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Tamil Nadu HSE Second Year Biochemistry study materials
Tamil Nadu HSE Second Year Biochemistry study resources

Tamil Nadu HSE Second Year Biochemistry practice questions

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Question 1Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Glucose and fructose both have the molecular formula C6H12O6 but differ in the arrangement of atoms around the carbonyl group. Glucose is classified as an aldose while fructose is a ketose. Which statement correctly describes the carbonyl group present in each sugar?

  1. Glucose has an aldehyde group at C-6, while fructose has a keto group at C-1.
  2. Glucose has an aldehyde group at C-1, while fructose has a keto group at C-2. Correct answer.
  3. Glucose has a keto group at C-2, while fructose has an aldehyde group at C-1.
  4. Both sugars have an aldehyde group at C-1 and differ only in ring size.

Explanation

Glucose is an aldohexose; its carbonyl carbon is C-1 and carries an aldehyde function. Fructose is a ketohexose with the carbonyl at C-2, which is why it is described as a ketose. This difference in the carbonyl position is the structural basis for their classification as aldose and ketose.

Question 2Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Sucrose is a disaccharide formed from glucose and fructose. Which pair of statements correctly describes the glycosidic linkage in sucrose and its reducing behaviour?

  1. It is an alpha-1,6-glycosidic bond and sucrose is a reducing sugar.
  2. It is a beta-1,4-glycosidic bond and sucrose is a non-reducing sugar.
  3. It is an alpha-1,2-glycosidic bond and sucrose is a non-reducing sugar. Correct answer.
  4. It is an alpha-1,4-glycosidic bond and sucrose is a reducing sugar.

Explanation

In sucrose the anomeric carbons of glucose (C-1) and fructose (C-2) are both involved in the linkage, giving an alpha-1,2-glycosidic bond. Because no free anomeric hydroxyl remains, sucrose cannot open to expose an aldehyde or keto group, so it is non-reducing.

Question 3Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A student hydrolyses a polysaccharide completely and finds only D-glucose units joined by alpha-1,4 bonds with occasional alpha-1,6 branch points. Which polysaccharide best matches this description?

  1. Amylopectin Correct answer.
  2. Cellulose
  3. Glycogen
  4. Chitin

Explanation

Amylopectin is a plant storage polysaccharide of D-glucose with alpha-1,4 linear segments and alpha-1,6 branch points at intervals of roughly 24 to 30 residues. Cellulose uses beta-1,4 bonds and is unbranched, chitin contains N-acetylglucosamine, and glycogen is far more highly branched.

Question 4Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

In the cyclic forms of D-glucose, the anomeric carbon is the one that carries both the ring oxygen and a hydroxyl group. Which carbon atom is the anomeric carbon in D-glucopyranose?

  1. C-1 Correct answer.
  2. C-6
  3. C-4
  4. C-2

Explanation

When the open-chain aldehyde at C-1 of glucose reacts with the hydroxyl at C-5, a six-membered pyranose ring forms. The carbonyl carbon becomes the anomeric carbon, so in D-glucopyranose it is C-1, the carbon that determines alpha or beta configuration.

Question 5Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Two samples of glucose are dissolved separately in water and allowed to stand. The optical rotation of each solution slowly changes until both reach the same value. Which property of glucose does this observation illustrate?

  1. Racemisation
  2. Mutarotation-independent epimerism
  3. Mutarotation Correct answer.
  4. Enolisation

Explanation

Freshly dissolved alpha-D-glucose and beta-D-glucose each show a different initial rotation, but equilibration through the open-chain form produces the same alpha-beta mixture. The gradual change of rotation to a common equilibrium value is called mutarotation.

Question 6Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A fatty acid is described as an omega-3 fatty acid. Which structural feature defines this designation?

  1. The first double bond counted from the methyl end lies between carbon 3 and carbon 4. Correct answer.
  2. The chain has three hydroxyl groups attached to it.
  3. The chain contains exactly three double bonds.
  4. The first double bond counted from the carboxyl end lies between carbon 3 and carbon 4.

Explanation

The omega nomenclature counts carbon atoms from the methyl terminus, which is designated omega-1. An omega-3 fatty acid therefore has its first double bond starting at the third carbon from the methyl end, as in alpha-linolenic acid.

Question 7Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Triglycerides and phospholipids are both lipids, but they differ in their behaviour at an oil-water interface. Which structural difference explains why phospholipids form bilayers while triglycerides do not?

  1. Phospholipids have a hydrophilic phosphate-containing head group and two hydrophobic tails, whereas triglycerides are entirely hydrophobic. Correct answer.
  2. Triglycerides contain a phosphate group, whereas phospholipids contain glycerol.
  3. Triglycerides contain a charged head group, whereas phospholipids are neutral.
  4. Phospholipids contain three fatty acid chains, whereas triglycerides contain only two.

Explanation

A phospholipid is amphipathic: the phosphate and alcohol head is polar and the two fatty acyl tails are nonpolar, so the molecule self-assembles into bilayers in water. A triglyceride has glycerol esterified to three fatty acids with no polar head, making it wholly hydrophobic.

Question 8Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

The iodine value of a fat is a measure of its degree of unsaturation. A sample with a high iodine value compared with another sample of the same mass indicates which property?

  1. A greater number of carbon-carbon double bonds in its fatty acid chains. Correct answer.
  2. A greater number of ester bonds in the glycerol backbone.
  3. A longer average chain length of saturated fatty acids.
  4. A higher melting point and more solid consistency.

Explanation

Iodine adds across carbon-carbon double bonds, so the mass of iodine absorbed per 100 g of fat rises with the number of double bonds present. A high iodine value therefore reflects greater unsaturation, which is typical of oils rather than hard fats.

Question 9Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A triglyceride is subjected to alkaline hydrolysis and the products are analysed. Which set of products is obtained?

  1. Glycerol and the sodium or potassium salts of fatty acids Correct answer.
  2. Glycerol and free fatty acids only
  3. Glycerol, fatty acids and phosphoric acid
  4. Fatty acids and glucose

Explanation

Alkaline hydrolysis of a triglyceride, called saponification, cleaves the three ester bonds and releases glycerol plus the salts of the fatty acids, which are soaps. Free fatty acids appear only if the hydrolysis is carried out with water under acidic conditions.

Question 10Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Amino acids are amphoteric because they contain both an acidic and a basic group. At a pH equal to the isoelectric point of a neutral amino acid such as glycine, which species predominates in solution?

  1. The fully protonated cation with both groups protonated
  2. The zwitterion with a protonated amino group and a deprotonated carboxyl group Correct answer.
  3. The fully deprotonated anion with both groups ionised
  4. An uncharged molecule with neither group ionised

Explanation

At its isoelectric point a neutral amino acid exists mainly as the dipolar zwitterion, in which the carboxyl group has lost a proton and the amino group has gained one. The net charge is zero even though separate charged centres are present.

Question 11Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which amino acid is classified as a sulphur-containing amino acid and contains a thiol group in its side chain?

  1. Serine
  2. Cysteine Correct answer.
  3. Lysine
  4. Methionine

Explanation

Cysteine carries a sulfhydryl or thiol group in its side chain, which can be oxidised to form disulphide bonds between polypeptide chains. Methionine also contains sulphur but as a thioether in a side chain that cannot form disulphide bridges.

Question 12Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A protein loses its biological activity when heated to 80 degrees Celsius, although its primary sequence is unchanged. Which explanation best accounts for this loss of function?

  1. Disulphide bonds are the only bonds broken, leaving the rest of the structure intact.
  2. Non-covalent interactions maintaining the secondary and tertiary structure are disrupted, unfolding the protein. Correct answer.
  3. Peptide bonds are hydrolysed, breaking the polypeptide into short fragments.
  4. The amino acid sequence is rearranged into a new primary structure.

Explanation

Heating supplies kinetic energy that breaks the weak hydrogen bonds, ionic interactions and hydrophobic contacts holding the folded conformation together. The chain unfolds and the active site is lost, which is denaturation; the peptide bonds and primary sequence remain intact.

Question 13Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

In the alpha-helix of a protein, hydrogen bonds stabilise the structure. Between which atoms do these hydrogen bonds form?

  1. The carbonyl oxygen of one peptide bond and the amide nitrogen of a peptide bond four residues away Correct answer.
  2. The side chains of two cysteine residues
  3. The carbonyl oxygen and the amide nitrogen of the same peptide bond
  4. The side-chain carboxyl groups of adjacent acidic residues

Explanation

The alpha-helix is held by hydrogen bonds running roughly parallel to the helix axis between the C=O of one residue and the N-H of the residue four positions further along the chain. Side chains project outward and do not participate in this regular bonding pattern.

Question 14Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A polypeptide is treated with a reagent that breaks disulphide bonds but leaves peptide bonds intact. Which level of protein structure is most directly affected by this treatment?

  1. The sequence of amino acids
  2. The planarity of the peptide bond
  3. Primary structure
  4. Tertiary structure Correct answer.

Explanation

Disulphide bridges between cysteine residues are covalent cross-links that hold distant parts of a folded chain together, so they contribute to tertiary structure. Breaking them can unfold the protein without altering the linear order of residues.

Question 15Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which bond links two monosaccharide units together in a disaccharide such as maltose?

  1. A phosphodiester bond between two hydroxyl groups
  2. A peptide bond between the carbonyl and amino groups of the two sugars
  3. An ester bond between two carboxyl groups
  4. A glycosidic bond between the anomeric carbon of one sugar and a hydroxyl oxygen of the other Correct answer.

Explanation

In a disaccharide the anomeric carbon of one monosaccharide is joined through an oxygen atom to a hydroxyl group of the second sugar, forming a glycosidic bond. In maltose this is specifically an alpha-1,4 linkage between two glucose units.

Question 16Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A nucleotide consists of three components. Which combination correctly identifies them?

  1. A pentose sugar, a fatty acid and a phosphate group
  2. A purine, a pyrimidine and a phosphate group only
  3. A nitrogenous base, a pentose sugar and one or more phosphate groups Correct answer.
  4. A nitrogenous base, a hexose sugar and an amino acid

Explanation

A nucleotide is formed when a nitrogenous base is attached to a pentose sugar to give a nucleoside, which is then phosphorylated. The three components are therefore a base, a pentose such as ribose or deoxyribose, and phosphate.

Question 17Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

In a double-stranded DNA molecule, adenine pairs with thymine and guanine pairs with cytosine. If a DNA sample contains 30 percent adenine, what percentage of its bases is guanine?

  1. 40 percent
  2. 70 percent
  3. 20 percent Correct answer.
  4. 30 percent

Explanation

Chargaff's rules require that adenine equals thymine and guanine equals cytosine. If adenine is 30 percent, thymine is also 30 percent, leaving 40 percent for guanine plus cytosine; halving this gives 20 percent guanine.

Question 18Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which structural feature distinguishes RNA from DNA in the sugar-phosphate backbone?

  1. RNA lacks phosphate groups in its backbone.
  2. RNA contains deoxyribose, whereas DNA contains ribose.
  3. RNA contains ribose with a 2-prime hydroxyl group, whereas DNA contains deoxyribose lacking that hydroxyl. Correct answer.
  4. RNA contains thymine in place of uracil.

Explanation

The pentose of RNA is ribose, which carries a hydroxyl at the 2-prime position, while DNA uses 2-deoxyribose with a hydrogen at that position. This 2-prime hydroxyl makes RNA more susceptible to alkaline hydrolysis and is a key chemical difference between the two nucleic acids.

Question 19Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

The two strands of the DNA double helix are held together mainly by hydrogen bonds between complementary bases. Which statement correctly describes these base pairs?

  1. Purines pair with purines and pyrimidines pair with pyrimidines.
  2. Adenine pairs with thymine through three hydrogen bonds, and guanine pairs with cytosine through two.
  3. Adenine pairs with thymine through two hydrogen bonds, and guanine pairs with cytosine through three. Correct answer.
  4. Adenine pairs with guanine through three hydrogen bonds, and thymine pairs with cytosine through two.

Explanation

The Watson-Crick pairing rules place the purine adenine opposite the pyrimidine thymine with two hydrogen bonds, and guanine opposite cytosine with three. Each pair thus contains one purine and one pyrimidine, keeping the helix diameter uniform.

Question 20Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A sample of DNA is hydrolysed with a nuclease that cleaves only phosphodiester bonds. Which statement describes the expected products?

  1. Nucleosides only, with no phosphate attached
  2. Intact DNA, because phosphodiester bonds are not part of DNA
  3. Free nitrogenous bases only, because the glycosidic bonds are also cleaved
  4. Nucleotides, because the phosphodiester backbone is broken between sugar residues Correct answer.

Explanation

Phosphodiester bonds link the 3-prime carbon of one sugar to the 5-prime phosphate of the next, forming the backbone. Cleaving them separates the chain into nucleotides while leaving the base-sugar glycosidic bond and the phosphate-ester attachments intact.

Question 21Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which observation most directly supports the conclusion that the two strands of DNA run in opposite directions?

  1. Both strands contain equal amounts of adenine and thymine.
  2. One strand has a free 5-prime phosphate end and the other has a free 3-prime hydroxyl end at the same end of the molecule. Correct answer.
  3. The sugar-phosphate backbones lie on the outside of the helix.
  4. The two strands are held together by hydrogen bonds.

Explanation

Antiparallel strands mean that one chain runs 5-prime to 3-prime in one direction while its partner runs 5-prime to 3-prime in the opposite direction. This is shown by the free ends: at a given end of the duplex one strand presents a 5-prime phosphate and the other a 3-prime hydroxyl.

Question 22Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A protein is found to contain 120 amino acid residues. Assuming the polypeptide has no signal peptide or post-translational cleavage and is synthesised from a single continuous coding sequence, what is the minimum number of nucleotides in the coding region of the mRNA?

  1. 360 Correct answer.
  2. 480
  3. 120
  4. 240

Explanation

The genetic code is a triplet code, so each amino acid is specified by three nucleotides. A polypeptide of 120 residues therefore requires 120 multiplied by 3, which equals 360 nucleotides in the coding region, excluding any stop codon.

Question 23Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

In a polypeptide, the peptide bond has partial double-bond character. Which consequence follows from this property?

  1. The peptide bond carries a full positive charge at all pH values.
  2. The peptide bond is readily hydrolysed at neutral pH without enzymes.
  3. The six atoms of the peptide group lie in a rigid plane, restricting rotation about the C-N bond. Correct answer.
  4. The peptide bond rotates freely, giving the backbone complete flexibility.

Explanation

Resonance between the carbonyl oxygen and the amide nitrogen gives the C-N bond partial double-bond character, so rotation about it is restricted. As a result the carbonyl carbon, oxygen, nitrogen, hydrogen and the two adjacent alpha carbons are coplanar, which constrains the possible folds of the chain.

Question 24Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A mixture of amino acids is subjected to paper electrophoresis at a pH above the isoelectric point of every component. Which amino acid migrates fastest towards the anode?

  1. The most basic amino acid in the mixture
  2. The most acidic amino acid in the mixture Correct answer.
  3. The amino acid with the largest side chain
  4. The amino acid with the highest isoelectric point

Explanation

At a pH above their isoelectric points all the amino acids carry a net negative charge and move towards the anode. The most acidic amino acid has the lowest isoelectric point and therefore the greatest excess of negative charge, so it migrates fastest.

Question 25Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A lipid sample is treated with bromine water and the bromine colour is rapidly discharged. Which conclusion about the sample is most reasonable?

  1. It contains glycerol but no fatty acids.
  2. It contains phospholipids but no triglycerides.
  3. It contains unsaturated fatty acid chains that add bromine across their double bonds. Correct answer.
  4. It consists only of saturated fatty acids that cannot react with bromine.

Explanation

Bromine adds across carbon-carbon double bonds, so rapid decolourisation indicates the presence of unsaturated acyl chains. Saturated fats lack such double bonds and react very slowly or not at all under these conditions.

Question 26Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which monosaccharide is a ketohexose and is commonly found in fruits and honey?

  1. Galactose
  2. Glucose
  3. Fructose Correct answer.
  4. Ribose

Explanation

Fructose is a ketohexose because it contains a ketone group at carbon 2 and has six carbon atoms; it is the main sugar in fruits and honey. Glucose and galactose are aldohexoses, and ribose is an aldopentose.

Question 27Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

Which statement best explains why a solution of a simple protein has a characteristic isoelectric pH at which its net charge is zero?

  1. At the isoelectric pH, the protein contains only non-polar amino acids.
  2. At the isoelectric pH, the protein loses its secondary structure completely.
  3. At the isoelectric pH, the number of positive and negative charges on the protein molecule is equal. Correct answer.
  4. At the isoelectric pH, all peptide bonds are hydrolysed.

Explanation

The isoelectric point (pI) is the pH at which the number of protonated basic groups equals the number of deprotonated acidic groups, giving zero net charge. At this pH the protein does not migrate in an electric field and is least soluble.

Question 28Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A polysaccharide is composed entirely of α-D-glucose units linked mainly by α(1→4) glycosidic bonds with occasional α(1→6) branch points. Which polysaccharide is described?

  1. Cellulose
  2. Amylose
  3. Glycogen Correct answer.
  4. Chitin

Explanation

Glycogen is a highly branched storage polysaccharide of α-D-glucose with α(1→4) links in the main chains and α(1→6) links at branch points. Amylose is unbranched, cellulose has β(1→4) links, and chitin contains N-acetylglucosamine.

Question 29Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

In a sample of double-stranded DNA, 30% of the total nucleotides are adenine. What percentage of the total nucleotides is guanine?

  1. 30%
  2. 40%
  3. 20% Correct answer.
  4. 70%

Explanation

By Chargaff's rules, adenine equals thymine, so thymine is also 30%. Together A + T = 60%, leaving 40% for G + C; since guanine equals cytosine, guanine is 20% of the total nucleotides.

Question 30Biomolecules: carbohydrates, lipids, amino acids and proteins, nucleic acids

A dipeptide is treated with phenyl isothiocyanate (Edman reagent), which reacts with the free α-amino group. Which statement best explains why only the N-terminal amino acid is released as a phenylthiohydantoin derivative?

  1. The Edman reagent reacts only with the α-amino group of the N-terminal residue, and subsequent acid treatment cleaves only the first peptide bond. Correct answer.
  2. The Edman reagent reacts only with the C-terminal carboxyl group, releasing the last amino acid.
  3. The Edman reagent hydrolyses all peptide bonds simultaneously.
  4. The Edman reagent reacts with the side chains of all amino acids, releasing each as a derivative.

Explanation

Phenyl isothiocyanate selectively labels the free α-amino group at the N-terminus; under acidic conditions the labelled residue cyclises and is cleaved as a phenylthiohydantoin without breaking the remaining peptide bonds. This allows sequential determination of the N-terminal sequence.

Question 31Enzymes: classification, kinetics, inhibition and regulation

According to the IUBMB system, an enzyme that catalyses the transfer of a phosphate group from ATP to glucose is placed in which class?

  1. Transferase Correct answer.
  2. Oxidoreductase
  3. Hydrolase
  4. Isomerase

Explanation

The IUBMB classification groups enzymes by the type of reaction catalysed. Hexokinase moves a phosphate group from ATP onto glucose, which is a group-transfer reaction, so the enzyme belongs to class 2, the transferases.

Question 32Enzymes: classification, kinetics, inhibition and regulation

An enzyme shows maximum activity at pH 7.4 and loses activity sharply on either side of this value. What is the most likely explanation for the loss of activity at pH 4?

  1. Complete hydrolysis of all peptide bonds in the backbone
  2. Alteration of ionisable groups in the active site and overall protein charge Correct answer.
  3. Denaturation caused by the absence of a coenzyme
  4. Irreversible precipitation of the enzyme as a metal sulphide

Explanation

Extreme pH changes the protonation state of acidic and basic side chains, including those in the active site, and alters the overall charge of the protein. This disrupts substrate binding and catalysis, so activity falls on either side of the optimum.

Question 33Enzymes: classification, kinetics, inhibition and regulation

In Michaelis-Menten kinetics, what does the Km value represent?

  1. The substrate concentration at which the reaction proceeds at half of Vmax Correct answer.
  2. The fraction of enzyme molecules present in the ES complex at equilibrium
  3. The rate constant for conversion of product back to substrate
  4. The maximum velocity reached when all enzyme is saturated

Explanation

Km is defined as the substrate concentration giving half the maximum velocity. It is derived from the Michaelis-Menten equation when v equals Vmax/2, and it is used as an inverse measure of enzyme affinity for substrate.

Question 34Enzymes: classification, kinetics, inhibition and regulation

Which statement correctly describes the effect of a competitive inhibitor on enzyme kinetics?

  1. Vmax decreases while Km is unchanged
  2. Both Vmax and Km increase without limit
  3. Both Vmax and Km decrease proportionally
  4. Vmax is unchanged while the apparent Km increases Correct answer.

Explanation

A competitive inhibitor binds reversibly at the active site, so it competes with substrate. Higher substrate concentration can overcome the inhibition, restoring Vmax, but more substrate is needed for half-maximal rate, so the apparent Km rises.

Question 35Enzymes: classification, kinetics, inhibition and regulation

An enzyme is inactive when synthesized but becomes active only after a specific peptide bond is cleaved. Which type of regulation does this illustrate?

  1. Allosteric activation by a positive effector
  2. Zymogen activation by limited proteolysis Correct answer.
  3. Covalent modification by phosphorylation
  4. Feedback inhibition by an end product

Explanation

Zymogens are inactive precursors that gain activity after limited proteolysis removes a masking peptide. Digestive enzymes such as trypsinogen are activated in this way, which prevents premature digestion of the tissue that makes them.

Question 36Enzymes: classification, kinetics, inhibition and regulation

Which coenzyme is required by dehydrogenases that transfer hydride ions from a substrate to an acceptor?

  1. Thiamine pyrophosphate
  2. Pyridoxal phosphate
  3. Biotin
  4. NAD+ Correct answer.

Explanation

NAD+ is a diffusible coenzyme that accepts a hydride ion during oxidation of a substrate, being reduced to NADH. Dehydrogenases depend on it for the two-electron oxidation step they catalyse.

Question 37Enzymes: classification, kinetics, inhibition and regulation

An enzyme has Km of 5 mM and Vmax of 100 micromoles per minute. At a substrate concentration of 5 mM the observed rate is 50 micromoles per minute. What is the rate at 20 mM substrate, assuming standard Michaelis-Menten behaviour?

  1. 20 micromoles per minute
  2. 100 micromoles per minute
  3. 50 micromoles per minute
  4. 80 micromoles per minute Correct answer.

Explanation

Using v = Vmax[S]/(Km+[S]), with [S] equal to four times Km the rate is 100 x 20/(5+20). This gives 80 micromoles per minute, showing that the rate approaches but does not reach Vmax at finite substrate concentration.

Question 38Enzymes: classification, kinetics, inhibition and regulation

A Lineweaver-Burk plot of an inhibited enzyme shows lines that intersect the y-axis at the same point but have a steeper slope than the uninhibited enzyme. Which inhibitor is present?

  1. An irreversible inhibitor that destroys the active site
  2. A competitive inhibitor Correct answer.
  3. A non-competitive inhibitor
  4. A mixed inhibitor that lowers Vmax and Km equally

Explanation

On a double-reciprocal plot the y-intercept equals 1/Vmax and the slope equals Km/Vmax. A shared y-intercept means Vmax is unchanged, while a steeper slope signals a higher apparent Km, which is the pattern of competitive inhibition.

Question 39Enzymes: classification, kinetics, inhibition and regulation

Which pairing of enzyme class with the reaction type it catalyses is correct?

  1. Oxidoreductase and transfer of an amino group between molecules
  2. Lyase and non-hydrolytic removal of a group to form a double bond Correct answer.
  3. Isomerase and joining of two molecules with ATP cleavage
  4. Ligase and hydrolysis of an ester bond

Explanation

Lyases cleave bonds without water by elimination, often producing a double bond, as in fumarase acting on malate. The other classes are defined by hydrolysis, ATP-dependent joining, or electron transfer respectively.

Question 40Enzymes: classification, kinetics, inhibition and regulation

An enzyme catalyses a reaction with a large negative standard free energy change, yet the uncatalysed reaction in a test tube is extremely slow. What does the enzyme change?

  1. It lowers the activation energy while leaving the equilibrium constant unchanged Correct answer.
  2. It makes the overall free energy change more negative
  3. It shifts the equilibrium position towards products
  4. It increases the energy released per mole of substrate

Explanation

A catalyst provides an alternative pathway of lower activation energy, so the rate increases. The free energy difference between substrate and product, and hence the equilibrium constant, is a thermodynamic property that catalysis does not alter.

Question 41Enzymes: classification, kinetics, inhibition and regulation

A student measures enzyme activity at several substrate concentrations and finds that a plot of v against [S] is a rectangular hyperbola. Which conclusion is justified?

  1. The enzyme is subject to substrate inhibition at all concentrations tested
  2. The reaction is first order with respect to substrate across the whole range
  3. The enzyme follows saturation kinetics consistent with Michaelis-Menten behaviour Correct answer.
  4. The enzyme is allosteric and shows cooperative binding

Explanation

A rectangular hyperbola is the diagnostic shape of simple saturation kinetics, in which the enzyme becomes saturated at high substrate concentration. It is not sigmoidal, so cooperativity is not indicated.

Question 42Enzymes: classification, kinetics, inhibition and regulation

Which statement about the active site of an enzyme is correct?

  1. It binds substrate covalently in every enzyme-catalysed reaction
  2. It is a three-dimensional cleft formed by residues that may be far apart in the primary sequence Correct answer.
  3. It is composed only of residues that are adjacent in the linear sequence
  4. It always occupies the entire surface of the enzyme molecule

Explanation

Folding brings together residues from different parts of the polypeptide chain to form the active site. This cleft provides specificity and catalytic groups, and its geometry explains why denaturation abolishes activity.

Question 43Enzymes: classification, kinetics, inhibition and regulation

An enzyme is inhibited by its end product binding to a site distinct from the active site, reducing the enzyme's affinity for substrate. What is this regulation called?

  1. Competitive inhibition at the active site
  2. Irreversible suicide inhibition
  3. Zymogen activation
  4. Allosteric feedback inhibition Correct answer.

Explanation

Binding of the end product at a regulatory site away from the active site changes the conformation of the enzyme and lowers substrate affinity. This allosteric feedback control prevents overproduction of the end product.

Question 44Enzymes: classification, kinetics, inhibition and regulation

Which feature distinguishes an irreversible inhibitor from a reversible one?

  1. It increases Vmax at high substrate concentration
  2. It binds the enzyme only when substrate is absent
  3. It forms a stable covalent bond with the enzyme that dialysis does not readily reverse Correct answer.
  4. It always binds at an allosteric site rather than the active site

Explanation

Irreversible inhibitors modify the enzyme covalently, so removing free inhibitor by dialysis does not restore activity. Reversible inhibitors associate and dissociate non-covalently, so their effect can be diluted out.

Question 45Enzymes: classification, kinetics, inhibition and regulation

An enzyme converts substrate to product with a very high turnover number. Which statement best explains the biological importance of a high kcat value?

  1. The reaction equilibrium is shifted further towards product
  2. The enzyme requires a higher activation energy than an uncatalysed reaction
  3. A small amount of enzyme can convert a large amount of substrate per unit time Correct answer.
  4. The enzyme has a lower affinity for its substrate than a low kcat enzyme

Explanation

kcat is the number of substrate molecules converted per enzyme molecule per unit time when the enzyme is saturated. A high value means less enzyme is needed to achieve a given flux, which is metabolically economical.

Question 46Enzymes: classification, kinetics, inhibition and regulation

In a reaction catalysed by an enzyme that follows ping-pong kinetics, which observation supports the mechanism?

  1. The reaction rate is independent of the concentration of the second substrate
  2. All lines intersect at a single point on the x-axis regardless of inhibitor present
  3. Parallel lines are obtained in a Lineweaver-Burk plot when the second substrate is varied at several fixed concentrations of the first Correct answer.
  4. The enzyme shows no saturation with the first substrate

Explanation

Ping-pong mechanisms involve a substituted enzyme intermediate, so varying one substrate at fixed levels of the other gives parallel double-reciprocal lines. Sequential mechanisms instead give intersecting lines because both substrates must bind before reaction.

Question 47Enzymes: classification, kinetics, inhibition and regulation

An enzyme from a thermophilic bacterium remains active at 70 degrees Celsius while its mesophilic counterpart is denatured. Which explanation is most consistent with this observation?

  1. The thermophilic enzyme lacks an active site and therefore cannot unfold
  2. The thermophilic enzyme has more stabilising interactions such as salt bridges and a more rigid folded structure Correct answer.
  3. The thermophilic enzyme has a much lower activation energy for the reaction at all temperatures
  4. The thermophilic enzyme is protected by a covalently attached competitive inhibitor

Explanation

Thermostability generally arises from increased structural rigidity, achieved through additional ionic interactions, hydrogen bonds, and compact packing. These features raise the temperature at which unfolding occurs without necessarily altering catalytic efficiency at moderate temperature.

Question 48Enzymes: classification, kinetics, inhibition and regulation

An enzyme is treated with a reagent that reacts specifically with the sulfhydryl group of a cysteine residue, and activity is lost. Addition of excess substrate during treatment protects against inactivation. What is the best interpretation?

  1. The enzyme is a metalloenzyme requiring zinc at the active site
  2. The reagent denatures the entire enzyme by disrupting all hydrogen bonds
  3. A cysteine residue essential for catalysis or substrate binding lies in the active site Correct answer.
  4. The cysteine residue is required only for maintaining the quaternary structure

Explanation

Protection by substrate indicates that the modified residue lies at or near the substrate-binding site. Specific chemical modification with loss of activity is classic evidence for an essential active-site group.

Question 49Enzymes: classification, kinetics, inhibition and regulation

A metabolic pathway is regulated by feedback inhibition in which the end product inhibits the first enzyme. Which property of this control makes it effective in conserving cellular resources?

  1. The inhibition is irreversible, permanently shutting down the pathway
  2. The end product activates the first enzyme to increase flux
  3. Only the final enzyme of the pathway is inhibited, allowing intermediates to build up
  4. The first committed step is inhibited, so no intermediates accumulate when the end product is plentiful Correct answer.

Explanation

Regulating the first committed step prevents the wasteful synthesis of all downstream intermediates when the end product is already sufficient. Inhibition of a later step would allow intermediates to accumulate without benefit.

Question 50Enzymes: classification, kinetics, inhibition and regulation

Two isoenzymes catalyse the same reaction but differ in their Km values. Which statement best describes the physiological advantage of this difference?

  1. The isoenzyme with higher Km has a higher Vmax by definition
  2. Both isoenzymes must have identical regulatory properties to function in different tissues
  3. The isoenzyme with lower Km cannot be saturated by substrate
  4. The isoenzyme with lower Km operates efficiently at low substrate concentration while the other works when substrate is abundant Correct answer.

Explanation

Different Km values allow isoenzymes to function over different ranges of substrate concentration. A low-Km form is effective when substrate is scarce, while a high-Km form responds when substrate rises, matching the metabolic needs of different tissues.