Topic 1: Biological Molecules — Edexcel A-Level Biology B
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Topic 1: Biological Molecules explained
Topic 1: Biological Molecules covers 8 specification points, including Carbohydrates i Know the difference between… and Lipids i Understand how a triglyceride is….
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1.1 Carbohydrates i Know the difference between monosaccharides, disaccharides and polysaccharides. ii Know the structure of the hexose glucose (alpha and beta) and the pentose ribose. iii Understand how monosaccharides (glucose, fructose, galactose) join to form disaccharides (sucrose, lactose and maltose) and polysaccharides (starch formed from amylose and amylopectin; glycogen) through condensation reactions forming glycosidic bonds, and how these can be split through hydrolysis reactions. iv Understand how the structure of glucose, starch, glycogen and cellulose relates to their function.
Monosaccharides are single sugar units; disaccharides are two joined units; polysaccharides are many. Glucose is a hexose drawn as a ring, with alpha and beta forms differing only in the position of the hydroxyl group on carbon 1. Ribose is a pentose. Two monosaccharides join by condensation, releasing water and forming a glycosidic bond: glucose plus fructose gives sucrose, glucose plus galactose gives lactose, and two glucose units give maltose. Starch is amylose plus amylopectin, and glycogen is a highly branched glucose polymer. Hydrolysis reverses condensation by adding water and splitting the bond. Structure links to function: glucose is soluble and easily respired; starch and glycogen are compact storage molecules; cellulose is straight-chain and structural.
1.2 Lipids i Understand how a triglyceride is synthesised, including the formation of ester bonds during condensation reactions between glycerol and three fatty acids. ii Know the differences between saturated and unsaturated lipids. iii Understand how the structure of lipids relates to their role in energy storage, waterproofing and insulation. iv Understand how the structure and properties of phospholipids relate to their function in cell membranes.
A triglyceride forms when glycerol reacts with three fatty acids in condensation reactions, each releasing water and creating an ester bond. Saturated fatty acids have no carbon-carbon double bonds in their hydrocarbon chains, while unsaturated fatty acids have at least one, producing kinks. Lipids store energy densely because their hydrocarbon chains are reduced and release much energy on oxidation; they are also waterproof and provide thermal insulation. Phospholipids have a glycerol backbone, two fatty acids and a phosphate group. The phosphate head is hydrophilic and the fatty acid tails are hydrophobic, so phospholipids form a bilayer in cell membranes that acts as a barrier to water-soluble substances.
1.3 Proteins i Know the structure of an amino acid (structures of specific amino acids are not required). ii Understand the formation of polypeptides and proteins (as amino acid monomers linked by peptide bonds in condensation reactions). iii Understand the role of ionic, hydrogen and disulfide bonding in the structure of proteins. iv Understand the significance of the primary, secondary, tertiary and quaternary structure of a protein in determining the properties of fibrous and globular proteins, including collagen and haemoglobin. v Understand how the structure of collagen and haemoglobin are related to their function.
Amino acids possess a central alpha carbon bonded to an amine group (-NH2), a carboxyl group (-COOH), a hydrogen atom (-H), and a variable R group. They link via condensation reactions between the carboxyl group of one amino acid and the amine group of another, forming peptide bonds to create polypeptides. A protein's primary structure is its amino acid sequence. Secondary structure involves hydrogen bonding between N-H and C=O groups of the polypeptide backbone, forming alpha-helices or beta-pleated sheets. Tertiary structure is the overall 3D folding stabilised by ionic, hydrogen, and disulfide bonds between R groups. Quaternary structure involves multiple polypeptide subunits. These structural levels determine whether a protein is fibrous (like collagen, which provides tensile strength) or globular (like haemoglobin, which transports oxygen).
1.4 DNA and protein synthesis i Know the structure of DNA, including the structure of the nucleotides (purines and pyrimidines), base pairing, the two sugar-phosphate backbones, phosphodiester bonds and hydrogen bonds. ii Understand how DNA is replicated semi-conservatively, including the role of DNA helicase, polymerase and ligase. iii Know that a gene is a sequence of bases on a DNA molecule coding for a sequence of amino acids in a polypeptide chain. iv Know the structure of mRNA including nucleotides, the sugar phosphate backbone and the role of hydrogen bonds. v Know the structure of tRNA, including nucleotides, the role of hydrogen bonds and the anticodon. vi Understand the processes of transcription in the nucleus and translation at the ribosome, including the role of sense and anti-sense DNA, mRNA, tRNA and the ribosomes. vii Understand the nature of the genetic code, including triplets coding for amino acids, start and stop codons, degenerate and non-overlapping nature, and that not all the genome codes for proteins. viii Understand the term gene mutation as illustrated by base deletions, insertions and substitutions. ix Understand the effect of point mutations on amino acid sequences, as illustrated by sickle cell anaemia in humans.
Nucleotides consist of a pentose sugar, phosphate group, and nitrogenous base. DNA features two sugar-phosphate backbones with phosphodiester bonds and complementary bases linked by hydrogen bonds. It replicates semi-conservatively via DNA helicase, polymerase, and ligase. A gene is a base sequence coding for a polypeptide, though not all the genome codes for proteins. Transcription in the nucleus builds single-stranded mRNA from anti-sense DNA. Translation at ribosomes uses tRNA, folded via internal hydrogen bonds to present an anticodon, to assemble amino acids. The genetic code uses non-overlapping, degenerate triplets with start and stop codons. Point mutations can alter polypeptides, causing sickle cell anaemia.
1.5 Enzymes i Know the structure of enzymes as globular proteins. ii Understand the concepts of specificity and the induced fit hypothesis. iii Understand that enzymes are catalysts that reduce activation energy. iv Understand how temperature, pH, substrate and enzyme concentration affect the rate of enzyme activity.
Enzymes are globular proteins: their polypeptide chains fold so that hydrophilic groups face outwards and hydrophobic groups cluster inside, producing a compact, roughly spherical shape with a specific tertiary structure. A small region, the active site, is formed by particular amino acid residues, and its shape and chemistry make the enzyme specific, usually to one substrate. The induced fit hypothesis explains this: the active site is not a rigid lock but moulds slightly around the substrate as the enzyme–substrate complex forms, straining bonds and lowering the activation energy needed for the reaction. Enzymes are therefore biological catalysts, unchanged at the end. Rate of activity rises with temperature until denaturation, has an optimum pH, and increases with substrate or enzyme concentration until a plateau when the other factor becomes limiting.
1.6 Inorganic ions i Understand the role in plants of: ● nitrate ions – to make DNA and amino acids ● calcium ions – to form calcium pectate for the middle lamellae ● magnesium ions – to produce chlorophyll ● phosphate ions – to make ADP and ATP.
Plants absorb inorganic ions from soil water, and each ion has a distinct structural or metabolic role. Nitrate ions supply nitrogen for building amino acids, which are joined into proteins, and for the nitrogenous bases of DNA, so growth and cell division depend on them. Calcium ions combine with pectin in the primary cell wall to form calcium pectate, the gel-like middle lamella that cements adjacent plant cells together. Magnesium ions sit at the centre of each chlorophyll molecule, so a shortage reduces chlorophyll and causes chlorosis. Phosphate ions are components of ADP and ATP, the molecules that carry and release energy in cells.
1.7 Water i Understand the importance of the dipole nature of water leading to hydrogen bonding and the significance of the following to organisms: ● high specific heat capacity ● polar solvent ● surface tension ● incompressibility ● maximum density at 4 °C.
In a water molecule the oxygen atom carries a slight negative charge and each hydrogen a slight positive charge, so water is a dipole. Opposite charges on neighbouring molecules attract, forming hydrogen bonds. These bonds explain water's biological importance. A large energy input is needed to raise its temperature, giving a high specific heat capacity that buffers organisms against sudden temperature change. Its polarity lets it dissolve ions and polar molecules, acting as a polar solvent for transport and reactions. Hydrogen bonding at the surface creates surface tension, supporting small organisms and capillary movement. Water is almost incompressible, so it can act as a hydrostatic skeleton and a shock-absorbing fluid. It is also most dense at 4 °C, so ice floats and insulates water below.
CORE PRACTICAL 1: Investigate a factor affecting the initial rate of an enzyme– controlled reaction. v Understand how the initial rate of enzyme activity can be measured and why this is important. vi Understand how enzymes can be affected by competitive, non-competitive and end-product inhibition. vii Know that enzymes catalyse a wide range of intracellular reactions as well as extracellular ones.
Core Practical 1 investigates one factor, such as temperature, pH or substrate concentration, affecting the initial rate of an enzyme-controlled reaction. The initial rate is the rate at the very start, measured as the gradient of the tangent to the steepest part of a progress curve, or from the time taken for a set change early in the reaction. It matters because it reflects the true catalytic activity before substrate depletion, product accumulation or enzyme denaturation slow the reaction. Enzymes can be inhibited competitively, when a similar molecule blocks the active site, non-competitively, when binding elsewhere changes active site shape, or by end-product inhibition, a negative feedback mechanism. Enzymes catalyse intracellular reactions, such as those in glycolysis, and extracellular ones, such as digestion.
Your focus
- Distinguish monosaccharides, disaccharides and polysaccharides by the number of sugar units they contain.
- Describe the ring structures of alpha glucose, beta glucose and ribose, including the carbon 1 difference.
- Explain condensation and hydrolysis in the formation and breakdown of named disaccharides and polysaccharides.
Show all 28 objectives
- Relate the structures of glucose, starch, glycogen and cellulose to their biological functions.
- Describe the condensation reactions that synthesise a triglyceride from glycerol and three fatty acids.
- Compare saturated and unsaturated lipids in terms of carbon-carbon double bonds.
- Explain how lipid structure supports energy storage, waterproofing and insulation.
- Relate phospholipid structure and properties to their role in cell membranes.
- Identify the components of an amino acid and describe how they link to form polypeptides.
- Explain how different types of bonding determine the secondary, tertiary, and quaternary structures of proteins.
- Relate the structural features of globular and fibrous proteins to their specific biological functions.
- Describe the structure of DNA and RNA nucleotides, including the pentose sugar, phosphate, and base, and how they form polynucleotide strands.
- Explain the roles of specific enzymes in semi-conservative DNA replication.
- Detail the stages of transcription in the nucleus and translation at the ribosome, including the structure of mRNA and tRNA and the role of hydrogen bonds.
- Explain the impact of point mutations on the resulting polypeptide, using sickle cell anaemia as an example, and recognise that not all the genome codes for proteins.
- Describe the globular protein structure of enzymes and the formation of the active site.
- Explain specificity using the induced fit hypothesis and enzyme–substrate complex formation.
- Explain how temperature, pH, substrate concentration and enzyme concentration affect the rate of enzyme activity.
- State the role of nitrate ions in making DNA and amino acids in plants.
- Describe how calcium ions form calcium pectate for the middle lamellae.
- Explain that magnesium ions are needed to produce chlorophyll.
- Identify phosphate ions as components of ADP and ATP.
- Describe how the dipole nature of water leads to hydrogen bonding.
- Explain how high specific heat capacity and polar solvent behaviour benefit organisms.
- Relate surface tension, incompressibility and maximum density at 4 °C to named biological roles.
- Describe how to measure the initial rate of an enzyme-controlled reaction and explain its importance.
- Explain competitive, non-competitive and end-product inhibition in terms of active site binding and shape.
- Give examples of enzymes catalysing intracellular and extracellular reactions.
Topic 1: Biological Molecules flashcards
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Monosaccharide
Topic 1: Biological Molecules exam tips
Marking Points
- Monosaccharides are single sugar units, disaccharides contain two monosaccharide units and polysaccharides contain many.
- Glucose is a hexose; alpha and beta glucose differ in the position of the hydroxyl group on carbon 1 of the ring.
- Ribose is a pentose sugar with five carbon atoms.
- Condensation joins two monosaccharides, releases water and forms a glycosidic bond; hydrolysis adds water and breaks that bond.
- Glucose plus fructose forms sucrose, glucose plus galactose forms lactose, and two glucose units form maltose.
- Starch is formed from amylose and amylopectin; glycogen is a branched glucose polymer.
- Structure relates to function: glucose is soluble for respiration; starch and glycogen are compact storage; cellulose is straight-chain and structural.
- A triglyceride is synthesised from one glycerol and three fatty acids.
- Each condensation reaction between glycerol and a fatty acid forms an ester bond and releases water.
- Saturated lipids have no carbon-carbon double bonds in their fatty acid chains; unsaturated lipids have at least one.
- Lipids store energy because their hydrocarbon chains are reduced and release much energy when oxidised.
- Lipids are waterproof and provide thermal insulation.
- Phospholipids contain glycerol, two fatty acids and a phosphate group.
- The hydrophilic phosphate head and hydrophobic fatty acid tails allow phospholipids to form a bilayer in cell membranes.
- Amino acids join via condensation reactions, releasing a water molecule and forming a peptide bond between the carboxyl group of one monomer and the amine group of another.
- Disulfide bonds are strong covalent links between the R groups of cysteine amino acids, which help to stabilise the tertiary structure of a protein.
- Haemoglobin is a globular protein with a quaternary structure comprising four polypeptide chains, each containing an iron-containing haem group to bind oxygen.
- Collagen is a fibrous protein consisting of three polypeptide chains wound in a triple helix, providing high tensile strength for structural support in tissues.
- A nucleotide is composed of a pentose sugar, a phosphate group, and a nitrogenous base; phosphodiester bonds link them into a sugar-phosphate backbone.
- DNA replication is semi-conservative: DNA helicase breaks hydrogen bonds, DNA polymerase joins adjacent nucleotides, and DNA ligase joins fragments.
- During transcription in the nucleus, RNA polymerase builds a single-stranded mRNA molecule with a ribose sugar-phosphate backbone, complementary to the anti-sense DNA strand.
- Hydrogen bonds play a crucial role in mRNA function by temporarily linking its bases to the DNA template during transcription and to tRNA anticodons during translation.
- tRNA molecules have a cloverleaf structure stabilised by internal hydrogen bonds, featuring a specific anticodon and an amino acid attachment site.
- A base substitution may not change the amino acid sequence due to the degenerate genetic code, whereas insertions or deletions cause a frameshift; not all the genome codes for proteins.
- In sickle cell anaemia, a point mutation changes a single base, resulting in a different amino acid that alters haemoglobin's structure and function.
- Enzymes are globular proteins whose tertiary structure creates a compact shape with an active site formed by specific amino acid residues.
- Specificity arises because the active site's shape and chemical properties complement a particular substrate.
- The induced fit hypothesis states that the active site changes shape slightly as the substrate binds, forming an enzyme–substrate complex.
- Enzymes are catalysts that lower the activation energy of a reaction and are not used up.
- Temperature increases rate up to an optimum, then denaturation destroys the tertiary structure and the active site.
- pH affects ionic and hydrogen bonding in the tertiary structure, so each enzyme has an optimum pH.
- Increasing substrate concentration raises rate until all active sites are occupied; increasing enzyme concentration raises rate if substrate is not limiting.
- Nitrate ions provide nitrogen used to synthesise amino acids, which are polymerised into proteins.
- Nitrate ions also provide nitrogen for the bases of DNA, supporting replication and cell division.
- Calcium ions form calcium pectate with pectin, and this cements the middle lamella between adjacent plant cell walls.
- Magnesium ions are a structural component of chlorophyll, so deficiency reduces photosynthesis and causes chlorosis.
- Phosphate ions are incorporated into ADP and ATP, which transfer energy in cellular reactions.
- Water is a dipole because oxygen is slightly negative and hydrogen slightly positive, so hydrogen bonds form between molecules.
- The high specific heat capacity means a large energy change causes only a small temperature change, stabilising organism temperature and aquatic habitats.
- As a polar solvent, water dissolves ions and polar molecules, allowing transport in blood or sap and providing a medium for reactions.
- Hydrogen bonding produces surface tension, which supports small organisms and contributes to capillary action.
- Water is incompressible, so it can act as a hydrostatic skeleton and protect tissues from shock.
- Water reaches maximum density at 4 °C, so ice is less dense and floats, insulating the water beneath.
- The initial rate is measured from the gradient at the start of a progress curve or from the time for an early, defined change.
- Initial rate is important because it avoids the effects of substrate depletion, product build-up and enzyme denaturation that reduce rate later.
- Competitive inhibition involves an inhibitor with a similar shape to the substrate binding to the active site, which can be overcome by more substrate.
- Non-competitive inhibition involves an inhibitor binding away from the active site, changing its shape so substrate cannot bind; extra substrate does not overcome it.
- End-product inhibition is negative feedback in which the final product inhibits an earlier enzyme in the pathway.
- Enzymes catalyse intracellular reactions, for example in respiration and metabolism, and extracellular reactions, for example digestion.
Examiner Tips
- 💡Learn each disaccharide as a pair of named monosaccharides so you can answer quickly and accurately.
- 💡When asked about structure and function, name the structural feature first, then state the functional advantage it gives.
- 💡Use the words condensation, hydrolysis and glycosidic bond precisely in multiple-choice options.
- 💡Sketch alpha and beta glucose side by side and label carbon 1 so the difference is clear.
- 💡Count the bonds and water molecules: three fatty acids mean three ester bonds and three water molecules released.
- 💡Use the terms hydrophilic and hydrophobic when explaining bilayer formation.
- 💡Link each lipid property to a named role, such as energy storage, waterproofing, insulation or membrane function.
- 💡Check whether a question asks about triglycerides or phospholipids before choosing an answer.
- 💡When describing protein structure, explicitly state that tertiary structure bonding occurs between the variable R groups of the amino acids.
- 💡Contrast the solubility of proteins by noting that globular proteins like haemoglobin are soluble due to outward-facing hydrophilic R groups, while fibrous proteins like collagen are insoluble.
- 💡Remember that a change in the primary structure can alter the position of bonds in the tertiary structure, potentially changing the protein's three-dimensional shape and function.
- 💡Clearly distinguish between purines (adenine and guanine) and pyrimidines (cytosine, thymine, and uracil) when describing complementary base pairing.
- 💡When explaining translation, specify that hydrogen bonds form temporarily between the mRNA codon and the complementary tRNA anticodon at the ribosome.
- 💡Use sickle cell anaemia as a concrete example of how a single base substitution can alter a protein's primary, tertiary, and quaternary structure.
- 💡Link every rate statement to the underlying molecular cause, such as more enzyme–substrate complexes or loss of active site shape.
- 💡When describing graphs, name the limiting factor at each region rather than only describing the shape.
- 💡Use the term 'denaturation' precisely and state that the tertiary structure of the active site is changed.
- 💡Link each ion to one clear product: nitrate to amino acids and DNA, calcium to calcium pectate, magnesium to chlorophyll, phosphate to ADP and ATP.
- 💡Use the exact product names in answers, because vague wording such as 'helps the plant grow' earns no credit.
- 💡When a question gives a deficiency symptom, work backwards: chlorosis points to magnesium, weak cell adhesion to calcium.
- 💡Check whether the question asks about plants specifically, since these roles are stated for plants in this specification point.
- 💡Always link a property to its consequence for an organism, for example high specific heat capacity to stable enzyme-controlled reactions.
- 💡Use the phrase 'hydrogen bonds' rather than 'bonds' alone, so the examiner sees the correct intermolecular attraction.
- 💡For surface tension, give a named context such as pond skaters or capillary rise in xylem.
- 💡For incompressibility, name a use such as a hydrostatic skeleton in earthworms or shock absorption around the brain.
- 💡State clearly which variable is changed, which is measured and which are controlled in the practical.
- 💡When explaining inhibition, always say where the inhibitor binds and what happens to the active site.
- 💡Use examples of both intracellular and extracellular enzyme action to show breadth.
Common Mistakes
- Thinking all polysaccharides are identical: correction is that starch is amylose plus amylopectin, glycogen is highly branched, and cellulose has straight chains with beta glucose.
- Saying condensation adds water: correction is that condensation releases water, while hydrolysis uses water to split a glycosidic bond.
- Confusing alpha and beta glucose: correction is that they differ only in the hydroxyl group position on carbon 1.
- Mixing up the disaccharide components: correction is sucrose from glucose and fructose, lactose from glucose and galactose, maltose from two glucose units.
- Saying a triglyceride contains three ester bonds but forgetting water is released: correction is that each condensation releases one water molecule as an ester bond forms.
- Treating saturated and unsaturated as about energy content only: correction is that the key difference is the presence or absence of carbon-carbon double bonds in the fatty acid chains.
- Describing phospholipids as having three fatty acids: correction is that phospholipids have two fatty acids plus a phosphate group.
- Saying the phospholipid bilayer blocks all substances: correction is that it is a barrier to water-soluble substances while allowing lipid-soluble substances through.
- Confusing secondary and tertiary structure bonding. Secondary structure relies on hydrogen bonds between N-H and C=O groups of the polypeptide backbone, whereas tertiary structure involves ionic, hydrogen, and disulfide bonds between R groups.
- Stating that all proteins have a quaternary structure. Quaternary structure only applies to proteins composed of more than one polypeptide chain, such as haemoglobin or collagen.
- Describing peptide bond formation as a hydrolysis reaction. Peptide bonds form through condensation reactions releasing water; hydrolysis breaks them.
- Stating that DNA polymerase breaks hydrogen bonds during replication. Correction: DNA helicase breaks the hydrogen bonds; DNA polymerase forms phosphodiester bonds between adjacent nucleotides.
- Forgetting the components of a nucleotide. Correction: always specify that a nucleotide contains a pentose sugar, a phosphate group, and a nitrogenous base.
- Believing that mRNA contains internal hydrogen bonds like tRNA. Correction: mRNA is a linear molecule with a ribose sugar-phosphate backbone; it only forms temporary hydrogen bonds with DNA or tRNA.
- Confusing the sense and anti-sense strands of DNA. Correction: the anti-sense strand acts as the template during transcription, so the resulting mRNA has the same sequence as the sense strand, with uracil replacing thymine.
- Thinking the active site is rigid and exactly complementary like a lock and key; the induced fit hypothesis says it moulds around the substrate as it binds.
- Saying enzymes are 'killed' by high temperature; they are denatured because the tertiary structure and active site are disrupted, and they were never alive.
- Believing rate keeps rising indefinitely with substrate concentration; it plateaus when active sites are saturated, so enzyme concentration becomes limiting.
- Confusing the effect of pH with temperature; pH alters charges and bonding in the active site rather than adding kinetic energy.
- Thinking nitrate ions are used directly as an energy source; correction: nitrate nitrogen is built into amino acids and DNA bases, while energy transfer involves ATP.
- Confusing calcium with magnesium; correction: calcium forms calcium pectate in the middle lamella, whereas magnesium is central to chlorophyll.
- Believing magnesium is used to make DNA; correction: magnesium produces chlorophyll, and nitrate supplies the nitrogen for DNA and amino acids.
- Assuming phosphate ions only build DNA; correction: phosphate is also a component of ADP and ATP, so it is central to energy transfer.
- Saying water molecules are held together by covalent bonds between separate molecules; correction: the attraction between molecules is hydrogen bonding, while covalent bonds hold each water molecule's atoms together.
- Confusing high specific heat capacity with high heat content; correction: it means a large energy input produces only a small temperature rise.
- Claiming ice is denser than liquid water; correction: water is most dense at 4 °C, so ice floats and insulates the water below.
- Describing water as a universal solvent; correction: it is a polar solvent that dissolves polar and ionic substances, not all substances.
- Measuring rate over the whole reaction rather than the initial period; later slowing due to substrate depletion or product accumulation gives a misleading value.
- Saying competitive inhibition cannot be overcome; increasing substrate concentration can outcompete the inhibitor at the active site.
- Confusing non-competitive inhibition with competitive inhibition; non-competitive inhibitors bind away from the active site and change its shape.
- Describing end-product inhibition as the product activating the pathway; it inhibits an earlier enzyme to switch the pathway off.