CSIR-NET Life Sciences Unit 13 PYQs: Methods in Biology Questions with Solutions

NMR detects transitions between nuclear spin states in a strong magnetic field. The energy gap is tiny, so the resonance frequencies are in the radio-frequency range, hundreds of MHz (for example 600 MHz for ¹H on a 14.1 T magnet). Microwaves are used in ESR/EPR, which detects electron spins.

Lateral resolution follows the Abbe/Rayleigh limit, d = 0.61λ/NA. Smaller is better, so you want the shortest wavelength and the highest numerical aperture. Magnification does not set resolution. 480 nm with NA 1.4 gives d ≈ 209 nm, better than the other three combinations.

Answer: (2) Intensity of light

Resolving power depends on the wavelength (the colour of light), the half-angle of light the objective accepts, and the refractive index of the medium. The last two together make up the numerical aperture. The intensity of light changes brightness and contrast, but not the smallest distance that can be resolved.

Answer: (2) dSTORM – stochastic optical reconstruction microscopy

dSTORM, like PALM, depends on single-molecule localisation. Photoswitchable or photoactivatable fluorophores are made to blink so that only a few, well-separated molecules are on in each frame. Their centres are found to nanometre precision and built up into an image. STED and SIM reach super-resolution by shaping the illumination and work with ordinary dyes.

Answer: (3) TIRF microscope

In TIRF microscopy, light totally internally reflected at the coverslip produces an evanescent field that reaches only about 100 nm into the sample. Out-of-focus fluorescence from the rest of the cell is never excited, so the background is low enough to see single fluorophores near the membrane.

Answer: (2) Visible and Infrared

Oxyhaemoglobin and deoxyhaemoglobin absorb differently in visible red light (about 660 nm), where deoxyhaemoglobin absorbs more, and in near-infrared (about 940 nm), where oxyhaemoglobin absorbs more. The oximeter compares the pulsatile part of the absorbance at the two wavelengths to work out saturation.

Answer: (2) hc(Y−X) / XY

Photon energy is E = hc/λ. Light is absorbed at X and emitted at the longer wavelength Y, so the energy lost in between (the Stokes shift, passed to the solvent) is hc/X − hc/Y = hc(Y − X)/XY.

Answer: (1) Mass spectrometry

Acetylation adds exactly 42.011 Da to a lysine. Mass spectrometry, usually tandem MS of a digest, detects that mass shift and finds the modified residue from the fragment ions. (It is told apart from trimethylation, +42.047 Da, by high mass accuracy.) SDS-PAGE, light scattering and CD cannot see a 42 Da change at a single site.

Answer: (2) sucrose

Rate-zonal centrifugation separates particles by how fast they sediment through a shallow, pre-formed density gradient, usually sucrose (for example 5–20%). The gradient stabilises the bands against convection. The run is stopped before anything reaches equilibrium. Isopycnic separations use CsCl instead.

Answer: (1) L mol⁻¹cm⁻¹

From the Beer–Lambert law, A = εcl. Absorbance has no units, so ε = A/(c × l), with c in mol L⁻¹ and l in cm. That gives L mol⁻¹ cm⁻¹ (M⁻¹ cm⁻¹).

Answer: (4) Passive agglutination

In the classical comparison of immunoassay sensitivities (Kuby), passive agglutination, with antigen-coated red cells or latex beads, detects antibody at about 0.006–0.06 µg/mL. Immunofluorescence detects about 1 µg/mL, rocket electrophoresis about 2, and immunoelectrophoresis needs 20–200. Among these options, passive agglutination is the most sensitive.

Answer: (3) RAPD (Randomly amplified polymorphic DNA)

RAPD amplifies genomic DNA with a single short (about 10-mer) arbitrary primer. Wherever two copies of it bind close together on opposite strands, a band forms, so polymorphisms show up as bands that are present or absent. SSR and SCAR use specific primer pairs, and AFLP uses adaptor-specific primers after restriction digestion and ligation.

Answer: (3) Unpaired t-test

A parametric test assumes a distribution, usually normal, and compares its parameters, such as means. The unpaired t-test is parametric. Kruskal–Wallis and the Wilcoxon signed-rank test work on ranks, and Fisher’s exact test works on counts in a contingency table, so all three are non-parametric.

Answer: (2) They are a way of measuring the precision of the estimate of the mean.

A 95% confidence interval measures the precision of the estimated mean. If sampling were repeated many times, 95% of intervals built this way would contain the true population mean. It does not describe where 95% of individual observations fall (that is spread, measured by the SD). It also says nothing about where the sample mean falls, since the interval is centred on it.

Answer: (2) y = 3x + 2

Every step of 2 in x raises y by 6, so the slope is 6/2 = 3. At x = 0, y = 2, which is the intercept. All the points lie exactly on y = 3x + 2 (check: x = 14 gives 44).

Answer: (3) 160 ± 20 mg/dL of blood cholesterol concentration in human athletes

For a variable that cannot go negative, a normal distribution is plausible only if the mean lies several SDs above zero. For 160 ± 20 the mean is 8 SD from zero, so the data can be symmetric. In the other three the SD is large relative to the mean (190/296, 45/58, 8/4), so the data must be skewed to the right.