Plasma bicarbonate has a pKa of 6.1, yet it is the most important extracellular buffer at pH 7.4. Which feature best explains its effectiveness?

CO₂ is exhaled or retained by the lungs and HCO₃⁻ adjusted by the kidneys, so the pair's ratio is regulated

Carbonic anhydrase in red cells shifts the pKa of carbonic acid up to 7.4, so the pair buffers at its own pKa

Bicarbonate binds H⁺ irreversibly, so its buffering does not depend on how close the pH is to the pKa

CO₂ is exhaled or retained by the lungs and HCO₃⁻ adjusted by the kidneys, so the pair's ratio is regulated

Its plasma concentration is so low that even a little added acid restores the ratio of the pair to 20:1

A protein of pI 5.2 is to be captured by ion-exchange chromatography from a lysate buffered at pH 7.5. Which matrix will bind it, and why?

DEAE-cellulose, because above its pI the protein carries a net negative charge

CM-cellulose, because above its pI the protein carries a net positive charge

DEAE-cellulose, because above its pI the protein carries a net negative charge

CM-cellulose, because its carboxymethyl groups are positively charged at pH 7.5

DEAE-cellulose, because its diethylaminoethyl groups are negatively charged at pH 7.5

On a calibrated gel-filtration column in native buffer, a purified enzyme elutes at the same volume as a 150 kDa globular standard. Reducing SDS-PAGE of the enzyme shows a single band at 50 kDa. Assuming the enzyme is globular, what is the most reasonable conclusion?

It is a trimer of polypeptide chains of about 50 kDa each

It is a single 150 kDa chain that SDS hydrolyses into three pieces

It is a dimer of 75 kDa chains that run faster in the SDS gel

It is a 50 kDa monomer that also binds two molecules of matrix

It is a trimer of polypeptide chains of about 50 kDa each

Two DNA samples have the same length and sequence, but one was made by cells grown on ¹⁵NH₄Cl and the other on ¹⁴NH₄Cl. Which method separates them?

Equilibrium (isopycnic) centrifugation in a CsCl gradient, which separates them by buoyant density

Rate-zonal centrifugation on a 5–20% sucrose gradient, which separates them by sedimentation coefficient

Agarose gel electrophoresis, which separates them by the charge carried on the phosphate backbone

Differential centrifugation at rising RCF, which pellets the heavier sample at the lower force

In SDS-PAGE, what is the main role of sodium dodecyl sulphate?

It unfolds the chains and coats them with negative charge in proportion to length, so mass sets their mobility

It reduces the disulphide bonds between cysteine residues, so that subunits held by S–S bridges separate on the gel

It cross-links the acrylamide chains into a mesh whose pore size is set by the detergent concentration

It unfolds the chains and coats them with negative charge in proportion to length, so mass sets their mobility

It holds every protein at its isoelectric point during the run, so that the proteins separate according to pI

Which of the following statements about electrophoretic methods are correct?

In isoelectric focusing, a protein stops migrating where the pH of the gradient equals its pI

In a standard 2-D gel, proteins are separated by pI in the first dimension and by mass in the second

Native PAGE separates proteins by polypeptide mass alone, because the gel pores sieve them

In isoelectric focusing, a protein stops migrating where the pH of the gradient equals its pI

Agarose gels resolve DNA strands differing by a single nucleotide better than polyacrylamide gels do

In a standard 2-D gel, proteins are separated by pI in the first dimension and by mass in the second

In an EMSA, a labelled 30-bp promoter probe gives a slower-migrating band when incubated with nuclear extract. Which further result shows that the binding is sequence-specific?

The shifted band is lost with a 100-fold excess of unlabelled wild-type probe but not with excess unlabelled probe whose site is mutated

The shifted band is lost with excess unlabelled probe whose site is mutated but not with a 100-fold excess of unlabelled wild-type probe

The shifted band is lost with a 100-fold excess of unlabelled wild-type probe but not with excess unlabelled probe whose site is mutated

The shifted band grows stronger as more nuclear extract is added, while the band of free probe grows weaker

The shifted band disappears when the nuclear extract is boiled for ten minutes before it is added to the probe

In a yeast two-hybrid screen, cells carrying the bait (protein X fused to the Gal4 DNA-binding domain) together with an empty activation-domain vector already grow on medium lacking histidine (HIS3 reporter). What does this mean for colonies obtained with library preys?

Their growth is not evidence of interaction, because the bait switches on the reporter by itself

Each growing colony carries a genuine partner of X, because the bait clearly reaches the nucleus

The HIS3 reporter is defective, so repeating the screen with a lacZ reporter will fix the problem

X must be a membrane protein, since membrane proteins give the strongest two-hybrid signals

Their growth is not evidence of interaction, because the bait switches on the reporter by itself

Which of the following statements about methods for studying macromolecular interactions are correct?

In ChIP, formaldehyde cross-linking in living cells captures protein–DNA contacts as they exist in vivo

In DNase I footprinting, the protein-binding site appears as a gap of missing or weakened bands in the cleavage ladder

In ChIP, formaldehyde cross-linking in living cells captures protein–DNA contacts as they exist in vivo

Co-immunoprecipitation of Y with an antibody to X proves that X and Y touch each other directly, with no protein bridging them

In DNase I footprinting, the protein-binding site appears as a gap of missing or weakened bands in the cleavage ladder

In a GST pull-down, the GST-tagged bait and its bound partners are captured on beads that carry Ni²⁺–NTA

When a ligand binds a protein that has a single tryptophan, the emission maximum (excitation at 295 nm) moves from 352 nm to 336 nm and the intensity rises. The most reasonable interpretation is that the tryptophan

has moved into a more nonpolar environment shielded from the solvent

has become more exposed to water, as happens when a protein unfolds

has moved into a more nonpolar environment shielded from the solvent

passes its excitation energy to the ligand by FRET, quenching itself

is now excited through tyrosine residues, which emit at shorter wavelengths

A plasmid DNA preparation gives A₂₆₀/A₂₈₀ = 1.5. What does this most likely indicate?

Contamination with protein or phenol, which absorb strongly near 280 nm

Contamination with RNA, which pulls the ratio below that of pure DNA

Partial melting of the DNA, which lowers its absorbance at 260 nm

A pure sample, since about 1.5 is the expected ratio for duplex DNA

Contamination with protein or phenol, which absorb strongly near 280 nm

Which of the following statements about absorbance and fluorescence measurements are correct?

A solution with an absorbance of 2.0 transmits 1% of the incident light

With one-photon excitation, fluorescence is emitted at a longer wavelength than the exciting light

Doubling the path length at constant concentration doubles the transmittance of the solution

A solution with an absorbance of 2.0 transmits 1% of the incident light

With one-photon excitation, fluorescence is emitted at a longer wavelength than the exciting light

NAD⁺ absorbs at 340 nm about as strongly as NADH, so the two cannot be told apart at that wavelength

Buffers, Chromatography, Centrifugation, Electrophoresis, Interaction Assays and Spectroscopy

This chapter answers Section XL1.4 of XL1 Biochemistry, one of the five optional sections of the Life Sciences (XL) paper. The heading, “Analytical Techniques”, reads: “Buffering against pH changes in biological systems, the Henderson-Hasselbalch equation; Ion exchange, size exclusion and affinity chromatography, Thin layer and paper chromatography, centrifugation; Characterization of biomolecules by electrophoresis; Analytical methods to study DNA-protein and protein–protein interactions; Lambert-Beer law, UV-visible, and fluorescence spectroscopy; 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.” This first chapter covers the physical methods up to spectroscopy: buffers, chromatography, centrifugation, electrophoresis, interaction assays and absorbance and fluorescence; the second chapter takes the DNA and immunological methods. The numericals are buffer pH and composition, blood acid–base arithmetic, Rf, relative centrifugal force, FRET efficiency and Beer–Lambert concentrations and enzyme activities.

1. Buffers and the Henderson–Hasselbalch equation

A buffer is a weak acid (HA) with its conjugate base (A⁻) that resists pH change: added H⁺ is taken up by A⁻ and added OH⁻ by HA. Taking logarithms of Ka = [H⁺][A⁻]/[HA] gives the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA], pH = pKa, and there buffer capacity is greatest; the useful range is about pKa ± 1, where the ratio lies between 1:10 and 10:1. The ratio sets the pH while the total concentration sets the capacity, so ideal dilution leaves pH unchanged but weakens the buffer.

Worked example (composition). For 0.10 M phosphate buffer at pH 7.20, taking pKa₂ (H₂PO₄⁻/HPO₄²⁻) = 6.86: log([HPO₄²⁻]/[H₂PO₄⁻]) = 0.34, so the ratio is 10^0.34 = 2.19. The base fraction is 2.19/3.19 = 0.686, giving 68.6 mM HPO₄²⁻ and 31.4 mM H₂PO₄⁻. The same logic fixes how much HCl to add to Tris base, since each mole of HCl converts one mole of Tris to TrisH⁺.

Buffers in the body. Inside cells, phosphate (pKa₂ near 6.9 at cellular ionic strength, 7.2 in dilute solution) and proteins dominate; histidine's imidazole group makes haemoglobin a major blood buffer. In plasma the main buffer is bicarbonate: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻, used clinically as pH = 6.1 + log([HCO₃⁻]/(0.03 × pCO₂)), with pCO₂ in mmHg and concentrations in mM. Normal values, 24 mM and 40 mmHg, give a ratio 24/1.2 = 20 and pH 7.40. A pKa 1.3 units from 7.4 should buffer poorly, but the system is open: the lungs set pCO₂ within minutes and the kidneys set HCO₃⁻ over hours to days, so the ratio itself is regulated.

Disorder	Primary change	Typical cause	Compensation
Metabolic acidosis	HCO₃⁻ falls	diabetic ketoacidosis, lactic acidosis, renal failure	hyperventilation lowers pCO₂
Metabolic alkalosis	HCO₃⁻ rises	prolonged vomiting (loss of HCl)	hypoventilation raises pCO₂
Respiratory acidosis	pCO₂ rises	hypoventilation: lung disease, depressed breathing	kidneys retain HCO₃⁻
Respiratory alkalosis	pCO₂ falls	hyperventilation: anxiety, high altitude	kidneys excrete HCO₃⁻

⚠️ Choosing a laboratory buffer

Pick a buffer whose pKa lies within one unit of the working pH. Tris (pKa about 8.1 at 25 °C) is temperature-sensitive: its pKa rises on cooling, so a Tris buffer set at room temperature is more alkaline in the cold room. Good's buffers such as MES, MOPS and HEPES were designed for biology: pKa near neutrality, little metal binding and poor membrane permeability.

2. Chromatography: ion exchange, size exclusion, affinity, paper and TLC

In every chromatography a molecule partitions between a stationary phase and a mobile phase; the more time it spends bound to or inside the stationary phase, the slower it travels. The methods differ in the property that decides the partition.

Method	Separates by	Typical matrix and elution
Ion exchange	net charge at the buffer pH	cation exchangers carry negative groups (carboxymethyl, CM; sulphopropyl, S); anion exchangers carry positive groups (diethylaminoethyl, DEAE; quaternary ammonium, Q); elute with a rising salt gradient or a pH shift
Size exclusion (gel filtration)	size and shape (hydrodynamic radius)	porous beads of cross-linked dextran, agarose or polyacrylamide; one buffer throughout, largest molecules first
Affinity	specific binding to an immobilised ligand	His₆-tag on Ni²⁺–NTA, eluted with imidazole; GST-tag on glutathione, eluted with free glutathione; IgG on protein A or G; glycoproteins on lectins; antigen on bound antibody
Paper	partition between water held by cellulose and a moving solvent	amino acids revealed with ninhydrin
Thin layer (TLC)	mainly adsorption on silica or alumina	fast, small samples; UV, iodine or ninhydrin detection

Ion exchange. A protein is positive below its pI and negative above it, so at pH < pI it binds a cation exchanger and at pH > pI an anion exchanger. Salt elutes it by competing for the charged groups: weakly charged proteins leave at low salt, highly charged ones later. Size exclusion. Molecules too large for the pores travel only between the beads and emerge at the void volume V₀; the smallest explore every pore and emerge near the total volume Vt. Kav = (Ve − V₀)/(Vt − V₀) lies between 0 and 1, and for globular proteins Ve falls linearly with log M, so standards give the native mass. Dividing it by the subunit mass from SDS-PAGE gives the number of subunits; an elongated protein elutes early and looks too large. Affinity chromatography uses one specific binding pair and can purify a protein greatly in a single step.

Paper and TLC report a retention factor, Rf = (distance moved by the spot's centre)/(distance moved by the solvent front), both measured from the origin line. Rf lies between 0 and 1 and depends on solvent, temperature and plate, so unknowns are run beside standards. With a less polar mobile solvent on paper, hydrophobic leucine travels further (higher Rf) than polar aspartate. A spot 3.5 cm from the origin with the front at 10.0 cm has Rf 0.35.

3. Centrifugation: RCF, sedimentation coefficient and gradients

A particle spun at angular velocity ω at radius r feels a field ω²r, quoted as a multiple of g: RCF = ω²r/g = 1.118 × 10⁻⁵ × r × N², with r in cm and N in rpm. At 10,000 rpm and r = 10 cm, RCF = 1.118 × 10⁻⁵ × 10 × 10⁸ ≈ 11,200 × g. Because RCF depends on r, protocols are written in × g, not rpm; to repeat a run in a rotor of another radius keep r·N² constant, so N₂ = N₁√(r₁/r₂).

The sedimentation coefficient s = v/(ω²r) is velocity per unit field, measured in svedbergs (1 S = 10⁻¹³ s). It rises with mass and with particle density relative to the solvent, and falls with the frictional coefficient, so an elongated molecule sediments more slowly than a compact one of equal mass. Hence s values are not additive: 30S and 50S subunits form a 70S ribosome, and 40S and 60S subunits an 80S one.

Differential centrifugation spins a homogenate at rising RCF and collects pellets: about 600–1,000 × g brings down nuclei and unbroken cells, about 10,000–20,000 × g mitochondria, lysosomes and peroxisomes, and about 100,000 × g microsomes (vesicles of endoplasmic reticulum and plasma membrane); the last supernatant is the cytosol. Fractions are crude and are checked with marker enzymes, such as succinate dehydrogenase for mitochondria and glucose-6-phosphatase for endoplasmic reticulum.
Rate-zonal centrifugation layers the sample on a shallow preformed gradient, often 5–20% sucrose, that stabilises the bands. Particles move as zones at speeds set by s, and the run is stopped before they reach the bottom; it separates ribosomal subunits, polysomes and complexes.
Isopycnic (equilibrium) centrifugation uses a gradient steep enough, such as CsCl, that each particle stops where the medium's density equals its buoyant density, however long the spin continues. It separates by density, not size: DNA by GC content or by ¹⁵N versus ¹⁴N label, as in the 1958 Meselson–Stahl experiment that showed semiconservative replication.

4. Electrophoresis: SDS-PAGE, native gels, IEF, 2-D and agarose

A charged molecule in a field E moves at v = qE/f, so its mobility v/E = q/f rises with charge and falls with frictional drag; a gel adds molecular sieving. Polyacrylamide (acrylamide cross-linked by bis-acrylamide, polymerised with ammonium persulphate and TEMED) has small, tunable pores for proteins and short DNA, and resolves DNA strands differing by one nucleotide; agarose has larger pores for DNA and RNA from about a hundred bases to tens of kilobases.

Method	Separates by	Key feature	Tells you
SDS-PAGE	polypeptide mass	SDS binds about 1.4 g per g protein, unfolds chains and swamps their own charge; β-mercaptoethanol or DTT breaks disulphides	subunit mass (log M linear with distance), purity
Native PAGE	charge, size and shape together	no denaturant; complexes and activity survive; enzymes seen by activity staining	oligomeric state, isozymes such as the five lactate dehydrogenase forms
Isoelectric focusing (IEF)	pI	pH gradient from carrier ampholytes or an immobilised gradient; a protein stops where its net charge is zero	pI, charge variants such as phosphorylated forms
2-D electrophoresis	pI, then mass	IEF strip laid across an SDS gel; spots identified by mass spectrometry	proteome maps, modified forms
Agarose gel	nucleic-acid length	phosphate backbone gives near-constant charge per unit length; ethidium bromide or SYBR stain	fragment size against a ladder, plasmid forms, RNA integrity

⚠️ Two SDS-PAGE traps

Compare reduced and non-reduced lanes: IgG runs near 150 kDa without reducing agent but as 50 kDa heavy and 25 kDa light chains with it, which exposes disulphide-linked subunits. And the Laemmli gel is discontinuous: in the stacking gel (pH 6.8, large pores) glycine is nearly uncharged and trails chloride, squeezing proteins into a sharp band; in the resolving gel (pH 8.8) glycine becomes anionic, overtakes the proteins, and sieving takes over.

5. Methods for DNA–protein and protein–protein interactions

Binding is studied in vitro with purified components, which gives affinity and the exact site, or in vivo, which shows that the interaction happens in the cell. Each method has a characteristic control that examiners ask about.

Method	Principle and readout	Key control or limit
EMSA (gel shift)	labelled DNA probe with protein on a native gel; bound probe runs slower than free probe	excess unlabelled specific probe competes the shift away, a mutant probe does not; an antibody to the protein gives a supershift
DNase I footprinting	end-labelled DNA lightly cut with and without protein; the bound region is a gap in the ladder	locates the exact site; hydroxyl-radical footprinting is finer
ChIP	formaldehyde cross-links proteins to DNA in living cells; chromatin is sheared, the protein immunoprecipitated, cross-links reversed and DNA read by qPCR or sequencing (ChIP-seq)	shows occupancy in vivo; needs a specific antibody
Yeast two-hybrid	bait fused to a DNA-binding domain (as of Gal4), prey to an activation domain; binding rebuilds an activator that switches on HIS3 or lacZ reporters	must occur in the yeast nucleus; a bait that activates alone gives false positives
Co-immunoprecipitation	antibody to X pulls X from a lysate; partner Y detected by western blot	endogenous complex, but contact may be indirect
Pull-down	tagged bait (such as GST fusion on glutathione beads) captures partners	with purified proteins it shows direct binding
Surface plasmon resonance (SPR)	ligand on a gold sensor chip; bound analyte changes the refractive index at the surface, read in real time without labels	gives kon, koff and KD = koff/kon
FRET	excited donor fluorophore transfers energy to a nearby acceptor without emitting; E = 1/[1 + (r/R₀)⁶]	works over about 1–10 nm; usable in living cells

Worked examples. SPR: kon = 1 × 10⁵ M⁻¹ s⁻¹ and koff = 1 × 10⁻³ s⁻¹ give KD = 10⁻⁸ M = 10 nM. FRET: at r = R₀ the efficiency is 0.5; at r = 0.5R₀ it is 1/(1 + 1/64) = 0.98. The sixth power makes FRET a sharp molecular ruler.

⚠️ Read the result for what it proves

A shifted EMSA band proves binding in vitro, not regulation in the cell; a ChIP signal proves occupancy, not function; a co-IP shows membership of one complex, not direct contact; a two-hybrid hit needs a second method to confirm it.

6. Lambert–Beer law, UV-visible and fluorescence spectroscopy

Light of intensity I₀ entering a solution leaves as I. Transmittance T = I/I₀ and absorbance A = log(I₀/I) = −log T = εcl, with ε the molar absorption coefficient (M⁻¹ cm⁻¹), c the molar concentration and l the path length (usually 1 cm). A = 1 transmits 10% of the light and A = 2 only 1%. Absorbances of independent absorbers add. The law holds for dilute solutions; at high absorbance, with scattering or stray light, or when molecules associate, the plot of A against c bends.

UV-visible spectroscopy. Proteins absorb at 280 nm mainly through tryptophan and tyrosine, and strongly below 230 nm through peptide bonds; nucleic acids absorb at 260 nm through their bases. Conventions: A₂₆₀ = 1.0 (1 cm) corresponds to about 50 µg mL⁻¹ double-stranded DNA, 40 µg mL⁻¹ RNA and 33 µg mL⁻¹ single-stranded DNA; pure DNA gives A₂₆₀/A₂₈₀ near 1.8 and pure RNA near 2.0, and a lower ratio flags protein or phenol. Melting DNA raises A₂₆₀ (the hyperchromic effect) because unstacked bases absorb more. NADH and NADPH absorb at 340 nm (ε = 6220 M⁻¹ cm⁻¹) while NAD⁺ and NADP⁺ do not, so dehydrogenases, and other enzymes through coupled assays, are followed at 340 nm.

Worked example. DNA diluted 1 in 20 reads A₂₆₀ = 0.40, so its concentration is 0.40 × 50 × 20 = 400 µg mL⁻¹. In an enzyme assay, A₃₄₀ falling by 0.0622 per minute (1 cm) means NADH is used at 0.0622/6220 = 1.0 × 10⁻⁵ M min⁻¹, or 10 µM min⁻¹; multiply by the reaction volume for µmol min⁻¹ (units) and divide by the protein mass for specific activity.

Fluorescence. An excited molecule loses some energy by vibrational relaxation and then emits a photon of longer wavelength (the Stokes shift) within nanoseconds; the quantum yield is photons emitted per photon absorbed. Measured against a dark background, fluorescence is far more sensitive than absorbance, but it is linear only at low absorbance and needs a standard curve. Tryptophan is the main intrinsic fluorophore: excited at 295 nm (to avoid tyrosine) it emits near 350 nm in water but at shorter wavelength (a blue shift) when buried in a nonpolar core, so folding and ligand binding can be followed. Quenchers, extrinsic dyes, fluorescent proteins such as GFP, and FRET extend the method.

Key takeaways

pH = pKa + log([A⁻]/[HA]): capacity is greatest at pH = pKa and useful over pKa ± 1; the ratio sets pH, the total concentration sets capacity. Blood pH = 6.1 + log([HCO₃⁻]/(0.03 pCO₂)); 24 mM over 1.2 mM gives 7.40, and lungs and kidneys regulate both terms.
Ion exchange separates by net charge (below pI a protein binds a cation exchanger, above pI an anion exchanger); gel filtration elutes the largest first and gives native mass from log M; affinity uses a specific ligand; Rf is measured from the origin line.
RCF = 1.118 × 10⁻⁵ r N² (r in cm, N in rpm); s is in svedbergs and is not additive. Differential centrifugation collects pellets, rate-zonal separates by s and is stopped early, isopycnic separates by buoyant density at equilibrium.
SDS-PAGE gives subunit mass, native PAGE keeps complexes and activity, IEF separates by pI, 2-D gels by pI then mass, agarose separates DNA by length.
EMSA, footprinting and ChIP probe DNA–protein binding; two-hybrid, co-IP, pull-down, SPR (KD = koff/kon) and FRET (sixth-power distance law over 1–10 nm) probe protein–protein binding, each with its own control.
A = εcl; proteins at 280 nm, nucleic acids at 260 nm (A₂₆₀ 1.0 ≈ 50 µg mL⁻¹ dsDNA), NADH at 340 nm (ε 6220 M⁻¹ cm⁻¹). Fluorescence is emitted at longer wavelength, is more sensitive, and a tryptophan blue shift means a more nonpolar environment.

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.

Plasma bicarbonate has a pKa of 6.1, yet it is the most important extracellular buffer at pH 7.4. Which feature best explains its effectiveness?
1. Carbonic anhydrase in red cells shifts the pKa of carbonic acid up to 7.4, so the pair buffers at its own pKa
2. Bicarbonate binds H⁺ irreversibly, so its buffering does not depend on how close the pH is to the pKa
3. CO₂ is exhaled or retained by the lungs and HCO₃⁻ adjusted by the kidneys, so the pair's ratio is regulated
4. Its plasma concentration is so low that even a little added acid restores the ratio of the pair to 20:1
Show answer
Answer: C — CO₂ is exhaled or retained by the lungs and HCO₃⁻ adjusted by the kidneys, so the pair's ratio is regulated
A closed buffer pair works best at its pKa, and 7.4 is 1.3 units away from 6.1. Bicarbonate succeeds because it is an open system: dissolved CO₂ is set by breathing and HCO₃⁻ by renal reabsorption and excretion, so both terms of the ratio are controlled. Carbonic anhydrase is the tempting choice, but an enzyme only speeds the CO₂ ⇌ H₂CO₃ interconversion and cannot change an equilibrium constant or a pKa. H⁺ binding is reversible, and a low concentration would weaken buffering, not explain it.
Arterial blood from a patient with diabetic ketoacidosis shows [HCO₃⁻] = 12 mM and pCO₂ = 25 mmHg. Using pH = 6.1 + log([HCO₃⁻]/(0.03 × pCO₂)), where 0.03 mM mmHg⁻¹ is the solubility of CO₂, the blood pH (to two decimal places) is ____.
Numerical answer — type the value.
Show answer
Answer: 7.30
Dissolved CO₂ = 0.03 × 25 = 0.75 mM. Ratio = 12/0.75 = 16, and log 16 = 1.204. pH = 6.1 + 1.204 = 7.30. The low HCO₃⁻ is the primary metabolic acidosis; the low pCO₂ (normal 40 mmHg) is respiratory compensation by hyperventilation. Without it, pCO₂ = 40 mmHg would give 6.1 + log(12/1.2) = 7.10.
To make a Tris buffer, 0.100 mol of Tris base is dissolved in water and 1.00 M HCl is added before making up to 1 L. Taking the pKa of TrisH⁺ as 8.06, assuming each mole of HCl converts one mole of Tris to TrisH⁺ and ignoring activity effects, the volume of HCl (in mL, to one decimal place) needed for pH 7.80 is ____.
Numerical answer — type the value.
Show answer
Answer: 64.5
log([Tris]/[TrisH⁺]) = 7.80 − 8.06 = −0.26, so base:acid = 10^−0.26 = 0.550. The protonated fraction is 1/(1 + 0.550) = 0.645, so 0.645 × 0.100 mol = 0.0645 mol HCl, which is 64.5 mL of 1.00 M HCl. Using the base fraction (0.355, giving 35.5 mL) is the usual slip: the target pH is below the pKa, so more than half the Tris must be protonated.
A protein of pI 5.2 is to be captured by ion-exchange chromatography from a lysate buffered at pH 7.5. Which matrix will bind it, and why?
1. CM-cellulose, because above its pI the protein carries a net positive charge
2. DEAE-cellulose, because above its pI the protein carries a net negative charge
3. CM-cellulose, because its carboxymethyl groups are positively charged at pH 7.5
4. DEAE-cellulose, because its diethylaminoethyl groups are negatively charged at pH 7.5
Show answer
Answer: B — DEAE-cellulose, because above its pI the protein carries a net negative charge
Above its pI a protein is net negative. DEAE (diethylaminoethyl) groups are protonated and positive, so DEAE-cellulose is an anion exchanger and binds the protein. CM (carboxymethyl) groups are negative, making CM-cellulose a cation exchanger that would bind this protein only below pH 5.2. The tempting distractor reverses the sign rule: proteins are positive below, not above, their pI.
On a calibrated gel-filtration column in native buffer, a purified enzyme elutes at the same volume as a 150 kDa globular standard. Reducing SDS-PAGE of the enzyme shows a single band at 50 kDa. Assuming the enzyme is globular, what is the most reasonable conclusion?
1. It is a single 150 kDa chain that SDS hydrolyses into three pieces
2. It is a dimer of 75 kDa chains that run faster in the SDS gel
3. It is a 50 kDa monomer that also binds two molecules of matrix
4. It is a trimer of polypeptide chains of about 50 kDa each
Show answer
Answer: D — It is a trimer of polypeptide chains of about 50 kDa each
Gel filtration in native buffer reports the mass of the intact assembly (150 kDa), whereas reducing SDS-PAGE reports the mass of each separated polypeptide (50 kDa); 150/50 = 3 subunits. SDS denatures and coats proteins but does not cleave peptide bonds, so one chain cannot become three bands. A 75 kDa chain would run at 75 kDa, and a protein that sticks to the matrix elutes later and looks smaller, not larger.
On a TLC plate the origin line is drawn 1.5 cm above the bottom edge. After development, the solvent front is 13.5 cm and the centre of an amino-acid spot is 6.3 cm above the bottom edge. The Rf of the amino acid (to two decimal places) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.40
Both distances are measured from the origin line, not from the plate edge: spot = 6.3 − 1.5 = 4.8 cm, front = 13.5 − 1.5 = 12.0 cm, so Rf = 4.8/12.0 = 0.40. Measuring from the edge gives 6.3/13.5 = 0.47, the usual error.
A protocol pellets mitochondria at 12,000 rpm in a rotor of effective radius 7.2 cm. The same relative centrifugal force (RCF proportional to r × N², with N in rpm) must be reached in a rotor of radius 5.0 cm. The required speed, in rpm to the nearest whole number, is ____.
Numerical answer — type the value.
Show answer
Answer: 14400
Equal RCF means r₁N₁² = r₂N₂², so N₂ = N₁√(r₁/r₂) = 12,000 × √(7.2/5.0) = 12,000 × √1.44 = 12,000 × 1.2 = 14,400 rpm. Both runs give about 1.118 × 10⁻⁵ × 7.2 × 12,000² ≈ 11,600 × g. A smaller radius needs a higher speed; scaling N in direct proportion to r (17,280 rpm) forgets that N is squared.
Two DNA samples have the same length and sequence, but one was made by cells grown on ¹⁵NH₄Cl and the other on ¹⁴NH₄Cl. Which method separates them?
1. Equilibrium (isopycnic) centrifugation in a CsCl gradient, which separates them by buoyant density
2. Rate-zonal centrifugation on a 5–20% sucrose gradient, which separates them by sedimentation coefficient
3. Agarose gel electrophoresis, which separates them by the charge carried on the phosphate backbone
4. Differential centrifugation at rising RCF, which pellets the heavier sample at the lower force
Show answer
Answer: A — Equilibrium (isopycnic) centrifugation in a CsCl gradient, which separates them by buoyant density
¹⁵N makes DNA slightly denser without changing its length or charge. In a CsCl gradient each molecule settles where the medium's density equals its own, so heavy, light and hybrid DNA form separate bands, the basis of the Meselson–Stahl experiment. In a sucrose rate-zonal run the tiny mass difference changes s negligibly; agarose separates by length, identical here; differential centrifugation cannot resolve so small a difference.
In SDS-PAGE, what is the main role of sodium dodecyl sulphate?
1. It reduces the disulphide bonds between cysteine residues, so that subunits held by S–S bridges separate on the gel
2. It cross-links the acrylamide chains into a mesh whose pore size is set by the detergent concentration
3. It unfolds the chains and coats them with negative charge in proportion to length, so mass sets their mobility
4. It holds every protein at its isoelectric point during the run, so that the proteins separate according to pI
Show answer
Answer: C — It unfolds the chains and coats them with negative charge in proportion to length, so mass sets their mobility
SDS binds about 1.4 g per g of protein, unfolds the chain and swamps its own charge, giving a near-constant charge-to-mass ratio, so the gel sieves by mass. Breaking disulphides is the job of β-mercaptoethanol or DTT, not SDS; that is the tempting distractor. Bis-acrylamide is the cross-linker, and separation by pI is isoelectric focusing.
Which of the following statements about electrophoretic methods are correct?
1. Native PAGE separates proteins by polypeptide mass alone, because the gel pores sieve them
2. In isoelectric focusing, a protein stops migrating where the pH of the gradient equals its pI
3. Agarose gels resolve DNA strands differing by a single nucleotide better than polyacrylamide gels do
4. In a standard 2-D gel, proteins are separated by pI in the first dimension and by mass in the second
Show answer
Answer: B — In isoelectric focusing, a protein stops migrating where the pH of the gradient equals its pI; D — In a standard 2-D gel, proteins are separated by pI in the first dimension and by mass in the second
At its pI a protein has zero net charge and stops moving in the pH gradient, and the standard 2-D gel runs IEF first and SDS-PAGE second. Native PAGE is wrong because without SDS mobility depends on charge, size and shape together. Agarose is wrong because its pores are too large for single-nucleotide resolution; that needs denaturing polyacrylamide, as in sequencing gels.
In an EMSA, a labelled 30-bp promoter probe gives a slower-migrating band when incubated with nuclear extract. Which further result shows that the binding is sequence-specific?
1. The shifted band is lost with excess unlabelled probe whose site is mutated but not with a 100-fold excess of unlabelled wild-type probe
2. The shifted band is lost with a 100-fold excess of unlabelled wild-type probe but not with excess unlabelled probe whose site is mutated
3. The shifted band grows stronger as more nuclear extract is added, while the band of free probe grows weaker
4. The shifted band disappears when the nuclear extract is boiled for ten minutes before it is added to the probe
Show answer
Answer: B — The shifted band is lost with a 100-fold excess of unlabelled wild-type probe but not with excess unlabelled probe whose site is mutated
Specific binding is competed by excess of the same sequence but not by the sequence with its site mutated, because the protein binds the mutant weakly. Dose-dependence and loss on boiling show only that a heat-labile protein binds the probe, which a non-specific DNA-binding protein would also do. The reversed competition pattern would mean the protein prefers the mutant, not the promoter site.
In a yeast two-hybrid screen, cells carrying the bait (protein X fused to the Gal4 DNA-binding domain) together with an empty activation-domain vector already grow on medium lacking histidine (HIS3 reporter). What does this mean for colonies obtained with library preys?
1. Each growing colony carries a genuine partner of X, because the bait clearly reaches the nucleus
2. The HIS3 reporter is defective, so repeating the screen with a lacZ reporter will fix the problem
3. X must be a membrane protein, since membrane proteins give the strongest two-hybrid signals
4. Their growth is not evidence of interaction, because the bait switches on the reporter by itself
Show answer
Answer: D — Their growth is not evidence of interaction, because the bait switches on the reporter by itself
The reporter should switch on only when bait–prey binding brings an activation domain to the promoter. If the bait with an empty AD vector already activates HIS3, X has its own transactivation activity (autoactivation), and growth with any prey is uninterpretable; the usual fix is a fragment of X lacking that activity or a more stringent reporter. The reporter is working, not defective, so a lacZ reporter would also be switched on; and membrane proteins are poor baits because the assay needs nuclear entry.
Which of the following statements about methods for studying macromolecular interactions are correct?
1. In ChIP, formaldehyde cross-linking in living cells captures protein–DNA contacts as they exist in vivo
2. Co-immunoprecipitation of Y with an antibody to X proves that X and Y touch each other directly, with no protein bridging them
3. In DNase I footprinting, the protein-binding site appears as a gap of missing or weakened bands in the cleavage ladder
4. In a GST pull-down, the GST-tagged bait and its bound partners are captured on beads that carry Ni²⁺–NTA
Show answer
Answer: A — In ChIP, formaldehyde cross-linking in living cells captures protein–DNA contacts as they exist in vivo; C — In DNase I footprinting, the protein-binding site appears as a gap of missing or weakened bands in the cleavage ladder
ChIP fixes contacts inside cells before lysis, and in footprinting the bound protein shields its site from DNase I, leaving a gap in the ladder. A co-IP shows only that X and Y are in one complex; a third protein may bridge them. GST binds glutathione, not Ni²⁺–NTA, which captures His-tagged proteins.
A donor–acceptor pair has a Förster distance R₀ = 5.0 nm. A conformational change moves the two dyes from 5.0 nm to 7.5 nm apart. Using E = 1/[1 + (r/R₀)⁶], the FRET efficiency changes from 0.50 to
1. about 0.08
2. about 0.33
3. about 0.31
4. about 0.23
Show answer
Answer: A — about 0.08
r/R₀ = 7.5/5.0 = 1.5 and 1.5⁶ = 11.4, so E = 1/(1 + 11.4) = 0.081. The steep sixth-power dependence is why FRET reports changes of a few nanometres so sharply. 0.33 assumes E falls in proportion to distance, and 0.31 and 0.23 use the second and third powers instead of the sixth.
Lactate dehydrogenase is assayed by following NADH oxidation at 340 nm (ε = 6220 M⁻¹ cm⁻¹, path length 1.0 cm) in a 3.0 mL reaction containing 0.020 mg of enzyme protein. A₃₄₀ falls by 0.124 per minute. The specific activity, in µmol min⁻¹ mg⁻¹ to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.99
Rate of concentration change = ΔA/(εl) = 0.124/6220 = 1.994 × 10⁻⁵ M min⁻¹ = 19.94 µM min⁻¹. In 3.0 mL this is 19.94 µmol L⁻¹ min⁻¹ × 0.0030 L = 0.0598 µmol min⁻¹. Divided by 0.020 mg it gives 2.99 µmol min⁻¹ mg⁻¹, that is 2.99 units per mg. Forgetting the 3.0 mL volume is the usual slip.
When a ligand binds a protein that has a single tryptophan, the emission maximum (excitation at 295 nm) moves from 352 nm to 336 nm and the intensity rises. The most reasonable interpretation is that the tryptophan
1. has become more exposed to water, as happens when a protein unfolds
2. has moved into a more nonpolar environment shielded from the solvent
3. passes its excitation energy to the ligand by FRET, quenching itself
4. is now excited through tyrosine residues, which emit at shorter wavelengths
Show answer
Answer: B — has moved into a more nonpolar environment shielded from the solvent
Tryptophan emission depends on its surroundings: exposed to water it emits near 350–355 nm, buried in a nonpolar pocket it emits at shorter wavelength (a blue shift), often with higher quantum yield. Unfolding causes the opposite, a red shift. Energy transfer to the ligand would lower, not raise, the intensity, and excitation at 295 nm is chosen precisely to avoid exciting tyrosine.
A plasmid DNA preparation gives A₂₆₀/A₂₈₀ = 1.5. What does this most likely indicate?
1. Contamination with RNA, which pulls the ratio below that of pure DNA
2. Partial melting of the DNA, which lowers its absorbance at 260 nm
3. A pure sample, since about 1.5 is the expected ratio for duplex DNA
4. Contamination with protein or phenol, which absorb strongly near 280 nm
Show answer
Answer: D — Contamination with protein or phenol, which absorb strongly near 280 nm
Pure DNA gives about 1.8 and pure RNA about 2.0, so a ratio of 1.5 means something that absorbs near 280 nm is present, typically protein (aromatic residues) or residual phenol. RNA would raise the ratio, and melting increases A₂₆₀ (the hyperchromic effect), which would also raise it.
Which of the following statements about absorbance and fluorescence measurements are correct?
1. Doubling the path length at constant concentration doubles the transmittance of the solution
2. A solution with an absorbance of 2.0 transmits 1% of the incident light
3. With one-photon excitation, fluorescence is emitted at a longer wavelength than the exciting light
4. NAD⁺ absorbs at 340 nm about as strongly as NADH, so the two cannot be told apart at that wavelength
Show answer
Answer: B — A solution with an absorbance of 2.0 transmits 1% of the incident light; C — With one-photon excitation, fluorescence is emitted at a longer wavelength than the exciting light
A = −log T, so A = 2.0 gives T = 10⁻² = 1%; and vibrational relaxation before emission makes one-photon fluorescence red-shifted (the Stokes shift). Doubling the path doubles the absorbance, which squares the transmittance (for example 0.5 becomes 0.25) rather than doubling it. Only the reduced forms NADH and NADPH absorb at 340 nm; that difference is why dehydrogenase assays are read there.