12 cards · Core A-level concepts

Biological molecules: active recall

Name the molecule, identify its bonds and explain how its structure helps it perform its role. Keep monomers, polymers and other large molecules distinct.

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1. How do condensation and hydrolysis change a bond between biological subunits?

Condensation joins subunits with formation of a bond and release of water. Hydrolysis breaks a bond using water. For a linear chain, count the bonds being formed or broken rather than assuming that the number of water molecules equals the number of subunits.

2. Distinguish a monosaccharide, a disaccharide and a polysaccharide.

A monosaccharide is one sugar unit, such as glucose. A disaccharide contains two joined sugar units, such as sucrose. A polysaccharide contains many sugar units. A disaccharide is not classified as a polysaccharide merely because it contains more than one unit.

3. How do amylose and amylopectin differ within starch?

Amylose consists of unbranched chains of alpha-glucose joined by alpha-1,4 glycosidic bonds and can coil. Amylopectin also has alpha-1,4 chains, with alpha-1,6 bonds at branch points. Branches provide more ends at which enzymes can act.

4. Why is glycogen suitable as a carbohydrate store?

It is a compact, branched alpha-glucose polymer with many ends for enzymatic removal of glucose units. It is relatively insoluble, reducing its effect on water potential compared with many separate dissolved glucose molecules. Storage still depends on regulated synthesis and breakdown.

5. How does cellulose structure support its mechanical role?

Beta-glucose units form unbranched beta-1,4 chains, with alternate units oriented differently. Hydrogen bonding between adjacent chains helps form strong microfibrils. This arrangement supports tensile strength in plant cell walls rather than making cellulose an animal energy store.

6. What forms a triglyceride, and why is it not described as a polymer?

One glycerol molecule is joined to three fatty acids through ester bonds. It is a large biological molecule, but it is not a chain of many repeating monomer units of the kind that defines a polymer. Its largely non-polar structure also makes it poorly soluble in water.

7. How does a phospholipid differ from a triglyceride?

In the typical structure used here, glycerol carries two fatty-acid tails and a phosphate-containing polar head group. The head can interact with water while the tails avoid it. This amphipathic arrangement helps explain bilayer formation.

8. How does an unsaturated fatty-acid chain differ from a saturated one?

An unsaturated chain contains one or more carbon–carbon double bonds. Cis double bonds introduce bends that can reduce close packing. A saturated chain has no such double bonds. The resulting packing influences the physical properties of lipids containing these chains.

9. Distinguish a protein’s primary and tertiary structure.

Primary structure is its amino-acid sequence, joined by peptide bonds. Tertiary structure is the overall three-dimensional folding of a polypeptide, maintained by interactions among side chains and the surrounding environment. Sequence influences which interactions and shapes are possible.

10. Does denaturation normally mean that every peptide bond has been broken?

No. Denaturation disrupts the folded structure and its stabilising interactions, often changing function, while the primary amino-acid sequence can remain intact. Hydrolysing peptide bonds is a different process from merely disrupting the protein’s folding.

11. What are the components of a nucleotide, and what differs between DNA and RNA nucleotides?

A nucleotide contains a phosphate group, a pentose sugar and a nitrogenous base. DNA uses deoxyribose and usually the bases A, T, C and G. RNA uses ribose and A, U, C and G. Base sequence and sugar–phosphate linkage are different parts of the structure.

12. How do hydrogen bonds help explain water’s biological importance?

Hydrogen bonding between water molecules contributes to cohesion and its relatively high specific heat capacity and latent heat of vaporisation. These properties help support continuous water columns, temperature stability and cooling by evaporation. Water’s polarity also supports interactions with ions and polar solutes.

Original recall material. AQA specification · OCR A specification resources.