How do the structures of lipids, proteins and nucleic acids equip them for their biological jobs?
2.1.2 Biological molecules: the structure and function of triglycerides and phospholipids; the structure of amino acids, the formation of peptide bonds and the four levels of protein structure; the structure of nucleotides, DNA and RNA; the biochemical tests for lipids (emulsion test) and proteins (biuret test).
A focused answer to the OCR H420 2.1.2 dot point on lipids, proteins and nucleic acids. Covers triglycerides and phospholipids, amino acids and the four levels of protein structure, nucleotide and DNA and RNA structure, and the emulsion and biuret tests.
Reviewed by: AI editorial process; not yet individually human-reviewed
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What this dot point is asking
OCR wants you to relate the structure of triglycerides and phospholipids to their roles, describe amino acids and the four levels of protein structure with the bonds that stabilise each, outline nucleotide, DNA and RNA structure, and carry out the emulsion and biuret tests.
The answer
Lipids
A triglyceride is one glycerol joined to three fatty acids by ester bonds, each formed in a condensation reaction (three condensations, three waters released). Fatty acids may be saturated (no carbon-carbon double bonds, solid fats) or unsaturated (one or more double bonds, liquid oils). Triglycerides are excellent energy stores: they are insoluble (so do not affect water potential), and release more than twice the energy per gram of carbohydrate; they also provide thermal insulation and, around organs, protection.
A phospholipid replaces one fatty acid with a phosphate group, giving a hydrophilic phosphate head and two hydrophobic fatty-acid tails. In water, phospholipids form a bilayer with heads facing the water and tails inward, which is the basis of every cell membrane.
Proteins
Amino acids share a central carbon bonded to an amino group, a carboxyl group, a hydrogen and a variable R group that gives each its identity. Two amino acids join by a condensation reaction, forming a peptide bond and releasing water. The four levels of structure:
- Primary: the sequence of amino acids, set by the gene.
- Secondary: local folding into alpha-helices or beta-pleated sheets, held by hydrogen bonds along the backbone.
- Tertiary: the overall three-dimensional fold of one polypeptide, stabilised by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions between R groups.
- Quaternary: two or more polypeptide chains (plus any prosthetic groups, such as the haem in haemoglobin) assembled together.
Globular proteins (for example haemoglobin and enzymes) fold into compact, soluble shapes with hydrophilic R groups outward and play metabolic and transport roles. Fibrous proteins (for example collagen and keratin) form long, insoluble, strong strands for structure and support.
Nucleic acids
A nucleotide is a pentose sugar, a phosphate group and a nitrogenous base. In DNA the sugar is deoxyribose and the bases are adenine, thymine, cytosine and guanine; in RNA the sugar is ribose and uracil replaces thymine. Nucleotides join by phosphodiester bonds (condensation) to form a polynucleotide. DNA is a double helix of two antiparallel strands held by hydrogen bonds between complementary base pairs: adenine with thymine (2 hydrogen bonds) and cytosine with guanine (3 hydrogen bonds). RNA is usually single-stranded and shorter.
The biochemical tests
- Lipids (emulsion test). Dissolve the sample in ethanol, then add water. A white emulsion forms if lipid is present.
- Proteins (biuret test). Add biuret reagent (sodium hydroxide then copper(II) sulfate). A colour change from blue to purple/lilac confirms peptide bonds (protein).
Examples in context
Example 1. Membranes. The amphipathic nature of phospholipids makes them assemble into the bilayer that forms every cell and organelle membrane, linking this topic to membrane transport later in the course.
Example 2. Haemoglobin and oxygen transport. Haemoglobin's quaternary structure (four polypeptides, each with a haem group) underpins cooperative oxygen binding, which you meet again in the oxygen dissociation curve in Module 3.
Try this
Q1. Name the bond formed between two amino acids and the reaction that forms it. [2 marks]
- Cue. A peptide bond, formed by a condensation reaction (releasing water).
Q2. Explain why phospholipids form a bilayer in water. [2 marks]
- Cue. They are amphipathic; the hydrophilic phosphate heads face the water and the hydrophobic tails point inward away from water.
Q3. State the positive result of the biuret test for protein. [1 mark]
- Cue. A colour change from blue to purple (lilac).
Exam-style practice questions
Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
OCR H420/01 20196 marksDescribe the four levels of structure of a protein, and explain how the structure of a globular protein such as haemoglobin differs from a fibrous protein such as collagen. [This question is marked using levels of response.]Show worked answer →
A Level-of-Response answer needs the four levels accurately, then a structured comparison.
Primary structure: the sequence of amino acids in the polypeptide, joined by peptide bonds. Secondary structure: local folding into alpha-helices or beta-pleated sheets, held by hydrogen bonds between the backbone. Tertiary structure: the overall three-dimensional fold of one polypeptide, stabilised by hydrogen bonds, ionic bonds, disulfide bridges and hydrophobic interactions between R groups. Quaternary structure: two or more polypeptide chains (and any prosthetic groups) assembled together.
Comparison: Haemoglobin is globular: it folds into a compact, roughly spherical shape with hydrophilic R groups outward, so it is soluble and can transport oxygen via its four haem groups (quaternary structure). Collagen is fibrous: three polypeptide chains wind into a rope-like triple helix with cross-links, giving an insoluble molecule of high tensile strength for support.
Top marks need accurate bonding at each level and at least two clear structure-function links in the comparison.
OCR H420/01 20203 marksDescribe how you would test a food sample for the presence of lipids, and state the positive result.Show worked answer →
The emulsion test, three marks for method plus result.
- Add the sample to ethanol and shake to dissolve any lipid, then decant into water.
- A positive result is a white (cloudy) emulsion forming, because the lipid comes out of solution as tiny droplets when mixed with water.
- A control with no lipid stays clear.
Markers reward the ethanol then water sequence and the white emulsion as the positive result.
Related dot points
- 2.1.2 Biological molecules: the properties of water and their importance to living organisms; the structure of monosaccharides, the formation of glycosidic bonds by condensation, and the structure and function of starch, glycogen and cellulose; the biochemical tests for reducing and non-reducing sugars and for starch.
A focused answer to the OCR H420 2.1.2 dot point on water and carbohydrates. Covers the biologically important properties of water, monosaccharides and condensation, the structure and function of starch, glycogen and cellulose, and the Benedict's and iodine tests.
- 2.1.4 Enzymes: the role of enzymes as biological catalysts in metabolic reactions; the mechanism of enzyme action including the lock-and-key and induced-fit models; the effects of temperature, pH, enzyme and substrate concentration on the rate of reaction; the action of competitive and non-competitive inhibitors; the roles of cofactors, coenzymes and prosthetic groups.
A focused answer to the OCR H420 2.1.4 dot point on enzymes. Covers enzymes as catalysts, the lock-and-key and induced-fit models, activation energy, the effects of temperature, pH and concentration, competitive and non-competitive inhibition, and cofactors.
- 2.1.3 Nucleotides and nucleic acids: the semi-conservative replication of DNA and the roles of DNA helicase, DNA polymerase and the complementary base pairing rule; the nature of the genetic code as a triplet code that is degenerate and non-overlapping; the roles of mRNA and tRNA in protein synthesis.
A focused answer to the OCR H420 2.1.3 dot point on DNA replication and the genetic code. Covers semi-conservative replication, the roles of DNA helicase and DNA polymerase, the Meselson-Stahl evidence, and the triplet, degenerate, non-overlapping code with transcription and translation.
- 2.1.1 Cell structure: the ultrastructure of eukaryotic and prokaryotic cells, the function of organelles including the role of the rough endoplasmic reticulum and Golgi apparatus in producing and secreting proteins; the use, calibration and resolution of light and electron microscopes.
A focused answer to the OCR H420 2.1.1 dot point on cell structure and microscopy. Covers every required eukaryotic and prokaryotic organelle, the protein secretory pathway, the three microscopes, eyepiece-graticule calibration and the magnification equation.
- 6.1.1 Cellular control: the nature of gene mutations and their effects on proteins; the control of gene expression at the transcriptional level, including operons (the lac operon) and transcription factors; the role of homeobox (Hox) genes in body plan development; and the role of apoptosis (programmed cell death).
A focused answer to the OCR H420 6.1.1 dot point on cellular control. Covers gene mutations and their effects, the control of transcription by the lac operon and transcription factors, the role of homeobox (Hox) genes in body plan development, and apoptosis.
Sources & how we know this
- OCR A Level Biology A (H420) Specification — OCR (2023)