Biology Paper 2 November 2005

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1. AAll of the DNA was hybrid, with none fully light.
2. BHalf of the DNA was hybrid and half was fully light.
3. CA quarter of the DNA was hybrid and three-quarters was fully light.
4. DAll of the DNA was fully light, with no hybrid DNA remaining.

1. AHalf of the DNA was hybrid and half was fully light, the same ratio as in the 2nd generation before this round of replication.
2. BAll of the DNA was fully light, because the original heavy strands could not survive a third round of replication in light medium.
3. CA quarter of the DNA was hybrid and three-quarters was fully light.
4. DAn eighth of the DNA was hybrid and seven-eighths was fully light, following the same halving pattern carried one generation further.

1. ADisulfide bond, formed by oxidation of two cysteine sulfhydryl groups, common in extracellular and fibrous proteins.
2. BPeptide bond, formed by a condensation reaction between the carboxyl and amino groups.
3. CIonic bond, formed by electrostatic attraction between oppositely charged R groups on separate amino acid side chains.
4. DHydrogen bond, formed by weak attraction between an —OH group and a carbonyl oxygen on a nearby residue.

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1. ABiuret test: add sodium hydroxide solution followed by dilute copper sulphate solution; a purple/violet colour indicates peptide bonds are present.
2. BIodine test: add iodine solution directly to the sample; a blue-black colour would indicate peptide bonds are present, since this test targets starch granules.
3. CSudan III test: shake the sample with Sudan III stain; a red-stained layer would indicate peptide bonds are present, since this test targets lipids.
4. DBenedict's test: heat the sample gently with Benedict's reagent; a brick-red precipitate would indicate peptide bonds are present, since this test targets reducing sugars.

1. AIt binds permanently to the active site and does not detach again, so the enzyme cannot bind substrate at all for the rest of the reaction.
2. BIt binds to an allosteric site away from the active site, changing the enzyme's overall shape so that the active site itself no longer fits the substrate.
3. CIt binds reversibly to the active site itself, blocking the substrate from entering and reducing the frequency of enzyme-substrate complex formation.
4. DIt reacts directly with free substrate molecules in solution, converting them chemically into a form the enzyme's active site cannot recognise.

1. AThe inhibitor gradually breaks down in the presence of excess substrate molecules, releasing free active enzyme back into solution as substrate concentration rises.
2. BThe inhibitor increases the enzyme's binding affinity for substrate molecules, so a higher substrate concentration actually speeds up the loss of enzyme activity.
3. CThe inhibitor binds at a site separate from the active site, changing the enzyme's shape so the active site can no longer bind substrate effectively, whatever the substrate concentration.
4. DThe inhibitor binds directly within the active site itself, so more substrate molecules present simply outcompete the inhibitor for that same site, exactly as happens with a competitive inhibitor.

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