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Continuity and change - Homeostasis

Grade 11IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Homeostasis is the maintenance of a constant internal environment within physiological tolerance limits, such as body temperature (37∘C37^{\circ}C in humans), blood pHpH (around 7.47.4), and blood glucose concentration.

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Negative Feedback Loops: The primary mechanism of homeostasis where a change in a variable (stimulus) triggers a response that opposes the initial change to return the system to its set point (Stimulus→Receptor→ControlCenter→Effector→ResponseStimulus \rightarrow Receptor \rightarrow Control Center \rightarrow Effector \rightarrow Response).

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Thermoregulation: The hypothalamus acts as a thermostat. If body temperature T>37∘CT > 37^{\circ}C, effectors trigger vasodilation and sweating. If T<37∘CT < 37^{\circ}C, effectors trigger vasoconstriction and shivering.

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Blood Glucose Regulation: Controlled by the endocrine pancreas. β\beta-cells secrete insulin to stimulate glucose uptake and glycogenesis when blood glucose is high; α\alpha-cells secrete glucagon to stimulate glycogenolysis when blood glucose is low.

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Osmoregulation: The control of water potential (Ψ\Psi) in the blood. High solute concentration triggers the release of Antidiuretic Hormone (ADH) from the posterior pituitary, increasing the permeability of the collecting ducts in the kidney to water.

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Positive Feedback: A physiological mechanism that amplifies a change, moving the system further away from the equilibrium (e.g., oxytocin release during childbirth or blood clotting cascades).

📐Formulae

Ψ=Ψs+Ψp\Psi = \Psi_s + \Psi_p

Q10=(R2R1)10T2−T1Q_{10} = \left(\frac{R_2}{R_1}\right)^{\frac{10}{T_2 - T_1}}

Rate of Diffusion∝Surface Area×Concentration GradientDiffusion Distance\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Gradient}}{\text{Diffusion Distance}}

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

💡Examples

Problem 1:

A patient has a blood glucose level of 140 mg/dL140\text{ mg/dL} after a meal. If the normal set point is 90 mg/dL90\text{ mg/dL}, calculate the deviation ΔG\Delta G and explain the homeostatic response.

Solution:

The deviation is calculated as: ΔG=140 mg/dL−90 mg/dL=50 mg/dL\Delta G = 140\text{ mg/dL} - 90\text{ mg/dL} = 50\text{ mg/dL} High glucose levels are detected by β\beta-cells in the Islets of Langerhans. They secrete insulin, which travels to the liver and muscles to convert glucose into glycogen (glycogenesis).

Explanation:

This is a classic example of negative feedback where the effector (insulin/liver) acts to reduce the stimulus (high glucose) back toward the set point.

Problem 2:

Calculate the temperature coefficient Q10Q_{10} for a metabolic reaction if the rate R1=5 units/sR_1 = 5\text{ units/s} at T1=20∘CT_1 = 20^{\circ}C and R2=10 units/sR_2 = 10\text{ units/s} at T2=30∘CT_2 = 30^{\circ}C.

Solution:

Using the formula: Q10=(105)1030−20Q_{10} = \left(\frac{10}{5}\right)^{\frac{10}{30 - 20}} Q10=(2)1010=21=2Q_{10} = (2)^{\frac{10}{10}} = 2^1 = 2

Explanation:

A Q10Q_{10} value of 22 indicates that the metabolic rate doubles for every 10∘C10^{\circ}C increase in temperature, emphasizing why thermoregulation is critical for maintaining stable metabolic rates.