Interpret yeast temperature and gas-rate data
Calculate rates and identify the fastest tested condition.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
Before you begin
- Recall cell processes, enzymes and the distinction between gas exchange and breathing.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which calculation gives a gas-volume rate?
What you will learn
- Use consistent volume and time
- Describe the rise and fall
Interpret yeast temperature and gas-rate data
| Feature | Meaning |
|---|---|
| 10 °C | 4 cm³ / 4 min = 1 cm³/min |
| 20 °C | 12 / 4 = 3 cm³/min |
| 30 °C | 20 / 4 = 5 cm³/min |
| 40 °C | 28 / 4 = 7 cm³/min |
| 60 °C | 4 / 4 = 1 cm³/min |
Use consistent volume and time
The invented data show gas collected over four minutes: 4, 12, 20, 28 and 4 cm³ at 10, 20, 30, 40 and 60 °C respectively. Calculate the mean gas-production rate by dividing collected volume by elapsed time. Keep the units as cm³/min; a larger volume is not automatically a larger rate if the collection periods differ.
These observations give rates of 1, 3, 5, 7 and 1 cm³/min. The rate increases over the tested 10–40 °C range and decreases strongly by 60 °C. Forty degrees is the fastest tested condition in this data set. An exact optimum is not established because temperatures between the chosen points were not sampled, and the result is not a universal yeast optimum.
Describe the rise and fall
Respiration involves enzyme-catalysed reactions. Within a suitable lower range, warming can increase reaction rate as particles move more rapidly and effective interactions become more frequent. At sufficiently high temperatures, enzyme structure can be disrupted and yeast cells can be damaged, reducing gas production. A falling high-temperature rate is consistent with this explanation under the controlled method.
Check the method before assigning cause: a leak, depleted glucose or inconsistent oxygen conditions can also alter collection. Collected gas is a proxy and can be influenced by dissolution and temperature-dependent gas behaviour. Do not call cold yeast necessarily dead, or describe enzymes as living organisms that die. A single graph describes the evidence supplied, not every strain or laboratory condition.
Sources for this lesson
Pause and explain
What can the invented data establish about 40 °C?
Worked example
At 30 °C, 20 cm³ is collected in four minutes. Find the rate.
Show the worked solution
- Use volume divided by time.
- 20 ÷ 4 = 5.
Answer 5 cm³/min.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the rates at 20 and 40 °C using 12 and 28 cm³ in four minutes.
Give me a hint
Divide each volume by four.
Compare my reasoning
- 12 ÷ 4 = 3.
- 28 ÷ 4 = 7.
20 °C: 3 cm³/min; 40 °C: 7 cm³/min.
Look for these in your work
- Calculate both rates
- Use cm³/min
Describe the whole observed temperature pattern and name the fastest tested point.
Give me a hint
Include the high-temperature decrease.
Compare my reasoning
- Rates rise from 10 to 40 °C.
- The rate falls at 60 °C; 40 °C is highest among the tested points.
Gas-production rate rises over 10–40 °C, then falls by 60 °C; 40 °C is fastest tested, not an established universal optimum.
Look for these in your work
- Describe both portions
- Qualify the optimum claim
A supplied trial collects 18 cm³ in six minutes and another collects 16 cm³ in four minutes. Which is faster?
Give me a hint
Compare rates rather than volumes.
Compare my reasoning
- 18 ÷ 6 = 3 cm³/min.
- 16 ÷ 4 = 4 cm³/min.
The second is faster: 4 versus 3 cm³/min, despite collecting less total gas.
Look for these in your work
- Normalise by time
- Compare the calculated rates
Common mistakes
- Using volume without elapsed time.
- Claiming exactly 40 °C is optimal for all yeast.
What can you explain now?
What can the invented data establish about 40 °C?
Compare with the explanation
It is the fastest tested temperature
Closer sampling is needed for the exact optimum.
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Make it stick
Calculate a rate and qualify the fastest-tested-temperature conclusion.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.