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Question 1
Correct
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A new blood test developed to screen individuals for cardiac failure was performed on 500 patients. The results were positive for 40 out of 50 patients with echocardiography-established heart failure. However, the test was also positive for 20 patients with no signs of heart failure. What is the positive predictive value of the test?
Your Answer: 0.66
Explanation:Positive predictive value = TP (true positives) / [TP + FP (false positives)] = 40 / (40 + 20) = 0.66
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This question is part of the following fields:
- Clinical Sciences
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Question 2
Correct
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A study is developed to assess a new mandible advancement device designed to reduce snoring. A 10 point scale was used to assess the severity of snoring before and after applying of the device by the respective partner. The number of the people involved in the study was 50. What test would you apply in this particular study?
Your Answer: Wilcoxon signed-rank test
Explanation:Steps required in performing the Wilcoxon signed rank test:
1 State the null hypothesis and, in particular, the hypothesized value for comparison
2 Rank all observations in increasing order of magnitude, ignoring their sign. Ignore any observations that are equal to the hypothesized value. If two observations have the same magnitude, regardless of sign, then they are given an average ranking
3 Allocate a sign (+ or -) to each observation according to whether it is greater or less than the hypothesized value (as in the sign test)
4 Calculate:
R+ = sum of all positive ranks
R- = sum of all negative ranks
R = smaller of R+ and R-
5 Calculate an appropriate P value What makes this test the most appropriate for this study is that the data is non-parametric, paired and comes from the same population. -
This question is part of the following fields:
- Clinical Sciences
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Question 3
Correct
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A 60-year-old male presents with dyspnoea and an urgent chest X-ray is scheduled. Sputum cultures reveal pneumonia and he receives treatment with erythromycin. What is the mechanism of action of this drug?
Your Answer: Inhibit 50S subunit of ribosomes
Explanation:Erythromycin is a bacteriostatic antibiotic. This means it stops the further growth of bacteria rather than directly destroying it. This is achieved by inhibiting protein synthesis. Erythromycin binds to the 23S ribosomal RNA molecule in the 50S subunit of the bacterial ribosome. This causes a blockage in the exiting of the peptide chain that is growing. Given that humans have 40S and 60S subunits, and do not have 50S subunits, erythromycin does not affect protein synthesis in human tissues.
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This question is part of the following fields:
- Clinical Sciences
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Question 4
Incorrect
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Choose the correct statement regarding the standard error of the mean:
Your Answer: Is the square root of standard deviation
Correct Answer: Gets smaller as the sample size increases
Explanation:When statistically comparing data sets, researchers estimate the population of each sample and examine them to see whether they are identical. The standard error of the mean (SEM) – not the standard deviation (SD), which represents the variation in the sample – is used to estimate the population mean. Via this process, researchers conclude that the sample used in their studies appropriately represents the population within the error range specified by the pre-set significance level.
The SEM is smaller than the SD, as the SEM is estimated usually with the SD divided by the square root of the sample size. For this reason, researchers are tempted to use the SEM when describing their samples. It is acceptable to use either the SEM or SD to compare two different groups if the sample sizes of the two groups are equal; however, the sample size must be stated in order to deliver accurate information. For example, when a population has a large amount of variation, the SD of an extracted sample from this population must be large. However, the SEM will be small if the sample size is deliberately increased. -
This question is part of the following fields:
- Clinical Sciences
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Question 5
Incorrect
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A 13-year-old girl is brought by her mother to the A&E with breathlessness, fatigue and palpitations. Anamnesis does not reveal any syncope or chest pain in the past. on the other hand, these symptoms were present intermittently for a year. Clinical examination reveals a pan-systolic murmur associated with giant V waves in the jugular venous pulse. Chest auscultation and resting ECG are normal. 24 hour ECG tape shows a short burst of supraventricular tachycardia. What is the most probable diagnosis?
Your Answer:
Correct Answer: Ebstein's anomaly
Explanation:Ebstein’s anomaly is characterised by apical displacement and adherence of the septal and posterior leaflets of the tricuspid valve to the underlying myocardium, thereby displacing the functional tricuspid orifice apically and dividing the right ventricle into two portions. The main haemodynamic abnormality leading to symptoms is tricuspid valve incompetence. The clinical spectrum is broad; patients may be asymptomatic or experience right-sided heart failure, cyanosis, arrhythmias and sudden cardiac death (SCD). Many Ebstein’s anomaly patients have an interatrial communication (secundum atrial septal defect (ASD II) or patent foramen ovale). Other structural anomalies may also be present, including a bicuspid aortic valve (BAV), ventricular septal defect (VSD), and pulmonary stenosis. The morphology of the tricuspid valve in Ebstein anomaly, and consequently the clinical presentation, is highly variable. The tricuspid valve leaflets demonstrate variable degrees of failed delamination (separation of the valve tissue from the myocardium) with fibrous attachments to the right ventricular endocardium.
The displacement of annular attachments of septal and posterior (inferior) leaflets into the right ventricle toward the apex and right ventricular outflow tract is the hallmark finding of Ebstein anomaly. -
This question is part of the following fields:
- Clinical Sciences
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Question 6
Incorrect
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Question 7
Incorrect
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A study is carried out to assess the efficacy of a rapid urine screening test developed to detect Chlamydia. The total number of people involved in the study were 200. The study compared the new test to the already existing NAAT techniques. The new test was positive in 20 patients that were Chlamydia positive and in 3 patients that were Chlamydia negative. For 5 patients that were Chlamydia positive and 172 patients that were Chlamydia negative the test turned out to be negative. Choose the correct value regarding the negative predictive value of the new test:
Your Answer:
Correct Answer: 172/177
Explanation:The definition of negative predictive value is the probability that the individuals with truly negative screening test don’t have Chlamydia. The equation is the following: Negative predictive value = Truly negative/(truly negative + false negative) = 172 / (172 + 5) = 172 / 177
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This question is part of the following fields:
- Clinical Sciences
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Question 8
Incorrect
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A 68-year-old male patient presents with central chest pain and associated flushing. He claims the pain is crushing in character. ECG reveals T wave inversion in II, III and AVF. Blood exams are as follows: Troponin T = 0.9 ng/ml. Which substance does troponin bind to?
Your Answer:
Correct Answer: Tropomyosin
Explanation:Troponin T is a 37 ku protein that binds to tropomyosin, thereby attaching the troponin complex to the thin filament.
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This question is part of the following fields:
- Clinical Sciences
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Question 9
Incorrect
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Which one of the following statements best describes a type II statistical error?
Your Answer:
Correct Answer: The null hypothesis is accepted when it is false
Explanation:In statistical hypothesis testing there are 2 types of errors:
– type I: the null hypothesis is rejected when it is true – i.e. Showing a difference between two groups when it doesn’t exist, a false positive.
– type II: the null hypothesis is accepted when it is false – i.e. Failing to spot a difference when one really exists, a false negative. -
This question is part of the following fields:
- Clinical Sciences
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Question 10
Incorrect
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Primarily, funnel plots are used to do what?
Your Answer:
Correct Answer: Demonstrate the existence of publication bias in meta-analyses
Explanation:Funnel plots are graphical tools to assess and compare clinical performance of a group of care professionals or care institutions on a quality indicator against a benchmark. Incorrect construction of funnel plots may lead to erroneous assessment and incorrect decisions potentially with severe consequences.
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This question is part of the following fields:
- Clinical Sciences
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Question 11
Incorrect
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Choose the wrong statement regarding hypocalcaemia:
Your Answer:
Correct Answer: Chvostek's sign is more sensitive and specific than Trousseau's sign
Explanation:Chvostek and Trousseau signs can be elicited in patients with hypocalcaemia. Chvostek sign is the twitching of the upper lip with tapping on the cheek 2 cm anterior to the earlobe, below the zygomatic process overlying the facial nerve. Trousseau sign (a more reliable sign present in 94% of hypokalaemic individuals and only 1% to 4% of healthy people) is the presence of carpopedal spasm observed following application of an inflated blood pressure cuff over systolic pressure for 3 minutes in hypokalaemic patients.
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This question is part of the following fields:
- Clinical Sciences
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Question 12
Incorrect
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Which one of the following congenital infections is most characteristically associated with chorioretinitis?
Your Answer:
Correct Answer: Toxoplasma gondii
Explanation:The common congenital infections encountered are rubella, toxoplasmosis and cytomegalovirus. Cytomegalovirus is the most common congenital infection in the UK. Maternal infection is usually asymptomatic.
Congenital toxoplasmosis is associated with fetal death and abortion, and in infants, it is associated with neurologic deficits, neurocognitive deficits, and chorioretinitis. -
This question is part of the following fields:
- Clinical Sciences
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Question 13
Incorrect
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You are a ST1 doctor working on a medical ward. You are struggling to cope with the workload and often leave the ward late. Who is the most appropriate action to take?
Your Answer:
Correct Answer: Speak to your consultant
Explanation:Speaking to your consultant is the most appropriate first action to take in this scenario. They are best placed to be able to take action to try and amend the situation. The consultant is also ultimately responsible for patient care and hterefore have a right to know if you are struggling, as this may affect patient care.
Arriving early and taking time off sick do not address the problem. -
This question is part of the following fields:
- Clinical Sciences
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Question 14
Incorrect
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Choose the most important stimulator of the central chemoreceptors:
Your Answer:
Correct Answer: Decrease in pH
Explanation:Central chemoreception refers to the detection of changes in CO2/H+ within the brain and the associated effects on breathing. In the conscious animal the response of ventilation to changes in the brain’s interstitial fluid (ISF) pH is very sensitive. Note that a small change in cerebrospinal fluid (CSF) pH from 7.30 to 7.25 is associated with a doubling of alveolar ventilation; it is a very sensitive reflex response. Note also that the relationship of alveolar ventilation to ISF pH is essentially the same for both types of stimulation, metabolic acid-base disorders and primary CO2 stimulation.
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This question is part of the following fields:
- Clinical Sciences
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Question 15
Incorrect
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A woman with severe renal failure undergoes a kidney transplant. However, after a few hours, she develops fever and anuria. The doctors are suspecting hyperacute organ rejection. Which are the cells primarily responsible for hyperacute organ rejection?
Your Answer:
Correct Answer: B Cells
Explanation:Hyperacute rejection appears in the first minutes following transplantation and occurs only in vascularized grafts. This very fast rejection is characterized by vessel thrombosis leading to graft necrosis. Hyperacute rejection is caused by the presence of antidonor antibodies existing in the recipient before transplantation. These antibodies induce both complement activation and stimulation of endothelial cells to secrete Von Willebrand procoagulant factor, resulting in platelet adhesion and aggregation. The result of these series of reactions is the generation of intravascular thrombosis leading to lesion formation and ultimately to graft loss. Today, this type of rejection is avoided in most cases by checking for ABO compatibility and by excluding the presence of antidonor human leukocyte antigen (HLA) antibodies by cross-match techniques between donor graft cells and recipient sera. This type of rejection is also observed in models of xenotransplantation of vascularized organs between phylogenetically distant species when no immunosuppressive treatment is given to the recipients.
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This question is part of the following fields:
- Clinical Sciences
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Question 16
Incorrect
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A 24-year-old male is admitted with worsening shortness of breath with signs of left ventricular failure. He has a known genetic condition. On examination, there is an ejection systolic murmur loudest over the aortic area radiating to the carotids, bibasal crepitations and pitting oedema to the knees bilaterally. On closer inspection of the patient, you note a wide vermillion border, small spaced teeth and a flat nasal bridge. The patient also has a disinhibited friendly demeanour. What is the likely precipitating valvular issue?
Your Answer:
Correct Answer: Supravalvular aortic stenosis
Explanation:Supravalvular aortic stenosis, is associated with a condition called William’s syndrome.
William’s syndrome is an inherited neurodevelopmental disorder caused by a microdeletion on chromosome 7. The most common symptoms of Williams syndrome are heart defects and unusual facial features. Other symptoms include failure to gain weight appropriately in infancy (failure to thrive) and low muscle tone. Individuals with Williams syndrome tend to have widely spaced teeth, a long philtrum, and a flattened nasal bridge.
Most individuals with Williams syndrome are highly verbal relative to their IQ, and are overly sociable, having what has been described as a cocktail party type personality. -
This question is part of the following fields:
- Clinical Sciences
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Question 17
Incorrect
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Which one of the following is true regarding Escherichia coli infection?
Your Answer:
Correct Answer: E coli is an important cause of neonatal meningitis
Explanation:Escherichia coli (also known as E. coli) is a gram-negative, facultatively anaerobic, rod-shaped bacterium commonly found in the lower intestine of warm-blooded organisms. Most E. coli strains are harmless, but some serotypes can cause serious food poisoning in their hosts, and are occasionally responsible for product recalls due to food contamination. The harmless strains are part of the normal flora of the gut, and can benefit their hosts by producing vitamin K2, and preventing colonization of the intestine with pathogenic bacteria. Virulent strains can cause gastroenteritis, urinary tract infections, and neonatal meningitis.
The most common causes of neonatal meningitis is bacterial infection of the blood, known as bacteremia (specifically Group B Streptococci (GBS; Streptococcus agalactiae), Escherichia coli, and Listeria monocytogenes). Although there is a low mortality rate in developed countries, there is a 50% prevalence rate of neurodevelopmental disabilities in E. coli and GBS meningitis -
This question is part of the following fields:
- Clinical Sciences
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Question 18
Incorrect
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You want to compare a new oral hypoglycaemic drug with an existing treatment, which would also lower HbA1c. Which study design would you choose?
Your Answer:
Correct Answer: Superiority trial
Explanation:When the aim of the randomized controlled trial (RCT) is to show that one treatment is superior to another, a statistical test is employed and the trial (test) is called a superiority trial (test).
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This question is part of the following fields:
- Clinical Sciences
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Question 19
Incorrect
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Intracellular proteins tagged with mannose-6-phosphate are destined to which of the following organelles?
Your Answer:
Correct Answer: Lysosome
Explanation:Lysosomal hydrolases are synthesized in the rough endoplasmic reticulum and specifically transported through the Golgi apparatus to the trans-Golgi network, from which transport vesicles bud to deliver them to the endosomal/lysosomal compartment. The explanation of how the lysosomal enzymes are accurately recognized and selected over many other proteins in the trans-Golgi network relies on them being tagged with a unique marker: the mannose-6-phosphate (M6P) group, which is added exclusively to the N-linked oligosaccharides of lysosomal soluble hydrolases, as they pass through the cis-Golgi network. Generation of the M6P recognition marker depends on a reaction involving two different enzymes: UDP-N-acetylglucosamine 1-phosphotransferase and α-N-acetylglucosamine-1-phosphodiester α-N-acetylglucosaminidase.
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This question is part of the following fields:
- Clinical Sciences
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Question 20
Incorrect
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A case-control study is developed to assess passive smoking as a risk factor for the development of asthma in children. The total number of patients recruited for this study is 200. 40 out of the 200 patients report at least one parent smoking in the house when they were younger. 200 more people without asthma are recruited and 20 out of them report that at least one parent smoked in the house when they were younger. What is the odds ratio of patients with asthma having been exposed to passive smoking during their childhood?
Your Answer:
Correct Answer: 2.25
Explanation:An odds ratio (OR) is a measure of association between an exposure and an outcome. The OR represents the odds that an outcome will occur given a particular exposure, compared to the odds of the outcome occurring in the absence of that exposure. Odds ratios are most commonly used in case-control studies, however they can also be used in cross-sectional and cohort study designs as well (with some modifications and/or assumptions). Where
a = Number of exposed cases
b = Number of exposed non-cases
c = Number of unexposed cases
d = Number of unexposed non-cases
OR=(a/c) / (b/d) = ad/bc
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This question is part of the following fields:
- Clinical Sciences
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Question 21
Incorrect
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A study was developed to assess a new oral antithrombotic drug on the chance of stroke in high-risk patients, compared to warfarin. The total number of patients receiving the new drug were 200 compared to 600 who were receiving warfarin. From the ones receiving the new drug, 10 patients had a stroke within 3 years, compared to 12 patients who were receiving warfarin and had a stroke. What is the relative risk of having a stroke within 3 years for patients receiving the new drug?
Your Answer:
Correct Answer: 2.5
Explanation:Relative Risk = (Probability of event in exposed group) / (Probability of event in not exposed group)
Experimental event rate, EER = 10 / 200 = 0.05Control event rate, CER = 12 / 600 = 0.02
Therefore the relative risk = EER / CER = 0.05 / 0.02 = 2.5
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This question is part of the following fields:
- Clinical Sciences
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Question 22
Incorrect
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A 52-year-old man presents with numbness and tingling in his left hand. On examination he has weakness of elbow extension, metacarpophalangeal joint flexion and extension and distal interphalangeal joint flexion. All other movements and reflexes are normal. Sensation is normal apart from reduced pin-prick sensation over the medial aspect of the hand. An MRI scan of the cervical spine is performed due to suspicion of a nerve lesion. Which of the following pathologies is most likely to be found on the scan based on the clinical findings?
Your Answer:
Correct Answer: Disc herniation between C7 and T1
Explanation:The C8 nerve forms part of the radial and ulnar nerves via the brachial plexus, and therefore has motor and sensory function in the upper limb. It originates from the spinal column from below the cervical vertebra 7 (C7).
The C8 nerve receives sensory afferents from the C8 dermatome. This consists of all the skin on the little finger, and continuing up slightly past the wrist on the palmar and dorsal aspects of the hand and forearm.
The other options available correspond to the C6 or C7 roots and these are unaffected as evidenced by normal elbow flexion and thumb sensation (C6) and normal sensation over the middle finger (C7). Elbow extension is weak as it has roots from both C7 and C8 and so cannot be used alone to decide between the two levels clinically.
The C8 nerve contributes to the motor innervation of many of the muscles in the trunk and upper limb. Its primary function is the flexion of the fingers, and this is used as the clinical test for C8 integrity, in conjunction with the finger jerk reflex.Trunk:
– Pectoralis major – Medial and lateral pectoral nerves (C5, C6, C7, C8, T1)
– Pectoralis minor – Medial pectoral nerve (C5, C6, C7,C8, T1)
– Latissimus dorsi – Thoracodorsal nerve (C6, C7, C8)
Upper arm:
– Triceps brachii – Radial nerve (C6, C7,C8)
Forearm
– Flexor carpi ulnaris – Ulnar nerve (C7, C8, T1)
– Palmaris longus – Median nerve (C7,C8)
– Flexor digitorum superficialis – Median nerve (C8, T1)
– Flexor digitorum profundus – Median and Ulnar nerves (C8, T1)
– Flexor pollicis longus – Median nerve (C7,C8)
– Pronator quadratus – Median nerve (C7,C8)
– Extensor carpi radialis brevis – Deep branch of the radial nerve (C7,C8)
– Extensor digitorum – Posterior interosseous nerve (C7,C8)
– Extensor digiti minimi – Posterior interosseous nerve (C7,C8)
– Extensor carpi ulnaris – Posterior interosseous nerve (C7,C8)
– Anconeus – Radial nerve (C6, C7,C8)
– Abductor pollicis longus – Posterior interosseous nerve (C7,C8)
– Extensor pollicis brevis – Posterior interosseous nerve (C7,C8)
– Extensor pollicis longus – Posterior interosseous nerve (C7,C8)
– Extensor indicis – Posterior interosseous nerve (C7,C8)
Hand
– Palmaris brevis – Superficial branch of ulnar nerve (C8, T1)
– Dorsal interossei – Deep branch of ulnar nerve (C8, T1)
– Palmar interossei – Deep branch of ulnar nerve (C8, T1)
– Adductor pollicis – Deep branch of ulnar nerve (C8, T1)
– Lumbricals – Deep branch of ulnar, Digital branches of median nerve
– Opponens pollicis – Recurrent branch of median nerve (C8, T1)
– Abductor pollicis brevis – Recurrent branch of median nerve (C8, T1)
– Flexor pollicis brevis – Recurrent branch of median nerve (C8, T1)
– Opponens digiti minimi – Deep branch of ulnar nerve (C8, T1)
– Abductor digiti minimi – Deep branch of ulnar nerve (C8, T1)
– Flexor digiti minimi brevis – Deep branch of ulnar nerve (C8, T1) -
This question is part of the following fields:
- Clinical Sciences
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Question 23
Incorrect
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Each one of the following statements regarding atrial natriuretic peptide are true, except:
Your Answer:
Correct Answer: Secreted mainly by the left atrium
Explanation:Atrial natriuretic peptide (ANP) is a 28-amino acid peptide that is synthesized, stored, and released by atrial myocytes in response to atrial distension, angiotensin II stimulation, endothelin, and sympathetic stimulation (beta-adrenoceptor mediated). ANP is synthesized and secreted by cardiac muscle cells in the walls of the atria in the heart. The main physiological actions of natriuretic peptides is to reduce arterial pressure by decreasing blood volume and systemic vascular resistance. It causes a reduction in expanded extracellular fluid (ECF) volume by increasing renal sodium excretion.
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This question is part of the following fields:
- Clinical Sciences
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Question 24
Incorrect
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Which of following does not promote the release of endothelin?
Your Answer:
Correct Answer: Prostacyclin
Explanation:Prostacyclin (PGI2) generated by the vascular wall is a potent vasodilator, and the most potent endogenous inhibitor of platelet aggregation so far discovered. Prostacyclin inhibits platelet aggregation by increasing cyclic AMP levels. Prostacyclin is a circulating hormone continually released by the lungs into the arterial circulation. Circulating platelets are, therefore, subjected constantly to prostacyclin stimulation and it is via this mechanism that platelet aggregability in vivo is controlled.
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This question is part of the following fields:
- Clinical Sciences
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Question 25
Incorrect
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A case-control study is being designed to look at the relationship between eczema and a new vaccine for yellow fever. What is the usual outcome measure in a case-control study?
Your Answer:
Correct Answer: Odds ratio
Explanation:A case–control study (also known as case–referent study) is a type of observational study in which two existing groups differing in outcome are identified and compared on the basis of some supposed causal attribute. Case–control studies are often used to identify factors that may contribute to a medical condition by comparing subjects who have that condition/disease (the cases) with patients who do not have the condition/disease but are otherwise similar (the controls).
An odds ratio (OR) is a statistic that quantifies the strength of the association between two events, A and B. The odds ratio is defined as the ratio of the odds of A in the presence of B and the odds of A in the absence of B or vice versa. -
This question is part of the following fields:
- Clinical Sciences
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Question 26
Incorrect
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Choose the correct definition regarding the standard error of the mean:
Your Answer:
Correct Answer: Standard deviation / square root (number of patients)
Explanation:The SEM is an indicator of how close the sample mean is to the population mean. In reality, however, only one sample is extracted from the population. Therefore, the SEM is estimated using the standard deviation (SD) and a sample size (Estimated SEM). The SEM computed by a statistical program is an estimated value calculated via this process.
Estimated Standard Error of the Mean (SEM)=SDn√
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This question is part of the following fields:
- Clinical Sciences
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Question 27
Incorrect
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Which one of the following statements regarding epidemiological measures is correct?
Your Answer:
Correct Answer: Cross-sectional surveys can be used to estimate the prevalence of a condition in the population
Explanation:The incidence rate is the number of new cases per population at risk in a given time period. For example, if a population initially contains 1,000 non-diseased persons and 28 develop a condition over two years of observation, the incidence proportion is 28 cases per 1,000 persons per two years, i.e. 2.8% per two years.
Prevalence is the proportion of a particular population found to be affected by a medical condition (typically a disease or a risk factor such as smoking or seat-belt use). It is derived by comparing the number of people found to have the condition with the total number of people studied, and is usually expressed as a fraction, as a percentage, or as the number of cases per 10,000 or 100,000 people.
Incidence should not be confused with prevalence, which is the proportion of cases in the population at a given time rather than rate of occurrence of new cases. Thus, incidence conveys information about the risk of contracting the disease, whereas prevalence indicates how widespread the disease is. -
This question is part of the following fields:
- Clinical Sciences
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Question 28
Incorrect
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Which one of the following statements regarding nitric oxide is incorrect?
Your Answer:
Correct Answer: Promotes platelet aggregation
Explanation:Nitric oxide, known as an endothelium-derived relaxing factor (EDRF), is biosynthesized endogenously from L-arginine, oxygen, and NADPH by various nitric oxide synthase (NOS) enzymes and is a signaling molecule in many physiological and pathological processes in humans.
One of the main enzymatic targets of nitric oxide is guanylyl cyclase. The binding of nitric oxide to the haem region of the enzyme leads to activation, in the presence of iron. -
This question is part of the following fields:
- Clinical Sciences
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Question 29
Incorrect
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Choose the correct stage in the cell cycle that vincristine acts on:
Your Answer:
Correct Answer: M
Explanation:Vincristine is part of the antimitotic agents, cell cycle specific (M phase). It binds to microtubules in the spindle apparatus and prevents their proper function, finally arresting mitosis.
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This question is part of the following fields:
- Clinical Sciences
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Question 30
Incorrect
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A 22-year-old man is investigated for visual loss and diagnosed with Leber's optic atrophy. Given the mitochondrial inheritance of this condition, which one of the following relatives is most likely to be also affected?
Your Answer:
Correct Answer: Sister
Explanation:The human cell has two type of DNA: Nuclear DNA and Mitochondrial DNA (MtDNA). A MtDNA copy is passed down entirely unchanged, through the maternal line. Males cannot pass their MtDNA to their offspring although they inherit a copy of it from their mother. Mitochondrial inheritance therefore has the following characteristics:
– Inheritance is only via the maternal line
– All children of affected males will not inherit the disease
– All children of affected females will inherit it -
This question is part of the following fields:
- Clinical Sciences
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