Human Physiology — Diagnostic Tests

Unit Tests

UT-1: Action Potential and Nerve Impulse Propagation

Question: A neuron has a resting membrane potential of $-70\ \text{mV}$ and the threshold potential is $-55\ \text{mV}$. The equilibrium potential for $\text{Na}^+$ is $+40\ \text{mV}$ and for $\text{K}^+$ is $-90\ \text{mV}$. Explain the ionic basis of: (a) the resting potential, (b) depolarisation, (c) repolarisation, and (d) the refractory period. Calculate the Nernst potential for $\text{K}^+$ given $[\text{K}^+]_{\text{in}} = 150\ \text{mmol dm}^{-3}$ and $[\text{K}^+]_{\text{out}} = 5\ \text{mmol dm}^{-3}$ at $37\ ^\circ\text{C}$.

Solution:

(a) Resting potential: The resting potential of $-70\ \text{mV}$ is close to the $\text{K}^+$ equilibrium potential ($-90\ \text{mV}$) because the membrane is predominantly permeable to $\text{K}^+$ at rest (via leak channels). The $\text{Na}^+/\text{K}^+$ pump maintains concentration gradients ($\text{Na}^+$ high outside, $\text{K}^+$ high inside). Some $\text{Na}^+$ leak inward slightly depolarises the membrane from the pure $\text{K}^+$ equilibrium potential.

(b) Depolarisation: When a stimulus reaches threshold ($-55\ \text{mV}$), voltage-gated $\text{Na}^+$ channels open. $\text{Na}^+$ rushes inward (down its electrochemical gradient), making the inside more positive. Positive feedback: depolarisation opens more $\text{Na}^+$ channels. The membrane potential approaches but does not quite reach the $\text{Na}^+$ equilibrium potential ($+40\ \text{mV}$), peaking at approximately $+30\ \text{mV}$.

(c) Repolarisation: The $\text{Na}^+$ channels inactivate (cannot reopen immediately) and voltage-gated $\text{K}^+$ channels open (with a slight delay). $\text{K}^+$ rushes outward (down its electrochemical gradient), making the inside negative again. The membrane briefly becomes more negative than the resting potential (hyperpolarisation to about $-80\ \text{mV}$) before the delayed $\text{K}^+$ channels close and the membrane returns to resting potential.

(d) Refractory period: During the absolute refractory period ($\sim 1\ \text{ms}$), the $\text{Na}^+$ channels are inactivated and cannot reopen regardless of stimulus strength. During the relative refractory period ($\sim 2$--$4\ \text{ms}$), some $\text{Na}^+$ channels have recovered but the membrane is hyperpolarised ( requiring a larger-than-normal stimulus to reach threshold. This ensures unidirectional propagation of the action potential.

Nernst calculation:

$E_{\text{K}} = \frac{RT}{zF}\ln\frac{[\text{K}^+]_{\text{out}}}{[\text{K}^+]_{\text{in}}} = \frac{(8.314)(310)}{(1)(96485)}\ln\frac{5}{150}$

$= 0.02669\ln(0.0333) = 0.02669 \times (-3.401) = -0.0908\ \text{V} = -90.8\ \text{mV}$

This matches the given equilibrium potential of $-90\ \text{mV}$, confirming the calculation.

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UT-2: Synaptic Transmission

Question: Describe the sequence of events at a cholinergic synapse from the arrival of an action potential at the presynaptic membrane to the generation of a postsynaptic potential. Explain how acetylcholinesterase contributes to synaptic function and why its inhibition is lethal.

Solution:

1. Action potential arrives at the presynaptic terminal, depolarising the membrane.
2. Voltage-gated $\text{Ca}^{2+}$ channels open. $\text{Ca}^{2+}$ flows inward down its electrochemical gradient.
3. Increased intracellular $\text{Ca}^{2+}$ causes synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane (exocytosis).
4. ACh is released into the synaptic cleft and diffuses across to the postsynaptic membrane.
5. ACh binds to nicotinic acetylcholine receptors (ligand-gated $\text{Na}^+$ channels) on the postsynaptic membrane.
6. The receptor channels open, allowing $\text{Na}^+$ to flow inward, causing depolarisation of the postsynaptic membrane (excitatory postsynaptic potential, EPSP).
7. Acetylcholinesterase (AChE) in the synaptic cleft rapidly hydrolyses ACh to choline and acetate, terminating the signal.
8. Choline is reuptaken by the presynaptic terminal via active transport and used to resynthesise ACh (with acetate from acetyl-CoA, catalysed by choline acetyltransferase).

Role of acetylcholinesterase: AChE ensures that the postsynaptic response is brief and precisely timed. Without AChE, ACh would remain in the synaptic cleft, continuously stimulating the postsynaptic receptors. This would cause sustained depolarisation (depolarisation block), preventing further action potential generation. The synapse becomes non-functional.

Inhibition is lethal because: (1) sustained muscle stimulation leads to continuous contraction (tetany), including the diaphragm -- the individual cannot breathe (respiratory failure); (2) excessive parasympathetic stimulation causes bradycardia, excessive salivation, bronchoconstriction, and miosis. Organophosphate nerve agents (e.g., sarin) and certain pesticides (e.g., malathion) are lethal precisely because they irreversibly inhibit AChE.

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UT-3: Hormone Action vs Nerve Impulse

Question: Compare and contrast the endocrine and nervous systems with respect to: (a) speed of response, (b) duration of effect, (c) specificity of target, (d) chemical nature of signals. Use adrenaline (epinephrine) and the fight-or-flight response to illustrate how both systems can work together.

Solution:

Feature	Nervous System	Endocrine System
Speed	Milliseconds (electrical impulses)	Seconds to hours (hormone transport in blood)
Duration	Very brief (ms to s)	Prolonged (minutes to weeks)
Specificity	Very specific (precise neural pathways to specific targets)	Moderately specific (hormone reaches all cells, only target cells with receptors respond)
Signal	Electrical impulses and neurotransmitters (at synapses)	Hormones (chemical messengers in blood)

Fight-or-flight response -- combined action:

When a threat is perceived:

1. Nervous system (immediate): The hypothalamus activates the sympathetic nervous system. Preganglionic neurons release ACh; postganglionic neurons release noradrenaline directly onto target organs. Effects within milliseconds: increased heart rate, bronchodilation, pupil dilation, redirection of blood to muscles.

2. Endocrine system (sustained): The sympathetic nervous system also stimulates the adrenal medulla to release adrenaline (and noradrenaline) directly into the bloodstream. Adrenaline reaches target organs within seconds and produces effects lasting several minutes: glycogenolysis in the liver (raising blood glucose), lipolysis in adipose tissue, further increase in heart rate and cardiac output, and inhibition of non-essential functions (digestion, immune response).

3. Hypothalamic-pituitary-adrenal (HPA) axis (long-term): The hypothalamus releases CRH, stimulating the anterior pituitary to release ACTH, which stimulates the adrenal cortex to release cortisol. Cortisol maintains elevated blood glucose over hours through gluconeogenesis and suppresses the immune system.

This illustrates the principle of dual control: the nervous system provides the immediate, rapid response while the endocrine system sustains and amplifies it over a longer timescale.

Integration Tests

IT-1: Kidney Function and Homeostasis (with Cell Biology)

Question: The kidneys maintain blood pH between 7.35 and 7.45. Explain the roles of: (a) the bicarbonate buffer system in the blood, (b) the proximal convoluted tubule in bicarbonate reabsorption, and (c) the intercalated cells in the distal tubule and collecting duct. A patient has a blood pH of 7.25. Calculate the ratio of $[\text{HCO}_3^-]/[\text{H}_2\text{CO}_3]$ if $\text{p}K_a = 6.10$ at body temperature. Is this acidosis or alkalosis?

Solution:

(a) Bicarbonate buffer system: $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$. The weak acid ($\text{H}_2\text{CO}_3$) and its conjugate base ($\text{HCO}_3^-$) buffer against pH changes. Since $\text{p}K_a = 6.10$ and blood pH $\approx 7.40$, the ratio $[\text{HCO}_3^-]/[\text{H}_2\text{CO}_3]$ is approximately 20:1, meaning the buffer is operating at a point where there is far more conjugate base than acid -- it is most effective at resisting acidification.

(b) Proximal convoluted tubule (PCT): The PCT reabsorbs approximately 85% of filtered bicarbonate. The mechanism: (1) $\text{Na}^+/\text{H}^+$ antiporter on the apical membrane secretes $\text{H}^+$ into the filtrate; (2) the secreted $\text{H}^+$ combines with filtered $\text{HCO}_3^-$ to form $\text{H}_2\text{CO}_3$; (3) carbonic anhydrase on the brush border converts $\text{H}_2\text{CO}_3$ to $\text{CO}_2 + \text{H}_2\text{O}$; (4) $\text{CO}_2$ diffuses back into the cell; (5) intracellular carbonic anhydrase converts $\text{CO}_2 + \text{H}_2\text{O}$ back to $\text{H}_2\text{CO}_3$, which dissociates to $\text{H}^+$ (recycled to step 1) and $\text{HCO}_3^-$; (6) $\text{HCO}_3^-$ exits across the basolateral membrane via the $\text{Na}^+/\text{HCO}_3^-$ cotransporter.

(c) Intercalated cells: Type A intercalated cells secrete $\text{H}^+$ (via $\text{H}^+$-ATPase) and reabsorb $\text{HCO}_3^-$ -- they respond to acidosis. Type B intercalated cells secrete $\text{HCO}_3^-$ and reabsorb $\text{H}^+$ -- they respond to alkalosis. This fine-tunes blood pH beyond what the PCT can achieve.

Calculation:

Using Henderson-Hasselbalch: $\text{pH} = \text{p}K_a + \log\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}$

$7.25 = 6.10 + \log\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}$

$\log\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 1.15$

$\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{1.15} = 14.1$

Normal ratio is approximately 20:1. The reduced ratio (14.1:1) confirms this is acidosis (blood pH below 7.35).

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IT-2: Immune System and Molecular Recognition (with Molecular Biology)

Question: Explain the process by which a B cell produces antibodies specific to a particular antigen, including the roles of: (a) antigen-presenting cells (APCs), (b) helper T cells, (c) plasma cells, and (d) memory cells. Explain how monoclonal antibodies are produced using the hybridoma technique.

Solution:

(a) APCs (dendritic cells, macrophages): When a pathogen enters the body, APCs engulf it by phagocytosis, process the antigens (proteolytically digest pathogen proteins into short peptide fragments), and present these peptide antigens on MHC class II molecules on their cell surface. The APC migrates to the nearest lymph node.

(b) Helper T cells ($\text{CD4}^+$): In the lymph node, the APC presents the antigen-MHC II complex to a helper T cell with a complementary T-cell receptor (TCR). The TCR binding triggers activation of the helper T cell, which requires a second signal (co-stimulation via CD28-B7 interaction). Once activated, the helper T cell proliferates (clonal expansion) and differentiates, then releases cytokines (interleukins, e.g., IL-4, IL-5, IL-6) that stimulate B cells.

(c) Plasma cells: A B cell whose surface immunoglobulin (IgM/IgD) binds the same specific antigen internalises it, processes it, and presents it on MHC II. The activated helper T cell provides the necessary cytokine signals for full B cell activation. The B cell undergoes clonal expansion and differentiates into plasma cells, which secrete large quantities of specific antibodies (up to 2000 molecules per second). These antibodies circulate in the blood and lymph, binding to the antigen and marking pathogens for destruction (opsonisation, complement activation, neutralisation).

(d) Memory cells: Some activated B cells differentiate into memory B cells instead of plasma cells. These persist for years or decades with the same surface antibody specificity. Upon re-exposure to the same antigen, memory B cells mount a faster, stronger secondary immune response (higher antibody titre, faster response, predominantly IgG class switching).

Hybridoma technique (Kohler and Milstein, 1975):

1. Immunise a mouse with the target antigen to stimulate B cell production.
2. Extract B cells from the mouse spleen (producing the desired antibody).
3. Fuse these B cells with myeloma (cancer) cells using polyethylene glycol (PEG). The myeloma cells provide immortality.
4. Select hybridomas (fused cells) using HAT medium: unfused myeloma cells die (they lack HGPRT enzyme), unfused B cells die naturally (limited lifespan). Only hybridomas survive.
5. Screen individual hybridoma clones for antibody production specific to the target antigen.
6. Culture the selected hybridoma indefinitely to produce unlimited quantities of identical (monoclonal) antibodies.

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IT-3: Cardiovascular Function and Gas Exchange (with Ecology)

Question: At sea level, atmospheric pressure is $101.3\ \text{kPa}$ and the partial pressure of $\text{O}_2$ in dry air is $21.2\ \text{kPa}$. Calculate the partial pressure of $\text{O}_2$ in: (a) the trachea (accounting for humidification), (b) the alveoli (accounting for $\text{CO}_2$ and water vapour), and (c) the pulmonary artery (mixed venous blood with $\text{O}_2$ saturation of 75%). At an altitude of $5000\ \text{m}$, atmospheric pressure drops to $54.0\ \text{kPa}$. Calculate the alveolar $\text{P}_{\text{O}_2}$ and explain why this causes altitude sickness.

Solution:

(a) Trachea: Air is humidified in the upper respiratory tract. Water vapour pressure at body temperature ($37\ ^\circ\text{C}$) $= 6.3\ \text{kPa}$.

$P_{\text{O}_2}(\text{trachea}) = (101.3 - 6.3) \times 0.21 = 95.0 \times 0.21 = 20.0\ \text{kPa}$

(b) Alveoli: Alveolar gas equation: $P_{\text{O}_2}(\text{alv}) = P_{\text{O}_2}(\text{inspired}) - \frac{P_{\text{CO}_2}(\text{alv})}{R}$

$P_{\text{O}_2}(\text{inspired}) = (101.3 - 6.3) \times 0.21 = 20.0\ \text{kPa}$

With $P_{\text{CO}_2}(\text{alv}) = 5.3\ \text{kPa}$ and respiratory quotient $R = 0.85$:

$P_{\text{O}_2}(\text{alv}) = 20.0 - \frac{5.3}{0.85} = 20.0 - 6.2 = 13.8\ \text{kPa}$

(c) Pulmonary artery (mixed venous): With 75% saturation, the $P_{\text{O}_2}$ is read from the oxyhaemoglobin dissociation curve. At 75% saturation, $P_{\text{O}_2} \approx 5.3\ \text{kPa}$ (about $40\ \text{mmHg}$).

At $5000\ \text{m}$:

$P_{\text{O}_2}(\text{inspired}) = (54.0 - 6.3) \times 0.21 = 47.7 \times 0.21 = 10.0\ \text{kPa}$

$P_{\text{O}_2}(\text{alv}) = 10.0 - \frac{5.3}{0.85} = 10.0 - 6.2 = 3.8\ \text{kPa}$

At sea level, alveolar $P_{\text{O}_2} = 13.8\ \text{kPa}$; at $5000\ \text{m}$, it drops to $3.8\ \text{kPa}$. This severely reduced $P_{\text{O}_2}$ places the blood on the steep portion of the oxyhaemoglobin dissociation curve, causing a dramatic drop in haemoglobin saturation (from $\sim 98\%$ to $\sim 70\%$). This hypoxia causes altitude sickness symptoms: headache, nausea, fatigue, and in severe cases, pulmonary or cerebral oedema. The body partially compensates through hyperventilation (reducing $P_{\text{CO}_2}$), increased heart rate, and eventually increased red blood cell production (polycythaemia).