Ch 2 · 8.2 Transport of oxygen and carbon dioxide · AS & A Level Biology 9700

Topic 2 of 3 in Transport in mammals — AS & A Level Biology 9700 — 58 lessons.

Introduction

Oxygen is poorly soluble in blood plasma, so mammals rely on haemoglobin inside red blood cells to carry almost all the oxygen transported from the lungs to respiring tissues. The way haemoglobin binds and releases oxygen, and the several ways carbon dioxide is carried in the opposite direction, are central to gas exchange physiology. (Syllabus section 8.2.)

Key ideas — Haemoglobin and oxygen transport

Haemoglobin is a globular protein with a quaternary structure of four polypeptide chains, two alpha and two beta chains in adult haemoglobin, each associated with a haem group containing an mathrmFe^2+ ion. Each mathrmFe^2+ ion can bind one oxygen molecule, so one haemoglobin molecule can carry up to four oxygen molecules, forming oxyhaemoglobin. Binding of the first oxygen molecule changes the tertiary structure of the polypeptide chains, making it easier for further oxygen molecules to bind, known as cooperative binding, and equally making it easier for oxygen to be released once one molecule has left.

🧑‍🏫 Teacher's explanation

Think of haemoglobin like a highly efficient, four-seater taxi. Each seat is a 'haem group' containing an iron ion (mathrmFe^2+). The magic of this taxi is that once the first passenger (oxygen) gets in, the seats automatically adjust their position to make it much easier for the next three passengers to hop in. This is called 'cooperative binding'.

📖 Definitions

🔍 Worked example — Easy

A student observes that at a pO_2 of 40 mmHg, haemoglobin is 75\% saturated, whereas at pO_2 of 100 mmHg, it is 98\% saturated. Calculate the percentage increase in oxygen saturation when moving from the tissues to the lungs.

🔍 Worked example — Standard

Describe the structural changes in a haemoglobin molecule when it transitions from a low-affinity state to a high-affinity state.

🔍 Worked example — Exam-style

Explain why the oxygen dissociation curve for adult haemoglobin is sigmoidal (S-shaped) rather than a simple hyperbola.

⚠️ Common errors

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🎯 Exam question

Key ideas — The oxygen dissociation curve

The oxygen dissociation curve plots the percentage saturation of haemoglobin with oxygen against the partial pressure of oxygen. It has a sigmoid (S-shaped) form because of cooperative binding: at very low partial pressure, little oxygen binds because the first oxygen molecule binds only with difficulty; as partial pressure rises, binding becomes progressively easier and saturation rises steeply; at high partial pressure, haemoglobin approaches full saturation and the curve levels off. At the high partial pressure of oxygen found in the lungs, haemoglobin is close to fully saturated; at the lower partial pressure found in respiring tissues, a much smaller percentage saturation means a large amount of oxygen is unloaded to the tissues.

🧑‍🏫 Teacher's explanation

Imagine a crowded bus where people are trying to find seats. At first, it is hard to get anyone to sit down because the bus looks empty and uninviting. However, once a few people sit down, it becomes much easier for others to find a spot because the 'pattern' of sitting is established. This is like haemoglobin; once the first oxygen molecule binds, the protein's shape changes to make it easier for the next ones to join. This is called 'cooperative binding'.

📖 Definitions

🔍 Worked example — Easy

A sample of haemoglobin has a partial pressure of oxygen (pO_2) of 40 ext mmHg and is 75\% saturated. Calculate the percentage of haemoglobin that is NOT saturated.

🔍 Worked example — Standard

Explain why the oxygen dissociation curve for adult haemoglobin has a sigmoid (S-shaped) shape.

🔍 Worked example — Exam-style

Compare the oxygen saturation of haemoglobin in the lungs versus in respiring tissues, assuming pO_2 in lungs is 100 ext mmHg and pO_2 in tissues is 40 ext mmHg.

⚠️ Common errors

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🎯 Exam question

(a) Describe how a decrease in blood pH affects the affinity of haemoglobin for oxygen. [1] (b) Explain why this change is advantageous for oxygen delivery to metabolically active tissues. [2] [3 marks]

Key ideas — The Bohr effect

A rise in carbon dioxide concentration, and the resulting fall in pH, shifts the oxygen dissociation curve to the right; this is the Bohr shift (Bohr effect). It happens because H⁺ ions from carbonic acid combine with haemoglobin, forming haemoglobinic acid and changing its shape so it holds oxygen less tightly. At a given partial pressure of oxygen, haemoglobin then has a lower affinity for oxygen and releases more of it. Actively respiring tissues produce more carbon dioxide, lowering local pH; the Bohr shift means haemoglobin unloads more oxygen exactly where it is needed most, matching oxygen supply to metabolic demand.

🧑‍🏫 Teacher's explanation

Imagine you are carrying a heavy backpack (oxygen) through a crowded room. If the room is calm and quiet (low CO_2), you hold onto the backpack tightly. But if the room becomes hot, loud, and hectic (high CO_2 and low pH), you find it much harder to hold on, and the backpack slips off more easily.

📖 Definitions

🔍 Worked example — Easy

A student observes that at a partial pressure of oxygen (pO_2) of 40 ext mmHg, the haemoglobin saturation is 75\%. In a highly active muscle, the pO_2 drops and CO_2 increases, causing a Bohr shift. If the saturation drops to 50\% at the same pO_2 due to this shift, calculate the percentage decrease in oxygen loading efficiency at this specific pressure.

🔍 Worked example — Standard

Describe the effect of an increase in pCO_2 on the oxygen dissociation curve of adult haemoglobin.

🔍 Worked example — Exam-style

Explain why the Bohr shift is physiologically advantageous in actively respiring tissues.

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🎯 Exam question

Key ideas — Carbon dioxide in red blood cells

Carbon dioxide from respiring cells diffuses into the plasma and into red blood cells. Inside the red blood cell:

🧑‍🏫 Teacher's explanation

Think of red blood cells as specialized delivery trucks. While most CO₂ travels in the plasma as dissolved bicarbonate, a significant portion is processed inside the cell to make the transport more efficient. Imagine a sponge soaking up water; the cell uses the enzyme carbonic anhydrase to turn CO₂ into a soluble form (bicarbonate) so it can be carried easily without changing the blood's pH too drastically.

📖 Definitions

🔍 Worked example — Easy

State the chemical equation for the formation of carbonic acid from carbon dioxide and water, and name the enzyme involved.

🔍 Worked example — Standard

Explain why the formation of haemoglobinic acid (HHb) is beneficial for oxygen transport in respiring tissues.

🔍 Worked example — Exam-style

A red blood cell enters a tissue with high PCO_2. Describe the sequence of events regarding CO₂ transport within the cell.

⚠️ Common errors

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🎯 Exam question

Key ideas — The chloride shift and the plasma

The hydrogencarbonate ions made in the red blood cell diffuse out into the plasma. To balance the loss of negative charge, chloride ions (Cl⁻) diffuse into the red blood cell from the plasma: this is the chloride shift. It matters because it keeps the red blood cell electrically neutral, so hydrogencarbonate can keep leaving, which keeps the concentration of HCO₃⁻ inside low and lets carbonic anhydrase keep converting CO₂.

🧑‍🏫 Teacher's explanation

Think of a crowded room where people are constantly passing heavy boxes (carbon dioxide) out of the room. If the room gets too full of boxes, people stop passing them. To keep the flow going, we need a way to move those boxes into a 'buffer' area. In your blood, the red blood cell is the room, and the plasma is the hallway.

Ch 2 · 8.2 Transport of oxygen and carbon dioxide · AS & A Level Biology 9700

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