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9700 · 14.1

Homeostasis in mammals — practice questions

Practice and worked examples for 9700 Homeostasis in mammals. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Explain how the body responds to a significant drop in core body temperature, leading to its normalisation. (6 marks)

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  1. A drop in core body temperature is detected by thermoreceptors in the skin (peripheral) and hypothalamus (central).
  2. This information is relayed to the thermoregulatory centre in the hypothalamus.
  3. The hypothalamus coordinates responses to increase heat production and decrease heat loss.
  4. Vasoconstriction occurs: arterioles supplying the skin capillaries constrict, reducing blood flow to the surface and minimising heat loss by radiation, convection, and conduction.
  5. Shivering is initiated: rapid, involuntary contractions of skeletal muscles generate metabolic heat.
  6. Hair erector muscles contract, causing hairs to stand erect and trap an insulating layer of air close to the skin, reducing heat loss. These combined actions raise the core body temperature back to the set point, illustrating negative feedback.

Worked example 2

A person with a total blood volume of 5.0 dm³ has a fasting blood glucose concentration of 90 mg per 100 cm³. After a sugary meal, their blood glucose rises to 140 mg per 100 cm³. Assuming glucose is evenly distributed in the blood plasma, which constitutes 55% of blood volume, calculate the total mass of glucose (in grams) that must be removed from the blood to return the concentration to the fasting level.

Show solution outline

  1. Calculate plasma volume:

    • Total blood volume = 5.0 dm³
    • Plasma volume = 55% of 5.0 dm³ = 0.55 × 5.0 dm³ = 2.75 dm³

  2. Convert concentrations to g/dm³:

    • Fasting concentration = 90 mg / 100 cm³ = 90 mg / 0.1 dm³ = 900 mg/dm³ = 0.9 g/dm³
    • Post-meal concentration = 140 mg / 100 cm³ = 140 mg / 0.1 dm³ = 1400 mg/dm³ = 1.4 g/dm³

  3. Calculate total mass of glucose at each concentration:

    • Mass at post-meal level = Concentration × Volume = 1.4 g/dm³ × 2.75 dm³ = 3.85 g
    • Mass at fasting level = Concentration × Volume = 0.9 g/dm³ × 2.75 dm³ = 2.475 g

  4. Calculate the mass of glucose to be removed:

    • Mass to remove = Mass at post-meal level - Mass at fasting level
    • Mass to remove = 3.85 g - 2.475 g = 1.375 g

    Final Answer: 1.375 g of glucose must be removed from the blood plasma to restore the fasting concentration.

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