OxfordAQA International GCSE Biology: Homeostasis, temperature control and blood glucose (9201)

This study guide is aligned to the OxfordAQA International GCSE Biology (9201) specification, Version 5.2, for exams from May/June 2018 onwards. It covers specification sections 3.4.2 (Homeostasis), 3.4.4 (Temperature control) and 3.4.5 (Control of blood glucose). The qualification is untiered, so every statement applies to every candidate. Water and ion balance (3.4.3) is not covered here.

Use this guide first. Then use the revision notes for quick recall and the practice questions to test yourself. Course pages: the OxfordAQA IGCSE Biology hub and the printable checklist. To find weak spots, try the free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do
3.4.2 a, b	State that automatic control systems keep internal conditions relatively constant. Name the parts: receptors, coordination centres, effectors.
3.4.2 c	Name where receptors are found and what each detects.
3.4.2 d, e	Name coordination centres. List the internal conditions that are controlled.
3.4.4 a, b	Explain the roles of the thermoregulatory centre and of skin temperature receptors.
3.4.4 c, d	Explain the responses to a core temperature that is too high or too low.
3.4.5 a	Explain how the pancreas monitors glucose, and how glucose is stored.
3.4.5 b, c	Explain the actions of insulin and glucagon.
3.4.5 d, e	Describe and compare Type 1 and Type 2 diabetes and how each is controlled.

The specification says that knowledge of the structure and function of sense organs such as the eye and ear is not required. It also says the name of the temperature centre in the brain (hypothalamus) is not required, so “thermoregulatory centre” is the term to use.

3.4.2 Homeostasis: control systems

Cells work best when their surroundings stay within narrow limits. Automatic control systems in the body keep conditions inside it relatively constant. This is called homeostasis. The word “relatively” matters: conditions move a little above and below a normal level, but the control system limits the change.

Every control system has three kinds of part.

Part	Job	Example
Receptor	Detects a stimulus, which is a change in conditions	Cells in the pancreas that detect blood glucose
Coordination centre	Receives and processes information from receptors	Brain, spinal cord, pancreas
Effector	Brings about a response	Muscles and glands

Where receptors are found (3.4.2 c)

Organ	Receptors are sensitive to
Eyes	Light
Ears	Sound, and changes in position (which helps balance)
Tongue and nose	Chemicals (taste and smell)
Skin	Touch, pressure, pain and temperature changes
Brain	Blood temperature and the concentration of water in the blood
Pancreas	The concentration of glucose in the blood

Notice that the brain and pancreas hold receptors for internal conditions. The skin holds receptors for what is happening at the body surface.

Conditions that are controlled (3.4.2 e)

Four are listed: temperature, the water content of the body, the ion content of the body and blood glucose levels. This guide covers temperature and blood glucose.

Worked example 1: naming the parts

Name the receptor, coordination centre and effector in each case.

Situation	Receptor	Coordination centre	Effector and response
A swimmer’s hand touches a very hot pool heater	Pain and temperature receptors in the skin	Spinal cord	Arm muscles contract and pull the hand away
Blood glucose rises after a sugary drink	Glucose-sensitive cells in the pancreas	Pancreas	Cells take up glucose after the pancreas releases insulin
Blood flowing through the brain becomes warmer	Temperature receptors in the brain	Thermoregulatory centre in the brain	Sweat glands release more sweat

Write the chain in order: stimulus, then receptor, then coordination centre, then effector, then response.

3.4.4 Temperature control

The thermoregulatory centre in the brain monitors and controls body temperature. It contains receptors that are sensitive to the temperature of the blood flowing through the brain, so it measures core body temperature directly. Temperature receptors in the skin send impulses to the same centre. They give information about skin temperature, and so about the outside world.

Core temperature is normally close to 37 °C. The centre compares what it detects with the normal level and sends signals to effectors. Those effectors are the sweat glands, the muscles and the blood vessels supplying the skin.

Core temperature too high (3.4.4 c, d)

• Vasodilation. The blood vessels supplying the skin capillaries dilate (widen). More blood flows through the skin capillaries, so more energy is transferred from the skin to the environment.
• More sweating. Sweat glands release more sweat. The sweat cools the body as it evaporates.
• Replacing fluid. More water is lost when it is hot, so more fluid has to be taken in through drink or food to balance the loss.

Core temperature too low (3.4.4 d)

• Vasoconstriction. The blood vessels supplying the skin capillaries constrict (narrow). Less blood flows through the skin capillaries, so less energy is transferred to the environment.
• Shivering. Muscles contract and relax rapidly. The contractions need respiration, and respiration transfers some energy that warms the body.

Worked example 2: a hot walk

A hiker walks for 4 hours in hot weather. Sweat loss per hour and core temperature were recorded. She drank 1.5 dm³ during the walk.

Hour	1	2	3	4
Sweat lost (dm³)	0.30	0.55	0.80	0.80
Core temperature at end of hour (°C)	37.6	37.9	38.0	38.0

Her core temperature was 37.0 °C at the start. (a) Calculate her total sweat loss. (b) Calculate how much more she lost than she drank, as a percentage of the total lost. (c) Explain the pattern in her temperature.

(a) 0.30 + 0.55 + 0.80 + 0.80 = 2.45 dm³.

(b) Shortfall = 2.45 − 1.5 = 0.95 dm³. Percentage = 0.95 ÷ 2.45 × 100 = 39% (to 2 significant figures).

(c) Exercise releases energy from respiration, so her core temperature rose at first (0.6 °C in hour 1). The rise slowed as sweat loss rose, because evaporating sweat cools the body, and her skin vasodilated so more energy was transferred to the surroundings. By hours 3 and 4 the temperature stayed at 38.0 °C, so heat loss matched heat gain.

Worked example 3: shivering

A volunteer sits still at 25 °C, then in a cold room at 5 °C. Oxygen used by the body is measured as a sign of respiration. It is 0.25 dm³ per minute at 25 °C and 0.60 dm³ per minute at 5 °C. In the cold room the volunteer shivers.

(a) Calculate the percentage increase in oxygen use. (b) Calculate the extra oxygen used in 30 minutes. (c) Explain the link to temperature.

(a) Increase = 0.60 − 0.25 = 0.35. Percentage = 0.35 ÷ 0.25 × 100 = 140%.

(b) 0.35 × 30 = 10.5 dm³.

(c) Shivering is rapid muscle contraction. The contractions need energy from respiration, so more oxygen is used. Some of the energy is transferred as heat, which warms the body and helps to keep core temperature near 37 °C.

3.4.5 Control of blood glucose

The pancreas monitors and controls the concentration of glucose in the blood. Much of the glucose is stored as glycogen in the liver and muscles. When these stores are full, excess glucose is stored as lipid.

Insulin (3.4.5 b)

If blood glucose is too high, the pancreas produces the hormone insulin. Insulin allows glucose to move from the blood into the cells. Blood glucose then falls, and the glucose taken in can be stored as glycogen.

Glucagon (3.4.5 c)

When blood glucose falls, the pancreas produces a second hormone, glucagon. Glucagon makes stored glycogen change back into glucose, and the glucose enters the blood. Blood glucose then rises.

	Blood glucose too high	Blood glucose too low
Hormone from pancreas	Insulin	Glucagon
Effect on glycogen	Glucose is stored as glycogen	Glycogen changes back into glucose
Effect on blood glucose	Falls	Rises

Worked example 4: a person with Type 1 diabetes and a meal

On two days a person with Type 1 diabetes eats the same meal at hour 0. On day A there is no insulin injection. On day B insulin is injected at hour 0. Blood glucose is in mmol/dm³.

Hour	0	1	2	3	4
Day A, no insulin	8.0	13.5	15.5	15.0	14.0
Day B, insulin	8.0	10.5	9.0	8.0	7.5

(a) Calculate the percentage rise to the peak on each day. (b) Explain the difference.

(a) Day A: 15.5 − 8.0 = 7.5, and 7.5 ÷ 8.0 × 100 = 94%. Day B: 10.5 − 8.0 = 2.5, and 2.5 ÷ 8.0 × 100 = 31%.

(b) After the meal, glucose enters the blood. With no injection, the person’s pancreas does not make enough insulin, so little glucose moves into cells and the level stays high. The injected insulin allows glucose to move from the blood into the cells, so the rise is smaller and the level returns towards 8.0.

Worked example 5: how long do the stores last?

Assume a liver holds 90 g of glycogen and the body uses glucose from it at 6.0 g per hour between meals.

Time = 90 ÷ 6.0 = 15 hours.

In a real fast, falling glucose stimulates the pancreas to release glucagon. Glucagon makes the glycogen change back into glucose, which enters the blood and slows the fall in blood glucose.

Diabetes (3.4.5 d, e)

	Type 1	Type 2
What goes wrong	The pancreas does not produce enough insulin	The body does not respond to its own insulin
Effect	Blood glucose may be too high	Blood glucose may be too high
Risk factor stated in the specification	None stated	Obesity is a significant factor
Control	Careful diet, exercise and injecting insulin	Careful diet, exercise and drugs that help the cells to respond to insulin

Insulin is given by injection in Type 1 diabetes because the problem is a shortage of the hormone. Drugs that help cells respond to insulin suit Type 2 diabetes because insulin is present but cells respond poorly.

Worked example 6: Type 2 diabetes in cells

Muscle cells from a healthy person (A) and a person with Type 2 diabetes (B) are given different insulin concentrations. Glucose uptake is in arbitrary units per hour.

Insulin (arbitrary units)	0	10	20	40
Uptake, A	2	10	17	20
Uptake, B	2	4	7	12

(a) Compare uptake at insulin 20. (b) Calculate each person’s percentage increase in uptake from 0 to 40. (c) Explain what the data suggest.

(a) 17 ÷ 7 = about 2.4, so A’s cells take up about 2.4 times as much glucose as B’s.

(b) A: (20 − 2) ÷ 2 × 100 = 900%. B: (12 − 2) ÷ 2 × 100 = 500%.

(c) B’s cells take up less glucose at every insulin concentration above zero. This fits Type 2 diabetes, where cells do not respond properly to their own insulin. Insulin is present, but the response is weaker.

Common errors

• Writing that insulin “breaks down” glucose. Insulin allows glucose to move from the blood into cells.
• Mixing up the two hormones. Insulin lowers blood glucose. Glucagon raises it.
• Saying that blood vessels “move” towards the skin. Their diameter changes.
• Writing that sweat cools the body because it is cold. The cooling happens as sweat evaporates.
• Saying Type 2 diabetes means no insulin. The body does not respond to its own insulin.
• Dividing by the final value in a percentage change. Divide by the starting value.

Official syllabus

The OxfordAQA International GCSE Biology (9201) specification, Version 5.2, for exams May/June 2018 onwards, published by OxfordAQA.