Question 1

Protein property – What is the approximate isoelectric pH (pI) of cytochrome c as commonly reported for the classic horse heart protein?

Accepted Answer

Introduction / Context: The isoelectric point (pI) of a protein is the pH at which its net charge is zero. Cytochrome c, a small heme protein of the electron transport chain, has a relatively high pI because it is rich in basic residues. Knowing approximate pI values helps in designing electrophoresis and chromatography protocols, especially isoelectric focusing and ion-exchange separations. Given Data / Assumptions: Reference is the well-studied horse heart cytochrome c used as an analytical standard. “Approximate” means recognizing the basic nature and order of magnitude rather than an exact universal constant. Buffers and post-translational modifications can shift observed pI slightly. Concept / Approach: Cytochrome c is lysine-rich and binds to negatively charged partners on the mitochondrial inner membrane. Its basic composition yields a pI around 10. The reported values in textbooks typically range near 10 to a little above 10 depending on measurement method, making 10.05 a reasonable representative option among the choices provided. Step-by-Step Solution: Recall that pI reflects the balance of acidic and basic side chains.Recognize cytochrome c’s abundance of lysine residues and positive charge at neutral pH.Associate this with a high pI close to 10.Select the value closest to ~10 from the options, i.e., 10.05. Verification / Alternative check: Empirical isoelectric focusing gels and database entries list cytochrome c pI around 10–10.5 for common species, supporting the selection of 10.05 as an approximate value. Why Other Options Are Wrong: 7.5 and 8.5: too acidic for cytochrome c’s known basicity. 11.05: higher than most reported values for typical preparations. 9.0: still lower than the widely cited basic pI near 10. Common Pitfalls: Assuming one exact pI for all species and conditions; minor sequence or modification differences can shift pI modestly, but not enough to move it down near neutrality for cytochrome c. Final Answer: 10.05

Question 2

Two-dimensional electrophoresis – What property is used in the first-dimension separation before SDS-PAGE in standard 2D gels?

Accepted Answer

Introduction / Context: Two-dimensional (2D) gel electrophoresis resolves complex protein mixtures by separating proteins according to two independent properties. The first dimension uses isoelectric focusing (IEF), which separates proteins by their isoelectric point (pI). The second dimension applies SDS-PAGE to separate proteins by molecular mass. This orthogonality greatly enhances resolving power for proteomics and differential expression studies. Given Data / Assumptions: Proteins carry different net charges at different pH values. In IEF, proteins migrate in a stabilized pH gradient until net charge is zero (at pI). SDS-PAGE then masks charge with SDS, making separation primarily size-based. Concept / Approach: Because pI and molecular mass are largely independent, combining IEF with SDS-PAGE spreads proteins across a 2D map: horizontal location reflects pI, vertical position reflects size. Staining (e.g., Coomassie, silver) or fluorescent dyes visualize the spots. Spots can be excised for mass spectrometric identification. Step-by-Step Solution: Perform IEF in immobilized pH gradient strips to focus proteins at their pI.Equilibrate strips with SDS, reducing and alkylating disulfides.Run strips atop SDS-PAGE gels to separate by size in the second dimension.Visualize and analyze protein spots across pI and mass coordinates. Verification / Alternative check: Proteins with identical mass but different pI (e.g., isoforms or PTM variants) resolve horizontally in IEF, confirming that the first dimension is based on isoelectric point rather than size or solubility. Why Other Options Are Wrong: Molecular mass: used in the second dimension, not the first. Solubility, folding, hydrophobicity: not the primary variables exploited in standard 2D electrophoresis. Common Pitfalls: Confusing IEF-first with blue native PAGE workflows; in classic 2D gels, the first dimension is always IEF by pI. Final Answer: Isoelectric point (pI) via isoelectric focusing)

Question 3

Signal processing in spectroscopy – How do you convert a time-domain NMR signal (free induction decay) into a frequency-domain spectrum?

Accepted Answer

Introduction / Context: Modern NMR and many spectroscopies rely on pulsed excitation and detection in the time domain. The recorded free induction decay (FID) contains all the spectral information, encoded as exponentially decaying sinusoids. To interpret chemical shifts and coupling patterns, we must convert this time-domain signal into a frequency-domain spectrum using the Fourier transform (FT). This is the basis of FT-NMR. Given Data / Assumptions: After an RF pulse, nuclear magnetization precesses and decays, producing the FID. Sampling is uniform, and digitization obeys the Nyquist criterion for the spectral width. Spectral processing may include apodization and zero-filling to improve appearance. Concept / Approach: The Fourier transform decomposes the FID into constituent frequencies, revealing resonance positions (chemical shifts) and line shapes. Phase correction aligns absorptive and dispersive components. Apodization (windowing) trades resolution for signal-to-noise in a controlled way, and zero-filling increases digital resolution without adding information. Step-by-Step Solution: Acquire FID after excitation pulse.Apply Fourier transform to convert time data to frequency spectrum.Perform phase and baseline correction; optionally apply window functions.Interpret chemical shifts, integrals, and multiplets in the frequency domain. Verification / Alternative check: Compare spectra processed with and without the FT step: only after FT do distinct peaks at specific frequencies (ppm) appear, confirming that Fourier transformation is essential. Why Other Options Are Wrong: Peak-area measurement is an analysis step, not a transformation. Michelson interferometers are used in FT-IR optics; the mathematical conversion is still a Fourier transform, and NMR does not use a Michelson interferometer. Wavelet averaging or “none” cannot replace FT for standard NMR spectra. Common Pitfalls: Confusing hardware used in other spectroscopies with the mathematical transformation required; in NMR, FT is the universal route from time to frequency domain. Final Answer: Fourier transformation

Question 4

Microscopy methods – In which technique is “negative staining” classically used to enhance contrast for very small biological particles?

Accepted Answer

Introduction / Context: Negative staining is a sample preparation method that enhances image contrast by surrounding the specimen with an electron-dense stain rather than staining the specimen itself. It is most strongly associated with transmission electron microscopy (TEM) for visualizing viruses, small protein complexes, and thin filaments at nanometer-scale resolution. Given Data / Assumptions: Common negative stains include uranyl acetate, phosphotungstic acid, and ammonium molybdate. The stain penetrates spaces around the specimen but not its interior, rendering the object light against a dark background. Samples are typically adsorbed onto carbon-coated grids. Concept / Approach: Because biological macromolecules scatter electrons weakly, surrounding them with a heavy-metal stain increases contrast. The stain forms a thin film; the specimen excludes the stain, appearing brighter. This quick-prep method provides shape and size information and is invaluable for rapid screening prior to higher-resolution cryo-EM work. Step-by-Step Solution: Place sample on a TEM grid and allow adsorption.Apply a drop of electron-dense stain; blot excess.Dry the grid; the stain forms a high-contrast background.Image under TEM; particles appear as bright silhouettes against dark stain. Verification / Alternative check: Compare identical specimens imaged with and without negative stain; the stained preparation reveals particles that are otherwise nearly invisible due to low intrinsic contrast. Why Other Options Are Wrong: Gel electrophoresis and standard light microscopy use different staining chemistries that color the specimen, not the background in an electron-dense manner. Immunocytochemistry emphasizes antigen–antibody localization; while immunogold EM exists, “negative staining” refers to TEM contrast, not immunofluorescence. AFM relies on mechanical probing rather than staining. Common Pitfalls: Confusing negative staining with dark-field light microscopy; despite similar visual outcomes (bright objects on dark), the physics and methods are distinct. Final Answer: Electron microscopy

Question 5

Scanning electron microscopy (SEM) – What is standard specimen preparation before imaging for conductive coating?

Accepted Answer

Introduction / Context: SEM images surfaces by scanning with a focused electron beam and detecting secondary or backscattered electrons. Biological specimens are inherently nonconductive and beam-sensitive. Standard preparation minimizes charging and damage by chemical fixation, dehydration, drying, and coating with a thin conductive metal film (e.g., gold, gold–palladium, platinum). Given Data / Assumptions: We consider conventional high-vacuum SEM of biological samples. Conductive coating provides a path for electrons to dissipate, reducing charging artifacts. Fixatives (e.g., glutaraldehyde) preserve ultrastructure prior to drying and coating. Concept / Approach: Charging distorts SEM images and reduces signal stability. Thin metal coatings (a few nanometers) improve conductivity and secondary electron yield, enhancing surface detail. While many coating metals are also transition metals, the key concept tested is the presence of a thin heavy-metal coating after proper fixation and drying, not the periodic table classification term itself. Step-by-Step Solution: Fix sample to stabilize structure.Dehydrate and critical-point dry or use alternative drying methods.Sputter-coat with a thin conductive metal layer.Image under SEM to obtain high-contrast, low-charging micrographs. Verification / Alternative check: Comparing uncoated versus coated samples shows dramatic reduction in charging artifacts and improved resolution, validating the necessity of metallic coating in standard protocols. Why Other Options Are Wrong: “Not fixed”: risks collapse and artifacts; fixation is standard. “Transition metal” wording is imprecise to the concept; the point is conductive heavy-metal coating after fixation. “None of the above” and uncoated room-pressure scanning do not describe conventional SEM workflows (except in variable-pressure or cryo-SEM special cases). Common Pitfalls: Confusing SEM with TEM preparation or assuming coating thickness should be large; overly thick coatings obscure fine surface details. Final Answer: A specimen is fixed and then coated with a thin layer of a heavy metal

Question 6

Spin physics – For a nucleus with spin quantum number I = 1/2 placed in an external magnetic field, how many distinct Zeeman energy orientations are possible?

Accepted Answer

Introduction / Context: Nuclear spin states split in a magnetic field into discrete energy levels. The number of allowed orientations (and hence energy levels) is determined by the spin quantum number I. This quantization underlies the basic two-level system exploited by NMR for many common nuclei such as 1H and 13C. Given Data / Assumptions: We consider a static magnetic field B0 and an isolated spin-1/2 nucleus. Magnetic quantum numbers mI take on values from +I to −I in integer steps. Selection rules allow transitions with ΔmI = ±1 under RF excitation. Concept / Approach: For I = 1/2, the allowed mI values are +1/2 and −1/2, yielding two Zeeman levels. The energy difference is proportional to the magnetic field and the gyromagnetic ratio (ΔE = ħ * γ * B0). This two-level system leads to a single resonance frequency at the Larmor frequency, broadened and shifted by chemical environment (chemical shift) and interactions (e.g., J-coupling). Step-by-Step Solution: Identify I = 1/2 → possible mI = +1/2, −1/2.Conclude there are exactly two orientations/levels.Recognize transitions between these levels produce the NMR signal.Extend to other spins (I > 1/2) to appreciate additional orientations and quadrupolar effects. Verification / Alternative check: NMR of 1H and 13C shows single-quantum transitions characteristic of two-level systems; contrast with quadrupolar nuclei (e.g., 14N, I = 1) that exhibit additional relaxation behavior due to more levels and electric quadrupole moments. Why Other Options Are Wrong: 3, 4, 6: correspond to spins I = 1 (three levels) or higher; not applicable to I = 1/2. 1: would imply no resonance; contradicts fundamental NMR behavior. Common Pitfalls: Assuming more lines automatically means more Zeeman levels; multiplets in 1H NMR arise from J-coupling, not additional intrinsic Zeeman states for I = 1/2 nuclei. Final Answer: 2

Question 7

FT-NMR excitation – In Fourier-transform NMR experiments, how are nuclear spins typically excited to generate a broadband free induction decay (FID)?

Accepted Answer

Introduction / Context: Fourier-transform NMR revolutionized spectroscopy by replacing slow frequency-sweep excitation with pulsed methods. A single short, intense radio-frequency pulse excites a broad swath of resonances simultaneously. The decaying signal captured afterward in the time domain (FID) is Fourier transformed to yield the frequency-domain spectrum containing all excited resonances at once. Given Data / Assumptions: The pulse duration determines excitation bandwidth (shorter pulses excite wider bandwidths). The spectrometer digitizes the FID at high sampling rates over a defined spectral width. Phase cycling and pulse sequences tailor coherence pathways for multidimensional NMR as needed. Concept / Approach: An intense, brief RF pulse generates transverse magnetization from the equilibrium longitudinal magnetization. Because the pulse has a broad frequency content, many chemical environments (chemical shifts) are excited at once. After the pulse, spins precess at their characteristic Larmor frequencies, producing the FID that encodes all resonance information for subsequent Fourier transformation. Step-by-Step Solution: Apply a broadband RF pulse (e.g., 90° pulse) to tip magnetization into the transverse plane.Record the FID as transverse magnetization decays.Fourier transform the FID to obtain the spectrum.Process (phase, baseline) to finalize peak positions and intensities. Verification / Alternative check: Compare with frequency-swept continuous-wave (CW) NMR: FT-NMR provides higher sensitivity via signal averaging (Fellgett advantage) and multiplex detection, confirming why pulsed, broadband excitation is standard. Why Other Options Are Wrong: Slow frequency sweeps describe CW NMR, not modern FT-NMR. Mechanical pressure or rapid heating are unrelated to coherent RF excitation of nuclear spins. “None” is incorrect because the pulse method is well established. Common Pitfalls: Equating “wide range of frequencies” with noise; the pulse is precisely timed and phase-controlled to create coherent excitation across the desired bandwidth. Final Answer: By a short, intense RF pulse that contains a wide range of frequencies

Question 8

Immunofluorescence microscopy — where are the fluorescent tags attached?  In cell biology and diagnostic microscopy, immunofluorescence relies on fluorescent compounds (fluorophores). These labels are covalently attached to which component in order …

Accepted Answer

Introduction / Context: Immunofluorescence is a cornerstone technique in cell and molecular biology for localizing proteins, pathogens, and organelles. The method uses antibodies coupled to fluorescent dyes (fluorophores) to target antigens with high specificity. Understanding exactly which molecule is labeled is key to interpreting images correctly and avoiding false conclusions in experimental design or diagnostics. Given Data / Assumptions: The goal is to visualize a defined subcellular target (antigen). Fluorescent compounds can be chemically linked to proteins. Antibody–antigen recognition provides specificity in situ. Concept / Approach: In direct immunofluorescence, the primary antibody carrying a covalently attached fluorophore binds the antigen. In indirect immunofluorescence, an unlabeled primary antibody binds the antigen first; then a fluorophore-labeled secondary antibody binds the primary antibody, amplifying signal. In both cases, the fluorescent tag is on an antibody species, not the antigen itself, which is native to the cell or tissue and typically not modified. Labelling the antibody preserves antigen structure and allows flexible multiplexing with different dyes. Step-by-Step Solution: Define the target: the antigen is the cellular molecule of interest.Select the binder: antibodies recognize the antigen with high affinity and specificity.Attach fluorophore: link dye to antibody (primary or secondary).Image acquisition: excite fluorophore and collect emission from the bound antibody at the target site. Verification / Alternative check: Controls such as secondary-only incubations show minimal background, confirming that fluorescence arises from antibody-mediated binding, not intrinsic antigen fluorescence. Co-localization with known markers further validates targeting. Why Other Options Are Wrong: Antigen labeling (B) is rarely practical in fixed samples and would alter native structures. Pre-bound complexes (C) misstate the workflow; the complex forms during staining. Random proteins (E) do not confer specificity. “None of these” (D) ignores standard practice. Common Pitfalls: Using poorly validated antibodies or mismatched secondary antibodies, leading to off-target fluorescence; not quenching autofluorescence; or over-fixation masking epitopes. Final Answer: An antibody specific for the subcellular structure

Question 9

NMR spectroscopy — meaning of the rotating frame of reference  In nuclear magnetic resonance (NMR), what does the term “rotating frame of reference” mean when describing magnetization dynamics at the Larmor frequency?

Accepted Answer

Introduction / Context: The rotating frame is a conceptual tool in NMR used to simplify the description of precessing nuclear magnetization. By transforming from the laboratory (static) frame to a frame that rotates at or near the Larmor frequency, the apparent motion of spins slows or stops, making radiofrequency (RF) pulse effects and relaxation easier to analyze. Given Data / Assumptions: Nuclear magnetic moments precess at the Larmor frequency around the static magnetic field B0. Analyses often track transverse magnetization behavior under RF fields B1 applied near resonance. A change of reference frame does not physically move hardware; it is a mathematical convenience. Concept / Approach: In the rotating frame, we mathematically rotate the coordinate system at the Larmor frequency (or the RF carrier frequency). In this frame, a resonant spin’s transverse magnetization appears stationary, and off-resonance effects are seen as slow nutations around an effective field. This simplifies Bloch equation solutions and pulse sequence design (e.g., 90° and 180° pulses, spin echoes, and decoupling). Step-by-Step Solution: Start with laboratory-frame precession at angular frequency omega0.Define a rotating coordinate system spinning at omegaRF ≈ omega0.Express magnetization vectors in the rotating frame; resonant spins become nearly static.Analyze RF interactions and relaxation with respect to the effective field. Verification / Alternative check: Bloch equations transform cleanly into the rotating frame, yielding intuitive nutation around B1 and straightforward visualization of phase evolution and off-resonance effects. Why Other Options Are Wrong: Physical spinning of samples (A) is MAS in solid-state NMR, not the rotating frame concept. Detectors and magnets do not rotate (C, E). “None of the above” (D) ignores the standard definition. Common Pitfalls: Confusing the rotating frame with mechanical spinning or assuming it changes measured frequencies rather than simplifying the mathematical description. Final Answer: Imagining the laboratory frame rotating at the Larmor frequency so that individual magnetic moment vectors appear stationary in space

Question 10

Determining protein primary structure — complementary strategies  Which of the following strategies are used together to determine the sequence of amino acids (primary structure) of a protein?

Accepted Answer

Introduction / Context: Protein sequencing integrates chemical and enzymatic methods to reconstruct the order of amino acids. Classic workflows combine end-group analysis, selective cleavage, chromatographic separation, and fragment sequencing (Edman degradation or tandem mass spectrometry). Understanding why multiple complementary steps are required ensures reliable reconstruction of the full sequence, including disulfide mapping. Given Data / Assumptions: Proteins are often too large to sequence intact by one method. Terminal residue identification constrains sequence assembly. Fragmentation at defined sites (e.g., trypsin, CNBr) generates overlapping peptides. Concept / Approach: Terminal analyses (N- and C-termini) provide anchors for assembling fragments. Controlled cleavage yields overlapping sets that can be aligned to deduce order. Modern tandem MS adds speed and sensitivity but still benefits from strategic proteolysis and knowledge of termini. Therefore, a comprehensive approach uses all listed strategies rather than relying on a single technique in isolation. Step-by-Step Solution: Determine N-terminus by derivatization and identification (e.g., FDNB or Edman).Determine C-terminus via carboxypeptidase digestion or specific tags.Generate fragments by enzymatic or chemical cleavage.Sequence fragments and align overlaps to reconstruct the full polypeptide. Verification / Alternative check: Concordance between independent fragment maps and termini assignments validates the final sequence; discrepancies trigger re-analysis for modifications or proteolysis artifacts. Why Other Options Are Wrong: Choosing only one technique risks ambiguities. Mass spectrometry alone (E) is powerful but typically paired with digestion and database/terminal constraints for de novo cases. Common Pitfalls: Ignoring post-translational modifications, disulfide bonds, or N-terminal blockage that require tailored strategies (e.g., reduction/alkylation, enzymatic deglycosylation). Final Answer: All of the above

Question 11

NMR terminology — comparing precession, transition, and Larmor frequencies  In magnetic resonance, the terms “frequency of precession,” “transition frequency,” and “Larmor frequency” are best regarded as:

Accepted Answer

Introduction / Context: Nuclear spins in a static magnetic field precess at the Larmor frequency, which also corresponds to the resonance (transition) frequency for absorption/emission of RF energy. Clarifying this equivalence helps students connect classical precession with quantum transitions in NMR and MRI. Given Data / Assumptions: The Larmor frequency is gamma * B0, where gamma is the gyromagnetic ratio and B0 the static field. Resonance occurs when the RF field is at this frequency, enabling transitions between Zeeman energy levels. Classical and quantum descriptions are consistent for spin dynamics. Concept / Approach: Classically, magnetization precesses about B0 at omega0 (Larmor). Quantum mechanically, the energy gap between spin states corresponds to a photon of frequency nu0 = omega0 / (2 * pi). Thus, the precession frequency and transition frequency are the same underlying quantity expressed in consistent frameworks. Different terms reflect perspective, not different values. Step-by-Step Solution: Relate classical precession to Larmor frequency (omega0 = gamma * B0).Relate quantum transition energy to frequency (DeltaE = h * nu0).Equate nu0 with omega0 / (2 * pi) at resonance.Conclude that all terms refer to the same frequency. Verification / Alternative check: Experimental resonance peaks occur exactly at the Larmor frequency computed from gamma and B0, validating equality of transition and precession frequencies. Why Other Options Are Wrong: Options C, D, and E incorrectly imply distinct quantities or no relationship. Option B is redundant wording; option A precisely states the equivalence. Common Pitfalls: Confusing angular frequency (omega) with linear frequency (nu); they differ by 2 * pi but describe the same resonance. Final Answer: Different terms for the same frequency

Question 12

FT-IR optics — functional role of the Michelson interferometer  In Fourier-transform infrared (FT-IR) spectroscopy, what is the primary function of the Michelson interferometer within the instrument?

Accepted Answer

Introduction / Context: FT-IR instruments differ from dispersive spectrometers by encoding spectral information in the time domain. The Michelson interferometer creates an interferogram by varying the optical path difference between two arms, translating high optical frequencies into a manageable modulation pattern that detectors can measure with high signal-to-noise ratio. Given Data / Assumptions: The IR source is broadband (polychromatic). Moving mirror changes path length continuously. Fourier transformation reconstructs the spectrum from the interferogram. Concept / Approach: The beam is split by a beamsplitter, reflected by fixed and moving mirrors, and recombined at the detector. As the moving mirror scans, constructive/destructive interference generates an intensity vs. time (or mirror position) signal known as the interferogram. After passing through the sample, the interferogram encodes the sample’s spectral absorptions, which are retrieved by a Fourier transform to yield the frequency-domain spectrum. Step-by-Step Solution: Split the source into two paths at the beamsplitter.Introduce a variable delay by moving one mirror.Recombine beams to form an interferogram at the detector.Apply a Fourier transform to obtain the spectrum. Verification / Alternative check: Interferogram features (central burst at zero path difference) and throughput advantages (Jacquinot/Fellgett) confirm modulation and multiplex benefits compared to dispersive devices. Why Other Options Are Wrong: Option A describes prisms/gratings, not interferometers. Option B is filtering, not modulation. Option E is an optical alignment function but not the defining role. “None of the above” ignores the central FT-IR principle. Common Pitfalls: Equating interferometers with wavelength splitters; forgetting that spectral resolution comes from maximum path difference, not slit width. Final Answer: Modulate the broadband IR signal into a low-frequency interferogram (time-domain signal) detectable by the sensor

Question 13

Spin-1/2 absorption in NMR — what changes upon RF excitation?  When a spin-1/2 nucleus in a magnetic field absorbs resonant radiation energy, which change best describes its magnetic moment relative to the applied field B0?

Accepted Answer

Introduction / Context: In NMR, nuclei with spin-1/2 (e.g., 1H, 13C, 19F) occupy one of two Zeeman energy levels in a static magnetic field B0: a lower-energy state aligned with the field and a higher-energy state opposed to it. Absorption of radiofrequency (RF) energy at the Larmor frequency induces transitions between these states and tips the net magnetization away from equilibrium. Given Data / Assumptions: B0 defines the quantization axis and causes Larmor precession. At thermal equilibrium, more spins populate the lower-energy (parallel) state. RF pulses at resonance transfer population to the higher-energy (anti-parallel) state and rotate the net magnetization. Concept / Approach: Absorption corresponds to promotion from the parallel to anti-parallel state for spin-1/2 nuclei. Classically, this appears as the magnetic moment vector tipping to oppose B0 (for a 180-degree inversion) or being rotated into the transverse plane (for a 90-degree pulse). The Larmor frequency itself is set by gamma and B0 and does not permanently increase; mechanical spin speed is not what changes—quantum state populations do. Step-by-Step Solution: Apply RF at the Larmor frequency to meet resonance conditions.Spins absorb energy and transition to the higher-energy orientation (anti-parallel).Macroscopic magnetization rotates according to pulse flip angle.Relaxation (T1/T2) returns the system toward equilibrium after excitation. Verification / Alternative check: Inversion recovery and saturation experiments demonstrate controllable population changes (and signal phase), consistent with flips between Zeeman levels rather than changes in B0 or intrinsic mechanical spin. Why Other Options Are Wrong: Option A confuses excitation with altered Larmor frequency. Option B misinterprets quantum transitions as changes in mechanical spin rate. Option E is physically incorrect; B0 is constant. “None” (D) is unnecessary because C is correct. Common Pitfalls: Equating precession frequency shifts with excitation; overlooking the distinction between state populations and classical analogies. Final Answer: The angle of precession flips so that the nuclear magnetic moment opposes the applied field (transition to the higher-energy state)

Question 14

FT-IR data domains — what does the instrument record?  An FT-IR spectrometer natively records the measured signal in which domain before applying a Fourier transform to obtain the spectrum?

Accepted Answer

Introduction / Context: Fourier-transform infrared (FT-IR) spectroscopy acquires an interferogram produced by a Michelson interferometer. This raw signal is a function of mirror position or time and must be mathematically transformed to reveal absorption intensity versus frequency (or wavenumber). Recognizing the native acquisition domain clarifies why digital processing is integral to FT-IR. Given Data / Assumptions: FT-IR uses a moving mirror to vary optical path difference continuously. The detector measures intensity changes as a function of time (or path difference). A computational Fourier transform converts the interferogram to the frequency (wavenumber) spectrum. Concept / Approach: The recorded interferogram encodes all spectral components through interference. Only after transforming from the time (or optical delay) domain does the familiar absorbance vs. wavenumber plot emerge. Therefore, the instrument records in the time domain and later presents results in the frequency/wavenumber domain. Step-by-Step Solution: Generate interferogram using a Michelson interferometer.Acquire detector signal vs. time or mirror displacement.Apply Fourier transform to compute spectral intensities vs. wavenumber.Display spectrum for interpretation and quantitative analysis. Verification / Alternative check: Central burst at zero path difference and apodization steps in software confirm time-domain acquisition. Resolution correlates with maximum optical path difference, not slit width as in dispersive instruments. Why Other Options Are Wrong: Option B confuses the processed domain with the recorded one. Option C is incorrect because acquisition and display occur sequentially, not simultaneously. Option E is a frequency-domain representation, obtained post-transform. “None” is unnecessary. Common Pitfalls: Assuming the detector directly measures absorbance vs. wavenumber; overlooking phase correction, apodization, and zero-filling steps in digital processing. Final Answer: Time domain

Question 15

NMR relaxation — naming the process from environmental magnetic field interactions  What is the name for the relaxation process driven by interactions between an excited nucleus and fluctuating magnetic fields from surrounding molecules (the “lattic…

Accepted Answer

Introduction / Context: NMR signals decay because excited nuclear magnetization returns to equilibrium through relaxation. Two principal mechanisms are T1 (spin–lattice) and T2 (spin–spin). Distinguishing them is pivotal for designing pulse sequences, interpreting contrasts in MRI, and understanding molecular dynamics that modulate local magnetic fields. Given Data / Assumptions: T1 processes exchange energy between spins and their surroundings (“lattice”). Fluctuating local fields at or near the Larmor frequency enable transitions. T2 processes dephase transverse magnetization via spin–spin interactions without necessarily exchanging energy with the lattice. Concept / Approach: The description specifically points to environmental magnetic fields produced by molecular motion that drive recovery of the longitudinal magnetization (Mz) back to equilibrium. This is the definition of spin–lattice relaxation (T1). Spin–spin relaxation (T2) concerns mutual interactions among spins that cause phase dispersion of Mxy and loss of coherence, not primarily energy exchange with the surroundings. Step-by-Step Solution: Identify the role of the lattice: a reservoir that absorbs/emits energy.Link fluctuating fields at the Larmor frequency to relaxation transitions.Map process to Mz recovery characterized by T1 constants.Select spin–lattice relaxation as the correct term. Verification / Alternative check: Inversion-recovery experiments measure T1 by tracking Mz recovery; results depend on molecular tumbling rates, confirming lattice-driven relaxation. Why Other Options Are Wrong: Spin–spin (B) is dephasing of transverse magnetization (T2), not energy exchange. Option C is not a standard term. “None” (D) is incorrect because T1 is well defined. Common Pitfalls: Assuming T2 processes restore longitudinal magnetization or conflating chemical exchange with T1 without considering spectral density at the Larmor frequency. Final Answer: Spin - lattice relaxation

Question 16

NMR signal averaging — why sum many FIDs?  In Fourier-transform NMR, why is it advantageous to acquire many free induction decays (FIDs) from the same sample and add them together before Fourier transforming?

Accepted Answer

Introduction / Context: FT-NMR relies on measuring weak time-domain signals (FIDs) that decay due to relaxation. Random electronic and thermal noise is uncorrelated from scan to scan, whereas the true NMR signal is coherent. Repeating the experiment and averaging the results is a universal strategy to enhance sensitivity without changing sample concentration or magnet strength. Given Data / Assumptions: Noisy FIDs contain both coherent signal and random noise. Successive scans are performed under identical conditions with adequate relaxation delays. Signal averaging follows root-N improvement for random noise. Concept / Approach: When N scans are summed, the coherent signal adds linearly (proportional to N), while uncorrelated noise adds as the square root of N. Therefore, the signal-to-noise ratio (SNR) improves approximately by sqrt(N), enabling detection of weak resonances and better quantitative integration. This is fundamental to multidimensional NMR and trace analyte detection. Step-by-Step Solution: Acquire FID #1 with defined pulse sequence and recycle delay.Repeat acquisition N times under identical conditions.Sum the FIDs point-by-point to form an averaged interferogram.Fourier transform the averaged FID to obtain a spectrum with improved SNR. Verification / Alternative check: Plotting SNR versus number of scans reveals near sqrt(N) dependence until systematic noise or drift dominates. This empirical behavior underpins standard NMR acquisition parameters (NS). Why Other Options Are Wrong: Not every nucleus must be excited individually (A); RF pulses excite ensembles. Field fluctuations (B) are managed by shimming and lock systems; averaging does not primarily fix B0 drift. T1 (E) is an intrinsic property and is not altered by averaging. Common Pitfalls: Using too short a recycle delay so spins do not fully relax, causing saturation and non-linear signal growth; ignoring drift that can limit averaging benefits. Final Answer: To increase sensitivity by improving the signal-to-noise ratio through signal averaging

FT IR Spectroscopy Questions