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Solution for Onion Plasmolysis Investigation

(a) (i) Dilution Calculations and Table:

The volume of stock sucrose solution S required is calculated using the formula: \( V_1 = \frac{C_2 \times V_2}{C_1} \), where \( C_1 = 1.0 \text{ mol dm}^{-3} \) and \( V_2 = 10.0 \text{ cm}^3 \).

• For 0.8 mol dm^-3: \( 8.0 \text{ cm}^3 \) of S + \( 2.0 \text{ cm}^3 \) of W
• For 0.6 mol dm^-3: \( 6.0 \text{ cm}^3 \) of S + \( 4.0 \text{ cm}^3 \) of W
• For 0.4 mol dm^-3: \( 4.0 \text{ cm}^3 \) of S + \( 6.0 \text{ cm}^3 \) of W
• For 0.2 mol dm^-3: \( 2.0 \text{ cm}^3 \) of S + \( 8.0 \text{ cm}^3 \) of W

(a) (ii) Results Table:

A well-constructed results table must have complete grid lines and clear headers with appropriate slash-separated units. Example of a standard data layout with typical biological trends:

• 0.2 mol dm^-3: 0 plasmolysed / 20 total cells counted = 0% plasmolysis
• 0.4 mol dm^-3: 3 plasmolysed / 20 total cells counted = 15% plasmolysis
• 0.6 mol dm^-3: 12 plasmolysed / 20 total cells counted = 60% plasmolysis
• 0.8 mol dm^-3: 19 plasmolysed / 20 total cells counted = 95% plasmolysis
• 1.0 mol dm^-3: 20 plasmolysed / 20 total cells counted = 100% plasmolysis

(a) (iii) Explanation:

As sucrose concentration increases, the solute potential of the external solution decreases, which lowers its water potential. This creates a water potential gradient where the water potential inside the vacuole is higher than the external water potential. Consequently, water moves out of the vacuole and cytoplasm by osmosis down the water potential gradient across the selectively permeable cell membrane. This causes the protoplast to shrink away from the cell wall, increasing the percentage of plasmolysed cells observed.

(a) (iv) Errors and Improvements:

• Error: Subjectivity in deciding whether a cell is 'just' starting to plasmolyse. Improvement: Establish a standard reference set of photomicrographs showing clear criteria for a plasmolysed cell.
• Error: Epidermal cells obtained from different parts of the onion bulb scale may have different pre-existing water potentials. Improvement: Take all tissue strips from the exact same scale and layer of a single onion bulb.

(a) (v) Modification:

Prepare a much narrower range of sucrose concentrations around the concentration that yielded approximately 50% plasmolysis (e.g., 0.45, 0.50, 0.55, 0.65 mol dm^-3). Count a larger sample size of cells (at least 100 cells per slide) to accurately determine the concentration that causes 50% plasmolysis (incipient plasmolysis), at which point the internal solute potential of the cells is equal to the external osmotic potential. Ensure temperature is controlled using a water bath.

(a) (i) [3 marks]
- 1 mark: Correctly calculates volumes of stock S and distilled water W to make all 4 concentrations.
- 1 mark: Formulates volumes to total exactly 10.0 cm³ for each concentration.
- 1 mark: Presents all calculated volumes consistently to 1 decimal place (e.g., 8.0 cm³, not 8 cm³).

(a) (ii) [6 marks]
- 1 mark: Table drawn with clear ruled borders and distinct columns.
- 1 mark: Headers with correct units presented as: 'Sucrose concentration / mol dm^-3'.
- 1 mark: Records raw count data for at least 5 different sucrose concentrations.
- 1 mark: Raw counts of cells are whole numbers.
- 1 mark: Correct calculation of percentage plasmolysis for all rows.
- 1 mark: Percentage values expressed consistently to either the nearest whole number or 1 decimal place.

(a) (iii) [4 marks]
- 1 mark: Higher sucrose concentration leads to a lower/more negative external water potential.
- 1 mark: Water moves out of the cell vacuole down a water potential gradient.
- 1 mark: Movement occurs by osmosis across a selectively permeable cell membrane.
- 1 mark: Protoplast shrinks away from the cell wall, resulting in a higher proportion of plasmolysed cells.

(a) (iv) [3 marks]
- 1 mark: Identifies one valid systematic or random error (e.g., subjective boundary of plasmolysis, cell variation).
- 1 mark: Identifies a second valid error.
- 1 mark: Proposes an appropriate, feasible improvement corresponding to at least one of the identified errors.

(a) (v) [4 marks]
- 1 mark: Uses a narrower interval/range of sucrose concentrations.
- 1 mark: Replicates and counts a larger number of cells (minimum 50-100 cells per concentration) to reduce random error.
- 1 mark: Identifies the precise point of incipient plasmolysis (where 50% of the cells are plasmolysed).
- 1 mark: Mentions controlling temperature of the solutions using a thermostatic water bath.

Solution for Catalase Floating Disc Investigation

(a) (i) Dilution Calculations:

To find the volume of stock H (3.0%) needed to prepare 20 cm³ of a desired concentration: \( V_1 = \frac{C_2 \times 20.0}{3.0} \).

• To make 2.5%: \( (2.5 / 3.0) \times 20.0 = 16.7 \text{ cm}^3 \) of H + \( 3.3 \text{ cm}^3 \) of W
• To make 2.0%: \( (2.0 / 3.0) \times 20.0 = 13.3 \text{ cm}^3 \) of H + \( 6.7 \text{ cm}^3 \) of W
• To make 1.5%: \( (1.5 / 3.0) \times 20.0 = 10.0 \text{ cm}^3 \) of H + \( 10.0 \text{ cm}^3 \) of W
• To make 1.0%: \( (1.0 / 3.0) \times 20.0 = 6.7 \text{ cm}^3 \) of H + \( 13.3 \text{ cm}^3 \) of W
• To make 0.5%: \( (0.5 / 3.0) \times 20.0 = 3.3 \text{ cm}^3 \) of H + \( 16.7 \text{ cm}^3 \) of W

(a) (ii) Results Table:

The table must have complete grid lines and headers with units, e.g., 'Hydrogen peroxide concentration / %', 'Time, t / s', 'Mean time / s', and 'Rate / s^-1'. Raw times should be whole numbers. Typical trends show that as hydrogen peroxide concentration increases, time decreases and rate increases:

• 2.5%: Trial 1 = 9s, Trial 2 = 11s, Mean = 10.0s, Rate = 0.100 s^-1
• 1.5%: Trial 1 = 18s, Trial 2 = 22s, Mean = 20.0s, Rate = 0.050 s^-1
• 0.5%: Trial 1 = 58s, Trial 2 = 62s, Mean = 60.0s, Rate = 0.017 s^-1

(a) (iii) Controlled Variables:

• Yeast volume / concentration on the disc: Keep the disc size identical by using the same paper hole-punch, and use a standardized blotting duration (e.g., 3 seconds on each side on a paper towel) to ensure uniform yeast volume.
• Temperature: Place all solution tubes and yeast suspension in a water bath maintained at a constant temperature (e.g., 25°C) and check with a thermometer.

(a) (iv) Control Experiment:

Boil the yeast suspension for 10 minutes to denature the catalase enzyme, cool it to room temperature, and repeat the procedure. The discs should remain at the bottom of the test-tubes and not rise, confirming that active enzyme is required for oxygen production and disc flotation.

(a) (v) Modification for pH:

Keep the concentration of hydrogen peroxide constant (e.g., 1.5%). Prepare a range of buffer solutions (e.g., pH 4, 5, 6, 7, 8). Mix a fixed volume of each buffer with the hydrogen peroxide before placing the yeast-soaked discs. Keep other factors like temperature and yeast concentration constant.

(a) (i) [4 marks]
- 1 mark: Correctly calculates volume of stock H for at least three dilution values.
- 1 mark: Correctly calculates volume of distilled water W for all dilutions.
- 1 mark: Total volume of each dilution sums to exactly 20.0 cm³.
- 1 mark: Records all calculated volumes consistently to 1 decimal place.

(a) (ii) [6 marks]
- 1 mark: Table constructed with clean ruled grid lines and clear headers.
- 1 mark: Appropriate units listed in column headers, e.g., 't / s', 'Rate / s^-1'.
- 1 mark: Records raw data for 2 trials across 5 different concentrations.
- 1 mark: Raw stopwatch times recorded as whole numbers (no decimal places for manual timing).
- 1 mark: Correct calculation of mean times from the raw values.
- 1 mark: Rate values calculated as 1/mean time and expressed to 3 significant figures.

(a) (iii) [4 marks]
- 1 mark: Identifies size/material of the filter paper disc as a variable to control.
- 1 mark: Explains control method (e.g., use the same paper grade, same punch tool).
- 1 mark: Identifies temperature of reaction as a variable to control.
- 1 mark: Explains control method (e.g., use of a water bath monitored with a thermometer).

(a) (iv) [3 marks]
- 1 mark: Uses denatured / boiled and cooled yeast suspension (or distilled water instead of yeast).
- 1 mark: Keeps all other physical conditions and volumes identical.
- 1 mark: States the expected result: the paper disc does not rise to the surface.

(a) (v) [3 marks]
- 1 mark: Keeps the hydrogen peroxide concentration constant.
- 1 mark: Uses at least 5 different pH buffer solutions.
- 1 mark: Describes mixing the buffer solution with the reaction mixture before starting timing.