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Transport in Animals - Composition of blood

Grade 12IGCSEBiology

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

🔑Concepts

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Blood consists of four main components: Red blood cells (erythrocytes), White blood cells (leukocytes), Platelets (thrombocytes), and Plasma.

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Red Blood Cells: Specialized for O2O_2 transport. They lack a nucleus to maximize space for Hemoglobin (HbHb) and have a biconcave shape to increase the surface area to volume ratio (SA:VSA:V).

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White Blood Cells: Phagocytes engulf and digest pathogens via phagocytosis. Lymphocytes produce specific antibodies to target antigens on the surface of pathogens.

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Platelets: Small cell fragments involved in blood clotting. They trigger the conversion of soluble fibrinogen into insoluble fibrin threads to form a mesh.

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Plasma: The liquid matrix (mostly H2OH_2O) that transports dissolved CO2CO_2, glucose, amino acids, urea, hormones, and heat energy.

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Oxygen Transport: Hemoglobin binds reversibly with oxygen to form Oxyhemoglobin: Hb+4O2⇌Hb(O2)4Hb + 4O_2 \rightleftharpoons Hb(O_2)_4.

📐Formulae

Hb+4O2⇌Hb(O2)4Hb + 4O_2 \rightleftharpoons Hb(O_2)_4

CO2+H2O⇌H2CO3⇌H++HCO3−CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-

Concentration of RBC=Number of CellsVolume of Blood (mm3)\text{Concentration of RBC} = \frac{\text{Number of Cells}}{\text{Volume of Blood (mm}^3\text{)}}

💡Examples

Problem 1:

A blood sample contains 5×1065 \times 10^6 red blood cells per mm3mm^3. If an average adult has 55 liters of blood, calculate the total number of red blood cells in the body using scientific notation.

Solution:

2.5×10132.5 \times 10^{13} cells

Explanation:

First, convert liters to mm3mm^3: 1 liter=1,000 cm3=1,000,000 mm31 \text{ liter} = 1,000 \text{ cm}^3 = 1,000,000 \text{ mm}^3. Therefore, 5 L=5×106 mm35 \text{ L} = 5 \times 10^6 \text{ mm}^3. Total cells = (5×106 cells/mm3)×(5×106 mm3)=25×1012=2.5×1013(5 \times 10^6 \text{ cells/mm}^3) \times (5 \times 10^6 \text{ mm}^3) = 25 \times 10^{12} = 2.5 \times 10^{13}.

Problem 2:

Explain how the structure of a Red Blood Cell is adapted to its function regarding the diffusion of O2O_2.

Solution:

Biconcave shape and lack of nucleus.

Explanation:

The biconcave shape increases the SAV\frac{SA}{V} ratio, allowing for faster diffusion of O2O_2 across the membrane. The absence of a nucleus provides more space for the protein Hemoglobin (HbHb), increasing the oxygen-carrying capacity per cell.

Composition of blood - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology