Donat Sciences and MathsBiology · Virtual lab worksheet

School: ........................................ Name: ................................ Class: ........ Date: ............

Population growth: yeast in a flask, impala and lions

Aim: To investigate the growth of a population in a closed environment and identify the phases of the growth curve, to estimate population size by sampling with a counting chamber, and to describe how predator and prey populations interact.

Apparatus (real laboratory): dried yeast, sugar solution, flasks, microscope, haemocytometer, methylene blue.

What to do (with the simulation on www.donatstem.co.rw)

1. Choose yeast, 30 °C, 5% sugar, closed flask. Tap Start. A sample is counted every 6 hours and added to the table.
2. Check one count yourself: cells counted ÷ 16 squares ÷ 0.25 = million cells per cm³.
3. Name the four phases on your curve. When does the sugar run out?
4. Repeat at 20 °C, at 10% sugar, and with fresh sugar added every day. Compare the curves.
5. Choose impala and lions. Start with 7 lions and run 40 years. Describe how the two populations change.
6. Try more grass (higher carrying capacity), no lions, and 10% poaching. What happens to each population?

Results

Questions

1. Name the four phases of a population growth curve in a closed flask. [2]
2. What is meant by carrying capacity? [2]
3. Why does the yeast population fall at the end? [2]
4. 80 yeast cells are counted in 16 small squares, each with a volume of 0.00025 mm³. Estimate the number of cells per cm³. [2]
5. Why do the lion peaks come after the impala peaks? [2]
6. Give two density-dependent factors that limit a population. [2]
7. Sketch and label a sigmoid population growth curve for yeast in a closed flask, and explain the cause of each phase. [8]
8. Explain how a predator population and its prey population affect each other over time. [4]
9. Suggest two reasons why the number of impala in a park may stay below the carrying capacity. [2]
10. A student counts yeast cells in 4 samples: 18, 22, 20 and 24 cells in 16 squares. Calculate the mean and explain why several samples are counted. [3]

© Donat Sciences and Maths · www.donatstem.co.rw

Donat Sciences and MathsBiology · Virtual lab worksheet

School: ........................................ Name: ................................ Class: ........ Date: ............

Marking scheme: Population growth: yeast in a flask, impala and lions

Expected results

Yeast model: growth rate up to 0.30 per hour at 32 °C (slower in the cold, falling fast above 32 °C), a 5-hour lag, growth limited by sugar and crowding, sugar used in proportion to growth; deaths rise once the sugar has been gone for about 10 hours. At 30 °C and 5% sugar: about 1 million cells/cm³ at 12 h, 18 at 24 h, a peak of about 22 at 30–36 h, then a slow decline to about 4 at 96 h. Counting chamber: each small square is 0.00025 mm³ (2.5 × 10⁻⁷ cm³), so cells per square × 4 = million cells per cm³; counts vary at random like real samples. Park model (logistic prey + predator): impala growth 0.5 per year, carrying capacity set by the grass, predation 0.01 per lion per year, lion growth 0.02 × predation, lion death 0.15 per year. With 7 lions and 3 000 capacity the populations cycle with a period of about 25 years. Real Akagera numbers differ: the model shows the pattern, not a forecast.

Answers

1. Name the four phases of a population growth curve in a closed flask. [2]

   Lag phase (slow growth while cells adapt), log or exponential phase (rapid growth), stationary phase (birth rate = death rate) and death or decline phase (deaths exceed births).

2. What is meant by carrying capacity? [2]

   The maximum population size that an environment can support with its available food, space and other resources.

3. Why does the yeast population fall at the end? [2]

   The sugar runs out and toxic wastes such as ethanol build up, so the death rate becomes greater than the birth rate.

4. 80 yeast cells are counted in 16 small squares, each with a volume of 0.00025 mm³. Estimate the number of cells per cm³. [2]

   80 ÷ 16 = 5 cells per square; 5 ÷ 0.00025 = 20 000 cells per mm³ = 2 × 10⁷ (20 million) cells per cm³.

5. Why do the lion peaks come after the impala peaks? [2]

   Lions need time to breed after food becomes plentiful; when there are many lions the impala fall, then the lions run short of food and fall too.

6. Give two density-dependent factors that limit a population. [2]

   Food supply, predation, disease, competition for space (any two).

7. Sketch and label a sigmoid population growth curve for yeast in a closed flask, and explain the cause of each phase. [8]

   Correct S-shaped curve with axes (2); lag – cells adapt, making enzymes (1–2); log – plenty of food, rapid division (1–2); stationary – limiting factors, births = deaths (1); decline – food used up / toxins (1).

8. Explain how a predator population and its prey population affect each other over time. [4]

   When prey increase, predators have more food and increase (1); more predators eat more prey, so prey decrease (1); with less food the predators decrease (1); fewer predators let the prey increase again, so both oscillate with the predator peak after the prey peak (1).

9. Suggest two reasons why the number of impala in a park may stay below the carrying capacity. [2]

   Predation by lions (1); poaching / disease / drought reducing grass (1).

10. A student counts yeast cells in 4 samples: 18, 22, 20 and 24 cells in 16 squares. Calculate the mean and explain why several samples are counted. [3]

    Mean = 84 ÷ 4 = 21 cells (1); cells are not spread evenly, so one sample may be unrepresentative (1); a mean of several samples is more reliable (1).

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