Lab Methods: Centrifugation, Spectroscopy & Assays

Lab Methods: Centrifugation, Spectroscopy & Assays

Part 1 of 3 — Centrifugation (Separating by Density & Size)

Where this fits. The Lab Methods: Separations & Electrophoresis lesson sorted molecules on columns and gels. This lesson adds centrifugation for particles and organelles, spectroscopy and Beer's law for measuring what you isolated, and the blots, PCR and ELISA that detect specific molecules.

Centrifugation spins samples to generate a force that drives denser/larger particles outward (to the bottom = pellet). The rest stays in the supernatant.

$\text{RCF} = 1.118 \times 10^{-5} \; r \; (\text{RPM})^2$

RCF (relative centrifugal force, in × g) depends on rotor radius $r$ (cm) and rotation speed.

Sedimentation Logic

$v_{sed} \;\propto\; (\rho_{particle} - \rho_{medium}) \, r_{particle}^2 \, \omega^2$

where $\omega$ is the rotor's angular speed (proportional to RPM).

• Larger and DENSER particles sediment faster.
• A particle stops moving when its density equals the surrounding medium's density (isopycnic point).

Two Major Techniques

Differential Centrifugation of a Cell Lysate (Order of Pelleting)

$\text{Low speed} \to \text{nuclei} \to \text{mitochondria} \to \text{microsomes/ER} \to \text{ribosomes (ultracentrifuge)}$

• Increase speed in steps; pellet the largest organelles first, re-spin the supernatant faster.
• Microsomes are small vesicles formed from fragments of the ER when the cells are homogenized.

Svedberg Units (S)

• S is a sedimentation coefficient — it is NOT additive because it depends on shape and density, not mass alone.
• Example: the 70S prokaryotic ribosome = 50S + 30S subunits (70 ≠ 50 + 30).

Centrifugation 🎯

Worked Examples — Centrifugation

Problem 1. You want to isolate mitochondria from liver cells without nuclear contamination, using differential centrifugation.

• (a) How should the cells be broken open?
• (b) What should the first, low-speed spin remove, and which fraction do you keep?
• (c) How do you then collect the mitochondria?
• (d) What is left in the final supernatant?

(a) Homogenize. Homogenize the cells in isotonic buffer, which preserves the organelles.

(b) Low-speed spin (~1000 g). It pellets the NUCLEI and debris, the biggest contaminant: discard that pellet and keep the supernatant.

(c) Medium-speed spin (~10,000 g). Spinning the post-nuclear supernatant from (b) faster pellets the MITOCHONDRIA: keep this pellet. The logic: remove the biggest contaminant first at low force, then collect the target at the next-higher force.

(d) What remains. The remaining supernatant holds the microsomes (ER fragments) and the cytosol.

Problem 2. In the Meselson–Stahl experiment, DNA labeled with heavy ${}^{15}N$ is shifted to ${}^{14}N$ medium. After one round of replication, a CsCl gradient shows a single band of intermediate density.

• (a) What property does a CsCl gradient separate DNA by?
• (b) What does a single intermediate band say about each duplex?
• (c) Which replication model does this result rule out?
• (d) Does this one result prove semiconservative replication?

(a) Buoyant density. A CsCl gradient is isopycnic: each DNA duplex bands where its buoyant density matches the medium's.

(b) Half heavy, half light. A single INTERMEDIATE band means every duplex has the same density, halfway between heavy and light, as if each held one old and one new strand.

(c) Conservative is out. Conservative replication would give two bands, one fully heavy and one fully light, so a single intermediate band rules it out and fits semiconservative replication.

(d) Not on its own. Dispersive replication would also give one intermediate band after one round. The second round, with one intermediate and one light band, rules dispersive out; together the two results are the classic density-gradient proof of semiconservative replication.

Problem 3. A rotor of radius $r = 8$ cm spins at 10,000 RPM. Use $\text{RCF} = 1.118 \times 10^{-5}\, r\, (\text{RPM})^2$.

• (a) What is the RCF, in × g?
• (b) What happens to the RCF if the speed doubles to 20,000 RPM?
• (c) Why should a protocol quote the force in × g rather than the RPM?

(a) RCF. $\text{RCF} = 1.118 \times 10^{-5} \times 8 \times (10000)^2 = 1.118 \times 10^{-5} \times 8 \times 10^{8}$

$= 8944 \times g \approx 8.9 \times 10^{3} \; g$

(b) Doubling the speed. RCF scales with the SQUARE of RPM, so doubling the RPM quadruples the force: $4 \times 8944 \approx 35{,}776 \times g$.

(c) Quote g. The same RPM gives a different force on a rotor of a different radius, so RPM alone is meaningless; the g-force is what reproduces the spin.

Key Takeaways — Part 1

• Sedimentation rate. Larger particles, and particles denser than the medium, sediment faster.
• Differential pelleting order. Rising speeds pellet nuclei, then mitochondria, then microsomes, then ribosomes.
• Isopycnic vs. rate-zonal. In a CsCl gradient each particle bands at its own density; a sucrose rate-zonal run sorts by S value.
• RCF ∝ r · RPM². A passage supplies the constant; RCF grows with rotor radius and RPM², so doubling RPM quadruples it.

Common traps

• Quoting RPM. RPM means nothing without the rotor radius; report the force in × g.
• Adding Svedberg units. S values are not additive: 50S and 30S subunits form a 70S ribosome, not 80S.
• Size sets the band. In an isopycnic gradient a band's position depends on density only, not on size.
• Keeping the wrong fraction. To isolate mitochondria, discard the low-speed nuclear pellet and re-spin its supernatant faster.

Lab Methods: Centrifugation, Spectroscopy & Assays

Part 2 of 3 — Spectroscopy & Beer's Law

Spectroscopy measures how molecules absorb/emit electromagnetic radiation. Different energies probe different transitions.

What Each Method Probes

Beer–Lambert Law

$A = \varepsilon \, l \, c$

• $A$ = absorbance (unitless, log scale)
• $\varepsilon$ = molar absorptivity $(\text{M}^{-1}\text{cm}^{-1})$ — an intrinsic property at a given wavelength
• $l$ = path length (cm)
• $c$ = concentration (M)

Linearity: absorbance is DIRECTLY proportional to concentration (and to path length) — the basis of quantitative assays. The relationship breaks down at high absorbance (A > ~1) due to stray light and deviations.

Absorbance vs. Transmittance

$A = -\log_{10}(T) = -\log_{10}\!\left(\frac{I}{I_0}\right)$

• $T = 1$ (100% transmitted) → $A = 0$.
• $T = 0.1$ (10%) → $A = 1$. $T = 0.01$ → $A = 2$.

Biochemistry Hooks

• $A_{260}$: nucleic acids (aromatic bases). $A_{280}$: proteins (Trp, Tyr).
• $A_{260}/A_{280}$ ratio ≈ 1.8 → pure DNA; ≈ 2.0 → pure RNA; lower → protein contamination.

Spectroscopy & Beer's Law 🎯

Worked Examples — Spectroscopy

Problem 1. A dye has $\varepsilon = 5000$ $\text{M}^{-1}\text{cm}^{-1}$ at 500 nm. In a 1 cm cuvette its absorbance is 0.75.

• (a) What is the dye's concentration?
• (b) Can this reading be trusted, and what would you do if A had been above 1?

(a) Concentration. $c = \dfrac{A}{\varepsilon l} = \dfrac{0.75}{5000 \times 1} = 1.5 \times 10^{-4} \text{ M}$

(b) Linear range. Yes: $A = 0.75$ lies in the linear range (≈ 0.1–1.0), and the stated path length is 1 cm (always check it). At $A > 1$, dilute the sample and remeasure.

Problem 2. Standards give A = 0.10, 0.20 and 0.30 at c = 2, 4 and 6 µM (path 1 cm). An unknown reads A = 0.25.

• (a) What is the slope of the standard curve?
• (b) What is $\varepsilon$, in $\text{M}^{-1}\text{cm}^{-1}$?
• (c) What is the unknown's concentration?
• (d) Why use a standard curve rather than a tabulated $\varepsilon$?

(a) Slope. The data are linear: A/c = 0.10/2 = 0.05 per µM, and that slope equals $\varepsilon l$.

(b) Molar absorptivity. With slope $= 0.05$ from (a) and $l = 1$ cm, $\varepsilon = 0.05$ µ$\text{M}^{-1}\text{cm}^{-1}$ = $5\times 10^{4}$ $\text{M}^{-1}\text{cm}^{-1}$.

(c) Unknown. $c = A/\text{slope} = 0.25/0.05 = 5$ µM.

(d) Why a standard curve. It captures $\varepsilon l$ empirically for your exact instrument and conditions, so concentration is read directly off the line.

Problem 3. A sample transmits 10% of the incident light. Use $A = -\log_{10}(T)$.

• (a) What is its absorbance?
• (b) What would the absorbance be at 1% transmittance?
• (c) What does each extra unit of absorbance mean for the transmitted light?

(a) 10% transmitted. $T = 0.10 \Rightarrow A = -\log_{10}(0.10) = 1.0$

(b) 1% transmitted. $T = 0.01 \Rightarrow A = -\log_{10}(0.01) = 2.0$

(c) A log scale. Absorbance is logarithmic: each unit of A means a 10× drop in transmitted light, so $A = 2$ lets only 1% through.

Key Takeaways — Part 2

• Beer–Lambert law. $A = \varepsilon l c$: absorbance is linear in concentration and path length, up to about $A = 1$.
• Absorbance is logarithmic. $A = -\log_{10} T$: 10% transmitted gives $A = 1$, and 1% gives $A = 2$.
• Match spectrum to question. UV-Vis: conjugation and concentration; IR: functional groups; NMR: H and C environments; MS: mass.
• $A_{260}/A_{280}$ purity. Nucleic acids absorb at 260 nm, proteins at 280 nm; pure DNA ≈ 1.8, pure RNA ≈ 2.0.

Common traps

• Reading A as linear in T. Each unit of absorbance is a tenfold drop in transmitted light, not a fixed step.
• Trusting A above 1. Beyond about $A = 1$ the line bends; dilute the sample and remeasure.
• Low ratio means RNA. A ratio below 1.8 signals protein contamination; RNA would push it up toward 2.0.
• UV-Vis for a carbonyl. UV-Vis reports conjugation; the C=O stretch near 1700 $\text{cm}^{-1}$ is read on IR.

Lab Methods: Centrifugation, Spectroscopy & Assays

Part 3 of 3 — Molecular & Immunological Assays + Review

This part covers the detection/amplification methods that follow separations, plus a method-selection review.

Nucleic Acid & Protein Detection (Know the Blots)

Mnemonic — SNoW DRoP: Southern = DNA, Northern = RNA, Western = Protein.

PCR Logic

Copies after $n$ cycles of ideal doubling: $\text{Copies} = N_0 \times 2^{n}$. Steps per cycle:

1. Denature (~95 °C): separate strands.
2. Anneal (~50–65 °C): primers bind.
3. Extend (~72 °C): heat-stable Taq polymerase synthesizes.

qPCR quantifies in real time; reverse-transcription (RT-PCR) starts from RNA (via complementary DNA, cDNA).

ELISA & Antibody Specificity

• Direct ELISA: antigen bound, enzyme-linked primary antibody → color.
• Sandwich ELISA: capture antibody traps antigen; detection antibody reports it (high specificity).
• Signal (color/absorbance via Beer's law) ∝ amount of antigen.

Method-Selection Cheat Sheet

SDS-PAGE (sorts proteins by size) and affinity chromatography (binds one tagged molecule) are from the Lab Methods: Separations & Electrophoresis lesson.

Molecular Assays Review 🎯

Worked Examples — Molecular Assays

Problem 1. A lab wants to know whether a gene is being TRANSCRIBED (made into mRNA) in a tumor sample, and to measure how much.

• (a) Which molecule must the assay measure?
• (b) Which method measures it quantitatively?
• (c) Why would a Western blot not answer the question?

(a) The molecule. Transcription produces mRNA, so the assay must measure RNA. Map the molecule of interest (DNA / RNA / protein) to the assay before anything else.

(b) RT-qPCR. Reverse-transcribe the mRNA to complementary DNA (cDNA), then run quantitative PCR. A Northern blot also detects RNA, but qPCR quantifies it precisely.

(c) Not a Western. A Western blot measures PROTEIN, not mRNA. A gene could be transcribed while its protein is degraded, so a Western answers a different question.

Problem 2. A single template molecule is to be amplified to over one million copies, assuming ideal doubling every cycle.

• (a) What inequality must the cycle number $n$ satisfy?
• (b) About how many cycles is that?
• (c) Will a real reaction reach the ideal copy number?

(a) The inequality. Starting from one copy, $n$ cycles give $2^n$ copies, so $2^n > 10^6$.

(b) Cycles. $n > \log_2(10^6) = \dfrac{6}{\log_{10}2} \approx \dfrac{6}{0.301} \approx 20$, so about 20 cycles exceed one million copies ($2^{20} \approx 1.05 \times 10^6$).

(c) Real PCR. No: real PCR plateaus as reagents deplete and efficiency drops below 2× per cycle, so practical yields fall short of the ideal.

Problem 3. A sandwich ELISA gives absorbances of 0.2, 0.4 and 0.6 for antigen standards of 10, 20 and 30 ng/mL. A patient sample reads 0.5.

• (a) What is the slope of the standard curve?
• (b) What is the antigen concentration in the patient sample?

(a) Slope. The standards are linear: 0.02 absorbance units per ng/mL (0.2/10).

(b) Patient. With the slope of 0.02 from (a): $c = 0.5 / 0.02 = 25$ ng/mL. This mirrors the Beer's-law standard curve in Part 2 of this lesson: ELISA turns antigen amount into a colored signal read by a spectrophotometer.

Key Takeaways — Part 3 (and Lesson Review)

• SNoW DRoP. Southern blots detect DNA, Northern blots RNA, and Western blots protein.
• PCR doubles each cycle. Copies $= N_{0} \times 2^{n}$ after $n$ denature, anneal and extend cycles; RT-PCR starts from RNA.
• Match molecule to method. MW: SDS-PAGE; protein: Western or ELISA; DNA: PCR; mRNA level: RT-qPCR; concentration: UV-Vis.
• Signal tracks amount. ELISA and UV-Vis read concentration off a standard curve, because signal is proportional to amount.

Common traps

• Western for transcription. mRNA is measured by RT-qPCR or a Northern blot; a Western blot measures protein.
• Linear PCR growth. Five cycles multiply the copies by $2^5 = 32$, not by $2 \times 5 = 10$.
• Ideal yield in real PCR. Real reactions plateau as reagents run out, so they fall short of $2^{n}$.
• Unbound antibody adds color. Unbound detection antibody is washed away; only antibody bound to antigen gives signal.