Biological Molecules
Biological molecules is the A-level Biology topic covering the structure and function of water, carbohydrates, lipids, proteins and enzymes in living organisms. It opens the Year 12 course (AQA 7402 and equivalent specifications) because later topics on cells, exchange and genetics all depend on it. Required practicals include food tests and enzyme rate investigations.
Before you start
No specific prerequisites - this is a good place to start.
Method
- Learn the monomer and polymer for each molecule type, and name the bond formed by condensation (glycosidic for carbohydrates, ester for lipids, peptide for proteins) and broken by hydrolysis.
- Link structure to function for each molecule: for example, starch's coiled, branched chains suit storage, while cellulose's straight, hydrogen-bonded chains suit structural strength.
- Practise the four qualitative food tests (iodine for starch, Benedict's for reducing sugar, Biuret for protein, emulsion test for lipid): know the reagent, the method and the exact positive colour change for each.
- For enzyme questions, explain rate changes using the active site and the induced-fit model, and state whether a factor (temperature, pH, concentration) increases collision frequency or denatures the enzyme.
- When a question gives a graph or table of rate data, read the values carefully before calculating, and check whether the enzyme or the substrate is limiting/saturating the reaction.
- For calculation questions (e.g. relative molecular mass, percentage change), write out every step of your working so partial credit is available even if the final answer is wrong.
Worked example
A triglyceride forms from one glycerol molecule (relative molecular mass, Mr = 92) and three identical fatty acid molecules (Mr = 284 each). Calculate the Mr of the triglyceride formed, and state the total mass lost as water during the reaction.
- Add the Mr of the starting molecules: 92 + (3 x 284) = 92 + 852 = 944.
- Three ester bonds form (one between glycerol and each fatty acid), each by a condensation reaction that releases one water molecule, so 3 water molecules are lost in total.
- Calculate the mass lost as water: 3 x 18 = 54.
- Subtract the mass of water lost from the total starting mass: 944 - 54 = 890.
- Final answer: the triglyceride has an Mr of 890; a total mass of 54 is lost as three water molecules.
Practice questions
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Q1Explain what it means to say that a water molecule is polar.Show answer
Answer: A molecule with an unequal (uneven) distribution of charge, having a slightly positive region and a slightly negative region.
Q2Name the type of reaction that joins two monosaccharides together, and name the bond formed.Show answer
Answer: Condensation reaction; forms a glycosidic bond.
Q3A food sample forms a white, cloudy emulsion when shaken with ethanol and then added to water. Which type of biological molecule does this confirm?Show answer
Answer: Lipid (positive emulsion test).
Q4State the term for a protein structure formed when two or more polypeptide chains are held together.Show answer
Answer: Quaternary structure.
Q5The initial rate of an enzyme reaction rose from 4 to 9 arbitrary units as temperature increased from 20 to 35 degrees C, then fell to 2 units at 65 degrees C. Give the reason for the fall in rate at 65 degrees C.Show answer
Answer: Denaturation - heat breaks the bonds holding the enzyme's tertiary structure, changing the shape of the active site so substrate can no longer bind.
Q6A triglyceride has an Mr of 794. It forms from glycerol (Mr = 92) and three identical fatty acids, releasing 3 water molecules (Mr = 18 each). Calculate the Mr of one fatty acid.Show answer
Answer: 252 (794 + 54 = 848 before water loss; 848 - 92 = 756 for the fatty acids; 756 / 3 = 252)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
Two amino acids join to form a dipeptide. Describe how the peptide bond forms, and name the type of reaction involved.
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A student investigated the effect of temperature on the initial rate of an enzyme-catalysed reaction from 10 degrees C to 70 degrees C, with all other variables controlled. State and explain the shape of the graph obtained when rate of reaction is plotted against temperature.
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The initial rate of an enzyme-catalysed reaction was measured at increasing substrate concentrations, with enzyme concentration and temperature kept constant. Results: substrate concentration (mmol/dm3): 2, 4, 8, 16, 32; initial rate (arbitrary units): 6, 11, 18, 22, 22. Calculate the percentage increase in rate between substrate concentrations of 4 and 8 mmol/dm3, and explain why the rate does not increase further between 16 and 32 mmol/dm3.
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Free printable worksheet
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