7.1.7Unit 3: Practical Skills in Biology I
Core Practical 15: Investigation of Respiration in Yeast — IAL Edexcel Linear Revision Notes & Topical Questions
5 min read 5 parts Lesson 24 of 27 Unit quiz
Curriculum Equation Key معادلة التنفس الخلوي الهوائي الموزونة - Balanced Cellular Respiration EquationFast Exam Solution
C₆H₁₂O₆ + 6O₂ ➔ 6CO₂ + 6H₂O + 36-38 ATP
Word Equation & Conditions: Glucose + Oxygen -> Carbon Dioxide + Water + Energy (36-38 ATP) | جلوكوز + أكسجين ← ثاني أكسيد كربون + ماء + طاقة
Location & Specifications: الميتوكوندريا (Mitochondria) والسيتوسول.
Part 1 of 5Principle of the Method
Redox indicator
- Redox indicator DCPIP and methylene blue are blue when oxidised and colourless when they gain hydrogen and become reduced.
Dehydrogenase
- Dehydrogenase Enzymes remove hydrogen from substrates during respiration, normally passing it to NAD.
Dye as acceptor
- Dye as acceptor The dye takes the hydrogen in place of NAD, so its decolourisation shows how fast hydrogen is being released.
Rate
- Rate The faster the respiration, the shorter the time to decolourise, so rate = 1 ÷ time with units s⁻¹.
Conditions
- Conditions Dye is added to a yeast suspension in glucose and the aim is to compare rates of anaerobic respiration.
Part 2 of 5Method
Either temperature or substrate concentration may be the independent variable.
1
Place a measured volume of yeast suspension and glucose solution in a test tube, then equilibrate it for five minutes in a water bath at the chosen temperature.
2
Add a measured volume of the indicator, start a stopwatch immediately and mix gently.
3
Stop the clock when the blue colour has gone and record the time to the nearest second, with the same person judging the end point each time.
4
Repeat at a range of temperatures such as 30, 35, 40 and 45 °C, or at different glucose concentrations, keeping all other factors the same.
5
Repeat each condition several times and calculate a mean time before converting it to a rate.
Part 3 of 5Controls and Interpretation
Controls and Interpretation
Variables to control
- Variables to control Keep constant the volumes of yeast and dye, the glucose concentration or temperature, the type of substrate and the pH.
Control tube
- Control tube Use a tube of yeast without dye to compare against, because the yeast suspension itself has a slight yellow colour.
Expected trend
- Expected trend Up to the optimum, higher temperature gives faster decolourisation because enzymes and substrate have more kinetic energy.
High temperatures
- High temperatures Enzymes denature above their optimum, so the dye may stay blue and the rate falls.
Subjectivity
- Subjectivity Judging the end point by eye is a source of error that a colorimeter would reduce.
Part 4 of 5Worked Example
Question
- A yeast suspension takes 120 s to decolourise DCPIP at 30 °C and 60 s at 40 °C.
- Calculate the rate at each temperature and the ratio of the rates.
- Rate = 1 ÷ time.
- At 30 °C: 1 ÷ 120 = 0.00833 s⁻¹.
- At 40 °C: 1 ÷ 60 = 0.0167 s⁻¹.
- Ratio = 0.0167 ÷ 0.00833 = 2.0.
Answer
- 0.0083 s⁻¹ at 30 °C and 0.017 s⁻¹ at 40 °C;
- the rate doubles.
Part 5 of 5Exam Checklist & Tips
7.1.7 – Key Points for the Exam
- Redox indicator A dye that changes colour when it is oxidised or reduced.
- Dehydrogenase An enzyme that removes hydrogen from a substrate in respiration.
- Anaerobic respiration Release of energy from glucose without oxygen, as in yeast.
- Rate of reaction 1 divided by the time taken, in s⁻¹.
Common Mistakes
- Starting the timer before adding the dye, which includes mixing time and temperature changes.
- Treating a longer time as a faster rate; rate is inversely related to time.
- Forgetting to equilibrate each tube at temperature before the dye is added.
- Failing to use a no-dye control, making the end point harder to judge.
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