Flashcards · 9700 · AS Topic 2

Biological molecules (AS Level) flashcards

27 cards on AS Topic 2, subtopics 2.1 to 2.4. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 27 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

2.1 Testing for biological molecules

Benedict's test: what it detects, method and positive result: Reducing sugars (glucose, fructose, maltose). Add an equal volume of Benedict's solution and heat in a water bath near boiling. Positive: green → yellow → orange → brick-red precipitate. Negative: stays blue.
Iodine test for starch: Add a few drops of iodine in potassium iodide solution; no heating. Positive: blue-black. Negative: stays orange-brown.
Emulsion test for lipids: Shake the sample with ethanol, then pour into water. Positive: a cloudy white emulsion (lipid comes out of solution as tiny droplets). Ethanol must come first.
Biuret test for proteins: Add biuret reagent (or potassium hydroxide then dilute copper sulfate); no heating. Positive: purple (lilac). Negative: stays blue. It detects peptide bonds.
Test for a non-reducing sugar: Benedict's first: stays blue. Boil a fresh sample with dilute HCl (hydrolyses glycosidic bonds). Cool, neutralise with sodium hydrogencarbonate. Benedict's again: a colour change to green–brick-red shows a non-reducing sugar.
How can a Benedict's test estimate the concentration of a reducing sugar?: Make at least five known concentrations; use the same volumes, Benedict's in excess, one water bath. Use the time to first colour change (graph) or colour standards, then compare the unknown. Shorter time = higher concentration.
Convert 0.30 mol dm⁻³ glucose to % (w/v): 0.30 × 180 = 54 g dm⁻³. 54 ÷ 10 = 5.4 % (w/v), since 1 % (w/v) = 1 g in 100 cm³.

2.2 Carbohydrates and lipids

α-glucose vs β-glucose: They differ at carbon 1. α-glucose: the OH on C1 is below the ring. β-glucose: the OH on C1 is above the ring.
Glycosidic bond formation and breakage: Formed by condensation (a water molecule is removed) between two sugars. Broken by hydrolysis (water is added). Maltose = α-glucose + α-glucose (α-1,4); sucrose = α-glucose + fructose.
Amylose vs amylopectin vs glycogen: All α-glucose. Amylose: unbranched helix, α-1,4 only. Amylopectin: branched, α-1,4 and α-1,6. Glycogen: like amylopectin but more branched and compact; the animal and fungal store.
Why are starch and glycogen good energy stores?: Compact, so much glucose in a small space. Insoluble, so no effect on water potential. Branches give many chain ends for enzymes to release glucose quickly.
How does the structure of cellulose suit it to the cell wall?: β-glucose units, alternate ones rotated 180°, give straight chains. Parallel chains are linked by many hydrogen bonds into microfibrils and fibres, giving high tensile strength so the wall resists bursting.
Structure of a triglyceride: One glycerol + three fatty acids joined by three ester bonds, formed by condensation (releasing 3 H₂O). Non-polar, hydrophobic and insoluble in water. It is not a polymer.
Saturated vs unsaturated fatty acid: Saturated: no C=C double bonds, straight chain. Unsaturated: one or more C=C double bonds, each causing a kink.
Phospholipid structure: Glycerol + two fatty acids (ester bonds) + a phosphate group. The phosphate head is charged, polar and hydrophilic; the fatty acid tails are non-polar and hydrophobic.

2.3 Proteins

General structure of an amino acid: A central carbon bonded to an amine group (–NH₂), a carboxyl group (–COOH), a hydrogen atom and an R group. Only the R group differs between amino acids.
α-helix: Secondary structure: the N–H of one amino acid hydrogen bonds to the C=O of the amino acid four places earlier; R groups point outwards.
Secondary vs tertiary structure: Secondary uses hydrogen bonds between backbone C=O and N–H only. Tertiary uses interactions between R groups (hydrophobic interactions, hydrogen bonds, ionic bonds, disulfide bonds).
Disulfide bond vs hydrogen bond in a protein: A disulfide bond is covalent and forms only between the sulfur atoms of two cysteine R groups. Hydrogen bonds are weak and form between many polar groups.
Structure of haemoglobin: Globular protein with quaternary structure: two α-globin and two β-globin chains. Each chain has a haem group with Fe²⁺, so up to 4 O₂ per molecule. Hydrophilic R groups outside make it soluble.
How does the structure of collagen give it high tensile strength?: Three helical chains wound into a triple helix (glycine every third position). Molecules lie parallel with staggered ends, linked by covalent cross-links, forming fibrils and fibres.
How many peptide bonds are in a protein of two chains of 50 and 30 amino acids?: (50 − 1) + (30 − 1) = 78 peptide bonds. Count each chain separately.

2.4 Water

Why is water polar, and how do hydrogen bonds form between water molecules?: Oxygen attracts electrons more strongly than hydrogen, so O is δ− and each H is δ+ (bent molecule). A hydrogen bond is the weak attraction between a δ+ H of one molecule and the δ− O of another.
Solvent action of water: explanation and roles: Polar water molecules surround ions and form hydrogen bonds with polar molecules, so they dissolve. Roles: transport in blood plasma, xylem and phloem; medium for metabolic reactions.
Specific heat capacity: The energy needed to raise the temperature of 1 kg (or 1 g) of a substance by 1 °C. About 4.2 kJ kg⁻¹ °C⁻¹ for water.
Why does water's high specific heat capacity matter to organisms?: Energy is used to break hydrogen bonds before the molecules move faster, so water changes temperature slowly. Aquatic habitats and bodies stay stable, keeping enzymes near their optimum.
Latent heat of vaporisation and its role: The energy needed to change a liquid into vapour without a change in temperature. Water needs a lot, because hydrogen bonds must break, so evaporating sweat or transpiration removes much heat for little water.