Physiology

Question 1

Define lung compliance

Answer 1

The volume displaced within the lung per unit pressure change i.e. the gradient of the pressure-volume curve

Specific lung compliance is the same, but indexed to total lung volume

Compliance is higher in expiration

Question 2

What forces are balanced at the end of expiration?

Answer 2

The elastic recoil of the lungs (causes passive expiration) and the elastic recoil of the chest wall (opposes the lung’s collapse)

Question 3

Give factors that increase and decrease lung compliance

Answer 3

Compliance is increased with age, surfactant, and emphysema

Compliance is decreased by high pulmonary venous pressures, fibrotic lung disease, and alveolar oedema. Compliance is also lower at extremes of lung volumes.

Question 4

What is a Reynold’s number?

Answer 4

The ratio between inertial and viscous forces, and a number which can therefore be used to predict the switch from laminar to turbulent flow

At low rates of flow, viscous forces predominate i.e. a gas has an inherent resistance to deformation at a given rate. At higher rates of flow this inherent resistance is overcome, so inertial forces predominate, which leads to more complex fluid dynamics with eddies etc. which is turbulent flow.

A Reynolds number describes the balance of these forces within a given liquid or gas, and so can be used to predict the change in flows.

Question 5

In which airways is turbulent flow more likely?

Answer 5

Larger airways, or those with bifurcations/ changes in diameter.

Larger airways have higher flow (Hagen-Pouiseulle equation) so inertial forces will trump viscous ones leading to turbulence (i.e. the Reynolds number will be higher)

Question 6

What effect would xenon have on flow through an airway?

Answer 6

More laminar flow

Xenon has a very low density, and density is proportional to the Reynolds number. Since turbulent flow begins at a threshold Reynolds number, the lower the density of gas, the more laminar flow.

Question 7

When is density relevant to gas flow rates through airways?

Answer 7

In turbulent flow

In laminar flow, the Hagen-Poiseulle equation governs, and it is not affected by density but viscosity

Question 8

Explain closing capacity

Answer 8

Closing capacity is the maximum volume at which collapse of small airways begins in the dependent parts of the lung. When lung is inflated, the stretch keeps small airways open which accounts for the improvement in compliance. Loss of elasticity of lung tissue will increase the residual capacity at which this small airway collapse starts. The dependent area of the lung is affected first because the effect of gravity on the lung above opposes the elastic forces. It is important because it leads to poor denitrogenation for anaesthesia and atelectasis of trapped gas. The atelectasis leads to low V/Q and accounts for a lot of the hypoxia during anaesthesia and associated with age.

The factors affecting it are:
- Increased expiratory effort or pressure. The pressure is transmitted to the small airways and acts to collapse them
- Small airways disease e.g. asthma/COPD which makes them narrowed with muscle or mucous so they close more easily
- Parenchymal lung disease as it reduces elasticity
- Surfactant deficiency as it promotes alveolar collapse which robs neighbouring small airways of the elastic force.
- Age, also reduces elasticity

Question 9

Define:

Dead space

Physiological dead space

Answer 9

Dead space is the volume of inspired air that does not take part in gas exchange (~30%)

Dead space consists of apparatus, anatomical, and alveolar

Physiological dead space discounts the apparatus

Question 10

How does Fowler’s method work?

Answer 10

A vital capacity breath of pure O2 is taken, after which the forced expiration is analysed for nitrogen. The volume exhaled between the start, and halfway up the slope of nitrogen detection is the anatomical dead space, as there was nitrogen present in the lung but pure oxygen in the anatomical dead space.

Question 11

Give an overview of perfusion vs. ventilation in a healthy lung

Answer 11

Zone 1/ Apex: Doesnt exist in physiological states but arises during IPPV or hypotension. Higher resting volume due to less compression secondary to gravity, which means it doesn’t ventilate that well as it can’t expand that much more (lower compliance). Comparatively poorly perfused due to gravity, in cases of IPPV or hypovolaemia, intrapleural pressure may exceed capillary pressure and splint them shut.
PA>Pa>Pv

Zone 2/ Mid: well balanced, capillaries aren’t splinted shut, venous pressure isn’t high enough to affect perfusion, and ventilation is okay. Perfusion depends on the arterial-alveolar gradient.
Pa>PA>Pv

Zone 3/ Base: lower resting volume due to compression from the above lung/ surrounding tissue, so ventilation is better because it has further capacity to expand than the apex (higher compliance). Higher capillary pressures because of the advantage of gravity, so perfusion is good, and even outstrips ventilation. Venous pressure may exceed alveolar pressure. Perfusion depends on the arterio-venous gradient.
Pa>Pv>PA

Question 12

Summarise body plethysmography, helium dilution, and nitrogen washout

Answer 12

Body plethysmography uses an airtight box and measures change in pressure within the box to extrapolate volumes during respiration

Nitrogen washout measures FRC - the subject breathes pure oxygen on a closed circuit and de-nitrogenates their lungs. The total amount of nitrogen washed out is then measured which gives the FRC, assuming the FRC was full of atmospheric air at the start (79% nitrogen)

Question 13

What is the most influential driver of ventilation?

Answer 13

PaCO2 stimulating Central chemoceptors by making CSF more acidic (CO2 + H2O -> HCO3 + H), this accounts for 80% of response to high CO2, though is slow at 1-3 minutes; peripheral chemoceptors generate a faster response. CSF has less protein than blood which gives it less buffering capacity so pH changes are greater for a given PaCO2 than in blood.

It’s unknown how the chemoceptors detect pH changes

Question 14

Why is the central response to hypercarbia blunted over time?

Answer 14

increased bicarbonate in CSF, either through active transport or passive diffusion - it’s unclear

Question 15

What is the difference in sensitivity between the carotid and aortic bodies?

Answer 15

Carotid bodies respond to hypoxia, hypercarbia, and pH. Aortic bodies do not respond to pH.

Aortic bodies have a lesser blood supply and so rely more on haemoglobin for their oxygen delivery i.e. are sensitive to the total oxygen content of the blood.

Conversely the carotid bodies have such a good blood supply they can be oxygenated purely by oxygen dissolved in solution and so will respond mostly just to PaO2.

Question 16

Describe the role of peripheral chemoceptors in responding to changes in gas tensions

Answer 16

Hypoxia is sensed breath to breath by the carotid bodies, which are very sensitive and have a massive blood flow through them. This response is increased by hypercarbia and acidosis, and they generate much faster responses than central chemoceptors.

Question 17

Describe the early features that compensate for cellular hypoxia

Answer 17

Local:
- pH drops due to anaerobic metabolism which shifts the Hb curve to the right allowing O2 to unload into the tissue more easily. Metabolites cause local vasodilation and improve perfusion

Ventilatory:
- Hypoxia and hypercarbia at peripheral chemoceptors stimulates an increase in minute ventilation

Cardiovascular:
- Hypoxia and hypercarbia at peripheral chemoceptors leads to vasoconstriction and tachycardia to improve perfusion

Question 18

Which of the following is true of adrenoceptors?

A) Voltage gated potassium channels trigger noradrenaline release
B) Amitryptilline activates NET
C) Most of the noradrenaline released is metabolised in the synapse by MAO
D) Cocaine increases synaptic noradrenaline concentration through MAO inhibition
E) Ephedrine works by exchanging for noradrenaline

Answer 18

E) Ephedrine works by exchanging for noradrenaline

Ephedrine has a modest direct agonism of post-synpatic α1 receptors, but mostly works as it is structurally similar enough to be transported across the pre-synpatic membrane by NET, and into noradrenaline vesicles by VMAT. This displaces, and exchanges with, noradrenaline which is pushed into the synapse. Because of this, ephedrine’s sfficacy is reduced in patients with depleted catecholamine reserves e.g. sepsis, or in patient’s taking inhibitors of NET (e.g. TCAs, SNRI).

For the other options:
- Noradrenaline release is triggered by voltage-gated calcium channel release
- Amitryptilline inhibits NET, and so reduces ephedrine efficacy as it can’t cross the pre-synaptic membrane
- Most of the noradrenaline release into the synapse is recycled into pre-synaptic vesicles via NET and VMAT, with only 25% being metabolised by MAO (and this takes place pre-synaptically rather than in the synapse). Outside of noradrenergic nerve terminals, noradrenaline may akso be metabolised to normetanephrine by COMT.
- Cocaine inhibits NET and EMT to prevent catecholamine reuptake

Question 19

Which two NANC (non-adrenergic, non-cholinergic) substances augment the action of noradrenaline?

Answer 19

ATP and neuropeptide Y

Question 20

Which of the following is true of α adrenoceptors?

A) In the gut, α receptors stimulate voltage gated calcium channel opening
B) Clonidine acts exclusively on α2 receptors
C) Activation has minimal effect on the vasculature of the heart and brain
D) Noradrenaline does not stimulate α2 receptors
E) The majority of the body’s α2 receptors are found pre-synaptically

Answer 20

C) Activation has minimal effect on the vasculature of the heart and brain

Clonidine exerts most of its action at α2 receptors, but also activates α1 and so may cause transient hypertension. Dexmedetomidine is 10 times more selective selective α2 than clonidine.

α1 receptors have an unusual activity in the gut - they do not activate calcium channels but instead allow potassium to cross the cell membrane. This hyperpolarises the membrane and inhibits action potentials that would cause contraction of gut smooth muscle.

Noradrenaline does stimulate α2 receptors - this is the natural negative feedback mechanism. The majority of α2 receptors are found outside the synapse.

Question 21

Which of the following is false of β adrenoceptors?

A) β2 receptors act to relax smooth muscle
B) Noradrenaline has a lesser affinity for β than α receptors
C) They stimulate NO-mediated splanchnic vasodilation
D) β1 receptors stimulate glucose and free fatty acid production from stores
E) β2 receptors are more important in sick hearts

Answer 21

B) Noradrenaline has a lesser affinity for β than α receptors

Contrary to popular belief, adrenaline is more potent at α and β2, and noradrenaline is more potent at β1, with a much lower affinity at β2. Both adrenaline and noradrenaline have greater affinity for β than α receptors, even though noradrenaline primarily acts as a vasoconstrictor and adrenaline as an inotrope. This is mostly explained by adrenaline’s significantly greater action at β2 which acts to reduce SVR and offsets its α1 activity. Furthermore, β2 receptors are particularly concentrated in the atria and SA node of the heart, hence stimulation of β2 causes more chonotropy.

β2 receptors account for only 20% of beta adrenoceptors in healthy hearts, but in heart failure there is de-coupling and de-population of β1 receptors such that β2 receptors make up ~50% of the beta adrenoceptor complement.

Question 22

What is the generic structure of a catecholamine?

Answer 22

A catechol ring (benzene ring with two hydroxyl groups) with a terminal amine group

Question 23

Which of the following is true regarding catecholamine production?

A) Dopamine is synthesised from noradrenaline
B) Tyrosine is made from phenylalanine in the adrenal medulla
C) Phenylalanine hydroxylase deficiency leads to profound deficiency of tyrosine
D) Dopamine exerts positive feedback on tyrosine hydroxylase
E) Tyrosine hydroxylase catalyses the rate-limiting step of catecholamine production

Answer 23

E) Tyrosine hydroxylase catalyses the rate-limiting step of catecholamine production

Tyrosine is the starting point for catecholamine production and can be acquired in the diet, or via conversion of phenylalanine in the liver, after which it is concentrated in chromaffin cells in the adrenaline medulla. Tyrosine undergoes hydroxylation to L-dopa, and then decarboxylation to dopamine, which is converted to noradrenaline (by dopamine-B hydroxylase), which is converted to adrenaline (by PNMT). Dopamine and noradrenaline exert negative feedback to tyrosine hydroxylase.

Question 24

Which of the following would cause a cell to depolarise?

A) Sodium channel opening
B) Potassium channel opening
C) Increased Na/K/ATPase pump activity
D) Increased frequency of chloride channel opening
E) Decreased intracellular cAMP

Answer 24

A) Sodium channel opening

Question 25

Which ion channel acts to prolong depolarisation in cardiac myocytes? A) Magnesium channels B) Chloride channels C) Sodium channels D) Potassium channels E) L-type calcium channels

Answer 25

E) L-type calcium channels After initial depolarisation (mediated by sodium and T-type calcium channel opening), potassium channels open as they would in a standard neuron to repolarise the membrane through potassium efflux. However the opening of L-type calcium channels and subsequent calcium influx prevents repolarisation of the membrane - this is the absolute refractory period. This is one of the heart's mechanisms to ensure the whole ventricle is depolarised together. As a result the cardiac myocyte action potential is 200-300 times longer than a standard neuron's. These cells also display prolonged refractory periods mediated by ion channel inactivation, to prevent tetanic contraction that would prevent ventricular filling.

Question 26

Rank these from lowest to highest trough membrane potential (i.e. most negative to least negative) Cardiac myocyte Cardiac pacemaker cell Standard neuron

Answer 26

Cardiac myocyte: -85mV Standard neuron: -70mV Cardiac pacemaker cell: -60mV

Question 27

Which of the following is false of cardiac pacemaker cells? A) Main depolarisation is not mediated by sodium B) They have no resting phase/ membrane potential C) Tachycardia is mediated by reducing membrane permeability to sodium D) There is continuous membrane permeability to sodium E) They don't display the same depolarisation plateau as myocytes

Answer 27

C) Tachycardia is mediated by reducing membrane permeability to sodium Cardiac pacemaker cells are found in the SA node in the RA, and throughout the heart's conducting system. Their membrane potential decays spontaneously towards a threshold potential of -40mV due to continuous permeability of their membrane to sodium, and progressive impermeability to potassium. Firing rate is modified by increasing membrane permeability to sodium (tachycardia) or potassium (bradycardia) during the pre-potential (phase 4). Once the threshold potential is met, L-type calcium channels open and are the main drivers of depolarisation of the cell. The membrane is repolarised by potassium efflux, but there is no resting phase in the cycle. T-type calcium channels also play a role in decay towards threshold potential alongside sodium channels.

Question 28

Describe the three main factors that govern resting membrane potential

Answer 28

Resting membrane potential (i.e. the potential difference in electrical charge across the membrane of a quiescent cell) is governed by: - Activity of the Na/K/ATPase pump which pumps 3 sodium out and 2 potassium in, and is responsible for 70% of energy consumption in the average neuron - Differential permeability of the membrane to sodium versus potassium. At rest, the cell membrane is 100 times more permeable to potassium than sodium. - The Donnan effect: charged particles that cannot cross the membrane such as proteins (like albumin) and phosphate that therefore keep the intracellular charge more negative irrespective of the above.

Question 29

Summarise CICR

Answer 29

Action potentials move along t-tubules (deep invaginations in the cell membrane) causing opening of voltage-gated L-type calcium channels. The calcium moves intracellularly and activates ryanodine receptors on the sarcoplasmic reticulum, causing calcium efflux into the cell. The influx of calcium through L-type channels is modulated by adrenaline - this is the mechanism of positive inotropy.

Question 30

Describe the action of surfactant

Answer 30

Surfactant is a phospholipid that acts to interrupt surface tension. Surface tension is a result of asymmetry of the normal attractive forces between fluid molecules because of an interface with gas. The alveolus is lined interiorly with fluid, and the net effect of surface tension here will be to crumple and collapse the alveolus (P=2T/R). This also increases the pressure required to increase lung volume i.e. compliance, which has been illustrated in experiments with saline-filled lungs having higher compliance. Surfactant interrupts surface tension and so prevents alveolar collapse. It becomes more effective at lower lung volumes as the hydrophobic tails come closer together, increasing hydrostatic repulsion. Hence surfactant makes the biggest difference to compliance at low lung volumes.

Question 31

Which of the following is false of surfactant? A) It is produced in type 2 pneumocytes from free fatty acids B) It has a hydrophilic head and hydrophobic tails C) It opposes Laplace's law D) It increases alveolar fluid transudation E) It has the greatest effect at low lung volumes

Answer 31

D) It increases alveolar fluid transudation Surfactant forms a hydrophobic layer on the inside of alveoli, and so reduces fluid moving into the alveoli.

Question 32

Which of the following is false? A) Airway resistance is overcome passively in expiration using stored elastic energy B) Lower RR and higher Vt minimise work of breathing in obstructive defects C) Area under the hysteresis curve equates to compliance D) The integral of the hysteresis curve equates to work done E) The 2 components of work of breathing are elastic and friction

Answer 32

C) Area under the hysteresis curve equates to compliance It equates to work done. The two components of work of breathing are elastic and resistive. Elastic forces are applied by the lung parenchyma as it is stretched, and so change with volume. Resistive forces are airway resistance and friction, and so are more constant with changed in lung volume, though the total work done will increase with RR.

Question 33

Which of the following regarding V/Q is false? A) Atelectasis causes shunt B) Increased dead space will increase EtCO2 C) A shunt will cause equilibration in gas composition between alveolus and capillary D) A normal 45 year-old lying supine will develop a basal shunt E) Where V/Q is ♾, PAO2 will tend towards PiO2

Answer 33

B) Increased dead space will increase EtCO2 The best example of dead space is cardiac arrest - none of the lung is perfused so there is almost no EtCO2. A normal 45 year old lying supine will generally have a closing capacity that impinges on their FRC, as FRC decreases by ~15% in the supine position.

Question 34

Give the four classifications of hypoxia

Answer 34

Hypoxic Anaemic Circulatory Cytotoxic

Question 35

Describe how a shunt develops as a consequence of general anaesthetic

Answer 35

FRC is reduced under GA due to alveolar collapse, this is due to loss of intercostal muscle tone and supine/ head down positioning). Additionally the inhaled gas mix contains more oxygen and less nitrogen than air, which promotes atelectasis as oxygen is more soluble than nitrogen. Reduction in FRC shifts the lung down the compliance curve - the volumes are now lower which shifts the apices onto a steeper segment i.e. better compliance, and the bases onto a shallower segment i.e. worse compliance. The bases are still better perfused, but now have worse compliance and so less ventilation - this is a shunt.

Question 36

How does hypoventilation cause hypoxaemia?

Answer 36

Not generally through a lack of oxygen. Hypoventilation causes hypercarbia, so the PaCO2 rises which reduces the PaO2 (alveolar gas equation). Accordingly supplemental oxygen can compensate for hypoxia due to hypoventilation by increasing the PaO2.

Question 37

Explain why the hysteresis loop exists

Answer 37

Some energy is dissipated during inspiration without being recovered in expiration, therefore higher pressure (i.e. more work done) is required to inflate the lung and account for inefficiency.

Question 38

Which of the following is true regarding CO2 transport in blood? A) Most CO2 is converted to bicarbonate by plasma carbonic anhydrase B) Carbonic anhydrase is relatively slow-acting C) Carbamino carriage in plasma accounts for 1/3 of the AV CO2 difference D) RBCs actively co-transport HCO3- into the cell with Cl- E) Buffering power of haemoglobin is mostly a result of histidine AA content

Answer 38

E) Buffering power of haemoglobin is mostly a result of histidine AA content The imidazole group of the amino acid histidine makes it the only AA that can be an effective buffer at physiological pH.

Question 39

Explain the two parts of the Haldane Effect

Answer 39

The Haldane Effect refers to the increased capacity of haemoglobin to transport CO2 when deoxygenated First and most significant: - deoxygenation induces a conformational change (allosteric modulation) which increases CO2 affinity for binding sites on haemoglobin. The total amount of CO2 carriage this way is much smaller than via bicarbonate, but this accounts for 1/3 of the total A/V difference in CO2 carriage. Secondly, deoxygenation of haemoglobin increases its buffering capacity - The dissociation of carbamic acid into carbamate ions within histidine AAs in haemoglobin is heavily influenced by whether the haemoglobin is oxygenated or not. Deoxygenated haemoglobin causes imidazole groups in histidine to become alkaline, so they buffer protons. This tips the equilibrium of the bicarbonate reaction in the direction of forming more bicarbonate, effectively storing more CO2.

Question 40

How do RBCs maintain favourable gradients for bicarbonate production?

Answer 40

1) Protons are buffered by histidine in haemoglobin (especially when deoxygenated) 2) Bicarbonate is actively transported from the cell by a membrane protein 'Band 3' which 'ping-pongs' chloride ions in for bicarbonate ions out. This is called the Hamburger shift Interestingly Band 3 is also an important structural protein that articulates closely with the actin cytoskeleton - a defect in this protein causes hereditary spherocytosis.

Question 41

Give the main causes of MR and AS

Answer 41

MR - degeneration - inflammation e.g. rheumatic heart disease - infection - connective tissue disease - ischaemia AS - bicuspid valve - calcification - inflammation e.g. rheumatic heart disease

Question 42

Describe the effects of AS on cardiac function

Answer 42

Diastolic: - Reduced relaxation of the hypertrophied ventricle so higher end diastolic LV pressure - This reduces compliance and passive filling of the ventricle, making the atrial kick more important - This pressure is transmitted to the coronaries, reducing the perfusion gradient and therefore blood supply to the heart muscle - This in combination with a hypertrophic ventricle that has a higher oxygen requirement leaves the heart vulnerable to ischaemia Systolic: - The ventricle hypertrophies in response to the increased pressure gradient through the valve - Compensates initially but will decompensate if demand for cardiac output increases e.g. exercise - In severe cases a drop in preload will decompensate the LV and it will be unable to make up for this with contractility.

Question 43

Describe the sequence of cardiovascular events that occur when standing from lying

Answer 43

1) Blood pools in the dependent capacitance veins of the legs due to gravity. This reduces preload which reduces stroke volume, so cardiac output falls. 2) The drop in CO reduces MAP. The drop in pressure is sensed by baroceptors in the heart and carotid and aortic sinuses which decreases their firing rate via afferent fibres (IX nerve for carotid, vagus for heart and aortic) to the pressor centre in the venterolateral medulla. 3) The response is an increase in sympathetic outflow. This causes vasoconstriction to boost BP, venoconstriction to mobilise volume from capacitance vessels and improve preload, and an increase in heart rate and contractility. 4) MAP is restored and the firing rate of baroceptors increases again.

Question 44

Which organs have post-ganglionic sympathetic cholinergic transmission?

Answer 44

Sweat glands, piloeretor muscles, and skeletal muscle blood vessels

Question 45

Describe the response to haemorrhage

Answer 45

Early: - Hypotension sensed by baroceptors, acidaemia due to lactate sensed by peripheral chemoceptors. Both increase sympathetic outflow - Vasoconstriction improves BP. Preferential redirection of arterial flow to core, bypassing mesenteric and renal circulation - Venoconstriction improves preload by mobilising blood from capacitance vessels in skin, lungs, splanchnic circulation, skeletal muscle. - Drop in intravascular hydrostatic pressure creates a gradient for Starling forces (net of hydrostatic and oncotic pressure) which recruits up to 0.25ml/kg/min fluid. - Other mechanisms for volume recruitment and vasoconstriction include activation of RAAS triggered by renal renin secretion in response to poor perfusion, which triggers ATII (vasoconstrictor) and aldosterone (Na and water retention) release. Finally, ADH is released by the neurohypophysis in reaction to reduced baroceptor stimulation +/- increased toxicity. This acts to vasoconstrict and retain water in kidney collecting ducts. Late: - Release of reticulocytes from bone marrow - Increase in plasma protein synthesis in the liver - Increased EPO release from kidneys

Question 46

List factors that would affect CC and FRC

Answer 46

Increase CC: - In neonates, upright 45 year-olds, and supine 65 year-olds, closing capacity is equal to FRC - Increase in intrathoracic pressure e.g. asthma - Smoking Decrease FRC: - Supine or head down positioning - Obesity/ abdominal mass/ abdominal pressure - GA especially with NMB - Restrictive lung disease - Young children - Female sex Increase FRC: - Obstructive lung disease - Increased age - PEEP

Question 47

Which of the following regarding V/Q is true? A) Hypoxia due to low V/Q ratio will respond to oxygen B) PACO2 will be lowest in zone 3 C) PAO2 is the same throughout the lungs D) A shunt will produce hypercarbia E) Hypoxia due to a low V/Q ratio will respond to hyperventilation

Answer 47

A) Hypoxia due to low V/Q ratio will respond to oxygen

Question 48

What are the physiological examples of a true shunt?

Answer 48

Bronchial arteries draining into pulmonary veins, and thebesian veins draining directly into coronary sinus

Question 49

Explain low and high V/Q ratio and their resulting gas tensions

Answer 49

Low/ Shunt: - Alveolar units are poorly ventilated so end-capillary blood tends towards mixed venous blood - Arterial PO2 falls and PCO2 rises, triggering an increase in respiratory stimulation from the CNS - This increase in ventilation can normalise CO2 but not O2 and the reason lies in the oxygen dissociation curve and the oxygen content equation - Hyperventilating well perfused alveolar units that were already well ventilated can increase the gradient for CO2 which will increase diffusion out of the blood. However increasing ventilation will barely increase the oxygen content of this blood as its Hb will already be close to saturation. - The poorly ventilated units can therefore not be compensated for by hyperventilation of normal units, leading to hypoxia without hypercarbia - Note that if a patient in this situation (e.g. asthma as they ventilate poorly but perfuse fine) fatigues the their compensatory hyperventilation will slow and they will become hypercapnic High/ Dead Space: - There are volumes of lung that are being ventilated but no longer take any part in gas exchange - This effectively decreases the alveolar minute ventilation - Alveolar ventilation is inversely proportional to PaCO2 and, as production of CO2 remains constant, this causes PaCO2 to rise.

Question 50

Which of the following is false of control of breathing? A) Central mechanisms govern most of the change in minute ventilation, and respond in 1-3 minutes B) Central chemoceptors are important in the response to hypoxia C) The carotid bodies receive 40 times the blood flow of the brain indexed to mass D) Sustained hypercapnia is compensated for by raising bicarb in CSF E) Hypocapnia significantly blunts the respiratory response to hypoxia

Answer 50

B) Central chemoceptors are important in the response to hypoxia Central chemoceptors in the medulla are sensitive to pH. CO2 diffuses across the BBB and is hydrated, then dissociated to produce protons. This increased acidity stimulates the chemoceptors and increases minute ventilation. The BBB doesn't allow protons to cross from the blood, and CSF has less protein and therefore buffering capacity than blood, which mean the dissolved CO2 has a proportionately larger effect on pH in CSF than in blood. Central chemoceptors are responsible for ~80% of control over minute ventilation, but are slower to react than peripheral chemoceptors in the carotid and aortic bodies which can detect hypoxia. The carotid bodies have an enormous blood flow at 2L/min/100g tissue and this extremely high flow allows them to detect changes in blood very quickly. This also means they can absorb all the oxygen they need from what is dissolved in solution - as a result they respond mainly to changes in PaO2. Conversely the flow to aortic bodies is lower and so they respond more normally i.e. to changes in total O2 content which is more dependent on Hb saturation. This also means the aortic bodies will respond to anaemic hypoxia where the carotid bodies will not. Input regarding PaCO2, PaO2, and pH is integrated to produce a respiratory response. Hypercapnia and hypoxia potentiate their respective responses, but equally hypocapnia will blunt the hypoxic response. If CO2 is very low, the response to hypoxia may not be activated until PaO2 is around 5, though interestingly the response to hypercapnia is better preserved even at high PaO2. The carotid bodies are so important in the response to hypoxia that animals without them die at altitude because they don't respond adequately to hypoxia.

Question 51

Give T/F for the following A) Stored blood loses 2,3 DPG which reduces its oxygenating efficiency B) Most deoxyhaemoglobin has one O2 bound C) 2,3-DPG causes a conformational change in alpha global chains to increase O2 affinity D) Haemoglobin is produced in the mitochondria E) The Bohr effect describes the reduction in haem affinity for O2 with acidosis F) Sickle cell, HbF, and CO poisoning shift the dissociation curve to the right

Answer 51

A - T B - F C - F D - T E - T F - F 2,3-DPG is an organic phosphate produced from one of the products of glycolysis. It binds to the beta chains of deoxyhaemoglobin and induces a conformational change that reduces its affinity for oxygen, thereby improving oxygen unloading. Haemoglobin is produced in the mitochondria, mostly only loses one oxygen molecule (venous sats of 75%), and all of the mentioned factors move the dissociation curve to the left which reduces oxygen unloading in tissues

Question 52

Which of the following is false of calcium homeostasis? A) PTH has a far shorter half life than 1,25-OH2VitD B) PTH acts directly in the gut to increase calcium absorption C) 1,25-hydroxycholecalciferol increases bone resorption D) Calcitonin reduces bone resorption and serum calcium level E) Hypocalcaemia in hypoventilation is caused by increased protein binding

Answer 52

B) PTH acts directly in the gut to increase calcium absorption This effect is exerted by PTH through hydroxylation of 25-hydroxycholecalciferol in the kidney, which produces activated vitamin D which stimulates absorption. PTH has a 4 minute half-life whereas activated Vit D has a half-life of 4 hours. Vitamin D increases serum calcium through gut absorption, kidney reabsorption, and bone de-mineralisation. Calcitonin acts to inhibit osteoclast activity, preventing bone resorption and reducing serum calcium. Hypocalcaemia during hypoventilation is a consequence of alkalosis. The alkalosis increases the protein binding of calcium.

Question 53

What would you expect the venous PCO2 to be in a subject breathing 100% oxygen?

Answer 53

Around 8kPa The explanation essentially centers around oxygen being very insoluble. A PaO2 of 100kPa will not increase the oxygen content much because of how insoluble it is. Thinking of the oxygen content equation (([Hb] x1.34xSO2 as a fraction) + (PaO2 x 0.225) = mlO2/L blood) it's clear that breathing pure oxygen will only add 20ml/L of oxygen in the average adult. Average oxygen extraction is around 25% which would reduce the content to 170ml/L, which would reduce the PaO2 to around 8.

Question 54

Explain the derivation of the Henderson-Hasselbalch equation

Answer 54

We start with the dissociation constant which describes the degree of dissociation for a given compound Ka = [H+]x[A-]/[HA] We then apply log10 which turns Ka into pKa and [H+] into pH. logKa = log[H+] + log[A-]/[HA] Rearranging: -log[H+] = -logKa + log[A-]/[HA] pH = pKa + log[A-]/[HA] For bicarbonate, the carbonic acid can be substituted for PaCO2 on the basis that one molecule of CO2 combines with one of H20 to form once H2CO3. This reaction is in equilibrium in the healthy subject and so PaCO2 should equal [H2CO3].

Question 55

Briefly explain the alveolar gas equation

Answer 55

The alveolar gas equation starts with basic input/output models to produce equations for the volume of CO2 produced and the volume of O2 consumed. There are several assumptions made and the important one here is that the respiratory quotient is 0.8. This allows us to compare the equations for oxygen consumption and CO2 production. This produces an equation that is almost the AGE. There are two final tweaks: PACO2 is substituted for PaCO2 because another assumption is that gas exchange is intact and so PCO2 will equilibrate totally between capillary and alveolus. The other assumption is that Dalton's law is valid, which just means the fractional percentage is equal to the partial pressure.

Question 56

Which of the following is false regarding air flow? A) Adding helium to an inhaled gas mixture would reduce its Reynolds number B) Gas density increases airway resistance in laminar flow C) Airway resistance decreases in inspiration D) Resistance to air flow is greatest in medium sized airways and negligible in small airways E) Adding helium to an oxygen mixture would reduce work of breathing

Answer 56

B) Gas density increases airway resistance in laminar flow It is viscosity rather than density that increases airway resistance in laminar flow. Viscosity is a product of attraction/friction between layers of a fluid in motion. Adding helium to a gas mixture would increase the likelihood of laminar flow as density is proportional to the Reynolds number, a lower number means laminar flow is more likely, and xenon has a low density. Airway resistance decreases in inspiration because the airways are tethered to the kung parenchyma. As the parenchyma expands during inspiration, there is radial traction of the airways increasing their diameter, which reduces the resistance. Despite expectations, experiments have shown that airways resistance is greatest in medium-sized airways and very small in small airways. This is because initially as airways divide and the radius decreases, airway resistance increases, peaking around the 5th generation. However as the number of branches increases exponentially (there are ~50,000 terminal bronchioles) the gas flow dwindles markedly, to the point where there is no flow bar passive diffusion from the 16th division onwards. This very low flow means the resistance is very low also.

Question 57

What is the formula for Reynolds number?

Answer 57

Re = 2 x velocity x density x radius/ viscosity It is essentially a ratio of kinetic energy to viscous damping

Question 58

Describe the benefit of IPPV in acute LV failure

Answer 58

Though unhelpful for the RV, and detrimental to cardiac output through decreasing preload, IPPV is actually helpful for the LV. There are two ways IPPV helps the LV: 1) The transmural pressure is reduced. I.e. in spontaneous respiration the negative intrapleural pressure opposes the contraction of the LV and essentially adds to its afterload by increasing the wall tension required for contraction. Positive pressure does the opposite and squeezes the LV, aiding in ejection. 2) Similarly, compression of the intrathoracic aorta creates a pressure gradient between it and the extrathoracic circulation which aids flow from the LV, reducing afterload. https://derangedphysiology.com/main/cicm-primary-exam/respiratory-system/Chapter-523/effects-positive-pressure-ventilation-cardiovascular-physiology

Question 59

Which of the following is false when comparing turbulent and laminar flow? A) Turbulent flow rate is proportional to the root of pressure gradient B) In turbulent flow the gas moves more as a solid column, than as a cone as in laminar C) Turbulent flow rate is independent of viscosity D) Turbulent flow rate is proportional to gas density

Answer 59

D) Turbulent flow rate is proportional to gas density In turbulent flow it is required pressure gradient/ resistance that is proportional to gas density, hence the utility of helium mixtures in divers to reduce work of breathing. https://partone.litfl.com/resistance.html

Question 60

Which of the following changes would make the resting potential of a neuron less negative? A) Decreased extracellular potassium B) Decreased extracellular sodium C) Increased extracellular chloride D) Increased extracellular potassium E) Increased extracellular sodium

Answer 60

D) Increased extracellular potassium The fucking giant squid axon question Potassium is the most influential determinant of resting membrane potential. This is determined by the neuronal membrane's greater permeability to potassium than other ions (100 times more for potassium than sodium). The electrochemical equilibrium of potassium has a concentration gradient moving out of the cell, balanced by an electrical gradient moving into the cell. Increasing the extracellular potassium lessens the concentration gradient which reduces the electrical gradient that equilibrium will be reached at to balance it. This is quantified in the Goldman-Hodgkin-Katz equation. The permeability of the membrane to other ions is so low that they exert little effect on the resting membrane potential.

Question 61

Regarding the nephron, are the following true or false? A) The size limit for molecules that are filtered in the glomerulus is ~30,000 Da B) Most sodium that is filtered is reabsorbed in the PCT C) Paracellular junctions in the PCT are leaky, allowing chloride movement which helps passive sodium movement D) Alterations in serum sodium/potassium are achieved in the Loop of Henle E) Principal cells control acid-base balance F) Potassium is less reabsorbed, and can be more maximally secreted, than sodium

Answer 61

A - F B - T C - T D - F E - F F - T

Question 62

Briefly explain the function of the Loop of Henle

Answer 62

The Loop of Henle's first purpose is to produce a hypertonic interstitium, and a hypotonic tubular fluid. This provides a hypertonic medulla to facilitate water absorption in the collecting ducts. This is only provided by the long loops of Henle, which comprise about 15% of all loops. Along the descending limb, water moves by osmosis into the interstitium. This leads to progressively more hypertonic tubular fluid, and also suggests more fluid will be lost early in the loop than late, as the gradient diminishes. Throughout the ascending limb ions are pumped into the interstitium, and as the end osmolarity is lower than the start, this would suggest more solute than water is excreted into the interstitium.