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Yeast cells can respire aerobically or anaerobically - AQA - A-Level Biology - Question 3 - 2018 - Paper 1

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Yeast cells can respire aerobically or anaerobically. A student used the apparatus shown in Figure 3 to measure the rate of respiration in yeast. She: - positioned... show full transcript

Worked Solution & Example Answer:Yeast cells can respire aerobically or anaerobically - AQA - A-Level Biology - Question 3 - 2018 - Paper 1

Step 1

Suggest one reason why it was important that the student left the apparatus for one hour after the yeast culture reached a constant temperature.

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Answer

Leaving the apparatus for one hour ensures that the temperature remains stable. This allows for a consistent environment for the yeast cells, which is crucial for obtaining accurate measurements of the respiration rate.

Step 2

Explain why it moved to the right.

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Answer

The coloured liquid moved to the right due to the consumption of oxygen by the yeast for respiration. As yeast cells respire, they absorb oxygen, causing a reduction in pressure within the apparatus. This pressure decrease draws the liquid in the capillary tubing towards the flask.

Step 3

Calculate the volume of gas produced in cm³ hour⁻¹. Show your working.

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Answer

To calculate the volume of gas produced, we first need to find the radius of the capillary tubing. The diameter is given as 1 mm, thus the radius (r) is:

r=1 mm2=0.5 mm=0.05 cmr = \frac{1 \text{ mm}}{2} = 0.5 \text{ mm} = 0.05 \text{ cm}

Next, we calculate the volume of gas produced in 24 hours using the given length (l) of the capillary tubing which can be calculated from the movement of the liquid. Since the liquid moved 1.5 cm in 24 hours, we have:

l=1.5 cml = 1.5 \text{ cm}

Now we can use the volume formula for a cylinder:

V=πr2l=3.14×(0.05)2×1.5V = \pi r^2 l = 3.14 \times (0.05)^2 \times 1.5

Calculating this gives:

V=3.14×0.0025×1.5=0.011775 cm3V = 3.14 \times 0.0025 \times 1.5 = 0.011775 \text{ cm}^3

To find the volume produced per hour, divide by 24:

0.01177524=0.000490625 cm3hour1\frac{0.011775}{24} = 0.000490625 \text{ cm}^3\text{hour}^{-1}

Thus, the final volume of gas produced is approximately equal to 0.00049 cm³ hour⁻¹.

Step 4

Explain why a log scale is used to record the number of cells.

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Answer

A log scale is used to record the number of cells because it allows for easier visualization and comparison of large variations in cell numbers. Cell populations can grow exponentially, and a log scale compresses this growth into a manageable graph format, making it easier to interpret.

Step 5

Suggest one reason why.

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Answer

Many yeast cells may die during the death phase due to the depletion of nutrients. As resources become scarce, the remaining cells struggle to survive, leading to an increase in cell death.

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