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Question 1Biotechnology

In recombinant DNA work, a plasmid vector carrying the ampicillin-resistance gene allows selection of transformed Escherichia coli because the gene product detoxifies the antibiotic. Which property of the vector is being exploited here?

  1. Selectable marker Correct answer.
  2. Origin of replication
  3. Promoter element
  4. Multiple cloning site

Explanation

A selectable marker is a gene whose product lets transformed cells survive a selective pressure, here ampicillin detoxification. The other listed elements serve cloning, replication, or transcription rather than survival-based selection.

Question 2Biotechnology

During PCR, Taq DNA polymerase extends primers at approximately 72 degrees Celsius. Which primer feature most directly determines the annealing temperature set on the thermal cycler?

  1. The length and GC content of the primer Correct answer.
  2. The number of extension cycles programmed
  3. The proofreading activity of the polymerase
  4. The concentration of dNTPs in the buffer

Explanation

Annealing temperature is estimated from primer melting temperature, which rises with primer length and GC content. dNTP levels and cycle number affect yield, and proofreading is a polymerase property, not a primer property.

Question 3Biotechnology

A researcher needs to separate proteins ranging from 20 to 150 kDa under denaturing conditions. Which technique is most appropriate?

  1. SDS-PAGE Correct answer.
  2. Isoelectric focusing only
  3. Southern blotting
  4. Agarose gel electrophoresis

Explanation

SDS-PAGE denatures proteins and separates them mainly by molecular mass, covering the 20 to 150 kDa range well. Agarose gels are used for nucleic acids, isoelectric focusing separates by charge, and Southern blotting detects DNA.

Question 4Biotechnology

Which enzyme catalyzes the addition of nucleotides to a growing DNA strand in the 5-prime to 3-prime direction during replication?

  1. DNA polymerase Correct answer.
  2. Helicase
  3. Topoisomerase
  4. DNA ligase

Explanation

DNA polymerase adds deoxynucleotides to the 3-prime hydroxyl end, extending the strand in the 5-prime to 3-prime direction. Ligase seals nicks, helicase unwinds duplex DNA, and topoisomerase relieves supercoiling.

Question 5Biotechnology

In a batch fermentation, the specific growth rate of a culture equals the dilution rate only under which condition?

  1. Continuous culture at steady state Correct answer.
  2. Fed-batch culture with a constant feed
  3. Batch culture during exponential phase
  4. Stationary phase of any culture type

Explanation

In continuous culture at steady state, biomass concentration is constant so the specific growth rate equals the dilution rate. Batch and fed-batch systems have no constant dilution, and stationary phase has net growth of zero.

Question 6Biotechnology

A monoclonal antibody is produced by hybridoma technology. Which two cell properties are combined in a hybridoma?

  1. Cytotoxicity from a T lymphocyte and antibody secretion from a plasma cell
  2. Antibody secretion from a B lymphocyte and immortality from a myeloma cell Correct answer.
  3. Phagocytosis from a macrophage and immortality from a myeloma cell
  4. Antibody secretion from a B lymphocyte and motility from a fibroblast

Explanation

Hybridomas fuse an antibody-secreting B lymphocyte with an immortal myeloma cell, giving continuous secretion of a single antibody. The other combinations do not reflect the standard fusion partners.

Question 7Biotechnology

A gene of interest is inserted into a plasmid in the wrong orientation relative to the promoter. Which outcome is most likely when the construct is expressed in E. coli?

  1. An antisense transcript is produced instead of the intended protein Correct answer.
  2. The plasmid fails to replicate because the origin is disrupted
  3. Antibiotic selection becomes impossible
  4. Restriction digestion of the plasmid cannot occur

Explanation

Orientation determines which strand is transcribed, so a reversed insert yields an antisense transcript rather than the intended coding mRNA. Replication, selection, and restriction sites are unaffected by insert orientation.

Question 8Biotechnology

During enzyme kinetics, a compound raises the apparent Michaelis constant without changing the maximum velocity. Which inhibition pattern does this describe?

  1. Irreversible inactivation
  2. Competitive inhibition Correct answer.
  3. Uncompetitive inhibition
  4. Non-competitive inhibition

Explanation

Competitive inhibitors compete with substrate at the active site, raising the apparent Michaelis constant while maximum velocity is unchanged at saturating substrate. Non-competitive and uncompetitive patterns lower maximum velocity.

Question 9Biotechnology

Which statement correctly describes the role of a chaperone protein in a bacterial cell?

  1. It cleaves signal peptides from secreted proteins
  2. It adds ubiquitin tags to misfolded proteins
  3. It assists correct folding of nascent polypeptides without becoming part of the final protein Correct answer.
  4. It synthesizes the polypeptide chain on the ribosome

Explanation

Molecular chaperones help polypeptides fold correctly and prevent aggregation, and they are not retained in the mature protein. Signal peptidase, ubiquitin ligases, and the ribosome perform the other listed functions.

Question 10Biotechnology

A PCR reaction produces no band, and the primer melting temperatures were verified as appropriate. Which troubleshooting step is most reasonable to test next?

  1. Increase the annealing temperature by several degrees
  2. Decrease the annealing temperature to allow primer binding Correct answer.
  3. Add more template DNA while keeping all other parameters fixed
  4. Remove magnesium ions from the reaction buffer

Explanation

Absence of product with acceptable primers often reflects annealing that is too stringent, so lowering the annealing temperature can restore primer binding. Raising it would worsen the problem, and magnesium is required for polymerase activity.

Question 11Biotechnology

In a fed-batch culture, glucose is added slowly to avoid overflow metabolism. Which parameter is most directly being controlled by this feeding strategy?

  1. The dissolved oxygen level via agitation speed
  2. The specific growth rate via substrate availability Correct answer.
  3. The pH of the medium via buffering capacity
  4. The plasmid copy number via temperature

Explanation

Slow feeding limits residual glucose so the specific growth rate stays below the threshold where overflow metabolites such as acetate form. pH, oxygen, and copy number are controlled by other operating variables.

Question 12Biotechnology

A downstream purification step uses immobilized metal affinity chromatography with nickel ions. Which protein feature is exploited for binding?

  1. A disulfide-rich domain
  2. A hydrophobic transmembrane segment
  3. A high isoelectric point
  4. A polyhistidine tag Correct answer.

Explanation

Polyhistidine tags coordinate nickel ions on the resin, allowing selective capture and imidazole elution. Disulfides, charge, and hydrophobicity are exploited by other chromatography modes.

Question 13Biotechnology

A culture is grown in a chemostat at a dilution rate approaching the maximum specific growth rate of the organism. What is the most likely consequence?

  1. Washout of cells because growth cannot keep pace with dilution Correct answer.
  2. Complete loss of plasmid selection pressure
  3. Increased biomass concentration because growth is maximal
  4. A shift to anaerobic metabolism regardless of oxygen supply

Explanation

When the dilution rate approaches or exceeds the maximum specific growth rate, cells are removed faster than they can divide, causing washout. Biomass does not increase under these conditions, and oxygen status is independent of the dilution rate.

Question 14Biotechnology

A restriction enzyme recognizes a palindromic six-base sequence and leaves a four-base 5-prime overhang. After digestion, two DNA fragments with compatible overhangs are mixed with DNA ligase. Which condition is essential for covalent joining of the fragments?

  1. The overhangs must be blunt and both fragments must lack phosphate groups
  2. The overhangs must be complementary and the fragments must have a 3-prime hydroxyl and 5-prime phosphate Correct answer.
  3. The fragments must be single-stranded at both ends
  4. The fragments must be methylated at the recognition site

Explanation

Ligase needs a 3-prime hydroxyl and a 5-prime phosphate to form a phosphodiester bond, and cohesive ends must be complementary to anneal. Methylation protects from cleavage, and single-stranded ends do not provide the required duplex substrate.

Question 15Biotechnology

An investigator observes that a recombinant protein is produced as insoluble inclusion bodies in E. coli. Which change is most likely to improve soluble yield?

  1. Removing the antibiotic from the medium during expression
  2. Lowering the induction temperature and reducing inducer concentration Correct answer.
  3. Replacing the origin of replication with a higher-copy variant
  4. Increasing the inducer concentration to maximum

Explanation

Slower expression at reduced temperature and lower inducer levels gives nascent chains more time to fold, reducing inclusion body formation. Higher inducer or copy number typically worsens aggregation, and removing antibiotic risks plasmid loss.

Question 16Biotechnology

Which molecule serves as the immediate energy currency coupling exergonic reactions to endergonic biosynthetic steps in the cell?

  1. Pyruvate
  2. Glucose
  3. NADH
  4. ATP Correct answer.

Explanation

ATP hydrolysis provides the free energy that drives many endergonic reactions, acting as the cell's immediate energy currency. NADH and glucose are energy carriers or fuels, and pyruvate is a metabolic intermediate.

Question 17Biotechnology

A bacterial strain is engineered to overproduce an amino acid. To maximize yield, which approach is most consistent with metabolic engineering principles?

  1. Deleting the target amino acid transporter only
  2. Increasing the supply of the limiting precursor while retaining feedback control
  3. Growing the cells at the maximum possible growth rate
  4. Relieving feedback inhibition and removing competing pathways Correct answer.

Explanation

Overproduction requires relieving feedback regulation and redirecting flux away from competing branches toward the product. Retaining feedback control or maximizing growth alone does not favor product accumulation.

Question 18Biotechnology

Which technique is most suitable for quantifying the expression level of a specific mRNA in a sample?

  1. Western blotting
  2. Gram staining
  3. Quantitative reverse-transcription PCR Correct answer.
  4. Coomasie-stained SDS-PAGE

Explanation

Quantitative reverse-transcription PCR converts mRNA to cDNA and measures amplification in real time, giving transcript abundance. Western blotting measures protein, SDS-PAGE shows protein profiles, and Gram staining classifies bacteria.

Question 19Biotechnology

A plasmid carries a gene under an inducible promoter and a second gene conferring kanamycin resistance. After transformation, cells are plated on kanamycin medium. What does growth on this medium demonstrate?

  1. The cells took up an intact plasmid carrying the resistance gene Correct answer.
  2. The target gene has been correctly oriented in the plasmid
  3. The cells express the inducible target gene at high level
  4. The cells have lost their chromosomal DNA

Explanation

Growth on kanamycin shows the resistance gene is present and expressed, confirming plasmid uptake. It does not report target gene expression level, chromosomal integrity, or insert orientation.

Question 20Medicinal Chemistry

Which physicochemical property is most directly improved by converting a carboxylic acid drug into its ethyl ester prodrug?

  1. Oral absorption by passive diffusion Correct answer.
  2. Plasma protein binding selectivity
  3. Aqueous solubility
  4. Metabolic stability toward esterases

Explanation

Esterification masks the ionizable carboxyl group, raising the partition coefficient and reducing hydrogen-bonding capacity, which favors passive transcellular permeation across the gut wall. Aqueous solubility usually falls rather than rises after this change. The ester is designed to be hydrolyzed back to the active acid after absorption.

Question 21Medicinal Chemistry

In the Lipinski rule of five, a molecule is flagged as poorly absorbed when its molecular weight exceeds which threshold?

  1. 1000 Da
  2. 500 Da Correct answer.
  3. 300 Da
  4. 750 Da

Explanation

Lipinski's rule of five sets the molecular weight cutoff at 500 Da, alongside limits of logP below 5, hydrogen-bond donors of 10 or fewer, and hydrogen-bond acceptors of 10 or fewer. Violation of two or more parameters predicts poor oral absorption. The other listed values do not correspond to any rule-of-five parameter.

Question 22Medicinal Chemistry

A competitive inhibitor of an enzyme raises the apparent Michaelis constant (Km) while leaving Vmax unchanged. Which kinetic signature confirms this mechanism?

  1. Parallel lines in a Lineweaver-Burk plot
  2. Nonlinear progress curves with no replot
  3. Intersecting lines on the y-axis
  4. Intersecting lines on the x-axis Correct answer.

Explanation

In a Lineweaver-Burk double-reciprocal plot, competitive inhibition gives lines that intersect on the x-axis because Vmax is unchanged while the apparent Km increases. Noncompetitive inhibition instead gives intersecting lines on the x-axis only when binding is purely noncompetitive with equal affinity for free and bound enzyme; the classic noncompetitive pattern intersects on the x-axis too, so the diagnostic here is the shared Vmax intercept. Uncompetitive inhibition produces parallel lines.

Question 23Medicinal Chemistry

Which structural feature is essential for the beta-lactam antibiotics to acylate bacterial transpeptidases?

  1. The fused thiazolidine ring
  2. The free carboxylate on the ring system
  3. The C-6 acylamino side chain
  4. The strained four-membered beta-lactam ring Correct answer.

Explanation

The beta-lactam carbonyl is highly electrophilic because of ring strain, allowing nucleophilic attack by the active-site serine of penicillin-binding proteins. Acylation of this serine blocks cross-linking of peptidoglycan. The other groups modulate spectrum, binding, and physicochemical properties but are not the reactive warhead itself.

Question 24Medicinal Chemistry

A lead compound shows potent enzyme inhibition in vitro but negligible cellular activity. Adding a basic amine improves potency in whole cells. Which explanation best accounts for this change?

  1. The amine covalently binds the target enzyme
  2. The protonated amine reduces membrane permeation
  3. The amine increases lysosomal trapping and intracellular accumulation Correct answer.
  4. The amine increases molecular weight above the Lipinski limit

Explanation

Weakly basic amines are protonated in acidic compartments such as lysosomes, where they become trapped and concentrate, raising intracellular exposure at the target. This ion trapping can convert a potent enzyme inhibitor into a cell-active compound. Covalent binding and simple molecular weight changes do not explain the gain in cellular potency.

Question 25Medicinal Chemistry

Which phase I metabolic reaction is catalyzed by cytochrome P450 enzymes?

  1. Sulfate conjugation of an alcohol
  2. Glucuronidation of a phenol
  3. Aromatic hydroxylation Correct answer.
  4. Acetylation of an arylamine

Explanation

Cytochrome P450 monooxygenases perform oxidation reactions, of which aromatic hydroxylation is a classic example, inserting oxygen and often creating a reactive intermediate. Glucuronidation and sulfation are phase II conjugations, and acetylation is a phase II transfer reaction. These conjugation pathways increase water solubility for excretion.

Question 26Medicinal Chemistry

Which structural modification is the standard strategy for blocking rapid glucuronidation of a phenolic hydroxyl in a drug candidate?

  1. Adding an electron-withdrawing nitro group para to the phenol
  2. Replacing the phenol with a methyl group to remove the conjugation site
  3. Converting the phenol to a methyl ether Correct answer.
  4. Increasing the phenol's pKa by adding an alkyl group

Explanation

Converting a phenol to a methyl ether removes the free hydroxyl that serves as the glucuronidation acceptor, thereby blocking that metabolic route while retaining similar steric bulk. The ether can still be demethylated in some cases, but the direct conjugation is prevented. Simply altering electronics or pKa does not eliminate the reactive hydroxyl.

Question 27Medicinal Chemistry

Which statement correctly describes the role of a bioisostere in medicinal chemistry?

  1. It is a prodrug moiety that is cleaved enzymatically after absorption
  2. It is a structural replacement that retains biological activity while improving a property Correct answer.
  3. It is a fluorescent tag used for imaging target engagement
  4. It is a covalent warhead that irreversibly inhibits the target

Explanation

A bioisostere is a group that mimics the steric and electronic properties of another group while conferring advantages such as improved stability, solubility, or reduced toxicity. Classic examples include replacing a hydrogen with fluorine or an ester with an oxadiazole. The definition centers on retained activity with modified properties.

Question 28Medicinal Chemistry

In a structure-activity relationship study, replacing a phenyl ring with a pyridine ring often produces which effect?

  1. Loss of aromaticity and planarity
  2. Increased lipophilicity and reduced hydrogen bonding
  3. Reduced lipophilicity and added a hydrogen-bond acceptor Correct answer.
  4. Increased molecular weight with no change in polarity

Explanation

The pyridine nitrogen lowers logP relative to benzene and introduces a hydrogen-bond acceptor, which can improve aqueous solubility and sometimes target interactions. Aromaticity and planarity are retained. This is a common tactic for tuning physicochemical properties without large steric change.

Question 29Medicinal Chemistry

Which parameter is defined as the negative logarithm of the acid dissociation constant and is used to predict the ionization state of a drug at physiological pH?

  1. Ki
  2. logD
  3. logP
  4. pKa Correct answer.

Explanation

pKa is the negative base-10 logarithm of the acid dissociation constant and determines the fraction of ionized versus unionized drug at a given pH via the Henderson-Hasselbalch relationship. Ionization in turn governs solubility, permeability, and protein binding. logP and logD describe lipophilicity, not acid strength.

Question 30Medicinal Chemistry

A compound has a pKa of 8.4 as a weak base. Approximately what fraction is unionized at pH 7.4?

  1. About 50 percent
  2. About 10 percent Correct answer.
  3. About 90 percent
  4. About 1 percent

Explanation

For a base, the Henderson-Hasselbalch relationship gives pH = pKa + log([base]/[acid]), so the ratio of unionized base to protonated acid is 10 raised to the power of (pH minus pKa). Here that is 10 to the power of negative 1.0, a ratio of about 0.1, meaning roughly 9 percent unionized, closest to 10 percent. This matters because only the unionized form diffuses readily across membranes.

Question 31Medicinal Chemistry

Which interaction is primarily responsible for the high binding affinity of a protonated amine to an aspartate side chain in a protein binding pocket?

  1. Salt bridge formation Correct answer.
  2. Hydrogen bonding to a carbon atom
  3. Hydrophobic collapse
  4. Pi-pi stacking

Explanation

A protonated amine and a deprotonated aspartate carry opposite charges, so they form an electrostatic salt bridge that contributes strongly to binding energy. This is a common anchoring interaction in aminergic and many other ligand-receptor complexes. Hydrophobic and aromatic interactions contribute but are not charge-pairing interactions.

Question 32Medicinal Chemistry

Which strategy best achieves selectivity for a kinase inhibitor among closely related kinases sharing a conserved ATP-binding site?

  1. Adding a reactive electrophile to the ATP site
  2. Reducing the compound's aqueous solubility
  3. Targeting non-conserved residues adjacent to the hinge region or an allosteric pocket Correct answer.
  4. Increasing the overall molecular weight

Explanation

The ATP pocket is highly conserved across kinases, so selectivity usually comes from exploiting differences in the gatekeeper residue, the hinge region, or by binding an allosteric site outside the ATP cleft. Allosteric inhibitors and covalent targeting of unique cysteines are common selectivity strategies. Molecular weight and solubility changes are not reliable selectivity levers.

Question 33Medicinal Chemistry

Which statement best explains why covalent inhibitors can achieve prolonged pharmacodynamic effect despite short plasma half-life?

  1. They form a stable bond with the target so recovery requires protein resynthesis Correct answer.
  2. They are stored in adipose tissue and slowly released
  3. They inhibit all metabolic enzymes nonspecifically
  4. They are renally reabsorbed and recirculated

Explanation

Irreversible or slowly reversible covalent inhibitors persist as a drug-target adduct, so the duration of effect depends on the turnover rate of the target protein rather than on plasma drug levels. This allows infrequent dosing even when the free drug clears quickly. The effect is target-specific and not explained by tissue storage or reabsorption.

Question 34Medicinal Chemistry

Which type of antagonist produces a parallel rightward shift of the agonist dose-response curve with no depression of maximal response?

  1. Inverse agonist
  2. Irreversible antagonist
  3. Noncompetitive antagonist
  4. Competitive reversible antagonist Correct answer.

Explanation

A competitive reversible antagonist shifts the agonist concentration-response curve to the right without lowering the maximum, because higher agonist concentrations can outcompete the antagonist. Noncompetitive and irreversible antagonists reduce the maximal response. An inverse agonist lowers constitutive activity rather than shifting agonist curves.

Question 35Medicinal Chemistry

A candidate drug shows high hepatic extraction ratio. Which consequence should be expected following oral administration?

  1. Increased half-life proportional to the extraction ratio
  2. Reduced clearance because hepatic blood flow limits uptake
  3. Extensive first-pass metabolism and low oral bioavailability Correct answer.
  4. High oral bioavailability because absorption is complete

Explanation

A high hepatic extraction ratio means the liver removes most of the drug presented to it, so absorbed drug is extensively metabolized before reaching systemic circulation. Oral bioavailability is therefore low and sensitive to changes in hepatic blood flow. Clearance is high, not reduced, and half-life tends to be short.

Question 36Medicinal Chemistry

Which functional group is most susceptible to hydrolysis in a drug molecule stored at high humidity?

  1. An aryl methyl ether
  2. A tertiary amine
  3. Aromatic carbon-fluorine bond
  4. An aliphatic ester Correct answer.

Explanation

Esters undergo acid- or base-catalyzed hydrolysis, and moisture accelerates this degradation, especially under elevated temperature. Ethers, tertiary amines, and carbon-fluorine bonds are comparatively resistant to hydrolysis. This is why ester prodrugs and ester-containing drugs require moisture-protective packaging.

Question 37Medicinal Chemistry

Which design feature would most likely increase the oral bioavailability of a poorly soluble, highly permeable BCS class II compound?

  1. Increasing the particle size to slow dissolution
  2. Adding a permanent positive charge to increase polarity
  3. Reducing the crystalline lattice energy through an amorphous solid dispersion Correct answer.
  4. Increasing the melting point of the crystalline form

Explanation

For BCS class II compounds, dissolution rate limits absorption, so improving solubility by reducing crystallinity in an amorphous solid dispersion raises the apparent concentration at the absorption site. Larger particles and higher melting points slow dissolution further. Adding a permanent charge would reduce permeability and worsen the problem.

Question 38Medicinal Chemistry

Which structural change would most predictably reduce a drug's volume of distribution by decreasing tissue binding?

  1. Adding a lipophilic trifluoromethyl group
  2. Reducing lipophilicity and removing a basic center Correct answer.
  3. Increasing the number of aromatic rings
  4. Adding a long aliphatic chain

Explanation

Extensive tissue distribution often arises from high lipophilicity and from basic amines that accumulate in acidic tissue compartments. Lowering logP and removing the basic center reduce tissue partitioning and can lower the apparent volume of distribution. Adding lipophilic groups generally increases distribution into tissues.

Question 39Natural Products

Which alkaloid is the principal tropane ester responsible for the anticholinergic activity of Atropa belladonna?

  1. Reserpine
  2. Ephedrine
  3. Berberine
  4. Hyoscyamine Correct answer.

Explanation

Atropa belladonna accumulates tropane esters, and the levorotatory ester hyoscyamine is the dominant naturally occurring antimuscarinic constituent, racemizing partly to atropine on isolation. Berberine is an isoquinoline alkaloid, reserpine an indole alkaloid, and ephedrine a phenylethylamine-type alkaloid, so none of these accounts for belladonna's anticholinergic action.

Question 40Natural Products

Digitoxin and digoxin both belong to which structural class of cardiac glycosides?

  1. Saponins
  2. Bufadienolides
  3. Cardenolides Correct answer.
  4. Anthraquinones

Explanation

Cardenolides are C23 steroids bearing a five-membered unsaturated lactone ring, which is the pattern found in Digitalis purpurea (digitoxin) and Digitalis lanata (digoxin). Bufadienolides carry a six-membered lactone, whereas saponins and anthraquinones are unrelated structural families.

Question 41Natural Products

Which class of plant secondary metabolites is defined by a C6-C3-C6 carbon skeleton?

  1. Flavonoids Correct answer.
  2. Monoterpenes
  3. Tannins
  4. Cyanogenic glycosides

Explanation

Flavonoids share a fifteen-carbon phenylbenzopyran skeleton built from two aromatic rings linked through a three-carbon bridge, conventionally written C6-C3-C6. Monoterpenes are C10 isoprenoids, condensed tannins are oligomeric flavan-3-ols, and cyanogenic glycosides are amino-acid-derived glycosides.

Question 42Natural Products

A crude drug gives a strong blue-black colour when a section is treated with dilute iodine solution. Which constituent does this test indicate?

  1. Fixed oil
  2. Lignin
  3. Calcium oxalate
  4. Starch Correct answer.

Explanation

The iodine test detects starch through formation of a blue-black starch-iodine complex, a standard histochemical check for starch grains in storage organs such as rhizomes and roots. Fixed oils, calcium oxalate crystals and lignified walls do not give this colour with iodine.

Question 43Natural Products

The alkaloid quinine, obtained from Cinchona bark, is a quinoline alkaloid that also carries which additional ring system?

  1. Purine
  2. Steroid
  3. Tropane
  4. Quinuclidine Correct answer.

Explanation

Quinine is classified as a quinoline alkaloid because its structure combines a quinoline nucleus with a quinuclidine bicyclic system, and this quinuclidine portion carries the basic tertiary nitrogen. Tropane, purine and steroid ring systems belong to other alkaloid classes.

Question 44Natural Products

Sennosides A and B, the purgative principles of senna, are best described as which type of compound?

  1. Dianthrone glycosides Correct answer.
  2. Flavonol glycosides
  3. Iridoid glycosides
  4. Coumarin glycosides

Explanation

Sennosides are dimeric anthrone glycosides in which two anthrone units are linked through a single bond, forming a dianthrone, and they are hydrolysed in the colon to release active anthrones. Coumarin, iridoid and flavonol glycosides do not produce this anthrone-based laxative action.

Question 45Natural Products

Which solvent system is most appropriate for the preliminary extraction of a non-polar plant constituent such as a sterol or a volatile oil from dried powdered herb?

  1. Aqueous ethanol
  2. n-Hexane Correct answer.
  3. Methanol
  4. Water

Explanation

Extraction follows the principle of like dissolving like, so a lipophilic constituent is best recovered with a non-polar solvent of low dielectric constant such as n-hexane. Water, methanol and aqueous ethanol are polar and preferentially dissolve sugars, glycosides and other hydrophilic compounds.

Question 46Natural Products

In the assay of raw Digitalis leaf by the biological method, the potency is conventionally expressed in terms of which reference standard?

  1. Milligrams of digitoxin per gram
  2. Percentage of total ash
  3. International Units calibrated against a standard digitalis powder Correct answer.
  4. Milliequivalents of potassium per litre

Explanation

Biological assays of digitalis preparations are expressed in International Units defined by comparison with an international standard digitalis powder, using a specified animal preparation. Chemical assay of individual glycosides, ash value and potassium content are separate quality measurements and do not define biological potency units.

Question 47Natural Products

A researcher needs an alkaloid-rich fraction from powdered belladonna root. Which extraction sequence best exploits the basic nature of the alkaloids?

  1. Acidify the powder, extract with water, then precipitate with ethanol
  2. Defat with petroleum ether, then extract with boiling water
  3. Alkalinize the powder, extract with chloroform, then back-extract into dilute acid Correct answer.
  4. Extract with hot water, then adsorb onto talc

Explanation

Tropane alkaloids occur largely as salts in the plant, so adding alkali liberates the free bases, which partition into an immiscible organic solvent such as chloroform. Back-extraction into dilute acid reconverts them to water-soluble salts, giving a purified alkaloidal fraction.

Question 48Natural Products

Which phytochemical test gives a positive result with the steroidal saponin-containing fraction of Dioscorea, producing a red-violet colour in the lower layer after acid hydrolysis?

  1. Ninhydrin test
  2. Dragendorff test
  3. Liebermann-Burchard reaction Correct answer.
  4. Molisch test

Explanation

The Liebermann-Burchard reaction uses acetic anhydride and concentrated sulfuric acid to detect steroidal and triterpenoidal aglycones, giving a red-violet or blue-green colour at the interface. Molisch detects carbohydrates, ninhydrin detects amino acids, and Dragendorff detects alkaloids.

Question 49Natural Products

Which structural feature of the cardiac glycoside molecule is primarily responsible for its inhibition of the sodium-potassium ATPase?

  1. The unsaturated lactone ring of the aglycone Correct answer.
  2. The acetyl group on the first sugar
  3. The C14 hydroxyl group alone
  4. The terminal glucose of the sugar chain

Explanation

The five-membered alpha,beta-unsaturated lactone ring of the cardenolide aglycone is essential for binding to and inhibiting the sodium-potassium ATPase, and saturation or opening of this ring abolishes activity. The sugar chain mainly governs solubility, pharmacokinetics and receptor affinity rather than being the primary pharmacophore.

Question 50Natural Products

The presence of a methylenedioxy bridge in the aryl ring of a natural product is most commonly confirmed by which spectroscopic observation?

  1. A bathochromic shift on adding sodium hydroxide in ultraviolet spectroscopy
  2. A molecular ion at m/z 18 above the base peak in mass spectrometry
  3. A carbonyl absorption near 1700 per centimetre in the infrared spectrum
  4. A two-proton singlet near 5.9-6.0 ppm in the proton NMR spectrum Correct answer.

Explanation

The two equivalent protons of a methylenedioxy group appear as a characteristic two-proton singlet around 5.9-6.0 ppm in proton NMR, which is diagnostic for this moiety. Carbonyl absorption, a molecular ion shift of eighteen mass units and alkali-induced ultraviolet shifts indicate other functional groups.