-
Question 1
Incorrect
-
One litre of water at 0°C and a pressure of 1 bar is in a water-bath. A 1 kW element is used in heating it.
Given that the specific heat capacity of water is 4181 J/(kg°C) or J/(kg K), how long will it take to raise the temperature of the water by 10°C?Your Answer: 7 minutes
Correct Answer: 42 seconds
Explanation: -
This question is part of the following fields:
- Physiology
-
-
Question 2
Incorrect
-
The SI unit of measurement is kgm2s-2 in the System international d'unités (SI).
Which of the following derived units of measurement has this format?Your Answer: Velocity
Correct Answer: Energy
Explanation:The derived SI unit of force is Newton.
F = m·a (where a is acceleration)
F = 1 kg·m/s2The joule (J) is a converted unit of energy, work, or heat. When a force of one newton (N) is applied over a distance of one metre (Nm), the following amount of energy is expended:
J = 1 kg·m/s2·m =
J = 1 kg·m2/s2 or 1 kg·m2·s-2The unit of velocity is metres per second (m/s or ms-1).
The watt (W), or number of joules expended per second, is the SI unit of power:
J/s = kg·m2·s-2/s
J/s = kg·m2·s-3Pressure is measured in pascal (Pa) and is defined as force (N) per unit area (m2):
Pa = kg·m·s-2/m2
Pa = kg·m-1·s-2 -
This question is part of the following fields:
- Physiology
-
-
Question 3
Incorrect
-
A patient on admission is given an infusion of 1000 mL of 10% glucose and 500 mL of 20% lipid over a 24 hour period.
Which of these best approximates to the energy input over this time period?Your Answer: 1600 kcal
Correct Answer: 1300 kcal
Explanation:1% solution contains 1 g of substance per 100 mL.
A solution of 10% glucose is 10 g/100mL. Therefore 1000 mL of this glucose solution will contain 100 g.
1 g of glucose yields about 4 kcal of energy. One litre of 10% glucose will therefore release approximately 4x100g = 400 kcal of energy.
A solution of 20% fat is 20 g/100mL. Therefore 1000 mL of this fat solution will have 200 g and 500 mL will contain 100 g.
1 g of fat yields approximately 9 kcal. 500 mL of 20% fat therefore has the potential to yield 900 kcal of energy.
The total energy input over this 24 hour period is approximately 400kcal + 900kcal = 1300 kcal.
-
This question is part of the following fields:
- Physiology
-
-
Question 4
Incorrect
-
Which of the following statement is true regarding hypoxic pulmonary vasoconstriction (HPV)?
Your Answer: The predominant stimulus is a low PO2 in the pulmonary arterial blood
Correct Answer: 20 parts per million (ppm) of nitric oxide will reduce hypoxic pulmonary vasoconstriction
Explanation:Hypoxic Pulmonary vasoconstriction (HPV) reflects the constriction of small pulmonary arteries in response to hypoxic alveoli (.i.e.; PO2 below 80-100mmHg or 11-13kPa).
These blood vessels become independent of the nerve stimulus, when blood with a high PO2 flows through the lung which contains a low alveolar PO2.
Thus a low PO2 within the alveoli has been shown to impact on hypoxic pulmonary vasoconstriction (HPV) more than a low PO2 within the blood.
HPV results in the blood flow being directed away from poorly ventilated areas of the lung and helps to reduce the ventilation/perfusion mismatch (not increase).
In animals, volatile anaesthetic agents can diminish HPV, while in adults, the evidence proves less persuading, in spite of the fact that it certainly doesn’t strengthen the effects.
HPV response will be suppressed by 20 parts per million (ppm) of nitric oxide.
-
This question is part of the following fields:
- Physiology
-
-
Question 5
Correct
-
You're summoned to the emergency room, where a 39-year-old man has been admitted following a cardiac arrest. He was rescued from a river, but little else is known about him.
CPR is being performed on the patient, who has been intubated. He's received three DC shocks and is still in VF. A rectal temperature of 29.5°C is taken with a low-reading thermometer.
Which of the following statements about his resuscitation is correct?Your Answer: No further DC shocks and no drugs should be given until his core temperature is greater than 30°C
Explanation:The guidelines for the management of cardiac arrest in hypothermic patients published by the UK Resuscitation Council differ slightly from the standard algorithm.
In a patient with a core temperature of less than 30°C, do the following:
If you’re on the shockable side of the algorithm (VF/VT), you should give three DC shocks.
Further shocks are not recommended until the patient has been rewarmed to a temperature of more than 30°C because the rhythm is refractory and unlikely to change.
There should be no drugs given because they will be ineffective.In a patient with a core temperature of 30°C to 35°C, do the following:
DC shocks are used as usual.
Because they are metabolised much more slowly, the time between drug doses should be doubled.Active rewarming and protection against hyperthermia should be given to the patient.
Option e is false because there is insufficient information to determine whether resuscitation should be stopped.
-
This question is part of the following fields:
- Physiology
-
-
Question 6
Incorrect
-
Question 7
Incorrect
-
During exercise, muscle blood flow can increase by 20 to 50 times.
Which mechanism is the most important for increased blood flow?Your Answer:
Correct Answer: Local autoregulation
Explanation:Skeletal muscle blood flow is in the range of 1-4 ml/min per 100 g when at rest. Blood flow can reach 50-100 ml/min per 100 g during exercise. With maximal vasodilation, blood flow can increase 20 to 50 times.
The adrenal medulla releases catecholamines and increases neural sympathetic activity during exercise. Normally, alpha-1 and alpha-2 would cause vasoconstriction in the muscle groups being used, but vasodilatory metabolites override these effects, resulting in a so-called functional sympathectomy. Local hypoxia and hypercarbia, nitric oxide, K+ ions, adenosine, and lactate are some of the stimuli that cause vasodilation.
However, the splanchnic and cutaneous circulations, which supply inactive muscles, vasoconstrict.
Sympathetic cholinergic innervation of skeletal muscle arteries is found in some species (such as cats and dogs, but not humans). Vasodilation is induced by stimulating smooth muscle beta-2 adrenoreceptors, but at rest, the alpha-adrenoreceptor effects of adrenaline and noradrenaline predominate. During exercise, the skeletal muscle pump promotes venous emptying, but it does not necessarily increase blood flow.
-
This question is part of the following fields:
- Physiology
-
-
Question 8
Incorrect
-
A 30-year old female athlete was brought to the Emergency Room for complaints of light-headedness and nausea. Clinical chemistry studies were done and the results were the following:
Na: 144 mmol/L (Reference: 137-144 mmol/L)
K: 6 mmol/L (Reference: 3.5-4.9 mmol/L)
Cl: 115 mmol/L (Reference: 95-107 mmol/L)
HCO3: 24 mmol/L (Reference: 20-28 mmol/L)
BUN: 9.5 mmol/L (Reference: 2.5-7.5 mmol/L)
Crea: 301 µmol/l (Reference: 60 - 110 µmol/L)
Glucose: 3.5 mmol/L (Reference: 3.0-6.0 mmol/L)
Taking into consideration the values above, in which of the following ranges will his osmolarity fall into?Your Answer:
Correct Answer: 300-313
Explanation:Osmolarity refers to the osmotic pressure generated by the dissolved solute molecules in 1 L of solvent. Measurements of osmolarity are temperature dependent because the volume of the solvent varies with temperature. The higher the osmolarity of a solution, the more it attracts water from an opposite compartment.
Osmolarity can be computed using the following formulas:
Osmolarity = Concentration x number of dissociable particles; OR
Plasma osmolarity (Posm) = 2([Na+]) + (glucose in mmol/L) + (BUN in mmol/L)Posm = 2 (144) + 3.5 + 9.5 = 301 mOsm/L
Suppose there is electrical neutrality, the formula will double the cation activity to account for the anions.
Plasma osmolarity (Posm) = 2([Na+] + [K+]) + (glucose in mmol/L) + (BUN in mmol/L)
Posm = 2 (144 + 6) + 3.5 + 9.5 = 313 mOsm/L
-
This question is part of the following fields:
- Physiology
-
-
Question 9
Incorrect
-
Concerning forced alkaline diuresis, which of the following statements is true?
Your Answer:
Correct Answer: Can be used in a barbiturate overdose
Explanation:In situations of poisoning or drug overdose with acid dugs like salicylates and barbiturates, forced alkaline diuresis may be used.
With regards to overdose with alkaline drugs, forced acid diuresis is used.
By changing the pH of the urine, the ionised portion of the drug stays in the urine, and this prevents its diffusion back into the blood. Charged molecules do not readily cross biological membranes.
The process involves the infusion of specific fluids at a rate of about 500ml per hour. This requires monitoring of the central venous pressure, urine output, plasma electrolytes, especially potassium, and blood gas analysis.
The fluid regimen recommended is:
500ml of 1.26% sodium bicarbonate (not 200ml of 8.4%)
500ml of 5% dextrose and
500ml of 0.9% sodium chloride. -
This question is part of the following fields:
- Physiology
-
-
Question 10
Incorrect
-
A 61-year-old woman with myasthenia gravis is admitted to the ER with type II respiratory failure. There is a suspicion of myasthenic crisis.
She is in a semiconscious state. Her blood pressure is 160/90 mmHg, pulse is 110 beats per minute, temperature is 37°C, and oxygen saturation is 84 percent.
With a PaCO2 of 75 mmHg (10 kPa) breathing air, blood gas analysis confirms she is hypoventilating.
Which of the following values is the most accurate representation of her alveolar oxygen tension (PAO2)?Your Answer:
Correct Answer: 7.3
Explanation:The following is the alveolar gas equation:
PAO2 = PiO2 − PaCO2/R
Where:
PAO2 is the partial pressure of oxygen in the alveoli.
PiO2 is the partial pressure of oxygen inhaled.
PaCO2 stands for partial pressure of carbon dioxide in the arteries.
The amount of carbon dioxide produced (200 mL/minute) divided by the amount of oxygen consumed (250 mL/minute) equals R = respiratory quotient. With a normal diet, the value is 0.8.By subtracting the partial pressure exerted by water vapour at body temperature, the PiO2 can be calculated:
PiO2 = 0.21 × (100 kPa − 6.3 kPa)
PiO2 = 19.8Substituting:
PAO2 = 19.8 − 10/0.8
PAO2 = 19.8 − 12.5
PAO2 = 7.3k Pa -
This question is part of the following fields:
- Physiology
-
-
Question 11
Incorrect
-
A 27-year-old woman takes part in a study looking into the effects of different dietary substrates on metabolism. She receives a 24-hour ethyl alcohol infusion.
A constant volume, closed system respirometer is used to measure CO2 production and consumption. The production of carbon dioxide is found to be 200 mL/minute.
Which of the following values most closely resembles her anticipated O2 consumption at the conclusion of the trial?Your Answer:
Correct Answer: 300 mL/minute
Explanation:The respiratory quotient (RQ) is the ratio of CO2 produced by the body to O2 consumed in a given amount of time.
CO2 produced / O2 consumed = RQ
CO2 is produced at a rate of 200 mL per minute, while O2 is consumed at a rate of 250 mL per minute. An RQ of around 0.8 is typical for a mixed diet.
The RQ will change depending on the energy substrates consumed in the diet. Granulated sugar is a refined carbohydrate that contains 99.999 percent carbohydrate and no lipids, proteins, minerals, or vitamins.
Glucose and other hexose sugars (glucose and other hexose sugars):
RQ=1Fats:
RQ = 0.7Proteins:
Approximately 0.9 RQEthyl alcohol is a type of alcohol.
200/300 = 0.67 RQ
For complete oxidation, lipids and alcohol require more oxygen than carbohydrates.
When carbohydrate is converted to fat, the RQ can rise above 1.0. Fat deposition and weight gain are likely to occur in these circumstances.
-
This question is part of the following fields:
- Physiology
-
-
Question 12
Incorrect
-
The following is normally higher in concentration extracellularly than intracellularly
Your Answer:
Correct Answer: Sodium
Explanation:The ions found in higher concentrations intracellularly than outside the cells are:
ATP
AMP
Potassium
Phosphate, and
Magnesium Adenosine diphosphate (ADP)Sodium is a primarily extracellular ion.
-
This question is part of the following fields:
- Physiology
-
-
Question 13
Incorrect
-
Which of the following, at a given PaO2, increases the oxygen content of arterial blood?
Your Answer:
Correct Answer: A reduced erythrocyte 2,3-diphosphoglycerate level
Explanation:The oxygen content of arterial blood can be calculated by the following equation:
(10 x haemoglobin x SaO2 x 1.34) + (PaO2 x 0.0225).
This is the sum of the oxygen bound to haemoglobin and the oxygen dissolved in the plasma.Oxygen content x cardiac output = The amount of oxygen delivered to the tissues in unit time which is known as the oxygen flux.
Any factor that increases the metabolic demand will encourage oxygen offloading from the haemoglobin in the tissues and this causes the oxygen dissociation curve (ODC) to shift to the right. This subsequently reduced the oxygen content of arterial blood.
Conditions like fever, metabolic or respiratory acidosis lowers the oxygen content and shifts the ODC to the right.
A low level of 2,3 diphosphoglycerate (2,3-DPG) is usually related to an increased oxygen content as there is less offloading, and so the ODC is shifted to the left.So for a given PaO2, a high blood oxygen content is related to any factors that can shift the ODC to the left and not to the right.
A low haematocrit usually means that there is a decreased haemoglobin concentration, and therefore is associated with decreased oxygen binding to haemoglobin.
-
This question is part of the following fields:
- Physiology
-
-
Question 14
Incorrect
-
The following statement is true with regards to the Nernst equation:
Your Answer:
Correct Answer: It is used to calculate the potential difference across a membrane when the individual ions are in equilibrium
Explanation:The Nernst equation is used to calculate the membrane potential at which the ions are in equilibrium across the cell membrane.
The normal resting membrane potential is -70 mV (not + 70 mV).
The equation is:
E = RT/FZ ln {[X]o
/[X]i}Where:
E is the equilibrium potential
R is the universal gas constant
T is the absolute temperature
F is the Faraday constant
Z is the valency of the ion
[X]o is the extracellular concentration of ion X
[X]i is the intracellular concentration of ion X. -
This question is part of the following fields:
- Physiology
-
-
Question 15
Incorrect
-
A transport ventilator connected to a size CD oxygen cylinder has a setting of air/oxygen entrainment ratio of 1:1 and a minute volume set at 10 litres/minute.
Which value best approximates to the FiO2?Your Answer:
Correct Answer: 0.6
Explanation:A nominal volume of 2 litres is contained in a CD cylinder. It has a pressure of 230 bar when full and contains litres 460 L of useable oxygen at STP.
For every 1000 mL 100% oxygen there will be an entrainment of 1000 mL or air (20% oxygen) in an air/oxygen mix.
The average concentration is, therefore, 120/2=60% or 0.6.
-
This question is part of the following fields:
- Physiology
-
-
Question 16
Incorrect
-
A 25-year old man needs an emergency appendicectomy and has gone to the operating room. During general anaesthesia, ventilation is achieved using a circle system with a fresh gas flow (FGF) of 1L/min, with and air/oxygen and sevoflurane combination. The capnograph trace is normal.
Changes to the end tidal and baseline CO2 measurements at 10 and 20 mins respectively are seen on the capnograph below:
10 minutes 20 minutes
End-tidal CO2 4.9 kPa 8.4 kPa
Baseline end-tidal CO2 0.2 kPa 2.4 kPa
The other vitals were as follows:
Pulse 100-105 beats per minute
Systolic blood pressure 120-133 mmHg
O2 saturation 99%.
The next most important immediate step is which of the following?Your Answer:
Correct Answer: Increase the FGF
Explanation:This scenario describes rebreathing management.
Changes is exhaustion of the soda lime and a progressive rise in circuit deadspace is the most likely explanation for the capnograph.
It is important that the soda lime canister is inspected for a change in colour of the granules. Initially fresh gas flow should be increased and then if necessary, replace the soda lime granules. Other strategies include changing to another circuit or bypassing the soda lime canister after the fresh gas flow is increased.
Any other causes of increased equipment deadspace should be excluded.
Intraoperative hypercarbia can be caused by:
1. Hypoventilation – Breathing spontaneously; drugs which include anaesthetic agents, opioids, residual neuromuscular blockade, pre-existing respiratory or neuromuscular disease and cerebrovascular accident.
2. Controlled ventilation- circuit leaks, disconnection, miscalculation of patient’s minute volume.
3. Rebreathing – Soda lime exhaustion with circle, inadequate fresh gas flow into Mapleson circuits, increased breathing system deadspace.
4. Endogenous source – Tourniquet release, hypermetabolic states (MH or thyroid storm) and release of vascular clamps.
5. Exogenous source – Absorption of CO2 absorption from the pneumoperitoneum. -
This question is part of the following fields:
- Physiology
-
-
Question 17
Incorrect
-
Which of the following statements is true with regards to 2,3-diphosphoglycerate (2,3-DPG)?
Your Answer:
Correct Answer: Production is increased in heart failure
Explanation:During glycolysis, 2,3-diphosphoglycerate (2,3-DPG) is
created in erythrocytes by the Rapoport-Luebering shunt.The production of 2,3-DPG increases for several conditions
in the presence of decreased peripheral tissue O2 availability.
Some of these conditions include hypoxaemia, chronic lung
disease anaemia, and congestive heart failure. Thus,
2,3-DPG production is likely an important adaptive mechanism.High levels of 2,3-DPG cause a shift of the curve to the right.
Low levels of 2,3-DPG cause a shift of the curve to the left,
as seen in states such as septic shock and hypophosphatemia. -
This question is part of the following fields:
- Physiology
-
-
Question 18
Incorrect
-
Regarding bilirubin, which one of the following statement is true?
Your Answer:
Correct Answer: Conjugated bilirubin is stored in the gall bladder
Explanation:Bilirubin is the tetrapyrrole and a catabolic product of heme. 70-90% of bilirubin is end product of haemoglobin degradation in the liver.
Bilirubin circulates in the blood in 2 forms; unconjugated and conjugated bilirubin.
Unconjugated bilirubin is insoluble in water. It travels through the bloodstream to the liver, where it changes from insoluble into a soluble form (i.e.; unconjugated into conjugated form).
This conjugated bilirubin travels from the liver into the small intestine and the gut bacteria convert bilirubin into urobilinogen and then into urobilin (not urobilin to urobilinogen). A very small amount passes into the kidneys and is excreted in urine.
-
This question is part of the following fields:
- Physiology
-
-
Question 19
Incorrect
-
The solutions that contains the most sodium is?
Your Answer:
Correct Answer: 3500 mL 0.9% N saline
Explanation:Sodium concentration for different fluids
3% N saline 513 mmol/L
5% N saline 856 mmol/L
0.9% N saline 154 mmol/L
Hartmann’s solution 131 mmol/L
0.45% N saline with 5% glucose 77 mmol/LThis means that:
500 mL 5% N saline contains 428 mmol of sodium
1000 mL 3% N saline contains 513 mmol of sodium
3500 mL 0.9% N saline contains 539 mmol of sodium
4000 mL Hartmann’s contains 524 mmol of sodium
6000 mL 0.45% N saline with 5% glucose contains 462 mmol of sodium. -
This question is part of the following fields:
- Physiology
-
-
Question 20
Incorrect
-
Left ventricular afterload is mostly calculated from systemic vascular resistance.
Which one of the following factors has most impact on systemic vascular resistance?Your Answer:
Correct Answer: Small arterioles
Explanation:Systemic vascular resistance (SVR), also known as total peripheral resistance (TPR), is the amount of force exerted on circulating blood by the vasculature of the body. Three factors determine the force: the length of the blood vessels in the body, the diameter of the vessels, and the viscosity of the blood within them. The most important factor that determines the systemic vascular resistance (SVR) is the tone of the small arterioles.
These are otherwise known as resistance arterioles. Their diameter ranges between 100 and 450 µm. Smaller resistance vessels, less than 100 µm in diameter (pre-capillary arterioles), play a less significant role in determining SVR. They are subject to autoregulation.
Any change in the viscosity of blood and therefore flow (such as due to a change in haematocrit) might also have a small effect on the measured vascular resistance.
Changes of blood temperature can also affect blood rheology and therefore flow through resistance vessels.
Systemic vascular resistance (SVR) is measured in dynes·s·cm-5
It can be calculated from the following equation:
SVR = (mean arterial pressure − mean right atrial pressure) × 80 cardiac output
-
This question is part of the following fields:
- Physiology
-
-
Question 21
Incorrect
-
Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your Answer:
Correct Answer: Left shift of the oxygen dissociation curve
Explanation:Smoking is a major risk factor associated with perioperative respiratory and cardiovascular complications. Evidence also suggests that cigarette smoking causes imbalance in the prostaglandins and promotes vasoconstriction and excessive platelet aggregation. Two of the constituents of cigarette smoke, nicotine and carbon monoxide, have adverse cardiovascular effects. Carbon monoxide increases the incidence of arrhythmias and has a negative ionotropic effect both in animals and humans.
Smoking causes an increase in carboxyhaemoglobin levels, resulting in a leftward shift in which appears to represent a risk factor for some of these cardiovascular complications.
There are two mechanisms responsible for the leftward shift of oxyhaemoglobin dissociation curve when carbon monoxide is present in the blood. Carbon monoxide has a direct effect on oxyhaemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2,3-DPG by inhibiting glycolysis in the erythrocyte. Nicotine, on the other hand, has a stimulatory effect on the autonomic nervous system. The effects of nicotine on the cardiovascular system last less than 30 min.
-
This question is part of the following fields:
- Physiology
-
-
Question 22
Incorrect
-
A mercury barometer can be used to determine absolute pressure. A mercury manometer can be used to check blood pressure. The SI units of length(mm) are used to measure pressure.
Why is pressure expressed in millimetres of mercury (mmHg)?Your Answer:
Correct Answer: Pressure is directly proportional to length of the mercury column and is variable
Explanation:A mercury barometer can be used to determine absolute pressure. A glass tube with one closed end serves as the barometer. The open end is inserted into a mercury-filled open vessel. The mercury in the container is pushed into the tube by atmospheric pressure exerted on its surface. Absolute pressure is the distance between the tube’s meniscus and the mercury surface.
Pressure is defined as force in newtons per unit area (F) (A).
Mass of mercury = area (A) × density (ρ) × length (L)
Pressure = ((A × ρ × L) × 9.8 m/s2)/A
Pressure = ρ × L x 9.8
Pressure is proportional to LThe numerator and denominator of the above equation, area (A), cancel out. The constants are density and the gravitational acceleration value.
The length is proportional to the applied pressure.
-
This question is part of the following fields:
- Physiology
-
-
Question 23
Incorrect
-
Which of the following statements best describes adenosine receptors?
Your Answer:
Correct Answer:
Explanation:Adenosine receptors are expressed on the surface of most cells.
Four subtypes are known to exist which are A1, A2A, A2B and A3.Of these, the A1 and A2 receptors are present peripherally and centrally. There are agonists at the A1 receptors which are antinociceptive, which reduce the sensitivity to a painful stimuli for the individual. There are also agonists at the A2 receptors which are algogenic and activation of these results in pain.
The role of adenosine and other A1 receptor agonists is currently under investigation for use in acute and chronic pain states.
-
This question is part of the following fields:
- Physiology
-
-
Question 24
Incorrect
-
Which of the following is true in the Kreb's cycle?
Your Answer:
Correct Answer: Alpha-ketoglutarate is a five carbon molecule
Explanation:Krebs’ cycle (tricarboxylic acid cycle or citric acid cycle) is a sequence of reactions to release stored energy through oxidation of acetyl coenzyme A (acetyl-CoA). Some of the products are carbon dioxide and hydrogen atoms.
The sequence of reactions, known collectively as oxidative phosphorylation, only occurs in the mitochondria (not cytoplasm).
The Krebs cycle can only take place when oxygen is present, though it does not require oxygen directly, because it relies on the by-products from the electron transport chain, which requires oxygen. It is therefore considered an aerobic process. It is the common pathway for the oxidation of carbohydrate, fat and some amino acids, required for the formation of adenosine triphosphate (ATP).
Pyruvate enters the mitochondria and is converted into acetyl-CoA. Acetyl-CoA is then condensed with oxaloacetate, to form citrate which is a six carbon molecule. Citrate is subsequently converted into isocitrate, alpha-ketoglutarate, succinyl-CoA, succinate, fumarate, malate and finally oxaloacetate.
The only five carbon molecule in the cycle is Alpha-ketoglutarate.
-
This question is part of the following fields:
- Physiology
-
-
Question 25
Incorrect
-
In a normal healthy adult breathing 100 percent oxygen, which of the following is the most likely cause of an alveolar-arterial (A-a) oxygen difference of 30 kPa?
Your Answer:
Correct Answer: Atelectasis
Explanation:The ‘ideal’ alveolar PO2 minus arterial PO2 is the alveolar-arterial (A-a) oxygen difference.
The ‘ideal’ alveolar PO2 is derived from the alveolar air equation and is the PO2 that the lung would have if there was no ventilation-perfusion (V/Q) inequality and it was exchanging gas at the same respiratory exchange ratio as real lung.
The amount of oxygen in the blood is measured directly in the arteries.
The A-a oxygen difference (or gradient) is a useful measure of shunt and V/Q mismatch, and it is less than 2 kPa in normal adults breathing air (15 mmHg). Because the shunt component is not corrected, the A-a difference increases when breathing 100 percent oxygen, and it can be up to 15 kPa (115 mmHg).
An abnormally low or abnormally high V/Q ratio within the lung can cause an increased A-a difference, though the former is more common. Atelectasis, which results in a low V/Q ratio, is the most likely cause of an A-a difference in a healthy adult breathing 100 percent oxygen.
Hypoventilation may cause an increase in alveolar (and thus arterial) CO2, lowering alveolar PO2 according to the alveolar air equation.
The alveolar PO2 is also reduced at high altitude.
Healthy people are unlikely to have a right-to-left shunt or an oxygen transport diffusion defect.
-
This question is part of the following fields:
- Physiology
-
-
Question 26
Incorrect
-
A participant of a metabolism study is to be fed only granulated sugar and water for 48 hours. What would be his expected respiratory quotient at the end of the study?
Your Answer:
Correct Answer: 1
Explanation:The respiratory quotient is the ratio of CO2 produced to O2 consumed while food is being metabolized:
RQ = CO2 eliminated/O2 consumed
Most energy sources are food containing carbon, hydrogen and oxygen. Examples include fat, carbohydrates, protein, and ethanol. The normal range of respiratory coefficients for organisms in metabolic balance usually ranges from 1.0-0.7.
Granulated sugar is a refined carbohydrate with no significant fat, protein or ethanol content.
The RQ for carbohydrates is = 1.0
The RQ for the rest of the compounds are:
Fats RQ = 0.7
The chemical composition of fats differs from that of carbohydrates in that fats contain considerably fewer oxygen atoms in proportion to atoms of carbon and hydrogen.Protein RQ = 0.8
Due to the complexity of various ways in which different amino acids can be metabolized, no single RQ can be assigned to the oxidation of protein in the diet; however, 0.8 is a frequently utilized estimate. -
This question is part of the following fields:
- Physiology
-
-
Question 27
Incorrect
-
A healthy 27-year old male who weighs 70kg has appendicitis. He is currently in the operating room and is being positioned to have a rapid sequence induction.
Prior to preoxygenation, the compartment likely to have the best oxygen reserve is:Your Answer:
Correct Answer: Red blood cells
Explanation:The following table shows the compartments and their relative oxygen reserve:
Compartment Factors Room air (mL) 100% O2 (mL)
Lung FAO2, FRC 630 2850
Plasma PaO2, DF, PV 7 45
Red blood cells Hb, TGV, SaO2 788 805
Myoglobin 200 200
Interstitial space 25 160Oxygen reserves in the body, with room air and after oxygenation.
FAO2-alveolar fraction of oxygen rises to 95% after administration of 100% oxygen (CO2 = 5%)
FRC- Functional residual capacity – (the most important store of oxygen in the body) – 2,500-3,000 mL in medium sized adults
PaO2-partial pressure of oxygen dissolved in arterial blood (80 mmHg breathing room air and 500 mmHg breathing 100% oxygen)
DF -dissolved form (0.3%)
PV-plasma volume (3L)
TG-total globular volume (5L)
Hb-haemoglobin concentration
SaO2-arterial oxygen concentration (98% breathing air and 100% when preoxygenated) -
This question is part of the following fields:
- Physiology
-
-
Question 28
Incorrect
-
The following is true about the extracellular fluid (ECF) in a normal adult woman weighing 60 kg.
Your Answer:
Correct Answer: Has a total volume of about 12 litres
Explanation:Total body water (TBW) is about 50% to 70% in adults depending on how much fat is present. ECF is relatively contracted in an obese person.
The simple rule is 60-40-20. (60% of weight = total body water, 40% of body weight is ICF and 20% is ECF)
For this woman, the total body water is 36 litres (0.6 × 60). ECF is 12 litres (1/3 of TBW) and 24 litres (2/3 of TBW) is intracellular fluid .
Sodium concentration is approximately 135-145 mmol/L in the ECF.
The ECF is made up of both intravascular and extravascular fluid and plasma proteins is found in both.
-
This question is part of the following fields:
- Physiology
-
-
Question 29
Incorrect
-
The action potential in a muscle fibre is initiated by which of these ions?
Your Answer:
Correct Answer: Sodium ions
Explanation:The cardiac action potential has several phases which have different mechanisms of action as seen below:
Phase 0: Rapid depolarisation – caused by a rapid sodium influx.
These channels automatically deactivate after a few msPhase 1: caused by early repolarisation and an efflux of potassium.
Phase 2: Plateau – caused by a slow influx of calcium.
Phase 3 – Final repolarisation – caused by an efflux of potassium.
Phase 4 – Restoration of ionic concentrations – The resting potential is restored by Na+/K+ATPase.
There is slow entry of Na+into the cell which decreases the potential difference until the threshold potential is reached. This then triggers a new action potentialOf note, cardiac muscle remains contracted 10-15 times longer than skeletal muscle.
Different sites have different conduction velocities:
1. Atrial conduction – Spreads along ordinary atrial myocardial fibres at 1 m/sec2. AV node conduction – 0.05 m/sec
3. Ventricular conduction – Purkinje fibres are of large diameter and achieve velocities of 2-4 m/sec, the fastest conduction in the heart. This allows a rapid and coordinated contraction of the ventricles
-
This question is part of the following fields:
- Physiology
-
-
Question 30
Incorrect
-
The typical fluid compartments in a normal 70kg male are:
Your Answer:
Correct Answer: intracellular>extracellular
Explanation:Body fluid compartments in a 70kg male:
Total volume=42L (60% body weight)
Intracellular fluid compartment (ICF) =28L
Extracellular fluid compartment (ECF) = 14LECF comprises:
Intravascular fluid (plasma) = 3L
Extravascular fluid = 11LExtravascular fluids comprises:
Interstitial fluid = 10.5L
Transcellular fluid = 0.5L -
This question is part of the following fields:
- Physiology
-
00
Correct
00
Incorrect
00
:
00
:
00
Session Time
00
:
00
Average Question Time (
Mins)