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Question 1
Correct
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If a large volume of 0.9% N. saline is administered during resuscitation, it is most likely to cause?
Your Answer: Hyperchloremic metabolic acidosis
Explanation:Crystalloids recommended for fluid resuscitation include 0.9% N saline and Hartmann’s solution(a physiological solution). 0.9% N. saline is not a physiological solution for the following reasons:
Compared with the normal range of 98-102 mmol/L, its chloride concentration is high (154 mmol/L)
It lacks calcium, magnesium, glucose and potassium
It does not have bicarbonate or bicarbonate precursor buffer necessary to maintain plasma pH within normal limitsThere is a difference in the activity (concentration) of strong ions at a physiological pH. This imbalance can explain abnormalities of acid base balance. A normal strong ion difference (SID) is in the order of 40.
SID = ([Na+] + [K+] + [Ca2+] + [Mg2+]) – ([Cl-] + [lactate] + [SO42-])
This imbalance is made up with the weaker anions to maintain electrical neutrality.
Administration of a large volume of 0.9% normal saline during resuscitation results in excessive chloride administration and this impairs renal bicarbonate reabsorption. The SID of 0.9% normal saline is 0 (Na+ = 154mmol/L and Cl- = 154mmol/L = 154 – 154 = 0). A large volume of NS will decrease the plasma SID causing an acidosis.Other causes of a hyperchloremic acidosis are:
Diabetic ketoacidosis
Total Parenteral Nutrition
Overdose of ammonium chloride and hydrochloric acid
Gastrointestinal losses of bicarbonate like in diarrhoea and pancreatic fistula
Proximal renal tubular acidosis with failure of bicarbonate reabsorption -
This question is part of the following fields:
- Physiology
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Question 2
Incorrect
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Regarding the treatment of bladder cancer, a study concerned with the usage of a combined or monotherapy was conducted. A forest plot was used for the visual representation of the data.
Which of the following is true regarding forest plots?Your Answer: Forest plots are typically used in qualitative studies
Correct Answer: Forest plots can present data from multiple studies
Explanation:Being the part of a meta analysis, forest plots are more valued as evidence then randomised control trials.
The notion that forest plots can only be used if the results are substantial is not true. They are good indicators of the significance of the data. If the diamond intersects the central line, the data is rendered significant. It also aggregates means and confidence intervals from studies conducted in the past which makes the study much more reliable as errors associated with individual studies tend to have less of an impact in this way.
The suggestion that forest plots are primarily used for qualitative data is factually incorrect. Forest plots require numerical values to function.
All in all, forest plots help us in determining whether or not there is a significant trend in that particular field of study.
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This question is part of the following fields:
- Statistical Methods
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Question 3
Correct
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The tip of a pulmonary artery flotation catheter becomes wedged when threaded through the chambers of the heart and the pulmonary artery.
Which of the following options best describes the sequence of pressures measured at the catheter's tip during its passage through a normal patient's pulmonary artery?Your Answer: 0-12 mmHg, 2-25 mmHg, 12-25 mmHg and 8-12 mmHg
Explanation:The tricuspid valve allows the tip of a pulmonary artery catheter to pass through the right atrium and into the right ventricle.
The balloon will be inflated before crossing the pulmonary valve and entering the pulmonary artery, where it will eventually wedge or occlude the artery, providing an indirect measure of left atrial pressure.
0-12 mmHg in the right atrium
2-25 mmHg in the right ventricle
12-25 mmHg in the pulmonary artery
8-12 mmHg is the occlusion pressure -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 4
Correct
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Following a lumbar epidural for foot surgery, a patient has residual weakness on hip flexion but normal knee flexion/extension.
Which nerve roots are most likely implicated?Your Answer: L1/2/3
Explanation:Iliacus, and Psoas major are the most important muscles that produce flexion at the hip.
They are collectively called the iliopsoas muscle. The iliacus muscle originates from the ilium while the psoas major muscle takes its origin from the lumbar vertebrae and sacrum. Their insertion is the lesser trochanter of the femur. They work together to produce flexion and external rotation of the hip. The nerve supply is from branches of the lumbar plexus (L1, 2, 3) femoral nerve (L2, 3, 4) and short direct muscular branches (T12, L1, L2, L3 and L4).
Sartorius, Rectus femoris, Tensor fasciae latae, and Pectineus muscles are two-joint muscles acting at the knee and having less influence on hip flexion:
Rectus femoris and sartorius are involved in extension of the knee. They are supplied by branches of the femoral nerve.
Myotomes associated with key movement of the lower limb are:
L1/L2 – Hip flexion
L2/L3/L4 – Hip adduction, quadriceps (knee extension)
L4/L5 – Hip abduction
L5 – Great toe dorsiflexion.Since knee extension is not affected, L2, L3 and L4 are still intact.
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This question is part of the following fields:
- Anatomy
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Question 5
Correct
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A randomized study aimed at finding out the efficacy of a novel anticoagulant, in preventing stroke in patients suffering from atrial fibrillation, relative to those already available in the market was performed. A 59 year old woman volunteered for it and was randomised to the treatment arm. A year later, following findings were reported:
165 out of 1050 patients who were prescribed the already prevalent medicine had a stroke while the number of patients who had a single stroke after using the new drug was 132 out of 1044.
In order to avoid one stroke case, what is the number of patients that need to be treated?Your Answer: 32
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (165/1050)-(132/1044)
ARR= (0.157-0.126)
ARR= 0.031
NNT= 1/0.031
NNT=32.3
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This question is part of the following fields:
- Statistical Methods
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Question 6
Incorrect
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What statement about endotoxins is true?
Your Answer: Are produced mainly by Gram positive bacteria
Correct Answer: Can often survive autoclaving
Explanation:Endotoxins are the lipopolysaccharides found in the outer cell wall of Gram-negative bacteria. They are responsible for providing the structure and stability of the cell wall.
They cannot be destroyed by normal sterilisation as they are heat stable molecules. They require the use of certain sterilant such as superoxide, peroxide and hypochlorite to be neutralised.
They stimulate strong immune responses, but can only be destroyed partially by specific antibodies. Repeat infections occur as memory T cells cannot be formed.
It can cause septicaemia and associated symptoms such as fever, shock, hypotension and nausea.
It activates the alternative complement pathway and the coagulation pathway using secreted cytokines.
It is not involved in botulism as clostridium botulinum, the responsible organism, secretes a neurotoxic exotoxin.
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This question is part of the following fields:
- Pathophysiology
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Question 7
Incorrect
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Bioelectric potentials that have been measured have an optimum bandwidth and typical frequency.
For a standard 12-lead ECG, which of the following bandwidth and voltage combinations is the best?Your Answer: Bandwidth 1-50 Hz, voltage 10-200 microvolts
Correct Answer: Bandwidth 0.05-150 Hz, voltage 100-4000 microvolts
Explanation:The potential difference (amplitude) and bandwidth frequencies of bioelectric signals are typical.
These are the following:
ECG: A bandwidth of 0.5-50 Hz is usually sufficient in monitoring mode, but a typical diagnostic bandwidth is 0.05-150 Hz (up to 200 Hz) with a typical voltage range of 0.1-4 millivolts (100-4000 microvolts).
EEG has a frequency range of 0.5-100 Hz and a voltage range of 0.5-100 microvolts.
EMG has a frequency range of 0.5 to 350 Hz and a voltage range of 0.5 to 30 millivolts.Prior to display, these small signals will need to be amplified and processed further.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 8
Correct
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A drug with a 2-hour half-life and a first-order kinetics of elimination is administered intravenously. The initial plasma concentration is calculated to be 12 mcg/mL and plasma concentrations is measured hourly.
At 6 hours, how much drug will be left?Your Answer: 1.5 mcg/mL
Explanation:In first order kinetics the rate of elimination is proportional to plasma concentration.
Rate of elimination is described by the following equation:
C = C0. e^-kt
Where:
C=drug concentration,
C0= drug concentration at time zero (extrapolated),
k = rate constant and
t = time.The initial concentration of this drug is 12 mcg/ml therefore:
The plasma concentration will have halved to 6 mcg/ml at 2 hours.
The plasma concentration will have halved to 3 mcg/ml at 4 hours and
The plasma concentration will have halved to 1.5 mcg/ml t 6 hours. -
This question is part of the following fields:
- Pharmacology
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Question 9
Correct
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A current flows through a simple electric circuit.
Which of the following electrical component configurations has the greatest potential difference?Your Answer: Two 5 ohm resistors in series with a passing current of 10 ampere
Explanation: -
This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 10
Incorrect
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Which statement is true about the autonomic nervous system?
Your Answer: Post ganglionic synapses use Acetylcholine as the neurotransmitter with nicotinic receptors in the parasympathetic system
Correct Answer: Preganglionic synapse utilise Acetylcholine as the neurotransmitter in both parasympathetic and sympathetic systems
Explanation:With regards to the autonomic nervous system (ANS)
1. It is not under voluntary control
2. It uses reflex pathways and different to the somatic nervous system.
3. The hypothalamus is the central point of integration of the ANS. However, the gut can coordinate some secretions and information from the baroreceptors which are processed in the medulla.With regards to the central nervous system (CNS)
1. There are myelinated preganglionic fibres which lead to the
ganglion where the nerve cell bodies of the non-myelinated post ganglionic nerves are organised.
2. From the ganglion, the post ganglionic nerves then lead on to the innervated organ.Most organs are under control of both systems although one system normally predominates.
The nerves of the sympathetic nervous system (SNS) originate from the lateral horns of the spinal cord, pass into the anterior primary rami and then pass via the white rami communicates into the ganglia from T1-L2.
There are short pre-ganglionic and long post ganglionic fibres.
Pre-ganglionic synapses use acetylcholine (ACh) as a neurotransmitter on nicotinic receptors.
Post ganglionic synapses uses adrenoceptors with norepinephrine / epinephrine as the neurotransmitter.
However, in sweat glands, piloerector muscles and few blood vessels, ACh is still used as a neurotransmitter with nicotinic receptors.The ganglia form the sympathetic trunk – this is a collection of nerves that begin at the base of the skull and travel 2-3 cm lateral to the vertebrae, extending to the coccyx.
There are cervical, thoracic, lumbar and sacral ganglia and visceral sympathetic innervation is by cardiac, coeliac and hypogastric plexi.
Juxta glomerular apparatus, piloerector muscles and adipose tissue are all organs under sole sympathetic control.
The PNS has a craniosacral outflow. It causes reduced arousal and cardiovascular stimulation and increases visceral activity.
The cranial outflow consists of
1. The oculomotor nerve (CN III) to the eye via the ciliary ganglion,
2. Facial nerve (CN VII) to the submandibular, sublingual and lacrimal glands via the pterygopalatine and submandibular ganglions
3. Glossopharyngeal (CN IX) to lungs, larynx and tracheobronchial tree via otic ganglion
4. The vagus nerve (CN X), the largest contributor and carries ¾ of fibres covering innervation of the heart, lungs, larynx, tracheobronchial tree parotid gland and proximal gut to the splenic flexure, liver and pancreasThe sacral outflow (S2 to S4) innervates the bladder, distal gut and genitalia.
The PNS has long preganglionic and short post ganglionic fibres.
Preganglionic synapses, like in the SNS, use ACh as the neuro transmitter with nicotinic receptors.
Post ganglionic synapses also use ACh as the neurotransmitter but have muscarinic receptors.Different types of these muscarinic receptors are present in different organs:
There are:
M1 = pupillary constriction, gastric acid secretion stimulation
M2 = inhibition of cardiac stimulation
M3 = visceral vasodilation, coronary artery constriction, increased secretions in salivary, lacrimal glands and pancreas
M4 = brain and adrenal medulla
M5 = brainThe lacrimal glands are solely under parasympathetic control.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 11
Correct
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Which of the following is a feature of a central venous pressure waveform?
Your Answer: An a wave due to atrial contraction
Explanation:The central venous pressure (CVP) waveform depicts changes of pressure within the right atrium. Different parts of the waveform are:
A wave: which represents atrial contraction. It is synonymous with the P wave seen during an ECG. It is often eliminated in the presence of atrial fibrillation, and increased tricuspid stenosis, pulmonary stenosis and pulmonary hypertension.
C wave: which represents right ventricle contraction at the point where the tricuspid valve bulges into the right atrium. It is synonymous with the QRS complex seen on ECG.
X descent: which represents relaxation of the atrial diastole and a decrease in atrial pressure, due to the downward movement of the right ventricle as it contracts. It is synonymous with the point before the T wave on ECG.
V wave: which represents an increase in atrial pressure just before the opening of the tricuspid valve. It is synonymous with the point after the T wave on ECG. It is increased in the background of a tricuspid regurgitation.
Y descent: which represents the emptying of the atrium as the tricuspid valve opens to allow for blood flow into the ventricle in early diastole.
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This question is part of the following fields:
- Pathophysiology
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Question 12
Correct
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Prophylactic antibiotics are required for which of the following procedures?
Your Answer: Caesarean section
Explanation:Staphylococcus aureus infection is the most likely cause.
Surgical site infections (SSI) occur when there is a breach in tissue surfaces and allow normal commensals and other pathogens to initiate infection. They are a major cause of morbidity and mortality.
SSI comprise up to 20% of healthcare associated infections and approximately 5% of patients undergoing surgery will develop an SSI as a result.
The organisms are usually derived from the patient’s own body.Measures that may increase the risk of SSI include:
-Shaving the wound using a single use electrical razor with a disposable head
-Using a non iodine impregnated surgical drape if one is needed
-Tissue hypoxia
-Delayed prophylactic antibiotics administration in tourniquet surgery, patients with a prosthesis or valve, in clean-contaminated surgery of in contaminated surgery.Measures that may decrease the risk of SSI include:
1. Intraoperatively
– Prepare the skin with alcoholic chlorhexidine (Lowest incidence of SSI)
-Cover surgical site with dressingIn contrast to previous individual RCT’s, a recent meta analysis has confirmed that administration of supplementary oxygen does not reduce the risk of wound infection and wound edge protectors do not appear to confer benefit.
2. Post operatively
Tissue viability advice for management of surgical wounds healing by secondary intentionUse of diathermy for skin incisions
In the NICE guidelines the use of diathermy for skin incisions is not advocated. Several randomised controlled trials have been undertaken and demonstrated no increase in risk of SSI when diathermy is used. -
This question is part of the following fields:
- Physiology And Biochemistry
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Question 13
Incorrect
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An 80-year old lady has a background history of a previous myocardial infarction which has left permanent damage to her heart's conduction system. The part of the conduction system with the highest velocities is damaged, and this has resulted in desynchronisation of the ventricles. The part of the heart that conducts the fastest is which of the following?
Your Answer: Atrial muscle
Correct Answer: Purkinje fibres
Explanation:The electrical conduction system of the heart starts with the SA node which generates spontaneous action potentials.
This is conducted across both atria by cell to cell conduction, and occurs at around 1 m/s. The only pathway for the action potential to enter the ventricles is through the AV node in a normal heart.
At this site, conduction is very slow at 0.05ms, which allows for the atria to completely contract and fill the ventricles with blood before the ventricles depolarise and contract.The action potentials are conducted through the Bundle of His from the AV node which then splits into the left and right bundle branches. This conduction is very fast, (,2m/s), and brings the action potential to the Purkinje fibres.
Purkinje fibres are specialised conducting cells which allow for a faster conduction speed of the action potential (,2-4m/s). This allows for a strong synchronized contraction from the ventricle and thus efficient generation of pressure in systole.
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This question is part of the following fields:
- Physiology And Biochemistry
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Question 14
Correct
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A 77-year-old woman is scheduled for day case cataract surgery under local anaesthesia. She has no cardiac or respiratory problems. Lisinopril is being used to treat her hypertension, which is under control.
Which of the following preoperative investigations are the most appropriate for this patient?Your Answer: No investigations
Explanation:Because the patient has mild systemic disease, he is ASA 2 and the procedure will be performed under local anaesthesia.
The following factors should be considered when requesting preoperative investigations:
Indications derived from a preliminary clinical examination
Whether or not a general anaesthetic will be used, the possibility of asymptomatic abnormalities, and the scope of the surgery.No special investigations are needed if the patient has no history of significant systemic disease and no abnormal findings on examination during the nurse-led assessment.
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This question is part of the following fields:
- Clinical Measurement
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Question 15
Correct
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What is the primary purpose of funnel plots?
Your Answer: Demonstrate the existence of publication bias in meta-analyses
Explanation:Funnel plot is essentially a scatterplot of the effect of treatment against a particular measure of study precision. Its primal purpose is to serve as a visual aid and help in detection of bias or systematic heterogenity.
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This question is part of the following fields:
- Statistical Methods
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Question 16
Incorrect
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Which of the given statements is true about standard error of the mean?
Your Answer: Confidence intervals cannot be applied to the standard error of the mean
Correct Answer: Gets smaller as the sample size increases
Explanation:The standard error of the mean (SEM) is a measure of the spread expected for the mean of the observations – i.e. how ‘accurate’ the calculated sample mean is from the true population mean. The relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size.
SEM = SD / square root (n)
where SD = standard deviation and n = sample size
Therefore, the SEM gets smaller as the sample size (n) increases.
If we want to depict how widely scattered some measurements are, we use the standard deviation. For indicating the uncertainty around the estimate of the mean, we use the standard error of the mean. The standard error is most useful as a means of calculating a confidence interval. For a large sample, a 95% confidence interval is obtained as the values 1.96×SE either side of the mean.
A 95% confidence interval:
lower limit = mean – (1.96 * SEM)
upper limit = mean + (1.96 * SEM)
Results such as mean value are often presented along with a confidence interval. For example, in a study the mean height in a sample taken from a population is 183cm. You know that the standard error (SE) (the standard deviation of the mean) is 2cm. This gives a 95% confidence interval of 179-187cm (+/- 2 SE).
Hence, it would be wrong to say that confidence levels do not apply to standard error of the mean.
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This question is part of the following fields:
- Statistical Methods
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Question 17
Correct
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A 30-year-old man has been diagnosed with a direct inguinal hernia.
A direct inguinal hernia passes through Hesselbach's triangle. What forms the medial edge of this triangle?Your Answer: Rectus abdominis muscle
Explanation:The inguinal triangle of Hesselbach’s is an important clinical landmark on the posterior wall of the inguinal canal. It has the following relations:
Inferiorly – medial third of the inguinal ligament
Medially – lower lateral border of the rectus abdominis
Laterally – inferior epigastric vesselsDirect inguinal hernia is when the bowel bulges directly through the abdominal wall. These hernias usually protrude through Hesselbach’s triangle
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This question is part of the following fields:
- Anatomy
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Question 18
Incorrect
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What is the order of the anatomical components of the tracheobronchial tree from proximal to distal?
Your Answer: Bronchioles, respiratory bronchioles, terminal bronchioles, alveolar ducts, alveolar sacs
Correct Answer: Bronchioles, terminal bronchioles, respiratory bronchioles, alveolar ducts, alveolar sacs
Explanation:The tracheobronchial tree is subdivided into the conducting and the respiratory zones.
The zones from proximal to distal are:
Trachea
Bronchi
Bronchioles
Terminal bronchioles
Respiratory bronchioles
Alveolar ducts
Alveolar sacsfrom the trachea to terminal bronchioles are the conducting zone while the respiratory zone is from the respiratory bronchioles to the alveola sacs
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This question is part of the following fields:
- Anatomy
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Question 19
Incorrect
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Which of the following statements is true about oxygen face masks?
Your Answer: Variable performance devices are colour coded and are marked with the recommended oxygen flow rate
Correct Answer: The oxygen concentration delivered by high air flow oxygen enrichment devices is not dependent on the respiratory pattern of the patient
Explanation:The normal peak inspiratory flow in healthy individuals is 20-30 L/min during each normal tidal ventilation. This is expected to increase with greater respiratory rate and deeper inspiration.
Face masks are used to facilitate the delivery of oxygen from a breathing system to a patient. Face masks can be divided into two types: fixed performance or variable performance devices.
In fixed performance devices (also known as high air flow oxygen enrichment or HAFOE), fixed inspired oxygen concentration is delivered to the patent, independent and greater than that of the patient’s peak inspiratory flow rate (PIFR). No random entrainment is expected to occur at the time of PIFR, hence, the oxygen concentration in HAFOE devices is not dependent on the patient’s respiratory pattern.
Moreover, in HAFOE masks, the concentration of oxygen at a given oxygen flow rate is determined by the size of the constriction; a device with a greater entrainment aperture delivers a lower oxygen concentration. Therefore, a 40% Venturi device will have lesser entrainment aperture when compared to a 31% Venturi. Venturi masks allow relatively fixed concentrations of supplemental oxygen to be inspired e.g. 24%, 28%, 31%, 35%, 40% and 60% oxygen. These are colour coded and marked with the recommended oxygen flow rate.
Variable performance devices deliver variable inspired oxygen concentration to the patient, and is dependent on the PIFR. The PIFR can often exceed the flow rate at which oxygen or an oxygen/air mixture is supplied by the device, depending on a patient’s inspiratory effort. In addition, these masks allow expired air to be released through the holes in the sides of the mask. Thus, with increased respiratory rate, rebreathing of alveolar gas from inside the mask may occur.
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This question is part of the following fields:
- Anaesthesia Related Apparatus
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Question 20
Correct
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A 72-year old farmer is hospitalized with acute respiratory failure and autonomic dysfunction. Suspected organophosphate poisoning.
Which one is the best mechanism for acute toxicity caused by organophosphates?Your Answer: Inhibition of acetylcholinesterase
Explanation:The toxicity of organophosphorus (OP) nerve agents is manifested through irreversible inhibition of acetylcholinesterase (AChE) at the cholinergic synapses, which stops nerve signal transmission, resulting in a cholinergic crisis and eventually death of the poisoned person. Oxime compounds used in nerve agent antidote regimen reactivate nerve agent-inhibited AChE and halt the development of this cholinergic crisis.
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This question is part of the following fields:
- Physiology
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Question 21
Incorrect
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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: Power
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
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Question 22
Correct
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A peripheral nerve stimulator is used to stimulate the ulnar nerve at the wrist to indicate the degree of neuromuscular blockade.
Which single muscle or group of muscles of the hand supplied by the ulnar nerve is best for monitoring the twitch function during neuromuscular blockade?Your Answer: Adductor pollicis
Explanation:In anaesthesia, adductor pollicis neuromuscular monitoring with ulnar nerve stimulation is commonly used. It is the gold standard for measuring the degree of block and comparing neuromuscular blocking drugs and their effects on other muscles.
Electrodes are usually placed over the ulnar nerve at the wrist to monitor the adductor pollicis.
Neuromuscular blocking drugs have different sensitivity levels in different muscle groups.
To achieve the same level of blockade, the diaphragm requires 1.4 to 2 times the amount of neuromuscular blocking agent as the adductor pollicis muscle. The small muscles of the larynx and the ocular muscles are two other respiratory muscles that are less resistant than the diaphragm (especially corrugator supercilii).
The abdominal muscles, Orbicularis oculi, peripheral muscles of the limbs, Geniohyoid, Masseter, and Upper airway muscles are the most sensitive to neuromuscular blocking agents.
The C8-T1 nerve roots, which are part of the medial cord of the brachial plexus, form the ulnar nerve. It enters the hand via the ulnar canal, superficial to the flexor retinaculum, after following the ulnar artery at the wrist.
The nerve then splits into two branches: superficial and deep. The palmaris brevis is supplied by the superficial branch, which also provides palmar digital nerves to one and a half fingers. The dorsal surface of the medial/ulnar 1.5 fingers, as well as the corresponding skin over the hand, are also supplied by it (as well as the palmar surface).
The ulnar nerve’s deep branch runs between the abductor and flexor digiti minimi, which it supplies. It also innervates the opponens, and with the deep palmar arch, it curves around the hook of the hamate and laterally across the palm. All of the interossei, the medial two lumbricals, the adductor pollicis, and, in most cases, the flexor pollicis brevis are supplied there.
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This question is part of the following fields:
- Anatomy
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Question 23
Incorrect
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A 27-year old man loses 1000ml blood after being stabbed on his thigh.
The most impactful physiological response occurring at the start to combat the decline in the intravascular blood volume of this man is?Your Answer: Arteriolar constriction
Correct Answer: Venoconstriction
Explanation:In contrast to the arterial system, which contains 15% of the circulating blood volume, the body’s veins contain 70% of it.
In severe haemorrhage, when sympathetic stimulation causes venoconstriction, venous tone is important in maintaining the return of blood to the heart.
Because the liver receives about 30% of the resting cardiac output, it is a very vascular organ. The hepatic vascular system is dynamic, which means it can store and release blood in large amounts – it acts as a reservoir within the general circulation.
In a normal situation, the liver contains 10-15% of total blood volume, with the sinusoids accounting for roughly 60% of that. The liver dynamically adjusts its blood volume when blood is lost and can eject enough blood to compensate for a moderate amount of haemorrhage.
In the portal venous and hepatic arterial systems, sympathetic nerves constrict the presinusoidal resistance vessels. More importantly, sympathetic stimulation lowers the portal system’s capacitance, allowing blood to flow more efficiently to the heart.
Net transcapillary absorption of interstitial fluid from skeletal muscle into the intravascular space compensates for blood loss effectively during haemorrhage. The decrease in capillary hydrostatic pressure (Pc), caused by reflex adrenergic readjustment of the ratio of pre- to postcapillary resistance, is primarily responsible for fluid absorption. Within a few hours of blood loss, these fluid shifts become significant, further diluting haemoglobin and plasma proteins.
Albumin synthesis begins to increase after 48 hours.
The juxtamedullary complex releases renin in response to a drop in mean arterial pressure, which causes an increase in aldosterone level and, eventually, sodium and water resorption. Increased antidiuretic hormone (ADH) levels also contribute to water retention.
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This question is part of the following fields:
- Physiology
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Question 24
Correct
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An 18-year old female was brought into the emergency room because of active seizures. The informant reported that it has been more than 5 minutes since the patient started seizing. The attending physician gave an initial diagnosis of status epilepticus.
According to the paramedics who brought in the patient, 10 mg of diazepam was given rectally. Upon physical examination, she was normotensive at 120/80 mmHg; tachycardic at 138 beats per minute; tachypnoeic at 24 breaths per minute; and well-saturated at 99% on high flow oxygen. Her random blood glucose level was normal at 7.0 mmol/L.
Given this situation and an initial diagnosis of status epilepticus, what would be the best initial anti-epileptic drug to administer to the patient?Your Answer: Lorazepam
Explanation:Lorazepam is an intermediate-acting benzodiazepine that binds to the GABA-A receptor subunit to increase the frequency of chloride channel opening and facilitate membrane hyperpolarization. It is the preferred treatment for status epilepticus, although Diazepam can also be used as an alternative.
Lorazepam has a longer duration of action than Diazepam, and binds with greater affinity to the GABA-A receptor subunit.
Phenobarbital is a barbiturate that acts on the GABA-A receptor site to increase the duration of chloride channel opening. Barbiturates, particularly phenobarbital, is considered the drug of choice for seizures in infants.
Phenytoin is a sodium-channel blocker that is given for generalized tonic-clonic seizures, partial seizures, and status epilepticus. Phenytoin is preferred in prolonged therapy for status epilepticus because it is less sedating.
Propofol or thiopentone is preferred when airway protection is required.
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This question is part of the following fields:
- Pharmacology
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Question 25
Incorrect
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A young male is operated on for an open inguinal hernia repair. During the procedure, the cord is mobilized, and the deep inguinal ring is located.
What structure forms the lateral wall of the deep inguinal ring?Your Answer: Conjoint tendon
Correct Answer: Transversalis fascia
Explanation:The deep inguinal ring is the entrance of the inguinal canal. It is an opening in the transversalis fascia around 1 cm above the inguinal ligament. Therefore, the superolateral wall is made by the transervalis fascia.
The inferior epigastric vessels run medially to the deep inguinal ring forming its inferomedial border.
The inguinal canal extends obliquely from the deep inguinal ring to the superficial inguinal ring.
An indirect inguinal hernia arises through the deep inguinal ring lateral to the inferior epigastric vessels. -
This question is part of the following fields:
- Anatomy
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Question 26
Correct
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A pre-operative evaluation for a trans-sphenoidal pituitary adenectomy is being performed on a 57-year-old woman. Her vision is causing her problems.
A macroadenoma compressing the optic chiasm is visible on MRI.
What is the most likely visual field defect to be discovered during an examination?Your Answer: Bitemporal hemianopia
Explanation:The pituitary gland plays a crucial role in the neuro-endocrine axis. It is located at the base of the skull in the sella turcica of the sphenoid bone. It is connected superiorly to the hypothalamus, third ventricle, and visual pathways, and laterally to the cavernous sinuses, internal carotid arteries, and cranial nerves III, IV, V, and VI.
Pituitary tumours make up about 10-15% of all intracranial tumours. The majority of adenomas are benign. Over-secretion of pituitary hormones (most commonly prolactin, growth hormone, or ACTH), under-secretion of hormones, or localised or generalised pressure effects can all cause symptoms.
Compression of the optic chiasm can result in visual field defects, the most common of which is bitemporal hemianopia. This is caused by compression of the nasal retinal fibres, which carry visual impulses from temporal vision across the optic chiasm to the contralateral sides before continuing to the optic tracts.
The interruption of the visual pathways distal to the optic chiasm causes a homonymous visual field defect. The loss of the right or left halves of each eye’s visual field is referred to as homonymous hemianopia. It’s usually caused by a middle or posterior cerebral artery territory stroke that affects the occipital lobe’s optic radiation or visual cortex.
Binasal hemianopia is a condition in which vision is lost in the inner half of both eyes (nasal or medial). It’s caused by compression of the temporal visual pathways, which don’t cross at the optic chiasm and instead continue to the ipsilateral optic tracts. Binasal hemianopia is a rare complication caused by the internal carotid artery impinging on the temporal (lateral) visual fibres.
A monocular visual loss (that is, loss of vision in only one eye) can be caused by a variety of factors, but if caused by nerve damage, the damage would be proximal to the optic chiasm on the ipsilateral side.
A central scotoma is another name for central visual field loss. Every normal mammalian eye has a scotoma, also known as a blind spot, in its field of vision. The optic disc is a region of the retina that lacks photoreceptor cells and is where the retinal ganglion cell axons that make up the optic nerve exit the retina. When both eyes are open, visual signals that are absent in one eye’s blind spot are provided for the other eye by the opposite visual cortex, even if the other eye is closed.
Scotomata can be caused by a variety of factors, including demyelinating disease such as multiple sclerosis, damage to nerve fibre layer in the retina, methyl alcohol, ethambutol, quinine, nutritional deficiencies, and vascular blockages either in the retina or in the optic nerve.
Bilateral scotoma can occur when a pituitary tumour compresses the optic chiasm, causing a bitemporal paracentral scotoma, which then spreads out to the periphery, causing bitemporal hemianopsia. A central scotoma in a pregnant woman could be a sign of severe pre-eclampsia.
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This question is part of the following fields:
- Pathophysiology
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Question 27
Correct
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With a 10-day history of severe vomiting, a 71-year-old man with a gastric outlet obstruction is admitted to the surgical ward.
The serum biochemical results listed below are available:
Sodium 128 mmol/L (137-144)
Potassium 2.6 mmol/L (3.5-4.9)
Chloride 50 mmol/L (95-107)
Urea 12 mmol/L (2.5-7.5)
Creatinine 180 µmol/L (60-110)
Which of the following do you think you are most likely to encounter?Your Answer: The standard base excess will be higher than actual base excess
Explanation:Hydrochloric acid is lost when you vomit for a long time (HCl). As a result, the following can be expected, in varying degrees of severity:
Hypokalaemia
Hypochloraemia
Increased bicarbonate to compensate for chloride loss and metabolic alkalosisThe alkalosis causes potassium to move from the intracellular to the extracellular compartment at first. Long-term vomiting and dehydration cause potassium to be excreted by the kidneys in order to conserve sodium. Dehydration can cause urea and creatinine levels to rise.
The actual base excess is always greater than the standard base excess.
The actual base excess (BE) is a measurement of a base’s contribution to a blood gas picture’s metabolic component. It’s the amount of base that needs to be added to a blood sample to bring the pH back to 7.4 after the respiratory component of a blood gas picture has been corrected (PaCO2 of 40 mmHg or 5.3 kPa). The BE has a normal range of +2 to 2. A large positive BE indicates a severe metabolic alkalosis, while a large negative BE indicates a severe metabolic acidosis. As a result, the actual BE in vitro is unaffected by CO2.
In vivo, however, standard BE is not independent of pCO2 because blood with haemoglobin acts as a better buffer than total ECF.
As a result, it is impossible to tell the difference between compensating for a respiratory disorder and compensating for the presence of a primary metabolic disorder.
The differences between in vitro and in vivo behaviour can be mostly eliminated if the BE is calculated for a haemoglobin concentration of 50 g/L (the ‘effective’ or virtual value of Hb if it was distributed throughout the extracellular space) rather than the actual haemoglobin. Because haemoglobin has a lower buffering capacity, the standard BE is higher than the actual BE. It reflects the BE better in the extracellular space rather than just the intravascular compartment.
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This question is part of the following fields:
- Pathophysiology
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Question 28
Incorrect
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The whole water content of the body is calculated by multiplying body mass with 0.6. This water is diffused into distinct compartments.
Which fluid compartment can be measured indirectly?Your Answer: Plasma volume
Correct Answer: Intracellular volume
Explanation:The total body water content of a 70kg man is (70 × 0.6) = 42 litres. For a woman, the calculation is (70 × 0.55) = 38.5 litres.
For a man, it is subdivided into:
Extracellular fluid (ECF) = 14L (1/3)
Intracellular fluid (ICF) = 28L (2/3).The ECF volume is further divided into:
Interstitial fluid = 10.5 litres
Plasma = 3 litres
Transcellular fluid (CSF/synovial fluid) = 0.5 litres.Directly measured fluid compartments:
Heavy water (deuterium) can be used to measure total body water content, which is freely distributed.
Albumin labelled with a radioactive isotope or using a dye called Evans blue can be used to measure Plasma volume . They do not diffuse into red blood cells.
Radiolabelled (Cr-51) red blood cells can be used to measure total erythrocyte volume.
Inulin as the tracer can be used to measure ECF volume as it circulate freely in the interstitial and plasma volumes.Indirectly measured fluid compartments:
Total blood volume can be calculated with the level of haematocrit and the volume of total circulating red blood cells.
ICF volume can be calculated by subtracting ECF volume from total blood volume. -
This question is part of the following fields:
- Basic Physics
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Question 29
Incorrect
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Which of the following statement is false regarding dopamine?
Your Answer: Part of dose is converted to Noradrenaline in sympathetic nerve terminals
Correct Answer: Urine output decreases due to inhibition of proximal tubule Na+ reabsorption
Explanation:Dopamine (DA) is a dopaminergic (D1 and D2) as well as adrenergic ? and?1 (but not ?2 ) agonist.
The D1 receptors in renal and mesenteric blood vessels are the most sensitive: i.v. infusion of a low dose of Dopamine dilates these vessels (by raising intracellular cAMP). This increases g.f.r. In addition, DA exerts a natriuretic effect by D1 receptors on proximal tubular cells.
Moderately high doses produce a positive inotropic (direct?1 and D1 action + that due to NA release), but the little chronotropic effect on the heart.
Vasoconstriction (?1 action) occurs only when large doses are infused.
At doses normally employed, it raises cardiac output and systolic BP with little effect on diastolic BP. It has practically no effect on nonvascular ? and ? receptors; does not penetrate the blood-brain barrier—no CNS effects.
Dopamine is less arrhythmogenic than adrenaline
Regarding dopamine part of the dose is converted to Noradrenaline in sympathetic nerve terminals.
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This question is part of the following fields:
- Pharmacology
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Question 30
Incorrect
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A 54-year-old man weighing 70kg, underwent mesh repair for inguinal hernia under general anaesthesia. He was given intravenous co-amoxiclav (Augmentin) following which the patient developed widespread urticarial ras, became hypotensive (61/30 mmHg), and showed clinical signs of bronchospasm. Anaphylaxis is suspected in this patient.
Which one of the following is considered as best initial pharmacological treatment for this condition?Your Answer: Intramuscular adrenaline 0.5 mg
Correct Answer: Intravenous adrenaline 50 mcg
Explanation:The drug of choice for the treatment of anaphylaxis is adrenaline. It has an intravenous route of administration. Since the patient already has intravenous access, the intramuscular route is not appropriate.
Second-line pharmacological intervention includes the use of chlorpheniramine 10mg intravenous, Hydrocortisone 200mg.
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This question is part of the following fields:
- Pharmacology
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