
What are the requirements of a surface to be highly efficient for gaseous exchange? How has this efficiency increased from fish to mammals?
Answer
498k+ views
Hint: The process by which oxygen and carbon dioxide move between the bloodstream and the lungs is known as gas exchange. This is the primary function of the respiratory system, and it is critical for maintaining a constant supply of oxygen to tissues while also removing carbon dioxide to prevent its accumulation.
Complete answer:
Diffusion across the outer surface of large animals is insufficient to maintain gas exchange. They created a variety of respiratory surfaces, all of which increase the surface area for exchange, allowing for larger bodies. Thin, moist epithelial cells cover the respiratory surface, allowing oxygen and carbon dioxide to exchange. Only when those gases are dissolved in water can they pass through cell membranes.
Respiratory surfaces are thin and have a large surface area, which are structural adaptations that maximize the gas exchange rate. To keep their plasma membranes intact, all living cells must be immersed in water. The respiratory surfaces of both terrestrial and aquatic animals are moist, and after dissolving in water, $O_2$ and $CO_2$ diffuse across these surfaces. The respiratory surface must supply oxygen and expel carbon dioxide for the entire body and several solutions to the issue of providing a large enough surface have emerged.
An endotherm has a larger respiratory surface area than an ectotherm of comparable size.
Note: Some small multicellular organisms, such as flatworms, can also exchange enough gas across the skin or cuticle that surrounds their bodies. However, in most larger organisms with low surface-area-to-volume ratios, specialized structures with convoluted surfaces such as gills, pulmonary alveoli, and spongy mesophyll provide the large area required for effective gas exchange.
Complete answer:
Diffusion across the outer surface of large animals is insufficient to maintain gas exchange. They created a variety of respiratory surfaces, all of which increase the surface area for exchange, allowing for larger bodies. Thin, moist epithelial cells cover the respiratory surface, allowing oxygen and carbon dioxide to exchange. Only when those gases are dissolved in water can they pass through cell membranes.
Respiratory surfaces are thin and have a large surface area, which are structural adaptations that maximize the gas exchange rate. To keep their plasma membranes intact, all living cells must be immersed in water. The respiratory surfaces of both terrestrial and aquatic animals are moist, and after dissolving in water, $O_2$ and $CO_2$ diffuse across these surfaces. The respiratory surface must supply oxygen and expel carbon dioxide for the entire body and several solutions to the issue of providing a large enough surface have emerged.
An endotherm has a larger respiratory surface area than an ectotherm of comparable size.
Note: Some small multicellular organisms, such as flatworms, can also exchange enough gas across the skin or cuticle that surrounds their bodies. However, in most larger organisms with low surface-area-to-volume ratios, specialized structures with convoluted surfaces such as gills, pulmonary alveoli, and spongy mesophyll provide the large area required for effective gas exchange.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

