BW Abitur Biology Bilingual EN study materials
BW Abitur Biology Bilingual EN study resources

BW Abitur Biology Bilingual EN: Stoffwechsel und Energieumsatz / Metabolism and bioenergetics

BW Abitur Biology Bilingual EN · Work through your exam, one topic at a time.

All guide sections

Stoffwechsel und Energieumsatz / Metabolism and bioenergetics

Metabolism items combine pathways with experiments: light-intensity and CO2 curves, the Hill reaction with DCPIP, respirometer readings, and inhibition kinetics. You should be able to place each reaction in its compartment, name its electron carrier, and justify a rate change by identifying the limiting factor.

Photosynthesis

  • Light reactions occur at the thylakoid membrane: water is split at photosystem II, supplying electrons and protons; electrons pass via plastoquinone, cytochrome b6f and plastocyanin to photosystem I, then via ferredoxin to NADP+ reductase. The O2 released comes from water, not CO2.
  • Chemiosmosis: protons accumulate in the thylakoid lumen; ATP synthase uses their return flow to make ATP. Products: ATP, NADPH, O2.
  • Calvin cycle in the stroma: RuBisCO fixes CO2 onto RuBP, giving 2 × 3-phosphoglycerate, which is reduced using ATP and NADPH to G3P; most G3P regenerates RuBP, a small fraction is exported. Three turns give one net G3P, six turns one glucose.
  • Photorespiration: at high temperature and closed stomata, RuBisCO binds O2 instead of CO2, wasting carbon. C4 plants separate initial fixation from the Calvin cycle spatially, CAM plants temporally.

Cellular respiration

  • Glycolysis (cytosol): glucose → 2 pyruvate, net 2 ATP and 2 NADH.
  • Link reaction and Krebs cycle (matrix): per glucose 6 NADH, 2 FADH2, 2 ATP, 2 CO2-rich steps releasing CO2.
  • Oxidative phosphorylation at the inner mitochondrial membrane: electrons pass down the chain, protons are pumped, and ATP synthase uses the return flow; O2 is the terminal electron acceptor. Exam convention: about 30–32 ATP per glucose.
  • Fermentation regenerates NAD+ so glycolysis can continue without O2.

Worked example

An enzyme assay gives Vmax = 100 units/min and Km = 2 mmol/L. With inhibitor A present: Vmax = 100, Km = 6 mmol/L. Reasoning: Vmax is unchanged, so the inhibitor does not remove active enzyme; Km is tripled, so more substrate is needed for half saturation - the inhibitor competes for the active site and its effect is overcome by raising substrate concentration. Inhibitor B (Vmax = 40, Km = 2) would instead be non-competitive.

Traps that cost marks

Saying O2 comes from CO2; writing NADH where the Calvin cycle needs NADPH; describing ATP production by the chain without chemiosmosis; treating the 10% or ATP-yield figures as exact rather than conventional; forgetting that light is not the limiting factor once the curve plateaus.