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Eduqas A-level Biology (A400QS) Β· Core Concepts (assessed across all three components)
Mini-Lesson Β· A-level

Core Concepts

The Eduqas Core Concepts are fundamental and may be assessed in any of the three components. They are: biological compounds, cell structure and organisation, cell membranes and transport, enzymes, and nucleic acids and their functions.

biological compounds cells & membranes enzymes & nucleic acids three strands you must be able to link together

Work through each screen, answer the questions as you go β€” several are A-level calculations β€” and collect ⭐ stars. Press Start when you are ready.

Core Concept 1 Β· biological compounds

Water, carbohydrates, lipids and proteins

Water is polar and hydrogen bonds to itself: hence its role as a solvent, its high specific heat capacity (temperature buffering), its high latent heat of vaporisation (cooling by evaporation), and its cohesion and surface tension.

Carbohydrates: monosaccharides join by condensation, forming a glycosidic bond. Starch and glycogen are branched, coiled, insoluble stores of Ξ±-glucose; cellulose is straight chains of Ξ²-glucose hydrogen-bonded into microfibrils.

Lipids: a triglyceride is glycerol + three fatty acids joined by three ester bonds. Saturated fatty acids have no C=C; unsaturated ones do. A phospholipid has a hydrophilic phosphate head and hydrophobic tails.

Proteins: amino acids join by peptide bonds (condensation). Primary β†’ secondary (Ξ±-helix, Ξ²-pleated sheet; hydrogen bonds) β†’ tertiary (hydrogen, ionic and disulfide bonds and hydrophobic interactions) β†’ quaternary. Fibrous proteins (collagen) are structural and insoluble; globular proteins (enzymes, haemoglobin) are soluble and functional.

Qualitative reagents: iodine β†’ blue-black for starch; Benedict’s (heated) β†’ brick-red for a reducing sugar; Biuret β†’ purple for protein; emulsion test β†’ white emulsion for lipid.

Match it

Which bond?

Tap the molecules being joined, then the bond that joins them.

Molecules joined
Bond
Calculate

Your turn β€” peptide bonds

1A single polypeptide chain contains 124 amino acids. Calculate the number of peptide bonds in the chain.
peptide bonds
Hint: Each bond joins two amino acids, so a chain of n amino acids has n βˆ’ 1 bonds.
Core Concept 2 Β· cell structure

Cell structure, organisation and microscopy

Eukaryotic cells: nucleus (envelope, pores, nucleolus), rough ER (protein synthesis and transport), smooth ER (lipid synthesis), Golgi (modification and packaging; makes lysosomes), mitochondria (cristae, matrix), lysosomes, 80S ribosomes; plants also have a cellulose wall, chloroplasts and a vacuole.

Prokaryotic cells: circular DNA, plasmids, 70S ribosomes, a murein (peptidoglycan) wall, sometimes a capsule and flagellum β€” and no membrane-bound organelles.

Levels of organisation: organelle β†’ cell β†’ tissue β†’ organ β†’ organ system β†’ organism.

magnification = image size Γ· actual sizerearranged: actual size = image size Γ· magnification. Convert both to the SAME unit first!

Magnification vs resolution: magnification is how much bigger the image is; resolution is the smallest distance at which two points can still be distinguished. Resolution is limited by wavelength β€” which is why the electron microscope (resolving ~0.1 nm) beats the light microscope (~200 nm), and why simply magnifying a light image further just gives a bigger blur.

Calculate

Your turn β€” magnification

2A drawing of an Amoeba measures 50 mm across. The actual organism is 25 Β΅m across. Calculate the magnification of the drawing. (1 mm = 1000 Β΅m)
Γ—
Hint: 50 mm = 50 000 Β΅m. Magnification = 50 000 Γ· 25.
Core Concept 3 Β· membranes and transport

The fluid-mosaic membrane, transport and water potential

The fluid-mosaic model: a phospholipid bilayer with intrinsic (channel and carrier) proteins, extrinsic proteins, cholesterol regulating fluidity, and glycoproteins and glycolipids for recognition.

  • Diffusion β€” small non-polar molecules straight through the bilayer, down the gradient. Passive.
  • Facilitated diffusion β€” polar molecules and ions through channel or carrier proteins, down the gradient. Passive.
  • Osmosis β€” water down the water potential gradient through a partially permeable membrane (largely through aquaporins).
  • Active transport β€” a carrier protein plus ATP, moving a substance against its gradient.
  • Endocytosis / exocytosis β€” bulk transport in vesicles; both require ATP.
ψ = ψs + ψppure water: ψ = 0. Adding solute makes ψ negative. Water moves from HIGHER (less negative) to LOWER (more negative) water potential.

Plant cell: in a dilute solution it becomes turgid (the wall pushes back, so ψp rises); in a concentrated one, plasmolysed. An animal cell has no wall, so in pure water it bursts. This is why the practical determines water potential by finding the concentration at which there is no change in mass or length.

Calculate

Your turn β€” water potential

3A cell has a solute potential (ψs) of βˆ’1200 kPa and a pressure potential (ψp) of +400 kPa. Calculate its water potential.
kPa
Hint: ψ = ψs + ψp = (βˆ’1200) + (+400).
Sort it

Does it need ATP?

Tap a process, then tap the group it belongs to.

➑️ Passive (no ATP)

⚑ Active (needs ATP)

πŸ’§ Water only

Core Concept 4 Β· enzymes

Enzymes and inhibition

Enzymes are globular proteins that lower the activation energy of a reaction. The active site is complementary to the substrate; its shape comes from the tertiary structure. The induced fit model says the active site changes shape slightly as the substrate binds, straining the substrate’s bonds.

  • Temperature β€” rate rises with kinetic energy; past the optimum, the bonds holding the tertiary structure break and the enzyme is denatured.
  • pH β€” extremes disrupt the ionic and hydrogen bonds of the tertiary structure, changing the active site.
  • Substrate concentration β€” the rate rises, then plateaus when all the active sites are saturated.
  • Competitive inhibitor β€” similar shape to the substrate; binds the active site; effect overcome by adding more substrate.
  • Non-competitive inhibitor β€” binds elsewhere and distorts the active site; not overcome by more substrate.

Immobilised enzymes (trapped in alginate beads) can be reused, do not contaminate the product, and are more stable to changes in pH and temperature β€” which is why industry uses them.

Calculate

Your turn β€” rate of reaction

4An enzyme-catalysed reaction produces 12 cmΒ³ of gas in 90 seconds. Calculate the mean rate of reaction in cmΒ³ per minute.
cm³ min⁻¹
Hint: 90 s = 1.5 minutes. Rate = 12 Γ· 1.5.
Quick check

Diagnose the inhibitor

?An inhibitor slows an enzyme reaction. Adding much more substrate restores the original maximum rate. Which type of inhibitor is it?
Core Concept 5 Β· nucleic acids

DNA, RNA and replication

A nucleotide = pentose sugar + phosphate + nitrogenous base. Nucleotides join by phosphodiester bonds to form a sugar–phosphate backbone.

DNA β€” an antiparallel double helix. A–T (two hydrogen bonds) and C–G (three). RNA β€” single-stranded, ribose, uracil instead of thymine.

Semi-conservative replication: DNA helicase breaks the hydrogen bonds; free nucleotides pair with the exposed bases; DNA polymerase joins them, always working 5′β†’3′ (hence a leading and a lagging strand). Each new molecule has one original and one new strand β€” as Meselson and Stahl proved using ¹⁡N and ¹⁴N.

Core Concept 5 Β· protein synthesis

Transcription and translation

Transcription (nucleus): RNA polymerase reads the DNA template strand and builds a complementary mRNA. In eukaryotes the primary transcript is spliced: non-coding introns are removed and the coding exons joined. The mature mRNA leaves through a nuclear pore.

Translation (ribosome): each mRNA codon is recognised by the complementary anticodon of a tRNA carrying a specific amino acid. The ribosome catalyses peptide bond formation and moves along one codon at a time until it reaches a stop codon.

The genetic code is triplet, non-overlapping, degenerate (so some base substitutions are silent) and effectively universal β€” which is why a human gene can be expressed in a bacterium.

ATP is the universal energy currency: hydrolysis of ATP β†’ ADP + Pi releases a readily usable quantity of energy, and the phosphate group can be transferred to another molecule to make it more reactive (phosphorylation). ATP is small, soluble, and easily and rapidly regenerated.

Quick check

Where does it happen?

?In a eukaryotic cell, where does translation take place, and what is the role of tRNA?
Quick check

Membrane structure

?Why is the model of the membrane described as "fluid mosaic"?
Recap

The big ideas to know

Water: polar; hydrogen bonding gives it a high specific heat capacity, a high latent heat of vaporisation, cohesion and surface tension, and makes it an excellent solvent.

Carbohydrates: Ξ±-glucose β†’ starch/glycogen (store); Ξ²-glucose β†’ cellulose (structure). Glycosidic bonds.

Lipids: triglyceride = glycerol + 3 fatty acids, ester bonds. Phospholipids form the bilayer.

Proteins: primary (peptide bonds) β†’ secondary (H-bonds, Ξ±-helix/Ξ²-sheet) β†’ tertiary (H-, ionic, disulfide bonds, hydrophobic interactions) β†’ quaternary.

Cells: eukaryotic ultrastructure (nucleus, rER, Golgi, mitochondria, lysosomes, chloroplasts); prokaryotic (70S ribosomes, murein wall, no membrane-bound organelles). Magnification = image Γ· actual size.

Membrane: fluid-mosaic. Transport: diffusion, facilitated diffusion, osmosis, active transport, endocytosis and exocytosis. ψ = ψs + ψp.

Enzymes: globular proteins; lower activation energy; specificity from the tertiary structure of the active site. Competitive inhibition is overcome by more substrate; non-competitive is not.

Nucleic acids: DNA is an antiparallel double helix (A–T, C–G); replication is semi-conservative; transcription β†’ mRNA β†’ translation on the ribosome.

You have covered all five Eduqas Core Concepts. Press Finish to see your score.

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