A qPCR standard curve of Ct against log₁₀(starting copies) has a slope of −3.45. Using efficiency = 10^(−1/slope) − 1, the amplification efficiency is closest to

95%, slightly less than perfect doubling

100%, perfect doubling in every single cycle

110%, slightly more than doubling

87%, well short of perfect doubling

95%, slightly less than perfect doubling

In a TaqMan (hydrolysis-probe) qPCR, why does reporter fluorescence increase as the target is amplified?

The 5′→3′ exonuclease of Taq cleaves the annealed probe, separating the reporter from the quencher

The reporter dye fluoresces only after it intercalates into the double-stranded DNA formed each cycle

The 3′→5′ proofreading exonuclease of Taq removes the quencher from the probe's mismatched 3′ end

Taq extends the probe from its 3′ end, which carries the quencher too far from the reporter dye

In whole-plasmid site-directed mutagenesis, the PCR product is treated with DpnI before transformation into *E. coli*. What is the purpose of this step?

To digest the Dam-methylated parental plasmid while sparing the unmethylated, newly made mutant DNA

To digest the unmethylated, newly made DNA while sparing the Dam-methylated parental plasmid

To seal the nicks left in the newly made mutant strands so that the circular plasmid can be transformed

To digest the Dam-methylated parental plasmid while sparing the unmethylated, newly made mutant DNA

To trim the single 3′ A overhangs left by the polymerase so that the plasmid ends become blunt

Which of the following statements about PCR and site-directed mutagenesis are correct?

Taq polymerase survives the repeated denaturation steps but lacks 3′→5′ proofreading activity

A mutagenic primer carries its mismatch near the middle, with matched sequence on both sides

Amplifying an mRNA target by PCR requires reverse transcription into cDNA first

Taq polymerase survives the repeated denaturation steps but lacks 3′→5′ proofreading activity

A mutagenic primer carries its mismatch near the middle, with matched sequence on both sides

SYBR Green fluorescence by itself distinguishes the intended product from primer dimers, with no melt curve

Amplifying an mRNA target by PCR requires reverse transcription into cDNA first

Pyrosequencing and ion-semiconductor sequencing both make more errors than reversible-terminator sequencing in stretches such as AAAAAAA. What is the reason?

A whole run of identical bases is added in one nucleotide flow, so its length must be judged from the signal size

Their polymerases stall on runs of identical bases and leave part of each run unread until the next cycle

A whole run of identical bases is added in one nucleotide flow, so its length must be judged from the signal size

Light and pH signals are generated only when a mismatched base, rather than a matched one, is incorporated opposite the template

Reversible terminators bind runs of adenine more tightly, so those runs give a stronger signal per base

Digestion of an IgG molecule with papain gives

two monovalent Fab fragments and one Fc fragment

one bivalent F(ab′)₂ fragment and small Fc peptides

two heavy and two light chains freed from each other

two Fc fragments and one bivalent binding fragment

On an Ouchterlony plate, antiserum is placed in the central well and antigens X and Y in adjacent outer wells. The precipitin lines formed by X and Y cross each other completely without fusing. This shows that

X and Y share no epitope that the antiserum recognises

X and Y are the same molecule carrying identical epitopes

X carries all the epitopes of Y plus at least one more

X and Y share no epitope that the antiserum recognises

X and Y are at exactly the concentration of the antibody

A laboratory must measure a small drug molecule (hapten-sized, with a single epitope) in serum using one monoclonal antibody. Which assay format is appropriate, and how does its signal behave?

Competitive ELISA, in which the signal falls as the drug concentration rises

Sandwich ELISA, in which the signal rises as the drug concentration rises

Competitive ELISA, in which the signal falls as the drug concentration rises

Indirect ELISA, in which the signal reports the patient's antibodies to the drug

Competitive ELISA, in which the signal rises as the drug concentration rises

In a radioimmunoassay for insulin, a fixed amount of ¹²⁵I-insulin and a limited amount of anti-insulin antibody are incubated with standards or patient serum. As the insulin concentration in the sample increases, the radioactivity in the antibody-bound fraction

decreases, because unlabelled insulin takes the limited sites from labelled insulin

increases, because more insulin in total is now bound to the antibody

stays constant, because the amount of antibody is the same in every tube

increases, because sample insulin cross-links the antibodies into a larger, more heavily labelled lattice

decreases, because unlabelled insulin takes the limited sites from labelled insulin

During hybridoma production, why do unfused myeloma cells die in HAT medium?

Aminopterin blocks de novo nucleotide synthesis, and lacking HGPRT they cannot salvage purines

They are not immortal, so they stop dividing and die within a few days of being placed in culture

Hypoxanthine is toxic to every cell that has lost the ability to secrete its own antibody

Thymidine in the medium inhibits the HGPRT enzyme on which the myeloma cells depend for growth

Serum immunoelectrophoresis of a patient with bone pain shows a dense, locally bowed arc in the γ region when developed with anti-IgG and with anti-κ antisera, while the arc with anti-λ antiserum is faint and normal in shape. Which interpretation fits best?

A monoclonal IgG-κ immunoglobulin made by one expanded plasma-cell clone, as in multiple myeloma

A polyclonal rise in IgG, as in chronic infection, in which κ- and λ-bearing antibodies rise together

A monoclonal IgG-λ immunoglobulin, because the λ arc is the one that has kept its normal shape

A monoclonal IgG-κ immunoglobulin made by one expanded plasma-cell clone, as in multiple myeloma

A deficiency of IgG, as in an inherited antibody deficiency, which leaves only the λ arc visible

Which of the following statements about flow cytometry are correct?

Forward scatter mainly reflects cell size, and side scatter internal complexity or granularity

A fluorescence-activated cell sorter can collect a chosen population of cells alive

Cells must be lysed before analysis so that their surface antigens can reach the laser beam

Forward scatter mainly reflects cell size, and side scatter internal complexity or granularity

Each laser can excite only one fluorochrome, so every additional colour needs its own laser

A fluorescence-activated cell sorter can collect a chosen population of cells alive

Which of the following statements about monoclonal antibodies are correct?

All molecules of one monoclonal antibody have the same paratope and bind the same epitope

Humanised antibodies reduce the human anti-mouse antibody response provoked by mouse antibodies

All molecules of one monoclonal antibody have the same paratope and bind the same epitope

One monoclonal antibody readily forms a precipitin lattice with a monomeric protein carrying one copy of its epitope

Humanised antibodies reduce the human anti-mouse antibody response provoked by mouse antibodies

In a hybridoma, HGPRT is supplied by the myeloma partner and immortality by the spleen B cell

PCR, Sequencing and Mutagenesis; Antibody–Antigen Interaction, Immunoassays and Monoclonal Antibodies

This chapter completes Section XL1.4 of XL1 Biochemistry, “Analytical Techniques”, with the second half of the heading: “next-generation DNA sequencing, PCR, and site-directed mutagenesis. Antibody-antigen interaction; Immunological techniques: immunodiffusion, immune-electrophoresis, RIA and ELISA, flow cytometry; monoclonal antibodies and their applications.” It teaches the PCR cycle and its arithmetic, real-time PCR, site-directed mutagenesis, how short-read and long-read sequencers work, antibody structure and binding, the gel, radio- and enzyme-immunoassays, flow cytometry, and hybridoma technology with the clinical uses of monoclonal antibodies. The numericals are PCR copy numbers and cycles, Ct and ΔΔCt fold changes, standard-curve quantitation, sequencing coverage and Phred scores.

1. PCR: components, cycle and exponential amplification

The polymerase chain reaction copies the stretch of DNA between two primers. A reaction holds the template, two primers (one for each strand, pointing towards each other), the four dNTPs, Mg²⁺ (the polymerase's cofactor) and a thermostable DNA polymerase, classically Taq from Thermus aquaticus, which survives repeated heating. Each cycle has three steps: denaturation at about 94–95 °C separates the strands; annealing, typically 50–65 °C and a few degrees below the primers' Tm, lets the primers pair with their targets; extension at 72 °C adds nucleotides to the primers' 3′ ends.

Arithmetic. Ideally each cycle doubles the target, so N = N₀ × 2ⁿ; with efficiency E (0 to 1), N = N₀(1 + E)ⁿ. From 100 copies, 25 perfect cycles give 100 × 2²⁵ ≈ 3.4 × 10⁹ copies, but at E = 0.9 only about 9.3 × 10⁸. Products of exactly primer-to-primer length first appear in cycle 3 and soon dominate. Later the reaction slows to a plateau as primers and dNTPs run down and the enzyme loses activity, so the end-point amount of product is a poor measure of the starting amount.

Primer Tm for short oligonucleotides is often estimated by the Wallace rule, Tm = 2(A + T) + 4(G + C) °C; the two primers are matched in Tm and checked so they cannot pair with each other (primer dimers).
Fidelity. Taq lacks 3′→5′ proofreading exonuclease and leaves a single 3′ A overhang (used in TA cloning); proofreading enzymes such as Pfu from Pyrococcus furiosus are preferred when the product will be cloned or expressed.
RT-PCR first copies RNA into cDNA with reverse transcriptase, so mRNAs and RNA viruses can be amplified. Because PCR detects a few molecules, a no-template control is always run to reveal contamination.

2. Real-time quantitative PCR (qPCR)

qPCR reads fluorescence in every cycle. SYBR Green fluoresces when bound to any double-stranded DNA, so it is cheap but also reports primer dimers; a melt curve after the run checks that a single product formed. A TaqMan (hydrolysis) probe carries a reporter dye at its 5′ end and a quencher at its 3′ end, which absorbs the reporter's energy while the probe is intact. The probe anneals between the primers, and the 5′→3′ exonuclease of Taq cleaves it during extension, freeing the reporter, so fluorescence rises only for the intended target.

The threshold cycle (Ct) is the cycle at which fluorescence crosses a threshold set in the exponential phase; it falls as the starting amount rises. At 100% efficiency a difference of 1 Ct is two-fold and 3.32 Ct is ten-fold. A standard curve of Ct against log₁₀N₀ has slope −3.32 at 100% efficiency, and efficiency = 10^(−1/slope) − 1. In the ΔΔCt method, ΔCt = Ct(target) − Ct(reference gene) in each sample, ΔΔCt = ΔCt(treated) − ΔCt(control), and fold change = 2^(−ΔΔCt), valid when both genes amplify at close to 100% efficiency.

🎯 Worked example

Target Ct 22.0 and reference (housekeeping gene) Ct 16.0 in treated cells; 25.0 and 16.5 in control cells. ΔCt = 6.0 and 8.5, ΔΔCt = −2.5, fold change = 2^2.5 ≈ 5.7-fold induction. Normalising to the reference corrects for unequal amounts of input RNA. Clinically, RT-qPCR measures viral load, and a lower Ct means more viral RNA.

3. Site-directed mutagenesis

Site-directed mutagenesis changes a chosen base, codon or short stretch of a cloned gene, so the role of a single residue can be tested. The change is carried by a synthetic mutagenic primer with the mismatch near its middle and matched sequence on each side, so that it still anneals. In the common whole-plasmid method, two complementary mutagenic primers and a high-fidelity polymerase copy the entire plasmid; the product is then treated with DpnI, which cuts only methylated GATC sites, so the Dam-methylated parental plasmid isolated from E. coli is destroyed while the unmethylated, newly made mutant DNA survives and is transformed into E. coli. Overlap-extension PCR places a mutation inside a fragment by joining two overlapping PCR products. Every clone is confirmed by sequencing.

Uses. Replacing a catalytic residue (for example the active-site serine of a serine protease with alanine) tests a proposed mechanism; alanine scanning maps the residues needed for binding; mutations can add or remove a glycosylation or phosphorylation site, raise stability, or introduce a cysteine for labelling.

⚠️ Loss of activity is not proof

A mutant may be inactive because it misfolds, not because the residue is catalytic. Show that the mutant is folded (a similar CD spectrum, normal gel-filtration behaviour or retained substrate binding) before concluding, and prefer conservative substitutions such as Asp to Asn or Ser to Ala, which change one property at a time.

4. Next-generation sequencing

Sanger sequencing, the reference method, reads one template per reaction: a polymerase extends a primer in the presence of fluorescent dideoxynucleotides, which lack a 3′-OH and terminate the chain, and capillary electrophoresis orders the fragments, giving reads of about 800–1,000 bases. Next-generation sequencing (NGS) reads millions of fragments in parallel. A library is made by fragmenting DNA and ligating adapters, with index barcodes so samples can be pooled; fragments are clonally amplified, by bridge amplification into clusters on a flow cell or by emulsion PCR on beads, and then read cycle by cycle.

Chemistry	Signal	Reads	Typical weakness
Reversible-terminator sequencing by synthesis (Illumina)	fluorescent nucleotides with removable blocks: one base added, imaged and unblocked per cycle	short, about 50–300 bases, often paired-end	low error, mostly substitutions; repeats are hard to place
Pyrosequencing	released pyrophosphate is turned into ATP by ATP sulfurylase, and ATP drives luciferase light	short to moderate	homopolymer length judged from light intensity
Ion semiconductor	H⁺ released on incorporation changes the pH in a microwell	short to moderate	homopolymer errors, as above
Single-molecule real-time (PacBio)	one polymerase in a nanoscale well; each fluorescent nucleotide is seen as it is incorporated	long, often above 10 kb	higher raw error, reduced by reading one molecule many times
Nanopore	ionic current changes as a DNA strand passes through a protein pore	very long, tens to hundreds of kb	higher raw error; needs no amplification

Coverage and quality. Mean depth of coverage C = N × L/G, where N is the number of reads, L the read length and G the genome size: 400 million reads of 150 bases on a 3.0 × 10⁹ bp genome give 20×. Each base call carries a Phred quality Q = −10 log₁₀P, where P is the probability that the call is wrong, so Q20 is one error in 100 and Q30 one in 1,000. Paired-end reads help place fragments across repeats. Applications include whole-genome and exome sequencing to diagnose inherited disease, RNA-seq for expression, ChIP-seq for binding sites, metagenomics, and screening cell-free fetal DNA in maternal blood.

5. Antibody–antigen interaction

An IgG molecule (about 150 kDa) has two identical heavy and two identical light chains joined by disulphide bonds into a Y. Each arm is a Fab region carrying one antigen-binding site, formed by the variable domains of one heavy and one light chain, whose hypervariable loops (CDRs) make the paratope; the stem is the Fc region, which engages Fc receptors and complement. Papain cuts above the hinge, giving two Fab and one Fc; pepsin cuts below it, giving one bivalent F(ab′)₂. The part of the antigen bound is the epitope, either linear (a continuous stretch of sequence) or conformational (residues brought together by folding, lost on denaturation).

Binding is non-covalent and reversible: hydrogen bonds, ionic attractions, van der Waals contacts and hydrophobic effects across closely fitting surfaces. Affinity is the strength of one paratope–epitope bond, Ka = [AbAg]/([Ab][Ag]); avidity is the overall strength of multivalent binding, which is why pentameric IgM binds cell surfaces well despite low affinity per site. Cross-reactivity arises when another antigen carries a similar epitope.

Precipitation and agglutination. A bivalent antibody and a multivalent antigen build a cross-linked lattice. In a precipitin curve the precipitate is greatest at equivalence; in antibody excess (prozone) or antigen excess (postzone) small soluble complexes form instead, so a very concentrated sample can test falsely weak. Cross-linking of particles such as red cells or bacteria is agglutination, the basis of blood grouping.

6. Immunodiffusion, immunoelectrophoresis, RIA, ELISA and flow cytometry

Technique	Principle	Readout and use
Single radial immunodiffusion (Mancini)	antigen diffuses from a well into gel containing antibody	a ring of precipitate whose diameter grows with antigen concentration, read against standards (serum IgG, complement)
Double immunodiffusion (Ouchterlony)	antigen and antibody diffuse towards each other from separate wells	lines that fuse (identity), fuse with a spur (partial identity) or cross (non-identity)
Immunoelectrophoresis	proteins are first separated by electrophoresis in gel; antiserum placed in a trough then diffuses towards them	precipitin arcs; abnormal serum proteins such as a monoclonal immunoglobulin (M-protein) in myeloma
Rocket immunoelectrophoresis	an electric field drives antigen into antibody-containing gel	a rocket-shaped peak whose height is proportional to antigen concentration
Counter-current immunoelectrophoresis	antigen and antibody are driven towards each other by the field	a precipitin line much faster than by diffusion alone

Radioimmunoassay (RIA) is a competition: a limited amount of antibody, a fixed amount of radiolabelled antigen (often ¹²⁵I) and the sample's unlabelled antigen compete for the binding sites. After bound and free antigen are separated, bound radioactivity falls as sample antigen rises, and concentrations are read from a standard curve. Developed for insulin by Rosalyn Yalow and Solomon Berson, it made hormones measurable at very low concentrations; its radiation hazard has led to wide replacement by enzyme and chemiluminescent assays.

ELISA replaces the radiolabel with an enzyme, usually horseradish peroxidase (with TMB substrate, read at 450 nm after an acid stop) or alkaline phosphatase (with p-nitrophenyl phosphate, read at 405 nm). Direct ELISA uses a labelled primary antibody; indirect ELISA detects antibodies in serum (antigen on the plate, patient antibody, then enzyme-labelled anti-human immunoglobulin), as in screening for infections; sandwich ELISA captures an antigen between two antibodies to different epitopes, so the signal rises with antigen; competitive ELISA suits small antigens with a single epitope, and its signal falls as antigen rises. Blocking protein prevents non-specific sticking and washes remove unbound reagent.

Flow cytometry passes cells in single file, by hydrodynamic focusing, through laser beams. Forward scatter reports mainly cell size and side scatter internal complexity or granularity; fluorochrome-labelled antibodies to surface markers (CD antigens) add channels, and gating selects populations, for example lymphocytes by scatter and then CD4⁺ cells. Uses include CD4 counts in HIV infection, immunophenotyping of leukaemias and lymphomas, and cell-cycle analysis with DNA dyes. A fluorescence-activated cell sorter (FACS) also charges and deflects droplets to collect chosen cells alive.

7. Monoclonal antibodies: hybridoma technology and applications

Serum after immunisation is polyclonal: antibodies from many B-cell clones against many epitopes, varying from batch to batch. A monoclonal antibody comes from one clone and recognises one epitope with one affinity. Georges Köhler and César Milstein produced them in 1975 by hybridoma technology (Nobel Prize, 1984). A mouse is immunised; its spleen B cells, which make antibody but die in culture, are fused (with polyethylene glycol) to myeloma cells, which are immortal but lack HGPRT and secrete no antibody of their own; the mixture is grown in HAT medium (hypoxanthine, aminopterin, thymidine).

Why HAT selects hybrids. Aminopterin inhibits dihydrofolate reductase and blocks de novo synthesis of purines and thymidylate, so cells must use the salvage pathway, which needs HGPRT to use hypoxanthine. Unfused myeloma cells (HGPRT⁻) die; unfused B cells die within days because they are not immortal; only hybrids, with HGPRT from the B cell and immortality from the myeloma, grow. Supernatants are screened, usually by ELISA, and positive wells are cloned by limiting dilution so that each culture descends from one cell.

Engineering and applications. Mouse antibodies given repeatedly provoke a human anti-mouse antibody (HAMA) response, so therapeutic antibodies are made chimeric (mouse variable, human constant regions; rituximab, anti-CD20), humanised (only mouse CDRs kept; trastuzumab, anti-HER2) or fully human (from phage display or transgenic mice; adalimumab, anti-TNF-α). Uses: diagnostics (home pregnancy tests detect hCG in a lateral-flow sandwich; ELISA kits; blood-group typing; CD markers in flow cytometry), immunoaffinity purification, imaging, and therapy, by blocking a ligand or receptor, recruiting complement or antibody-dependent cell-mediated cytotoxicity (ADCC), or delivering a drug as an antibody–drug conjugate.

Key takeaways

PCR cycles through denaturation (about 95 °C), annealing (50–65 °C) and extension (72 °C) with a thermostable polymerase; N = N₀(1 + E)ⁿ until the plateau. Taq has no proofreading; RT-PCR starts from RNA; a no-template control detects contamination.
In qPCR a lower Ct means more template: 1 cycle is two-fold and 3.32 cycles ten-fold at 100% efficiency; efficiency = 10^(−1/slope) − 1, and fold change = 2^(−ΔΔCt). SYBR Green binds any dsDNA; a TaqMan probe is cleaved by Taq's 5′→3′ exonuclease.
Site-directed mutagenesis uses mutagenic primers; DpnI removes the methylated parental plasmid; clones are sequenced, and a mutant's folding must be checked before a residue is called catalytic.
NGS builds an adapter library, amplifies clonally and reads in parallel: reversible terminators give short accurate reads, pyrosequencing and ion chemistry err in homopolymers, PacBio and nanopore give long reads. Coverage C = NL/G; Phred Q = −10 log₁₀P.
IgG has two Fab arms and an Fc stem (papain gives 2 Fab + Fc, pepsin F(ab′)₂); binding is non-covalent, with affinity per site and avidity overall. Lattices precipitate best at equivalence; immunodiffusion and immunoelectrophoresis read precipitin rings, lines, arcs and rockets.
In RIA and competitive ELISA the signal falls as antigen rises, in sandwich ELISA it rises; flow cytometry reads size (forward scatter), granularity (side scatter) and markers. Hybridomas survive HAT with HGPRT from the B cell and immortality from the myeloma; chimeric, humanised and human antibodies progressively reduce the HAMA response.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A PCR starts with 500 copies of a target sequence. Assuming perfect doubling in every cycle, the minimum number of cycles needed to obtain at least 1.0 × 10⁹ copies is ____.
Numerical answer — type the value.
Show answer
Answer: 21
500 × 2ⁿ ≥ 10⁹ requires 2ⁿ ≥ 2 × 10⁶, so n ≥ log₂(2 × 10⁶) = 20.93, and the smallest whole number is 21. Check: 20 cycles give 500 × 1,048,576 = 5.2 × 10⁸ (too few), 21 cycles give 1.05 × 10⁹. Rounding 20.93 down to 20 is the usual slip.
A qPCR standard curve of Ct against log₁₀(starting copies) has a slope of −3.45. Using efficiency = 10^(−1/slope) − 1, the amplification efficiency is closest to
1. 100%, perfect doubling in every single cycle
2. 110%, slightly more than doubling
3. 87%, well short of perfect doubling
4. 95%, slightly less than perfect doubling
Show answer
Answer: D — 95%, slightly less than perfect doubling
10^(1/3.45) = 10^0.290 = 1.949, so efficiency = 0.949, about 95%. Perfect doubling needs a slope of −3.32, since 10^(1/3.32) = 2.00; a steeper slope means more cycles per ten-fold step and so less than doubling. 110% would need a shallower slope near −3.1, and 87% a steeper one near −3.68.
In an RT-qPCR experiment, a target gene gives Ct = 24.6 and the reference gene Ct = 18.2 in drug-treated cells; in control cells the target gives Ct = 27.1 and the reference Ct = 18.7. Assuming 100% efficiency for both genes, the fold change in target expression in treated relative to control cells (by the 2^−ΔΔCt method) is ____.
Numerical answer — type the value.
Show answer
Answer: 4
ΔCt(treated) = 24.6 − 18.2 = 6.4; ΔCt(control) = 27.1 − 18.7 = 8.4; ΔΔCt = 6.4 − 8.4 = −2.0; fold change = 2^2.0 = 4, a four-fold induction. Comparing the raw target Ct values alone (27.1 − 24.6 = 2.5, giving 5.7-fold) ignores the different amounts of input RNA that the reference gene corrects for.
An RT-qPCR assay for a virus has the standard curve Ct = 38.0 − 3.32 log₁₀N₀, where N₀ is the number of target copies in the reaction. A patient sample gives Ct = 26.38. The number of target copies in the reaction (to the nearest whole number) is ____.
Numerical answer — type the value.
Show answer
Answer: 3162
log₁₀N₀ = (38.0 − 26.38)/3.32 = 11.62/3.32 = 3.50, so N₀ = 10^3.5 = 3,162 copies. The slope of −3.32 corresponds to 100% efficiency, so each 3.32 cycles below the intercept is a ten-fold step. Forgetting the intercept and using 26.38/3.32 gives an absurd 10^7.9.
In a TaqMan (hydrolysis-probe) qPCR, why does reporter fluorescence increase as the target is amplified?
1. The 5′→3′ exonuclease of Taq cleaves the annealed probe, separating the reporter from the quencher
2. The reporter dye fluoresces only after it intercalates into the double-stranded DNA formed each cycle
3. The 3′→5′ proofreading exonuclease of Taq removes the quencher from the probe's mismatched 3′ end
4. Taq extends the probe from its 3′ end, which carries the quencher too far from the reporter dye
Show answer
Answer: A — The 5′→3′ exonuclease of Taq cleaves the annealed probe, separating the reporter from the quencher
While the probe is intact the quencher, a few nanometres away, absorbs the reporter's energy. During extension Taq meets the annealed probe and its 5′→3′ exonuclease degrades it, releasing the reporter, so each new copy frees more reporter. Intercalation describes SYBR Green, not a probe. Taq has no 3′→5′ proofreading activity, and the probe's 3′ end is blocked so it cannot be extended.
In whole-plasmid site-directed mutagenesis, the PCR product is treated with DpnI before transformation into E. coli. What is the purpose of this step?
1. To digest the unmethylated, newly made DNA while sparing the Dam-methylated parental plasmid
2. To seal the nicks left in the newly made mutant strands so that the circular plasmid can be transformed
3. To digest the Dam-methylated parental plasmid while sparing the unmethylated, newly made mutant DNA
4. To trim the single 3′ A overhangs left by the polymerase so that the plasmid ends become blunt
Show answer
Answer: C — To digest the Dam-methylated parental plasmid while sparing the unmethylated, newly made mutant DNA
DpnI cuts GATC only when the adenine is methylated. Template plasmid purified from a dam⁺ E. coli strain is methylated, whereas DNA made in vitro by PCR is not, so DpnI removes the wild-type template and enriches the mutant. The reversed statement is the tempting error. Nicks are repaired by the host cell after transformation, not by DpnI, and a restriction enzyme does not trim A overhangs.
Which of the following statements about PCR and site-directed mutagenesis are correct?
1. Taq polymerase survives the repeated denaturation steps but lacks 3′→5′ proofreading activity
2. A mutagenic primer carries its mismatch near the middle, with matched sequence on both sides
3. SYBR Green fluorescence by itself distinguishes the intended product from primer dimers, with no melt curve
4. Amplifying an mRNA target by PCR requires reverse transcription into cDNA first
Show answer
Answer: A — Taq polymerase survives the repeated denaturation steps but lacks 3′→5′ proofreading activity; B — A mutagenic primer carries its mismatch near the middle, with matched sequence on both sides; D — Amplifying an mRNA target by PCR requires reverse transcription into cDNA first
Taq is thermostable but has no proofreading exonuclease; a central mismatch with matched flanks lets the mutagenic primer anneal and prime; and PCR polymerases copy DNA, so RNA must first be reverse-transcribed. SYBR Green binds any double-stranded DNA, so it cannot tell primer dimers from the product; a melt curve or a gel is needed.
Pyrosequencing and ion-semiconductor sequencing both make more errors than reversible-terminator sequencing in stretches such as AAAAAAA. What is the reason?
1. Their polymerases stall on runs of identical bases and leave part of each run unread until the next cycle
2. A whole run of identical bases is added in one nucleotide flow, so its length must be judged from the signal size
3. Light and pH signals are generated only when a mismatched base, rather than a matched one, is incorporated opposite the template
4. Reversible terminators bind runs of adenine more tightly, so those runs give a stronger signal per base
Show answer
Answer: B — A whole run of identical bases is added in one nucleotide flow, so its length must be judged from the signal size
These flow-based chemistries supply one kind of unblocked nucleotide at a time, so the polymerase fills an entire homopolymer in that flow; the light (pyrophosphate to ATP to luciferase) or H⁺ signal is proportional to the number added, and telling 7 from 8 is hard. Reversible terminators allow exactly one base per cycle, so a run is counted base by base. Signals come from correct incorporation, not mismatches, and stalling is not the cause.
A 4.6 Mb bacterial genome is to be sequenced to a mean coverage of 30× using 2 × 150 bp paired-end reads, so each read pair contributes 300 bases. Ignoring duplicate and unmapped reads, the number of read pairs required is ____.
Numerical answer — type the value.
Show answer
Answer: 460000
Coverage C = N × L/G, so N = C × G/L = 30 × 4.6 × 10⁶ / 300 = 460,000 read pairs. Counting 150 bases per pair instead of 300 doubles the answer to 920,000.
A sequencer estimates that a particular base call has a probability of 1 in 2,000 of being wrong. Using Q = −10 log₁₀P, its Phred quality score (to one decimal place) is ____.
Numerical answer — type the value.
Show answer
Answer: 33.0
P = 1/2,000 = 5 × 10⁻⁴, so Q = −10 log₁₀(5 × 10⁻⁴) = 10 × 3.301 = 33.0. As checks, Q30 is 1 error in 1,000 and Q40 one in 10,000, so a value between them is expected.
Digestion of an IgG molecule with papain gives
1. two monovalent Fab fragments and one Fc fragment
2. one bivalent F(ab′)₂ fragment and small Fc peptides
3. two heavy and two light chains freed from each other
4. two Fc fragments and one bivalent binding fragment
Show answer
Answer: A — two monovalent Fab fragments and one Fc fragment
Papain cleaves IgG on the N-terminal side of the hinge disulphides, releasing two separate Fab arms, each with one antigen-binding site, and the Fc stem. Pepsin cuts on the other side of the hinge, so the two Fab arms stay joined as F(ab′)₂ while Fc is degraded; that is the tempting distractor. Separating heavy from light chains needs reduction of disulphides, not a protease, and an IgG has only one Fc.
On an Ouchterlony plate, antiserum is placed in the central well and antigens X and Y in adjacent outer wells. The precipitin lines formed by X and Y cross each other completely without fusing. This shows that
1. X and Y are the same molecule carrying identical epitopes
2. X carries all the epitopes of Y plus at least one more
3. X and Y share no epitope that the antiserum recognises
4. X and Y are at exactly the concentration of the antibody
Show answer
Answer: C — X and Y share no epitope that the antiserum recognises
Each antigen forms its own line with the antibodies specific to it; if the two antigens share no epitope, neither line blocks the other's antibodies and the lines cross (non-identity). Identical antigens give lines that fuse smoothly into one arc (identity), and an antigen carrying extra epitopes gives fused lines with a spur (partial identity). The pattern reports epitope sharing, not concentration.
A laboratory must measure a small drug molecule (hapten-sized, with a single epitope) in serum using one monoclonal antibody. Which assay format is appropriate, and how does its signal behave?
1. Sandwich ELISA, in which the signal rises as the drug concentration rises
2. Competitive ELISA, in which the signal falls as the drug concentration rises
3. Indirect ELISA, in which the signal reports the patient's antibodies to the drug
4. Competitive ELISA, in which the signal rises as the drug concentration rises
Show answer
Answer: B — Competitive ELISA, in which the signal falls as the drug concentration rises
A sandwich needs two antibodies bound at the same time to two separate epitopes, which a molecule with one small epitope cannot provide. In a competitive ELISA the sample drug competes with a fixed amount of labelled (or plate-bound) drug for the limited antibody, so more drug in the sample means less label captured and a lower signal. Indirect ELISA measures antibodies, not the drug itself.
In a radioimmunoassay for insulin, a fixed amount of ¹²⁵I-insulin and a limited amount of anti-insulin antibody are incubated with standards or patient serum. As the insulin concentration in the sample increases, the radioactivity in the antibody-bound fraction
1. increases, because more insulin in total is now bound to the antibody
2. stays constant, because the amount of antibody is the same in every tube
3. increases, because sample insulin cross-links the antibodies into a larger, more heavily labelled lattice
4. decreases, because unlabelled insulin takes the limited sites from labelled insulin
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Answer: D — decreases, because unlabelled insulin takes the limited sites from labelled insulin
The sites are limited and fixed, so labelled and unlabelled insulin share them in proportion to their amounts. More sample insulin means a smaller share for the tracer, so bound counts fall and free counts rise; the standard curve is read in that inverse direction. Total bound insulin may stay near the binding capacity, but only the labelled part is counted.
During hybridoma production, why do unfused myeloma cells die in HAT medium?
1. Aminopterin blocks de novo nucleotide synthesis, and lacking HGPRT they cannot salvage purines
2. They are not immortal, so they stop dividing and die within a few days of being placed in culture
3. Hypoxanthine is toxic to every cell that has lost the ability to secrete its own antibody
4. Thymidine in the medium inhibits the HGPRT enzyme on which the myeloma cells depend for growth
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Answer: A — Aminopterin blocks de novo nucleotide synthesis, and lacking HGPRT they cannot salvage purines
Aminopterin, a folate antagonist, inhibits dihydrofolate reductase and so de novo purine and thymidylate synthesis; survival then needs the salvage enzymes, including HGPRT for hypoxanthine. The myeloma partner is chosen to be HGPRT-deficient, so alone it dies. Lack of immortality is why unfused B cells die, the tempting distractor. Hypoxanthine and thymidine are salvage substrates, not toxins or inhibitors.
Serum immunoelectrophoresis of a patient with bone pain shows a dense, locally bowed arc in the γ region when developed with anti-IgG and with anti-κ antisera, while the arc with anti-λ antiserum is faint and normal in shape. Which interpretation fits best?
1. A polyclonal rise in IgG, as in chronic infection, in which κ- and λ-bearing antibodies rise together
2. A monoclonal IgG-λ immunoglobulin, because the λ arc is the one that has kept its normal shape
3. A monoclonal IgG-κ immunoglobulin made by one expanded plasma-cell clone, as in multiple myeloma
4. A deficiency of IgG, as in an inherited antibody deficiency, which leaves only the λ arc visible
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Answer: C — A monoclonal IgG-κ immunoglobulin made by one expanded plasma-cell clone, as in multiple myeloma
A single plasma-cell clone secretes one heavy-chain class with one light-chain type, so its excess protein (the M-protein) distorts only the arcs formed with anti-IgG and anti-κ, and its restricted mobility makes the arc bow locally. A polyclonal rise involves both κ and λ and thickens the arcs evenly. The λ arc is normal, so the abnormal protein is not λ, and a dense IgG arc rules out IgG deficiency.
Which of the following statements about flow cytometry are correct?
1. Cells must be lysed before analysis so that their surface antigens can reach the laser beam
2. Forward scatter mainly reflects cell size, and side scatter internal complexity or granularity
3. Each laser can excite only one fluorochrome, so every additional colour needs its own laser
4. A fluorescence-activated cell sorter can collect a chosen population of cells alive
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Answer: B — Forward scatter mainly reflects cell size, and side scatter internal complexity or granularity; D — A fluorescence-activated cell sorter can collect a chosen population of cells alive
Scatter of light at low angles tracks size and at right angles tracks internal structure, and FACS deflects charged droplets carrying chosen cells into collection tubes without killing them. Intact cells are analysed one at a time; lysis would destroy the very thing being measured. One laser routinely excites several fluorochromes that share an excitation band but emit at different wavelengths, read through separate filters.
Which of the following statements about monoclonal antibodies are correct?
1. All molecules of one monoclonal antibody have the same paratope and bind the same epitope
2. One monoclonal antibody readily forms a precipitin lattice with a monomeric protein carrying one copy of its epitope
3. Humanised antibodies reduce the human anti-mouse antibody response provoked by mouse antibodies
4. In a hybridoma, HGPRT is supplied by the myeloma partner and immortality by the spleen B cell
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Answer: A — All molecules of one monoclonal antibody have the same paratope and bind the same epitope; C — Humanised antibodies reduce the human anti-mouse antibody response provoked by mouse antibodies
A monoclonal antibody is one molecular species, hence one paratope and one epitope; and humanising keeps only the mouse CDRs, so far less foreign protein provokes HAMA. A lattice needs a multivalent antigen: a monomer with one copy of the epitope can bind only one antibody, so no cross-linked precipitate forms. In a hybridoma the roles are reversed from that statement: HGPRT comes from the B cell and immortality from the myeloma.