Biology HL

1. 1.

   A linear polypeptide contains 120 amino-acid residues. Calculate the number of peptide bonds formed during its assembly, name the two types of functional group joined, and explain why changing one residue can alter function without changing chain length.

   [4 marks] · no calculator

   Worked explanation

   1. Count one fewer link than residues and distinguish peptide bonds from hydrogen bonds involved in folding.
   2. The backbone connection joins an amine and carboxyl group with water released.
   3. A same-length replacement can change charge or hydrophobicity, affecting shape or interactions without guaranteeing loss of function.

   Common mistake: A substitution does not necessarily shorten the protein or destroy every peptide bond; its effect depends on position and properties.

   Marking points

   • 119 peptide bonds join 120 residues in one linear chain.
   • A carboxyl group reacts with an amine group in condensation.
   • A replacement can change the chemical properties of an R-group at that position.
   • Altered interactions can change folding or binding and therefore function; not every substitution necessarily does so.

   Examiner tip: A substitution does not necessarily shorten the protein or destroy every peptide bond; its effect depends on position and properties.

2. 2.

   A soluble globular protein has many non-polar R-groups buried inside and charged R-groups exposed to water. Explain this arrangement, identify two other interactions stabilizing tertiary structure, and explain how a major pH change may disrupt the fold without necessarily hydrolysing its peptide backbone.

   [4 marks] · no calculator

   Worked explanation

   1. Separate interactions with the aqueous surroundings from covalent backbone assembly.
   2. Tertiary folding is maintained by several kinds of interaction, not just one generic bond type.
   3. Changes in protonation can disturb charge-dependent interactions; denaturation does not require complete hydrolysis.

   Common mistake: Denaturation usually changes folding and function; it does not mean all peptide bonds have been cut into separate amino acids.

   Marking points

   • Burial of non-polar groups shields them from water while polar/charged groups interact favourably with water.
   • Identify one further stabilizing interaction: hydrogen bonds, ionic interactions or covalent disulfide bonds between cysteine residues.
   • Identify a second distinct interaction from hydrogen bonds, ionic interactions and disulfide bonds; accept any two, without requiring hydrogen bonds.
   • pH can change R-group charges and disrupt interactions, changing conformation without necessarily breaking peptide bonds.

   Examiner tip: Denaturation usually changes folding and function; it does not mean all peptide bonds have been cut into separate amino acids.

3. 3.

   Insulin is a compact protein hormone, whereas collagen forms long fibres that resist tension. Explain how the different forms suit signalling in blood and mechanical support, and distinguish tertiary from quaternary structure in a protein containing multiple polypeptide chains.

   [4 marks] · no calculator

   Worked explanation

   1. Use shape and exposed chemistry to link insulin with transport and binding rather than describing it as an enzyme.
   2. Repeated alignment in collagen supports mechanical strength rather than transport of a soluble signal.
   3. Multiple chains introduce quaternary organization in addition to the folding of individual chains.

   Common mistake: A protein need not contain a non-protein prosthetic group to have quaternary structure; more than one chain is the key distinction.

   Marking points

   • Compact globular insulin has a surface enabling solubility in blood and specific receptor interaction.
   • Long fibrous collagen molecules assemble into strong fibres suited to resisting tensile forces.
   • Tertiary structure is the three-dimensional folding of one polypeptide chain.
   • Quaternary structure is the arrangement/association of multiple polypeptide chains; collagen and mature insulin contain multiple chains.

   Examiner tip: A protein need not contain a non-protein prosthetic group to have quaternary structure; more than one chain is the key distinction.

4. 4.

   A hypothetical globular enzyme loses activity when heated, but analysis shows its amino-acid sequence is unchanged. Explain how activity can be lost without peptide-bond hydrolysis, and why cooling does not guarantee restoration of activity.

   [3 marks] · no calculator

   Worked explanation

   1. Separate amino-acid order from the three-dimensional arrangement of that chain.
   2. Activity depends on a suitable active-site structure, not merely on retaining the same monomers.
   3. Some changes are reversible under suitable conditions, but irreversible aggregation or alternative folds make a guaranteed recovery claim unjustified.

   Common mistake: Denaturation does not necessarily break peptide bonds, and retaining primary structure does not guarantee an active enzyme.

   Marking points

   • Heat can disrupt interactions maintaining the folded secondary/tertiary structure without changing the primary sequence.
   • A changed active-site conformation can impair substrate binding or catalysis.
   • Misfolding or aggregation may prevent recovery, so cooling does not necessarily restore the original functional conformation.

   Examiner tip: Denaturation does not necessarily break peptide bonds, and retaining primary structure does not guarantee an active enzyme.

5. 5.

   Compare soluble insulin with collagen as examples of proteins whose shapes suit different roles. Explain why both can contain peptide bonds even though one acts as a signal and the other provides tensile strength.

   [3 marks] · no calculator

   Worked explanation

   1. Begin with a shared chemical feature, then compare higher-order structure rather than treating the proteins as different classes of monomer.
   2. For insulin, connect shape to specific molecular recognition; for collagen, connect elongated assemblies to mechanical support.
   3. A shared bond type does not require identical sequence, shape or biological activity.

   Common mistake: Do not equate all proteins with enzymes or assume all peptide-linked chains have the same shape and function.

   Marking points

   • Insulin is a relatively compact globular signalling protein whose specific shape enables receptor recognition.
   • Collagen forms fibrous triple-helical structures and assemblies that contribute tensile strength.
   • Both contain amino acids joined by peptide bonds; differences in sequence, folding and assembly produce different properties and roles.

   Examiner tip: Do not equate all proteins with enzymes or assume all peptide-linked chains have the same shape and function.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.