SME
Principles of Biology I · Problem type
How efficient respiration is
Use this for “How efficient is respiration if one glucose makes 32 ATP?”
Chart
This is an example. Type your own values to start a new problem.
η (%)
=
ATP
×
30.5
÷
2870
×
100
E
Energy kept in ATP
calculated · kJ/mol
Eglucose
Energy glucose releases
worked out (not typed) · kJ/mol
Formulas
E
=
ATP
×
30.5
976 = 32 × 30.5
Eglucose = 2,870
2,870 = 2,870
η
=
E
Eglucose
×
100
34
=
976
2,870
×
100
Assumptions
•
Glucose oxidized to CO₂ and H₂O releases 2870 kJ/mol; each ATP keeps 30.5 kJ/mol. Both energies are per mole of glucose.
•
Today’s count is 30–32 ATP per glucose; older books say 36–38, which the menu keeps for their questions.
•
Efficiency is the energy kept in ATP over the energy glucose releases; the rest leaves as heat.
Step-by-step
WE KNOW
ATP = 32
FIND
energy kept in ATP (E), energy glucose releases (Eglucose), efficiency (η)
Step 1 · Find energy kept in ATP (E)
Use E = ATP × 30.5
Each ATP made from one glucose keeps 30.5 kJ/mol.
E
=
32
×
30.5
E
=
976
kJ/mol
Step 2 · Find energy glucose releases (Eglucose)
Use Eglucose = 2,870
Each mole of glucose releases 2870 kJ when oxidized completely.
Eglucose = 2,870 kJ/mol
Step 3 · Find efficiency (η)
Use η = E ÷ Eglucose × 100
The share of glucose’s energy kept in ATP, as a percent.
η
=
976
2,870
×
100
η
=
0.3401
×
100
η
=
34%
Check
Put the numbers back in. Both sides should match.
976 = 32 × 30.5 ✓
2,870 = 2,870 ✓
34
=
976
2,870
×
100
✓
Topic check
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