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Calculate value X in Table 4 - AQA - GCSE Biology - Question 6 - 2010 - Paper 1

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Calculate value X in Table 4. X =

Worked Solution & Example Answer:Calculate value X in Table 4 - AQA - GCSE Biology - Question 6 - 2010 - Paper 1

Step 1

Calculate value X in Table 4

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Answer

To find the value of X in Table 4, we compare the surface area to volume ratio. Given the volumes of 3600 and 1200, we can set up the relationship:

  1. Calculate the surface area to volume ratio:

    • For the first jellyfish: Ratio = Surface Area / Volume = X / 3600
    • For the second jellyfish: Ratio = Surface Area / Volume = 3 / 1200
  2. Set the ratios equal to each other:

    • Thus, we have: X3600=31200\frac{X}{3600} = \frac{3}{1200}
  3. Cross-multiply to solve for X:

    • This leads to: 1200X=3×36001200X = 3 \times 3600
    • Simplifying gives us: 1200X=108001200X = 10800
  4. Finally, divide both sides by 1200: X=108001200=9X = \frac{10800}{1200} = 9

Thus, the value of X is 9.

Step 2

Describe the relationship between the size of a jellyfish and its surface area to volume ratio

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Answer

As the size of a jellyfish increases, the surface area to volume ratio decreases. This means that larger jellyfish have less surface area relative to their volume, which can affect their ability to absorb oxygen.

Step 3

Name one other substance that enters cells by diffusion

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Answer

One substance that enters cells by diffusion is carbon dioxide.

Step 4

Suggest two factors that affect the rate of diffusion of oxygen into a jellyfish

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Answer

  1. Concentration gradient: A higher concentration of oxygen outside the jellyfish compared to the inside will increase the rate of diffusion.
  2. Surface area: A larger surface area will facilitate a higher rate of oxygen diffusion.

Step 5

Name the organs where gas exchange takes place in fish

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Answer

Gas exchange in fish takes place in the gills.

Step 6

Explain how the human breathing system is adapted to maximise the rate of gas exchange

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Answer

The human breathing system has several adaptations that maximize the rate of gas exchange:

  1. Large number of alveoli: The lungs contain millions of alveoli, providing a vast surface area for gas exchange, which increases the efficiency of oxygen uptake and carbon dioxide removal.

  2. Thin walls: The alveoli and blood vessels (capillaries) have very thin walls, facilitating rapid diffusion of gases across the alveolar and capillary membranes.

  3. Proximity of alveoli and capillaries: Alveoli are closely associated with capillaries, which minimizes the distance that oxygen and carbon dioxide must diffuse.

  4. Good blood supply: The lungs have a rich supply of blood vessels to ensure that oxygen can be quickly transported away from the lungs and that carbon dioxide can be removed efficiently.

  5. Ventilation: The respiratory system is actively ventilated; inhalation brings fresh oxygen into the lungs, while exhalation removes carbon dioxide-rich air, maintaining the concentration gradient necessary for efficient diffusion.

  6. Capillary network: The extensive network of capillaries surrounding each alveolus increases the surface area available for gas exchange, further enhancing the rate at which gases can diffuse in and out.

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