start/higher-fidelity polymerase), not evidence that the reaction has somehow violated the basic chemistry of PCR.

(c) This is a direct exponential-amplification calculation.

Given. Initial template copy number $N_0 = 10$ (double-stranded DNA molecules); number of cycles $n = 10$.

Find. The number of DNA copies (and total single strands) expected after 10 PCR cycles.

Approach. Assume ideal, 100% amplification efficiency — i.e. every template strand present at the start of a cycle is fully copied during that cycle, so the number of target molecules exactly doubles each cycle. This is the standard simplifying assumption for a "textbook" PCR calculation at low-to-moderate cycle number (well before the reaction would begin to plateau from reagent depletion, which typically only becomes significant after ~25–35 cycles) and is explicitly stated as an assumption per the exam's instruction to record assumptions made.

  1. Exponential amplification formula. Under the 100%-efficiency assumption, the number of double-stranded copies after $n$ cycles starting from $N_0$ template copies is $$ N(n) = N_0 \cdot 2^{\,n} $$
  2. Substitute $N_0=10$, $n=10$. $$ N(10) = 10 \cdot 2^{10} = 10 \cdot 1024 = \boxed{10{,}240\ \text{double-stranded copies}} $$
  3. Convert to total single strands (if the question is read as asking for strand count rather than duplex-copy count). Each double-stranded copy is composed of two complementary single strands, so $$ \text{strands} = 2 \times N(10) = 2 \times 10{,}240 = \boxed{20{,}480\ \text{single strands}} $$
Question 5(c) — final results
QuantityValue
Double-stranded DNA copies after 10 cycles10,240
Total single DNA strands after 10 cycles20,480
Amplification factor$2^{10} = 1024\times$
Check — assumption

The calculation assumes 100% per-cycle amplification efficiency (every template molecule doubles every cycle) with no reagent-limited plateau, which is realistic at only 10 cycles but would systematically overestimate yield if extended to the higher cycle numbers used in part (b) (24 cycles), where efficiency typically falls below 100% as primers/dNTPs/polymerase become limiting.