201. Exchange region P has a normal thin membrane and a moderate partial-pressure gradient. Region Q has a larger gradient but a markedly thickened membrane. Which conclusion is most justified without additional rate data?
ⓐ. The larger gradient and thicker membrane have opposing effects
ⓑ. The larger gradient necessarily makes Q exchange faster
ⓒ. The thicker membrane necessarily makes Q exchange slower
ⓓ. The two effects necessarily cancel, giving equal exchange rates
Correct Answer: The larger gradient and thicker membrane have opposing effects
Explanation: A larger partial-pressure difference favours net diffusion by strengthening the driving force. Increased membrane thickness opposes transfer by lengthening the pathway that gas molecules must cross. Region Q contains one favourable and one unfavourable change. Without numerical information about their relative magnitudes, it is not valid to claim that either effect must dominate. Thickening does not necessarily stop all diffusion, and the presence of a gradient does not make thickness irrelevant. The safest biological conclusion recognises that respiratory-gas transfer depends on interacting factors. Predicting exchange from one variable alone can fail when another major determinant changes in the opposite direction. From alveolar air toward blood, the barrier is arranged as alveolar epithelium, basement substance and capillary endothelium. From the supplied evidence, the thinness of each component is an exchange adaptation rather than a mechanical support feature. Within this diffusion comparison, efficient transfer requires the diffusion barrier to remain thin while alveolar air and capillary blood maintain different partial pressures.
202. Four alveolar regions contain the same respiratory gas.
| Region | Partial-pressure gradient | Barrier | Capillary association |
|---|
| P | Large | Very thin | Extensive |
| Q | Large | Thickened | Extensive |
| R | Small | Very thin | Reduced |
| S | Absent | Very thin | Extensive |
Which region should show the greatest sustained gas transfer?
ⓐ. Region Q
ⓑ. Region P
ⓒ. Region S
ⓓ. Region R
Correct Answer: Region P
Explanation: Region P combines all three favourable conditions listed in the table. Its large partial-pressure gradient provides a strong directional drive, its very thin barrier supplies a short diffusion distance, and its extensive capillary association maintains close contact with blood. Region Q loses part of this advantage through membrane thickening. Region R has both a smaller gradient and reduced vascular association. Region S has an intact thin, vascular surface but lacks a partial-pressure difference, so it lacks a driving force for net transfer. All rows must be integrated because no single favourable feature identifies the best exchange region. P is the only region in which gradient, membrane structure and blood association all support sustained gaseous exchange. The same thin air–blood barrier carries both gases, with their separate gradients fixing opposite directions.
203. An alveolus remains well ventilated, its diffusion membrane is thin, and an initial oxygen gradient exists. Blood flow through its surrounding capillaries then stops. Which outcome is most likely as the condition persists?
ⓐ. Oxygen exchange increases since stationary blood remains beside the alveolus
ⓑ. Carbon dioxide solubility immediately restores normal capillary circulation
ⓒ. The alveolar epithelium begins transporting gases to distant tissues directly
ⓓ. Sustained exchange declines because blood no longer renews gases at the surface
Correct Answer: Sustained exchange declines because blood no longer renews gases at the surface
Explanation: A thin membrane and an initial pressure difference permit gas movement at first, but effective pulmonary exchange must also be linked to circulating blood. Flowing blood carries newly acquired oxygen away and brings carbon-dioxide-rich blood to the alveolar surface. When capillary flow stops, local blood is not continuously renewed. Oxygen accumulation and carbon-dioxide depletion near the membrane weaken the gradients that support continuing transfer. The alveolar epithelium cannot deliver gases directly to distant tissues, and gas solubility cannot restore blood flow. The case demonstrates why extensive vascularisation is a functional exchange adaptation rather than merely an anatomical association. Sustained gas exchange depends on both diffusion across the barrier and transport by the adjacent circulation. In the present context, the thinness of each component is an exchange adaptation rather than a mechanical support feature. Efficient transfer therefore requires both a short diffusion path and continuous renewal of blood that preserves the gas gradients.
204. Match each factor with its most direct contribution to respiratory-gas exchange. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Partial-pressure gradient | 1. Provides close contact with circulating blood |
| Q. Gas solubility | 2. Influences transfer per unit pressure difference |
| R. Membrane thinness | 3. Supplies direction and driving force for net diffusion |
| S. Extensive vascularisation | 4. Shortens the distance crossed by gases |
ⓐ. P-2, Q-3, R-1, S-4
ⓑ. P-4, Q-1, R-3, S-2
ⓒ. P-3, Q-2, R-4, S-1
ⓓ. P-1, Q-4, R-2, S-3
Correct Answer: P-3, Q-2, R-4, S-1
Explanation: A partial-pressure gradient establishes the direction and driving force for net diffusion, giving P-3. Solubility affects how readily a gas can transfer under a given pressure difference, producing Q-2. Membrane thinness shortens the diffusion pathway, so R maps to 4. Extensive vascularisation brings circulating blood into close contact with the alveolar surface and continually connects local exchange with whole-body transport, giving S-1. These factors perform related but non-identical roles. A thin membrane cannot create a gradient, and a gradient cannot replace the need for blood contact. Their combined operation makes the alveolar surface suitable for oxygen entry and carbon-dioxide removal. A thicker barrier lengthens the diffusion path and slows transfer even when area and pressure gradients are unchanged.
205. Assertion: Extensive vascularisation is an important adaptation of the alveolar exchange surface.
Reason: Carbon dioxide is approximately \(20\text{–}25\) times more soluble than oxygen.
ⓐ. Both statements are true, and the Reason correctly explains the Assertion
ⓑ. Both statements are true, but the Reason does not explain the Assertion
ⓒ. The Assertion is true, whereas the Reason is false and cannot explain it
ⓓ. The Assertion is false, whereas the Reason is true and cannot explain it
Correct Answer: Both statements are true, but the Reason does not explain the Assertion
Explanation: Extensive vascularisation brings a continuous supply of deoxygenated blood close to alveolar air, carries absorbed oxygen away and delivers carbon-dioxide-rich blood for pulmonary release. This close and renewed blood contact is therefore an important adaptation of the alveolar exchange surface. Carbon dioxide is indeed approximately \(20\text{–}25\) times more soluble than oxygen, which helps it diffuse efficiently despite a smaller partial-pressure gradient. However, solubility is a property of the gas, whereas vascularisation concerns the anatomical supply and removal of blood at the exchange surface. Both features improve gaseous exchange, but the solubility difference does not explain why alveoli require an extensive capillary network. The two statements are true without forming the proposed explanatory relation.
206. Across a stated volume of blood, a total of \(200\,\mathrm{mL}\) of oxygen is transported. Using the approximate transport fractions, how much is carried by red blood cells and how much remains dissolved in plasma?
ⓐ. \(6\,\mathrm{mL}\) by red blood cells and \(194\,\mathrm{mL}\) in plasma
ⓑ. \(140\,\mathrm{mL}\) by red blood cells and \(60\,\mathrm{mL}\) in plasma
ⓒ. \(190\,\mathrm{mL}\) by red blood cells and \(10\,\mathrm{mL}\) in plasma
ⓓ. \(194\,\mathrm{mL}\) by red blood cells and \(6\,\mathrm{mL}\) in plasma
Correct Answer: \(194\,\mathrm{mL}\) by red blood cells and \(6\,\mathrm{mL}\) in plasma
Explanation: The stated \(200\,\mathrm{mL}\) is the total transported oxygen, not the volume of blood. Under the approximate normal reference, red blood cells carry \(97\%\) of that oxygen, mainly through haemoglobin binding; the remaining \(3\%\) is physically dissolved in plasma. Calculate the two parts separately:
\[
0.97\times200\,\mathrm{mL}=194\,\mathrm{mL}
\]
\[
0.03\times200\,\mathrm{mL}=6\,\mathrm{mL}
\]
The amounts add to the supplied total:
\[
194\,\mathrm{mL}+6\,\mathrm{mL}=200\,\mathrm{mL}
\]
Red blood cells therefore account for the much larger oxygen-transport share, while dissolved oxygen contributes a small but non-zero share. The percentages describe how oxygen is carried in this sample; they do not say how much oxygen will be unloaded at a particular tissue. That delivery depends on local oxygen pressure and other conditions that favour haemoglobin dissociation. Keeping transport fraction separate from tissue delivery avoids assigning the \(97\%\) value to the wrong physiological question. A second check is to subtract the red-cell share from the total: \(200-194=6\,\mathrm{mL}\); this matches the independently calculated plasma share.
207. Compare the approximate transport patterns.
| Gas | RBC-associated fraction | Dissolved in plasma | Other major form |
|---|
| \(\mathrm{O_2}\) | \(97\%\) | \(3\%\) | None listed as dominant |
| \(\mathrm{CO_2}\) | \(20\text{–}25\%\) | \(7\%\) | \(70\%\) bicarbonate |
Which inference is supported?
ⓐ. Oxygen is mainly RBC-bound, whereas carbon dioxide is mainly bicarbonate
ⓑ. Both gases are transported mainly as substances freely dissolved in plasma
ⓒ. Carbon dioxide relies more heavily than oxygen on direct RBC association
ⓓ. Oxygen and carbon dioxide have identical transport distributions
Correct Answer: Oxygen is mainly RBC-bound, whereas carbon dioxide is mainly bicarbonate
Explanation: The oxygen row shows that approximately \(97\%\) of transported oxygen is associated with red blood cells and only \(3\%\) is dissolved in plasma. The carbon-dioxide row has a different dominant pathway: about \(70\%\) is carried as bicarbonate, while approximately \(20\text{–}25\%\) is associated with red blood cells and \(7\%\) is dissolved. The comparison requires identifying the largest category for each gas rather than assuming that both use the same route. Oxygen transport depends overwhelmingly on red-cell haemoglobin, whereas carbon dioxide is distributed mainly through chemical conversion to bicarbonate. The table demonstrates that sharing the same bloodstream does not require identical molecular forms of carriage. The dominant transport form reflects chemical conversion or binding, not simply physical dissolution in plasma. Approximate carbon-dioxide fractions may not sum to exactly \(100\%\) when the carbamino fraction is given as a range.
208. Consider the following statements about oxygen transport in blood.
I. Approximately \(97\%\) of oxygen is carried by red blood cells.
II. About \(3\%\) of oxygen remains dissolved in plasma.
III. Plasma-dissolved oxygen is the principal transport form.
IV. The listed percentages are approximate physiological values.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Red blood cells carry the overwhelming majority of blood oxygen, approximately \(97\%\), mainly through haemoglobin. A much smaller fraction, about \(3\%\), remains physically dissolved in plasma. Statements I and II correctly state this distribution. Statement III reverses the relative importance of the two forms and is inconsistent with the percentages. Statement IV is also valid, since transport fractions are physiological approximations rather than exact values identical in every person and condition. Statements I, II and IV identify the dominant oxygen carrier while preserving the approximate nature of the numerical values. Plasma contributes to oxygen transport, but it does not replace the much larger red-blood-cell-associated fraction. Transport percentages describe distribution among forms and do not by themselves specify the direction of diffusion.
209. Haemoglobin loses most of its ability to bind oxygen, but oxygen solubility in plasma remains unchanged. Which immediate transport effect is most likely?
ⓐ. Plasma oxygen transport compensates for nearly all lost haemoglobin binding
ⓑ. Most oxygen capacity is lost; the small dissolved fraction remains
ⓒ. Oxygen shifts mainly to carbamino-haemoglobin transport
ⓓ. Bicarbonate formation restores the lost oxygen-carrying capacity
Correct Answer: Most oxygen capacity is lost; the small dissolved fraction remains
Explanation: Approximately \(97\%\) of blood oxygen is normally transported through red blood cells, chiefly by reversible binding to haemoglobin. Only about \(3\%\) is dissolved directly in plasma. If haemoglobin loses most of its oxygen-binding ability, the dominant transport route is impaired. The unchanged plasma solubility preserves only the small dissolved component and cannot maintain the normal total oxygen-carrying capacity. Bicarbonate is a major transport form for carbon dioxide, not a substitute oxygen carrier. The altered condition affects oxygen transport immediately through loss of haemoglobin function, even though plasma can still contain some dissolved oxygen. The prediction follows from combining carrier function with the relative proportions of the two oxygen-transport forms. Red blood cells are central to both haemoglobin binding and rapid bicarbonate formation through carbonic anhydrase.
210. A blood volume transports \(1000\,\mathrm{mL}\) of carbon dioxide in total. Which set best represents the approximate amounts carried as bicarbonate, through red blood cells and dissolved in plasma?
ⓐ. \(700\,\mathrm{mL}\); \(200\text{–}250\,\mathrm{mL}\); \(70\,\mathrm{mL}\)
ⓑ. \(200\text{–}250\,\mathrm{mL}\); \(700\,\mathrm{mL}\); \(70\,\mathrm{mL}\)
ⓒ. \(700\,\mathrm{mL}\); \(70\,\mathrm{mL}\); \(200\text{–}250\,\mathrm{mL}\)
ⓓ. \(70\,\mathrm{mL}\); \(200\text{–}250\,\mathrm{mL}\); \(700\,\mathrm{mL}\)
Correct Answer: \(700\,\mathrm{mL}\); \(200\text{–}250\,\mathrm{mL}\); \(70\,\mathrm{mL}\)
Explanation: Approximately \(70\%\) of transported carbon dioxide is carried as bicarbonate. For a total of \(1000\,\mathrm{mL}\):
\[
1000\times\frac{70}{100}=700\,\mathrm{mL}
\]
The red-blood-cell-associated fraction is approximately \(20\text{–}25\%\):
\[
1000\times\frac{20\text{–}25}{100}=200\text{–}250\,\mathrm{mL}
\]
The dissolved fraction is about \(7\%\):
\[
1000\times\frac{7}{100}=70\,\mathrm{mL}
\]
The categories are approximate, and the range means their reported maximum values need not sum to exactly \(100\%\). The ordered amounts are \(700\,\mathrm{mL}\), \(200\text{–}250\,\mathrm{mL}\) and \(70\,\mathrm{mL}\). Within reversible oxygen carriage, most oxygen is carried reversibly by haemoglobin in red blood cells, leaving only a small fraction dissolved in plasma. Within the bicarbonate pathway, most carbon dioxide is transported as bicarbonate, with lower fractions associated with haemoglobin or dissolved in plasma. At the relevant gas interface, percentage calculations should be applied to the stated total gas quantity before comparing transport forms.
211. Match each transport form with its approximate fraction. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Carbon dioxide as bicarbonate | 1. \(70\%\) |
| Q. RBC-associated carbon dioxide | 2. \(20\text{–}25\%\) |
| R. Carbon dioxide dissolved in plasma | 3. \(7\%\) |
| S. RBC-associated oxygen | 4. \(97\%\) |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-4, Q-3, R-2, S-1
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Bicarbonate is the dominant carbon-dioxide transport form at approximately \(70\%\), giving P-1. About \(20\text{–}25\%\) of carbon dioxide is transported through red blood cells, mainly as carbamino-haemoglobin, so Q maps to 2. The dissolved carbon-dioxide fraction is about \(7\%\), producing R-3. Oxygen has a very different distribution, with approximately \(97\%\) carried by red blood cells, giving S-4. The completed mapping distinguishes gases as well as transport forms. A red-cell association is dominant for oxygen but secondary for carbon dioxide, whose largest fraction undergoes conversion to bicarbonate. Red-cell haemoglobin carries about \(97\%\) of transported oxygen, leaving only about \(3\%\) dissolved in plasma. Within the bicarbonate pathway, most carbon dioxide is transported as bicarbonate, with reduced fractions associated with haemoglobin or dissolved in plasma.
212. A student adds the upper values of the carbon-dioxide transport categories and obtains:
\[
70\%+25\%+7\%=102\%
\]
Which interpretation is most appropriate?
ⓐ. The three transport forms represent mutually exclusive blood samples
ⓑ. The bicarbonate estimate should be adjusted to \(68\%\)
ⓒ. The listed approximate ranges need not sum to exactly \(100\%\)
ⓓ. The dissolved fraction should be excluded when totals are compared
Correct Answer: The listed approximate ranges need not sum to exactly \(100\%\)
Explanation: The reported carbon-dioxide fractions are approximate physiological values. The red-blood-cell-associated category is given as a range of \(20\text{–}25\%\), while bicarbonate and dissolved fractions are also approximate. Selecting the maximum of the range and adding it to the other rounded values can produce a total slightly above \(100\%\). This does not imply that carbon dioxide uses only one form or that the dissolved fraction should be omitted. It also does not justify replacing a stated approximate value with an invented exact percentage. The correct interpretation preserves the approximation qualifier: the figures describe the relative importance of the transport routes rather than an exact partition that must sum perfectly at every chosen endpoint. In the present context, the dominant transport form reflects chemical conversion or binding, not simply physical dissolution in plasma. At the specified reference point, approximate carbon-dioxide fractions may not sum to exactly \(100\%\) when the carbamino fraction is given as a range.
213. Evaluate the following statements about carbon-dioxide transport.
I. Bicarbonate accounts for the largest approximate fraction.
II. About \(20\text{–}25\%\) is transported through red blood cells, mainly as carbamino-haemoglobin.
III. Approximately \(7\%\) remains dissolved in plasma.
IV. Carbamino-haemoglobin accounts for about \(70\%\) of total carbon-dioxide transport.
ⓐ. I, II and III only
ⓑ. I and IV only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Bicarbonate is the principal carbon-dioxide transport form and accounts for approximately \(70\%\) of the total. A smaller fraction, about \(20\text{–}25\%\), is transported through red blood cells, mainly as carbamino-haemoglobin, while roughly \(7\%\) remains dissolved in plasma. Statements I, II and III accurately represent this distribution. Statement IV transfers the dominant bicarbonate percentage to carbamino-haemoglobin and is not valid. The accepted combination distinguishes chemical conversion of carbon dioxide into bicarbonate from reversible association of carbon dioxide with haemoglobin. These routes operate simultaneously, but their approximate contributions are unequal; bicarbonate carries by far the largest share under normal conditions. The fractions describe transport forms in blood and do not represent separate sequential stages of gas movement.
214. The conversion pathway that normally produces bicarbonate from tissue carbon dioxide is strongly impaired, while dissolved transport and carbamino-haemoglobin formation remain unchanged. Which consequence is most likely?
ⓐ. The dissolved carbon-dioxide fraction is selectively reduced
ⓑ. The major bicarbonate transport route is selectively reduced
ⓒ. The carbamino route is reduced while bicarbonate transport remains normal
ⓓ. Diffusion from tissues is reduced before transport-form conversion
Correct Answer: The major bicarbonate transport route is selectively reduced
Explanation: About \(70\%\) of carbon dioxide is normally carried as bicarbonate, making it the dominant transport form. If the conversion pathway producing bicarbonate is strongly impaired, this largest component is directly reduced. The dissolved and carbamino-haemoglobin routes remain available, so some carbon-dioxide transport can continue through those smaller fractions. Their continued operation does not mean that the \(7\%\) dissolved portion automatically expands to replace the missing bicarbonate fraction. Tissue-to-blood diffusion occurs before or alongside conversion and is not abolished simply by naming a transport defect. The predicted outcome follows from identifying which route carries the greatest share and then applying the selective impairment to that route. The major transport route is determined by chemical conversion or molecular binding rather than by simple plasma dissolution.
215. Assertion: Formation of oxyhaemoglobin is reversible.
Reason: Once oxygen binds to haemoglobin, it remains permanently fixed to its iron-containing pigment.
ⓐ. Both statements are true, and the Reason correctly explains the Assertion
ⓑ. Both statements are true, but the Reason does not explain the Assertion
ⓒ. The Assertion is true, whereas the Reason is false and cannot explain it
ⓓ. The Assertion is false, whereas the Reason is true and cannot explain it
Correct Answer: The Assertion is true, whereas the Reason is false and cannot explain it
Explanation: Haemoglobin is an iron-containing red pigment present in red blood cells. It binds oxygen reversibly to form oxyhaemoglobin. Reversibility is essential to its transport role: oxygen must be loaded under favourable conditions in the lungs and released under the different conditions present in tissues. The Assertion is valid. The proposed Reason is incorrect since permanent fixation would prevent oxygen unloading and would make haemoglobin unsuitable as a reusable carrier. Binding does not destroy haemoglobin or lock oxygen to it indefinitely. The reversible association allows the same haemoglobin molecule to participate repeatedly in oxygen uptake, blood transport and tissue delivery. One haemoglobin molecule can bind a maximum of four oxygen molecules through four haem groups.
216. Red blood cells are first exposed to high \(\mathrm{pO_2}\) under conditions favouring oxygen association and later enter a tissue region where oxygen availability is lower and cellular demand continues. The observations show oxyhaemoglobin formation followed by oxygen release. This demonstrates that:
ⓐ. Bicarbonate is the main reversible carrier of oxygen
ⓑ. Haemoglobin loads oxygen in tissues and unloads it in alveoli
ⓒ. Oxygen release occurs only after haemoglobin is degraded
ⓓ. Haemoglobin binds oxygen reversibly for loading and unloading
Correct Answer: Haemoglobin binds oxygen reversibly for loading and unloading
Explanation: In the first condition, high oxygen partial pressure favours the association of oxygen with haemoglobin and the formation of oxyhaemoglobin. When the same red blood cells reach tissues with lower oxygen availability, the balance shifts toward dissociation and oxygen is released for use by cells. The carrier itself remains available for another transport cycle. This sequence demonstrates reversible binding rather than permanent attachment or destruction of haemoglobin. Bicarbonate is associated mainly with carbon-dioxide transport and does not replace haemoglobin as the dominant oxygen carrier. The case links two different environments to complementary carrier functions: loading in an oxygen-rich region and unloading where oxygen is needed. Reversibility allows the same haemoglobin molecules to alternate between pulmonary loading and tissue delivery without being consumed in the process.
217. Consider the following statements about haemoglobin.
I. It is an iron-containing pigment present in red blood cells.
II. One haemoglobin molecule can bind a maximum of four \(\mathrm{O_2}\) molecules.
III. A fully loaded haemoglobin molecule can bind a fifth \(\mathrm{O_2}\) molecule at sufficiently high \(\mathrm{pO_2}\).
IV. Its association with oxygen is reversible.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Haemoglobin is the iron-containing red respiratory pigment located within red blood cells. It binds oxygen reversibly to form oxyhaemoglobin, allowing oxygen to be loaded in the lungs and released in tissues. A single haemoglobin molecule can carry a maximum of four oxygen molecules. Once all four available binding positions are occupied, the molecule is fully loaded and cannot accept a fifth oxygen molecule merely by increasing \(\mathrm{pO_2}\). Statements I, II and IV correctly describe its location, capacity and reversible carrier function. Statement III exceeds the supported molecular maximum. The four-molecule limit distinguishes the carrying capacity of one haemoglobin molecule from the percentage saturation of an entire blood sample. The \(1:4\) molecule ratio must be applied before scaling to multiple haemoglobin molecules.
218. A sample contains \(250\) haemoglobin molecules, and every molecule reaches its maximum oxygen-carrying capacity. How many \(\mathrm{O_2}\) molecules are bound?
ⓐ. \(1000\) oxygen molecules
ⓑ. \(750\) oxygen molecules
ⓒ. \(500\) oxygen molecules
ⓓ. \(250\) oxygen molecules
Correct Answer: \(1000\) oxygen molecules
Explanation: Each haemoglobin molecule can bind a maximum of four oxygen molecules. The sample contains \(250\) haemoglobin molecules, each described as fully loaded. The total number of bound oxygen molecules is:
\[
250\times4=1000
\]
The multiplication uses haemoglobin molecules as the carrier count and four oxygen molecules as the capacity of each carrier. A result of \(250\) would count only one oxygen molecule per haemoglobin, while \(500\) and \(750\) would represent incomplete average loading. Under complete saturation, the \(250\) haemoglobin molecules collectively carry \(1000\) oxygen molecules. Full loading describes maximum occupancy; partial loading would require information about saturation that is not supplied. Counting oxygen molecules differs from counting oxygen atoms, so the requested unit must be retained. Haemoglobin binding greatly increases blood oxygen-carrying capacity above the amount dissolved in plasma alone. Within reversible oxygen carriage, one haemoglobin molecule can bind a maximum of four oxygen molecules through four haem groups.
219. A preparation initially contains \(120\) haemoglobin molecules. One-fourth of them become completely unable to bind oxygen, while the remaining molecules retain a maximum capacity of four \(\mathrm{O_2}\) molecules each. What is the new maximum oxygen load?
ⓐ. \(120\) \(\mathrm{O_2}\) molecules
ⓑ. \(240\) \(\mathrm{O_2}\) molecules
ⓒ. \(480\) \(\mathrm{O_2}\) molecules
ⓓ. \(360\) \(\mathrm{O_2}\) molecules
Correct Answer: \(360\) \(\mathrm{O_2}\) molecules
Explanation: One-fourth of \(120\) haemoglobin molecules become non-functional:
\[
120\times\frac{1}{4}=30
\]
The number remaining capable of oxygen binding is:
\[
120-30=90
\]
Each functional haemoglobin molecule can bind a maximum of four oxygen molecules, so the new maximum load is:
\[
90\times4=360\ \mathrm{O_2\ molecules}
\]
The original maximum of \(480\) would incorrectly include the \(30\) carriers that have lost all binding capacity. The calculation assumes complete saturation of the remaining functional molecules; a partial saturation value is not supplied. The requested unit is oxygen molecules, not oxygen atoms, so no additional factor of two is applied. The ratio must be applied after removing the non-functional carriers, because those molecules contribute no binding sites. Multiplying all \(120\) molecules by four and then reducing the oxygen count by only one-fourth leads to the same result here, but the carrier-first method makes the biological assumption explicit.
220. Match each haemoglobin state with the appropriate oxygen-load description. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Unbound haemoglobin | 1. Carries \(0\) \(\mathrm{O_2}\) molecules |
| Q. Haemoglobin carrying two \(\mathrm{O_2}\) molecules | 2. Partially oxygenated state |
| R. Fully loaded haemoglobin | 3. Carries \(4\) \(\mathrm{O_2}\) molecules |
| S. Proposed haemoglobin carrying five \(\mathrm{O_2}\) molecules | 4. Exceeds the maximum capacity |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-2, R-3, S-4
ⓒ. P-3, Q-4, R-1, S-2
ⓓ. P-4, Q-3, R-2, S-1
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Unbound haemoglobin carries no oxygen, giving P-1. A molecule carrying two oxygen molecules is neither unbound nor fully loaded, so it represents a partially oxygenated state and gives Q-2. Complete loading occurs when four oxygen molecules are bound, producing R-3. A proposed load of five exceeds the supported maximum for one haemoglobin molecule, so S maps to 4. The relationships separate actual loading states from percentage saturation of a larger blood sample. Haemoglobin can occur at intermediate levels of oxygen association, but its molecular carrying capacity cannot extend beyond four oxygen molecules. Scaling begins with the maximum relation of four oxygen molecules for each haemoglobin molecule. Maximum occupancy is stated explicitly; a partially loaded sample would require a supplied saturation value.
