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A student investigated the effect of surface area on osmosis in cubes of potato - AQA - A-Level Biology - Question 2 - 2018 - Paper 1

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A student investigated the effect of surface area on osmosis in cubes of potato. He cut two cubes of potato tissue, each with sides of 35 mm in length. He put one ... show full transcript

Worked Solution & Example Answer:A student investigated the effect of surface area on osmosis in cubes of potato - AQA - A-Level Biology - Question 2 - 2018 - Paper 1

Step 1

Describe the method the student would have used to obtain the results in Figure 3.

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Answer

To obtain the results in Figure 3, the student would have followed a systematic method that includes the following steps:

  1. Preparation of Potato Cubes: After cutting the large potato cube into eight smaller cubes, it is essential to ensure that all cut surfaces of the small cubes are exposed to the sucrose solution. This maximizes the surface area for osmosis to occur.

  2. Controlling Variables: The student should have controlled the temperature in the environment where the experiment was conducted. This is important because temperature can affect the rate of osmosis.

  3. Drying the Samples: Before weighing, the potato cubes should be dried with a paper towel to remove excess surface moisture. This ensures that the weight measurements reflect only the mass of the potato tissue and not any excess water from the solution.

  4. Measuring Mass Over Time: The water loss due to osmosis needs to be recorded at regular intervals. The mass of both the large and small cubes should be measured consistently, taking readings at specific time intervals, such as every 5 or 10 minutes, to show the change in mass over the duration of the experiment.

Step 2

Is the rate of osmosis per mm² per minute different between A and B during this time? Use appropriate calculations to support your answer.

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Answer

To compare the rate of osmosis between the large cube (A) and the small cubes (B), first we need to calculate the surface area for both types of cubes:

  • Surface Area Calculation:

    • For the large cube:

      extSurfaceArea(A)=6imes(35extmm)2=7350extmm2 ext{Surface Area (A)} = 6 imes (35 ext{ mm})^2 = 7350 ext{ mm}^2

    • For one small cube:

      extSurfaceArea(smallcube)=6imes(17.5extmm)2=1837.5extmm2 ext{Surface Area (small cube)} = 6 imes (17.5 ext{ mm})^2 = 1837.5 ext{ mm}^2

    • For eight small cubes combined:

      extTotalSurfaceArea(B)=8imes1837.5extmm2=14700extmm2 ext{Total Surface Area (B)} = 8 imes 1837.5 ext{ mm}^2 = 14700 ext{ mm}^2

  • Rate of Osmosis Calculation:

    • For A (0 to 40 mins): The mass loss is approximately 1.5 g.

      • Rate per minute:

      ext{Rate}_{A} = rac{1.5 ext{ g}}{40 ext{ min}} = 0.0375 ext{ g/min}

      • Rate per mm²:

      ext{Rate}_{ ext{A per mm}^2} = rac{0.0375 ext{ g/min}}{7350 ext{ mm}^2} = 5.1 imes 10^{-6} ext{ g/mm}^2/ ext{min}

    • For B (0 to 40 mins): The mass loss is approximately 5 g.

      • Rate per minute:

      ext{Rate}_{B} = rac{5 ext{ g}}{40 ext{ min}} = 0.125 ext{ g/min}

      • Rate per mm²:

      ext{Rate}_{ ext{B per mm}^2} = rac{0.125 ext{ g/min}}{14700 ext{ mm}^2} = 8.5 imes 10^{-6} ext{ g/mm}^2/ ext{min}

  • Comparison:

    • The rate of osmosis for B (small cubes) is approximately 8.5imes1068.5 imes 10^{-6} g/mm²/min, which is higher than that for A at approximately 5.1imes1065.1 imes 10^{-6} g/mm²/min, indicating that osmosis occurs more rapidly in the smaller cubes due to their greater surface area relative to volume.

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