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GATE Life Sciences Chemistry of Biomolecules Flashcards
59 question-and-answer cards covering Chemistry of Biomolecules as it is examined in GATE Life Sciences. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Chemistry of Biomolecules deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Define epimers and give a hexose example.
Epimers are diastereomers that differ in configuration at a single chiral carbon. For example, D-glucose and D-galactose are C-4 epimers; D-glucose and D-mannose are C-2 epimers.
What makes a sugar a 'reducing sugar', and which test detects it?
A reducing sugar has a free anomeric carbon (free aldehyde or ketone group capable of being oxidized). It gives a positive Benedict's, Fehling's, or Tollens' test; e.g. glucose and fructose are reducing sugars.
How many stereoisomers are possible for an aldohexose, and why?
An aldohexose has 4 chiral centers (C-2, C-3, C-4, C-5), so it has $2^{4} = 16$ stereoisomers (8 D-forms and 8 L-forms).
What is a triglyceride (triacylglycerol) chemically?
A triglyceride is an ester of glycerol (a trihydric alcohol) with three fatty acid molecules, formed by esterification of the three hydroxyl groups; it is the main storage form of fat.
Write the general reaction of triglyceride formation.
$$\ce{Glycerol + 3\,Fatty\ acids -> Triglyceride + 3\,H2O}$$ Each fatty acid carboxyl condenses with a glycerol hydroxyl to form an ester bond, releasing water.
What is saponification of a triglyceride?
Saponification is the alkaline (e.g. $\ce{NaOH}$/$\ce{KOH}$) hydrolysis of a triglyceride yielding glycerol and the salts of the fatty acids (soaps).
Define the saponification number of a fat or oil.
The saponification number is the number of milligrams of $\ce{KOH}$ required to saponify 1 gram of fat or oil; it is inversely related to the average molecular weight (chain length) of the fatty acids.
Define the iodine number (iodine value) of a fat.
The iodine number is the number of grams of iodine ($\ce{I2}$) that can be added to 100 grams of fat; it measures the degree of unsaturation (number of C=C double bonds) — higher value means more unsaturation.
Why do unsaturated triglycerides (oils) have lower melting points than saturated ones (fats)?
Cis double bonds introduce kinks that prevent tight packing of the fatty acid chains, weakening van der Waals interactions and lowering the melting point, so unsaturated triglycerides are liquid (oils) at room temperature while saturated ones are solid (fats).
On what property does ion-exchange chromatography separate molecules?
It separates molecules based on their net surface charge — by reversible electrostatic interaction between charged groups on the analyte and oppositely charged groups fixed on the stationary phase (resin).
Distinguish between a cation exchanger and an anion exchanger.
A cation exchanger carries negatively charged groups (e.g. $\ce{-SO3^-}$, carboxymethyl/CM) and binds positively charged (cationic) molecules. An anion exchanger carries positively charged groups (e.g. $\ce{-N(CH3)3^+}$, DEAE) and binds negatively charged (anionic) molecules.
How are bound molecules typically eluted in ion-exchange chromatography?
By increasing the salt concentration (ionic strength) of the buffer, which competes for the charged binding sites, or by changing the pH to alter the net charge of the bound molecules — usually applied as a gradient.
On what basis does gel filtration (size-exclusion) chromatography separate molecules?
It separates molecules according to their size (and shape / hydrodynamic radius / molecular weight), not charge — using a porous gel matrix that excludes large molecules and retains small ones.
In gel filtration, which molecules elute first and why?
Large molecules elute first because they are too big to enter the pores of the beads and pass through the void volume (the spaces between beads), traveling a shorter path; small molecules enter the pores, take a longer path, and elute later.
What do the void volume ($V_0$) and total volume ($V_t$) represent in gel filtration?
$V_0$ (void volume) is the volume of solvent between the gel beads, at which fully excluded (largest) molecules elute. $V_t$ is the total column volume; molecules small enough to fully penetrate all pores elute near $V_t$. The elution volume $V_e$ of a molecule depends on its size, with $V_0 < V_e < V_t$.
Name common matrix materials used in gel filtration chromatography.
Cross-linked dextran (Sephadex), agarose (Sepharose), and polyacrylamide (Bio-Gel P) beads of defined pore size.
State the Beer–Lambert law in equation form and define each term.
$$A = \varepsilon \, c \, l$$ where $A$ is absorbance, $\varepsilon$ is the molar absorptivity (extinction coefficient, $\text{M}^{-1}\text{cm}^{-1}$), $c$ is the molar concentration, and $l$ is the path length of the cuvette (cm).
How is absorbance ($A$) related to transmittance ($T$) and to incident/transmitted light intensity?
$$A = -\log_{10}(T) = \log_{10}\!\left(\frac{I_0}{I}\right)$$ where $I_0$ is the incident light intensity and $I$ is the transmitted intensity; transmittance $T = \frac{I}{I_0}$.
According to the Beer–Lambert law, what is the relationship between absorbance and concentration, and why is it useful?
Absorbance is directly proportional to concentration ($A \propto c$) at fixed path length and wavelength. This linear relationship allows the unknown concentration of a colored or UV-absorbing solute to be determined spectrophotometrically from a calibration (standard) curve.
Under what conditions does the Beer–Lambert law deviate from linearity?
At high concentrations (solute–solute interactions, refractive index changes), with non-monochromatic light, due to scattering by turbid/particulate samples, chemical changes (association/dissociation, fluorescence), or stray light — causing the $A$ vs $c$ plot to deviate from a straight line.
At what wavelength do proteins typically absorb in the UV, and which residues are responsible?
Proteins absorb maximally around $280\ \text{nm}$ due to the aromatic side chains of tryptophan and tyrosine (and weakly phenylalanine); the peptide bond itself absorbs around $190$–$220\ \text{nm}$.
At what wavelength do nucleic acids (DNA/RNA) absorb maximally, and why?
Nucleic acids absorb maximally at $260\ \text{nm}$ due to the aromatic purine and pyrimidine bases; the $A_{260}/A_{280}$ ratio is used to assess DNA/protein purity (pure DNA $\approx 1.8$).
What is the role of piperidine in Maxam–Gilbert sequencing?
After base-specific chemical modification, hot piperidine catalyzes cleavage of the DNA backbone at the modified (and thereby displaced) bases, producing the labeled fragments that are then size-separated.
Why is one peptide-bond plane considered rigid and planar?
The peptide ($\ce{C-N}$) bond has partial double-bond character due to resonance delocalization of the carbonyl/amide electrons, restricting rotation and keeping the six atoms of the peptide group ($\ce{C_\alpha, C, O, N, H, C_\alpha}$) coplanar; the trans configuration is generally favored.
What this deck covers
The Chemistry of Biomolecules deck follows the GATE Life Sciences Chemistry of Biomolecules syllabus — 6 chapters and 7 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 9.8 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 212 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Chemistry of Biomolecules flashcards FAQ
How many Chemistry of Biomolecules flashcards are in this GATE Life Sciences deck?
59 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these GATE Life Sciences flashcards free?
Yes. The preview here is free to read with no signup, and the full 59-card deck is free inside the Examius app.
What do the Chemistry of Biomolecules cards cover?
They follow the GATE Life Sciences Chemistry of Biomolecules syllabus — 6 chapters and 7 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.