krit.club logo

Adaptation and Survival - Adaptations to Climate and Habitat

Grade 9IB

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

🔑Concepts

•

Adaptation is the evolutionary process whereby an organism becomes better suited to its habitat through changes in its structure, behavior, or physiology.

•

Structural Adaptations are physical features of an organism's body, such as the streamlined body of a shark or the thick blubber layer in seals for insulation (Thermal ConductivityThermal\ Conductivity).

•

Behavioral Adaptations are the ways an organism acts to survive, such as migration, hibernation, or being nocturnal to avoid the daytime heat in deserts.

•

Physiological Adaptations are internal systematic responses to external stimuli, such as the production of concentrated urine by camels to conserve water or the presence of antifreeze proteins in the blood of Antarctic fish.

•

The Surface Area to Volume ratio (SA:VSA:V) is a critical factor in heat exchange. Animals in cold climates typically have a lower SA:VSA:V ratio to minimize heat loss, often following Bergmann's Rule (larger body size) and Allen's Rule (shorter extremities).

•

Abiotic factors influencing adaptation include temperature, rainfall (precipitationsprecipitations), sunlight intensity, and soil salinity.

•

Xerophytes are plants adapted to arid environments. They often possess thick waxy cuticles, sunken stomata, and succulent tissues to store water and reduce transpiration rates (EE).

📐Formulae

Surface Area to Volume Ratio=SAV\text{Surface Area to Volume Ratio} = \frac{SA}{V}

Surface Area of a Cube=6s2\text{Surface Area of a Cube} = 6s^2

Volume of a Cube=s3\text{Volume of a Cube} = s^3

Rate of Heat Loss∝Surface Area\text{Rate of Heat Loss} \propto \text{Surface Area}

💡Examples

Problem 1:

Compare the Surface Area to Volume ratio (SA:VSA:V) of two cubic organisms: Organism A with a side length of 2 cm2\text{ cm} and Organism B with a side length of 10 cm10\text{ cm}. Determine which organism is better adapted to survive in a polar environment based on heat retention.

Solution:

For Organism A (s=2 cms = 2\text{ cm}): SA=6×(22)=24 cm2SA = 6 \times (2^2) = 24\text{ cm}^2 V=23=8 cm3V = 2^3 = 8\text{ cm}^3 SA:V=248=3 cm−1SA:V = \frac{24}{8} = 3\text{ cm}^{-1}

For Organism B (s=10 cms = 10\text{ cm}): SA=6×(102)=600 cm2SA = 6 \times (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 (0.6 cm−10.6\text{ cm}^{-1}) compared to Organism A (3 cm−13\text{ cm}^{-1}). Since heat loss occurs through the surface, a lower ratio means less surface area relative to the volume of heat-generating tissue. Therefore, Organism B is better adapted for a polar environment as it will retain heat more efficiently.

Problem 2:

A desert plant has a total leaf surface area of 0.05 m20.05\text{ m}^2. If it reduces its surface area by 40%40\% through leaf curling to survive a drought, calculate the new surface area.

Solution:

Reduction=0.40×0.05=0.02 m2\text{Reduction} = 0.40 \times 0.05 = 0.02\text{ m}^2 New Surface Area=0.05−0.02=0.03 m2\text{New Surface Area} = 0.05 - 0.02 = 0.03\text{ m}^2

Explanation:

Leaf curling is a behavioral/structural adaptation to reduce the area exposed to dry air, thereby decreasing the rate of transpiration and conserving water during periods of high temperature or low water availability.