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Form and function - Adaptation to environment

Grade 11IBBiology

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

🔑Concepts

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Adaptations are inherited characteristics of an organism that enhance its survival and reproduction in specific environments. These are categorized into structural (morphological), physiological (biochemical), and behavioral adaptations.

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The Surface Area to Volume Ratio (SA:VSA:V) is a critical factor in adaptation. Smaller organisms have a higher SA:VSA:V ratio, leading to faster heat loss and higher metabolic rates to maintain body temperature.

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Xerophytes are plants adapted to arid environments. They utilize adaptations such as thick waxy cuticles, sunken stomata, and reduced leaf surface area to minimize the transpiration rate (EE).

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Hydrophytes are plants adapted to aquatic environments. They often possess aerenchyma (air-filled tissue) for buoyancy and gas exchange, and have minimal or no waxy cuticle because water conservation is not a priority.

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Thermoregulation adaptations include Bergmann's Rule (larger body size in colder climates to reduce heat loss via lower SA:VSA:V) and Allen's Rule (shorter appendages in cold climates to minimize surface area).

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Counter-current heat exchange is a physiological adaptation in many endotherms where heat is transferred from warm arterial blood to cold venous blood returning from the extremities, maintaining a core temperature TcoreT_{core}.

📐Formulae

SAcube=6s2SA_{cube} = 6s^2

Vcube=s3V_{cube} = s^3

SA:V=6s2s3=6sSA:V = \frac{6s^2}{s^3} = \frac{6}{s}

SAsphere=4πr2SA_{sphere} = 4 \pi r^2

Vsphere=43πr3V_{sphere} = \frac{4}{3} \pi r^3

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

💡Examples

Problem 1:

Compare the SA:VSA:V ratio of two cubic organisms, Organism A with side length s=2 cms = 2\text{ cm} and Organism B with side length s=10 cms = 10\text{ cm}. Explain which organism is better adapted to a cold environment based on heat retention.

Solution:

For Organism A (s=2s = 2): SA=6(2)2=24 cm2SA = 6(2)^2 = 24\text{ cm}^2 V=23=8 cm3V = 2^3 = 8\text{ cm}^3 SA:V=248=3.0 cm−1SA:V = \frac{24}{8} = 3.0\text{ cm}^{-1}

For Organism B (s=10s = 10): SA=6(10)2=600 cm2SA = 6(10)^2 = 600\text{ cm}^2 V=103=1000 cm3V = 10^3 = 1000\text{ cm}^3 SA:V=6001000=0.6 cm−1SA:V = \frac{600}{1000} = 0.6\text{ cm}^{-1}

Explanation:

Organism B has a much lower SA:VSA:V ratio (0.60.6) compared to Organism A (3.03.0). Because heat loss occurs at the surface, a lower ratio means less heat is lost relative to the body mass (volume) generating that heat. Therefore, Organism B is better adapted to conserve heat in a cold environment.

Problem 2:

A xerophyte has stomata located in deep pits. If the ambient humidity is 20%20\% and the humidity inside the pit is 80%80\%, while the humidity inside the leaf is 100%100\%, how does this structure affect the water potential gradient Δψ\Delta \psi?

Solution:

The pit creates a microenvironment where the humidity (80%80\%) is significantly higher than the external air (20%20\%). The gradient between the leaf interior and the pit is Δψint=100%−80%=20%\Delta \psi_{int} = 100\% - 80\% = 20\%, whereas the gradient between the leaf and external air would be Δψext=100%−20%=80%\Delta \psi_{ext} = 100\% - 20\% = 80\%.

Explanation:

By trapping moist air in the pit, the plant reduces the concentration gradient of water vapor between the inside of the leaf and the air immediately outside the stomata. According to Fick's Law, a smaller gradient results in a lower rate of transpiration, conserving water.