301. Three patients show different cardiac conditions.
| Patient | Observation |
|---|
| P | The heart pumps inadequately and pulmonary congestion develops |
| Q | The heartbeat ceases |
| R | Cardiac muscle is suddenly damaged after insufficient blood supply |
Which classification is correct?
ⓐ. P cardiac arrest, Q heart attack, R heart failure
ⓑ. P heart attack, Q heart failure, R cardiac arrest
ⓒ. P heart failure, Q cardiac arrest, R heart attack
ⓓ. P heart failure, Q heart attack, R cardiac arrest
Correct Answer: P heart failure, Q cardiac arrest, R heart attack
Explanation: Patient P has inadequate pumping with pulmonary congestion, which defines heart failure. The heart may continue beating, but its output is insufficient for the body’s needs. Patient Q shows cessation of heartbeat, which is cardiac arrest. Patient R has sudden damage to cardiac muscle following insufficient blood supply, which describes a heart attack. The defining events are inadequate pumping, cessation of heartbeat and sudden myocardial damage, respectively. Pump insufficiency, heartbeat cessation and sudden myocardial damage are distinct biological conditions. Their terms must not be used interchangeably. The mapping is P heart failure, Q cardiac arrest and R heart attack, with each label fixed by a different defining event. Using the presence or absence of heartbeat alone is insufficient for P and R, so the pumping and tissue-damage evidence must also be interpreted.
302. A patient has breathlessness associated with fluid congestion in the lungs. Cardiac recordings show that the heart is still beating, but its minute output is insufficient to meet body needs. No evidence shows sudden cessation of heartbeat. The most appropriate diagnosis is:
ⓐ. cardiac arrest because lung congestion always means the heart has stopped
ⓑ. angina because every breathing difficulty is acute myocardial pain
ⓒ. heart failure because pumping is inadequate despite continued heartbeat
ⓓ. hypertension because blood pressure must be above \(140/90\,\text{mm Hg}\)
Correct Answer: heart failure because pumping is inadequate despite continued heartbeat
Explanation: The case provides the defining functional evidence for heart failure: the heart continues beating but cannot pump enough blood to satisfy body requirements. Pulmonary congestion supports this interpretation; ineffective pumping can cause blood and fluid to accumulate in the lung circulation. Cardiac arrest is excluded by the continued heartbeat. Angina is primarily acute chest pain caused by insufficient oxygen reaching cardiac muscle, and the findings do not supply that symptom pattern. Hypertension requires repeated high-pressure readings, which are not provided. The diagnosis should follow the measured pumping failure and congestion rather than an unrelated assumption. Heart failure is distinct from both stopped cardiac activity and sudden myocardial damage. The minute-output measurement supplies direct functional evidence, while the lung finding provides a compatible consequence of the same sustained cardiac pumping defect.
303. Conditions P, Q and R are defined as follows.
P. Persistent elevation of arterial pressure
Q. Atherosclerotic narrowing of vessels supplying cardiac muscle
R. Inadequate pumping that may produce pulmonary congestion
Which classification is accurate?
ⓐ. P hypertension, Q coronary artery disease, R heart failure
ⓑ. P coronary artery disease, Q heart failure, R hypertension
ⓒ. P heart failure, Q hypertension, R coronary artery disease
ⓓ. P angina, Q cardiac arrest, R heart attack
Correct Answer: P hypertension, Q coronary artery disease, R heart failure
Explanation: Persistent elevation of arterial pressure identifies hypertension, particularly when repeated readings meet or exceed the stated threshold. Atherosclerotic narrowing of vessels supplying cardiac muscle identifies coronary artery disease, whose deposits reduce the lumen of coronary vessels. Inadequate pumping that fails to meet body needs and may produce pulmonary congestion identifies heart failure. The three conditions differ in their primary disturbed feature: pressure regulation, coronary-vessel structure and pump effectiveness. Angina is acute chest pain caused by insufficient oxygen reaching the myocardium, cardiac arrest is cessation of heartbeat, and heart attack is sudden cardiac-muscle damage from insufficient blood supply. The classification is P hypertension, Q coronary artery disease and R heart failure.
304. Four blood-derived fractions are compared.
| Fraction | Formed elements | Clotting factors | Albumins and globulins |
|---|
| P | Present | Present | Present |
| Q | Absent | Present | Present |
| R | Absent | Absent | Present |
| S | Concentrated | Only traces of surrounding fluid | Only traces of surrounding fluid |
Which identification is consistent with all the observations?
ⓐ. P is serum, Q is whole blood, R is plasma and S is formed elements
ⓑ. P is whole blood, Q is plasma, R is serum and S is formed elements
ⓒ. P is plasma, Q is serum, R is whole blood and S is formed elements
ⓓ. P is whole blood, Q is serum, R is plasma and S is lymph
Correct Answer: P is whole blood, Q is plasma, R is serum and S is formed elements
Explanation: Fraction P contains the fluid matrix, clotting factors and formed elements, so it represents whole blood. Fraction Q has no cells or cell fragments but retains clotting factors and major plasma proteins, identifying plasma separated from anticoagulated blood. Fraction R lacks formed elements and clotting factors while retaining albumins, globulins and other dissolved substances; this is serum rather than protein-free fluid. Fraction S contains the concentrated erythrocytes, leucocytes and platelets left after most surrounding plasma has been removed. The comparison integrates cellular content with the presence of clotting factors. Absence of cells alone cannot distinguish plasma from serum, since both are cell-free. The decisive distinction is that plasma retains clotting factors, whereas serum is the fluid remaining after those factors have been removed or consumed. This classification also explains why only whole blood simultaneously carries cells, dissolved transport substances and an intact coagulation system.
305. A clear blood-derived fluid contains albumins, globulins, mineral ions, glucose and other dissolved substances. It contains no formed elements and cannot supply the complete set of clotting factors required for normal fibrin formation. The fluid is:
ⓐ. whole blood, because dissolved substances are present with circulating formed elements
ⓑ. a formed-element concentrate, because erythrocytes and platelets dominate its composition
ⓒ. plasma, because fibrinogen and the other inactive clotting factors remain available
ⓓ. serum, because many plasma solutes remain although clotting factors are absent
Correct Answer: serum, because many plasma solutes remain although clotting factors are absent
Explanation: Serum is the fluid obtained after blood has clotted and the clot has been removed. It retains water, albumins, globulins, mineral ions, nutrients and many other dissolved substances, so it is not pure water and is not devoid of all proteins. Its defining distinction from plasma is the absence of clotting factors that have been removed or consumed during clot formation. Plasma would contain fibrinogen and other inactive clotting components, whereas whole blood would also include erythrocytes, leucocytes and platelets. A formed-element concentrate would be rich in those cells and cell fragments rather than being a clear fluid. The functional clue is decisive: the sample can still transport dissolved materials, but it cannot by itself restore the complete fibrin-forming system supplied by normal plasma.
306. Evaluate the following statements.
I. Whole blood contains plasma together with formed elements.
II. Erythrocytes, leucocytes and platelets belong to the formed-element fraction.
III. Serum is obtained when all plasma proteins are removed from blood.
IV. Serum lacks clotting factors but retains many other dissolved substances.
ⓐ. I, II and IV only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is valid since blood is organised into a fluid plasma fraction and suspended formed elements. Statement II correctly includes erythrocytes, leucocytes and platelets in the formed-element category, even though platelets are cell fragments rather than complete cells. Statement III is false. Serum is not produced by removing every plasma protein; it is plasma after clotting factors have been removed or consumed during coagulation. Albumins, globulins, ions, nutrients and many other dissolved materials remain. Statement IV is valid and supplies the important qualifier missing from Statement III. The correct set is I, II and IV. This distinction prevents serum from being confused with pure water and also separates it from plasma, which retains the clotting components needed for fibrin formation. Cellular content separates whole blood from its fluid fractions, while clotting-factor status separates plasma from serum.
307. Three formed-element populations are selectively removed from separate blood samples. Removal of Population P sharply reduces respiratory-gas carrying capacity. Removal of Population Q reduces phagocytosis by the most abundant leucocyte but leaves haemoglobin concentration and platelet count unchanged. Removal of Population R markedly delays coagulation. Population Q is most likely:
ⓐ. erythrocytes
ⓑ. platelets
ⓒ. neutrophils
ⓓ. lymphocytes
Correct Answer: neutrophils
Explanation: Population P must represent erythrocytes, since their haemoglobin content accounts for most respiratory-gas carrying capacity. Population R must represent platelets, whose loss delays coagulation. The remaining evidence identifies Population Q more specifically than merely calling it a leucocyte. Its removal reduces phagocytosis by the most abundant white blood cell, while erythrocyte and platelet measurements remain unchanged. Neutrophils form approximately \(60\text{–}65\%\) of circulating leucocytes and are active phagocytes, so their selective removal fits the observation. Monocytes also phagocytose foreign material, but they are agranulocytes present at only about \(6\text{–}8\%\). The experiment uses functional loss and relative abundance together, allowing neutrophils to be distinguished from another phagocytic population without relying on a single morphological clue. The preserved haemoglobin and platelet measurements act as controls that exclude erythrocytes and platelets from the identity of Population Q.
308. Match each formed element with the evidence that identifies it. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Erythrocyte | 1. Agranulocyte occurring as B and T forms |
| Q. Neutrophil | 2. Megakaryocyte-derived fragment involved in coagulation |
| R. Lymphocyte | 3. Biconcave, usually anucleate and rich in haemoglobin |
| S. Platelet | 4. Most abundant leucocyte and an active phagocyte |
ⓐ. P-4, Q-3, R-2, S-1
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-2, Q-1, R-4, S-3
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-3, Q-4, R-1, S-2
Explanation: Erythrocytes are biconcave, usually lack a nucleus in adult mammals and contain haemoglobin, giving P-3. Neutrophils are granulocytes that form the largest fraction of circulating leucocytes and perform phagocytosis, giving Q-4. Lymphocytes are agranulocytes present as B and T forms involved in immune responses, giving R-1. Platelets are fragments released from bone-marrow megakaryocytes and participate in coagulation, giving S-2. Each mapping uses a different identifying dimension: morphology and pigment, relative abundance and function, immune classification, or cellular origin. The complete pattern prevents two common confusions—treating platelets as nucleated leucocytes and assigning all phagocytosis to agranulocytes. The resulting relation is P-3, Q-4, R-1 and S-2. The mapping also distinguishes complete blood cells from the platelet fragments that share the formed-element category. Reading the four relations together links blood-cell structure, abundance, defence and haemostasis without reducing formed elements to one shared function.
309. An unknown formed element is nucleated, lacks haemoglobin, has no conspicuous cytoplasmic granules, forms approximately \(6\text{–}8\%\) of leucocytes and phagocytoses foreign organisms. Which nested relationship identifies it?
ⓐ. Formed element → erythrocyte → granulocyte → eosinophil
ⓑ. Formed element → platelet → agranulocyte → lymphocyte
ⓒ. Plasma component → leucocyte → granulocyte → basophil
ⓓ. Formed element → leucocyte → agranulocyte → monocyte
Correct Answer: Formed element → leucocyte → agranulocyte → monocyte
Explanation: The object is a formed element since it is one of the suspended cellular components of blood. Its nucleus and absence of haemoglobin place it among leucocytes rather than erythrocytes. Lack of conspicuous granules places it in the agranulocyte class. The differential range of approximately \(6\text{–}8\%\), together with phagocytosis, identifies a monocyte rather than a lymphocyte. The correct hierarchy narrows from the broad blood-component category to the cell class, then to the leucocyte subdivision and finally to the specific cell type. Platelets cannot fit the hierarchy since they are megakaryocyte-derived fragments, and plasma components are not formed elements. The nested relation preserves every supplied feature rather than selecting a cell from one function alone. The abundance range resolves the identity because both neutrophils and monocytes can phagocytose, but only monocytes fit the stated agranulocyte percentage.
310. A sensitised recipient has blood group A, Rh negative. Four erythrocyte donors are available:
P. A, Rh negative
Q. O, Rh negative
R. B, Rh negative
S. A, Rh positive
Which donors satisfy both ABO compatibility and Rh safety for this recipient?
ⓐ. P and Q only
ⓑ. P and S only
ⓒ. Q and R only
ⓓ. P, Q and S only
Correct Answer: P and Q only
Explanation: A group A recipient has anti-B antibody in plasma, so donor erythrocytes must not carry B antigen. Group A cells and group O cells satisfy that ABO requirement; group B cells do not. The recipient is Rh negative and already sensitised, meaning anti-Rh antibodies may react with Rh-positive donor erythrocytes. Donors must be Rh negative as well. Donor P is A, Rh negative and Donor Q is O, Rh negative, so each passes both filters. Donor R fails the ABO test through B antigen, while Donor S passes the ABO test but fails the Rh test. The case illustrates an intersection rule: a donor is acceptable only when compatibility is maintained simultaneously in the ABO and Rh systems. Passing only one compatibility test is insufficient, since either recipient anti-B or sensitised anti-Rh can make donor cells unsafe.
311. Let the antigen set on donor erythrocytes be compared with the antibody-target set of a sensitised group B, Rh-negative recipient. The recipient can react against \(\{\mathrm{A},\mathrm{Rh}\}\). Donor antigen sets are:
P. \(\{\mathrm{B}\}\)
Q. \(\{\mathrm{A}\}\)
R. \(\{\mathrm{B},\mathrm{Rh}\}\)
S. \(\varnothing\)
Which donor sets have an empty intersection with the recipient’s antibody targets?
ⓐ. Q and R only
ⓑ. P and S only
ⓒ. P, Q and S only
ⓓ. R and S only
Correct Answer: P and S only
Explanation: A safe donor set must satisfy:
\[
\mathrm{Donor\ antigens}\cap\{\mathrm{A},\mathrm{Rh}\}=\varnothing
\]
Set P contains only B antigen. A group B recipient does not carry anti-B, and the set lacks Rh antigen, so the intersection is empty. Set S contains neither A, B nor Rh antigens, making its intersection empty as well. Set Q contains A antigen, which is targeted by the recipient’s anti-A antibody. Set R contains Rh antigen, which is targeted after sensitisation even though its B antigen is ABO-compatible. The symbolic test combines the two compatibility systems without treating either as optional. P corresponds to B, Rh-negative erythrocytes and S to O, Rh-negative erythrocytes; both avoid every stated antibody target. The empty-intersection rule gives a compact way to express that no donor antigen may match any antibody capable of attacking it. The sensitised recipient has anti-A from group B status and anti-Rh from prior exposure. Donor sets containing A or Rh antigen are reactive, whereas B antigen alone is not targeted.
312. Proposed erythrocyte transfusions are listed below.
| Proposal | Recipient | Donor | Relevant condition |
|---|
| P | AB, Rh positive | B, Rh positive | No anti-A or anti-B in recipient plasma |
| Q | O, Rh positive | A, Rh positive | Recipient has anti-A |
| R | A, Rh negative | O, Rh negative | Recipient is sensitised |
| S | B, Rh negative | B, Rh positive | Recipient is sensitised |
Which proposals are compatible under the supplied ABO and Rh conditions?
ⓐ. P and Q only
ⓑ. Q and S only
ⓒ. P, Q and R only
ⓓ. P and R only
Correct Answer: P and R only
Explanation: Proposal P is compatible: an AB recipient lacks natural anti-A and anti-B antibodies, and the donor and recipient are both Rh positive. Proposal Q is ABO-incompatible since the O recipient has anti-A that can react with A antigen on donor erythrocytes. Proposal R is compatible under both systems. Group O donor cells lack A and B antigens, and the Rh-negative donor avoids reaction with anti-Rh in the sensitised Rh-negative recipient. Proposal S is ABO-compatible but Rh-incompatible, since the sensitised recipient may attack Rh-positive donor cells. Each proposal must pass both the ABO antigen-antibody condition and the stated Rh condition. P and R are the only proposals that satisfy both the ABO antigen-antibody condition and the supplied Rh condition. This row-by-row comparison shows why an ABO-compatible proposal may still be rejected after Rh status and sensitisation are considered.
313. Equal plasma preparations contain normal fibrinogen and \(\mathrm{Ca^{2+}}\), and injury-related factors produce thrombokinase normally. In Preparation P, prothrombin is removed; fibrin does not form until active thrombin is added. In Preparation Q, fibrinogen is removed; active thrombin still fails to produce fibrin. Which inference is best supported?
ⓐ. Fibrinogen lies upstream of prothrombin and generates thrombokinase after injury.
ⓑ. Prothrombin yields thrombin, which then acts on fibrinogen to form fibrin.
ⓒ. Thrombin acts before prothrombin and substitutes for fibrinogen in clot formation.
ⓓ. Prothrombin and fibrinogen perform the same interchangeable step in coagulation.
Correct Answer: Prothrombin yields thrombin, which then acts on fibrinogen to form fibrin.
Explanation: Preparation P lacks prothrombin, so the normal pathway cannot generate thrombin even though thrombokinase and calcium are available. Adding active thrombin bypasses that missing conversion and restores the next step, showing that prothrombin lies upstream as the inactive precursor of thrombin. Preparation Q lacks fibrinogen. Supplying thrombin cannot restore fibrin because the enzyme has no fibrinogen substrate to convert into insoluble fibrin threads. The paired results locate the two plasma proteins at different positions in the cascade: prothrombin is converted into thrombin, and thrombin then acts on fibrinogen. Their functions are not interchangeable. The experiment also separates an enzyme precursor from the substrate of the active enzyme. A fibrin network can form only when the pathway provides active thrombin and retains fibrinogen for the final conversion. The rescue pattern identifies pathway order by showing which missing reaction is bypassed and which indispensable downstream substrate cannot be replaced.
314. Three preparations receive the same vessel-injury signal. Preparation P contains platelet-rich plasma with normal clotting factors and \(\mathrm{Ca^{2+}}\). Preparation Q contains serum reconstituted with washed platelets to the same count as P and added \(\mathrm{Ca^{2+}}\). Preparation R contains platelet-rich plasma with normal clotting factors but no \(\mathrm{Ca^{2+}}\). Only P forms an effective fibrin clot. Which conclusion best explains the pattern?
ⓐ. Effective clotting requires plasma clotting factors and \(\mathrm{Ca^{2+}}\) along with platelet activation
ⓑ. Serum supplies every inactive clotting factor, whereas added \(\mathrm{Ca^{2+}}\) prevents prothrombin activation
ⓒ. Platelets alone generate thrombin and fibrin when plasma clotting factors and \(\mathrm{Ca^{2+}}\) are absent
ⓓ. \(\mathrm{Ca^{2+}}\) directly replaces fibrinogen and polymerises into the structural threads of the clot
Correct Answer: Effective clotting requires plasma clotting factors and \(\mathrm{Ca^{2+}}\) along with platelet activation
Explanation: Preparation P contains activated platelets, plasma clotting factors including fibrinogen, and \(\mathrm{Ca^{2+}}\). Following the common injury signal, this combination can support the linked reactions that produce thrombin; thrombin can then convert fibrinogen into an insoluble fibrin mesh. Preparation Q has been made by adding washed platelets and calcium to serum. Restoring platelet number and calcium does not restore the clotting factors lost or consumed when serum was produced, so Q cannot make an effective fibrin clot under the stated conditions. Preparation R retains platelet-rich plasma and its soluble precursors but lacks \(\mathrm{Ca^{2+}}\), which is needed for effective progression of coagulation. Thus the two failed preparations isolate different missing requirements: Q lacks the required plasma clotting components, whereas R lacks calcium. Platelets or calcium alone cannot replace fibrinogen and the other needed plasma factors. Only P retains the stated combination, making option A correct.
315. A plasma sample has normal albumin and globulin concentrations but lacks fibrinogen. Platelet activation, thrombokinase formation and thrombin production occur normally after vessel injury. Which functional profile is expected?
ⓐ. Osmotic balance fails, while defence and fibrin-mesh formation remain normal
ⓑ. Defence mechanisms fail, while osmotic balance and fibrin-mesh formation remain normal
ⓒ. Osmotic and defence functions remain supported, but an adequate fibrin mesh cannot form
ⓓ. All three functions fail because fibrinogen normally performs the roles of every plasma protein
Correct Answer: Osmotic and defence functions remain supported, but an adequate fibrin mesh cannot form
Explanation: Albumins contribute importantly to plasma osmotic balance, and globulins are primarily associated with defence. Since both protein groups remain normal, their principal functions can continue under the stated condition. Fibrinogen has a different role: it is the soluble plasma precursor converted by thrombin into insoluble fibrin threads. Normal platelet activation and normal thrombin production preserve the upstream stages of coagulation, but thrombin cannot build an adequate fibrin network when its specific substrate is absent. The developing haemostatic response may begin, yet it lacks the stable mesh that traps formed elements and strengthens the clot. This profile separates the functions of the three major plasma-protein groups instead of treating them as interchangeable. A defect in fibrinogen chiefly impairs fibrin formation rather than directly removing albumin-dependent osmotic support or globulin-associated defence.
316. Capillary walls in a tissue become temporarily impermeable to water and small dissolved substances, while the lymphatic vessels remain patent and continue carrying previously collected lymph. Which paired outcome is most likely at first?
ⓐ. Fresh tissue-fluid formation increases, and erythrocytes enter the intercellular spaces
ⓑ. Fresh tissue-fluid formation and capillary exchange decline, while existing lymph still returns toward veins
ⓒ. Plasma proteins are converted into lymphocytes inside the capillary wall
ⓓ. Tissue fluid becomes serum because all local clotting factors are consumed
Correct Answer: Fresh tissue-fluid formation and capillary exchange decline, while existing lymph still returns toward veins
Explanation: Tissue fluid is formed when water and many small dissolved substances move out of tissue capillaries while most formed elements and larger plasma proteins remain within the blood. If the capillary wall becomes impermeable to those normally filtered materials, little fresh tissue fluid can enter the intercellular spaces. Exchange between capillary blood and nearby cells consequently declines because the immediate extracellular medium is no longer being replenished normally. The lymphatic vessels perform a later and separate function: they collect excess tissue fluid and return it toward major veins. Since those vessels remain patent, lymph already present within them can continue moving even though formation of new tissue fluid has fallen. Erythrocytes do not transform into lymphocytes, and serum formation depends on clotting rather than loss of capillary permeability. The altered condition separates capillary formation of tissue fluid from lymphatic collection and return.
317. Three fluid compartments are compared.
| Compartment | Location | Formed elements | Large plasma proteins | Principal relation |
|---|
| P | Blood vessels | Abundant | Abundant | Circulating transport fluid |
| Q | Spaces around cells | Mostly absent | Much lower | Immediate exchange medium |
| R | Lymphatic vessels | Specialised lymphocytes present | Lower than blood | Returns excess fluid toward veins |
Which conclusion best integrates the table?
ⓐ. Q is whole blood that has entered tissues without selective filtration
ⓑ. R forms independently of Q and does not return to blood
ⓒ. P and Q are identical since both contain water and dissolved solutes
ⓓ. Q is capillary-derived fluid, and excess Q enters lymphatics as R
Correct Answer: Q is capillary-derived fluid, and excess Q enters lymphatics as R
Explanation: Compartment P is whole blood, identified by abundant formed elements and plasma proteins within vessels. Compartment Q lacks most formed elements and contains far fewer large proteins, showing that it forms through selective movement of water and small solutes across capillary walls rather than through unrestricted leakage of whole blood. Its position around cells makes it the immediate exchange medium. Excess Q enters lymphatic vessels and becomes Compartment R, which contains specialised lymphocytes and travels toward venous return. The table describes linked stages rather than three unrelated fluids. Shared water and solutes do not make the compartments identical; location, cellular content, protein content and route all differ. Selective capillary formation followed by lymphatic collection explains the transition from P to Q to R. The progression preserves continuity of body-fluid transport while explaining why the three compartments differ in composition and immediate function.
318. A person has markedly reduced plasma albumin and, independently, an obstruction of lymphatic drainage in one limb. Which combined effect is most likely in that limb?
ⓐ. Tissue-fluid accumulation from low albumin and blocked lymph drainage
ⓑ. Reduced tissue fluid since both changes force water into capillaries
ⓒ. Increased erythrocyte production within the lymphatic vessels
ⓓ. Conversion of tissue fluid into serum through loss of clotting factors
Correct Answer: Tissue-fluid accumulation from low albumin and blocked lymph drainage
Explanation: Albumin contributes importantly to the osmotic balance that helps retain water within plasma. A marked reduction weakens this retaining influence and favours greater persistence of fluid outside the capillaries. Lymphatic vessels provide the return route for excess tissue fluid. Obstruction of that route prevents efficient collection and venous return from the affected limb. The two changes act at different points in fluid balance: one alters the plasma-side osmotic condition, while the other blocks removal of fluid already present in tissue spaces. Their effects reinforce one another, making tissue-fluid accumulation more likely. Erythrocytes are produced in red bone marrow, not in obstructed lymphatics, and serum formation depends on clotting rather than tissue-fluid accumulation. The combined mechanism is reduced retention plus reduced drainage. This integration connects a plasma-protein function with a vessel-drainage function, showing two independent controls over intercellular fluid volume.
319. Use the heart arrangement described below. Chamber P is an upper chamber receiving the venae cavae. Blood passes through Valve Q into lower Chamber R. It then leaves through Valve S and enters an artery carrying blood toward the lungs. Which identification is correct?
ⓐ. P is the left atrium, Q is the mitral valve, R is the left ventricle and S is the aortic valve
ⓑ. P is the right atrium, Q is the pulmonary semilunar valve, R is the right ventricle and S is the tricuspid valve
ⓒ. P is the right atrium, Q is the tricuspid valve, R is the right ventricle and S is the pulmonary semilunar valve
ⓓ. P is the left atrium, Q is the tricuspid valve, R is the left ventricle and S is the pulmonary semilunar valve
Correct Answer: P is the right atrium, Q is the tricuspid valve, R is the right ventricle and S is the pulmonary semilunar valve
Explanation: The venae cavae return systemic venous blood to the right atrium, so Chamber P must be the right atrium. The opening from the right atrium to the right ventricle is guarded by the tricuspid valve, identifying Q. Lower Chamber R is the right ventricle. Blood leaves the right ventricle through the pulmonary semilunar valve, labelled S, and enters the pulmonary artery toward the lungs. The complete identification follows the permitted one-way route:
\[
\text{Venae cavae}\rightarrow\text{right atrium}\rightarrow\text{tricuspid valve}\rightarrow\text{right ventricle}\rightarrow\text{pulmonary semilunar valve}\rightarrow\text{pulmonary artery}
\]
The chamber positions, connected vessels and valve types all support the same right-sided pulmonary outflow pathway. Following every connection in order also rules out left-sided structures, which receive pulmonary venous blood and eject through the aorta.
320. Complete the pressure relations for normal valve movement. At the beginning of ventricular systole, atrioventricular valves close when \(X\). Later, semilunar valves open when \(Y\).
ⓐ. \(X:P_{\mathrm{A}}\gtP_{\mathrm{V}};\ Y:P_{\mathrm{artery}}\gtP_{\mathrm{V}}\)
ⓑ. \(X:P_{\mathrm{artery}}\gtP_{\mathrm{V}};\ Y:P_{\mathrm{A}}\gtP_{\mathrm{V}}\)
ⓒ. \(X:P_{\mathrm{V}}\gtP_{\mathrm{artery}};\ Y:P_{\mathrm{V}}\gtP_{\mathrm{A}}\)
ⓓ. \(X:P_{\mathrm{V}}\gtP_{\mathrm{A}};\ Y:P_{\mathrm{V}}\gtP_{\mathrm{artery}}\)
Correct Answer: \(X:P_{\mathrm{V}}\gtP_{\mathrm{A}};\ Y:P_{\mathrm{V}}\gtP_{\mathrm{artery}}\)
Explanation: Atrioventricular valves lie between atria and ventricles. When ventricular contraction begins, \(P_{\mathrm{V}}\) rises above \(P_{\mathrm{A}}\), reversing the filling gradient and pushing the tricuspid and mitral valves closed. Ventricular pressure continues increasing while the semilunar valves remain closed. Ejection begins only after \(P_{\mathrm{V}}\) exceeds pressure in the pulmonary artery or aorta, represented by \(P_{\mathrm{artery}}\). The corresponding semilunar valve then opens. The two thresholds occur in sequence:
\[
P_{\mathrm{V}}\gtP_{\mathrm{A}}\quad\text{before}\quad P_{\mathrm{V}}\gtP_{\mathrm{artery}}
\]
This ordering creates a brief interval with both valve groups closed and prevents backflow into the atria before arterial ejection starts. The required identities are the ventricular-over-atrial relation for \(X\) and the ventricular-over-arterial relation for \(Y\). The pressure symbols describe relative values at the same moment, not permanent pressure rankings throughout the entire cardiac cycle.
