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Question 1
Correct
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A 68-year-old man is to be operated.
His past history is significant for a stroke, and some residual neurological deficit. The cranial nerves are examined clinically. He is unable to rotate his head to the left side when resistance is applied. Moreover, there is tongue wasting on the right side. There are no unusual sensory signs and symptoms.
The most likely reason for these clinical findings is?Your Answer: Damage to hypoglossal (XII) and spinal accessory (XI) nerves
Explanation:The upper five cervical segments of the spinal cord give rise to the XI cranial nerve. They connect with a few smaller branches before exiting the skull through the jugular foramen. The sternomastoid and trapezius muscles get their motor supply from the accessory root. Except for the palatoglossus, the hypoglossal nerve supplies motor supply to all tongue muscles.
The inability to shrug the shoulder on the affected side and rotate the head to the side against resistance is caused by damage to the spinal accessory nerve. This is due to the trapezius and sternomastoid muscles’ weakness.
The hypoglossal nerve is damaged, resulting in tongue wasting and inability to move from side to side.
The stylopharyngeus receives motor supply from the glossopharyngeal nerve. It also carries taste sensory fibres from the back third of the tongue, as well as the carotid sinus, carotid body, pharynx, and middle ear.
Motor supply to the larynx, pharynx, and palate; parasympathetic innervation to the heart, lung, and gut; and sensory fibres from the epiglottis and valleculae are all provided by the vagus nerve.
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This question is part of the following fields:
- Pathophysiology
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Question 2
Correct
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Ondansetron is serotonin antagonist indicated for post-operative nausea and vomiting.
In which of the following anatomic structures does ondansetron produce its effects?Your Answer: Nucleus tractus solitarius
Explanation:Ondansetron is a serotonin antagonist at the 5HT3 receptor. 5HT3 receptors in the gastrointestinal tract and in the vomiting centre of the medulla participate in the vomiting reflex. They are particularly important in vomiting caused by chemical triggers such as cancer chemotherapy drugs.
The nucleus solitarius is the recipient of all visceral afferents, and an essential part of the regulatory centres of the internal homeostasis, through its multiple projections with cardiorespiratory and gastrointestinal regulatory centres. It participates in the reflexes of the nerves innervating the nucleus, so it mediates cough reflex, carotid sinus reflex, gag reflex, and vomiting reflex.
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This question is part of the following fields:
- Pharmacology
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Question 3
Incorrect
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With a cervical dilation of 9 cm, a 23-year-old term primigravida is in labour. She is otherwise in good health. She's been in labour for 14 hours and counting.
Early foetal pulse decelerations can be seen on the cardiotocograph, and a recent foetal scalp blood sample revealed a pH of 7.25.
Which of the following is true about this patient's care and management?Your Answer: The patient requires a category 2 caesarean section under spinal anaesthetic
Correct Answer: Monitor for downward trend in fetal scalp blood pH as caesarean section is not indicated at the present time
Explanation:Once the decision to deliver a baby by caesarean section has been made, it should be carried out with a level of urgency commensurate with the risk to the baby and the mother’s safety.
There are four types of caesarean section urgency:
Category 1 – Endangering the life of the mother or the foetus
Category 2 – Maternal or foetal compromise that is not immediately life threatening
Category 3 – Early delivery is required, but there is no risk to the mother or the foetus.
Category 4: Elective delivery at a time that is convenient for both the mother and the maternity staff.Caesarean sections for categories 1 and 2 should be performed as soon as possible after the decision is made, especially for category 1. For category 1 caesarean sections, a decision to deliver time of 30 minutes is currently used.
In most cases, Category 2 caesarean sections should be performed within 75 minutes of making the decision.
The condition of the woman and the unborn baby should be considered when making a decision for a quick delivery, as it may be harmful in some cases.
There is no evidence of foetal compromise in the example above (early foetal pulse decelerations and a pH of less than 7.25). Early foetal pulse decelerations are most likely caused by the uterus compressing the foetal head. The foetus is not harmed by these. A spinal anaesthetic is preferred over a general anaesthetic whenever possible.
If the foetal scalp blood pH is greater than 7.25, it’s a good idea to repeat the test later and look for any changes. When a foetus decelerates, the mother should be given oxygen, kept in a left lateral position, and kept hydrated to avoid the need for a caesarean section.
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This question is part of the following fields:
- Pathophysiology
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Question 4
Correct
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Which measure of central tendency is most useful for a continuous, non-skewed data?
Your Answer: Mean
Explanation:Mean, also known as the average, is the most common measure of central tendency. It is the sum of all observed values divided by the number of observation. It is not useful for skewed data, which has an abnormal distribution. It is useful, instead, for numerical data that have symmetric distribution. It reflects the contributions of each data in the group, and are sensitive to outliers.
The median is the value that falls in the middle position when the observations are ranked in order from the smallest to the largest. If the number of observations is odd, the median is the middle number. If it is even, the median is the average of the two middle numbers. Unlike the mean, the median is useful on skewed data, and can be used for ordinal or numerical data if skewed.
The mode is the value that occurs with the greatest frequency in a set of observations, and is utilized for bimodal distribution.
The variance and the standard deviation are not measures of central tendency, but of dispersion.
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This question is part of the following fields:
- Statistical Methods
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Question 5
Incorrect
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A 30-year-old man has been stabbed in an area of the groin that contains the femoral triangle. He will undergo explorative surgery.
Which of the following makes the lateral wall of the femoral triangle?Your Answer: Adductor longus
Correct Answer: Sartorius
Explanation:The femoral triangle is a wedge-shaped area found within the superomedial aspect of the anterior thigh. It is a passageway for structures to leave and enter the anterior thigh.
Superior: Inguinal ligament
Medial: Adductor longus
Lateral: Sartorius
Floor: Iliopsoas, adductor longus and pectineusThe contents include: (medial to lateral)
Femoral vein
Femoral artery-pulse palpated at the mid inguinal point
Femoral nerve
Deep and superficial inguinal lymph nodes
Lateral cutaneous nerve
Great saphenous vein
Femoral branch of the genitofemoral nerve -
This question is part of the following fields:
- Anatomy
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Question 6
Incorrect
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Which of the following best explains the association between smoking and lower oxygen delivery to tissues?
Your Answer: Increased closing volume
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.
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This question is part of the following fields:
- Physiology
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Question 7
Incorrect
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Which of the following is the most appropriate first-line pharmacologic treatment for status epilepticus?
Your Answer: Phenytoin
Correct 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 cause membrane hyperpolarization.
Lorazepam has emerged as the preferred benzodiazepine for acute management of status epilepticus. Lorazepam differs from diazepam in two important respects. It is less lipid-soluble than diazepam, with a distribution half-life of two to three hours versus 15 minutes for diazepam. Therefore, it should have a longer duration of clinical effect. Lorazepam also binds the GABAergic receptor more tightly than diazepam, resulting in a longer duration of action. The anticonvulsant effects of lorazepam last six to 12 hours, and the typical dose ranges from 4 to 8 mg. This agent also has a broad spectrum of efficacy, terminating seizures in 75-80% of cases. Its adverse effects are identical to those of diazepam. Thus, lorazepam also is an effective choice for acute seizure management, with the added possibility of a longer duration of action than diazepam.
Phenobarbitone is a long-acting barbiturate that binds to GABA-A receptor site and increase the duration of chloride channel opening. It also blocks glutamic acid neurotransmission, and, at high doses, can block sodium channels. It is considered as the drug of choice for seizures in infants.
Phenytoin is an anti-seizure drug that blocks voltage-gated sodium channels. It is preferred in prolonged therapy of status epilepticus because it is less sedating.
In cases wherein airway protection is required, thiopentone and propofol are the preferred drugs.
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This question is part of the following fields:
- Pharmacology
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Question 8
Incorrect
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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.
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This question is part of the following fields:
- Physiology
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Question 9
Correct
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A weakly acidic drug with a pKa of 8.4 is injected intravenously into a patient.
At a normal physiological pH, the percentage of this drug unionised in the plasma is?Your Answer: 90
Explanation:Primary FRCA is concerned with two issues. The first is a working knowledge of the Henderson-Hasselbalch equation, and the second is a working knowledge of logarithms and antilogarithms.
The pH at which the drug exists in 50 percent ionised and 50 percent unionised forms is known as the pKa.
To calculate the proportion of ionised to unionised form of a drug, use the Henderson-Hasselbalch equation.
pH = pKa + log ([A-]/[HA])
or
pH = pKa + log [(salt)/(acid)]
pH = pKa + log ([ionised]/[unionised])Hence, if the pKa − pH = 0, then 50% of drug is ionised and 50% is unionised.
In this example:
7.4 = 8.4 + log ([ionised]/[unionised])
7.4 − 8.4 = log ([ionised]/[unionised])
log −1 = log ([ionised]/[unionised])Simply put, the antilog is the inverse log calculation. In other words, if you know the logarithm of a number, you can use the antilog to find the value of the number. The antilogarithm’s definition is as follows:
y = antilog x = 10x
Antilog to the base 10 of 0 = 1, −1 = 0.1, −2 = 0.01, −3 = 0.001 and, −4 = 0.0001.
[A-]/[HA] = 0.1
Assuming that we can apply the approximation [A-] << [HA} then this means the acid is 0.1 x 100% = 10% ionised so the percentage of (non-ionized) acid will be 100% – 10% = 90%
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This question is part of the following fields:
- Pharmacology
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Question 10
Correct
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Which of the following combinations of signs seen in a patient would most likely confirm ingestion of substances with anticholinesterase effects?
Your Answer: Bradycardia and miosis
Explanation:An acetylcholinesterase inhibitor or anticholinesterase is a chemical that inhibits the cholinesterase enzyme from breaking down acetylcholine (ACh) therefore increasing the level and duration of action of the neurotransmitter acetylcholine(ACh).
ACh stimulates postganglionic receptors to produce the following effects:
Salivation
Lacrimation
Defecation
Micturition
Sweating
Miosis
Bradycardia, and
Bronchospasm.Since these effects are produced by muscarine, they are referred to as muscarinic effects, and the postganglionic receptors are called muscarine receptors.
SLUD (Salivation, Lacrimation, Urination, Defecation – and emesis) is usually encountered only in cases of drug overdose or exposure to nerve gases. It is a syndrome of pathological effects indicating massive discharge of the parasympathetic nervous system.
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This question is part of the following fields:
- Pathophysiology
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