Physiology MCQs ANZCA Flashcards
Tight junctions between cells:
A. impermeable to water and solutes
B. involved in active transport
C. permeable to water and solutes
D. permeability is NOT under hormonal control
E. permeable to large compounds (or something else wrong)
Tight junctions are one of the types of junction which connect adjacent cells. They are located mainly on the apical side of epithelial cells.
Functions
They have 2 main functions:
Fence function: Inhibits membrane protein moving from apical side to the basal side of the cell. This is essential in those cells in which various parts of the cell membrane have different membrane proteins for different membrane functions. For example, in proximal tubule cells:
Na-K ATPase (sodium pump)(coloured green in diagram below) is absent from the apical membrane but present in the basolateral membrane
Na-H exchanger (coloured red in diagram below) in apical membrane but not in basolateral membrane
Gate function: Tight junctions can have a low water permeability and thus contribute to the barrier function of the epithelium (eg in blood-brain barrier). In other epithelial cells the water permeability of the tight junctions can vary, eg in proximal tubule cells where the tight junctions control access from the tubule lumen to the intercellular space. The role of the tight junctions here is likened to a gate which can be closed or can open.
TightJunction.gif
Tight junctions & the blood-brain barrier
“The BBB plays a vital role in maintaining brain homeostasis. Composition of the brain interstitial fluid is controlled within a precise range, independent of fluctuations within the blood, allowing optimal neuronal function to occur. The BBB is situated at the endothelial tight junctions of the cerebral microvessels.”
“The cerebral endothelial cells form a continuous membrane with no fenestrations, unlike peripheral vessels. The endothelial cells of the BBB are connected via a network of tight junctions that create a rate-limiting barrier to paracellular diffusion of solutes. Structurally, tight junctions form a continuous network of parallel, interconnected, intramembrane protein strands, which are composed of an intricate combination of transmembrane and cytoplasmic proteins linked with the actin-based cytoskeleton, allowing the tight junction to form a seal while remaining capable of rapid modulation and regulation.”
ECG effects of hypokalaemia: A. Short PR interval B. Ventricular extrasystoles C. Elevated ST segments D. Long QRS interval E. Long QT interval F. Q waves
Answer - B
Hypokalaemia will result in: A. Prolonged QRS duration B. Prolonged QT interval C. Peaked T wave D. Hyperpolarisation of cell membrane E. Shortened PR interval
Unknown
Hypokalaemia: (Jul98 version) A. Hyperpolarises membrane B. Peaked T waves C. Prolonged QT D. VEBs E. ST elevation
Answer D, perhaps A
Hypokalaemia: A. Hyperpolarizes the membrane B. Shortens the QRS C. Shortens the PR interval D. Depresses the ST segment E. Prolongs the QT interval
Answer D, perhaps A
Hypokalemia
A. ST segment changes ("It did read changes")
B. P wave flattening
C. Shortened QT
D. No Q waveAnswer A
Alt version (Mar 05 & July 05): Hypokalemia A. P wave flattening B. ST segment depression C. Q wave D. Shortened PR E. Delta waves
Answer B ECG changes with hypokalaemia prolongation of the PR interval ST segment depression T wave: decreased T wave amplitude, late inversion prominent U waves If the T and U waves merge, the apparent QT interval is prolonged, but, if they are separated, the QT interval is seen to be normal. In addition, hypokalaemia: hyperpolarise the membrane causes ventricular extrasystoles Hypokalaemia does: NOT cause q waves NOT shorten QRS NOT prolong QT Does that mean both hypo/hyperkalemia prolong PR segment? As every other source I've read says increased K increases PR segment, and then Ganong says low K increases PR
The ion with lowest intracellular concentration is: A: Na+ B: HCO3- C: Ca+2 D: Mg+2 E: K+
Intracellular Concentration Na+ = 10mmol/L HCO3- = 10mmol/L Ca+2 = 100nmol/L (note: nanomoles/l) Mg+2 = 10mmol/L K+ = 150mmol/l Answer is C - Calcium has the lowest intracellular concentration
The rate of diffusion across semipermeable membrane:
A. is inversely proportional to thickness
B. is proportional to molecular weight
C. ?
D. ?
E. ?
Answer = A
Rate of diffusion = K. A. (P2-P1)/D
Fick’s law states that the rate of diffusion of a gas across a membrane is:
Constant for a given gas at a given temperature by an experimentally determined factor, K
Proportional to the surface area over which diffusion is taking place, A
Proportional to the difference in partial pressures of the gas across the membrane, P2 − P1
Inversely proportional to the distance over which diffusion must take place, or in other words the thickness of the membrane, D.
The exchange rate of a gas across a fluid membrane can be determined by using this law together with Graham’s law.
I looked at the formula and this is not what I got used to from J West book:
Diffusion through a tissue sheet= A x D x ( P1-P2)/ T where A - area D - diffusion constant ( P1- P2) the difference in partial pressure T - thickness
Fick’s Law of diffusion: - rate of diffusion of a gas through a tissue slice is proportional to the area but inversely proportional to the thikness. - diffusion rate is proportional to the partial pressure difference - diffusion rate is proportional to the solubility of the gas in the tissue but inversely proportional to the square root of the molecular weight.
So… A is correct, B is incorrect - diffusion rate inversely proportional to the square root of the MW
Set of blood gases with high pH, high HCO3 and high CO2. Options: A. Metabolic acidosis B. Acclimatisation to altitude C. COAD D. ? E. Prolonged vomiting
Correct Answer is E.
Gas shows a metabolic alkalosis with respiratory compensation.
A. Altitude causes hyperventilation because of hypoxia leading to a respiratory alkolosis which can last days (4 days?) with metabolic compensation. Even chronically may have high respiratory rate.
B. COAD would cause a respiratory acidosis with metabolic compensation.
C. Metabolic acidosis is wrong. pH is >7.4
D. ?
E. Prolonged Vomiting. Loss of acid+ will cause a metabolic alkalosis which will be compensated for by hypoventilation and rise in pCO2.
Boston Rules Strategy:
pH 7.48 PO2 70 pCO2 48 HCO3 35
Step 1 pH>7.44 = alkalosis
Step 2 pCO2 and HCO3 raised = metabolic alkalosis OR respiratory acidosis
Step 3 No clues
Step 4 Assess respiratory compensation- One and a Half Plus Eight Rule Expected pCO2 = 1.5[HCO3] + 8 = 1.5[35]+8 = 60.5»_space; 48 (but 58 about right)
Step 5 Metabolic alkalosis with appropriate respiratory compensation
Only plausible answer is vomiting (loss of acid), although note can lose bicard dependent on where vomiting came from (see brandis).
I think the wrong Bedside rule has been applied. It should be the Point Seven + Twenty Rule instead
Hence Expected pCO2 = 0.7(35)+20 = 44.5(+/-5)
Food for thought: Assuming this person is breathing room air & at sea level, there is an A-a gradient i.e. Expected pO2 = 150 - 48/0.8 = 90 Why???
–cos the formula you used was wrong!
The A-a gradient in this case is [0.21(760-47) - 48/0.8] - 70 = 19.73mmHg (which isn’t that nasty)
You used exactly the same formula! And yes the A:a gradient is raised.. we know nothing about the patient’s age, which will impact on it..
Base excess calculation from A. when PaCO2 is 40 mm Hg B. difference of measured HCO3 from standard HCO3 C. lower with higher HCO3 D. is an indicator of cellular buffers E. is negative when pH greater than 7.40
Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4
Edit - yes, but that’s not the same as saying assumes a CO of 40 is it? Doesn’t the base excess adjust the values to what they would be IF PCO2 were 40mmHg?
(1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement.
(2) As the measurement occurs “as if” ABG PCO2 was 40 mmHg, ‘A’ will be the “most correct” of the available answers.
AD18 [Feb12] version: The base excess on an arterial blood gas? A. Assumes a CO2 of 40mmHg B. Is measured at 20 degrees Centigrade C. ..Something about titratable acids... D. Same as plasma bicarbonate E. Measures respiratory acid base status
Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4
Edit - yes, but that’s not the same as saying assumes a CO of 40 is it? Doesn’t the base excess adjust the values to what they would be IF PCO2 were 40mmHg?
(1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement.
(2) As the measurement occurs “as if” ABG PCO2 was 40 mmHg, ‘A’ will be the “most correct” of the available answers.
BP01 [aqr] [Mar05] [Jul05]
Gap junctions:
A. Maintain cellular polarity
B. Occur at the apices of cells
C. Have corresponding connections between cells
D. Are formed by ridges on adjacent cells
E. Gives cells stability and strength
Gap junctions permit the transfer of ions and other molecules between cells via proteins called connexons and these form a channel when lined up with the corresponding connexon in the adjacent cell (C). As an example, gap junctions permit current flow and electrical coupling between myocardial cells.
Tight junctions occur on the apices of cells (B) and are formed by ridges on adjacent cells (D) and give cells stability and strength (E).
Polarity is due to enzymes in the apical cell membrane differing from those in the basolateral membrane.
Answer is C.
BP02 [Jul97]
Bulk flow:
A. Is related to concentration gradient
B. Is related to permeability coefficient
C. Depends on hydrostatic and oncotic pressure
D. ?
ANSWER
Concentration gradient - important for diffusion, NOT for bulk flow
Hydrostatic & oncotic pressures - important in filtration (which is bulk flow across a membrane).
Permeability coefficient is the diffusion constant in the membrane divided by membrane thickness (therefore related to diffusion rather than bulk flow).
I would say the answer is C Glomerular filtration is the bulk flow of fluid from glomerular capillaries into Bowman’s Capsule (Vander p17) Filtration is the process by which fluid is forced through a membrane or other barrier because of a difference in pressure on the two sides (Ganong p36) Although Kf is a factor in GFR, the Net Filtration Pressure is the major determinant of GFR
—
“Bulk Flow (ultrafiltration): a process whereby fluid moves from capillary to interstitial fluid by excess of hydrostatic over oncotic pressure.” Faunce. pg 7.
BP03 [gko] All of the following histamine effects are mediated by H2-receptors EXCEPT: A. Vasodilatation B. Bronchoconstriction C. Gastric acid secretion D. Tachycardia E. Increased contractility
Bronchoconstriction is due to stimulation of H1 receptors
Maconochie JG et al. Effects of H1- and H2-receptor blocking agents on histamine-induced bronchoconstriction in non-asthmatic subjects. Br J Clin Pharmacol. 1979; 7(3): 231-6. [1]
1 Two studies have been carried out to investigate the effect of H1- and H2-receptor blocking agents on histamine-induced bronchoconstriction in non-asthmatic subjects.
2 The H2-receptor blocker cimetidine administered orally had no effect on histamine-induced bronchoconstriction on any of the subjects tested. In three of four subjects, the H1-receptor blocker, chlorpheniramine given orally, inhibited the effect of the histamine in the lung.
3 The effects of intravenous chlorpheniramine and cimetidine, both alone and in combination, upon histamine-induced bronchoconstriction, were also studied. Chlorpheniramine inhibited the effect of the histamine and this was significantly dose related. This was not so with cimetidine and there was no evidence that the dose response curve to chlorpheniramine was affected by the additional administration of cimetidine.
4 The results show that histamine-induced bronchoconstriction in non-asthmatic subjects is not mediated by H2-receptors, but it is likely that H1-receptors are involved.
“ Histamine, like many other transmitters, mediates responses via receptors, which are
divided into three subtypes H1, H2 and H3.
- H1 receptors are found in the smooth muscle of the intestines, bronchi, and blood vessels.
Also found in nerve endings - activation causes itch &pain - The H2 receptor is found in gastric parietal cells and in the vascular and central nervous systems.
Also found in cardiac muscle and on mast cells for negative feedback mechanism. - H3 receptors are found in brain and in the periphery and regulate histamine release.
- from [2]
Choice A: Vasodilation via H2 (Vasoconstriction is via H1) ??Reference
Choice B: Pulmonary vasodilation via both H1 and H2 [p666, Ganong 21th ed]
Choice C: Histamine stimulates gastic acid secretion via H2 [p497, Ganong 21th ed]
[Ganong 21th ed, p600]
Vasoconstriction via is H1
Vasodilation is via H2
However, Rang and Dale 5th ed, p230 states
H1 receptor mediates vasodilation and bronchoconstriction
H2 mediates tachycardia and increased contractility
- Likewise, Katzung is a little vague.
H1 receptors act via increasing IP3/DAG, and intracellular Ca
H2 act via increased cAMP
It says that H2 receptors directly cause an increased contractility and tachycardia, but also that the cardiovascular response - including vasodilation I presume - can be blocked by a mixture of H1 and H2 antagonists (and then says that low dose H1 antagonists can block the CVS response!)
the vasodilation is probably mediated via nitric oxide released from endothelial cells
It says that H1 receptors mediate bronchoconstriction, and H2 gastric secretion
B sounds like it then - and vasodilation could be a mixture
BP04 [Feb00] [Jul09] The trace element that is an integral component of carbonic anhydrase, lactic dehydrogenase, and several other peptidases: A. Magnesium B. Manganese C. Zinc D. Cobalt E. Copper
They all contain Zinc.
BP05 [Jul04] [Mar05] An example of autoregulation is: A. Renin-angiotensin-aldosterone system B. Tubuloglomerular feedback C. Baroreceptors D. ? E. Increased tissue vascularity
Answer - B
Autoregulation - refers to the capacity of tissues to regulate their own blood flow.
Tubuloglomerular feedback - The macula densa senses the amount of sodium and chloride entering the distal convoluted tubule. An increased solute load results in adenosine release, which then causes vasoconstriction of the afferent arteriole, thus controlling renal blood flow and GFR. This is an intrinsic mechanism of the kidney.
BP06 [JUl04] [Jul05] Which is not essential for pain? A. Conscious awareness B. Actual tissue damage C. something like 'May be modulated over time'
think the best answer is B actual tissue damage. Think of phantom limb pain - no tissue damage there but pain persists. Also think of psychosomatic pain where it’s all in the mind so to speak!
From Acute Pain Management: scientific evidence 2005 p 1:
IASP definition of pain is “an unpleasant sensory and emotional experience associated with actual or potential tissue damage, or described in terms of such damage” and in addition notes that inability to communicate pain doesn’t negate the possibility that pain is present. From this it’s clear that ACTUAL tissue damage isn’t necessary hence this option is correct.
However the extension to the definition and the discussion that follows implies that unconscious patients can experience pain so this may also be a correct option (although perhaps not the MOST correct).
More comments…
It’s actually interesting the last comment - we’ve all seen that surgical incision causes a sympathetic response to pain that would also be seen in an awake patient. It also seems that if we prevent this (regionals, ketamine) then post operatively the patient does better.
However, does the “experience” of pain require consciousness?
(It does. The unconscious patient can respond to noxious stimuli, by a spinal reflex for example, or an autonomic response, but the “experience” of pain occurs when nociceptive stimuli reaches consciousness).
If a tree falls in the forest and no one is there to hear it, does it make a sound?
BP06b [Jul05] (Above MCQ remembered slightly differently: Whis is not true of pain pathways? A. Withdrawal pathways are involved B. Emotional pathways are involved C. Tissue damage must occur D. Requires conscious awareness
I think the best answer is and C - actual tissue damage. Think of phantom limb pain - no tissue damage there but pain persists. Also think of psychosomatic pain where it’s all in the mind so to speak!
BP08 [July-07] [Feb08] Giant Squid Axons are used to study action potentials because: A. They are large B. They only contain sodium channels C. ? D. ? E. ?
Answer. A
Giant squid axons are very large up to 1mm diameter allowing scientists to insert voltage clamp electrodes into the nerve to study the ionic mechanisms of the action potential.
Which is incorrect regarding the Kreb’s cycle:
A. Acetyl-CoA is metabolized to CO2 & H+
B. ?
C. Oxaloacetate is recycled
D. 12 ATP is generated
E. Cycle is continous during anaerobic metabolism but at slower rate
E - “The Citric Acid cycle requires oxygen and does not function under anaerobic conditions” - Ganong
This is because the ETC provides new NAD, NADP, FAD for accepting Hydrogens, and these will not be replenished without oxidative phosphorylation (although some NAD is replenished during anaerobic glycolysis by converting pyruvate to lactate)
However, C is also possibly wrong, given that technically, the citric acid cycle only directly produces one molecule of ATP per revolution (not counting the subsequent effect of oxidative phosphorylation) (Guyton)
If the ATP produced indirectly from the citric acid cycle (via H carriers NAD, etc and oxidative phosphorylation) then it does produce 12 ATP per revolution (and 24 per molecule of glucose)
Answers directly from Ganong 21st Ed page 289 - 291
A - Correct “Acetyl CoA is metabolised to CO2 and H atoms”
C - Correct “In a series of seven reactions, …… regenerating oxaloacetate”
D - Correct (in a way) - “24 ATPs are formed during the subsequent two turns of the citric acid cycle” - therefore 12 ATP per cycle —- although really, most of these ATPs are really off-shoots of the citric acid cycle entering oxidative phosphorylation rather than directly from the citric acid cycle itself
E - False (see above quote from Ganong)
Summary of Energy Production in Aerobic Metabolism
Glycolysis
Glucose metabolised to pyruvate (x 2 molecules)
4 ATP formed, 2 used
2 ATP net production
2 NADH produced (6 ATP via ETC)
ATP produced = 2 ATP + 6 ATP
in between
Pyruvate converted to acetyl coA
2 NADH formed from 2 pyruvates
2 NADH produced (6 ATP via ETC)
Citric Acid Cycle
Each molecule of pyruvate forms a molecule of Acetyl CoA for subsequent entry into the cycle - see above
Each revolution of cycle produces 1 ATP (?via GTP)
therefore, 2 revolutions per glucose molecule, so 2 ATP
2 revolutions produce 6 NADH (18 ATP via ETC), 2 FADH2 (4 ATP via ETC)
In total, for every glucose
Glycolysis = 2 ATP
Citric Acid Cycle = 2 ATP
ETC = 34 ATP (6 via glycolysis, 6 via pyruvate to acetyl coA, 22 via Citric Acid Cycle)
I hope that confuses everyone as much as it has me :)
For answer A it should be noted that in the addition of Acetyl CoA to Oxaloacetate to create the 6C citrate of the krebs cycle the subsequent loss of carboxyl groups comes from the oxaloacetate base not the Acetyl CoA addition. The actual carbons may in fact be shunted to other reactions outside the mitochondrion via malate. Though it is hard to argue against Ganong as this is obviously where the question comes from.
BP10 [Feb08] Cytochrome c oxidase catalyses: A. O2 + 2H+ -> H2O B. ? C. ? D. H+ + HCO3- -> H2CO3 E. None of the above
(Think this may have actually been asking about cytochrome a3)
If talking about cytochrome oxidase - then this is the last enzyme in the ETC
A (although chemical reaction is not balanced) - combines Oxygen with electrons and hydrogen ions = water
NB. Cytochrome c oxidase is otherwise known as Complex IV in the ETC. It contains cytochrome a3, cytochrome a, and 2 copper ions.
The question was therefore correct I believe…
Another direct Ganong lifted question. Last sentence in the section on Biologic oxidations: The final enzyme… cytochrome c ox… which transfers hydrogens to oxygen forming H20. It contains 2 Fe, 3 Cu and 13 subunits.
BP11 [Feb08][Jul09]
In regards to the Na+/K+ ATPase:
A. Three K+ out for every two Na+ pumped in
B. Stimulated by Oaubain
C. 3ATP broken down to ADP and P for every 3Na+ pumped in
D. Is inhibited by high extracellular concentrations of Na+
E. An electrogenic pump
Below based on first version above:
A - Wrong - 3x Na+ pumped OUT for every 2x K+ pumped IN
B - Wrong - Oaubain is a cardiac glycoside that inhibits pump (see see http://en.wikipedia.org/wiki/Ouabain)
C - Wrong as worded - Na pumped OUT, not in. Not sure how many ATP are used… anyone?
D - True? - High extra cellular Na levels might inhibit the pump, or just cause more Na+ leak back in… anyone?
E - True - The Na/K pump is an electrogenic pump (3 +ve charge out, for 2 +ve let in, therefore contributed to -ve intracellular potential)
— Power & Kam Pg 5 1st Ed Pump is electrogenic 1ATP = 3Na out + 2K in +ADP,phosphate
Na/K ATPase A. Is electrogenic B. Is impaired by low extracellular ECF conc C. 3 ATP used for each 3 na pumped D. 3 K+ in for 2 Na+ out E. ?
?
BP12 [Feb12]
Which of the following is not true regarding intracellular organelles:
A. The Endoplasmic reticulum is involved in protein synthesis
B. The Golgi apparatus ..?..
C. ..Maybe an option about gene transcription..?..
D. All cells contain a nucleus
E. ..something true about peroxisome?
A. True Rough ER is involved in protein synthesis
D. False Anucleate cells include RBCs, platelets, outer layer of epidermis.
FE01 [Mar96] [Mar98] [Jul98] [Apr01] [Jul01] [Mar03] [Jul03] [Mar05] [Jul05] [Feb06] Jul10 Aug14 ECG effects of hypokalaemia: A. Short PR interval B. Ventricular extrasystoles C. Elevated ST segments D. Long QRS interval E. Long QT interval F. Q waves
B
Aug14 remembered version: Hypokalaemia will result in: A. Prolonged QRS duration B. Prolonged QT interval C. Peaked T wave D. Hyperpolarisation of cell membrane E. Shortened PR interval
D, ?A
Hypokalaemia: (Jul98 version) A. Hyperpolarises membrane B. Peaked T waves C. Prolonged QT D. VEBs E. ST elevation
D ?A
Hypokalaemia: A. Hyperpolarizes the membrane B. Shortens the QRS C. Shortens the PR interval D. Depresses the ST segment E. Prolongs the QT interval
D ?A
Alt version: Hypokalemia
A. ST segment changes ("It did read changes")
B. P wave flattening
C. Shortened QT
D. No Q waveA
Alt version (Mar 05 & July 05): Hypokalemia A. P wave flattening B. ST segment depression C. Q wave D. Shortened PR E. Delta waves
B. ECG changes with hypokalaemia prolongation of the PR interval ST segment depression T wave: decreased T wave amplitude, late inversion prominent U waves If the T and U waves merge, the apparent QT interval is prolonged, but, if they are separated, the QT interval is seen to be normal. In addition, hypokalaemia: hyperpolarise the membrane causes ventricular extrasystoles Hypokalaemia does: NOT cause q waves NOT shorten QRS NOT prolong QT Does that mean both hypo/hyperkalemia prolong PR segment? As every other source I've read says increased K increases PR segment, and then Ganong says low K increases PR
FE02 [Mar97] [Jul04]
For two solutions separated by a semi-permeable membrane (Solution A: saline solution AND solution B: H2O): Which ONE of the following statements is true?
A. A hydrostatic pressure applied to A will stop osmotic pressure (?)
B. There will be bulk flow from A to B
C. The fluid level in B will go up
D. The NaCl concentration at A will remain the same
E. Water will move from A to B by diffusion
Firstly the idea of a semipermeable membrane is that the solvent water can cross the membrane but the solutes cannot.
A. - All the other options are wrong (see below) so by excluson this option is the correct one. However, the remembered wording is very poor and is wrong. What the option should say is something like: “A suitable hydrostatic pressure applied to A will stop the osmotic movement of water from B into A”. This afterall is essentially the definition of “osmotic pressure”. However the wording of “stop osmotic pressure” is wrong and this is probably because the idea of the option has been remembered rather than the exact wording.
B. - WRONG. Net water movement is by diffusion from B->A as determined by the osmotic gradient. This is not bulk flow but due to diffusion of water molecules.
C. - WRONG. Water moves from B->A (down its osmotic gradient) so the fluid level in B will drop.
D. - WRONG but “technically correct”. The [Na+] in A will decrease as water moves into the solution. However, strictly speaking, NaCl is TOTALLY ionised in water so there is in reality NO undissociated NaCl present in the solution, only Na+ and Cl- ions. So the [NaCl] = 0 throughout - it “remains the same”. However the examiners are unlikwely to be asking for your knowledge of this situation.
E. - WRONG but “technically correct”. Water moves BOTH ways across a semi-permeable membrane (by diffusion) but where there is an osmotic gradient (ie difference in water concentrations across the membrane) there will be a NET movement of water in one direction, in this case from B->A. So presumingtb this option really refers to net movement then it is incorrect.
Comment
Depends on what you think semi-permeable means- permeable to water, permeable to small cations? If you think the membrane will be permeable to Na/Cl - then i think B will be right.
Comment
Bulk flow refers to movement of water and solutes together down a pressure gradient. This is seen in the circulation of CSF. As solute movement does not occur in this scenario, bulk flow is not present.
Diffusion refers to the net movement of water down a concentration gradient due to the random movement of individual molecules. If the examiners are referring to net movement (in this admittedly poorly phrased question), then this option is potentially correct.
The osmotic pressure refers to the tendancy of water to diffuse across the membrane. It is defined by the hydrostatic pressure that must be applied to the recipient solution in order to prevent said movement. For a solution with osmolality of 287mOsmol/L, this is 5547mmHg. If the examiners are indeed referring to the hydrostatic pressure required to prevent the movement of water from B to A, this option is the most correct.
All we can be sure of is that the question will be better written on the day!
FE03 [Mar97] [Jul97] [Jul99]
Rapid (?ingestion/?infusion) of 2 litres of normal saline causes:
A. Increased ECF, increased ICF, decreased [Na+]
B. Increased ECF, unchanged ICF, increased [Na+]
C. Unchanged ECF, increased ICF, increased [Na+]
D. Increased ECF, unchanged ICF, unchanged [Na+]
N/saline is distributed throughout the ECF only as its [Na+] restricts its distribution.
In contrast, note that ‘pure water’ is distributed throughout the total body water so each compartment would obtain an amount of the infused fluid in proportion to its % contribution to total body water. (You would not want to infuse “pure water” as it could cause haemolysis but you could drink a couple of litres of pure water). Note that giving IV 5% dextrose solution is the equivalent of infusing ‘pure water’ as the glucose is rapidly taken up by cells.
Infusing N/saline would not cause osmotic shifts of water as its osmolality is the same as that of body fluids. (This may not be true if the person is either hyponatraemic or alternatively hypertonic for some reason but you can assume it is true for the normal situation). So ECF [Na+] would not change.
The distribution of the infused N/saline would be totally to the ECF. The ICF volume would not change.
So we would get increased ECF, unchanged ICF, unchanged [Na+] which is the correct option.
Comment
Technically you should get a slight rise in ECF [Na+] as [Na+] in N/Saline is 150 mmol/L, compared with 140 mmol/L in plasma
Question says ‘rapid’. In other words, kidneys have no time to compensate for the additional [Na+] load.
Thus expected increase in [Na+] = 20mmol/14L = 1.43 mmol/L
Further Comment:
Ref: The Australian and New Zealand Primary Exam by Faunce pg 119
Rapid infusion of 2L NS causes inc ECF, unchanged ICF, and UNCHANGED [Na+] due to plasma solids effect
Alternative view
As per Brandis (2nd edition), the plasma consists of 93% plasma water and 7% plasma solids (protein and lipid). However, measurement of Na+ by flame emission spectrophotmetry or ISE measure it as a whole. Hence, the measured plasma Na+ appears less than the actual amount of Na+ contained in the plasma.
Again as per Brandis “…the decrease in measured plasma Na+ due to the plasma solids effect is about the same as the increase in Na+ that occurs due to the Gibbs-Donnan effect…”, with the result being that measured plasma Na+ is about the same as measured ISF Na+. Hence, the postulated reduction in plasma [Na+] that is present as a result of the plasma solids effect is negated, and we can effectively treat the system (for measurement purposes only) as though the two do not exist.
Consequently, infusing a solution that contains 150mmol/L of NaCl into a compartment that contains 140mmol/L of NaCl will result in:
Expansion of ECF
No change in ICF (despite a measured osmolality of 300mOsmol/L, this is not true in-vivo)
Increase in measured [Na+]
The above relies on the fact that NaCl dissociates completely in solution and in vivo. However, this may not be the case as:
Serum osmolality is 285 mOsmol/L, and whilst 0.9% saline has an osmolality of 300mOsmol/L it is still considered isotonic;
The solution contains 154 mmol/L of Na+ and 154 mmol/L of Cl-, yet has an osmolality of 300mOsmol (should be 308mOsmol if complete dissolution were to occur).
Can anyone reconcile the above? Preferably within the next 7 days?
Hyperkalaemia: A. Causes a prolonged QT interval B. Prolongs the QRS duration C. Causes ST segment elevation D. Potentiates digoxin toxicity E. Causes loss of P wave
B (?before E, also correct)
Not A/C
Not D - may be a complication of digoxin toxicity, but hypokalemia potentiates digoxin toxicity
FE04b [Mar02] [Feb07] ECG changes in hyperkalaemia include: A. ST depression B. T wave inversion C. P wave flattening D. Sinus tachycardia E. ?
C
Feb07 options: A. Long PR B. Wide QRS C. Long QT D. ST elevation E. ST depression
B (although I have also read A in a few areas too – for e.g. Brandis page 15 under “Potassium”
ECG changes with hyperkalaemia (in order that they occur)
Tall peaked T waves
loss of the P wave
prolongation of the QRS complex
ventricular arrhythmias - tachycardias and fibrillation
asystole
FE05 [Mar98] [Apr01] [Jul04] Thoracic lymph contains: A. Clotting factors B. Higher protein content than plasma C. Similar composition to ISF D. Rarely contains fat E. ?
A. Clotting factors - Yes. All Lymph contains clotting factors (just low fibrinogen)
B. Higher protein content than plasma – WRONG: Always lower even in hepatic lymph
C. Similar composition to ISF – WRONG: Thoracic duct lymph contains more protein than usual ISF because of the contribution of hepatic lymph which has a relatively high protein concentration (as compared to lymph from the rest of the body).
D. Rarely contains fat – WRONG: Lymphatics from small bowel contains fat (chylomicrons) after meals – important in fat absorption (approx 90% fat absorbed this way)
E. ?
Thoracic duct lymph contains protein about 50g/L (hepatic lymph 60g/L), whereas lymph from muscle/skin contains 20g/L
Plasma contains about 70g/L protein.
Albumin 40g/L
Globulins 25g/L
Fibrinogen 5g/L
FE06 [Mar98] [Jul98] [Mar99] [Feb04] [09B] [11B] [12B] [14A]
Gibbs-Donnan effect leads to:
A. Non-diffusible ions between 2 sides will be equal
B. Diffusible ions between 2 sides will be equal
C. Equal concentrations of ions on both sides
D. Equal passive diffusion
E. Osmotic gradient
F. Important in the measurement of plasma oncotic pressure
Gibbs-Donnan leads to :
E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood.
Gibbs-Donnan effect
The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules.
Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. “About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone.” (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno).
I thought that ‘oncotic pressure’ a.k.a. ‘colloid osmotic pressure’ was osmotic pressure due to proteins alone, whereas ‘osmotic pressure’ refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me.
Re: Aug 2011 version:
A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium
B. (seems like an incomplete option)
C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells.
D. ?
E. ?
Best answer: ‘A’
Alternate view - Why couldn’t B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?
Gibbs-Donan equilibrium
A. explains the distribution of diffusable ions between intravascular and extravascular space
B. effects distribution of plasma proteins across capillary membrane
C. effects distribution of proteins across cell membrane
D. explains the distribution intracellular and extracellular ions
E. explains the distribution of non-diffusable ions across cell membrane
Gibbs-Donnan leads to :
E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood.
Gibbs-Donnan effect
The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules.
Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. “About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone.” (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno).
I thought that ‘oncotic pressure’ a.k.a. ‘colloid osmotic pressure’ was osmotic pressure due to proteins alone, whereas ‘osmotic pressure’ refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me.
Re: Aug 2011 version:
A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium
B. (seems like an incomplete option)
C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells.
D. ?
E. ?
Best answer: ‘A’
Alternate view - Why couldn’t B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?
[Aug 11]:
Gibbs-Donnan Effect:
A. explains distribution of charge between intra and extraVASCULAR space
B. explains difference in ionic concentration gradients
C. explains distribution of charge between intra and extracellular spaces
D. ?
E. ?
Gibbs-Donnan leads to :
E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood.
Gibbs-Donnan effect
The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules.
Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. “About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone.” (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno).
I thought that ‘oncotic pressure’ a.k.a. ‘colloid osmotic pressure’ was osmotic pressure due to proteins alone, whereas ‘osmotic pressure’ refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me.
Re: Aug 2011 version:
A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium
B. (seems like an incomplete option)
C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells.
D. ?
E. ?
Best answer: ‘A’
Alternate view - Why couldn’t B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?
[Feb14]:
The Gibbs Donnan Effect:
A. Explains distribution of charge between intra and extravascular space
B. Explains distribution in charge between intra and extracellular space
C. Explains differences in ionic concentration gradients
D. Explains the distributions of plasma proteins across cell membranes
E. Explains the distribution of all proteins across cell membranes
Gibbs-Donnan leads to :
E. Osmotic gradient- by altering the concentrations of diffusible ions, resulting in a small net increase in ions present in the plasma and maintenance of the plasma oncotic pressure in the blood.
Gibbs-Donnan effect
The observation that charged molecules starting on one side of a semipermeable membrane sometimes will not evenly distribute themselves by diffusion on both sides of the membrane. This effect is probably because there are other charged substances already present which cannot move through the membrane themselves and which are creating an electric field that influences the movement of the incoming charged molecules.
Regards option F: The Donnan effect increase the Plasma Oncotic pressure by about 50% via retention of Na+ attracted to -ve charged plasma proteins. Guyton 11th p 185-190 has a discussion of oncotic pressure. “About 19mmHg of the colloid osmotic pressure is due to the dissolved protein, but an aditional 9mmHg is due to the positively charged cations, mainly sodium, that bind to the negatively charged proteins. This is called the Donnan equilibrium effect, which causes the colloid osmotic pressure in the plasma to be about 50% greater than that produced by the protein alone.” (This is actually from the Pocket Companion of Guyton 11th ed p113..so the wording may differ from the original- I only read pocket manuals!! Wayno).
I thought that ‘oncotic pressure’ a.k.a. ‘colloid osmotic pressure’ was osmotic pressure due to proteins alone, whereas ‘osmotic pressure’ refers to osmotic pressure due to all solutes. Thus option F seems more like a confusing statement/red herring than a correct option to me.
Re: Aug 2011 version:
A. Correct - the Gibbs-Donan effect across the vascular wall explains why plasma has higher [Na+] and lower [Cl-] than the interstitium
B. (seems like an incomplete option)
C. Wrong - The Na-K-ATPase pump is the most important explanation. However, this does set up the dual Gibbs-Donan effects of intracellular proteins and extracellular Na+ nearly balance one another to prevent osmotic expansion of cells.
D. ?
E. ?
Best answer: ‘A’
Alternate view - Why couldn’t B be right? The point being that it affects ionic gradient everywhere, both between intravascular and extravascular, and intersitial/ICF?
FE07 [Jul98] [Mar05] [Jul05]
With decreased osmolality and hypovolaemia, you would see:
A. (?Decreased/increased) urine output
B. Decreased ADH secretion
C. Decreased aldosterone
D. Increased permeability of collecting ducts to water
E. Decreased renin
Decreased urine output due to increased ADH secretion.
Osmoreceptors are very sensitive to a decrease in osmolality and the decreased osmolality will result in a decreased ADH secretion.
However, the low pressure baroreceptors (volume receptors) in the right atrium and great veins though being less sensitive than the osmoreceptors respond far more powerfully. The peak [ADH] resulting from hypovolaemia is much higher than that produced by maximal response to osmoreceptor input.
So the knowledge this question is testing is whether you know that hypovolaemia over-rides the inhibition from the hypo-osmolality. Volume is maintained at the expense of a decreased osmolality.
But either way it is true that permeability of collecting duct to water is increased, therefore correct answer D.
Hartmann's solution contains: A. Potassium 2 mmol/l B. Calcium 3 mmol/l C. Magnesium 2 mmol/l D. Sodium 154 mmol/l E. Chloride ?131 ?154 mmol/l
Answers
1st wording - none correct
Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B
Hartmann’s solution contains Ca++ but no Mg++.
The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don’t give whole blood much anymore).
Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++.
The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this.
Hartmann’s Solution
Solute Concentration
Na+ 129 mmol/L
K+ 5 mmol/L
Ca2+ 2 mmol/L
Cl- 109 mmol/L
Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver)
pH 6.5
Osmolarity 279 mosm/L
Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.
FE08b [Apr01] [Mar05] Hartmann's solution contains no: A. Na+ B. Ca++ C. Mg++ D. Lactate E. Cl-
Answers
2nd wording - no magnesium
Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B
Hartmann’s solution contains Ca++ but no Mg++.
The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don’t give whole blood much anymore).
Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++.
The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this.
Hartmann’s Solution
Solute Concentration
Na+ 129 mmol/L
K+ 5 mmol/L
Ca2+ 2 mmol/L
Cl- 109 mmol/L
Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver)
pH 6.5
Osmolarity 279 mosm/L
Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.
FE08c [Aug 11] - (NB: Same as GP47) Hartmanns is \:A. Isotonic \:B. Contains Ca 2 mmol/L \:C. Contains lactate 5 mmol/L \:D. ? \:E. ?
Answers
3rd wording - both A and B are correct
Edit - dont think so, Hartmanns is actually hypotonic relative to plasma, I would go with B
Hartmann’s solution contains Ca++ but no Mg++.
The Ca++ content of Hartmanns is important as it can result in clotting in the tubing if blood is given slowly before or after it in the same line. Because of this problem the traditional nursing practice in the past was to give a 100mls of normal saline before and after each bag of blood. This clotting problem is less of a concern during anaesthesia if we fluids are given quickly (and since we don’t give whole blood much anymore).
Some people use ‘Plasmalyte 148’ which contains Mg++ but no Ca++.
The other difference between these 2 solutions is the slightly lower [Na+] in Hartmann’s solution: 130mmol/l compared to 140mmol/l in Plasmalyte. Some neuroanaesthetists prefer not to use Hartmanns because of this.
Hartmann’s Solution
Solute Concentration
Na+ 129 mmol/L
K+ 5 mmol/L
Ca2+ 2 mmol/L
Cl- 109 mmol/L
Lactate 29 mmol/L (converted into HCO3- via oxidation and gluconeogenesis by the liver)
pH 6.5
Osmolarity 279 mosm/L
Comment: Most sources (inc. Sasada & Smith) say Na-131mM, Cl-111mM.
FE09 [gij] [Mar99] [Feb00] [Jul00] [Jul08] The total osmotic pressure of plasma is: A. 25 mmHg B. 285 mOsm/l (or ?308mOsm/l) C. 5900 mmHg D. 300 kPa E. None of the above
One source gives a total plasma osmotic pressure of plasma is about 5562mmHg or 7.31 atmospheres of 733kPa.
This can be calculated using the van’t Hoff equation and the actual value in a person depends on the plasma osmolality. See the Fluid Physiology tutorial on this site (section 2.4.2) where a value of 5409 mmHg is calculated for a plasma osmolality of 280 mOsm/kg. Using this equation, the value of 5562mmHg quoted above is the value obtained when the osmolality is 288mOsm/kg.
The oncotic P is the component due to colloids is 25 to 28mmHg, or 0.5% of the total (ie 19.3 x plasma osmolality due to colloids)
Using Van’t Hoff in KB’s book a total plasma osmotic pressure of 5900mmHg would mean that the plasma osmolality would have to be 305 mOsm/kg. This sounds high but is actually the upper end of the physiological range quoted in Yentis A-Z (280-305 mOsm/kg). This could mean that answer C is close enough to be true.
NB:More likely is that the actual value on the paper was lower than the remembered
5,900mmHg value quoted in this reconstructured version of the MCQ. So option C is correct but
may have been somrthing like 5,500mmHg on the paper.
Surely given the figures above assuming that they are remembered correctly then E is the correct answer as working backwards 5900 mmHg gives a 305mOsm/kg which is above the normal or 288-293 mOsm/kg depending on the text you read. Obviously if option C had read 5500mmHg then you could say that is was correct, but at 5900mmHg its a bit high
FE10 [Mar99] [Jul05] Normal saline: A. Osmolality of 300-308 mOsm/l B. Has pH 7.35 to 7.45 C. ? D. ?
Normal saline
pH range 4.0 - 7.0
Osmolality 300mOsm
FE11 [Mar99] [Mar03] [Jul03] Mar10 Obligatory water loss from body: A. 400 mls in faeces B. 300 mls from lung C. Loss from skin & respiratory tract 700ml D. ??Insensible water loss E. 500 mls in urine
Comment: All figures seemed slightly off from standard text
Response: Thats because 4 of the options have to be wrong. The actual question will expect an
answer consistent with values in the ANZCA ‘Recommended texts’. Don’t grieve too much over
the values in this remembered version; just use this MCQ as a reminder to learn the standard
values so you can answer the question if it appears.
Daily Water Loss (Power & Kam)
900ml insensible loss from skin and lungs
50ml as sweat in normal climates
100ml in faeces
430ml urine - minimum volume to excrete the daily solute load (600mOsmol / 1400mosmol/kg)
It seems that 500ml for urine loss is probably the closest answer! (Comment - 500mls is the minimum daily urine, not normal daily loss, which is about 1.5L/day. Food for thought.)
Fluid Output (for a normal individual): (Guyton & Hall)
insensible-skin 350ml
insensible-lung 350ml
sweat 100ml
faeces 100ml
urine 1400ml
I think that 700ml from the lung and skin is therefore the best answer.
I think the wording of the 1st question as “obligatory” water loss would suggest that the answer is urine, rather than insensible losses from skin/lung (see link to fluid physiology notes below)
“An obligatory urine loss occurs because of the need to remove various solutes from the body”
Comment: totally agree. Obligatory refers to urine excretion which is obligatory to prevent toxic buildup of certain byproducts of metabolism. Maximum human urine osmolality is ~ 1400mOsm/L, and at that concentration we would have to produce 400-500ml urine at least to remove adequate waste products.
Normal amount of daily water loss in a 70kg man:
A. 300mls faeces
B. 500mls from urine
C. 700mls from lungs and skin (?insensible)
D. ?
E. None of the above
Comment: All figures seemed slightly off from standard text
Response: Thats because 4 of the options have to be wrong. The actual question will expect an
answer consistent with values in the ANZCA ‘Recommended texts’. Don’t grieve too much over
the values in this remembered version; just use this MCQ as a reminder to learn the standard
values so you can answer the question if it appears.
Daily Water Loss (Power & Kam)
900ml insensible loss from skin and lungs
50ml as sweat in normal climates
100ml in faeces
430ml urine - minimum volume to excrete the daily solute load (600mOsmol / 1400mosmol/kg)
It seems that 500ml for urine loss is probably the closest answer! (Comment - 500mls is the minimum daily urine, not normal daily loss, which is about 1.5L/day. Food for thought.)
Fluid Output (for a normal individual): (Guyton & Hall)
insensible-skin 350ml
insensible-lung 350ml
sweat 100ml
faeces 100ml
urine 1400ml
I think that 700ml from the lung and skin is therefore the best answer.
I think the wording of the 1st question as “obligatory” water loss would suggest that the answer is urine, rather than insensible losses from skin/lung (see link to fluid physiology notes below)
“An obligatory urine loss occurs because of the need to remove various solutes from the body”
Comment: totally agree. Obligatory refers to urine excretion which is obligatory to prevent toxic buildup of certain byproducts of metabolism. Maximum human urine osmolality is ~ 1400mOsm/L, and at that concentration we would have to produce 400-500ml urine at least to remove adequate waste products.
FE12 [Jul99] [Mar03] [Jul03] [Jul05]
Which ONE of the following statements about intravenous crystalloid solutions is TRUE?
A. Rapid infusion of (?one litre) Hartmann’s may cause lactic acidosis
B. Osmolality of Hartmann’s solution is 300-308 mosm/kg
C. 0.9% saline pH 7.35-7.45
D. Osmolality of N/saline is 300-308 mosm/kg
E. 0.9% sodium chloride has a pH 6.5-7.5
F. One litre of Hartmann’s solution contains 150 mmol of Na+
A. False
B. False - osmolality of Hartmann’s is 274mOsm
C. False - 0.9% saline pH range 4.0 - 7.0
D. True - 0.9% sailne osmolality 300mOsm
FE13 [Mar99] [Jul99] [Feb00] Water handling by the kidney (% reabsorption): A. 93% B. 94% C. 99% D. 99.4% E. 99.9%
Based on Ganong: GFR = 180 litres/day Urine = 1.0 litres/day Percent reabsorption = 99.4% Other figures calculated: 99.7% if assume 500ml obligate urine output 99.2% if assume 1500ml urine output) The 99.9% option is incorrect as the urine volume is below that required for the minimum urine volume to excrete the daily solute load (typically 600-700 mosmoles/day)
FE14 [j] [Jul04] [Mar05] [Jul05] [Feb06] Feb12
The ion with lowest intracellular concentration is:
A: Na+
B: HCO3-
C: Ca+2
D: Mg+2
E: K+
Intracellular Concentration Na+ = 10mmol/L HCO3- = 10mmol/L Ca+2 = 100nmol/L (note: nanomoles/l) Mg+2 = 10mmol/L K+ = 150mmol/l Answer is C - Calcium has the lowest intracellular concentration
FE15 [Apr01]
Total plasma osmolarity can be calculated via:
A Van Halen’s equation
B. Starling equation
C. P = nRT
D. (multiplying 19.2mmHg/mOsm/L by body Osm) (it worked out in the exam!)
E. None of the above
Total Plasma Osmolarity can be calculated by several formulae, one of which is:
Osmolarity (mOsmol/kg) = [2 x (Na + K)] + urea + glucose
according to fluid website ( link below): Each mOsm/kg of solute contributes about 19.32mmHg to the osmotic pressure (derived from van’t Hoff eqn) So maybe close enough for 19.20 to be correct?
Addition: the wording in the question is a bit confusing. If the answer is D the question must have been asking about total plasma osmotic pressure. Osmotic pressure can be measured (eg oncometer) or calculated using the Vant Hoff equation (ref Physiology viva, KB). Osmolarity can be measured (eg freezing point depression) or calculated: Osmolarity (mOsmol/kg) = [2 x (Na + K)] + urea + glucose
Answer: Depends on the wording of the question
If question says:
Osmolality: E - none of the above - osmolality can never be calculated
Osmolarity: E - none of the above - an appropraite formula may be = 2(Na+K) + glucose + urea, for example
Osmotic pressure: D - mOsm x 19.3 = osmotic pressure (in mmHg). Or could use Hoff’s Law: π = cRT JB2012
FE16 [Apr01] Which of the following will increase plasma potassium concentration A. Beta adrenergic receptor AGONIST B. Insulin C. Aldosterone D. hemolysis E. None of the above
Reduce plasma potassium
Beta adrenergic receptor AGONIST
Insulin
Aldosterone
Hyperglycaemia
Carbonic anhydrase inhibitors
Norepinephrine and epinephrine cause an initial rise via liver release and then a prolonged fall in plasma potassium due to skeletal muscle (via B2 adrenergic receptors) Ganong pg 360
Beta receptor agonism causes increased movement of potassium intracellularly, presumably due to increased Na/K ATPase activity, as well as increasing insulin release from the pancreas (which has a similar effect). Think salbutamol in hyperkalaemia.
Increase plasma potassium
Metabolic acidosis
http://www.aic.cuhk.edu.hk/web8/Potassium%20metabolism.htm
and haemolysis - think haemolysed sample K levels in U+E’s.
FE16b [Ju05] Which will increase plasma [K+]? A. Hyperglycaemia B. Aldosterone C. Metabolic acidosis D. diuretics E. Carbonic anhydrase inhibitors
Reduce plasma potassium
Beta adrenergic receptor AGONIST
Insulin
Aldosterone
Hyperglycaemia
Carbonic anhydrase inhibitors
Norepinephrine and epinephrine cause an initial rise via liver release and then a prolonged fall in plasma potassium due to skeletal muscle (via B2 adrenergic receptors) Ganong pg 360
Beta receptor agonism causes increased movement of potassium intracellularly, presumably due to increased Na/K ATPase activity, as well as increasing insulin release from the pancreas (which has a similar effect). Think salbutamol in hyperkalaemia.
Increase plasma potassium
Metabolic acidosis
http://www.aic.cuhk.edu.hk/web8/Potassium%20metabolism.htm
and haemolysis - think haemolysed sample K levels in U+E’s.
E17 [Apr01] Osmotic pressure in plasma is usually 1.6 mosmol/L more than ISF. This is because of: A. Plasma Proteins B. Plasma Oxygen Tension C. Plasma creatinine D. ? E. ?
This CANNOT be “osmotic pressure” as ISF and plasma have the same osmolality so must have the same osmotic pressure. The stem must be about “oncotic pressure”. Additionally, osmotic pressure has pressure units (eg mmHg) and NOT concentration units as here.
Disagree, the above statement is wrong.
Guyton and Hall “textbook of medical physiology” (11th ed 2006)- sites the difference between as being about 1mosmol/L and that plasma proteins is the cause of this.
States that it is responsible for about 20mmHg greater pressure in the capillaries than in the interstitial fluid.
FE18 [Apr01]- [Mar03]- [Jul03]- [Feb04]- [Jul04]- Aug14
(Responses to ?increased osmolarity)
A. ?Thirst and ADH from stimulation of osmoreceptors in posterior hypothalamus
B. ?Thirst via stimulation of SFO and OVLT via Angiotensin II in hypovolaemia
C. Baroreceptors afferents to the Posterior Pituitary
D. Increased ADH levels
E. Aldosterone
Clearly these options have been fairly poorly recalled. I think the answers should be: 1st version:
A) wrong (anterior hypothalamus not posterior)
B) true
C) false (baroreceptor afferents go to medullary vasomotor centre)
D) & E) false as ADH and aldosterone do not directly affect thirst although these are both TRUE if the question really is “response to osmolarity” as they are both responses to this
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
[03A] Increases in plasma osmolarity in a healthy young person produce:
A. Production of ADH from posterior pituitary
B. Thirst via ADH effect on paraventricular nuclei
C. ….. angiotensin?
D. ?
Alt version 03A
A) false - ADH produced from hypothal, secreted from post pituitary
B) see above
C) ?? true depending on wording. Thirst is in part regulated by angiotensin
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
In hypovolaemic shock, thirst is triggered via:
A. Angiotensin II acting on the circumventricular organs
B. ?
Alt version:
A) true (subfornical organ & OVLT are cirumventricular organs)
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
[04A] Thirst is stimulated by:
A. Release of angiotensin II
B. Supraoptic nuclei
Re 2004B version:
A) false
B) true
C) false
D) false (see above for all)
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
04B Thirst in hypovolaemia from:
A. Stimulation of baroreceptors which stimulate posterior pituitary
B. Angiotensin II stimulating SFO and OVLT
C. increased ADH levels
D. Aldosterone
(?), osmoreceptor stimulation not an option).
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
14B Thirst is stimulated by
A. reduction in volume
B. ANP
C. increased osmolarity sensed by posterior pituitary
D. Angiotensin II inhibiting the subfornical organ
Re Aug14 version: A - T B - F - C - F - sensed by hypothalamus D - F - ATII stimulate SFO and OVLT
Guyton & Hall referenced below, p325&326, identify a number of important factors in stimulating thirst:
Increased osmolarity is probably the most important. Stmiulates thirst by intracellular dehydration of cells in the thirst centre
Drop in blood volume, and blood pressure- as in haemorrhage, probably via baroreceptors. Note that the response exist independent of osmolarity changes
Increases in ADH
Dry mucous membranes
Along with thirst being stimulated by osmotic pressure of plasma and decreased ECF volume, Ganong also mentions thirst is stimulated by psychological factors, thought to underly “prandial drinking” (intake of liquids while eating), as well as the dryness of mucous membranes mentioned above.
It mentions increased plasma osmolality acts via osmoreceptor located in the anterior hypothalamus
Decreased extracellular fluid volume (ie hypovolaemia) acts via baroreceptors and angiotensin II. Angiotensin II acts via the subfornical organ and OVLT.
The central controller for water balance is the hypothalamus but there is no single anatomically defined ‘center’ which is solely responsible for producing an integrated response to changes in water balance. There are numerous pathways interconnecting the various centers or areas in the hypothalamus. The osmoreceptors are located in the area known as the AV3V (anteroventral 3rd ventricle). Lesions in the AV3V region in rats cause acute adipsia.
The thirst centre is located in the lateral hypothalamus. It receives input from osmoreceptors in the AV3V region and from the subfornical organ and the organum vasculosum of the lamina terminalis (OVLT) which are sites for angiotensin II action. The OVLT is in the AV3V region.
ADH is formed predominantly in the neurones of the supraoptic and paraventricular nuclei. These nuclei receive input from the osmoreceptors and also from ascending adrenergic pathways from the low and the high pressure baroreceptors. Aquaporin 4 has recently been identified in cells in the hypothalamus particularly in the paraventricular and supraoptic nuclei.
The key parts of the hypothalamus involved in water balance are:
Osmoreceptors
Thirst centre
OVLT & SFO (respond to angiotensin II)
Supraoptic & paraventricular nuclei (for ADH synthesis)
Regarding whether angiotensin II affects SFO or OVLT osmoreceptors or BOTH to stimulate thirst:
Ganong says there is evidence that OVLT, not just SFO, are stimulated by ATII to produce thirst
But this study [1] suggests it’s only SFO that stimulates thirst
However Ganong is a prescribed text and therefore correct in exam parlance
Also, regarding the primary thirst-stimulating mechanism in hyperosmolar states
Ganong states that thirst is due to (a) osmoreceptors and (b) ATII
The response due to osmolarity appears to be the primary response
The OVLT and SFO appear to be in the anterior and lateral but not posterior hypothalamus
Referenced from Ganong 21st ed p244
FE19 [Apr01] [Feb04]
Sweat in patients acclimatised to hot weather (as compared to patients in
a temperate climate) contains less Na+ because:
A. Takes longer for Na+ to be transported through sweat ducts
B. Aldosterone effect causing a reduction in Na+ in sweat
C. Increased intake of water causing a reduction in Na concentration
D. ?
E. ?
B
FE20 [Jul01]
Magnesium is required for:
A. To Depolarise excitable cell membranes B. Na-K ATPase C. Coagulation D. ? E. ?
FE20
From ‘The Physiology Viva’ by Kerry Brandis (p17):
Mg++ is the major intracellular divalent cation.
Intracellular function
Catalysing Mg++ dependent enzymes
all enzymes for phosphate transfer
all enzymes requiring thiamine phosphorylase as a co-factor (therefore Na+/K+ pump, oxidative phosphorylation, all reactions involving ATP)
acts as ‘plug’ in NMDA receptors ->voltage-dependent block of channel.
Extracellular function
Reduces nerve and membrane excitability (similar to but
Magnesium is essential for: A. muscle contraction B. cofactor in Na/K/ATPase C. something about bone D. ? E. ?
FE20b
B. Correct - cofactor in Na/K/ATPase
FE21 [Jul01] Intracellular ?osmolality is greater than interstitial ?osmolality because:
A. Proteins in plasma
B. Cells producing intracellular proteins
C. ?
D. ?
E. ?
B: Gibbs Donnan effect: ‘because of proteins in cells, there are more osmotically active particles in cells than in interstitial fluid, osmosis would make them swell and rupture if not for Na/K/ATPase pumping ions back out’ (p6, Ganong 20thed)
FE22 [Mar02] [Jul02] [Mar03] [Jul03] Mar10
During sweating in strenuous exercise, the [Na+] in sweat is:
A. Less than plasma
B. Equal to plasma
C. More than plasma
D. ?
E. ?
Alt stem:
Regarding sweat osmolality during exercise (repeat):
As far as I can find:
At rest loss of [Na] is approximately 11 mmols/l by sweating (physiology for the anaesthetist). However this can vary from 30 to 65 mmols/l depending on the level of acclimatisation. Aldosterone decreases the [Na+] of sweat.
This concentration is less than plasma [Na] concentration of 140 mmols/l Therefore A is correct
FE23 [Mar03] [Jul03] [Feb04] [Jul05]
Acute onset (4 hours) diabetes insipidus in an otherwise healthy person produces these biochemical changes
(“these numbers may not be exact”):
A. Na+ 130, K+ 3.0, Osm 260 B. Na+ 130, K+ 4.0, Osm 300 C. Na+ 150, K+ 3.0, Osm 260 D. Na+ 150, K+ 3.5, Osm 320 E. Na+ 160, K+ 3.0, Osm 320
“For the DI question there were a set of normal electrolytes as an option too - which is missing from the bank thus far (the problem with the discussion previously on the Bulletin Board ie the most ‘normal’ set of electrolytes were hyponatraemic/hypoosmotic - if your thirst mechanism is intact and you have access to water you have normal electrolytes but tend towards hypernatraemia/hyperosmolality. The actual numbers in the MCQ were Na 140 K 3.5 Osm 300. (One of the options also had a K of 6.0!)”
“I must admit I still answered normal electrolytes in spite of the ‘untreated’ bit in the Q since drinking is part of the normal physiology of DI (most DI’s are not the head injured ventilated pt’s we tend to see in ICU but rather the compensating DI’s in renal clinics)… ie I regarded drinking lots as normal physiology for DI, not a treatment for it.”
Comment Ganong 22nd ed pg 247 states 30% are neoplastic, 30% post-traumatic, 30% idiopathic, remainder vascular lesions, infections, systemic diseases eg sarcoidosis. Surgical removal of post hypothesis does not guarantee permanent DI as suprooptic and paraventricualr fibres can regrow.
Contrary View:
Plasma osmolality is high, Urine osmolality is low,Plasma sodium may be high, Dehydrated.
Case report notes K high: http://adc.bmj.com/cgi/content/full/80/6/548
Perhaps D?
- I agree that the blood results would likely be abnormal. In a normal physiological state, there is ADH, thirst, ATII, etc, etc, so I think even if you were able to drink water, without one of the other fine-controlling mechanisms you would have at least a partially abnormal end result
Oh´s Intensive Care Manual:
“The plasma osmolality measures in central DI is usually in the higher regions of the normal range or very slighltly supranormal” “Hyperosmolality or hypernatraemia suggest impaired sensation of thirst or inabilty to access water”–Kajuprim 20:52, 26 Feb 2010 (EST)
FE24 [Mar03] [Jul03]
Colligative properties:
A. Increase BP, decrease freezing point, decrease SVP
B. Other combinations: increase/ decrease…boiling point/FP/SVP
C. ?
The Colligative properties of a solution are those properties that depend ONLY on the particle concentration (number of solute particles per unit volume) - and NOT on the chemical properties of the substance or size of the particles.
(as concentration of particles increases)
These properties are:
Freezing Point Depression
Boiling Point Elevation
Vapour Pressure Depression – reduction of the solvent molecules ability to leave the solution
Osmotic pressure
The inorganic ion necessary in Na+-K+ ATPase is: A. ? B. ? C. Mg+2 D. PO4 E. SO4-2
Mg++ is an essential co-factor for the Na+-K+ ATPase (cell membrane sodium pump).
Perhaps options A or B were an ion of ATP?
FE26 [Jul04] [Mar05] [Jul05] A patient is given an infusion of 100 mL of 8.4% sodium bicarbonate solution. This represents an osmotic load of: A. 42 mosmol B 84 mosmol C 100 mosmol D 168 mosmol E 200 mosmol
Firstly: The MW of NaHCO3 is 84 so a one molar solution would contain 84 grams in a liter (ie 8.4 grams in a 100mls which is an 8.4% solution).
Secondly: Sodium bicarbonate splits in two ions on dissolving: Na+ and HCO3-.
So an 8.4% solution of sodium bicarbonate contains 1,000mmols of Na+ and 1,000mmoles of HCO3-. The osmolality is this 2 Osm/litre (or 2,000 mOsm/litre).
A 100mls of this solution is an osmotic load of 200 Mosmoles.
FE27 [Mar05]
Regarding the ECF concentrations of K+ and H+:
A. K+ rise causes pH rise
B. They move in the same direction
C. ?
D. Hypokalaemia inhibits renal H+ excretion
E. ?
FE27
Answer = B. (don’t think this is right- hyperkalaemia doesn’t cause alkalosis- can’t remember where I read from but it says something along the lines that in an attempt to correct acidosis, the cells will take H+ in in exchange for a K+ ion- hence acidosis is a/w high K+.)
Metabolic acidosis leads to hyperkalaemia due to compartment shifts. K+ shifts out of cells causing [K+] increase by 0.6mmol/L per each pH decrease by 0.1 (effect is greater with non-organic acids eg KCl - Cl is obligatory extracellular anion so as H+ taken up by cells, K+ must move out to maintain neutrality)
K+ depletion stimulates H+ secretion however hypokalaemia decreases aldosterone release which decreases H+ secretion. Combination of hypokalaemia and secondary hyperaldosteronism synergistically to stimulate H+ secretion causing metabolic alkalosis.
Ref Mark Finnis notes
Interrelation of potassium and hydrogen ion gradients in metabolic alkalosis Jared J. Grantham 1 and Paul R. Schloerb 1
To study the interrelationships of potassium and hydrogen ion intra- and extracellularly, as an extension of previous investigations on gastric alkalosis, metabolic alkalosis was produced in dogs by dietary K depletion, NaHCO3 loading, and deoxycorticosterone administration, with and without chloride depletion. Analysis of skeletal muscle and measurement of intracellular pH confirmed the development of cellular K depletion, with increases of intracellular sodium and hydrogen ion associated with extracellular alkalosis and hypokalemia. These changes were not influenced by concomitant dietary chloride restriction. The intracellular K concentration was linearly and inversely related to the intra-extracellular hydrogen ion gradient. The intra-extracellular K gradient was about 10 times greater than the corresponding hydrogen ion gradient. These respective gradients were linearly related, in this proportion, over the range of K depletion studied in this type of metabolic alkalosis as well as that resulting from gastric juice loss. Deviation from this relationship occurred only with the most severe K depletion. These experimental observations demonstrate in vivo that the potassium and hydrogen ion gradients change in the same direction and are nearly constantly related in at least two types of metabolic alkalosis.
Kerry says
Acidosis is commonly said to cause hyperkalaemia by a shift of potassium out of cells. The effect on potassium levels is extremely variable and indirect effects due to the type of acidosis present are much more important. For example hyperkalaemia is due to renal failure in uraemic acidosis rather than the acidosis. Significant potassium loss due to osmotic diuresis occurs during diabetic ketoacidosis and the potassium level at presentation is variable (though total body potassium stores are invariably depleted).
FE27b [Feb11] Regarding potassium and hydrogen: A. both go in the same direction. B. acidosis increases potassium loss. C. insulin affects the interaction between potassium and H+ D. hypokalaemia inhibits acid secretion E. ?
FE27b
A. Partly correct - plasma H+ and K+ are associated through an unknown mechanism, possibly modulating the cellular Na-K-ATPase, but this is not a direct linear correlation as implied in the option
B. Partly correct - acidosis increases plasma concentration and renal filtration of potassium, but low intracellular pH inhibits Principal cell capacity to secret K+ (paradoxical potassium retention) “These topics are fraught with difficulty because the effects are not consistently seen.”
C. Wrong - insulin increases cellular uptake of potassium (important for plasma potassium regulation with meal boluses), but does not have a direct affect on the interaction between potassium and H+
D. Wrong - hypokalaemia stimulates H+ excretion and HCO3- generation in tubular cells, itself generating a metabolic alkalosis
E. ?
This is a problematic topic, but my “most correct” answer would be: A
FE28 [Jul05] Hyperkalemia caused by: A. Metabolic acidosis B. Aldosterone excess C. ? D. ?
A is correct, B is false (aldosterone causes Na retention and K excretion)
From Vander p147:
Low plasma pH is usually associated with hyperkalaemia. There are 2 known reasons for the effects of acid base status on potassium.
changes in extracellular [H+] cause exchange of H+ with cellular cations (mainly K+). Acidosis results in H+ being taken up by cells which is balanced by efflux of K+ from the cell.
intracellular pH inhibits Na-K-ATPase contributing to K+ loss from cells and an inability of cells to uptake extracellular K+. Principal cells have their Na-K-ATPase pumps and luminal membrane K+ channels inhibited which results in paradoxical K+ retention
From Vander p141:
Aldosterone stimulates K+ secretion by the principal cells by 3 mechanisms:
activation of apical K+ channels (ROMK).
stimulation of basolateral membrane Na-K-ATPase pumps which increases intracellular K+ and the gradient driving K= secretion into the lumen.
increases the activity or number of luminal membrane Na channels which allows more Na to enter the principal cell to then be pumped out by the Na-K-ATPase basolaterally (K+ secretion is dependent on Na+ being pumped out so more Na+ entering the cell allows more K+ to leave)
FE29 [Feb06]
The rate of diffusion across semipermeable membrane:
A. is inversely proportional to thickness
B. is proportional to molecular weight
C. ?
D. ?
E. ?
Answer = A
Rate of diffusion = K. A. (P2-P1)/D
Fick’s law states that the rate of diffusion of a gas across a membrane is:
Constant for a given gas at a given temperature by an experimentally determined factor, K
Proportional to the surface area over which diffusion is taking place, A
Proportional to the difference in partial pressures of the gas across the membrane, P2 − P1
Inversely proportional to the distance over which diffusion must take place, or in other words the thickness of the membrane, D.
The exchange rate of a gas across a fluid membrane can be determined by using this law together with Graham’s law.
I looked at the formula and this is not what I got used to from J West book:
Diffusion through a tissue sheet= A x D x ( P1-P2)/ T where A - area D - diffusion constant ( P1- P2) the difference in partial pressure T - thickness
Fick’s Law of diffusion: - rate of diffusion of a gas through a tissue slice is proportional to the area but inversely proportional to the thikness. - diffusion rate is proportional to the partial pressure difference - diffusion rate is proportional to the solubility of the gas in the tissue but inversely proportional to the square root of the molecular weight.
So… A is correct, B is incorrect - diffusion rate inversely proportional to the square root of the MW
FE30a [Jul07] Infusion of 40ml/kg of 0.9% saline solution will cause: *new* A. Hypochloraemic metabolic acidosis. B. Hypochloraemic metabolic alkalosis. C. Hyperchloraemic metabolic acidosis. D. Hyperchloraemic metabolic alkalosis. E. No acid base disturbance.
c
N/saline infusion causes a hyperchloraemic metabolic acidosis.
FE30b ANZCA version [Apr08]
The use of large amounts of normal saline for patient resuscitation is associated with:
A. hyperchloraemic acidosis B. hyperchloraemic alkalosis C. hypernatraemic acidosis D. hypernataemic alkalosis E. serum hyperosmolarity
A
FE31 [Jul07]
Lymph flow:
A. greatest when skeletal muscle contracting
B. when interstitial pressure 1-2mmHg above atmospheric
C. approx. 1000ml per hour via thoracic duct
D. ?
E. ?
Answer:
?A Ganong states “assisted when muscles contract” (pg 593)
B: This is approximately the interstitial pressure and a higher pressure would reduce lymph formation by decreasing the hydrostatic pressure gradient across the capillary membrane.
C: Flow 100mls/hr (in ref)
120 ml/hr total lymph flow at rest includes 100 ml/hr of thoracic duct lymph
Interstitial hydrostatic pressure usually b/w 0 and -3mmHg - Guyton
FE32 [Jul07]
Post-thoracotomy the drain is leaking fluid with protein, fat, lymphocytes etc. What could be the cause?
A. Bleeding
B. Thoracic duct injury
C. sympathectomy
D. Pleural fluid
E. ?? “something like CHF or pulmonary oedema”
Alt stem: “Post-thoracotomy the drain is leaking fluid that contains protein, coagulation factors,
with a high fat & lymphocyte count”
B
FE33 [Feb08] Hyponatraemia is usually due to: A. Excess lipids B. Excess glucose C. Free water deficit D. Excess protein E. Free water excess
E. Free water excess
Both excess lipids and excess glucose can cause a pseudohyponatraemia but the most common cause is probably water excess.
Hypertonic fluid is used in resuscitation for:
A. increase in total body sodium
B. reduction in viscosity
C. improve coagulation
D. reduce intracellular oedema
E. rapid expansion of intravascular volume
Answer E
FE35 Feb12
Chronic hypokalaemia (3.0 mM) will cause which ECG changes?
A. Flat p waves
B. Flat T waves
C. Cardiac arrest in diastole
D. More prone to arrhythmia than acute hypokalaemia
E. Resting membrane potential will be higher?
A - wrong - not great reference but http://lifeinthefastlane.com/ecg-library/basics/hypokalaemia/
B - seems most correct of these options
C - wrong - this happens in hyperkalemia - Ganong pp566 Ed 21
D - probably wrong as chronic electrolyte changes are usually better tolerated than acute
E - wrong - “RMP in hyperkalemia decreases” Ganong
Generally, these questions are difficult to reference to source texts…
FE36 Feb12
With regards to chloride:
A. Hyperchloraemia leads to decreased plasma HCO3
B. Intracellular concentration is
FE36
A: not sure.. hyperchloremia usually ends up with a decreased serum bicarb - normal anion gap metabolic acidosis, but is it causal?
Under the Stewart acid-base system, hyperchloraemia ought to decrease the Strong Ion Difference of plasma (usually about 40 mEq/L), reducing bicarbonate concentrations (a weak ion), and so cause a metabolic acidosis. The problem is that this assumes an isolated hyperchloraemia, which is a rare event, and we don’t have complete blood chemistry to really comment. Hard call.
B: is true, value is 9 mM from Ganong Ed 21 pp 8
This is true, but higher in erythrocytes (~77 mM). Assuming this question was correctly recalled,
B seems the “most correct” answer.
With regards to chloride:
A. ? changes in direct proportion to bicarboate
B. it is the major cation extracellularly
C. is a weak base
D. ?
E. Intracellular concentration
FE36b A. Partly correct - per A above B. Wrong - Cl- is an anion (but is the major anion extracellularly) C. Wrong - subject to the acid-base definition being employed, it is a strong base (taken to be "H+ acceptor" under the Brønsted-Lowry definition) D. ? E. Partly correct - per B above "Most correct" answer: E References
FE37 [Feb12] A person with undiagnosed adrenocortical insufficiency will have the following electrolyte profile: A. Na 122 K 6.2 Cl 72 HCO3 40? B. Low Na, High K, Low Cl, Low HCO3 C. Low Na, High K, Low Cl, Normal HCO3 D. Low Na, High K, Low Cl, Raised HCO3 E. High Na, Low K ??
Not clear whether this MCQ had example values, or was just high,low, normal options
This is a table straight from Ganong: In adrenal insufficiency
Na low
Cl low (goes with Na presumably)
K high (hypoaldosteronism so Na wasted, K retained)
HCO3 unchanged
Ganong pp380, Ed 21
FE38 Feb12 Long PR interval, ST depression, T wave inversion and U wave is caused by which electrolyte abnormality? A. Hypokalaemia B. Ca++ C. Na+ D. ? E. ?
A
FE39 [Mar10]]
Which is true regarding colloids?
A. dextrans stay in the circulation for 6-8hrs
B. gelatins have a higher rate of anaphylaxis than starches
C. ?? has greater effect on coagulation than ??
D. do not depend on renal clearance for excretion
E. ?
Dextran 40 has been shown to improve microcirculation presumably by decreasing blood viscosity thereby improving laminar flow in the microcirculatory beds. Both Dextran 40 and Dextran 70 possess antiplatelet effects and may interfere with blood typing. (Nagelhout pg. 411, M&M pg. 694)
References
FE39b [Feb13] Probably a different MCQ on the same topic
Colloids:
A. ?
B. ?
C. HES is completely excreted by the kidney
D. Dextran 40 is used to improve micro-circulatory flow
E. ?
Dextran 40 has been shown to improve microcirculation presumably by decreasing blood viscosity thereby improving laminar flow in the microcirculatory beds. Both Dextran 40 and Dextran 70 possess antiplatelet effects and may interfere with blood typing. (Nagelhout pg. 411, M&M pg. 694)
References
FE40 [Aug14] Which organ has no lymphatic vessels? A. Brain B. Stomach C. Kidney D. Heart E. Lung
?
FE41 14A What is true for Magnesium (Mg++?: A. normal plasma level is 1.1 - 2.2 B. slows sinoatrial conduction C. increases uterine tone D. is a diuretic E. is a direct respiratory depressant
A - False - Normally 0.7-1.1mmol/L
B - True - Slow sinoatrial conduction/increases SA node recovery time [1]
C - False - Mg2+ decreases uterine tone
D - False - No diuretic effects
E - False - Respiratory depression at toxic doses (>3.0mmol/L) due to diaphragm paralysis from neuromuscular blockade (not direct)
AD01 [Mar96] [Apr01] [Mar05] [Jul05]
ABGs: pH 7.35, pCO2 60 mmHg, pO2 40 mmHg.
These blood gas results are consistent with:
A. Atelectasis
B. Morphine induced respiratory depression (OR: Acute morphine overdose)
C. Diabetic ketoacidosis
D. Patient with COAD
E. Lobar pneumonia (OR: bronchopneumonia)
F. Metabolic acidosis
(Alt version of the gas results: pH 7.35; pO2 45mmHg;
pCO2 60mmHg; HCO3- 34mmol/l)
Ans D.
I think answer D is incorrect.
PO2 of 40mmHg correlates with oxygen saturations of 75%. While this fits some patients who need oxygen therapy, it does not fit with the vast majority of COPD patients. I think the most correct answer is B (morphine overdose).
Nope, not B, there is a compensated respiratory acidosis, has to be longer term than morphine OD. Ans IS D
I agree that the answer is D, and I imagine that the HCO3 was probably included in the question. The fact that the pO2 is so low really rules out an acute morphine overdose(particularly if in conjunction with metabolic compensation HCO3 34).
Using the “4 for 10” rule for Chronic Resp Acidosis and the presumptive HCO3 of 34. Expected HCO3 = 24 + 4[(60-40/10)] = 32. Close to what we have.
I dont think B is correct. Using alveolar equation with Pco2 of 60, will give much higher Po2.
Assuming otherwise normal lung function, morphine OD alone will nit cause this picture
The ABGs of a 60yr old man who has overdosed on morphine would be:
A. paO2 60, paCO2 55, pH 7.29, HCO3 32, BE -1
B. paO2 40, paCO2 60, pH 7.37, HCO3 26, BE +5
C. ?
I think the correct answer is A.
Morphine overdose would cause acute hypoventilation with hypoxia (paO2 60) with an acute respiratory acidosis (pH7.29, paCO2 55). Renal compensation will take some time with increases in plasma HCO3. (BE within +/-1).
B is incorrect because it shows a fully corrected respiratory acidosis with a positive base excess which usually occurs in alkalosis not acidosis. This does follow the 1 in 10 rule i.e for each 10mmHg increase in CO2 there will be 1mmol increase in HCO3…but is probably still wrong because the base excess should be negative.
I disagree - both show a primary respiratory acidosis for every 10mmHg of CO2, the HCO3 will compensate by 1 acutely… thus in A, it has actually compesated by 8 which doesn’t make sense in B, the CO2 is 20mmHg above 40, so you’d expect the bicarb to rise by 2, which it has.
Therefore either the person in A already has some primary metabolic alkalosis before the event, or the numbers got remembered wrong (which is more likely)
Both of these stems seem incorrect to me. The base excess clearly has been remembered incorrectly as it should reflect the bicarb so for option A), the base excess should be about +7 and for option B) the base excess should be about 0 +/- 1. Option A) is incorrect as usually morphine overdose is acute so there should not be this much metabolic compensation by the kidneys as yet. Option B) is incorrect because although the patient should have an acute respiratory acidosis, we would expect the pH to be lower than this for a CO2 of 60 and the base excess doesn’t make any sense.
Answer should be: hypoxic with acute, uncompensated respiratory acidosis and a base excess that reflects the bicarb - so neither!!
Addit comment: If you calculate the Standardised Base Excess (SBE) from the data in option B it is:
SBE = 0.93[26]+13.77[7.37]-124.58 (wikpedia in mEq/L) = +1.08
Which is correct for the situation of acute respiratory acidosis. Option B therefore fits a whole lot better (for what its worth).
response to the 2nd comment above - a compensatory process does not move the pH into the normal range - in this case there must be a second acid-base disorder going on, and given the BE of +5, it’s likely that this person would also have an underlying metabolic alkalosis, allowing the pH to be 7.37 in the face of a respiratory acidosis.
Agree neither answer seems to fit. This should be an acute respiratory acidosis.
Power & Kam re: in vivo titration curves - for every increase pCO2 by 10mmHg, corresponding decrease in pH by 0.07 (gives 7.29 for CO2 55, 7.26 for CO2 60), increase in HCO3- by 0.08 (gives 25 for CO2 55, 26 for CO2 60). BE should be negative.
AD03 [Jul97] [Mar99]
Buffering of a bicarbonate infusion:
A. 60 to 70% occurs intracellularly
B. Exchanged for Cl- across the red cell membrane
C. Compensated for by increased respiratory rate.
D. Intracellular proteins
Please Note: I took this question to mean how a bicarbonate infusion would buffer a metabolic acidosis when given for that clinical indication. Apologies if I’ve misinterpreted it.
Notes from the Acid-Base Physiology section of Kerry Brandis’ Physiology Book, also on-line at: http://www.anaesthesiamcq.com/AcidBaseBook/ab2_2.php
Answer A: NO Metabolic acidosis: 57% of buffering occurs intracellularly and 43% occurs extracellularly (Of the 43% EC component, 80% is performed by the bicarbonate buffer system). Metabolic alkalosis: 30% IC, 70% EC Respiratory acidosis: 99% IC, 1% EC Respiratory alkalosis: 97% IC, 3% EC
Answer B: YES
Accounts for about 6% of buffering in metabolic acidosis
“Protein buffers in blood include haemoglobin (150g/l) and plasma proteins (70g/l). Buffering is by the imidazole group of the histidine residues which has a pKa of about 6.8. This is suitable for effective buffering at physiological pH. Haemoglobin is quantitatively about 6 times more important then the plasma proteins as it is present in about twice the concentration and contains about three times the number of histidine residues per molecule.” (sic)
Answer C: YES
The body’s ability to change alveolar ventilation (“physiological buffering”) rapidly is what makes the system such a good buffer despite the pKa being low at 6.1.
Answer D: “YES”
ICF buffering (57%) occurs by protein phosphate and bicarbonate buffers due to entry of H+ by:
Na+-H+ exchange 36%
K+-H+ exchange 15%
Other 6% (mostly Cl-)
Comment. An infusion of HCO3- will cause a metabolic alkalosis; there will be renal and respiratory compensation, as well as buffering. Compensation will take the form of renal excretion of HCO3- and a respiratory acidosis. Thus RR will decrease. Therefore C is incorrect. In addition, B is less correct than D, leaving D as the correct answer assuming that the question is remembered correctly. Reference: Brandis
Alternate View
If interpreted as the buffering for a bicarbonate load, i.e. metabolic alkalosis, then compensation is CO2 retention:
Hence, (A) ~70% will occur in ECF (B) Enters RBC through Cl- exchange and forms CO2 and H2O (C) Decreased respiratory rate and (D) Is incorrect.
If above is true then correct answer is B. Note that respiratory compensation is not the same as buffering to produce CO2.
Addit notes: Bicarbonate formed by CO2 passage into the RBC is 70% exchanged for Cl- with plasma. The buffering of Bicarbonate therefore happens intracellularly through the catalyst carbonic anyhydrase. Excess H is buffered by proteins and Hb and there is a significant pH change between venous and arterial blood, a process with is reversed when in the lungs and CO2 is lost to atm.
AD04 [Jul97] [Mar98] [Mar99] [Feb00] Mar10
Phosphate buffer system is an effective buffer intracellularly and in renal tubules because:
A. Its pKa is close to the operating pH
B. High concentration in distal tubule
C. High concentration intracellularly
D. All of the above
For a buffer to be effective in a given location you need:
A high enough concentration of it must be present
the buffer’s pKa to be within 1 pH unit of the local pH
The acid to be buffered cannot be itself
A high enough concentration of it must be present.
The phosphate concentration is high enough in urine and ICF to be an important buffer there. In comparison, in ECF, the phosphate concentration is too low for it to provide much buffering (despite meeting the other requirement for buffer effectiveness ie pKa within +/- 1 pH unit of prevailing pH)
The buffer’s pKa to be within 1 pH unit of the local pH
Actually this rule is for ‘closed’ buffer systems - which is the common situation. If there is an ‘open’ buffer system then the buffer can be effective wuith a larger ph-pKa difference - as occurs with the bicarbonate buffer which buffers in plasma with a pH of 7.4 despite having a pKa of 6.1.
For phosphate buffer system, the relevant reaction is the conversion between monohydrogen phosphate and dihydrogen phosphate. This reaction has a pKa of 6.8 and so is within the required range (of 7.4 +/- 1). In plasma and ISF however the phosphate concentration is just too low to actually contribute much buffering.
The acid to be buffered cannot be itself
For example, the bicarbonate buffer cannot buffer a respiratory acidosis because it cannot buffer itself. Consequently most of the buffering of respiratory acid-base disorders (97-99%) occurs intracellularly on protein and phosphate buffers.
Mar 2010 version:
Why is phosphate such a good buffer in ICF and urine?
A. ICF has lower pH than ECF (I’m pretty sure this was ICF has higher conc of phosphate than ECF actually)
B. tubular pH is low
C. pKa is close to pH
D. concentration is high in urine and ICF (don’t remember them asking about conc in the urine)
E. all of above
For a buffer to be effective in a given location you need:
A high enough concentration of it must be present
the buffer’s pKa to be within 1 pH unit of the local pH
The acid to be buffered cannot be itself
A high enough concentration of it must be present.
The phosphate concentration is high enough in urine and ICF to be an important buffer there. In comparison, in ECF, the phosphate concentration is too low for it to provide much buffering (despite meeting the other requirement for buffer effectiveness ie pKa within +/- 1 pH unit of prevailing pH)
The buffer’s pKa to be within 1 pH unit of the local pH
Actually this rule is for ‘closed’ buffer systems - which is the common situation. If there is an ‘open’ buffer system then the buffer can be effective wuith a larger ph-pKa difference - as occurs with the bicarbonate buffer which buffers in plasma with a pH of 7.4 despite having a pKa of 6.1.
For phosphate buffer system, the relevant reaction is the conversion between monohydrogen phosphate and dihydrogen phosphate. This reaction has a pKa of 6.8 and so is within the required range (of 7.4 +/- 1). In plasma and ISF however the phosphate concentration is just too low to actually contribute much buffering.
The acid to be buffered cannot be itself
For example, the bicarbonate buffer cannot buffer a respiratory acidosis because it cannot buffer itself. Consequently most of the buffering of respiratory acid-base disorders (97-99%) occurs intracellularly on protein and phosphate buffers.
AD05 [Jul97] [Apr01]
Arterial gases including pH 7.46 bicarbonate 31mmol/l PCO2 46mmHg indicate:
A. Metabolic alkalosis with respiratory compensation
B. Respiratory alkalosis
C. Respiratory acidosis with compensation
D. Metabolic acidosis with respiratory compensation
E. Mixed metabolic and respiratory alkalosis
.
(Apr 01: AB05 changed so 2 top stems read partially compensated then bottom stem was “none of the above”)
Correct Answer is A.
Metabolic alkalosis as pH is higher than 7.4 and HCO3 is high. There is some respiratory compensation as pCO2 > than 40mmHg.
OR
Alkalaemia as pH is greater than 7.4
Metabolic as HCO3 is greater than normal.
Rule of thumb 6. (0.7 x HCO3) + 20 = pCO2 (+/- 5mmHg) Here = 42
Within range of appropriate respiratory compensation. Therefore most likely = primary metabolic alkalosis with respiratory compensation.
The changed stem is more difficult. “Partial compensation” of A is as close as “none of the above”.
Alternative: Boston Rules:
pH 7.46 pCO2 46mmHg HCO3 31 mM/L
Step 1 pH>7.44, so alkalosis.
Step 2 Both pCO2 (35-45) and HCO3 (22-28) are high:
This suggests metabolic alkalosis OR respiratory acidosis
But, we already know it is an alkalosis.
Step 3 Look for clues - none provided (eg: AG, Crea, Glucose, etc.)
Step 4 Assess compensatory response using The Boston Rules:
Rule 6: The Point Seven Plus 20 Rule
Expect pCO2 = 0.7 [HCO3] + 20 (range: +/- 5)
= 0.7 x 31 + 20 = 41.7 +/- 5
= TRUE! Thus, respiratory compensation plausible.
Step 5 Metabolic alkalosis.
AD06 [Jul98] Metabolic acidosis is characterised by: A. Increased [H+] intracellularly B. Decreased production of bicarbonate C. hypoalbuminaemia D. ? E. ?
The answer is A
Metabolic acidosis -> exchange of H+ for K+ across the cell membrane -> increased intracellular H+. Bicarbonate production is INCREASED
also: hypoalbuminaemia can cause a metabolic alkalosis
UNLESS, of course, the acidosis is caused by a decreased HCO3 production in the first place.
AD07 [Jul99] [Feb00] [Jul00] [Jul03] [Feb04] Aug15
The bicarbonate system is the most important ECF buffer system because:
A. It has a pKa close to physiological pH
B. CO2 can be exchanged in lungs and HCO3 excreted in the kidneys
C. HCO3- occurs in such large amounts
D. ?
E. CO2 can be regulated by lung & HCO3 by the kidneys
For a buffer system [1] to be useful it ordinarily must possess two properties:
Its components must be present in a sufficient concentration
It must have a pKa in the range +/-1 of the prevailing pH
For example, if we assess the phosphate buffer system in the ECF against these 2 criteria:
Concentration is about 1 mmol/l
pKa is 6.8 (& ECF pH is 7.4)
So, the phosphate bufffer has a suitable pKa BUT its ECF concentration is much too low to be useful here. [2]
Now, consider the bicarbonate buffer:
Concentration is 24 mmol/l
pKa is 6.1 (& ECF pH is 7.4)
So, the concentration is OK (plenty of it here) BUT its pKa is more than i pH unit away from 7.4. We would predict that it is not a useful buffer in these circumstances.
This is correct if we assess its usefulness as a simple closed buffer system (that is as a simple chemical buffer system like the phosphate system). However, the bicarbonate buffer system is not closed but open. Indeed it is said to be “open at both ends”. This means that both components of this buffer system are regulated by physiological mechanisms: bicarbonate is maintained by the kidneys, and CO2 is excreted via the lungs. The effect of these physiological mechanisms is to overcome the limitation of having a pKa more than 1 pH unit away from ECF pH.
The main reasons why the bicarbonate system is the most important ECF buffer are:
Bicarbonate is present in significant (ie sufficient) concentrations in the ECF
The buffer system is open at both ends and this overcomes the limitation of its low pKa
There is a lack of other useful buffer systems in the ECF.
This final point is important as to be the “most important” it helps if there is little competition. For example:
The phosphate buffer is not important (concentration too low in ECF)
Plasma proteins provide the only other buffer system in the ECF but the concentration is lower than that of bicarbonate AND the protein concentration in the interstitial fluyid is even lower then in plasma
However, it is wise not to dismiss the “proteins” too lightly. The reasons for this are:
Firstly, the plasma proteins can buffer the H+ produced when bicarbonate is formed (Obviously, this cannot be buffered by the bicarbonate system as a buffer system cannot buffer itself).
Secondly, another important protein contributes very significantly to buffering of the H+ produced by bicarbonate formation. This is haemoglobin in the red cells. Now because of its intracellular location it is NOT really an extracellular buffer. However, as the bicarbonate (& its H+) are produced in the red cell before diffusing out into the plasma, the haemoglobin really does a significant amount of buffering of this H+. Thus the haemoglobin is sometimes said to be an important extracellular buffer EVEN THOUGH it is actually intracellular. [3]
Comment I think the answer is E - due to the open-ended nature of the system, and ability to regulate CO2 via the lungs, HCO3 via the kidneys.
quote from Ganong the system is one of the most effective buffer systems in the body because the amount of dissolved CO2 is controlled by respiration. In addition, the plasma concentration of HCO3– is regulated by the kidneys
B is not quite right, because it is not so much HCO3 excretion by the kidneys, as regulation (it mainly reabsorbs HCO3, not excretes)
Hmmmmm If the kidney doesn’t reabsorb a filtered load is it excreted? I thought so??. thats why it handles an alkilosis so easily? I think B is correct, but so is E and prob safer if they both pop up- surely they didn’t have both stems so similar?
It is the use of the word regulate: It is the lungs that exchange CO2, they do not regulate it, regulation is part of the control system mediated by the respiratory centre, chemoreceptors, efferent nerves ect. Though the excretion of HCO3?? I’d favour B.
Re: above Yes, the respiratory centre regulates the lung, but the actions/abilities/healthiness of the lung regulates CO2 levels. Does the speed of the motor regulate the wind or is it the speed of the fan blades? Much of a muchness.
AD07b Aug15 version
Bicarbonate buffer most important extracellular buffer system because:
A. Because CO2 and HCO3- are present in large concentrations
B. Because of carbonic anahydrase
C. Because CO2 is delt with by the lung and HCO3 by the kidneys
D. Of the Henderson-Hasselbalch equation
For a buffer system [1] to be useful it ordinarily must possess two properties:
Its components must be present in a sufficient concentration
It must have a pKa in the range +/-1 of the prevailing pH
For example, if we assess the phosphate buffer system in the ECF against these 2 criteria:
Concentration is about 1 mmol/l
pKa is 6.8 (& ECF pH is 7.4)
So, the phosphate bufffer has a suitable pKa BUT its ECF concentration is much too low to be useful here. [2]
Now, consider the bicarbonate buffer:
Concentration is 24 mmol/l
pKa is 6.1 (& ECF pH is 7.4)
So, the concentration is OK (plenty of it here) BUT its pKa is more than i pH unit away from 7.4. We would predict that it is not a useful buffer in these circumstances.
This is correct if we assess its usefulness as a simple closed buffer system (that is as a simple chemical buffer system like the phosphate system). However, the bicarbonate buffer system is not closed but open. Indeed it is said to be “open at both ends”. This means that both components of this buffer system are regulated by physiological mechanisms: bicarbonate is maintained by the kidneys, and CO2 is excreted via the lungs. The effect of these physiological mechanisms is to overcome the limitation of having a pKa more than 1 pH unit away from ECF pH.
The main reasons why the bicarbonate system is the most important ECF buffer are:
Bicarbonate is present in significant (ie sufficient) concentrations in the ECF
The buffer system is open at both ends and this overcomes the limitation of its low pKa
There is a lack of other useful buffer systems in the ECF.
This final point is important as to be the “most important” it helps if there is little competition. For example:
The phosphate buffer is not important (concentration too low in ECF)
Plasma proteins provide the only other buffer system in the ECF but the concentration is lower than that of bicarbonate AND the protein concentration in the interstitial fluyid is even lower then in plasma
However, it is wise not to dismiss the “proteins” too lightly. The reasons for this are:
Firstly, the plasma proteins can buffer the H+ produced when bicarbonate is formed (Obviously, this cannot be buffered by the bicarbonate system as a buffer system cannot buffer itself).
Secondly, another important protein contributes very significantly to buffering of the H+ produced by bicarbonate formation. This is haemoglobin in the red cells. Now because of its intracellular location it is NOT really an extracellular buffer. However, as the bicarbonate (& its H+) are produced in the red cell before diffusing out into the plasma, the haemoglobin really does a significant amount of buffering of this H+. Thus the haemoglobin is sometimes said to be an important extracellular buffer EVEN THOUGH it is actually intracellular. [3]
Comment I think the answer is E - due to the open-ended nature of the system, and ability to regulate CO2 via the lungs, HCO3 via the kidneys.
quote from Ganong the system is one of the most effective buffer systems in the body because the amount of dissolved CO2 is controlled by respiration. In addition, the plasma concentration of HCO3– is regulated by the kidneys
B is not quite right, because it is not so much HCO3 excretion by the kidneys, as regulation (it mainly reabsorbs HCO3, not excretes)
Hmmmmm If the kidney doesn’t reabsorb a filtered load is it excreted? I thought so??. thats why it handles an alkilosis so easily? I think B is correct, but so is E and prob safer if they both pop up- surely they didn’t have both stems so similar?
It is the use of the word regulate: It is the lungs that exchange CO2, they do not regulate it, regulation is part of the control system mediated by the respiratory centre, chemoreceptors, efferent nerves ect. Though the excretion of HCO3?? I’d favour B.
Re: above Yes, the respiratory centre regulates the lung, but the actions/abilities/healthiness of the lung regulates CO2 levels. Does the speed of the motor regulate the wind or is it the speed of the fan blades? Much of a muchness.
AD08 -[Jul00]-Jul05-14B
During infusion of an acidic solution (?HCl infusion) , which contributes MOST to buffering?
A. Phosphate buffer
B. Bicarbonate buffer
C. Intracellular buffers
D. Proteins (?intracellular proteins) (?Haemoglobin)
E. None of the above
Bicarbonate is the most important ECF buffer against metabolic acids (including hydrochloric acid).
I disagree… See http://www.anaesthesiamcq.com/AcidBaseBook/ab2_2b.php
Experiments in metabolic acidosis have shown that 57% of buffering occurs intracellularly and 43% occurs extracellularly. The processes involved in this buffering are:
ECF 43% (by bicarbonate & protein buffers)
ICF 57% (by protein phosphate and bicarbonate buffers) due to entry of H+ by: Na+-H+ exchange 36%, K+-H+ exchange 15%, Other 6%
[yet another disagreement] I think the [sneaky] answer is proteins, because collectively intracellular proteins and Hb (a protein) BOTH contribute synergistically to the buffering effect.
[yet another disagreement with the disagreement] Which finally brings us back to the original answer: that bicarbonate is the most important buffer because it is in BOTH ECF and ICF and is open ended - which makes it more effective than proteins in buffering metabolic acids!! Consider the maths: (a) intra cellular buffer most important (57%) BUT most of the extracellular buffering (43%) is by bicarbonate (b) Bicarbonate “less” intracellularly but still important e.g. in RBC, bicarbonate accounts for 18% buffering whilst Hb 35% (power and cam p.225), and so % buffering by bicarbonate can’t be that much less in other cells - the only question is whether the “other” cells have as much protein as Hb is in RBCs… probably not - so I still maintain bicarbonate as the most important… but I can see why protein looks like the right answer.
From Ganong 22nd ed, (where many of the MCQs have been lifted, from a single sentence), it says “in metabolic acidosis, only 15-20% of the acid load is buffered by the H2CO2-HCO3 system in the ECF, and most of the remainder is buffered in cells. In metabolic alkalosis, about 30-35% of the OH load is is buffered in cell, whereas in respiratory acidosis and alkalosis, almost all the buffering is intracellular” (page 733) ….. SInce infusion of acid is a metabolic acidosis, it would follow that the answer to the question is C, intracellular buffers.
I like C too. As stated above, “57% occurs intracellularly”. It doesn’t say by which buffers, so it includes HCO3, proteins/Hb, phosphate. Fantastic.
Brandis Page 36 suggests answer is B
- but he also says “assuming all the buffering is extracellular and is by bicarbonate. These assumptions are not really correct…”
The other question is how long has this infusion been going on for? if it is a slow infusion over long period of time… probably C is most likely. but if the duration is short. the infused HCl will probably be buffered by ECF HCO3 and breathed out before it is shifted into the cells! It is not exactly the same situation as endogenous metabolic acidosis where acid production is intracellular… I would probably still go for B!
The rate of infusion is not important. Even if it is dumped into the blood the actual buffering in the blood occurs either in proteins or Hb where Hb is 6 times more important (due to quantity of Hb and number of histidine residues - 38, ganong 22nd ed pg 732). To buffer with Hb the acid load dealt with in the ICF due to the presence of carbonic anhydrase in the (RBC).
Another view
As per Power & Kam (2nd Ed, Pg. 256), “…approximately 70% of non-carbonic acids produced by the body are buffered by plasma bicarbonate…”, whilst “…more than 90% of the capacity of the blood to buffer carbonic acid is contributed by haemoglobin…”.
Metabolic acidosis assumes that the H+ is produced intracellularly, and so the intracellular buffers will have first contact. As HCl is infused, rather than produced intracellularly, the correct answer is likely B - bicarbonate buffer.
D09 [Jul00] [Jul01] [Mar05] [Feb12] The same MCQ for 2001-2 exams but probably different for 2005 & 2012 exams.
In a patient with diabetic ketoacidosis, the following are true except:
A. ?
B. There is decreased PaCO2
C. There is decreased concentration of H+ intracellularly
D. Renal excretion of titratable acids will be increased
E. There is increased synthesis of bicarbonate
Re MCQ AD09: Diabetic ketoacidosis is a metabolic acidosis. Acid Base Physiology Case 2:A Sick Diabetic Patient
A. ?
B. Respiratory compensation (Kussmaul breathing) is hyperventilation which causes a decreased arterial pCO2
C. The keto-acids are produced in the liver and released into the ECF so only only the liver cells have a low pH (increased [H+]) due to intracellular acid production. However CO2 is very lipid soluble and crosses cell membranes easily. The low arterial pCO2 results in low intracellular CO2 -> decreased intracellular [H+])
D. The kidney excretion of titratable acids increases due partly to the fact that when the urine pH is low, keto-acids themselves contribute to “titratable acidity”.
Comment: I think that C is true - compensation for metabolic acidosis involves exchange of intracellular K+ for ECF H+ and so a rise in H+ in intracellular fluid. This is a more general effect in metabolic acidosis than ketone synthesis in the liver.
From KB Acid Base eBook: In diabetic ketoacidosis (DKA), the ketoacids are produced in the liver and not in every cell in the body. The intracellular alkalinising effect of the compensatory hypocapnia that occurs will however affect every cell and not just the hepatocytes. Does this mean that DKA produces an extracellular rise in [H+] but the opposite change in most tissues (excluding the liver) where the net effect is a fall in intracellular [H+] due to the compensatory hypocapnia? Ketoacids can enter most cells and be used as an energy substrate and this would initially cause a fall in intracellular [H+]. Intracellular pH may not be altered much once maximal respiratory compensation has been achieved because of these opposing effects. It is possible that though the maximal respiratory compensation does not fully correct the extracellular acidaemia, it may be sufficient to prevent much change in intracellular pH. This discussion is speculative and has not been fully investigated. The purpose here is merely to show that looking at acid-base disorders from the intracellular viewpoint can lead to ideas which are different from those of the conventional extracellular viewpoint.
From 2005:
Patient with metabolic acidosis with excess ketones, which of the following is true:
A. Decreased urinary NH4 excretion
B. Normal CO2 concentration
C. Hyperventilation can excrete non volatile acid
D. Increased intracellular H+
E. Increased urinary excretion of HCO3-
In the Mar 2005 version, I think D is correct, which suggests C is the answer for the 2000-01 version.
From 2012:
What is the compensation for a person with increased keto-acids?
A. Increased ammonium production (I thought it said decreased urinary NH4+ excretion?)
B. Hyperventilation to expel non-volatile acids
C. Decreased absorption of bicarbonate
D. Increased intracellular H+ concentration
E. No change in PaCO2?
Again, the only answer that makes sense is that H+ is increased intracellularly
Alternative:
Unlike previous years, the 2012 question as recalled here specifically asks for a compensatory mechanism.
Increase intracellular pH is only compensatory if you consider intracellular buffering to be a ‘compensation’, which is not a common usage of ‘compensation’.
Assuming that option A was correctly recalled as, “Increased ammonium production”, it would be the best answer available.
AD10 [Apr01] [Jul04] [Mar05] [Aug11]
A patient is draining 1 litre of fluid per day from a
pancreatic fistula while maintaining normal
volume status. The most likely acid-base disorder is:
A. Hyperchloraemic metabolic acidosis
B. Hypochloraemic metabolic acidosis
C. Metabolic acidosis with normal chloride
D. Hyperchloraemic metabolic alkalosis
E. Hypochloraemic metabolic alkalosis
Alt stem: Fluid loss from pancreatic fistula with normovolaemia:
Loss of a sufficient amount of bicarbonate rich pancreatic or biliary secretions will cause a hyperchloraermic (or normal anion gap) metabolic acidosis.
This is the ‘standard’ acid-base disorder associated with a pancreatic or biliary fistula.
A correct answer
AD11 [Apr01] ABG's in healthy young man with pneumothorax: A. pO2=50, pCO2=25 B. pO2=50, pCO2=46 C. pO2=90, pCO2=25 D. pO2=90, pCO2=46 Alt version: ABG of young male who develops total collapse of one lung postop: A. pO2 95mmHg pCO2 50 mmHg B. pO2 80mmHg pCO2 50mmHg C. pO2 90mmHg pCO2 25mmHg D. pO2 60mmHg pCO2 50mmHg
A healthy young male so assume lungs & heart otherwise noromal.
Pneumothorax-> sensation of dyspnoea -> hyperventilation -> decreased arterial pCO2 (resp alkalosis).
Now assume the arterial pCO2 is 25mmHg (as in the given options), what then would the arterial pO2 be?
Using the Alveolar gas equation for alveolar pO2:
Alveolar pO2 = inspired pO2 - (pCO2)/0.8
(as it is reasonable to assume respiratory exchange ratio = normal (0.8) and that alveolar pCO2 = arterial pCO2).
Substituting:
Alveolar pO2 = 149 - (25/0.8) = 149 -31 = 118 mmHg.
Now of course the arterial pO2 MUST be lower than this because of the shunt and V/Q mismatch. However, assuming one lung is totally collapsed-> assume minimal pulmonary blood flow here (thus shunt this blood to the ventilated lung), so pO2 would be lower than 118mmHg but not really low. The option of pO2 = 90mmHg would be a reasonable expectation.
So the option with pO2 =90 & pCO2 =25mmHg would seem realistic.
If pO2 is 90, pCO2 is 25, then there is neither hypoxic or hypercapnic resp drive -TRUE, neither hypoxaemia or hypercapnia will occur so NEITHER is important here.
For the following blood gas results, which clinical scenario fits best?
ABG results: pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15.
These are consistent with:
A. Mixed metabolic and respiratory acidosis
B. Acute respiratory alkalosis
C. Metabolic acidosis with compensated respiratory alkalosis
D. Chronic respiratory disease
E. Mountain climber after several weeks at altitude
F. Hyperventilating consistent with acclimatisation to altitude
G. Hyperventilation for 5 mins
This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above)
A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg).
Therefore there has been some metabolic ‘compensation’ to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1])
In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of ‘compensation’ for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l
(Ref: Acid-Base on-line tutorial at [2])
In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l
(Ref: Acid-Base on-line tutorial at [3])
If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula:
Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4])
Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn’t fit if 15 is the right number. I am happy to accept the above answers without the BE.
Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis’ acid-base tutorial has a good argument for not using BE which includes the point: “It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder.” For further info via acid-base calculators using BE see: [5] and [6]
Calculations:
pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15
Step 1 Alkalosis as pH > 7.44
Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis
Step 3 No clues
Step 4 Assess compensations:
If acute respiratory alkalosis, Boston Rule 3 predicts:
Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg
If chronic respiratory alkalosis, Boston Rule 4 predicts:
Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg
Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis.
Q. how long does it take for full renal compensation?
See Ganong Table 36-3 Values for respiratory alkalosis:
After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26
In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27
Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.
Arterial Blood Gas Result: pH 7.56, HCO3 43, pCO2 53. Most likely?
A. Mountaineer at altitude after a (week?)
B. ?
C. ?
D. Hyperventilation
E. Prolonged vomiting
This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above)
A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg).
Therefore there has been some metabolic ‘compensation’ to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1])
In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of ‘compensation’ for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l
(Ref: Acid-Base on-line tutorial at [2])
In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l
(Ref: Acid-Base on-line tutorial at [3])
If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula:
Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4])
Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn’t fit if 15 is the right number. I am happy to accept the above answers without the BE.
Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis’ acid-base tutorial has a good argument for not using BE which includes the point: “It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder.” For further info via acid-base calculators using BE see: [5] and [6]
Calculations:
pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15
Step 1 Alkalosis as pH > 7.44
Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis
Step 3 No clues
Step 4 Assess compensations:
If acute respiratory alkalosis, Boston Rule 3 predicts:
Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg
If chronic respiratory alkalosis, Boston Rule 4 predicts:
Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg
Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis.
Q. how long does it take for full renal compensation?
See Ganong Table 36-3 Values for respiratory alkalosis:
After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26
In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27
Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.
Aug2014 remembered version: An ABG given pH of 7.42, reduced bicarbonate, reduced CO2 (exact numbers not recalled). A. altitude B. COPD C. metabolic acidosis D. hyperventilation E. prolonged vomiting
This is a Compensated Respiratory Alkalosis. (NB: This comment and analysis refers to the 1st version above)
A CO2 this low (24) would be expected to produce a pH of about 7.53. (0.1 change for every 12mmHg of CO2 away from 40mmHg).
Therefore there has been some metabolic ‘compensation’ to reduce the pH to 7.48, which is consistent with a low HCO3 (19). (Ref: [1])
In an Acute Respiratory Alkalosis there you would expect drop in HCO3- by 2 mmol/l (from 24 mmol/l)for every 10mmHg decrease in pCO2 from the reference value of 40mmHg. The lower limit of ‘compensation’ for this process is 18mmol/l - so bicarbonate levels below that in an acute respiratory alkalosis indicate a co-existing metabolic acidosis. This would give us an expected HCO3- = 21 mmol/l
(Ref: Acid-Base on-line tutorial at [2])
In a Chronic Respiratory Alkalosis an average 5 mmol/l decrease in HCO3- (from 24 mmol/l) per 10mmHg decrease in pCO2 from the reference value of 40mmHg would be expected. This takes 2 to 3 days. The limit of compensation is a HCO3- of 12-15 mmol/l. This would give us an expected HCO3- = 16 (+/- 2) mmol/l
(Ref: Acid-Base on-line tutorial at [3])
If it was a metabolic acidosis with respiratory compensation the expected CO2 would be about 36-37, from the formula:
Expected pCO2 = 1.5 (Actual HCO3 ) + 8 mmHg (Ref: Acid-Base on-line tutorial at [4])
Can someone please explain why the BE would be positive then? Unless the number is remembered wrongly. Respiratory alkalosis or metabolic acidosis are associated with a negative base excess wherelese respiratory acidosis and metabolic alkalosis are associated with positive base excess which doesn’t fit if 15 is the right number. I am happy to accept the above answers without the BE.
Good point. I struggled with this when I researched the answer, and came to the conclusion that the number was probably wrong, as a BE of +15 would indicate a mixed respiratory and metabolic alkalosis which is not in the list of possible answers. Perhaps the answers are wrong. Kerry Brandis’ acid-base tutorial has a good argument for not using BE which includes the point: “It does not distinguish compensation for a respiratory disorder from the presence of a primary metabolic disorder.” For further info via acid-base calculators using BE see: [5] and [6]
Calculations:
pH 7.48, PCO2 24 (or 26), HCO3 19 mmol/l, BE 15
Step 1 Alkalosis as pH > 7.44
Step 2 Both pCO2 and HCO3 trending down - either respiratory alkalosis OR metabolic acidosis
Step 3 No clues
Step 4 Assess compensations:
If acute respiratory alkalosis, Boston Rule 3 predicts:
Expected [HCO3] = 24 - 2 [(40 - actual pCO2) / 10]= 24 - 2 [(40-24)/10] = 20.8 mmHg
If chronic respiratory alkalosis, Boston Rule 4 predicts:
Expected [HCO3] = 24 - 5 [(40 - actual pCO2) / 10] = 24 - 5 [(40-24)/10] = 16.0 mmHg
Step 5 Formulate diagnosis: Appears to be a partially compensated acute respiratory alkalosis.
Q. how long does it take for full renal compensation?
See Ganong Table 36-3 Values for respiratory alkalosis:
After 3 wks at 4000m altitude pH 7.48, HCO3- 18.7, pCO2 26
In voluntary hyperventilation pH 7.53, HCO3- 22, pCO2 27
Whilst not exactly correct CO2 values, I think this table makes a good argument for E in the first version and G in the second version.
A 19 year old is admitted unconscious. She has the with the following arterial blood gases:
PaO2 117mmHg, PaCO2 11mmHg, pH 7.1 Base excess -15
.
This is most consistent with:
A. Metabolic acidosis with respiratory compensation
B. Respiratory alkalosis with metabolic compensation
C. Mixed metabolic and respiratory acidosis
D. ?
The story and gases are typical of acute diabetic ketoacidosis. The low pH despite the very low pCO2 indicates a very low bicarbonate. This is a severe metabolic acidosis with respiratory compensation.
Additional Comment
pH = 6.1 + log ( [HCO3] / (0.03 x pCO2) ), Hence log ( [HCO3] / (0.03 x pCO2) ) must = 1, and ( [HCO3] / (0.03 x pCO2) ) = 10, As such, [HCO3] would = 3.3. Given pCO2 = 1.5 x [HCO3] + 8 +/- 2, which is 12.95 +/- 2, this is consistent with metabolic acidosis at maximal respiratory compensation, not complicated by other acid-base disorders. Also, PaO2 = FiO2 x (760-47) - ([CO2] / 0.8) = 135.98 mmHg. There is a slight A-a gradient (135.98 - 110) and this patient is unlikely to be receiving supplemental oxygen.
LOVE mathematics today said, likely answer is D. 1. pH = 6.1 + log ( [HCO3] / (0.03 x pCO2), PH=7.1, pCO2=11, aHCO3=3.3. 2. expected HCO3 in acute=24-2((40-pCO2)/10), pCO2=11, eHCO3=18.2
expected HCO3 in chronic=24-5((40-pCO2)/10), pCO2=11, eHCO3=9.5
3.aHCO3
AD15 [Mar05] A previously healthy man with this blood gas: pH 7.40 pCO2 50mmHg pO2 88mmHg
Must indicate
A. Breathing FIO2> 0.21 B. Acute respiratory acidosis C. Fully compensated metabolic acidosis D. HCO3 levels will be raised E. ?Mixed respiratory & metabolic acidosis
We are unable to reliably interpret a blood gas without having a history. We can however say whether something is consistent or not.
In this case, option B (acute respiratory acidosis) as a single disorder (ie a simple acid base disorder) is not possible as the bicarbonate (see calculation below) is too high. The 1 for 10 rule (Rule 1) would predict a HCO3 of 25.
Option C (fully compensated metabolic acidosis) is ridiculous. The compensation for a metabolic acidosis is hyperventilation to reduce the arterial pCO2 (below 40, to a value predicted by Rule 5 ie pCO2=1.5 Bicarb + 8). In any case the elevated bicarbonate excludes metabolic acidosis.
Option E (as remembered here) is also impossible. A mixed acidosis can NEVER result in a normal pH of 7.4.
OPtion A may be true but there is no sure way of knowing on the information given - so this CANNOT be the correct answer. If the pO2 was high (eg >140 mmHg say) then we could say that the person must be breathing a high pO2 but this is not the case here. (Technically though this need not be a high FIO2 as the FIO2 for example could be low but the inspired pO2 high if the pertson was in a hyperbaric chamber.)
The only option left is option D. Now this does not depend on knowing any clinical histoiry so does not violate the essential history requirement prior to interpreting a blood gas. This is not an interpretation, just a simple application of the Henderson-Hasselbalch equation.
THUS:
Application of Henderson-Hasselbach indicates that there MUST be an increase in bicarbonate in order for a normal pH to be maintained:
7.4 = 6.1 + log ([HCO3]/0.03 pCO2)
So: log ([HCO3]/1.5) = 1.3
And as: log 2 = 0.3 and log 10 = 1 (thus log 20 = 1.3)
So:
[HCO3] = 20 x 1.5 = 30 mmol/l
D is correct: Bicarbonate is elevated.
Comment
It is possible to roughly work out what the FiO2 is, using the alveolar gas equation (and assuming pAO2 is very close to paO2) - I think it roughly works out to room air FIO2, so A is wrong anyway
as for anything else, the pH is normal - so it doesn’t matter what the primary acid-base disorder is, if we know that paCO2 is raised, then HCO3 would be raised also (either as the primary or compensating mechanism)
AD16 [Mar05]
A man is air lifted up to 5000m and his arterial blood gas is taken after ½ hr.
He lives there and his blood gas is repeated after 1 week.
Compared to the first sample, the second blood gas shows:
A. No change in PaCO2 and PaO2 B. PaCO2 increase, PaO2 increase C. PaCO2 increase, PaO2 decrease D. PaCO2 decrease, PaO2 decrease E. PaCO2 decrease, PaO2 increase
Option E - CO2 decrease and O2 increase - is correct.
Traditional (and simple answer) is that ↑ in minute vol. in response to hypoxia at altitude is limited by the resulting respiratory alkalosis which tends to ↓ MV via central chemoreceptors. Then the renal compensation to respiratory alkalosis (HCO3 excretion - maximal at 3-4 days) results in HCO3 movement out of CSF and thus correcting CSF pH toward normal - ‘releasing the brakes’ on respiratory centre and allowing further increase in minute ventilation and further decrease in pCO2, and further improvement of pO2.
Real situation appears to be more complex (isn’t it always?). Has now been shown that changes in MV don’t actually parallel the compenstaion of respiratory alkalosis. Not well understood but observed changes as above probably due to a change in the ‘set-point’ of the carotid bodies repsonse to hypoxia, or changes in the hypoxic repsonse of the respiratory centre itself.
AD17 [Jul05] [Feb06]
ABG: pH 7.48, PaO2 70, HCO3 raised (~35mmHg), PaCO2 48 (OR 58).
This ABG could be explained by:
A. Acclimatisation to altitude B. COAD C. Metabolic acidosis D. ? E. Prolonged vomiting
Correct Answer is E.
Gas shows a metabolic alkalosis with respiratory compensation.
A. Altitude causes hyperventilation because of hypoxia leading to a respiratory alkolosis which can last days (4 days?) with metabolic compensation. Even chronically may have high respiratory rate.
B. COAD would cause a respiratory acidosis with metabolic compensation.
C. Metabolic acidosis is wrong. pH is >7.4
D. ?
E. Prolonged Vomiting. Loss of acid+ will cause a metabolic alkalosis which will be compensated for by hypoventilation and rise in pCO2.
Boston Rules Strategy:
pH 7.48 PO2 70 pCO2 48 HCO3 35
Step 1 pH>7.44 = alkalosis
Step 2 pCO2 and HCO3 raised = metabolic alkalosis OR respiratory acidosis
Step 3 No clues
Step 4 Assess respiratory compensation- One and a Half Plus Eight Rule Expected pCO2 = 1.5[HCO3] + 8 = 1.5[35]+8 = 60.5»_space; 48 (but 58 about right)
Step 5 Metabolic alkalosis with appropriate respiratory compensation
Only plausible answer is vomiting (loss of acid), although note can lose bicard dependent on where vomiting came from (see brandis).
I think the wrong Bedside rule has been applied. It should be the Point Seven + Twenty Rule instead
Hence Expected pCO2 = 0.7(35)+20 = 44.5(+/-5)
Food for thought: Assuming this person is breathing room air & at sea level, there is an A-a gradient i.e. Expected pO2 = 150 - 48/0.8 = 90 Why???
–cos the formula you used was wrong!
The A-a gradient in this case is [0.21(760-47) - 48/0.8] - 70 = 19.73mmHg (which isn’t that nasty)
You used exactly the same formula! And yes the A:a gradient is raised.. we know nothing about the patient’s age, which will impact on it..
AD18 [Feb06] [Feb08] Mar10 Jun10 Base excess calculation: A. When PaCO2 is 40 mm Hg B. Difference of measured HCO3 from standard HCO3 C. lower with higher HCO3 D. is an indicator of cellular buffers E. is negative when pH greater than 7.40
Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4
Edit - yes, but that’s not the same as saying assumes a CO of 40 is it? Doesn’t the base excess adjust the values to what they would be IF PCO2 were 40mmHg?
(1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement.
(2) As the measurement occurs “as if” ABG PCO2 was 40 mmHg, ‘A’ will be the “most correct” of the available answers.
AD18 [Feb12] version: The base excess on an arterial blood gas? A. Assumes a CO2 of 40mmHg B. Is measured at 20 degrees Centigrade C. ..Something about titratable acids... D. Same as plasma bicarbonate E. Measures respiratory acid base status
Base excess or deficit is the amount of acid or base required to titrate whole blood at 37 degrees celcius and PaCO2 of 40 mmHg to a pH of 7.4
Edit - yes, but that’s not the same as saying assumes a CO of 40 is it? Doesn’t the base excess adjust the values to what they would be IF PCO2 were 40mmHg?
(1) Base excess assesses the metabolic component by applying PCO2 40 mmHg during lab measurement.
(2) As the measurement occurs “as if” ABG PCO2 was 40 mmHg, ‘A’ will be the “most correct” of the available answers.
AD19 [Feb07]Aug15 Which of the following is a 'strong ion'? A. PO4 B. SO4-2 C. Cl D. ? E. ?
Stewart (in his quantitative acid-base analysis) said there were 3 independent variables that determined pH (and the other dependent variables required for acid-base analysis) in each body fluid compartment, namely:
pCO2
Strong ion difference (SID)
Total concentration of non-volatile weak acids (Atot)
will determine the status of body pH, HCO3, & other dependent variables in body fluid
Plasma SID
= (Sum of concentration of all the strong cations) - (Sum of the concentration of all the strong anions)
= about 40-42 mEq/l under normal conditions
Common Strong Cations: Na+,K+,Mg++,Ca++
Common Strong Anions: Cl-,Lactate-, Ketone bodies
Atot is the total concentration of non-weak volatile acid. (i.e. not CO2) in that compartment. In plasma this mainly consists of: Inorganic phosphate, serum proteins, and albumin
From Acid-base text:
In particular, the substances which affect acid-base balance in body fluids can all be classified into 3 groups
based on their degree of dissociation. This allows certain generalisations and simplifications which are useful
in understanding complex solutions.
Body fluids can be considered as aqueous solutions that contain:
- strong ions
- weak ions
- non-electrolytes
Strong ions in solution are those ions which are always fully dissociated.
They exist only in the charged form.
Weak ions in solution are only partially dissociated
Strong ions are mostly inorganic (eg Na+, Cl-, K+)
but some are organic (e.g. lactate anion).
In general, any substance which has a dissociation constant greater then 10-4 Eq/l
is considered as a strong ion because in the body it exists ONLY in the charged form,
and so does not participate in proton transfers
(this last thing being the essential difference from the weak ions and the non-electrolytes.)
The actual answer to this MCQ depends on the options on the actual MCQ and thge wording of the stem (which may be different from what has been remembered above). In this remembered version, there are 2 correct answers as both the chloride ion and the sulphate ion are ‘strong ions’. However sulphate concentration ([SO4–in plasma is low and it is not normally measured so does not figure much in calculations (though Peter Lloyd - see comments below - does use it in his calculator).
A significant problem in determining derived parameters for a Stewart analysis is that they involve mathematical manipulations of multiple ions, the determination of each of which has a measurement error. Consequently the total error is the sum of the errors of the individual measurements and so the range within which the correct values lies is large.
AD21 Which of the following sets of values are measured directly by ABG machine A. pCO2, paO2, pH B. paCO2, HCO3, pH C. paCO2, base excess, paO2 D. Something with base excess E. Something with HCO3
Answer is A Arterial blood gas machines measure PaO2, PaCO2 and pH. HCO3 is determined by substitution of the known values into the Henderson Hasselbach equation
AD22 [Feb08]
Person with these blood gas results: pH 7.33 CO2 58 HCO3 33
A. Acclimitization after several weeks at altitude
B. Person with chronic pulmonary disease
C. Diabetic ketoacidosis
D. Hyperventilation
E. Prolonged vomiting
B - chronic pulmonary disease.
pH is near normal, but mildly acidaemic. pCO2 is raised, so the primary disorder is a respiratory acidosis.
Using the 1 for 10 rule (acute RA) = HCO3 = 26 Using the 4 for 10 rule (chronic RA) = HCO3 = 32
I think that’s the best answer, so chronic respiratory acidosis caused by pulonary disease.
AD23 [Feb08]
Person with these blood gas results: pH 7.53 pCO2 27 HCO3 22
A. Acclimitization after several weeks at altitude
B. Person with chronic pulmonary disease
C. Diabetic ketoacidosis
D. Hyperventilation
E. Prolonged vomiting
Fits with an uncompensated acute respiratory alkalosis, so “Hyperventilation” it is.
AD24 [Feb11]|[Aug11]|[Feb12]|Aug15
Buffering by Hb better than by plasma proteins because
A. Hb has 38 carboxyl residues
B. great amount
C. plasma protein pKa near pH of Blood
D.
E.
Haemoglobin is an effective buffer because:
A. present in large concentrations
B. Has 38 carboxyl residues per globin molecule
C. pKa close to physiological pH
D. deoxyHb is more acidic than oxyHb?
E. Proteins have pKa 6.8
Haemoglobin is a better buffer than plasma proteins because
A. present in much greater quanitity
B. Hb contains 38 carboxyl residues
C. plasma proteins have pKa closer to physiological pH
D. ?
E. ?
NB: These 2 versions differ in that the first one is looking for a reason why Hb is better than PPs,
whereas the second is just about Hb as a buffer. The answer (as always) depends on the wording
of the actual question, not on what the remembered version says. The options here are so close
these remembered versions are certainly the same question.
Hb has 38 carboxyl residues: Incorrect Hb has got 38 Histidine (imidazole) residues which are important in buffering (i.e. not carboxyl residues)
Great amount: Correct “Hb is about six times more important quantitatively as a buffer compared to the plasma proteins because :
Its concentration is about twice as much[Hb] 150 gr/L versus [plasma proteins] 70 gr/L
Each haemoglobin molecule contains about three times more histidine residues than the average plasma protein.”
Ref: from p25 physiology viva Kerry Brandis
Plasma protein pKa near pH of Blood - The relevant group for buffering by proteins is the imidazole group of histidine residues. This has a pKa about 6.8 so as a ‘closed’ buffer system this provides effective buffering for a pH of 7.4 (i.e. by the pKa +/- 1 rule). However this applies equally to both plasma proteins and Hb.
In summary:
A. Correct - present in large concentrations
B. Wrong - has 38 HISTIDINE residues, and that in total rather than per globin molecule
C. Correct - pKa of ~6.8 is close to physiological pH 7.4 (note that Feb 2011 recalls ‘closer’, which would make this option more tenuous as the reason for a difference between Hb and plasma proteins)
D. Wrong - deoxyHb is a better H+ acceptor (base) than oxyHb
E. Not entirely correct - Proteins are zwitterions with multiple pKas. As physiological pH, only those proteins with imidazole groups have the a pKa 6.8
With two correct choices, one wonders whether the Feb 2011 option C is in fact the correctly recalled option with ‘closer’ pKa. If we were talking about the relative buffering power of Hb compared to plasma proteins, ‘A’ would be correct.
AD25 [Aug11] If pH is 7, then H+ concentration of pure water is: A. O B. 40 nmol/L C. 70 nmol/L D. 100nmol/L E. 1000nmol/L
f pH is 7, then [H+] is 10-7 = 100 nmol/L, therefore option “D”.
Note:
40 nmol/L for pH 7.4,
36 nmol/L for pH 7.44
46 nmol/L for pH 7.36
AD25 [Mar10] At pH 7.4, [H+] is (repeat)? A. 40nmol/L B. 40mmol/L C. 40mOsmol/L D. ? E. ?
f pH is 7, then [H+] is 10-7 = 100 nmol/L, therefore option “D”.
Note:
40 nmol/L for pH 7.4,
36 nmol/L for pH 7.44
46 nmol/L for pH 7.36
AD27 [Mar10] [Aug11] [Feb12] With pCO2 200mmHg, what else would you find A. Hyperkalemia B. Bradycardia C. ?Hyper/hypocalcaemia D. Hypokalaemia E. Hypermagnesimia
This is 1 MAC of CO2. Henderson hasselbach equation suggests the increased dissociation of carbonic acid will raise H+ concentration (ICF where carbonic anhydrase is) leading to exchange across the cellular membrane and consequent decrease of ICF K+ and therefor increase in ECF K+ to maintain electroneutrality.
No idea regarding the other answers.
Yep so A correct.
AD28 Mar10 In plasma, a 'strong ion': A. is usually a cation B. is usually an anion C. has its pKa close to 7.40 D. almost completely dissociates E. ?
Answer = D
By definition, a strong ion is essentially completely dissociated at physiological pH. JB2012
AD29 [Feb14]
The following ABG is suggestive of:
pH 7.56, HCO3 46, pC02 58
A. Mountain climber after several weeks at altitude B. Chronic pulmonary disease C. Diabetic coma D. Five minutes of hyperventilation E. Prolonged vomiting
There is a significant alkalaemia present (pH > 7.44), so there must be an underlying alkalosis (unless there is a measurement error).
The pattern of a high HCO3 and a high pCO2 indicate a metabolic alkalkosis.
The commonest causes of metabolic alkalosis (accounting for 90% of cases) are:
loss of gastric acid (prolonged vomiting, or prolonged NG suction, in the absence proton pump inhibitors of course), and
use of certain diuretics (thiazides, frusemide).
On this basis, a quick check of the options indicates option E as the correct option.
A is wrong: Hyperventilation occurs at (significant) altitude (due to low arterial pO2 stimulating the peripheral chemoreceptors), so arterial pCO2 is decreased. This increase in ventilation is initially limited by central chemoreceptor mediated inhibition of the respiratory centre (due to the low pCO2). With time at altitude, there is equilibration across the blood brain barrier and the pH around the central chemoreceptor returns towards normal (despite continuing low pCO2) so ventilation can increase further. This is a respiratory alkalosis. This helps maintain a higher arterial pO2 at altitude. The high pCO2 in this blood gas excludes this option.
B is wrong. The acid base disorder associated with chronic pulmonary disease is a respiratory one, not metabolic alkalosis.
C is wrong. “Diabetic coma” is due to diabetic ketoacidosis (though actual “coma” is uncommon). This is a metabolic acidosis. A hyperosmolar nonketotic state can occur in diabetics and this can also result in impaired consciousness (‘coma’) but without an acid base disorder; indirectly any coma could result in airway obstruction with a respiratory acidosis.
D is wrong. Five minutes of hyperventilation causes an acute respiratory alkalosis, that is the arterial pCO2 is low.
AD30 - 13B - 14A
If the pH of a solution changes from 7.4 to 7.1 what is happening to the H+ ion concentration?
A. Decrease by approximately 75%
B. Increase by approximately 150%
C. Increase by approximately 100%
D. Increase by approximately 20%
E. No increase due to the effect of a buffer
A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options).
This is easy to work out if you are familiar with a small trick.
pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving).
Thus:
If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles).
If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves).
You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l.
Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on).
The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+].
Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles.
.
Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get:
pH = 6.1 + log[24 / (0.03 x 40)]
pH = 6.1 + log[24/1.2]
pH = 6.1 + log(20)
Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So:
pH = 6.1 + 1.3 = 7.4 (as expected)
AD30b - From 15B paper, requiring the same knowledge. See also AD25
Hydrogen ion concentration in plasma with a pH of 7.1
A. 40 nmol/l
B. 80 nmol/l
C. 120 nmol/l
D. 0 nmol/l
E. ?
A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options).
This is easy to work out if you are familiar with a small trick.
pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving).
Thus:
If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles).
If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves).
You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l.
Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on).
The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+].
Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles.
.
Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get:
pH = 6.1 + log[24 / (0.03 x 40)]
pH = 6.1 + log[24/1.2]
pH = 6.1 + log(20)
Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So:
pH = 6.1 + 1.3 = 7.4 (as expected)
AD30c - Alt remembered version from 15B paper:
Pure water at ph 7 has hydrogen ion concentration of:
A. 0 nanomol/L
B. 40 nanomol/L
C. 70 nanomol/L
D. 100 nanomol/L
E. 1000 nanomol/L
A change in pH from 7.4 to 7.1 is a change of 0.3 and the solution becomes more acidic. The [H+] doubles, so this is an increase of 100% (in the terms of the options).
This is easy to work out if you are familiar with a small trick.
pH is a log to base 10 scale, and the key fact to know is that log102 = 0.3. So a change in pH of 0.3 represents a change in [H+] by a factor of 2, either increasing or decreasing depending on the direction of the change: more acidic is a doubling, and less acidic is a halving).
Thus:
If the pH is decreased by 0.3 (from any starting pH), then this is becoming more acidic (lower pH) so the [H+] increases by a factor of 2 (i.e. it doubles).
If the pH is increased by 0.3 (from any starting pH), then this is becoming less acidic (higher pH) so the [H+] decreases by a factor of 2 (i.e. it halves).
You should know that pH = 7.4 (the typical value for arterial pH) is a [H+] = 40 nmoles/l. So when the pH is 7.1, the [H+] = 80 nmoles/l, and a further 0.3 pH drop to 6.8 is a further doubling to 160 nmoles/l). Going ther other way, then a pH of 7.7 is a [H+] = 20 nmoles/l.
Another easy [H+] to remember is 100 nmoles/l at pH = 7.0 and so the [H+] at pH = 7.3 is 50 nmoles/l (0.3 pH units above 7.0), and [H+] at pH = 7.6 is 25 nmoles/l (and so on).
The reference range for pH in arterial plasma is 7.36 to 7.44 which is almost exactly a [H+] range of 44 to 36 nmoles/l. (That is, pH=7.44 is [H+]=36nmoles/l; and pH=7.36 is [H+]=44nmoles/l.) So now its easy to work out in your head what the [H+] is at pH = 7.06, and at pH = 7.14 (and so on) as these are 0.3 pH units away from pH values where you know the [H+].
Knowing this (log2 = 0.3) fact makes this question extremely easy to answer, and certainly a lot easier than than doing conversions from pH to [H+] by calculating it from first principles.
.
Another way this knowledge is useful is when considering the normal values in the Henderson-Hasselbalch equation for the bicarbonate buffer, and substituting normal values for arterial pCO2(=40) and bicarbonate(=24), and for pKa(=6.1),we get:
pH = 6.1 + log[24 / (0.03 x 40)]
pH = 6.1 + log[24/1.2]
pH = 6.1 + log(20)
Now 20 = 2 x 10 so log(20) log(2 x 10) = log(2) + log(10) [i.e. the log of a product is the sum of the numbers as you will remember from high school maths] and we know log2 = 0.3, and log10 = 1, so log(20) = 0.3 + 1.0 = 1.3. So:
pH = 6.1 + 1.3 = 7.4 (as expected)
AD31 14B The H+ production by body metabolism that has to be excreted by the kidney to prevent acidosis is? A. 32 nmol B. 32 mmol C. 0.68 mmol D. 6.8 mmol E. 68 mmol
Body metabolism every day produces a net excess of acid that in order to maintain acid base homeostasis has to be excreted by the body.
The largest amount of net ‘acid’ production due to metabolism is of carbon dioxide (or equivalently, carbonic acid). This is excreted by the lungs. As it evident here, it is CO2 that is excreted, not hydrogen ion. However it can be seen that the CO2 excretion is equivalent to the elimination of the same amount of carbonic acid. The question asks about excretion by the kidneys, so CO2 is not relevant.
The net “fixed acid” production by the body per day is 1.0-1.5 mmols/kg/day which for the standard 70kg person is 70-100 mmol/day. Strictly it is not the hydrogen ion that is excreted (so in that sense the question is wrongly worded) but in a less strict sense this excretion of fixed acid (by the kidney) is commonly spoken of as acid excretion (and thus moire loosely, as “hydrogen ion excretion”). Obviously the question expects you to understand this common use.
Of the options available (assuming they are correctly remembered), thern the correct answer os E. 68 mmol.
Technical Note
Assuming the remembered version are correct, then this question is VERY poorly structured. Why? Because the correct answer can be determined by a pattern analysis of the answers without any knowledge of acid-bases physiology at all.
Firstly, note that 4 options have mmol (millimol) and only one answer has nmol (nanomol), so eliminate A.
Secondly, 3 answers have 68 in them, so eliminate A and B (which have 32)
Thirdly, note that 3 of the options have larger numbers in them (32, 32 & 68) without a decimal point, so eliminate C & D.
The remaining option then is E, which is correct.
Quite a few years ago now the college appointed a Director of Education (now long gone) who was an educationalist, not medical at all. He reviewed the MCQs and noted that a significant number had internal structural clues to the correct answer, which would enable a person to guess the correct answer without any knowledge of the subject at all. Writing MCQs with this defect is a very common thing to do and most people are not aware they are doing it. Anyway, as a consequence the college then reviewed ALL the MCQs and removed some and reworded many others. This was of course a few years ago, and with change in staff then there is a risk of forgetting this lesson in MCQ writing - maybe this is what happened here, but of course this is just a ‘remembered’ version and it could be that the person remembering the question and knowing the correct answer was the person who unknowingly caused the pattern. If interested, see for flaws in MCQs: [1]
The answer of 68 mmol is of course not the same as 70-100 mmol as I stated above (or the same as 70 mmol as given in some texts) but is very close to the answer I state. But then why is it different? The answer is probably related to the rule that the examiners use when writing primary MCQs, namely that each question and answer must be referenced to one (or more) of the recommended books in the reference text list. My informed guess is that one or more of these sources uses the 68 mmol figure.
AD32 In blood gas analysis, the base excess: A. Is directly measured B. Assumes a pCO2 of 40mmhg C. Assesses respiratory acid base status D. Is the same as plasma bicarbonate E. Is independent of blood haemoglobin level
Base excess is not measured, it is calculated according to a formula programmed into the machine. To see the formula used just ask your pathology department to let you view the manual of the machine.
It is used to assess metabolic acid base status
It is NOT the same as bicarbonate
Answer:B
RE01 [Mar96] Which of the following is a normal characteristic of lung? A. 3,000,000 alveoli B. Alveolar diameter 3 mm C. External surface area: 10 m2 D. Alveolar surface area: 5 to 10 m2 E. None of the above
“The mean total number of alveoli has recently been established as 480 million” - Nunn’s applied respiratory physiology 2005. Not 3 million, so A. is incorrect.
Each has a diameter of 1/3mm not 3 mm so B. is incorrect(The size of the alveoli is proportional to lung volume but owing to gravity they are normally larger in the upper part of the lung, except at maximal inflation, when the vertical gradient in size disappears. At FRC the mean diameter is 0.2mm- (Ref: Nunn’s Applied Respiratory Physiology’ p19).
Alveolar surface area is 50-100m2
External surface area is much less than 10m2
Answer is E. None of the above
RE02 [Mar96] [Mar99] [Apr01] A young man collapses one lung. His ABGs on room air would be: A. pO2 80, pCO2 50 mmHg B. pO2 50, pCO2 80 mmHg C. pO2 50, pCO2 50 mmHg D. ?
Collapse of one lung will result in shunt. The paO2 will decrease as decreased O2 content in blood from the collapsed lung causes significant decline in pO2 (shifts point down on to steep part of Oxygen Hb Dissociaton curve).Increasing inhaled O2 concentration to 100% does not correct pO2 ,as the addition of a small amount of shunted blood with its low O2 concentration greatly reduces the p02 of arterial blood. (West-p61)
Hypoxic pulmonary vasoconstriction in the collapsed lung, and the increased pulmonary vascular resistance as a consequence of the loss of radial stretch on the vessels will both result in a decrease in the amount of blood shunting through the collapsed lung.
The collapsed lung will result in a sensation of increased difficulty with breathing (dyspnoea) AND impair pulmonary gas exchange. BOTH the dyspnoea, and any tendency for arterial pCO2 to rise will stimulate ventilation. This would tend to maintain at least a normal paCO2 (or even result in a lowered paCO2) as the decreased CO2 in blood from the normal lung will tend to counteract the increase in pCO2 in blood from the collapsed lung. This ability to “counteract” fairly well is due to both the linear shape of the CO2 dissociation curve and the ability of paCO2 to stimulate ventilation.
In summary, the shunt will impair both oxygenation and CO2 excretion, but the increased ventilation in this situation will overcome the CO2 excretion difficulty (linear dissociation curve) BUT will not be able to overcome the oxygenation problem. So, the answer will be the one with a low pO2 and a normal or low paCO2.
The logic is that:
PaO2 will be impaired by the shunt (
RE02 [Jul96] [Mar97] The ABGs in a healthy young 70kg male with one collapsed lung are: A. paO2 50 mmHg, pCO2 25 mmHg B. paO2 95 mmHg, pCO2 40 mmHg C. paO2 60 mmHg, pCO2 45 mmHg D. paO2 60 mmHg, pCO2 25 mmHg
Part B
A - unlikely to be correct pCO2 very low
B - definitely incorrect; pO2 too high
C - most likely
D - possible
Application of the Alveolar Gas Equation/A-a gradient
Assessing the proposed values above withe the above in light of a significant shunt may help to narrow down a more likley set of values. In Part A, both A and C are the most probable answers on face value. A results in a small A-a gradient (8mmHg)which seems unlikely, C results in an A-a gradient of 37.5mmHg which would more accurately reflect the situation.
In Part B, a similar application of the A-a gradient can narrow down the field by allowing you to exclude those answers with a normal A-a gradient. Most likely answers are C or D
For those like me that have forgotten so close to the exam!
A-a (O2) = (FiO2%/100) * (Patm - 47 mmHg) - (PaCO2/0.8) - PaO2
Thus, for pO2 50 and pCO2 50
A-a (O2) = 0.21(760-47) - (50/0.8) - 50 = 37.2mmHg
RE03 [Mar96] [Mar99] [Feb04] Pulmonary vascular resistance: A. Is minimal at FRC B. ?Increases/?decreases with increase in lung volume C. Increases with elevated CVP D. ?
Pulmonary vascular resistance is minimal at FRC, and increases with a change in lung volume in either direction, ie with increased or decreased lung volume. The increase in PVR with decreased lung volume is due to changes in the extra alveolar vessels, which are pulled open by radial traction at higher volumes, or close at lower volumes due to their smooth muscle tone. The increase in PVR at high lung volumes is due to the alveolar vessels, which are distorted and their calibre decreased.
PVR decreases with an increase in pulmonary arterial pressure, firstly due to recruitment of previously closed vessels (increasing the number of open capillaries as much as threefold), and then at higher pressures due to distension of vessels (by as much as twofold). PVR also decreases with increased pulmonary venous pressure, although not as much. Therefore an increased left atrial pressure should lead to a decrease in PVR.
It seems to me that probably an elevation in CVP would exert effects on PVR indirectly through effects on pulmonary artery pressure, although I haven’t yet found anything that specifically addresses this question.
“Accumulation of CO2 leads to a drop in pH in the area, and a decline in pH also produces vasoconstriction in the lungs, as opposed to the vasodilation it produces in other tissues” (Ganong p641). Therefore option B - hypocarbia and C - alkalosis are BOTH wrong.
Refs Guyton 11th ed pgs 483- 487, West’s Pulmonary Physiology and Pathophysiology pgs 86-92, Brandis p156-157
Above comments noted. ABOVE SAYS PVR dcrease as pulmonary venous pressure increase,it should be otherway around as the pulmonary venous pressure or the lt atrial pressure increase this will increase PVR as
PVR = PAP-PVP/ CO
so i think the answer is clearly PVR INCREASE IF PV PRESSURE OR THE LT ATRIAL PRESSURE INCREASE
I don’t think this above comment is correct. If you increase Pulmonary Venous Pressure, this will make the numerator in the above formula smaller and thus DECREASE pulmonary vascular resistance. The textbooks say that increasing pulmonary venous pressure will recruit non-patent pulmonary vessels and distend those that are already patent - thereby increasing their total surface area and decreasing resistance to flow.
The trouble with using this equation is that if PVP goes up, PAP and CO would change with it. PVP goes up, CO would probably come down a bit. So you can’t just apply this equation with an isolated change in PVP. In any case, yeah, it was applied incorrectly to start with. This whole concept of decreasing PVR by recruitment and distension is dodgy. If the PVR truly decreases when PAP goes up, which is what the books are implying is the cause of the r&d, then PAP wouldn’t have gone up in the first place. The fact that PAP went up, meant that the decrease in PVR was inadequate to compensate for whatever is putting up the pressure, so even though the total PVR decreases, the PVR per unit of CO actually increased. Of course West doesn’t mention this to make it sound neat. Or was it because he didn’t really understand it? Perhaps nobody does. And now you have a bunch of confused people trying to make sense of it all for an exam. — Hypercapnia has a slight pressor effect on PVR
Acidosis (metabolic or respiratory) augments HPV
Alkalosis (metabolic or respiratory) causes pulmonary vasodilation
Alkalosis (metabolic or respiratory) can reduce or even abolish HPV
Page 102 (Ch 7) Nunns 6th ed.
If CVP were to affect pul vascular resistance via pulm art pressure, a rise in JVP would likely cause rise in PA pressure, causing recruitment of pulm vasculature, thus reduction in overall resistance to flow.
To clarify the comment on change in PVR with change in lung volume: PVR is sum of intraalveolar vessel resistance and extra-alveolar resistance. As lung volume increases intraalveolar resistance increases and extraalveoar resistance falls, and visa-versa for a decrease in lung volume, (for the reasons prevously mentioned). The sum of the two is minimal at lung volume = FRC. Graph with three lines, for intra-alveolar, extra-alveolar, and total resistance, can demonstrate this.
Summary:
PVR increases with increased H+, pCO2, lung volume (> FRC), pulmonary pressure, arterial pressure. (Note fig 4-4 West: demonstrates decreasing PVR with increasing pulmonary aterial and venous pressures - due to distension and recruitment)
PVR decreases with decreasing lung volume (
RE03b [Jul00] [Feb12] Pulmonary vascular resistance is increased in : A. Increase in pulmonary arterial pressure B. Hypocarbia C. Alkalosis D. Increased left atrial pressure E. Head down tilt F. Hypoxic pulmonary vasoconstriction
Well I think that increasing left Atrial pressure will end up increasing pulmonary vascular resistance. As has been pionted out pressure gradient between PA pressure and LA pressure deterimines flow. with increase LA pressure there would be smaller pressure gradient. If flow has decreased then the resistance to the flow would have been increased. thus PVR must be increased since its one of the factors that resists blood flow. my reasoning may not be acceptable to some but you will find that most descriptions of pathological conditions that increase LAP eg mitral stenosis state that pulmonary hypertension is present. if indeed there were to be a decrease in PVR then pulm HPT would not be a problem in these groups of patients.
[2] emedicine mitral stenosis
[3] wikipedia Mitral stenosis
Comment:
This reasoning is incorrect with regards to PVR.
It is true to say that increasing LA pressure reduces lung perfusion pressure. That would reduce flow only if resistance is unchanged.
However, recruitment and distension of pulmonary capillaries will occur with increase of either pulmonary artery OR venous pressure. This is because the recruitment is acting against the alveolar gas pressure: even venous blood pressure will improve capillary engorgement. In this way, increased LA pressure will reduce PVR.
Increased LA pressure also leads to pulmonary hypertension. However hypertension does not equal resistance. The implication is that the is low pulmonary vascular resistance due to recruitment and distension, but a pulmonary hypertension due to cardiac valvular disease. The pulmonary capillary bed is thus compensating for the increased resistance within the left heart: low pulmonary vascular resistance compensating for high intra-cardiac resistance.
SO for the Feb 12 version, hypoxic pulmonary vasoconstriction as the best answer?
Yes, ‘A’.
RE03c [Aug 11] Which of the following increase pulmonary vascular resistance: A. Hypocarbia B. Alkalosis C. Raised pulmonary artery pressure D. Raised left atrial pressure E. None of the above
RE03c [Aug 11] Which of the following increase pulmonary vascular resistance:
A. Wrong - Hypocarbia, with respiratory alkalosis, causes pulmonary vasodilation. Hypercarbia has a pressor effect.
B. Wrong - Alkalosis causes pulmonary vasodilation
C. Wrong - Raised pulmonary artery pressure decreases PVR through recruitment and distension of capillaries
D. Wrong - Raised left atrial pressure decreases PVR through recruitment and distension of capillaries (even though perfusion pressure may have decreased)
E. None of the above
Most correct answer: E
RE04 [Mar96] [Jul97] [Jul02]
The greatest increase in (?physiological) dead space would be expected with:
A. Pulmonary embolism
B. Atelectasis (or: collapse of one lung)
C. Pneumothorax
D. Bronchoconstriction
E. Obesity
Physiological dead space is the part of tidal volume that does not take part in gas exchange. It comprises of both anatomical and alveolar dead space.
Anatomical dead space is the volume of the conducting airways (some sources list it as the total lung volume minus the volume of alveoli).
Alveolar dead space is where alveoli are ventilated but not (adequately) perfused.
- -
Pulmonary embolism leads to an increase of alveolar dead space (before any atelectasis) as previously perfused areas of the lung are now no longer perfused.
Atelectasis/collapse will mean the involved lung isn’t ventilated; it doesn’t receive any of the tidal volume.
A pneumothorax will decrease lung volume, and decrease how much it is ventilated.
Bronchoconstriction reduces ventilation, not perfusion. (If you constrict your bronchioles you decrease anatomical dead space)
Obesity shouldn’t cause a decrease in perfusion??? (although may cause atelectasis, see above) - Agreed - Pressure of abdominal contents, especially if supine, should decrease basal ventilation, potentially causing atelectasis. However, there would be Alveolar hypoxia causing increased vascular resistance in that lung region, with decreased perfusion to return V/Q towards 1.
So the best answer is A.
RE05 [Mar96] [Jul00] [Apr01] [Jul01] [Jul02] [Feb04]
As go from the top of the erect lung to the bottom:
A. Water vapour pressure remains constant
B. pN2 is higher at the apex than is at the base
C. pCO2 at apex is higher than at the base
D. pO2 at base is lower than at the apex
E: V/Q is higher at base than apex
F. Ventilation goes up as go up lung
G. Compliance is more at base than apex
A1. Correct - Water vapour pressure will remain constant throughout the lung.
B1. False - Apex 553mmHg, Base 582 mmHg - West says variation in effect default due to variations in others.
C1. False - pCO2 lower at apex
D1. True - pO2 higher at apex (130mmHg vs 88 Kam p 92)
E1. False - V/Q ratio is much greater at apex (>3.0)
F1. False - greater ventilation achieved at base because greater compliance at base
G1. True - compliance greater at base
V/Q is higher at the apex than the base. Both ventilation and perfusion decrease from base to apex, but perfusion decreases more. Compliance is greater at the base than the apex, because of the position of the lung on the pressure-volume curve. Therefore ventilation, when seen as a change in volume per unit of resting volume, is greater at the base than the apex (West Pulmonary physiology and pathophysiology, p 41-42).
It is the change in V/Q ratio which affects regional pCO2 and pO2 differences. If the alveoli were ventilated but not at all perfused, the alveolar gas would approach inspired gas composition (pO2 of 149mmHg and pCO2 close to 0). If the alveoli were perfused but not at all ventilated, the alveolar pO2 and pCO2 would approach venous values (pO2 of roughly 40mmHg and pCO2 of roughly 45mmHg). So therefore pO2 must be higher at the apex than the base, and pCO2 higher at the base than at the apex. pN2 is lower at the apex (553mmHg at apex compared to 582mmHg at base). The variation is by default as the total alveolar gas pressure is constant throughout the lung. [The Riley approach to V/Q; Nunn p.116]
Incidentally, i dimly remember something about TB granulomas being generally in the upper regions of the lungs because of the higher pO2, although I can’t remember where I read this(West p66)
West p41-42, Guyton 11th ed p500. West pg 66
A- Water Vapour pressure is constant throughout the lung- TRUE, as temperature and relative humidity are the factors that alter water vapour, both these are not variable conditions in the lung. The saturated water vapour is higher in warm than cool air ( 20C it is 17.5mmHg. at body temp 37C it is 47mmHg)
Decreases with increasing humidity, at 40% humidity at 20C it is 7mmHg.
C- False ,apex - high V/Q ration of 3, leading to high Po2 130mmHg, low Pco2 28mmHg.
D -false, Base- low V/Q ratio of 0.6 , with low pO2 88mmHg, high pCO2 42mmHg
F- false, ventilation is highest in the lower zones (West Fig 2-7, p 21)
The difference between the apical and basal alveoli in a erect lung:
A. Apical PaO2 Basal PaCO2
C. V/Q mismatch Apical Apical
REFERENCES in West:
Fig 5-8, p.65: V, Q, V/Q vs base-apex lung.
Fig 7-8, p.102: Explanation of regional differences in ventilation. Illustrates: 1) Why ventilation is greater at base than apex, and 2) why compliance (dV / dP) is greater at base than apex.
Fig 4-7, p.43: Explanation of regional differences in perfusion (ie: gravity).
Fig 5-10, p.66, Regional difference in gas exchange down lung with numbers for Vol, VA, Q, pN2, etc.
physi16.jpg
E1: If Ventilation much greater as stated in comment above, then ‘V/Q is higher at base than at apex’ should be true (see RE06) E1 is talking about the V/Q ratio, not “ventilation”. The base is relatively more perfused than ventilated (low V/Q) and the apex is relatively more ventilated than perfused (high V/Q). It’s the ratio not the actual amount of “ventilation” or gas exchange that goes on so E1 is INCORRECT.
Compliance is generally greater at the bottom of the lung compared with the top. However, compliance at the bottom of the lung can be smaller at low lung volumes. Therefore option G is not strictly correct.
In alternative version: Isn’t PaO2 referring to arterial pO2 and will reflect venous admixture in left atrium and therefore area of lung become irrelevant to Q?. Is PaO2 and PaCO2 referring to pulmonary end capillary tension or is it just a typo and meant to be PAO2 & PACO2?
http://www.kerrybrandis.com/wiki/mcqwiki/index.php?title=RE05
Alternative [Aug 11]
In the upright lung (or something like that):
A. The apical alveolar PCO2 is low (28mmHg)
B. The basal alveolar V/Q is high (approx 3)
C. The apical alveolar V/Q is low (approx 0.6)
D. ?
E. ?
E1: If Ventilation much greater as stated in comment above, then ‘V/Q is higher at base than at apex’ should be true (see RE06) E1 is talking about the V/Q ratio, not “ventilation”. The base is relatively more perfused than ventilated (low V/Q) and the apex is relatively more ventilated than perfused (high V/Q). It’s the ratio not the actual amount of “ventilation” or gas exchange that goes on so E1 is INCORRECT.
Compliance is generally greater at the bottom of the lung compared with the top. However, compliance at the bottom of the lung can be smaller at low lung volumes. Therefore option G is not strictly correct.
In alternative version: Isn’t PaO2 referring to arterial pO2 and will reflect venous admixture in left atrium and therefore area of lung become irrelevant to Q?. Is PaO2 and PaCO2 referring to pulmonary end capillary tension or is it just a typo and meant to be PAO2 & PACO2?
http://www.kerrybrandis.com/wiki/mcqwiki/index.php?title=RE05
RE06 [Mar96] [Mar99] [Jul01]
Distribution of pulmonary ventilation & perfusion in the erect position:
A. Gradient of change in ventilation is greater than that for perfusion
B. Ventilation increases as go up the lung
C. Perfusion increases as go up the lung
D. V:Q ratio at apex is greater than at base
E. None of the above
For 1996,1999,2001 version:
(A to E below refer to first version above):
A. WRONG - Perfusion gradient is steeper
B. WRONG - Ventilation is higher at the base
C. WRONG - Perfusion is higher at the base
D. TRUE -V/Q ratio is higher (3.3) than at the base (0.63).Ventilation is less at the top than the bottom,but the differences in blood flow are more marked.Consequently,the ventilation-perfusion ratio decreases down the lung,and all the differences in gas exchange follow from this(West p 66).
(These values for V/Q ratio are the standard ones from West “Respiratory Physiology”)
E. WRONG - Because D. is correct.
D is obviously correct. But whoever wrote this question has no strict mathematical understanding of the word “gradient”. The gradient changes depending on which direction you are plotting the graph. There is no reason why the graph must be plotted as shown in West. Replace the word “greater” with “steeper” and this would make more mathematical sense.
RE06b [Feb12] version Apex compared to base of lung A. Lower ventilation to perfusion ratio B. Higher perfusion than at the base C. Higher transmural pressures D. Intrapleural pressure is less negative
For Feb 12 version A is wrong, V/Q higher at apex B is wrong, less perfusion at apex C is correct, alveolar pressure is the same throughout the lung and intrapleural pressure is more negative at apex, hence higher transmural (probably meant to be transpulmonary) D is wrong
RE07 [Mar96] Oxygen unloading: A. Increases with increased paCO2 B. Decreases with increase in temperature C. Decreases with increase in 2,3 DPG D. Decreases with decrease in pH E. ?
A right shift of the haemoglobin oxygen dissociation curve (ODC) indicates a decreased oxygen affinity, that is increased O2 unloading in the capillaries as indicated by a lower saturation at any particular pO2 value.
The haemoglobin ODC is shifted to the right by:
increased PaCO2
increased temperature
increased H+ concentration (ie decreased pH)
increased 2,3 DPG.
The only correct option is A: Increases with increased paCO2
[NB: This MCQ has not resurfaced since asked in 1996; this is presumably because it was so easy, it had a poor discrimination index and was eliminated from the ANZCA Question Bank]
RE08 [Mar97] Alveolar dead space: A. Is less than physiological dead space B. Is decreased with mechanical ventilation C. Is increased with hypotension D. Is measured by Fowler’s method
A: Correct. Alveolar dead space is a component (subset) of physiological dead space
B: Incorrect. Mechanical ventilation typically creates/increases alveolar dead space
C: Correct. (Pulmonary) hypotension can increase alveolar dead space by increasing zone 1 of the lung
D: Fowler’s method measures Anatomical dead space.
Physiological dead space is the part of tidal volume which does not take part in gas exchange. It consist of anatomical and alveolar dead space. Alveolar dead space is where alveoli is ventilated but not perfused. Pleural effusion, pneumothorax and CCF leading to pulmonary oedema all compromise ventilation more than perfusion and decrease alveolar dead space. So for RE08b, A,B & C are FALSE
Hypotension and alveolar dead space is a reference to “zone 1” of the lung from p43 of West
“…zone 1 does not occur under normal conditions…but may be present if the arterial pressure is reduced (eg following severe hemorrhage) or if alveolar pressure is raised (during positive pressure ventilation). This ventilated but unperfused lung is useless for gas exchange and is called alveolar dead space”
Therefore alveolar dead space is increased with hypotension so for RE08b, D is correct
Further discussed in Nunn’s:
“Low cardiac output (?relate to hypotension -ed), regardless of the cause, results in pulmonary hypotension and failure of the uppermost parts of the lungs(Zone 1). During anaesthesia with controlled ventilation, sudden changes in end-expiratory CO2 therefore usually indicate changing alveolar dead space secondary to abrupt variations in cardiac output.” - Nunn’s
RE08b [Jul98] [Jul99] [Feb00] [Jul02] Alveolar dead space is increased with: A. Pleural effusion B. CCF C. Pneumothorax D. Hypotension E. None of the above
A: Correct. Alveolar dead space is a component (subset) of physiological dead space
B: Incorrect. Mechanical ventilation typically creates/increases alveolar dead space
C: Correct. (Pulmonary) hypotension can increase alveolar dead space by increasing zone 1 of the lung
D: Fowler’s method measures Anatomical dead space.
Physiological dead space is the part of tidal volume which does not take part in gas exchange. It consist of anatomical and alveolar dead space. Alveolar dead space is where alveoli is ventilated but not perfused. Pleural effusion, pneumothorax and CCF leading to pulmonary oedema all compromise ventilation more than perfusion and decrease alveolar dead space. So for RE08b, A,B & C are FALSE
Hypotension and alveolar dead space is a reference to “zone 1” of the lung from p43 of West
“…zone 1 does not occur under normal conditions…but may be present if the arterial pressure is reduced (eg following severe hemorrhage) or if alveolar pressure is raised (during positive pressure ventilation). This ventilated but unperfused lung is useless for gas exchange and is called alveolar dead space”
Therefore alveolar dead space is increased with hypotension so for RE08b, D is correct
Further discussed in Nunn’s:
“Low cardiac output (?relate to hypotension -ed), regardless of the cause, results in pulmonary hypotension and failure of the uppermost parts of the lungs(Zone 1). During anaesthesia with controlled ventilation, sudden changes in end-expiratory CO2 therefore usually indicate changing alveolar dead space secondary to abrupt variations in cardiac output.” - Nunn’s
RE09 [Mar97] [Jul97] [Mar99] [Jul00] [Jul01] If dead space is one third of the tidal volume and arterial pCO2 is 45 mmHg, what is the mixed expired pCO2? A. 20 mmHg B. 25 mmHg C. 30 mmHg D. 45 mmHg E. 60 mmHg
The above equation appears not to be working so the answer is: Bohr Equation Vd/Vt = (PaCO2-PeCO2)/PaCO2 ∴ 1/3 = (45-PeCO2)/45 ∴ 15 = 45 - PeCO2 ∴ PeCO2 = 30 Answer is C
RE10 [Mar97] [Jul98] [Mar99] [Jul00] [Jul01] [Mar03] [Jul03]
With constant FIO2, CO and VO2, an increase in mixed venous O2 content would be seen with:
A. Hypothermia
B. Increased paCO2
C. Decreased 2,3 DPG
D. Alkalosis
E. None of the above
Alt wording:
Without a change in body oxygen consumption or cardiac output, mixed venous oxygen
tension increases with:
Alt wording (March 03):
With constant FIO2 and cardiac output and no change in position of ODC, mixed venous
blood oxygen tension increases with:
(see also CV47 ??same Q)
Jul03: If CO constant and ODC unchanged, mixed venous oxygen tension is decreased in: A. Cyanide toxicity B. Anaemia C. Hypothermia D. Hypercarbia E. ?
RE10 The below answers all have an effect on the oxygen dissociation curve… Question possibly remembered incorrectly too!
A - ? correct; hypothermia would decrease metabolic requirements (although VO2 is the same ruling this out); hypothermia will cause a left shift in the haemoglobin desaturation curve meaning that less oxygen has been liberated at the tissues??
B - increased paCO2 will cause a right shift and a decrease in SvO2
C - decreased 2-3 DPG will cause a left shift
D - Alkalosis will also cause a left shift
E -
July 03
B - anaemia; decreased oxygen carrying capacity in the blood will mean that greater extraction has to take place from the blood delivered to the tissues and hence a lower SvO2; also remember O2 content = 1.34Sat%Hb + 0.003*pO2, so with lower Hb, there will be lower O2 content
(but stem now says TENSION which is partial pressure not CONTENT.) - comment - a lower mixed venous content will also result in a lower mixed venous pO2/saturation
The relevant formula is Fick’s law as applied to oxygen uptake in the lungs. When re-arranged this results in:
Mixed venous O2 content = Arterial O2 content - (VO2 /CO)
If VO2 & CO are constant, then a consideration of this equation shows that the ONLY way for mixed venous O2 content to increase is if there is an increase in arterial O2 content. — Now, considering the oxygen flux equation, the ONLY ways to increase arterial oxygen content (if CO constant) are:
an increase in [Hb],
an increase in saturation, or
an increase in dissolved O2.
The only way to increase dissolved O2 content is to increase arterial pO2. Now given that FIO2 is fixed this is very unlikely to occur. Also if FIO2 is constant than it is unlikely that arterial O2 saturation will change. So the only possible option then is to increase [Hb]. If this is an option in the “actual” version of the question then it would be correct. If it were not an option, then the answer would be the “none of the above” option.
Consideration of factors which affect the position of the ODC are TOTALLY IRRELEVANT. The ODC is essentially a graphical way to convert a pO2 value into a Saturation value. This question talks about O2 CONTENT, not pO2, so there is absolutely no sense in referring to the ODC. As mentioned above, the only way that pO2 comes into it is because “dissolved O2 content” is proportional to pO2 (as per Henry’s Law), so it is possible that a large increase in arterial pO2 can significantly increase arterial O2 content. As per the Fick’s Law equation then this could result in an increase in mixed venous pO2. However, the MCQ says “fixed FIO2” thus effectively excluding this possibility.
COMMENT re: above. I may be wrong but I think possibly an increased CO2 may be the answer in the first stem.
CvO2 = CaO2 - VO2/CO and we are holding VO2, CO and FiO2 all constant
Therefore the only thing that can change is CaO2 which equals Hb x sats x 1.34 + PaO2 x 0.003
Sats is unlikely to change because FiO2 is constant. Therefore an increase in CO2 will shift the ODC to the right. If we hold sats constant on that new curve, the new PaO2 will be slightly higher. This means that CaO2 will be a bit higher (although not much as have to multiply by 0.003)
So although it’s a silly question and I think maybe remembered incorrectly, perhaps increased CO2 is the answer??
Further comments: With a fall of temp you get increased gas solubility as per Henry’s law - shouldn’t this mean you get an increase content of oxygen - (Probably a small increase but an increase none the less)? It will be a fall in PO2, but an increase in dissolved O2.
Reply: Yes this is correct. BUT the temp difference would be small (a few degrees maybe) and the increased amount of dissolved O2 would be very small indeed. Probably too little to even be able to measure. Thus, this would not be a meaningful option in my opinion. ALSO, with hypothermia, VO2 would decrease but the question (in the top version anyway) says constant VO2. The idea with this question is to understand and be able to apply Fick’s Law. The correct answer will depend on the actual wording, and the options on the actual MCQ on the day.
Comment I think Jul 03 - answer is anaemia.
need greater oxygen extraction per unit volume blood due to decreased oxygen carrying capacity, therefore decreasing pvO2
I think the first question would be “none of the above”, as most of them need the position of the ODC to change or you would not be able to have a change in oxygen extraction if you could not change position of ODC for a given CO, oxygen uptake, etc
For the first question, the only thing that could change the mixed venous O2 content given the constants - would be arterial oxygen content (if oxygen uptake, and CO are constant, then by the fick principle, mixed venous content would only change for arterial content)
if the question uses the term “tension”, then B would increase the pvO2 for a given venous content - as it R shifts the curve (as with all the constants - venous content does not change)
Comment re Jul 03 Anaemia would not be correct because O2 tension is not affected by Hb. A person can have normal O2 tension with Hb of 5. However Anaemia would lead to a decrease O2 CONTENT (not partial pressure).
yes you are correct, anaemia does not alter partial pressure. But this only applies to arterial side. As oxygen content is decreased that is supplying the tissue, with no change in oxygen consumption, there will be much less content left in venous blood, which will result in decreased partial pressure in venous blood.
The PO2 is related to the CONTENT of O2 through the ODC which is unchanged (as given in the question). Lower CONTENT equals lower PO2 because the other unchanged variables means the Hb oxygen complex will have a lower %O2 and if you follow the ODC at a lower saturation: a lower PO2. Which makes sense really.
Comment re Jul 03
I am thinking maybe hypercarbia could cause it perhaps if due to increased metabolism (eg Malignant hyperthermia) then it stands to reason that there would be increased in Co2 production / o2 consumption and then this would indeed reduce mixed venous o2 sats. but i guess on the D day with all options available and correctly worded it would be hopefully easier to chose the one best answer! As it stands its probably B or D
Hypercarbia without change in VO2 or Hb would lead to an increase in the PO2 from the right shift of the ODC. Without changing the ODC (!) yeah right I’d like to try controlling for that in that lab.
IMHO these are not the same question and should be dissociated.
From Peter Kam’s NSW primary course the answer to the first question with oxygen content is E but it is annotated that the question was abandoned due to an unclear best answer.
RE11 [Jul97] [Jul01] With altitude: A. Increased 2,3 DPG B. Increased oxygen unloading in peripheries C. Increased oxygen uptake in the lungs D. ? E. ?
Not sure how acute this question is.
Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below).
In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content.
Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by “normalising” the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day.
The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR.
As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude.
This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries.
Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days.
In brief:
Ventilation increases at altitude.
Cardiac output and heart rate will increase (& this increases tissue O2 delivery)
Plasma and red cell volume both increase
Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30%
Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance
Polycythaemia will develop
Red cell 2,3 DPG will increase due to decreased breakdown
Pulmonary and cerebral oedema are known complications of ascent to altitude.
However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs.
High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent.
High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent.
Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means…
References
RE11b
In acclimatisation to altitude:
A. P50 is reduced, improving O2 uptake in the lungs
B. P50 is increased, improving O2 offloading in the tissues
C. 2,3 DPG levels are reduced, improving O2 offloading in the tissues
D. Alkalaemia reduces the affinity for O2, increasing p50
E. Increase in 2,3 DPG and a decrease in P50
Not sure how acute this question is.
Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below).
In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content.
Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by “normalising” the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day.
The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR.
As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude.
This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries.
Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days.
In brief:
Ventilation increases at altitude.
Cardiac output and heart rate will increase (& this increases tissue O2 delivery)
Plasma and red cell volume both increase
Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30%
Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance
Polycythaemia will develop
Red cell 2,3 DPG will increase due to decreased breakdown
Pulmonary and cerebral oedema are known complications of ascent to altitude.
However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs.
High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent.
High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent.
Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means…
References
RE11c With acute acclimitisation to altitude: A. Hypoventilation B. Decreased cardiac output C. Pulmonary oedema D. Polycythaemia E. Increase in 2,3 DPG
Not sure how acute this question is.
Initially at altitude, acute hypoxaemia leads to increased ventilation (acting via peripheral chemoreceptors in aortic and carotid bodies) resulting in decreased arterial pCO2 secondary to increased exhalation; this inhibits the increase in ventilation (acting via the central chemoreceptors). This means that initial Hypoxic Ventilatory Response (HVR) is blunted but increases over the next 2-3 days as the pH of arterial blood and CSF is normalized (see below).
In native lowlanders, the carotid bodies increase in size and weight due to increasing vascularity. In native highlanders, they increase through hyperplasia. They respond to the lower arterial oxygen tension, not the lower oxygen content.
Over the next 2 to 3 days, the respiratory centre of the brain stem loses around four fifths of its sensitivity to pCO2 and H+ concentrations, thereby removing much of the hypocapnoeic inhibition on the ventilation increase. This increased ventilation then limits the amount of hypoxaemia (at the expense of a very low arterial pCO2). This is achieved firstly by “normalising” the pH of CSF by increased transport of HCO3- across the blood brain barrier (out of the CSF), and secondly (slower) by renal compensation for low pCO2 by increased excretion of bicarbonate. So by the third day or so, someone at altitude should be ventilating much greater volumes than they were on the first day.
The action of acetazolamide to stimulate respiratory drive is by inhibiting carbonic anhydrase, so that H+ ions accumulate in the CSF, lowering its pH and stopping the alkalotic blunting of the HVR.
As part of acclimatisation to altitude, the concentration of red cell 2,3 DPG increases. This is caused by hypoxic induced alkalosis, which inhibits the enzyme 2,3-DPG Diphosphatase that is responsible for the breakdown of 2,3-DPG. Slight elevations of the 2,3-DPG concentration are seen after the first few hours of exposure to altitude.
This leads to a right shift of the oxygen-haemoglobin dissocation curve, therefore increased P50 and increased oxygen unloading in the peripheries. This right shift is often ablated by the respiratory alkalosis which shifts the ODC to the left. ACIDaemia reduces the affinity of haemoglobin for oxygen, ie also shifts the curve right, therefore increasing P50 with resulting increased oxygen unloading in the peripheries.
Tissue hypoxia causes the JuxtaGlomerular Apparatus (JGA) to secrete Erythropoietin which stimulates erythropoiesis (increased by about 30%) in the bone marrow, increasing the haemoglobin concentration. EPO secretion begins to rise in the first few hours and is maximal at about 48 hours of exposure to hypoxia. Increase of Hb concentration is one of the slowest parts of acclimatization to altitude. The red blood cells maintain their normal lifespan of 120 days.
In brief:
Ventilation increases at altitude.
Cardiac output and heart rate will increase (& this increases tissue O2 delivery)
Plasma and red cell volume both increase
Coronary blood flow decreases and myocardial oxygen extraction increases, both by about 30%
Hypoxic Pulmonary Vasoconstriction increases the Pulmonary Vascular Resistance
Polycythaemia will develop
Red cell 2,3 DPG will increase due to decreased breakdown
Pulmonary and cerebral oedema are known complications of ascent to altitude.
However, at extreme altitude, a left shift of the oxygen-haemoglobin dissociation curve is seen, secondary to extreme respiratory alkalosis. this leads to reduced P50 and improved O2 uptake in the lungs.
High altitude pulmonary oedema occurs acutely in unacclimatised people who ascend to altitude. They experience the usual symptoms of pulmonary oedema, and the condition may be fatal. Treatment is descent.
High altitude cerebral oedema mirrors high altitude pulmonary oedema. Features are confusion, ataxia, hallucinations, and eventually loss of consciousness and death; again, treatment is descent.
Notes: It would be helpful to know how rapidly the onset of polycythaemia occurs - this paper (http://jap.physiology.org/cgi/content/abstract/81/2/636) notes that there is no change within 13 days - so NOT an acute effect by any means…
References
RE12 [d] [Jul98] [Jul01] Central chemoreceptors: A. Bathed in CSF B. Respond to increase in CSF pH C. Bathed in ECF D. In medullary respiratory centre
Central chemoreceptors were thought to be in the medullary respiratory centre, but are now recognised as separate. They:
lie on the surface of the anterior medulla, a short distance away from the respiratory neurones of the medulla (respiratory centre).
are bathed in ECF, the composition of which is determined by CSF, local blood flow and metabolism.
respond to increased hydrogen ion concentration, although the precise mechanism by which pH change causes stimulation of these chemoreceptors is still unknown.
are mediated by muscarinic ACh receptors.
Some light seems to have been thrown on to the mechanism by which pH canges cause stimulation of chemoreceptors: (See: Am J Physiol Cell Physiol 287: C1493–C1526, 2004.) which suggests an interplay of intracellular Ca, gap junctions, oxidative stress, and neurotransmitters regulate the action of K+ and Ca2+ channels.
Answers:
A - false, bathed in ECF not CSF
B - false, respond to ECF pH not CSF pH
C - true
D - false, close to (but not in) the respiratory centre
Comment
This is a terrible question and I hope it’s been abandoned. True, the chemoreceptors are bathed in ECF. However this ECF is right next to CSF and according to West, CSF is “apparently the most important” factor in determining the ECF pH. Furthermore at the top of p107 he summarises the preceding section with three points, one of which is” “(Central Chemoreceptors:) respond to the change in pH of the ECF/CSF”.
This is one of those anal questions that has no application in clinical practice and is really testing us on whether or not we have read a prescribed text in detail.
In conclusion, B is true, however I agree that C is “more” true.
RE13 [d] [Jul98] [Mar99] [Apr01] [Jul01] [Jul02] [Mar03] [Jul03] [Feb04] - 15A The peripheral chemoreceptors: A. Have a nonlinear response to paO2 changes B. Have an intact response at 1MAC C. Respond to a fall in paCO2 D. Respond slowly to rise in paCO2 E. Respond to alkalaemia F. Respond only to ?incr-/decr-eased H+ G. Respond only to arterial hypoxaemia H. Innervated by glossopharyngeal nerve I. Low metabolic rate J. Stimulated by carbon monoxide K. Stimulated by cyanide L. Blood flow of 2 ml/gram/min (OR Blood flow of 200mls/G/min) M. Aortic body innervated by vagus N. Changes in arterial oxygen content O. Low O2 extraction (OR: Low A-V O2 difference P. Have glomus cells
A. they do have a nonlinear response to paO2 changes- in fact they register changes as high as pO2 500mmHg, but the response really gears up at around 70mmHg, and maximum response is below 50mmHg.
B. their response is eliminated by as little as 0.1 MAC- bad news for CO2 retainers
C. they respond to a rise in paCO2
D. they respond around five times more rapidly than the central chemoreceptors to an increase in paCO2
E. they respond to acidaemia (their response is to pO2, pH and pCO2).
NB:They also do respond to reduction in their perfusion rate (Nunn 63)
F. therefore they do not respond only to H+
G. or only to arterial hypoxaemia
H. they are innervated by Hering’s nerves -> Glossopharyngeal n -> dorsal respiratory centre of the medulla
I. they have a high metabolic rate
J. Probably not. Nunn (6th p65) states Cyanide and Carbon Monoxide are chemical stimulants due to interference with the cellular cytochrome system (Nicotine and acetylcholine are also stimulants, via sympathetic ganglia). However, while Ganong (19th p644) agrees CO is toxic to the tissue cytochromes, this is at an amount 1000x the lethal dose, so plays no role in clinical poisoning. Ganong goes on to say CO is not a stimulant of the peripheral chemoreceptors as the PaO2 remains normal. Also Kam (2nd p109): “The peripheral chemoreceptors are stimulated by low oxygen tension (PaO2), not low oxygen content (CaO2) of blood. Conditions with low blood oxygen content but relatively normal oxygen tension (anaemia, carboxyhaemoglobin) do not stimulate ventilation via the peripheral chemoreceptors.
K. Cyanide is a chemical stimulants due to interference with the cellular cytochrome system, decreasing ATP synthesis. Nunn (6th p65)
L. blood flow is 20 times their weight per minute- 20mL/g/min or 2,000mL/100g/min
M. the aortic bodies are innervated by the vagus -> dorsal medullary respiratory centre
N. No. See J.
O. depite their high metabolic rate, their small size and the enormous amount of blood flow results in very little oxygen being removed; therefore they have a very low A-V O2 difference, and respond to arterial rather than venous pO2
P. yes they have Type I and II glomus cells
RE13 - 15A version (See also RE76 - also from 15A)
Peripheral chemoreceptors:
A. Respond to decreased O2 saturation
B. Respond to increased arterial pH
C. Respond to decreased arterial CO2 tension
D. Nonlinear increase with arterial oxygen tension
E. Slow response to changes in arterial carbon dioxide tension
Feb 04 Version Peripheral chemoreceptors: A. In the carotid sinus B. Have glomus cells C. Low A-V difference D. Innervated by glossopharyngeal nerve E. Blood flow of 200mls/g/min
A. the carotid bodies are at the carotid bifurcation, near the carotid sinus (which contains the baroreceptors).
B. yes, contain glomus cells
C. yes, they have a low A-V difference
D. yes, they are innervated by the glossopharyngeal nerve (via Hering’s nerves)
E. see L above
RE13b [Feb04]
Carotid bodies (Similar to RE13)
A. Have glomus cells
B. Innervated by vagus
C. Blood flow of 200mls/g/min
D. High A-V difference
E. Afferents via CN IX
(See also:RE36)
A. true
B. false. the aortic bodies are innervated by the vagus, the carotid bodies by the glossopharyngeal nerve
C. true. See L of the first question (L states 20mls/g/min, C is false.)
D. false. they have a low A-V difference, despite high metabolic rate, because they have a very high blood flow
E. true. Ganong (18th p629) Afferents from the carotid body ascend to the medulla via the carotid sinus and IX nerves, fibres from the aortic bodies ascend in the vagi.
RE14 [d] [Jul98] [Jul99] [Jul00]
Surfactant:
A. Causes hysteresis (Or: Is the ONLY cause of hysteresis)
B. Is produced by type 1 pneumocytes
C. Is commonly deficient in term neonates
D. Acts like detergent in water
E. Reduces the amount of negative intrapleural pressure
F. Production is slow
G. Increases pulmonary compliance
Re RE14:
A. not sure- but the saline-filled lung pressure-volume curve doesn’t show hysteresis
Surfactant is a radially- dependent surface tension reducing agent - providing more reduction in surface tension with smaller radius ie it counteracts Laplace’s Law. Also Fig 7.7B West shows the lung extract (surfactant) change in surface tension with area follows the hysteresis curve. Can these be correlated?? My conclusion is that surfactant is at least involved with hysteresis.
B. produced by type II pneumocytes
C. is commonly deficient in premmies
D. Yes. Detergent molecules (hydrophobic) repel each other, reducing the attractive forces between water moleculres and thus surface tension. Surfactant does the same.
(NO! Look at Fig 7-7, p.101, West , 7th Ed. Detergent does reduce surface tension, but it is independent of area, whereas dipalmitoyl phosphatidylcholine (DPPC) curve is different [hysteresis]. Would appreciate comments!!)
E. Yes. Reduction of surface tension reduces static lung recoil pressure, therefore reduces the amount of negative IPP
F. production is fast, but so is turnover
G. true, it increases pumonary compliance
RE14b [Jul04]
Surfactant
A. Surface tension is inversely proportional to surfactant concentration
B. Lung compliance decreases with surfactant
C. Is produced by alveolar type 1 cells
D. Stabilises alveoli to allow smaller alveoli to empty into larger ones
E. Increases surface tension in smaller alveoli to promote stability
Re: RE14b:
A. true, surface tension is decreased with increasing surfactant concentration
B. lung compliance increases with surfactant
C. produced by type II cells
D. stabilises alveoli to prevent them emptying into bigger ones
E. decreases surface tension in smaller alveoli to promote stability
Hysteresis is shown by all other elastic bodies. It is due to intrisic elastic forces within the lung tissues. Nunn Page 30)
RE15 [Jul97] [Apr01] In quiet breathing, exhalation is: A. Passive due to elastic tissue alone B. Passive due to surface tension in the alveoli and elastic tissue recoil C. Active due to intercostal contraction D. ? E. ?
Work of breathing = Work to overcome elastic recoil + Work to overcome resistance forces
2 factors are responsible for the lung’s elastic recoil - surface tension + elastic tissues
During quiet breathing, expiration is passive in that no active muscular contraction occurs, the energy used for expiration is the stored elastic energy achieved by inspiration.
Thus the answer should be B. Passive due to surface tension in the alveoli and elastic tissue recoil.
RE16 [d] [Mar98] [Jul98] [Apr01] [Mar03] [Jul03] [Jul09]
The normal arterio-venous difference for CO2 is:
A. 2 ml/100ml
B. 4 ml/100ml
C. 6 ml/100ml
D. 8 ml/100ml
E. 10 ml/100ml
(Mixed venous blood contains 52 mlsCO2/100mls blood & arterial blood
contains 48 mlsCO2/100 mls blood.)
Correct answer : B Venous CO2 : 53.5ml/100m blood Arterial CO2: 49ml/100ml blood A-V difference : 4.5ml/100ml blood Refer arterial and mixed venous CO2 dissociation curves
Ganong (23rd ed, pg 611, table 36-1) gives slightly different values, but with a similar outcome:
arterial CO2 = 46.4 mL/dL
venous CO2 = 49.7 mL/dL
difference = 3.3 (close to 4)
RE17 [d] [Jul98] [Mar99] [Jul00] [Apr01] [Jul01] [Mar02] [Jul02] [Jul04] [Jul09] The lung: A. Removes/inactivates serotonin (5HT) B. Activates bradykinin C. Converts angiotensin II to I D. Inactivates aldosterone E. Takes up noradrenaline
Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs.
Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung.
Up to 80% of bradykinin is inactivated (by ACE).
Angiotensin I is converted to angiotensin II (by ACE).
aldosterone is unaffected.
Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO.
Vasopressin is unaffected.
ADH is unaffected.
The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites.
Prostaglandins E2 and F2alpha are removed.
Leukotrienes are removed.
prostaglandin A2 and prostacyclin are not affected.
Histamine is unaffected.
Surfactant has a fast turnover, but I’m not sure whether it’s actually degraded. Guyton: ‘surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells
Which of the following substances is removed (?inactivated) by the lungs? A. Serotonin B. Noradrenaline C. Angiotensin I D. Bradykinin E. All of the above
Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs.
Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung.
Up to 80% of bradykinin is inactivated (by ACE).
Angiotensin I is converted to angiotensin II (by ACE).
aldosterone is unaffected.
Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO.
Vasopressin is unaffected.
ADH is unaffected.
The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites.
Prostaglandins E2 and F2alpha are removed.
Leukotrienes are removed.
prostaglandin A2 and prostacyclin are not affected.
Histamine is unaffected.
Surfactant has a fast turnover, but I’m not sure whether it’s actually degraded. Guyton: ‘surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells’
References
Which of the following is inactivated in the lung: A: Angiotensin II B: Angiotensin I C: Bradykinin D: Vasopressin E: Noradrenaline
Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs.
Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung.
Up to 80% of bradykinin is inactivated (by ACE).
Angiotensin I is converted to angiotensin II (by ACE).
aldosterone is unaffected.
Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO.
Vasopressin is unaffected.
ADH is unaffected.
The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites.
Prostaglandins E2 and F2alpha are removed.
Leukotrienes are removed.
prostaglandin A2 and prostacyclin are not affected.
Histamine is unaffected.
Surfactant has a fast turnover, but I’m not sure whether it’s actually degraded. Guyton: ‘surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells’
References
Metabolic functions of the lung include which one of the following?
A. Inactivates ADH
B. Converts Angiotensin II to Angiotensin I
C. Activates bradykinin
D. Inactivate serotonin (5HT)
E. Activation of prostaglandins
Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs.
Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung.
Up to 80% of bradykinin is inactivated (by ACE).
Angiotensin I is converted to angiotensin II (by ACE).
aldosterone is unaffected.
Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO.
Vasopressin is unaffected.
ADH is unaffected.
The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites.
Prostaglandins E2 and F2alpha are removed.
Leukotrienes are removed.
prostaglandin A2 and prostacyclin are not affected.
Histamine is unaffected.
Surfactant has a fast turnover, but I’m not sure whether it’s actually degraded. Guyton: ‘surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells’
References
Which biologically active substances are partially ?degraded by the lung? A. Surfactant B. Histamine C. Angiotensin D. Noradrenaline E. ?all/?none of the above
Endothelium is the most metabolically active cell type in the lungs. It may activate, deactivate or remove hormones that pass through the lungs.
Serotonin is removed by the lung (through uptake and degradation) with 98% taken up in single pass through lung.
Up to 80% of bradykinin is inactivated (by ACE).
Angiotensin I is converted to angiotensin II (by ACE).
aldosterone is unaffected.
Up to 30% of noradrenaline is taken up into the endothelium where it is metabolised by MAO and COMT. Dopamine, isoprenaline, histamine and adrenaline are unchanged as they are NOT taken up into the endothelium where they can be metabolised by MAO.
Vasopressin is unaffected.
ADH is unaffected.
The lung is a major site of synthesis, uptake, metabolism and release of arachidonic acid metabolites.
Prostaglandins E2 and F2alpha are removed.
Leukotrienes are removed.
prostaglandin A2 and prostacyclin are not affected.
Histamine is unaffected.
Surfactant has a fast turnover, but I’m not sure whether it’s actually degraded. Guyton: ‘surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells’
References
RE18 [] [Mar98] [Jul98] Breathing oxygen : A. Causes pain on re-expansion of collapsed alveoli B. Reduces vital capacity C. ? D. ?
Breathing 100% Oxygen can cause absorption atelectasis therefore reduces vital capacity.
Question not specific enough in this form - I breathe oxygen everyday without any problems. me too. Count me in too.
I believe A is the correct answer. Nunn’s 5th Ed, pg 564. “ Voluntary maximal inspirations have been effective in clearing areas of absorption atelectasis in subjects who had been breathing oxygen near residual volume. The process imparted a distinctive tearing sensation in the chest, but rapidly restored the PO2 and CXR to Normal.”
RE19 [] [Mar98] [Jul98] [Feb00] [Mar02] [Jul02] Contribution to the increase in CO2 carriage as blood passes from artery into vein: Carbamino HCO3 Dissolved A. 5% 90% 5% B. 30% 60% 10% C. ? D. ? E. ?
CO2 is carried in the blood in three forms:
Bicarbonate
Carbamino compounds
Dissoved CO2
In arterial blood, the proportions of CO2 present in these 3 forms is as in option A in the question. The majority of CO2 (90%) is present as bicarbonate.
However, the question is asking about the percent contribution of these three forms of carriage to the additional CO2 picked up in systemic capillaries. The ‘standard’ textbook values are the figures in option B above.
Note the increased importance of carbamino compounds to the carriage of this extra 4mls CO2/dl of blood.
RE20 [Mar98] [Mar03]
Increased physiological dead space with:
A. Decreases with age
B. Anaesthesia
C. Supine position
D. Calculated from Bohr equation using end-tidal CO2
E. Calculated from endtidal CO2 and arterial CO2
F. Decreases with increase in anatomical dead space
G. Increases with PEEP
(see RE04 & RE08)
Physiological dead space is “the volume of gas not equilibrating with blood (wasted ventilation)” and comprises of the anatomical dead space plus the alveolar dead space. Anything that increases anatomical dead space (the volume of the conducting airways) or alveolar dead space (area of the lung where the alveoli are ventilated but not perfused) will increase physiological dead space.
Both B and G are true!
A. False: Physiological dead space increases by approx 1ml/yr from early adulthood.
B. Physiological dead space increases with anaesthesia - due to increased alveolar dead space the exact cause of which is unclear - it’s not a Zone 1. (Nunn’s 6th Ed. p310)
C. Supine position makes West Zone 1 unlikely (well, West Zones are in the upright lung, but the same principle applies - the effect due to gravity is less). [Clarification:”In the supine position differences in blood flow between apices and bases are replaced by differences between posterior and anterior regions”. That is, the V-Q relationship is much the same. And whilst “the VD/VT ratio decreases from a mean value of 34% in the upright position to 30% in the supine position”, this is actually “largely explained by differences in anatomical dead space”. [Nunn 6th Ed p114 and 121]] (Physiologic dead space consists of anatomic + alveolar dead space. Anatomic reduces from 150mL to 100mL when moving from sitting to lying - Nunn 6th Ed Pg 119)
D. The Bohr equation uses the pCO2 of mixed expired gas, not end-tidal pCO2.
E. See D. The Enghoff modification to the Bohr equation used arterial pCO2 as an estimate of average (ideal) alveolar pCO2.
F. False, see the definition above.
G. Yes, but only a slight increase in the short term in normal lungs. Long term PEEP can cause a greater increase in physiological dead space. (Nunn’s Respiratory Physiology 6th Ed. p 433). See below quote and also answer to B in RE20b below.
“The acute application of moderate amounts of PEEP causes only a slight increase in VD/VT ratio. The alveolar component of physiological dead space may be increased by ventilation in patients with lung injury or when mean intrathoracic pressure is high, such as with significant amounts of PEEP. Under the latter conditions, lung volume is increased to such an extent that not only does cardiac output fall but pulmonary vascular resistance rises as well. Perfusion to overexpanded alveoli is reduced and areas of lung with high V/Q ratios develop, which constitute alveolar dead space. In healthy lungs this is not seen until PEEP levels exceed 10-15cm H2O. However with IPPV in lung injury there is now good evidence that overdistension occurs in the relatively small number of functional alveoli and local perfusion to these lung units is likely to be impeded. There is indirect evidence that long term application of PEEP may cause a very large increase in the dead space, probably because of bronchiolar dilation.” - Nunn’s Respiratory Physiology p433
RE20b [Jul98] [Feb00] Physiological dead space increases with: A. Pulmonary hypertension B. Hypotension C. Atelectasis D. Pleural effusion E. None of the above
Here the best answer is B: Hypotension
A. Pulmonary hypertension: Pulmonary vascular obstruction can occur late in pulmonary hypertension, which would lead to an increase in alveolar dead space. (Decreased perfusion also increases this, so perhaps A is a possible answer)
B. Hypotension: This can lead to the development of West Zone 1 - where pA > pa > pv (i.e. alveolar pressure is greater than pulmonary arterial pressure, therefore no perfusion occurs) leading to an increase in alveolar dead space, (which by definition increases physiological dead space).
C. Atelectasis: This causes a region of lung to not be ventilated - therefore it can’t contribute to dead space.
D. Pleural effusion: occurs outside the lung, so won’t increase dead space directly.
RE21 [Mar98] [Mar99] [Feb00] Shunt can be calculated by knowing: A. Cardiac output B. Arterial oxygen content C. Mixed venous oxygen content D. End pulm. capillary oxygen content E. All of the above
To calculate shunt, you need all of the above.
shunt blood flow/total blood flow = (oxygen content in end capillary blood - oxygen content in arterial blood)/(oxygen content in end capillary blood - oxygen content in mixed venous blood)
note that oxygen CONTENT is in mL/L (not pO2 - but can figure it out from pO2 by using ODC) and flows are in L/min
total blood flow = cardiac output (L/min)
CcO2 - pulmonary end capillary oxygen content is assumed to be that of oxygen content that you would have with PAO2 as the blood at the end of the pulmonary capillaries has equilibrated with the alveoli.
summarised a little more succintly as
Qs/Qt = (CcO2 - CaO2)/(CcO2 - CvO2)
which i like to think of as
proportion of shunt flow to total flow = proportion of impairment of oxygenation to oxygenation that should be achieved
which makes it a little easier for me to remember
Alveolar pressure:
A. Is always negative throughout normal quiet breathing
B. Is zero (atmospheric pr) during pause between inspiration and expiration
C. Is greater than 5-6 cm H2O during quiet expiration
D. Is less than 5-6 cms H2O during quiet inspiration
During normal quiet breathing alveolar pressure oscillates between -1 to 1 cmH2O, and is zero during pauses between inspiration and expiration. So B and D are correct.
In the second version, D is correct.
Alveolar pressure during quiet breathing:
A. 5 cmH2O negative at inhalation
B. 5 cmH2O positive at expiration
C. Follows intrapleural pressure closely
D. Is atmospheric between inhalation & exhalation
During normal quiet breathing alveolar pressure oscillates between -1 to 1 cmH2O, and is zero during pauses between inspiration and expiration. So B and D are correct.
In the second version, D is correct.
RE23 [Mar99] [Apr01] [Jul03] [Feb04]
Patient with chronic airflow limitation:
A. Gradient maximal in effort independent part of flow volume loop
B. Will have increased total lung capacity
C. Has increased static compliance
D. ?
FIRST VERSION
Best answer is B
A: INCORRECT: Shallower gradient in effort dependent part of the loop
C: INCORRECT: increased volume means that the lung compliance curve is at the top shallower (less compliant) end.
Nunn’s (5th ed p48) says that ‘emphysema is unique in that static pulmonary compliance is increased’ - does this mean that C could also be true, or is it going to far to assume that they mean COPD/emphysema when they say chronic airflow limitation?
Comment: I believe that chronic airflow limitation (CAL) is an old acronym for what is now called COPD.
Primary MCQs-Aug2015 In chronic obstructive airways disease A. Increased total lung capacity B. increased slope of effort dependent portion C. increased vital capacity D. reduced compliance
2015 VERSION
A: CORRECT: secondary to hyperinflation from gas trapping
B:INCORRECT: shallower slope of effort dependent portion
C: INCORRECT: although total lung capacity increases, much of this is gas trapping so is part of RV rather than VC
D:INCORRECT: increased compliance
RE24 [Jul98] [Mar99] [Jul00]
One lung anaesthesia:
A. High FIO2 will completely correct paO2
B. CPAP will completely correct paO2
C. Supine position will give better VQ matching
D. Associated with hypercarbia
One lung anaesthesia creates a large shunt, hence A and B are incorrect. Lying ventilated lung side down would theoretically give better V/Q matching of ventilated lung - helping unventilated lung become West’s Zone 1. D would be correct, hyperventilation of the one lung may not completely abolish the hypercarbia.
PEEP - Ventilated lung should have mild PEEP to avoid atelectasis and maximise ventilation of this lung. However, PEEP to ventilated lung may exacerbate hypoxia by reducing cardiac output and forcing more blood through the uppermost, non-ventilated lung. It may be tried to remedy hypoxia or with a ‘sinking’ mediastinum, but the results are unpredictable. I think in the context of the question the answer is false to these parts. (A-Z under ‘one-lung anaesthesia’, Oxford Handbook ref. from Miller)
Initial version
a - high FiO2 will completely correct PaO2
incorrect
one lung ventilation creates an obligatory right to left transpulmonary shunt through the non-ventilated non-dependent lung. For the same FiO2 (as two lung ventilation), one lung ventilation results in a much larger PAO2-PaO2 gradient and lower PaO2 than two lung ventilation. – Miller’s 6th ed., One Lung Ventilation
i.e. Large shunt from non-ventilated lung means that PaO2 cannot be completely corrected by increased PAO2 in ventilated lung.
b - CPAP will completely correct PaO2
incorrect
applying PEEP to the ventilated lung has the risk of PEEP induced increase in lung volume which can cause compression of small dependent lung intra-alveolar vessels and increase this lungs PVR and therefore resulting in a shunt to non-ventilated lung –> increasing shunt and decreasing PaO2. PEEP is a trade off between positive effect of increased FRC and negative effect of increased PVR and shunt Millers 6th One lung ventilation section
However, mild CPAP to non-ventilated lung can be helpful in decreasing the shunt, but won’t completely correct it.
i.e. Large shunt from non-ventilated lung means that PEEP to ventilated lung cannot completely correct PaO2.
c - supine position will give better V/Q matching
incorrect
ventilated (dependent) lung has better V/Q matching in a lateral position, compared with supine, due to gravity effect of improving perfusion
d - associated with hypercarbia
correct
during one-lung ventilation, ventilated lung can eliminate enough carbon dioxide to compensate for non-ventilated lung. Overtime however retention of CO2 from blood traversing non-ventilated lung usually slightly exceeds increased elimination of CO2 from blood traversing ventilated lung, PaCO2 will usually slowly increase. – Miller’s 6th ed., One Lung Ventilation
July 2000 version:
With regards to hypoxia with one lung anaesthesia:
A: Oxygenation is better supine
B: Should have 10cm H2O PEEP to lower lung
C: Is usually associated with hypercarbia, (?can be associated with hypercarbia)
July 2010 version
a - oxygenation better in supine
false
oxygenation better with ventilated lung as the dependent lung in LDP (lateral decubitus position)
b - should have 20 cmH2O PEEP to lower lung
false
PEEP should be cautiously applied as above
c - usually/can be associated with hypercarbia
true
can be associated with hypercarbia as above
Just ventilate more?
The patient is anaesthetised and likely ventilated. Whether hypercarbic or not is completely dependent on how much you ventilate him. It’s more likely that you would over ventilate than under ventilate, to keep up PO2. So unless you are slack, shouldn’t this be usually associated with low PCO2?
In the normal surgical patient, this is essentially correct. However, in one-lung ventilation, there is a dramatic increase in shunt. There is a limit as to how much you can ventilate a patient, especially when you have only one lung available. Oxygenation is worse (but with SpO2 typically >90%), but increased ventilation is limited. Hence, since PaCO2 ~ PACO2 ~ VCO2/VA (i.e. CO2 arterial tension is proportional to production / alveolar ventilation), the decrease in ventilation will tend to result in hypercarbia.
Mild CPAP to the non-ventilated lung can also be helpful at maintaining oxygenation. (I think this is why there is an (incorrect) answer regarding CPAP to the ventilated lung.)
Ref: One-lung ventilation and arterial oxygenation. Current Opinion in Anaesthesiology. 24(1):24-31, February 2011.
RE25 [Jul98] [Mar99] [Mar03] [Jul03] [Jul07]
The partial pressure of oxygen in dry air at sea level:
A. 163 mmHg
B. 159 mmHg
C. 149 mmHg
D. 100 mmHg
E. ?
This can be calculated as follows: Assuming: barometric pressure (PB) = 760mmHg inspired O2 concentration (FIO2) = 0.21 dry air so pH2O = 0 mmHg then: pIO2 = (0.21)(760 - 0) = 159.6 mmHg
RE26 [Mar99] [Jul04] Cause of increased minute ventilation with exercise: A. Oscillation in paO2 & paCO2 B. Hypercarbia C. Hypoxaemia D. Acidosis E. None of the above
None of the above.
None of them become deranged enough to exert an effect. Something else causes an increase in ventilation before they can be affected. Maybe a CNS effect.
West discusses A. (page 133) but goes on to say that no current theory adequately explains observed increase in ventilation.
Agreed None of the above is the most suitable answer. Muscle mechanoreceptors & raised body temperature (via higher centres) thought to contribute. Faunce mentions a possible role of increased plasma K+ (via peripheral chemoreceptors) & of PCO2 fluctuations.
Johnson BD - Chest - 01-AUG-1999; 116(2): 488-503 does not directly answer the question however does provide an insight into minute ventilation changes during exercise.
However, Power&Kam say on pages 97 and 326.
The rapid response in minute ventilation at the onset of exercise result from “neural inputs to the respiratory centre from the motor cortex and proprioceptive receptors in the exercising muscle”.
The gradual response in minute ventilation that comes later may be related to fluctuations in arterial oxygen tension and hence the “oscillatory discharge of central chemoreceptors provides a potent respiratory stimulus in exercise”. Nunn 6thEd’ (p244) says that “peripheral chemoreceptors do contribute, in a small way, to exercise-induces hyperpnoea”.
Other factors are “the release of epinephrine and norepinephrine, and a rise in blood temperature”.
So “A” is possible?
The sentence after the above Nunn quote is “In spite of this caveat, it is difficult to avoid the conclusion that arterial blood gas tensions acting on the chemoreceptors cannot be the main factor in the increase of ventilation during exercise. Option E probably remains the best.
West 8th Ed P135: “One hypothesis is that oscillations in arterial PO2 and PCO2 may stimulate the peripheral chemoreceptors, even though the mean level remains unaltered.” So A is definitely possible.
Nunns applied physiology- 7th edition-page-266 Peripheral chemo receptors are responsible for exercise induced hyperopnoea- during the non steady state. This may not result from changes in Po2, but from oscillations in arterial Pco2. Humoral factors are important in heavy and severe exercise. Lactic acidocis contributes to excess ventilation during heavy and severe exercise. Slight additional respiratory drive may result from hyperthermia.
I agree with the answer A since it provides a possible mechanism as West suggested which makes sense. We know that hypoxaemia/hypercarbia/acidosis do not occur during exercise.
Comment: perhaps the answer lies in the difference between the words “cause” and “hypothesis”?
RE27 [Jul99] [Feb00] [Apr01] Work of breathing (as % of total VO2) in normal healthy adult:: A. 1% B. 5% C. 10% D. 20%
As worded, the question is not asking about efficiency, but about how much of basal energy expenditure is devoted to respiration - “…as % of total VO2…”
Work of breathing is ~3mL O2/min, or 1% of basal O2 consumption, as per Brandis page 148,
RE28 [Feb00] [Mar03] [Jul03] PEEP: A. Has a variable effect on FRC B. Reduces lung compliance C. Reduces lung water D. Reduces airway resistance E. No effect on lung compliance
PEEP:
Increases FRC
Increases lung compliance in lower dependant parts of the lung and reduces compliance in non-dependant parts (Nunn 6th Ed, pg 434) therefore also improving VQ matching
Does not reduce (total) lung water, but may redistribue extravascular lung water.
Reduces airway resistance
So answer should be D.
PEEP also improves V/Q inequalities (Morgan & Murray)
I agree with above statements, agree answer D and also add:
significant PEEP (raised intrathoracic pressure) can increase alveolar component of physiological dead space, can also occur with ventilation in lung injury (high V/Q ratios, which constitute alveolar dead space, don’t develop in healthy lungs until PEEP > 10-15cmH20) Nunn 6th Ed, p 432
PEEP doesn’t change total amount of lung water but greater proportion in extra-alveolar interstitial space and increased lymphatic drainage (in haemodynamic pulmonary oedema in dogs) Nunn p 392
animal models of pulmonary oedema indicate that increased lung volume increases capacity of interstitium to hold liquid Nunn p 392
increasing lung volume decreases airway resistance and helps to prevent gas trapping (achieved with CPAP in spontaneously breathing or PEEP in ventilated patient) Nunn p 44
At an atmospheric pressure of 247 mmHg, what is the moist inspired pO2? A. 200 mmHg B. 2 mmHg C. 40 mmHg D. 50 mmHg E. ?
Step 1: Assuming by “moist” inspired pO2 it means saturated air at 37C in the airways (so pH20 = 47mmHg), then:
pO2 = (247 -47) x 0.21
Step 2: Now approximating 0.21 to 0.2 (ie a one fifth so divide by 5 which is easy) then:
pO2 = 200/5 = 40mmHg.
Query: The MCQ is asking about moist ispired pO2. I think we don`t need to subtract 47 mmHg from 247 mmHg. So answer is D 50 mmHg?
Response: This is of course a ‘remembered’ version of the actual MCQ so the wording is not perfect. I think that 47 has to be subtracted for the following reasons:
If you think the question means the air is already ‘moist’, then that will only be for the ambient temperature (which is not stated) and in any case it doesn’t say ‘saturated’ so some value for pH2O will have to be subtracted but you have no way of knowing what. (SVP is temp dependent). So the assumption you make (if you followed it correctly) just renders the MCQ unanswerable.
Being ‘moist’ before inspiration still dilutes the ambient pO2 in any case but as you do NOT know the ambient temperature then again you do NOT know what value of pH2O to use to correct for this. The MCQ on your assumption is unanswerable.
The given value of 247mmHg suggests that the MCQ is trying to make the calculation easy as “(247-47)/5” is easy mental arithmetic. The exam afterall is not a test of mathematical skill and MCQs that require calculation are set up so the calculation is easy to do. The problem is knowing the correct formula - in this case:
pO2 = (pB - pIO2) x FIO2
where pB is barometric pressure.
So, once you know that calculation MCQs have to:
have easy calculations (Exception if you are supposed to ‘know’ the answer as a standard value)
AND
that MCQs have to have an answer (Exception if the examiners actually make a mistake -which happens)
THEN
the confusion is overcome.
The expectation is that the candidates who do not understand the topic very well will get confused about whether to subtract 47 or not, and where to subtract it from, or whether to use 0.2 or not and the distractors will be designed to exploit this confusion, for example:
some might think it is: (0.2 x 247)-47 = 2 (which is option B);
some will forget that air is only 21% O2 and just use 247-47=200 (option A)
some may try (247-47)-(40/0.8) = 150mmHg (possibly the missing option E), and – incorrect alveolar gas equation yeap, would be (247-47)x0.21 - (40/0.8) = an impossible answer
some will try 247 * 0.2 = about 50mmHg (option D).
RE30 [Feb00] Type II pneumocytes A. Develop from type I pneumocytes B. Are macrophages C. Are very flat and practically devoid of organelles D. ?Metabolise surfactant
Type I pneumocytes are flat with minimal organelles. They are easily damaged, and then replaced by proliferating type II pneumocytes which then differentiate into type I pneumocytes. Neither are macrophages.
Type II have microvilli whereas Type I are a flat thin sheet. Surfactant works by sitting on top of the water layer anyway, so would be no good bound to a brush border. Type II pneumocytes produce surfactant.
neumocito.gif Type II pneumocyte secreting surfactant Surfactant is cleared from alveolar fluid by alveolar macrophages and type II cell reuptake (Faunce p21). Therefore, answer RE30 is D. II is bigger than I, therefore type II pneumocytes are fatter than type I pneumocytes, and have more equipment to do more stuff. Additional information: from Nunn 6th edition p 21,22,26,27 type 2 (alveolar epithelial cells) are stem cells and don't function as gas exchange membranes type 1 cells develop from type 2 type 2 cells contain stored surfactant in striated osmiophiluc organelles (surfactant formed and released from type 2 cells) type 2 are involved in pulmonary defence mechanisms, secrete cytokines and contribute to pulmonary inflammation type 2 are resistant to oxygen toxicity and replace type 1 cells after prolonged exposure to high oxygen concentrations (type 1 are very sensitive to damage from high O2 concentrations) type 2 have SP-A receptors, stimulation of these receptors produces negative feedback on surfactant secretion from type 2 cells and increases the reuptake of surfactant components into the type 2 cells
“Type II pneumocytes are plump, cuboidal cells that make up about 16% of the cell population of the
alveolar region. They have microvilli on their outer edges (seen in the scanning EM on the right).
They cover a small amount of alveolar surface area - only ~5% (compared with 95% of surface area
covered by Type I cells).”
clip_image002_000.gif
“Under normal conditions, Type II cells synthesize surfactant, store it within cytoplasmic vacuoles
(electron dense structures called lamellar bodies) and then release that surfactant onto the alveolar
surface where it forms a thin film. Surfactant serves to lower surface tension and therefore it helps
alveoli remain expanded, especially during the expiratory phase of breathing. Following injuries,
Type II cells proliferate and replace damaged Type I cells.” -from [1]
dustcells.jpg
Pneumocytes are naughty… when no one is looking ‘2 become 1’
RE30b [Jul00] Type I pneumocytes A: Give rise to Type II pneumocytes B: Are flat & minimal organelles C: Bind surfactant (? receptors) on their brush border D. ?
Type I pneumocytes are flat with minimal organelles. They are easily damaged, and then replaced by proliferating type II pneumocytes which then differentiate into type I pneumocytes. Neither are macrophages.
Type II have microvilli whereas Type I are a flat thin sheet. Surfactant works by sitting on top of the water layer anyway, so would be no good bound to a brush border. Type II pneumocytes produce surfactant.
neumocito.gif Type II pneumocyte secreting surfactant Surfactant is cleared from alveolar fluid by alveolar macrophages and type II cell reuptake (Faunce p21). Therefore, answer RE30 is D. II is bigger than I, therefore type II pneumocytes are fatter than type I pneumocytes, and have more equipment to do more stuff. Additional information: from Nunn 6th edition p 21,22,26,27 type 2 (alveolar epithelial cells) are stem cells and don't function as gas exchange membranes type 1 cells develop from type 2 type 2 cells contain stored surfactant in striated osmiophiluc organelles (surfactant formed and released from type 2 cells) type 2 are involved in pulmonary defence mechanisms, secrete cytokines and contribute to pulmonary inflammation type 2 are resistant to oxygen toxicity and replace type 1 cells after prolonged exposure to high oxygen concentrations (type 1 are very sensitive to damage from high O2 concentrations) type 2 have SP-A receptors, stimulation of these receptors produces negative feedback on surfactant secretion from type 2 cells and increases the reuptake of surfactant components into the type 2 cells
“Type II pneumocytes are plump, cuboidal cells that make up about 16% of the cell population of the
alveolar region. They have microvilli on their outer edges (seen in the scanning EM on the right).
They cover a small amount of alveolar surface area - only ~5% (compared with 95% of surface area
covered by Type I cells).”
clip_image002_000.gif
“Under normal conditions, Type II cells synthesize surfactant, store it within cytoplasmic vacuoles
(electron dense structures called lamellar bodies) and then release that surfactant onto the alveolar
surface where it forms a thin film. Surfactant serves to lower surface tension and therefore it helps
alveoli remain expanded, especially during the expiratory phase of breathing. Following injuries,
Type II cells proliferate and replace damaged Type I cells.” -from [1]
dustcells.jpg
Pneumocytes are naughty… when no one is looking ‘2 become 1’
RE31 [Jul00]
Control (?inspiratory) of the diaphragm originates in:
A. Pneumotactic centre
B. Apneustic centre in pons
C. Dorsal medullary (?neurons of) respiratory centre
D. Ventral medullary (?neurons of) respiratory centre
A. Pneumotaxic centre in upper pons inhibits the inspiratory ramp, therefore regulates inspiration volume and time.
B. Apneustic centre in lower pons has an excitatory effect on inspiratory area of medulla tending to prolong ramp action potentials (?function in humans)
C. Inspiratory control originates from the dorsal medullary respiratory centre.
D. Ventral medullary respiratory centre is associated with expiration
ie C is correct
RE32 [Jul00] For a normal Hb-O2 dissociation curve, the most correct relationship is: A. PaO2 340mmHg, SaO2 99% B. PaO2 132mmHg, SaO2 98% C. PaO2 68mmHg SaO2 ? D. PaO2 60mmHg, SaO2 91% E. None of the above
D. Pa 60mmHg, SaO2 91% (“ICU point”)
RE33 [Jul00] Alveolar dead space ??? A. Measured by Fowler’s method B. ?? C. ? D. ?
Fowler’s method measures anatomical dead space.
The Bohr equation calculates physiological dead space.
Alveolar dead space is physiological dead space minus anatomical dead space.
RE34 [Jul00] [Jul09] [Mar10]
Oxygen toxicity:
A: Is caused by superoxide dismutase (OR: Increased by increased SOD)
B: Causes CNS toxicity at over 100kPa
C: Is caused by absorption atelectasis
D: Is due to formation of superoxide radicals
E: Prolonged ventilation at 50kPa causes pulmonary toxicity
F. Causes lipid peroxidation
A: False. Superoxide dismutase is an antioxidant that protects against oxygen toxicity
B: False
C: False. Causes absorption atelectasis
D: True
E: False
F: True
Addit: Pulmonary oxygen toxicity occurs after 16 hours at levels of 50kPa or more (~375 mmHg or 50% oxygen) So E is true
West p146 says CNS toxicity at PO2 exceeding 760mmHg. doesn’t it mean B is correct?
Central nervous system toxicity is caused by short exposure to high concentrations of oxygen at greater than atmospheric pressure. Pulmonary and ocular toxicity result from longer exposure to elevated oxygen levels at normal pressure. Pulmonary toxicity occurs with prolonged exposure of 16–24 hours or more to elevated concentrations of oxygen greater than 50% at normal atmospheric pressure. O2 toxicity causes oxidative damage to cell membranes, the collapse of the alveoli in the lungs, retinal detachment, and seizures.
Additional comment: According to Wikipedia, need 160kPa for CNS toxicity. Atmospheric pressure is about 100kPa. Also says need more than 50kPa for O2 toxicity. http://en.wikipedia.org/wiki/Oxygen_toxicity. Agree with either D or F.
RE35 [Jul00] [Apr01]
Pulmonary stretch receptors:
A. ?
B: Are only stimulated by maintained stretch
C: Show (?slow) adaptation
D: Cause an immediate decrease in tidal volume
E. ?
C= True hmmm……
Pulmonary stretch receptors are aka slowly adapting pulmonary stretch receptors. Lie within airway smooth muscle. They discharge in response to distension of the lung, activity is sustained with lung inflation therefore showing little adaptation. Impulses travel via Vagus nerve.
Reflex is to slow respiratory frequency due to an increase in in expiratory time = the Hering-Breuer inflation reflex. Inflation of the lungs tend to inhibit further insp muscle activity. opposite is also seen = the deflation reflex.
These reflexes are largely inactive in humans unless tidal vol exceeds 1 litre
RE36 [Jul00] The peripheral chemoreceptors are located: A. Carotid sinus B. Carotid bodies C. The vasomotor centre D. ?
B= True
The peripheral chemoreceptors are located in the carotid bodies and the aortic bodies.
There’s a good outline of peripheral chemoreceptors at question RE13.
RE37 [Apr01] [Mar03] [Jul03] Mixed venous blood: A. Higher haematocrit than arterial B. Saturation of 48% C. Higher pH than arterial Blood D. Can be sampled from the right atrium E. pO2 lower than coronary sinus blood F. Coronary sinus O2 saturation of 30%
A and F true
Option A: For each CO2 molecule which diffuses into a RBC either an HCO3 or chloride atom appears inside cell (the latter due to chloride shift when some HCO3- (out)exchanges for a Cl- (in). This results in the presence of one osmotically active particle for each CO2, which attracts H2O and causes the RBC to swell slightly. This together with a very small amount of fluid returning via lymphatics means that the haematocrit of venous blood is normally about 3% greater than arterial blood.
Saturation is 75% so option B is wrong.
Higher pCO2 = 46mmHg means lower pH so option C is wrong
Option D is incorrect because true mixed blood incorporates venous blood from all venous beds, therefore cannot be sampled from the right atrium as blood not adequately mixed here. can only be sampled from pulmonary artery.
Option E: pO2 is 40mmHg (when breathing room air) whereas coronary sinus blood has a pO2 of 20mmHg.
Re option F:
Just wondering if there is a reference for ‘F’?
Brandis says pO2 = 20mmHg - and I’m not sure where this correlates on the ODC
Calculating it using O2 content = (1.39)(Hb)(SO2) + (0.003)(pO2)
assuming Hb=15; SO2=97; and ignoring the dissolved O2 portion…
And AV O2 difference for heart muscle being 114ml/L
O2 content in venous blood is 86ml/L or 8.6ml/L
which working backwards gives SO2 of 40% (ish)
Thoughts?
Alternatively: Oxygen extraction ratio across the coronary circulation is typically 55-65% (i.e. high compared to systemic circulation as a whole) so starting from a nominal 97% this would predict a mived venous saturation of 32-42%. Extraction ration can get as high as 70% which would result in 27%. So the 30% in option E is a tad low but not unreasonable, though 35-40% would be a more comfortable answer in this option.
Of course there are 6 remembered options (instead of 5) so this question is wrong. We can assume the “real” question would be more clearcut.
Another way of calculating sats from pO2 of 20mmHg for coronary sinus blood is as follows:
We know CaO2 (arterial content of blood) is approx 20mls O2 / 100mls (Hb x sats x 1.34 = 15 x 1 x 1.34)
We know that coronary blood flow is about 250mls / min.
Therefore the coronaries are getting about 50mls O2 / min
They extract about 55%. Meaning 45% is returned to the coronary sinus (ie 45% of 50 is 22.5 mls O2 in 250mls)
22.5 in 250 is equal to 9mls O2 / 100mls = CvO2
Working back from CvO2 = Hb x sats x 1.34 where we assume Hb is about 15, we get sats = 45%
PS i doubt we would have to do all this on a single MCQ!! A simpler way would be that we know p50 on the ODC is about 26. So 20mmHg must be a bit less than sats 50%!
Re F: Has a nice interactive ODC curve where you set pH/temp/PCO2 and then can enter Sat/PO2 to calculate the other. With CO2 46 and pO2 20–> gives Sat 33%
addit: based on ODC then the real sats would be even lower than 33% since the lower pH causes a right shift
RE38 [Apr01] Carbon dioxide carriage: A. 10% dissolved B. 30% carbamino C. 85% bicarbonate D. 60% bicarbonate E. Unaffected by pO2
The ‘standard’ textbook values for CO2 carriage are:
Bicarbonate - 90%
Carbamino compounds - 5%
Dissolved CO2 - 5%
Also the degree of oxygenation of haemoglobin has an important affect on the carriage of CO2 (due to the Haldane effect)
As written above, none of the options are correct but 85% bicarbonate is the closest and may well be the figure used in some sources. The 60,30,10 values refer to proportions for carriage of the AV difference for CO2.
See also the comments for RE19
RE39 - [Apr01] Factors that favour formation of carbamino-haemoglobin include: A. Carbonic anhydrase B. A decrease in oxygen tension C. An increase in oxygen tension D. A decrease in pH E. None of the above
B. True
Deoxyhaemoglobin is 3.5 times more able to form carbamino compounds (as compared to oxyhaemoglobin).
This is important in CO2 transport and accounts for 70% of the magnitude of the Haldane effect.
Carbonic anhydrase catalyses the reaction CO2 + H2O - H2CO3, not formation of carbamino acid. Decrease in O2 tension favours formation of carbamino-Hb through the Haldane effect. Decrease in pH (increase H+) would decrease formation of carbamino-Hb also decreased pH means the HbO2 curve is shifted to the right and therefore mixed venous PO2 is higher, therefore less reduced Hb and less H+ buffering for formation of HCO3 and less CO2 carried as carbamino
A reduction in pH causes decrease formation of Carbamino-Hb due to the Left shift of the following chemical equation:
Hb.NH2 + CO2 Hb.NHCOOH Hb.NHCOO- + H+
I agree with above, but perhaps here is a spanner in the works… The decrease in oxygen tension, would have to be significant enough to cause de-saturation of Hb, for the top statement to be correct. If the paO2 drops from 400mmhg, to 200mmhg for eg, then Hb saturation would be unchanged at 100% (meaning E could be the “trick” answer). (However, I hope they wouldn’t be that nasty, and that I have just got too much time on my hands to think about things like this)
Not sure how well this question was memorised, but it appears tricky to me. I would have thought a decrease in pH shift the ODC to right, i.e. promoting offloading of O2 from Hb. More DeoxyHb = increased ability to carry CO2 via H+ liberated when carbonic acid dissociates and via formation of carbamino-Hb.
Disagree with the above comment regarding decrease in oxygen tension being a ‘trick’ answer because even if PaO2 reaches 760mmhg, once the Hb is saturated the rest are just dissolved O2, and PvO2 is not going to be too much higher than having a PaO2 of 100mg. But I guess it also really depends on what ‘decrease in oxygen tension’ really means here – does it mean by decrease in O2 saturation with Hb??
Additional: The interaction of oxygen with Hb is at best described not as an absolute but as a reaction constant or probability if you like. Therefore the increase of PO2 does actually increase the saturation of Hb but to an increasingly small amount (likelihood).
Mar 2010 remembered version Which would increase carbamino-Hb formation? A. decreased pH B. increased carbonic anhydrase C. decreased carbonic anhydrase D. decreased pO2 E. ?
B. True
Deoxyhaemoglobin is 3.5 times more able to form carbamino compounds (as compared to oxyhaemoglobin).
This is important in CO2 transport and accounts for 70% of the magnitude of the Haldane effect.
Carbonic anhydrase catalyses the reaction CO2 + H2O - H2CO3, not formation of carbamino acid. Decrease in O2 tension favours formation of carbamino-Hb through the Haldane effect. Decrease in pH (increase H+) would decrease formation of carbamino-Hb also decreased pH means the HbO2 curve is shifted to the right and therefore mixed venous PO2 is higher, therefore less reduced Hb and less H+ buffering for formation of HCO3 and less CO2 carried as carbamino
A reduction in pH causes decrease formation of Carbamino-Hb due to the Left shift of the following chemical equation:
Hb.NH2 + CO2 Hb.NHCOOH Hb.NHCOO- + H+
I agree with above, but perhaps here is a spanner in the works… The decrease in oxygen tension, would have to be significant enough to cause de-saturation of Hb, for the top statement to be correct. If the paO2 drops from 400mmhg, to 200mmhg for eg, then Hb saturation would be unchanged at 100% (meaning E could be the “trick” answer). (However, I hope they wouldn’t be that nasty, and that I have just got too much time on my hands to think about things like this)
Not sure how well this question was memorised, but it appears tricky to me. I would have thought a decrease in pH shift the ODC to right, i.e. promoting offloading of O2 from Hb. More DeoxyHb = increased ability to carry CO2 via H+ liberated when carbonic acid dissociates and via formation of carbamino-Hb.
Disagree with the above comment regarding decrease in oxygen tension being a ‘trick’ answer because even if PaO2 reaches 760mmhg, once the Hb is saturated the rest are just dissolved O2, and PvO2 is not going to be too much higher than having a PaO2 of 100mg. But I guess it also really depends on what ‘decrease in oxygen tension’ really means here – does it mean by decrease in O2 saturation with Hb??
Additional: The interaction of oxygen with Hb is at best described not as an absolute but as a reaction constant or probability if you like. Therefore the increase of PO2 does actually increase the saturation of Hb but to an increasingly small amount (likelihood).
RE40 [Apr01]
Carbon monoxide (CO) is diffusion-limited because:
A. Combines avidly with Hb
B. Partial pressure in blood increases as partial pressure in air increases
C. ?
CO forms a very tight bond with Hb, therefore a large amount can be taken up by the red cell with almost no increase in partial pressure. The amount of CO which diffuses into blood in the pulmonary capillaries is NOT limited by the amount of blood available (therefore A correct).
RE41 [Jul01] [Jul05] [Jul06]
Oxygen toxicity may be seen:
A. In CNS and lungs if breath 100% at 1 ATA (?) for 24 hours
B. In CNS and lungs if breath 30% at 1 ATA (?) for 24 hours
C. In CNS if breathe 100% oxygen for 48 hours
D. ?
E. CNS toxicity seen with O2 concs far greater than 760mmHg
A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher.
B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty.
C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure.
E - Correct As above.
Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.
Oxygen toxicity may be seen:
A. In CNS if breath 100% at 1 ATA for 24 hours
B. In lungs if breath 30% at 1 ATA for 48 hours
C. In CNS and lungs if breathe 100% oxygen for 48 hours
D. ?
E. CNS toxicity seen with O2 concs far greater than 760mmHg
I thought the question was CNS and Pulmonary toxicity are caused at:
A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher.
B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty.
C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure.
E - Correct As above.
Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.
Oxygen toxicity
A. CNS affected only if significantly above 760mmHg of PiO2
B. CNS and RS affected at 760mmHg PiO2 for 24 hours
C. RS affected at FiO2 30% and 1atm for 48hours
D. mediated by superoxide dismutase
E. involves lipid peroxidation
A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher.
B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty.
C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure.
E - Correct As above.
Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.
Oxygen toxicity:
A. CNS effects if 100% O2 for 24 hrs
B. Resp effects if FiO2>30% O2 for 24 hrs
C. both pulmonary and CNS toxicity at 760 mmHg
D. ?
E. CNS effects only if PO2 significantly higher than 760 mmHg
Ref: Nunn 6th edition p357
A. - False Inhalation of 80%-100% oxygen at 1 ATM for 8 hours or more produces irritation of respiratory passages with substernal distress, sore throat nasal congestion and coughing and decreased VC- i.e ? mild pulmonary toxicity. Will not cause CNS toxicity - CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher.
B. - False Abundant evidence exists that prolonged exposure to pO2 of below 255mmHg (33% at 1ATM) causes no pulmonary toxicty.
C. - False CNS convulsions (Paul Bert effect) not below 2 atmospheres pure oxygen, usually higher - irrespective of length of exposure.
E - Correct As above.
Administration of 100% oxygen at higher pressures than atmospheric accelerates the onset of oxygen toxicity with, in addition to tracheobronchial irritation, muscle twitching, tinnitis, dizziness, convulsions and coma. Speed of development of symptoms proportionate to the pressure (cf. concentration) at which oxygen is administered.
RE42 [Jul01]
Breathing 0.04% CO2 in one atmosphere for 30 minutes, you would see:
A. Periodic apnoeas (or: ‘periods of apnoea’)
B. Hyperpnoea
C. Signs of acidosis
D. Signs of alkalosis
E. No change
Correct answer is E- no change. 0.04% is what we actually breathe. see question 45.
see http://en.wikipedia.org/wiki/Carbon_dioxide
I don’t know, I often get tachypnoea when breathing 0.04% CO2… although I admit this is usually when I’m thinking about the exam
RE43 [Jul01] [Feb04]
In the lung, airway resistance
A Mainly in small airways
B Varies with change in lung volume
C Increased by stimulation of adrenergic receptors
D Can be measured by flow rate divided by pressure difference between mouth and alveolus
E Increased by breathing helium-oxygen mixture
(Q42 Jul 01)
A: False. Mainly in the medium airays (up to 7th generation)
B: True. Airway resistance decreases with increased lung volume due to radial traction increasing calibre of airways.
C: False. Decreased resistance with stimulation of beta2 adrenergic receptors.
D: False. Measured by pressure difference divided by flow rate.
E. False. Decreased by breathing helium-oxygen due to decreased density.
RE44 [Jul01] The effect of decreasing airway diameter has the following effect on airway resistance: A. 1/8 B. ¼ C. ½ D. 4 times E. 16 times
According to the equation R= (8nl)/(pi* r^4), with each unit decrease in radius, there would be a corresponding 4 time increase in airway resistance
It should be 16 times
E = true
I think its probably improperly worded, people state ‘a unit’ decrease will lead to 16 times inc in resistance, but from what i understand decreasing radius by a certain ratio will cause a increase in resistance by that ratio to the power 4. IE halving the radius will inc resistance 16 times.
Another view:
Nunn p42 states that laminar flow does not happen until the 11th airway generation. Predominantly turbulent flow occurs in the conducting airways. We know that resistance = deltaP/q and Davis and Kenny page 93 say that pressure is proportional to flow in laminar flow and pressure is proportional to the square of the flow in turbulent flow so assuming turbulent flow then the answer would be C. Comments please.
- a note about the above statement, you are right about the pressure drop being proportional to the square of the flow in turbulent flow, however the question is asking about a change in airway diameter.
It is my understanding, that in turbulent flow, the flow is proportional to the 5th power of the radius. (see Examiner’s Report on question about Factors affecting airway resistance from 1998 on this site)[[1]].
Therefore, it would be a 32-fold change, which is not mentioned as an answer, so E sounds most correct.
This question doesn’t make sense as written. It only changes the resistance by 16 times if the radius is halved or doubled. For instance, changing the radius from 5 units to 4 units will change the resistance by 2.43 times. The correct answer should be “altered by power of 4 of the radius” or something like that.
This is obviously an incompletely remembered question. The answer is obvious but the above discussion is lacking, hence my comment. The answer lies in the Hagen-Poiseuille equation:
Q=(P1-P2) x pi x r^4 divided by 8 x n x L
Ohm’s Law states that V=IR i.e. P1-P2=QxR
Rearranging the two formulae, R = (P1-P2)/Q = 8 x n x L divided by pi x r^4 as someone correctly stated above.
∴ If radius doubles, resistance decreases by a factor of 16
If radius is halved, resistance increases by a factor of 16.
RE45 [Mar02] [Jul02] [Mar03] [Jul03]
Gas composition of air? (%)
O2 CO2 N2 Other gases
A. 20.98 0.4 ?
B. 20.98 0.4 ?
C. 21 0.04 ?
D. 20.98 0.04 78.58 0.42
E. 20.98 0.04 78.2 0.98According to Guyton 11th ed p493, atmospheric air is :
PO2 159mmHg (20.84%)
PN2 597mmHg (78.62%)
PCO2 0.3mmHg (0.04%)
PH2O 3.7mmHg (0.50%)
Note that there is lots of room for PH2O to change with humidity, eg in a tropical summer with temperature of 37 degrees and 80% humidity. Guyton’s doesn’t give temperature or humidity with these figures.
Nunn 6th Ed, Page 4 (Table 1.1), indicates that atmospheric air is composed of:
O2 20.95%
N2 78.08%
CO2 0.037%
Argon 0.93%
The closest answer that approximates these figures is E.
A-Z Anaesthesia: N2 78.03 O2 20.00 CO2 0.03 Argon 0.93 Answer: E
I agree with E. This is from Ganong: “The composition of dry air is 20.98% O2, 0.04% CO2, 78.06% N2, and 0.92% other inert constituents such as argon and helium”
I guess if you can’t be bother to remember argon, just remember that O2 21% + N2 78% = 99%, knowing CO2 is 0.04%, the rest is going to be near 0.9%!
RE46 [Mar02] [Jul02] [Feb04]. What happens to lung function in COAD? A. Decreased static compliance B. Increased TLC C. Decreased airway resistance D. Increased FEV1 E. ??
A. False. Increases static compliance (but decreases dynamic compliance) [Do you have a reference??!]
Static compliance is increased in emphysema but unchanged in chronic bronchitis
B. True
C. False. Increases airway resistance
D. False. Decreases FEV1
Definitions
The static compliance of the lung is the change in volume for a given change in transpulmonary pressure with zero gas flow.
Dynamic compliance measurements are made by monitoring the tidal volume used, while intra thoracic pressure measurements are taken during the instance of zero air flow that occur at the end inspiritory and expiratory levels with each breath.
RE47 [Mar03] [Jul03] [Feb04] [Jul04] [Mar05] [Feb13]
The amount of oxygen dissolved in plasma is
A. 0.03 mlO2/100ml at PaO2 100mmHg
B. 6 mlO2/100ml breathing 100% O2 at 3 atmospheres
C. 6 mlO2/100ml breathing room air at 3 atmospheres
D. 0.3 mlO2/l breathing room air at 1 atmosphere
E. 6 mlO2/100ml breathing 100% O2
his remembered version does not indicate that the MCQ is talking about ARTERIAL blood,
but we will assume its arterial blood (or arterial plasma) that was in the real MCQ
A is incorrect - plasma contains 0.3mLO2/100ml at PaO2 100mmHg
B is correct -> see calculation below
C is incorrect -> paO2 will not be high enough.
D is incorrect - as it says per litre of blood (Would be 3 mlsO2/litre of blood)
E is incorrect - see B
(West’s Pulmonary Physiology and Pathophysiology, p134)
Calculations:
Dissolved O2 is 0.003 ml O2/mmHg pO2/100ml blood.
So for PaO2 of 100mHg -> 0.3 mlO2/100ml blood
NOW for a person breathing 100% at 3 atmospheres, the ALVEOLAR pO2
can be calculated using the alveolar gas equation:
PAO2 = PIO2 - (PACO2/RR)
= [(760x3)-47] - (40/0.8)
= 2233 - 50
= 2183 mmHg
Now the arterial pO2 will be lower than this but cannot be calculated.
So just to get a ball-park figure, lets use a guesstimate of 2100 mmHg
Amount of O2 dissolved in ARTERIAL blood at this pO2 is:
PaO2 = 0.003 x 2100 = 6.3 mlO2/100ml blood
This is pretty close to the 6 value given in the MCQ.
(Actually a value of 6 is correct for an arterial pO2 = 2,000mmHg).
Nunn (5th ed p262):
Normal arterial content O2 = 0.25-0.3 ml/100ml (NB: The exact value depends on the arterial pO2
Plasma dissolves 0.003ml O2 per mmHg pO2 per 100ml blood
100% O2 @ 1 atm = 2mL/100mL
100% O2 @ 3 atm = 6mL/100mL
If there is 6 mls O2 dissolved in arterial blood (and assuming CO & regional flows are the same and arterial-venous difference in O2 is 5mL/100mL, then breathing 100% O2 at 3 atmospheres means haemoglobin in mixed venous blood will be 100% saturated as all the body’s oxygen required can be supplied by dissolved O2. However breathing 3 ATM of O2 results in convulsions so this is only a very short-term situation.
Of course, Hb will desaturate in areas with high O2 extraction ratios eg the coronary circulation. Here at 3 ATMs, a coronary flow of 250 mls/min & extraction of 8 mls O2/100 mls of blood (as MVO2 = 7-9 mlsO2/min), the 6 mls dissolved will not be enough.
RE48 [Mar03] [Jul03] [Jul04]
Closing capacity (in young adults)
A. Increases with anaesthesia
B. 10% vital capacity
C. Decreases with age
D. Responsible for relative hypoxaemia in healthy adult patients under anaesthesia
E. The same as FRC in elderly supine patients
A. False - Decreases in parallel with FRC (Accord. to Nunn’s 6th Ed p304)
B. False - closing volume is about 10% of vital capacity in young normal subjects (West 7th ed p. 163)
C. False - increases with age. CC=FRC at mean age 44 supine, CC=FRC at mean age 66 erect
D. I’d say false, it is more the reduction in FRC and its reaching CC under anaesthesia that causes the atelectsis, not the CC itself. Although this may be the only right answer ???
E. False - unless 44 is considered “elderly”
I presumed that if it is equal to FRC supine at 44yo, then it will be at least equal to FRC supine in elderly, so E would be correct. Although not perfectly correct. Thoughts??] – in elderly, say 66 for arguments sake, CC is the same as FRC in erect position, but CC exceeds FRC in supine, so I’ll stick with False for E
p118 Brandis - Closing Capacity is the lung volume at which the small airways in the lung first start to close. This closure occurs first in the dependent parts of the lung. So in the erect person, the small airways in the base are the first to close. This airway closure impairs gas exchange as any blood flow through the non-ventilated areas is shunted blood. The closing capacity is the sum of the residual volume and closing volume CC = RV + CV
Firstly, quoting Kerry above does not answer the question as to which is the best option.
Finally, “The role of airway closure in causing relative hypoxaemia during anaesthesia has been confirmed and accounts for approximately 35% of the increase in venous admixture.”(Wahba, p. 1147).
Hence answer is D
Comment: I interpret the above statement as airway closure as a result of reduced FRC and the resulting shunts rather than ‘increased closing capacity/volume’ as such. My understanding of closing capacity has always been pathological or due to aging secondary to small airways disease. Appreciate comments.
RE49 [Mar03] [Jul03] [Feb04]
Measurement of Functional residual Capacity (FRC):
A. Helium dilution does not measure unventilated spaces on chest
B. Body plethysomography inaccurate if high FIO2 used
C. Helium used to decrease airflow viscosity
D. Body plethysomography requires oesophageal probe
E. ?
A is true
B - can’t think of a reason for this to be true
C Helium is used because very little is lost into blood as poorly soluble.
D is false
Just my 5c:
For comment re:B is this because the ratio of O2 used to CO2 produced is not 1:1. Therefore, would expect that the calculation becomes skewed especially at high FiO2s
another 2c:
?absorption atelectasis - therefore decreases your FRC, and end up with a reading that does not correlate to your original FRC….
Having said that, I like A as the answer
addit: 1. not sure how FiO2 would affect the result given that the test is measuring pressure and volume changes 2. like the idea of absorption atelectasis – but I guess it probably only matter in people with actual diseased, obstructed airway. probably not significant in younger person, yes? I support answer A
4th 2c: mcq technique – having the ability to ignore the unusual answers when there is a most correct answer screaming at you. now, move on please
RE50 [Mar03] [Jul03] [Feb04] The absolute humidity of air saturated at 37C: A. 760 mmHg B. 47 mmHg C. 100% D. 44g/m3 E. 17mg/m3
Absolute humidity = actual Water Vapour content
When air is saturated at 37C, the absolute humidity is 44 mgs/litre = 44 g/m3. (Note that the units are equivalent because 1,000 litres in a m3 and 1,000 mg in a gram).
44mg/L is equal to 44g/m3 (not 44mg/m3).
Answer is D
Absolute humidity is the mass of dissolved water vapour per unit volume of total moist air.
Absolute humidity of air saturated at 37°C is 44mg/L = 44g/m3
Absolute humidity of air saturated at 20°C is 17mg/L = 17g/m3
i.e. 27mg/L water vapour is added on inspiration from “room temperature” to body temperature.
RE51 [Jul03] [Feb04] [Jul04] (This question renumbered from CM31)
Surface Tension
A. Is inversely proportional to the concentration of surfactant molecules per unit area
B. Cause the small alveoli to collapse into the larger ones
C. ?
D. ?
RE14 includes the above options, and more. This question (RE51) is probably mis-remembered, imo.
Comments
SURFACTANT stabilises small alveoli to prevent them empty into the larger ones.
SURFACE TENSION causes small alveoli to collapse into the larger ones.
I still think both A and B are true, since the question is about surface tension rather than surfactant.
RE52 [Jul03] Atelectasis causes hypoxaemia because of: A. ? B. ? C. ? D. ? E. Increased shunt
Shunt & V/Q mismatch
RE53 [Feb04] Which of the following is closest value for mixed venous PO2 breathing 100% oxygen? A. 50 mmHg B. 75 mmHg C. 100 mmHg D. 600 mmHg E. 660 mmHg
Its about 48 to 50 mmHg (making all the “usual assumptions” ie CO is constant, no changes in organ blood flow distribution, normal O2 consumption, so a-v O2 content difference stays at 5 ml O2/dL blood).
“Mixed venous blood” is a mixture of blood from SVC, IVC and coronary sinus - this can only be obtained from one site in the body: the pulmonary artery.
The pO2 of mixed venous sample in a normal healthy person is about 40mmHg (and 75% oxygen saturation of Hb).
My boss today told me I was wrong when I spouted this fact in a tutorial. She insisted PvO2 would be 90% if breathing 100% oxygen. So, to prove the point:
For FiO2 of 21%:
PAO2 = 149 - 40/.08 + 2 = 101mmHg [let us assume A-a gradient of 4mmHg]
PaO2 = 97mmHg [let us assume SpO2 = 98%; and Hb = 120g/L, or 12g/dL]
CaO2 = 1.39 x 12 x .98 + 0.003 x 97 = 16.28 ml/100ml [let us assume 4.8ml/100ml extraction]
CvO2 = 16.28 - 4.8 = 11.48 ml/100ml
Now, this part is a bit dodgy mathematically, but: Because CvO2 = 1.39 x 12 x SvO2 + 0.003 x PvO2, for ease of math I’m going to assume 0.003 x PvO2 is insignificant;
Therefore SvO2 = 11.48/(1.39x120) = 68.8%
And if we refer to the O2 dissociation curve… PvO2 =~ 35mmHg
For FiO2 = 100%
PAO2 = 713 - 40/.08 + 2 = 663mmHg
PaO2 = 659mmHg [let us assume SpO2 = 100% with the higher FiO2, just for the benefit of my consultant]’
CaO2 = 1.39 x 12 x 1 + 0.003 x 663 = 18.67 ml/100ml [let us assume 4.8ml/100ml extraction]
CvO2 = 18.67 - 4.8 = 13.87 ml/100ml
Therefore SvO2 = 13.87/(1.39x120) = 83.2%
And if we refer to the O2 dissociation curve… PvO2 =~ 48mmHg
Presumably the difference between my results and the stated normal value for PvO2 is the rightward shift of the curve in venous blood, but this does prove the point that PvO2 will be only about 10mmHg higher in venous blood when breathing 100% oxygen. So, if you were in any doubt, there you go. Now, back to actual study…
RE54 [Feb04] [Jul04]
Which of the following is the best explanation for the different effects
on PaO2 and PaCO2 of VQ mismatch?
A. Different solubilities of O2 and CO2
B. Different dissociation curves
C. Effect of compensatory hyperventilation
B is true. WHY?????
paO2 is decreased in V/Q mismatch - why?
Say there are 3 alvoeli - Alveolus A has low V/Q, B has normal V/Q, C has high V/Q. The total O2 content would be the addition of O2 content in the end capillaries of A, B and C.
Alveolus A has low V/Q, low pO2, high pCO2. The low pO2 lies on steeper part of the ODC, therefore O2 content in end capillary A is significantly decreased (eg 16mL/dL as in the example in West)
Alveolus B has normal V/Q, O2 content is normal (19.5mL/dL)
Alveolus C has high V/Q, high pO2, low pCO2. The high pO2 means that it lies on the flattened part of the ODC, therefore O2 content in end capillary C is not significantly greater than that of normal (20mL/dL in West)
So when we add all 3 end capillary blood together to give us the final pulmonary venous blood, the low O2 content from the low V/Q alveoli depresses the total O2 content much greater than the high V/Q alveoli would increase total O2.
Therefore the effect of V/Q mismatch on PaO2 is because of the shape of the dissociation curve.
For CO2, the dissociation curve is almost linear in the physiological range, so the high V/Q and low V/Q should balance each other out and PaCO2 should not be affected.
Note: If there is severe hypoxaemia from V/Q mismatch eg in a large shunt, hyperventilation may occur due to peripheral chemoreceptors. In this situation the pCO2 will then by affected. However, in this question the best answer is still B. _____________________________________________________________________________
The dissociation curve for CO2 is linear in the physiological range however the contribution from areas of low VQ ratios is going to be more because of the higher blood flow in these units. I agree that for O2 the shape of the HbO2 dissociation curve produces the greatest effect, but in terms of CO2 I would argue that hyperventilation is more important. Can anyone comment on this?
The point is that the reason for the differing effects of hyperventilation on PaCO2 and PaO2 is the difference in their respective disocciation curves.
The way I understood what West said was that you are able to excessively ventilate the high VQ ratio units to remove the CO2 reducing content here (not altering O2 content) by an amount that compensates for the low VQ ratio units (which are not really able to be ventilated much more than they are already. The linear disscoiation curve of CO2 means you can do this. Hyperventilation itself doesn’t to me seem like a good answer as it is not the reason for the lower O2 is not able to be improved. Does that help at all? nr
The other clue in the question could be the use of “hyperventilation”. I would argue that if the ventilation needs to be increased to eliminate CO2 - to keep paCO2 in the normal range - then this is merely “adequate” ventilation, not hyperventilation.
RE55 [Feb04]
Functional Residual Capacity
A. Decreases with age
B. Decreases with obesity
From The Physiology Viva (Kerry Brandis 2007 ed)
FRC is about 30ml/kg in adults and children. It is almost fully established in the neonate at 60mins after birth.
Increased with:
increased height
changing from supine to erect (30% higher)
less elastic recoil (ie, emphysema)
Decreased with
obesity
pregnancy
supine positioning
anaesthesia (with or without paralysis)
pulmonary disease causing increased elastic recoil (pulmonary fibrosis)
It does not really change with age - however, the relationship between closing capacity and FRC does.
Functions of the FRC
Oxygen Store
Arterial pO2 buffer
minimizes airway resistance, PVR, V/Q mismatch, work of breathing
prevents atelectasis (also part of minimizing WOB)
improves positioning on compliance curve (also part of minimizing WOB)
FRC measurement
Body plethysmography
Gas dilution (nitrogen washout, Helium wash-in)
As PEEP improves FRC, you can see all the reasons above why a bit of PEEP is good here and there.
RE56 [Jul04] [Jul10]
Correction of hypoxaemia in anaesthetised patient:
A. Increase airway pressures between breaths
B. V/Q matching
C. Decrease dead space
D. ?
E. ?
I think A?
If hypoxaemia is related to atelectasis then, yes, increasing airway pressures between ventilation (PEEP) would correct this. (Brings FRC up). Isn’t this then V/Q matching also?
It is but the initiating problem is atelectasis treatable with PEEP. Hypoxaemia during anaesthesia occurs due to decreased FRC, below closing capacity, and atelectasis with consequent shunting. Hypoxic pulmonary vasoconstriction is impaired by volatile agents and shunting is more of a problem than in the awake patient with similar degree of atelectasis. Unfortunately there are not good interventions to improve HPV but PEEP (or recruitment manoeuvres) will reduce the atelectasis which is the original cause of the problem. For this reason I would choose A.
Addition: On the other hand… PEEP can create more West zone 1 thus increase alveolar dead space so i thinks it B or C
Add: But a shunt eg atelectasis is more likely to cause hypoxaemia than a small increase in dead space? I would go for A, would always reach for the peep knob instead of the adrenaline or fluid in a mild dropping sats in OT..
It’s a funny way to say PEEP, why don’t they just say ‘PEEP”? (NB: They may do so - this is just the partially remembered version which is both incomplete and possibly a bit misleading - KB) A is the best answer because that’s what you do in real life - increase FiO2, ensure adequate minute ventilation, then put on some PEEP. (while at the same time exploring sinister causes)Duncan 18:20, 5 Dec 2006 (EST)
if the patient hypotensive , then PEEP will kill him , B looks better
Applying some PEEP increases FRC or reduces atelectasis and hence improve V/Q matching, for patient in supine position, in the posterior part of the lungs. Intra-abdominal surgery/obesity/pregnancy/muscle paralysis are going to worsen V/Q matching by reduciing FRC, so I think applying PEEP is a sensible and obviously a common move to improve hypoxaemia. Anything related to increase West zone 1 and dead space is clearly less significant in contributing to hypoxaemia – that’s why a big PE doesn’t cause hypoxaemia via dead space but mainly via creating significant shunt, and hence giving O2 alone produces only small improvement. As for the comment regarding hypotension – I can only hope that we all watch at the monitor closely and not to dial the PEEP up too quickly… you may be surprised how much difference a PEEP of 2 cmH2O can make, and I don’t think a PEEP of 2 would result in so much haemodynamic instability that would stop you from using it completely. And I guess if the patient was hypotensive at the first instance with hypoxaemia, restore circulation first! Answer from me is A
RE57 [Jul04] [Feb12]
Lung compliance
A. Measurement requires a respiratory laboratory
B. dynamic greater than static (or other way round)
C. Static and dynamic same in emphysema
D. In healthy person, difference between static and dynamic due to airflow resistance
E. Due to surface tension
F. units cmH2O/ml
E Surface Tension- seems to the the closest, as in it has the greatest influence on Compliance, the others look wrong.
You need a respiratory lab when measuring static compliance as uses oesophageal balloon.
Compliance (both static & dynamic) is measured at points of “no flow” so airway resistance does NOT come into it.
Yes, but dynamic compliance is “related to airway resistance during the equilibration of gases throughout the lung at end-inspiration or expiration” (Yentis p130)
Add:
One can measure static (and dynamic) compliance in paralysed patient on appropriate ventilators eg in ICU, but then you are measuring static compliance of whole system including abdo and chest wall..
Add:
surface tension causes alveoli to collapse - surfactant increases compliance ie surface tension decreases compliance so
E is wrong
I think what you mean is that Surface Tension contributes significantly to the Elastic Recoil of the lung. Elastance is the reciprocal of Compliance. The Compliance of the Lung is very much related to Surfactant. The clinical correlation is a surfactant-depleted lung (such as lung disease of prematurity or even ARDS), where the lung without Surfactant becomes “stiff” and poorly Compliant. The Compliance of the normal lung is partially related to the presence of Surfactant, but also to the inherent elasticity of the lung parenchyma. E is a possible answer. [To my mind, B and C and D make little sense]. [Nunn Ch3]
So what’s the actual answer???? Sounds like it is resistance.
The actual answer is D. Difference between static and dynamic due to airflow resistance
Static Compliance
Measured after lung volume has been held at a fixed volume for as long as is practicable (in conscious subjects, obviously much easier in paralyzed victims). This means sufficient time should be given for functional respiratory units to equilibrate and contribute equally in the calculation of compliance.
Dynamic Compliance
Sacrifices this luxury for the sake of simplicity in measurement, whereby compliance is (hastily) assessed at points of no-flow (at the mouth – not within the lung) at end-inspiration and end-expiration. Therefore not allowing complete equilibration in the case of diseased lung where airway resistance in differing functional units will yield different time constants.
HOWEVER, doesn’t the answer A correct as well? I can’t see how this could be done outside the respiratory lab – moving the spirometry and oesophageal barometry devices to some other places doesn’t count!! —-> we measure compliance on the anaesthetic machine every day…
But I do agree that answer D is correct too. And, static compliance IS higher than dynamic compliance.
Edit: I don’t think D is right - a major component of the difference between static and dynamic compliance is stress relaxation of elastic bodies, which has nothing to do with AWR. The AWR contributes to part of the difference, i.e. heterogeneity of time constants, which itself is a function of both AWR and compliance. Think A has to be right, you can’t do lung compliance on its own without oesophageal monometry, an anaesthetic machine is measuring the respiratory system, careful.. not the lung.
Comment: I agree with the ‘edit’ above. ‘A’ is the “most correct” answer.
A - Correct. TOTAL respiratory compliance can be measured on a ventilator, and many can measure both dynamic and static compliance. However, LUNG compliance requires oesophageal manometry.
B - Wrong. Dynamic pressure > Static pressure for given volume, therefore Dynamic compliance
Barometric pressure is half that of sea level at: A. 550m B. 1500m C. 5500m D. 7000m E. 19500m
At 18,000ft, or 5486m, pressure is half that of sea level (Yentis)
Barometic pressure decreases roughly in a exponential manner. Highest permanent human habitation is 4900-5000m in the Andes. The summit of Mt Everest is at 8848m (around 29,000 ft) where PiO2 43mmHg, and Base Camp sits at 5400m, which is half of the barometric pressure at sea level. At 20,000m (60,000ft) barometric pressure is around 47mmHg, so FIO2 is zero.
When going underwater, barometric pressure increases by 1 atm roughly every 10m depth (or every 33 feet).
RE59 [Mar05] [Jul05]
Regarding O2 carriage in blood (or regarding red blood cells):
A. ?
B. ?
C. HbS less soluble than HbA
D. ?
E. MetHb has 85% the O2 carrying capacity of normal Hb
(Comment: A,B,D were all wrong I’m pretty sure so I put C as E is also wrong.
Another similar comment: A,B,D all involved changes in 2,3 DPG and all wrong)
C- correct, HbS where valine replaces glutamic acid on beta-chain causes critical loss of solubility if reduced Hb leading to polymerisation and “sickle” shape of Hb at PaO2 less than 40mmHg. (Nunn resp phys)
Methaemoglobin
Haemoglobin in which the iron has been oxidised (trivalent ferric form). metHb is unable to combine with oxygen but is slowly reconverted to haemoglobin (can be treated with ascorbic acid or methylene blue).
May arise from
OxyHb scavenges nitric oxide (physiological process to limit endogenous NO activity)
Drugs (prilocaine, benzocaine, nitrites, dapsone)
metHb —> Hb via four mechanisms
NADH-metHb reductase enzymes in erythrocytes (most important mechanism)
Ascorbic acid via direct chemical reaction (accounts for 16%)
Glutathione-based reductive enzymes
NADPH-dehydrogenase in erythrocytes (considered ‘reserve’ metHb reductase)
RE60 [Jul05]
The greatest increase in venous admixture is due to:
A. Hypoventilation
B. ?
Venous admixture is the adition of mixed venous blood to end alveolar capillary blood. It is the entrance of blood to the arterial circulation which has not been exposed to ventilated alveoli (ie shunted blood).
It accounts for the PaO2 being less than the PAO2 in healthy individuals at rest, and is due to blood from the bronchial veins and thebesian vein returning to the left side of the heart without undergoing gas exchange in the lungs.
Venous admixture can be calculated by the shunt equation as detailed in West; derived from the fact that total oxygen must equal the sum of end capillary oxygen and shunt oxygen (providing the shunted blood is mixed venous blood).
The shunted blood reduces the PaO2, even if the shunt is small. A small drop in oxygen content leads to a large drop in PaO2 due to the almost flat ODC.
Increasing Fi02 to correct hypoxaemia cannot compensate for a shunt of more than 30% of cardiac output. 100% oxygen has a less than expected impact on PaO2.
A shunt usually has no impact on PaCO2, as chemoreceptors detect an inrcease in CO2 content and respond by increasing ventilation.
Shunt fraction –> arterial content on top, venous content on bottom: Qs/Qt = CcO2-CaO2/CcO2-CvO2.
RE61 [Jul05] [Mar09] [Jul09] Static lung compliance A. Is change in pressure per unit volume B. Affected by airway resistance C. Is equal to pulmonary elastance D. Depends upon surface tension forces E. Combination of lung and chest wall compliance
Is typically 200ml/cmH2O (1mmHg=1.36cmH2O).Is measured in vivo with a conscious,relaxed,upright subject,at different degrees of deflation from near TLC when no air is flowing. Factors affecting static compliance:
lung vol (normally measured at FRC. If only 1 lung dV is half that for a given dP)
FRC variation due to body size (specific compliance 0.05cmH2O relates changes in dV & dP to FRC (dV/dP)/FRC )
pulmonary blood vol (comp v with congestion)
alveolar collapse
lung disease
Quite rhetorical, because if there were no surface tension, then surfactant would do nothing.
Physiologically speaking, the determinants of lung compliance are its intrinsic elasticity, and the surface tension (about 50% each).
RE62 [Jul05]
Gas solubilities with decreased temperature
(Also remembered as ‘Under a general anaesthetic, if a patient becomes hypothermic, you can expect to see:’)
A. Increased PACO2, Decreased PAO2
B. Increased PACO2, Increased PAO2
C. No change in PACO2 or PAO2 (OR: PAO2 no change, decreased PACO2)
D. Decreased PACO2, Increased PAO2
E. Decreased PACO2, Decreased PAO2
oth O2 and CO2 - increased solubility with hypothermia
Since partial pressure is measuring dissolve gases (Brandis p135), then both PaCO2 and PaO2 should increase.
I think you might have this backwards. The solubilty increases, which means the amount dissoved increases. If the system is closed then this means that the partial pressure must drop.
Henry’s law: gas content dissolved in soln is directly proportional to the sol coefficent x partial pressure in the gas phase.The solubility coefficent increases as temp deceases. pp = conc dissolved gas / sol coefficent shows the inverse relationship btw partial pressure and solubility coefficent when temp changing. Therefore if temp v, sol coeff ^,and pp v. In the case of a closed system the partial pressures falls with v temp as the solubilty increases.Note that this law applies only when there is no reaction with the solvent.(Mnemonic Henry Lawson was bit of a piss pot and the amount of the intoxicant he could dissolve was directly proportional to the pressures on him and his attempts to maintain a constant of proportion)
ref: faunce http://gasboys.net
It depends on how the question is asked - if the patient is hypothermic then the A-v difference in O2 would drop as well right? So it’s important whether that’s asking A (as in alveoli) or a (as in arterial)
Comment
Now let me get this right: Don’t matter what the temperature is, atmospheric pressure should be 760mmHg (PV = nRT only applies to a set volume….) Now does that mean that inspired, saturated alveolar pO2 will still be (760-47) x 0.21 - 40/0.8 = 100? –> nope, the point here is, 5% of CO2 carried in blood is dissolved normally, but now more can be dissolved for LESS overall CO2!! –> there is less VO2 hence less VCO2 when you are hypothermic (Brain VO2 definitely down 7% per defgree, but ?14% decrease in BMR for each decrease temperature! p.281 Ganong - says about inc temp, not other way) So PCO2 has to decrease. So if CO2 decreases, that means O2 has to increase because pCO2 is which means if PAO2 increases, PaO2 should increase too… (now this assumes that R is constant)
So my conclusion? decreased temperature –> (a) increased solubility (b) decreased CO2 production –> decreased arterial PCO2 partial pressure –> increased alveolar O2 partial pressure –> increased arterial O2 partial pressure. (Correct me if I’m wrong please!!!!!!)
Now, having said all that above…. I believe the above conclusion drawn does not answer this MCQ. My conclusion here basically means that there will be MORE OXYGEN CONTENT taken up into the body, i.e. increased arterial oxygen partial pressure due to increased availability of alveolar oxygen partial pressure (not to mention left shift of the ODC). However, if the question asks about the PaO2 for a GIVEN OXYGEN CONTENT when temperature decreases, then partial pressure of oxygen will DECREASE, as quoted from Kerry’s site (see below).
adding to the confusiong – hypothermia itself reduces VO2 as well, it sort of complicates the whole scenario of how the question was asked then because that could affect the PaO2 and PvO2. But I guess that’s just me being silly because when we put ppl under anaesthesia they are having surgery, that means surgical stress is applied which results in increased VO2… the conclusion is, this is a s*** question. Or a decent question being remembered badly.
Quote from http://www.anaesthesiamcq.com/EduResources/Gases/Gas005.php
“pO2 decreases with a decrease in temperature for reasons as for pCO2. There is an additional factor here as the haemoglobin oxygen dissociation curve shifts to the left with a decrease in temperature - this means Hb has a higher oxygen affinity at a lower temperature. These 2 factors (increased solubility & increased Hb affinity) make for a more complex situation then with CO2. If pO2 is high the Hb is already fully saturated and the change in affinity won’t have much effect. At lower pO2 levels, the increased affinity means some of the dissolved oxygen will bind to Hb causing a decrease in pO2. This is additional to the decrease in pO2 that occurs with the increased solubility of O2 at the lower temperature. Total oxygen content does not change.”
RE63 [Feb06] Anatomical dead space A. measured by carbon monoxide inhalation B. 2ml/kg in average adult. C. ? D. ?
Anatomical dead space is the volume of the conducting airways which is around 150ml or 2.2ml/kg for the average adult. It is influenced by size (17mls for every 10cm) age posture (sitting 157ml supine 101ml) the posistion of the head and neck,lung volume and intubation (increases apperatus dead space decreases anatomic). Fowler’s method is used to calculate anatomic dead space.(single insp of 100% O2 followed by measurement of expired N2. Increasing conc of expired N2 is plotted against expired volume). The Bohr equation (nmemonic ‘DAETA’)is used to calculate physiologic dead space. Anatomic dead sapce is mostly atmospheric air with a little CO2
RE64 [Feb06]
With regard to dead space:
A. Bohr equation can be used for anatomical dead space
B. Nitrogen washout can be used for alveolar dead space
C. Physiological dead space calculated from end-tidal CO2
D. Physiological dead space can be calculated from end-tidal CO2 and alveolar CO2.
Fowlers method is used to calculate anatomic dead space (single insp of 100% O2 followed by measurement of expired N2.The increasing conc of expired N2 is plotted against expired volume).
The Bohr equation (mnemonic ‘DAETA’)is used to calculate physiologic dead space.
A- false, the Bohr equation is used to measure physiological dead space
B- false,nitrogen washout, used in the Fowler method is used to measure anatomical dead space
C- Calculation of physiological/total dead Space(Vt) requires end-tidal pCO2 and alveolar pCO2.
{EDIT} Its not caculated from Endi-tidal, its Mixed Expired C02; they are all wrong.
By utilising and modifying Bohr equation, anatomical dead space can be calculated-by replacing alveolar with end-expiratory gas in the Bohr equation. Hence, Bohr equation can theoretically be used to calculate anatomical dead space. Ref: Nunn 6th ed., p121.
Physiological dead space = anatomical dead space + alveolar dead space
One can use Bohr’s equation and substitute mixed expired Pco2 with end-tidal Pco2 and that gives you the alveolar dead space. By calculating the physiological dead space using mixed expired Pco2 as well, using the above formula anatomical dead space is derived. By this mean you ARE using Bohr’s equation to measure anatomical dead space.
Unless the answers was remembered wrongly, e.g. D is Alveolar dead space can be calculated from end-tidal CO2 and alveolar CO2, or Physiological dead space can be calculated from mixed expired CO2 and alveolar CO2
RE65 [Feb06]
Regarding the work of lungs in breathing:
A. ?
B. Most work is to overcome airway resistance
C. Increased by increasing respiratory rate
D. ?
E. Work done is determined by integral of pressure volume loop
On a plot of respiratory rate vs work of breathing, as respiratory rate increases work to overcome elastic forces decreases and work to overcome airway resistance increases. The total work is minimal when both elastic work and airway resistance work contribute 50%. The total work vs respiratory rate curve is U-shaped and is minimal at rest respiratory rates. However this minimum work point occurs at higher rates in pathologies that increase elastic work (such as pulmonary fibrosis) and at lower rates in pathologies that increase airway resistance work (such as COPD).
work due to airway resistance depends on the respiratory rate. change in work due to increasing respiratory rate depends on what the respiratory rate was initially. E is definately true and is the best answer
Components that make up the work of breathing during quiet inspiration:
Nonelastic work (Viscous resistance (7%)Airway resistance (28%))
Elastic work (65%)
The higher the breathing rate the faster the flow velocity (which by using Reynold’s number - is more likely to cause turbulent flow, and require increasing pressure changes for the same volume) and the larger the work done in overcoming non elastic work. The work done in overcoming the elastic forces is decreased as respiratory rate increases.
Therefore, people with highly compliant lungs and high airways resistance - breath at higher volumes and mroe slowly (ie, emphysema)
people with less compliant lungs breath at faster rates, and more shallow volumes (ie, pulmonary fibrosis)
_______________________________________________________________________________________________________________ I do not believe E to be the correct answer. Assuming the question is correctly remembered, C is the best option. The integral of the pressure volume LOOP only encompasses the area within the loop itself ie work done to overcome non elastic forces of inhalation (active) and non elastic forces of exhalation (supplied by potential energy stored by inspiratory work done against compliance). The true amount of total work encompasses the integral of the entire loop, as well as the area between the y axis and the loop itself.
Also, I disagree with another line above, stating that “The work done in overcoming the elastic forces is decreased as respiratory rate increases.” Elastic forces are not related to resp rate, but to compliance ie surface tension and intrinsic lung or chest wall elasticity. Therefore, work done against elastic forces are increased with increasing tidal volumes, and work done against non-elastic forces increases with increasing respiratory rate.
I agree with pretty much everything else above however, especially this bit: Therefore, people with highly compliant lungs and high airways resistance - breath at higher volumes and mroe slowly (ie, emphysema)
people with less compliant lungs breath at faster rates, and more shallow volumes (ie, pulmonary fibrosis)
RE66 [Feb-06][Jul10] A-a gradient is increased with: A. Atelectasis B. Venous admixture C. Hypoventilation D. Reduced cardiac output E. Increased diffusion distance for oxygen
A-a gradient = PAO2 - PaO2
where:
PaO2 is partial pressure of O2 in the artery -obtained from the arterial blood gases.
PAO2 is partial pressure of O2 in the alveoli - obtained from the Alveolar Gas equation:
PAO2 = ( FiO2 * (PB - pH2O)) - (PaCO2 / R) Alveolar gas equation
where:
PB is atmospheric pressure (eg 760mmHg at sea level)
pH2O is SVP of water at body temp (eg 47mmHg at 37C)
R is the respiratory exchange ratio (In steady state or over any period of time, this equals RQ; a typical value on a typical diet is 0.8)
Thus (if breathing room air at sea level and if arterial pCO2 = 40):
PAO2
= (FiO2 * (PB - PH2O)) - (PaCO2 / R)
= (0.21 * (760-47)) - (40/0.8)
= 150 - 50
= 100 mmHg
The Aa gradient provides an assessment of gas exhange between alveolar and capillaries, and is used as an index of V/Q mismatch in the lung.
Aa gradients can be increased due to an increase in FiO2, respiratory disease/venous admixture. In room air the normal Aa gradient is
RE67 [Jul10]
What percentage of total blood volume is found in the pulmonary capillaries?
A. 1% B. 3% C. 9% D. 11% E. 15%
As:
Pulmonary capillary blood volume = 80 mls
Total blood volume = 5,000mls
So: % = 80 x 100 / 5000 = 1.6%
Did it really say pulmonary capillaries? Or pulmonary circulation? - Because the pulmonary circulation has the capacity to hold 400-500mL blood as a reservoir. Which would make it 8-10%.
Additional comment: P+K (p121 2nd ed) states that 6% of total blood volume is in the pulmonary capillaries. – Actually, on page 120 rather than 121, its says 6% is in the systemic capillary exchange vessels, and only 3% is in the pulmonary capillaries
RE68 (Feb11) Blood draining from an unventilated part of lung will have an O2 composition identical to A. Coronary sinus B. Pulmonary artery C. Bronchial artery D. Alveolar gas
Answer is B
Blood draining from an unventilated part of the lung will have no opportunity to gain O2 and should not lose any either. Therefore it should have a composition identical to the blood supplying that part of the lung, i.e. the pulmonary artery.
Consideration of the other options reveals
a. coronary sinus - Incorrect. The primary venous drainage of the myocardium, has a high O2 extraction and pO2 of ~20mmHg, far lower than mixed venous O2
b. pulmonary artery - correct answer - see above
c. bronchial artery - Incorrect. These arise from origins in the aorta and therefore have a normal systemic PaO2. (Typically ~100mmHg, allowing for a PAO2 of 105 - physiological Aa gradient.)
d. Alveolar Gas - Incorrect. PAO2 is described by the alveolar gas equation as (760-47 * 0.21) - pCO2 / respiratory quotient = ~150 - 45 = 105mmHg. Clearly impossibly high as higher than arterial O2
RE69 [Feb12] Most likely cause of hypoxaemia post abdominal surgery? A. Increased shunt B. Increased dead space C. Hypoventilation D. ? E. ?
A: Could be, I think most likely , from Miller (e-version no page sorry - chapter on PACU), “Atelectasis and alveolar hypoventilation are the most common causes of transient postoperative arterial hypoxemia in the immediate postoperative period.” So in other words, could be A or C
B - no
C - could be right see above, not sure we are supposed to definitively answer this, anyone?
RE70 [Feb07] [Feb12] The anatomical dead space is increased by: A. Intubation B. Chin tuck position C. Moving from supine to erect D. Moving from sitting to semi-recumbent E. Bronchospasm
A. Intubation - decrease (anatomical dead space as distinct from apparatus dead space)
B. Chin tuck position - decrease
C. Moving from supine to erect - increase - the correct answer
D. Moving from sitting to semi-recumbent - decrease
E. Bronchospasm - decrease
Supine 101mls, sitting 147mls, semi recumbent 124mls NUNN
Nunn, factors affecting anatomical dead space:
size of subject
age (decrease birth to 6yr; increases early adulthood onwards)
posture (supine = 2/3 sitting)
position neck and jaw (increases from neck flexion to extension)
lung volume end inspiration (increases with increasing volume)
intubation/LMA (apparatus dead space bypassing 1/2 anatomical dead space)
drugs (bronchodilators increase)
Vtidal + RR (reducing Vt reduces dead space as measured by Fowler’s method; due to air flow changes)
Answer: C
Anatomic dead space increases A. supine to erect B. erect to semi-reclining C. on intubating the patient D. when neck is flexed and chin is pushed down
A. Intubation - decrease (anatomical dead space as distinct from apparatus dead space)
B. Chin tuck position - decrease
C. Moving from supine to erect - increase - the correct answer
D. Moving from sitting to semi-recumbent - decrease
E. Bronchospasm - decrease
Supine 101mls, sitting 147mls, semi recumbent 124mls NUNN
Nunn, factors affecting anatomical dead space:
size of subject
age (decrease birth to 6yr; increases early adulthood onwards)
posture (supine = 2/3 sitting)
position neck and jaw (increases from neck flexion to extension)
lung volume end inspiration (increases with increasing volume)
intubation/LMA (apparatus dead space bypassing 1/2 anatomical dead space)
drugs (bronchodilators increase)
Vtidal + RR (reducing Vt reduces dead space as measured by Fowler’s method; due to air flow changes)
Answer: C
RE71 [Jul10] [Feb12] The VO2 max for a sedentary 40 year old male is about? A. 3ml/kg/min B. 11ml/kg/min C. 40ml/kg/min D. 90ml/kg/min E. 250ml/kg/min
Assuming:
70kg Fit young adult = 3 litres/min
70year = 2 litres/min
Sedentary = reduced by 50%
Middle age = ?? half way between 2 and 3 = 2.5 and sedentary = 1.25L/minute = 17ml/kg/minute
is there such thing as sedentary VO2max? I think answer should be 40ml/kg/min.
I also agree with C. The “most correct” answer given above would give a V02 MAX of 770ml/min - a MAXIMUM just over three times basal rate. I would be surprised if he could do up his shoelaces at this rate!
Nunn clearly states that VO2 MAX is the oxygen consumption of a subject when exercising as hard as possible for that subject which means that this sedentary persons VO2 MAX may be reduced compared to an active person but that is still the MAX for him (p335 5th Ed). So if we use Nunn’s definition VO2 MAX for this pork chop = 1.5L/min = 20ml.kg.min - which leaves me with no answer… dunno
I took this is meaning the VO2-max of an untrained 40 yr old as opposed to a marathron-running 40 year old. Both would have the same VO2-couch, but very different VO2-max. 40mL/kg/min was then an easy answer to circle.
I suggest, based upon this study, the ball park answer is not 11 but 40 ml/kg/min. Aerobic work capacity in sedentary men and active athletes in Israel
Agreed that this question is likely incorrectly remembered. There clearly is no such thing as sedentary VO2 max. The options would likely be:
what is VO2 max in a 40yr old male?
what is VO2 in a sedentary 40yr old male?
what is VO2 max in an untrained 40 yr old male:
option 1: VO2max in a 40yo male (with some degree of fitness) is likely to be approximately 40mls/kg/min
option 2: VO2 in a sedentary 40yo male (Basal metabolic rate ) 200-250mls /min or 2/8-3/8mls/kg/min for a 70kg individual
option 3: VO2 max in an untrained 40yo male ? between 28-34mls/kg/ min (number from the Israeli study above)
Disagree that this is incorrectly recalled. It is not the VO2max that is sedentary, it is the male subject! Measuring VO2max will excercise the slovenly, sedentary subject as hard as his jelly legs will take him. (As a sedentary 39 year-old, I’m allowed to write that.)
Nunn, “Maximal oxygen uptake refers to the oxygen consumption of a subject when exercising as hard as possible for that subject.”
Per the assumptions made earlier, his VO2max can be no better than a fit young adult (~42 mL/kg/min) and no worse than a sedentary 70 year-old (~17 mL/kg/min).
My guess is that the “most correct” answer is ‘C’.
I think the above discussions are making it a bit more complicated than it needs to be. Nunn says, a fit young person should be able to attain a V02max of 3L/min. Athletes should be able to get up to 5L/min. This “sedentary” 40 year old is thus not an athlete and would have the usual V02 max of 3L/min.
Assuming he’s 70kg, that would mean 3000L/min = 42ml/kg/min therefore C is the correct answer.
RE72 [Mar03] [Jul03] [Mar10] Respiratory exchange ratio: A. Always equals respiratory quotient B. Increases in strenuous exercise C. Decreases after payment of oxygen debt D. Is measured at steady state E. ?
[Comments for A-D in the 2nd version above
A. - Incorrect => “Not to be confused”
B - Incorrect - This is metabolic rate - R can be measured at any instant in time and does not require equilibrium to have been reached
C - Incorrect - Increases during severe exercise as CO2 increases - can rise to 2
D - Increases - Decreases whilst repaying oxygen debt - can fall to 0.5
Hence E must have been something correct or others remembered differently
The respiratory exchange ratio:
A. is the same as the respiratory quotient
B. is always measured at rest
C. decreases during severe exercise
D. increases when repaying an oxygen debt
E. ?
[Comments for A-D in the 2nd version above
A. - Incorrect => “Not to be confused”
B - Incorrect - This is metabolic rate - R can be measured at any instant in time and does not require equilibrium to have been reached
C - Incorrect - Increases during severe exercise as CO2 increases - can rise to 2
D - Increases - Decreases whilst repaying oxygen debt - can fall to 0.5
Hence E must have been something correct or others remembered differently
During normal tidal ventilation
A. Intrapleural pressures between -5 & -8mmHg
B. Alveolar pressures between -2 & +2 cmH2O
C. Tracheal flow is sinusoidal
D. Peak flow is 5L/s
E. ?intrapleural pressure curve is sinusoidal
a) correct p109 west, but isn’t tracheal flow sinusoidal. Peak flow is 0.5L/s. and technically alveolar pressures are between those values (-1 to +1).
intrapleural pressure curve is not sinusoidal.
Alt version:
In a normal healthy 75kg person:
A. Intrapleural pressure during tidal breathing is between -5cmH2O to -8cmH2O
B. Alveolar pressure during tidal breathing is between +5cmH2O to -5cmH2O
C. Tidal volume is 400ml
D. inspiration last 1 second, expiration last 4 secs
A. Correct - intrapleural pressure varies between -5 cmH2O to -8 cmH2O per (West 9th Ed, Fig 7-13 p 112)
B. Partly correct - Alveolar pressure during tidal breathing varies from +1 cmH2O to -1 cmH2O, which is within the range but a significantly different number. Making this clear is probably why the question was changed from +/-2 cmH2O to +/-5 cmH2O.
C. Wrong - Vtidal ~ 7 mL/kg which is about 525 mL in a healthy 75 kg person
D. Wrong - RR 12/min and I:E 4 is pathological
“Most correct” answer is ‘A’.
RE74 [Jul06] [Feb12] FEF 25-75% A. Includes the effort dependent part B. Measured during first half of expiration C. always related to FEV1 D. fastest / steepest in 1 sec? E. Increased in COPD
A - no, is effort independent - “It has the theoretical advantage of avoiding measurement during the more effort-dependent first quarter of the FVC.” Gibson, Clinical Tests of Respiratory Function, ED 2, pp46
B no - is the middle half of expiration of the FVC
C -
D - ? not sure..
E - should be decreased in airway obstruction
FEF 25-75%
A. ?
B. ? the same in restrictive & obstructive lung disease
C. Is independent of expiratory effort
D. Measured in the first half of expiration
E. Relates? to FEV1
Alternative:
A. Wrong - specifically designed to exclude effort dependent expiration
B. Wrong - measured in middle half
C. Partly correct - FEV1 and FEF both vary with obstruction/ restrictive disease
D. Wrong - time is not a consideration in a flow-volume relationship
E. Wrong - COPD will typically have some early dynamic airway closure, reducing FEF (which is also why they have a longer expiratory phase).
Most correct answer: ‘C’
RE75 [Feb12]
With regards to blood sampled from the distal lumen of a pulmonary artery catheter (when it is wedged)
A. PO2 will be the equal to mixed venous PO2
B. PO2 will be less than mixed venous PO2
C. PCO2 will be equal to mixed venous PCO2
D. PCO2 will be less than mixed venous PCO2
E. PCO2 will be more than mixed venous PCO2
Don’t understand this question - sampling from a pulmonary artery catheter is how you obtain mixed venous blood values isn’t it? So how could anything in it be different from mixed venous blood? So A and C would both be correct.
My reading is that this is about sampling blood distal to the wedge.
After aspirating an intermediate 5 mL sample, blood aspirated distal to the wedge should be similar to an ABG. This technique is used to confirm correct placement, hence validate PCWP measurements.
Hence “most correct” answer: ‘D’
RE76 15A
Regarding peripheral chemoreceptors:
A their blood flow is 3x their metabolic rate
B Something about response rate
C Type 1 fibres are incontact with glossopharyngeal nerve
D Type 2 fibres are . . .
E Aortic body is responsible for most respiratory responses
C may be the most correct - http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3919066/
RE77 - 15A Which respiratory parameter cannot be obtained using spirometry? A. Tidal volume B. Vital capacity C. Inspiratory capacity D. Residual volume E. Expiratory reserve capacity
Residual volume - and any capacity which includes RV eg FRC, TLC
In IPPV and PEEP, what changes would you expect:
A. Change in a systolic parameter
B. Change in a diastolic parameter
C. On echocardiography you would notice a left shift in inter ventricular septum
D. ?
E. ?
C
RE79 - 15A
In which form is the majority of CO2 carried in blood?
A. Carbamino groups bound to proteins
B. Carbamino groups bound to haemoglobin
C. Dissolved in plasma
D. Bicarbonate in red blood cells
E. Bicarbonate in plasma
Bicarbonate accounts for 90% of CO2 carriage in arterial blood, and for 60% of the additional 4mls/dl picked up after passage through the systemic capillaries. Bicarbonate is formed inside the red cell (due presence of carbonic anhydrase) but then exchanges for Cl- across the cell membrane to enter the plasma.
RE80 - Feb15
Severe hypercapnia is most likely to be associated with?
A. Increased catecholamines
B. Increased urine output
C. Increased myocardial contractility
D. ?…
E. ?…
Yentis states that hypercapnia is associated with increased sympathetic activity, thus increasing circulating catecholamine levels.
These catecholamines offset the direct myocardial depressant effect of CO2 itself.
RE81 Feb15
Regarding muscles of respiration
A. Diaphragm moves 1 cm in normal breathing
B. Diaphragm can be an accessory muscle of expiration
C. Internal intercostal muscles are inspiratory
D. 50% of normal breathing is due to intercostals
E. Sternocleidomastoid is an accessory muscle of inspiration that acts by raising the first rib
A seems to be correct.
RE82 - Feb15
Airway Resistance:
A. Decreases with decreasing viscosity
B. Increases with increasing lung volume
C. The pressure between the alveoli and mouth divided by flow
D. ?
E. Mediated by α receptors
ANSWER A and C
A: CORRECT: Resistance = (8Lη)/(πr4), therefore decreased viscosity -> decreased resistance
B: INCORRECT: decreases with increasing lung volume
C: CORRECT: Resistance = pressure/flow
D: ?
E: INCORRECT: β2
Dr mitta 06:34, 3 December 2015 (CST)
C may be more correct than A as the Hagen–Poiseuille equation is based on laminar flow. Most of the flow in lower airways is transitional.
CV01 [Mar96] [Jul97] [Mar99] [Feb00] [Jul01] [Jul02] [Feb04] In a normal cardiac cycle:
A. RA systole precedes LA systole
B. RV ejection precedes LV ejection in expiration
C. RV contraction precedes LV contraction in inspiration
D. Pulmonary valve closes after aortic in inspiration
E. Pulmonary valve closes before aortic in expiration
A - T (See below quote from Ganong) B - T C - F D - T E - F
Ganong (21 ed) p568 - “RA systole precedes LA systole and contraction of the RV starts after that of the LV. However, since pulmonary arterial pressure is lower than aortic pressure, RV ejection begins before LV ejection. During expiration, the pulmonary and aortic valves close at the same time; but during inspiration, the aortic valve closes slightly before the pulmonary.”
Physiological systole lasts from the start of isovolumic contraction to the peak of the ejection phase, so that physiological diastole commences as the LV pressure starts to fall (Braunwald Heart disease 6th 463-4)
Further comment:
Ganong 21ed pg 567. Figure 29-3. Defines atrial systole from the mid point of the p-wave to the closure of the atrioventricular valves. Ventricular systole is from the closure of the atrioventricular valve to the closure of the semilunar valves. Diastole is from the closure of the semilunar valvue until the onset of atrial systole.
When conflicts arise between texts I think the examiners would give more priority to texts such as Ganong so I would use this definition.
And further…
S2 splitting attributed to “lower impedance” of pulmonary circuit by Ganong (p.566, 22nd Ed) - whatever that means. Other texts suggest that it is due to increased right ventricular return and therefore delay in right pulmonary valve closing when inspiring as decreased intrathoracic pressure.
splitting can be attributed to the fact that the semilunar valves only close once the pressure in the respective ventricles fall below the pressures in the circuits they are perfusing. logically since the pressure in the pumonary circuit is lower than the systemic circulation, it would close at a lower pressure, causing splitting.
- aortic valve closes earlier due to the higher impedance or resistance of the systemic circulation compared to the pulmonary circulation and its effects on the pulmonary valve. Valve closure requires a reversal of pressure gradient, and amount of back flow I suppose, and this pressure depends on the resistance/impedance of the circulations. In expiration, there is slightly higher intrathoracic pressure which probably adds to the pulmonary circulation’s impedance - meaning PV closes earlier (same time as AV).
In summary: (pg 566 Ganong 22nd Edn Chp 29)
RA contract before LA
RV contract after LV but RV ejection before LV ejection
In inspiration Aortic valve closes before pulmonary valve
In expiration Aortic valve and Pulmonary valve close at the same time
Alt version 1:
In a normal cardiac cycle (same as CV01 but we remembered the options as)
A. RA ejection precedes LA ejection
B. RV contraction starts before LV contraction
C. LV ejection starts before RV ejection
D. Pulmonary valve closes before aortic valve
E. Aortic valve closes after pulmonary valve in ?expiration
Alt Ver 1 A - T (See below quote from Ganong) B - F C - F D - F E - F
In summary: (pg 566 Ganong 22nd Edn Chp 29)
RA contract before LA
RV contract after LV but RV ejection before LV ejection
In inspiration Aortic valve closes before pulmonary valve
In expiration Aortic valve and Pulmonary valve close at the same time
Alt version 2:
With respect to the cardiac cycle:
A. Right ventricle starts ejecting before left ventricle
B. Pulmonary valve closes before aortic valve
C. Right & left atrial systole occur simultaneously
D. Peak aortic blood flow coincides with jugular venous c wave
E. Right ventricular ejection precedes left ventricular ejection
(The above version is reported as accurate for the July 01 paper -
It was Q14 on the Physiol paper)
Alt Ver 2 A - T (See below quote from Ganong) B - F C - F D - F (coincides with systole but not peak flow) E - T
In summary: (pg 566 Ganong 22nd Edn Chp 29)
RA contract before LA
RV contract after LV but RV ejection before LV ejection
In inspiration Aortic valve closes before pulmonary valve
In expiration Aortic valve and Pulmonary valve close at the same time
CV01b [Jul04] Physiological systole is defined as: A. AV open to AV Close B. MV close to MV open C. MV close to AV Close D. AV open to MV open
CV01B
C - T
In summary: (pg 566 Ganong 22nd Edn Chp 29)
RA contract before LA
RV contract after LV but RV ejection before LV ejection
In inspiration Aortic valve closes before pulmonary valve
In expiration Aortic valve and Pulmonary valve close at the same time
CV02 [acd] [Jul98] [Jul99] [Apr01] [Jul02] [Feb04] [Jul04]
Version 1: Normal jugular venous pressure c waves occur:
A. Just prior to atrial systole
B. Just after atrial systole
C. During ventricular systole
D. During expiration
E. ?
C
From Ganong 21st Ed (p 571) “The c wave is the transmitted manifestation of the rise in atrial pressure produced by the bulging of the tricuspid valve into the atria during isovolumetric ventricular contraction.”
A wave coincides with atrial contraction C wave coincides with isovolumetric contraction - bulging of the tricuspid valve. V wave mirrors the rise in atrial pressure before the tricuspid valve opens in diastole. It reflects peripheral venous return filling the atrium. X decent occurs after the C wave because contraction of the right ventricle pulls the fibrous A-V ring that results increased compliance of the atrium (the atrium does not enlarge as written in text but capacity to expand for a given pressure improves). Y decent after the tricuspid valve opens and reflects the decreasing compliance of the right ventricle allowing blood to spill over into this cavity from the right atrium.
COMMENT
in Berne + Levy’s Wigger’s diagram, the c wave of the JVP corresponds with LV ejection, due to the time taken for the wave to travel there I presume
I think the C wave differs depending on whether you talk about atrial pressure waves or JVP waves
atrial pressure wave - caused by the bulging of the atroventricular valve into the atrium during isovolumetric contraction
JVP - coincides with rapid ejection phase due to time taken. Caused by above, but also (According to Levy & Pappano) by the pulsation in the carotid
Version 2: The ‘c’ wave in the JVP corresponds most closely with: A. Peak aortic flow B. Isovolumetric contraction C. Isovolumetric relaxation D. Closure of aortic valve E. Closure of mitral valve
B
From Ganong 21st Ed (p 571) “The c wave is the transmitted manifestation of the rise in atrial pressure produced by the bulging of the tricuspid valve into the atria during isovolumetric ventricular contraction.”
A wave coincides with atrial contraction C wave coincides with isovolumetric contraction - bulging of the tricuspid valve. V wave mirrors the rise in atrial pressure before the tricuspid valve opens in diastole. It reflects peripheral venous return filling the atrium. X decent occurs after the C wave because contraction of the right ventricle pulls the fibrous A-V ring that results increased compliance of the atrium (the atrium does not enlarge as written in text but capacity to expand for a given pressure improves). Y decent after the tricuspid valve opens and reflects the decreasing compliance of the right ventricle allowing blood to spill over into this cavity from the right atrium.
COMMENT
in Berne + Levy’s Wigger’s diagram, the c wave of the JVP corresponds with LV ejection, due to the time taken for the wave to travel there I presume
I think the C wave differs depending on whether you talk about atrial pressure waves or JVP waves
atrial pressure wave - caused by the bulging of the atroventricular valve into the atrium during isovolumetric contraction
JVP - coincides with rapid ejection phase due to time taken. Caused by above, but also (According to Levy & Pappano) by the pulsation in the carotid
CV03 [Mar96] [Mar99] [Feb04] In a normal heart at rest the LV end-systolic volume is: A. 10 to 30 ml B. 50 to 70 mls C. 120 to 150 ml D. ?80 - 100 ml
B
According to Ganong (p 568 21st Ed)(p 565 22nd Ed): the end-diastolic volume of each ventricle is about 130ml and end-systolic volume of each is about 50ml (SV 70-90ml at rest, EF about 65%).
CV03b [Mar97] [Jul00] [Apr01] Left ventricular end-diastolic volume is: A. 10-30 mls B. 30-50 mls C. 50-70 mls D. 70-120 mls E. 120-150 mls (Jul 00 & Apr 01 versions recalled as RV EDV rather than LV)
E
According to Ganong (p 568 21st Ed)(p 565 22nd Ed): the end-diastolic volume of each ventricle is about 130ml and end-systolic volume of each is about 50ml (SV 70-90ml at rest, EF about 65%).
CV04 [ad] [Jul98] [Jul01] In moderate exercise, the LV end-systolic volume is: A. 10 mls B. 30 mls C. 70 mls D. 120 mls E. 140 mls
In moderate exercise LV end systolic volume stays the same (Berne and Levy p274)
Power and Kam seems to disagree. I’d probably go with Power and Kam as they sit on the examination board. There is a non-linear increase in SV during excercise. This increase in SV occurs mainly in light to moderate excerice, with changes at maximal excercise contributing little. Most of the increase in SV is from an increase in end-diastolic volume (from increased VR) and from a decrease in end-systolic volume (caused by increased emptying due to increased sympathetic activity). Power and Kam pg 163
Answer will be B or C depending on which text you believe!
Question doesn’t state whether isotonic or isometric exercise…
New version of B&L: “Cardiovascular Physiology”, M. N. Levy & A. J. Pappano, 9th ed, Mosby, 2007. page 243 states SV increases 10-35% in mild to moderate exercise - so if 70ml is the upper limit of normal stroke volume (50-70 ml) I think 30ml is probably the best answer.
“Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 633 states with isometric exercise little change in SV, but with isotonic “marked increase in SV”.
Take your pick! Gray
Guyton p103 claims the LVESV can fall as low as 10-20mls with increased contraction - doesn’t classify the degree of exertion. Also states the LVEDV can increase to 150-180mls in the normal heart and the stroke volume output can double!
B&L states the SV increases by 10-35%. That give a SV ranges of 45-63mls if normal SV is 70 mls or 32.5-45 if normal SV is 50mls. Doesn’t comment on range of increase in LVEDV hence difficult to subtract down to LVESV.
I think I’d pick B (30 mls) as despite increase in LVEDP associated with moderate exertion I would anticipate an increased EF thus a less than normal LVESV.
I think in moderate exercise you would have a 10-30 % increase in SV, most of the increase in CO comes from increasing HR (like in L & P where they talk about exercising dogs - denervated hearts have increased SV but normally innervated hearts have more of an increase in HR). I’d pick B too.
I disagree: There are some misleading statements here. LVESV = LVEDV - SV. LVEDV and SV both increase in moderate exercise, whereas LVESV only decreases mildly if at all - not a massive drop to 30ml. That might occur but only in strenuous exercise.
CV05 [Mar96] [Feb00] [Jul00]
Effect of tilting table from flat to head up include:
A. Decreased activation of RAS
B. Changes to skin blood flow occur immediately
C. ?
D. ?
E. None of the above
The fall in venous return to the right heart, and the decrease in hydrostatic pressure relative to the heart (i.e above the heart due to gravity), causes an immediate fall in arterial pressure at the carotid and arch barroreceptors. This decreases their level of firing. This is projected to the NTS which decreases the firing of the neurons projecting from the NTS to the RVLM, (the main efferent vasomotor/cardiac centre). The normal signal from the NTS to the RVLM is inhibitory, so decreasing it’s firing results in decreased inhibition of the RVLM. Thus there is greater SNS efferent signal to the heart(increasing heart rate and contractility and thus cardiac output),and the resistance vessels,(increasing vasomotor tone and thus PVR). To a lesser degree there is also an increase in venomotor tone increasing venous return. Overall the increase in VR adds to the rise in CO, and this coupled with the rise in PVR, tends to restore the BP measured at the barroreceptors. The primary physiological aim is to maintain CPP. The increase in SNS output/vasomotor tone will decrease skin blood flow and this effect occurs very quickly, (effectively immediately) as the barroreceptor are responsible for response to rapid transient changes.
CV06 [Mar96] The best site to measure mixed venous pO2 is: A. SVC B. RA C. Pulmonary artery D. Pulmonary vein E. ?
Answer: C
The pulmonary artery is only site for sampling for measurement of mixed venous blood. Right atrial blood is not mixed sufficiently for consistent and reliable results.
Mixed venous oxygen saturation is a direct measurement of the blended blood in the right ventricle, a mix of blood from the inferior vena cava, the superior vena cava and the coronary circulation. Mixed venous oxygen saturation is the percentage of reduced hemoglobin left after tissue oxygen extraction. The normal value for SvO2 is 60 - 80%. The measurement can be made from a blood sample drawn from the ventricle using the pulmonary artery port of a Swan-Ganz catheter.
Reference: MIXED VENOUS OXYGEN SATURATION by David Kissin, BS, RRT
Berne & Levy - CVS Physiology book
Same answer in The Physiology Viva Q&A by KB p90 Mixed venous blood can only be obtained from the pulmonary artery (Providing that no shunts have been contaminated with blood). This is the only site in which adequate mixing of the 3 venous streams can be said to have occurred (venous blood from SVC, IVC and coronary sinus).
CV07 [ad] [Jul98] [Jul99] [Apr01] Changes with raised intracerebral pressure (ICP): A. BP increase, HR decrease, RR decrease B. BP increase, HR increase, RR decrease C. BP decrease, HR increase, RR increase D. BP increase, HR decrease, RR increase E. No change in BP or HR
Answer: A
“When ICP is elevated to more than 33mmHg over a short period, cerebral blood flow
is significantly reduced. The resultant ischaemia stimulates the vasomotor area and
systemic blood pressure rises. Stimulation of vagal outflow produces bradycardia,
and respiration is slowed. The BP rise, which was described by Cushing and is
sometimes called the Cushing reflex, helps to maintain the cerebral blood flow.”
- Ganong 20th ed p595
“When intracranial pressure is increased, the blood supply to the vaso motor area
is compromised, and the local hypoxia and hypercapnia increase its discharge…
…and over a considerable range the (resultant) blood pressure rise is proportional
to the increase in ICP. The rise in BP decreases heart rate via the arterial baroreceptor
reflex”. (Respiratory rate not mentioned).
“Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 609.
Mneumonic for causes of raised ICP: Trauma Hydrocephalus Infection Neoplasia Bleeding Benign Intracranial Hypertension (THIN BB) Template:Www.anaesthesiauk.com Initially described in patients with severely raised intracranial pressure, the phenomenon consisted of bradycardia, hypertension and cessation of respiration. Cushing H: Concerning a definite regulatory mechanism of the vasomotor centre which controls blood pressure during cerebral compression. Bull Johns Hopkins Hosp 12:290-292, 1901 Mohran CV Physiology describe bradycardia, hypertension and respiratory abnormality with increased ICP but not specific as to the respiratory effects
CV08 [Mar96] [Jul97] [Mar98] [Apr01] With increased heart rate: (OR: “With moderate tachycardia:”) A. Myocardial oxygen demand increases B. Ratio of systole to diastole increases C. Vascular filling is unchanged D. Prolonged action potential E. Decrease in diastolic filling F. Decrease in coronary perfusion G. None of the above
A - Correct (Ganong 22nd ed, p575)
B - Correct (Power and Kam, p110, Ganong (22nd ed) p566)
C - Incorrect
D - Incorrect (Shortens - Levy & Pappano p31)
E - ? (The diastolic filling TIME certainly decreases, and at high heart rates filling can be compromised, however tachycardia may be associated with increase diastolic filling in exercise)
F - Incorrect - coronary perfusion increases (Brandis Revised ed, pg 71 - shorter diastole means less time for coronary perfusion however with a normal coronary circulation autoregulation of coronary blood flow will fully compenstate for this and will meet the increased myocardial oxygen demand due to the higher heart rate - ie. coronary perfusion must increase due to the increased myocardial oxygen demand caused by higher HR)
Comment: do they really mean “vascular filling” or could it mean “ventricular filling”?? Ganong 22e says that up to 180bpm filling is adequate providing venous return is normal. So C could also be correct. Thoughts?
CV09 [Mar96] In exercising muscle, the major increase in blood flow is due to: A. Sympathetic vasodilatation B. Metabolic vasodilatation C. Muscle pumping D. ? E. ?
Best answer - B
A. The blood flow of resting muscle is known to double after sympathectomy, suggesting that some decrease in tonic vasoconstrictor discharge may contribute to any increase in blood flow. Exercising muscle however, can increase its blood flow as much as x30, hence sympathetic vasodilation is unlikely to be the major cause of increase in blood flow.
B. Local mechanisms maintaining a high blood flow in exercising muscle include:
Fall in tissue pO2
Rise in tissue pCO2
Accumulation of K+ and other vasodilator metabolites (?role of lactate)
Rise in temperature
This is the correct answer.
C. When a muscle contracts or “pumps”, blood flow actually ceases during the contraction. It is only between contractions that flow is increased and this is maintained by local mechanisms (refer B above).
Although metabolic vasodilation is the MAJOR source of increased blood flow in the excercins muscle, SYMPATHETIC VASODILATION may contribute to blood flow in the excercising muscle.
“In addition to their vasoconstrictor innervation, resistance vessels in skeletal muscles are innervated by vasodilator fibres, which although they travel with the sympathetic nerves, are cholinergic (Sympathetic Cholinergic Vasodilator System). There is no tonic activity in the vasodilator fibres, but the vasoconstrictor fibres to most vascular beds have some tonic activity. When the sympathetic nerves are cut (sympathectomy), the blood vessels dilate. In most tissues, vasodilation is produced by decreasing the rate of tonic discharge in the vasoconstrictor nerves, although in SKELETAL MUSCLE it can be produced by activating the SYMPATHETIC CHOLINERGIC VASODILATOR SYSTEM.
CV10 [Mar96] Which circulation has predominant metabolic control? A. Renal B. Liver C. Lung D. Splanchnic E. ?
CV10b
Skin blood flow is regulated mainly by the sympathetic nervous system (Berne and Levy Cardiovascular Physiology p241-2)
In lung hypoxia has the most important influence in pulmonary vasomotor tone (Berne and Levy Cardiovasc Phys p254)
In skeletal muscle local factors predominate in exercise, whilst at rest neural control predominates (Berne and Levy Cardiovasc Phys p247)
Renal blood flow is autoregulated via the myogenic mechanism in glomerular arterioles and tubuloglomerular feedbeck (Brandis p74)
In the liver the portal venous system (providing 60-80% or total hepatic blood flow) does not autoregulate (Berne and Levy Cardiovasc Phys p264)
Assuming the question refers to exercising skeletal muscle then the answer is C
Could the other way of thinking about it be - which of each organs autoregulatory mechanisms would “win” if pitted against each other?
eg, SNS regulation due to temperature regulation would “win” against local metabolic factors
in skeletal muscle, local autoregulation tends to predominate (during exercise) against the vasoconstrictive effects of the SNS
CV10
Autoregulation in intestinal circulation is not as well developed as other vascular beds, eg brain and kidney. The principal mechanism for autoregulation is metabolic although a myogenic mechanism probably also participates (Berne and Levy Cardiovascular Physiology p261)
I think the best answer may be D. Splanchic
In Ganong (p623, 22nd edition), it states “The blood flow to the mucosa…responds to chnages in metabolic activity”, but it also states “The intestinal circulation is capable of extensive autoregulation”
If metabolic autoregulation refers to things such as, adenosine, K, decreased pH, increased CO2 - then couldn’t lung also be seen as metabolically regulated? (just mainly opposite to the rest of the circulation - ie, these things cause vasoconstriction, at least hypoxia, hypercapnoea, and acidosis do anyway according to Ganong)… so maybe the best answer is the lungs (as this may be a greater effect than SNS stimulation, compared with the other answers)
Infact, Levy & Pappano says that “Hypoxia has the most important influence on pulmonary vasomotor tone”, which would tend to mean that this form of “metabolic autoregulation” outweights the importance of the autonomic nervous sytem and humoral factors.
But the hypoxia that controls pulmonary blood flow relates to inspired (alveolar) gas, and isn’t related to the metabolic rate of the lung itself - I’d go with splanchnic as the best of the poor choices here but hope that E was something more obvious.
CV10b [Mar02] [Jul02] [Jul03] Local metabolic control is most important in determining flow to: A. Skin B. Lung C. Skeletal muscle D. Kidney E. Liver
(Alt wording: Which tissues autoregulate blood flow prominently: )
(Alt wording in July 03: Metabolic regulation of arteriolar resistance is most important in)
(Comment received Jul03: “Options D & E were Lungs & Kidneys - two clearly
incorrect answers since their flow is varied according to factors important
to the body as a whole rather than for their own benefit ie oxygenation &
filtration (& temperature regulation, in the case of skin). “
Circulation in skeletal muscles match tightly with the metabolic demand. Studies reviewed that capillaries capable to co-ordination of muscle blood flow responses to changed muscle metabolism. Further research are done to showed there are links to direct arterials to communicate with up branching vessels. So not just appply the condition to exercising muscle but skeletal muscle as a whole would match its metabolic demand
CV11 [Mar96] Myocardial ischaemia in shock is due mainly to: A. Decreased coronary artery pressure B. Increased myocardial O2 demand C. Decreased myocardial O2 supply D. ?
Answer C Shock can be divided into: Distributive Hypovolaemic Cardiogenic Obstructive
General terms such as shock require general answer. Therefore the obvious must be stated – ischaemia is the relative decrease in oxygen supply to demand. So it can be caused by:
Coronary blood flow or oxygen carrying capacity declines relative to demand.
Demand for O2 increases without the corresponding increase in oxygen supply.
Ischaemia therefore results.
The best answer is then relative.
Comment: If the precipitating cause is shock….a low output/supply state then the cause is a supply issue. If someone has a heightened metabolic state then maybe the demand may be the primary issue. Sure, someone in shock becomes tachycardic and has adrenaline etc which increases myocardial o2 requirements, but it is the initial low supply (shock) that started it all. So C
I would argue that by definition, shock means lower blood pressure, therefore lower coronary perfusion pressure (being the cause of decreased oxygen supply). This is a quote from the AnaesthesiaUK site - “A vicious circle of poor coronary perfusion with worsening cardiac output occurs”
Having said that, here’s something to muddy the waters from Oh’s Intensive Care Manual - “decreased coronary blood flow is not the usual cause of myocardial dysfunction in severe sepsis as coronary sinus catheter studies showed normal coronary blood flow”
(I guess this was talking about myocardial dysfunction in sepsis, as opposed to ischaemia)
Sepsis is a special form of shock - the cardiac output is high. Therefore the coronary flow was normal.
Why not B? considering in shock your blood volume is preserved for vital organs like brain and heart in the expense of periphery… so your coronary blood flow should be only slightly reduced but the demand is largely increased all because of the significantly increased workload due to the increased SNS activity. relatively speaking the decreased supply probably less than the increased demand…
CV12 [Mar97] [Mar99] [Jul99] [Jul03] The atrial component of ventricular filling A. 5% B. 10% C. 30% D. 50% E. 80%
C. 30% -most textbooks
No simple answer really.
Increases with age. One paper (Am J Cardiol. 1987 May 1;59(12):1174-8) suggested that 20yo may be 12% while 80yo with out heart disease may be 45%.
Increases with heart rate. Suggestions of 10% at rest up to 40% at maximum HR
Increases with most forms of heart disease.
CV13 [Mar97] Skin perfusion decreases: A. With standing B. ? C. ? D. ?
Standing causes blood to pool in the legs which results in a sudden decrease in venous return/cardiac output. The baroreceptor reflex (predominantly carotid baroreceptor) causes simultaneous sympathetic activation and vagal inhibition. This causes an increase in the tone of resistance vessels resulting in decreased perfusion to areas such as the skin and will also increase heart rate.
Ganong 21st ed. p633
CV14 [Mar97] [Jul98] [Mar99] [Feb00] In a 70 kg man 2 metres tall with right atrial pressure of 2 mmHg & aortic root pressure 100 mmHg, the pressure in the dorsum of the foot is: A. 0 mmHg B. 2 mmHg C. 5 mmHg D. 30 mmHg E. >50 mmHg
Go for E
My assumption for this question is that the man is standing quietly, and that the question is referring to the pressure in a blood vessel in the dorsum of the foot. They haven’t specified artery or vein, but according to Ganong, in either case the answer is (E).
p588 states that in an upright adult human with MAP of 100 mmHg, the pressure in a large artery in the foot is 180 mm Hg.
p595 states that during quiet standing, venous pressure at the ankle is 85-90 mm Hg.
(NB: If the man is contracting his leg muscles, the venous pressure can be reduced to less than 30 mm Hg)
CV15 [Mar97] [Apr01] When moving from a supine to an erect position: A. Mean arterial pressure increases B. Skin perfusion immediately decreases C. Decreased renin-angiotensin II D. Cardiac output increases E. Increased ADH secretion F. TPR increases
Moving from supine to erect causes hypotension, increased TPR and increased ADH. (Ganong p608)
See excellent Fig 33-1, p. 557 “Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; (631 21Ed)
Additional to this: Cardiac output drops secondary to reduced venous return and subsequent stroke volume reduction, although this is partially compensated for by an increased heart rate. Minimal venoconstriction. Rapid increase in circulating renin and aldosterone (Ganong 21Ed p633; 23Ed p557).
CV15b [Mar98] Changes from supine to standing causes: A. Hypotension B. Adrenal gland activation C. ? D. ? E. (See also CV05)
Moving from supine to erect causes hypotension, increased TPR and increased ADH. (Ganong p608)
See excellent Fig 33-1, p. 557 “Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; (631 21Ed)
Additional to this: Cardiac output drops secondary to reduced venous return and subsequent stroke volume reduction, although this is partially compensated for by an increased heart rate. Minimal venoconstriction. Rapid increase in circulating renin and aldosterone (Ganong 21Ed p633; 23Ed p557).
CV16 [Mar97] [Jul99] [Mar03] [Jul03] The lowest intrinsic discharge activity resides in the: A. SA node B. AV node C. Bundle branches D. Purkinje fibres E. Ventricular fibres (see also CV28)
The myocardial fibres in the ventricle have the lowest intrinsic discharge rate, which is zero. These fibres have a stable phase 4 and are not pacemaker cells. Hence as Ganong (20th ed p530, 22nd ed p548) says, “atrial and ventricular muscles do not have pre-potentials, and they discharge spontaneously only when injured or abnormal.”
The bundle branches contain “latent pacemakers” and only take over setting the heart rate when the SA & AV nodes are depressed or conduction from them is blocked.
(does no intrinsic discharge mean lowest? or should we choose the one that actually has an intrinsic activity?)
Slowest conduction (velocity) occurs in: A. Atrium B. AV Node C. Bundle of His D. Purkinje Fibres E. Ventricular muscle
The slowest conduction velocity is in the SA & AV nodes. According to the table in Ganong (20th ed p530, 22nd ed p549): SA node: 0.05 m/s Atrial pathways: 1 m/s AV node: 0.05 m/s Bundle of His 1 m/s Purkinje system: 4 m/s Ventricular muscle: 1 m/s
CV17 [Mar97] [Jul98] [Apr01] The hepatic artery : portal vein blood flow ratio is: A. 1 : 10 B. 3 : 1 C. 2 : 1 D. 1 : 6 E. 1 : 3
Answer is E - see below. Also note that hepatic artery flow less than portal vein hence B and C automatically wrong
Different texts give different numbers.
Power & Kam p. 152 indicate that the ratio is 1:3. “Portal vein normally accounts for three quarters of the blood supply”
Figure 32-15 on page 624 of Ganong 22nd Edition shows hepatic artery average blood flow of 500mL/min and portal blood flow of 1000mL/min. This would give a ratio of 1 : 2.
SAQ examiner comment: 25% and 75% resp.
CV18 [Mar97] [Mar98] [Jul01]
CSF production & absorption:
{Diagram of CSF pressure versus flow with lines}
A. Unit for x-axis is mmCSF
B. A is shifted to A1 when paCO2 is 50mmHg
C. ?
D. B is shifted to B1 with hypothermia to 33°C
E. B is shifted to B1 with metabolic acidosis
Obviously without the diagram this question isn’t much use. The diagram in Ganong doesn’t outline the changes with temperature, hypothermia and acidosis. Those factors all modify cerebral blood flow, which according to Ganong, hovers at 11.2 mmCSF. Between 68mmCSF and 200mmCSF, CSF absorption increases linearly, while production remains the same. Increased cerebral blood flow will increase intracranial pressure and thus increase aborption of CSF slightly.
Comments
“Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. 23Ed; page 572 figure 34-3 is a diagram or CSF pressure versus flow showing absorption and formation. The units of the X-axis is mmCSF. I don’t know if this is the same diagram but it would fit.
CV19 [Jul97] [Mar03] [Jul03] [Jul04] Which ONE of the following causes vasodilatation (?vasoconstriction): A. TXA2 B. Serotonin (5HT) C. Endothelin D. Neuropeptide Y E. Angiotensin II F. VIP (Comment from July 2003: Question states vasodilatation)
VIP causes vasodilatation.
Neuropeptide Y causes vasoconstriction.
Vasoconstrictors:
TXA2 - promotes platelet aggregation and vasoconstriction
Endothelin - three types produced by endothelial cells, endothelin-1 is one of most potent vasoconstrictors
Angiotensin II - increased with elevated renin secretion when BP falls or ECF volume drops, works to maintain BP
Serotonin - has effects on CNS, GIT and vascular systems
Neuropeptide Y - augments vasoconstrictive effects of noradrenergic neurons, found in brain and autonomic nervous system
Vasodilators:
Prostacyclin - inhibits platelet aggregation and promotes vasodilation
VIP - found in nerves in GIT and circulating blood
ACh, Histamine via H1Rs, Bradykinin, Substance P and VIP act on endothelial system
Adenosin, Histamine via H2Rs produced relation of vascular smooth muscle independent of endothelium
Nitric Oxide
Reference
“Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 603 Table 31-2
Ganong 20th Edition page 575
cGRP - calcitonin gene related peptide is also a vasodilator
CV19b [Feb00] Which of the following is NOT a vasodilator? A. cGRP B. VIP C. Neuropeptide Y D. Bradykinin E. Acetylcholine
VIP causes vasodilatation.
Neuropeptide Y causes vasoconstriction.
Vasoconstrictors:
TXA2 - promotes platelet aggregation and vasoconstriction
Endothelin - three types produced by endothelial cells, endothelin-1 is one of most potent vasoconstrictors
Angiotensin II - increased with elevated renin secretion when BP falls or ECF volume drops, works to maintain BP
Serotonin - has effects on CNS, GIT and vascular systems
Neuropeptide Y - augments vasoconstrictive effects of noradrenergic neurons, found in brain and autonomic nervous system
Vasodilators:
Prostacyclin - inhibits platelet aggregation and promotes vasodilation
VIP - found in nerves in GIT and circulating blood
ACh, Histamine via H1Rs, Bradykinin, Substance P and VIP act on endothelial system
Adenosin, Histamine via H2Rs produced relation of vascular smooth muscle independent of endothelium
Nitric Oxide
Reference
“Review of Medical Physiology”, W. F. Ganong, 22nd ed, Lange/McGraw-Hill, 2006. page 603 Table 31-2
Ganong 20th Edition page 575
cGRP - calcitonin gene related peptide is also a vasodilator
CV20 [Jul97] [Feb04] Which ONE of the following causes vasoconstriction: A. Serotonin B. Prostacyclin C. Neuropeptide Y D. Substance P E. Alkalaemia F. cGRP G. Oxytocin
Locally released platelet serotonin and neuropeptide Y both cause vasoconstriction. Ref: Ganong 22nd Edition page 603 table 31-2
Note: Serotonin vasoconstricts in most tissue beds but vasodilates in skeletal muscle and heart. (Faunce Pg 75 and also Goodman’s and Gilman’s on ANZCA website)
Oxytocin can also cause vasoconstriction and hypertension [1]
CV20b [Mar99] Which ONE of the following is true? A. Neuropeptide Y secreted by vagus B. CGRP present in afferent nerves C. ?
Neuropeptide Y is found in postganglionic sympathetic nerves.
CGRP is found in sensory nerves near blood vessels. –> incomplete. Ganong:
Two forms in rats and thus probably humans: CGRPα and CGRPβ
CGRPα found in GI tract
CGRPβ found in: (a) primary afferent neurons (b) neurons projecting taste into thalamus (c) neurons in medial forebrain bundle (d) sensory nerves near blood vessels
Causes weak vasodilation and has other weak effects
Ref: Ganong 22nd Edition page 602, 114
