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13 Cards in this Set

  • Front
  • Back
A 29 year old female presents with a one year history of irregular periods, deteriorating hirsutism and weight
gain. Investigations reveal:
Serum testosterone 4 mmol/L (0.5-3)
Serum dehydroepiandosterone sulphate(DHEAS) 15 umol/L(0.3-9.3)
Which one of the following statements is most probable for this patient

1)Pituitary gonadotrophins are likely to become suppressed.
2)She is likely to develop acanthosis nigricans.
3)She has an increased risk of multiple pregnancies.
4)She is at increased risk of autoimmune disease.
5)She is at increased risk of ovarian carcinoma
Answer 2)

This patient has oligomennorhoea, weight gain and hirsutism. The investigations show a modest elevation of androgens
and support a diagnosis of polycystic ovarian syndrome. This condition is associated with insulin resistance and
acanthosis nigricans is a feature.
A letter published in a medical journal suggests that an established antidepressant may cause photosensitivity.
The manufacturer wishes to set up a study to determine rapidly and efficiently whether this is a true association.
Which one of the following techniques is most appropriate

1)case control study
2)dose ranging study
3)double blind, randomized, placebo controlled study
5)sequential trial
Answer 3)

The drug is an established one and the correct answer can be found by elimination. A "double-blind, randomized,
placebo controlled study" would be time consuming, expensive and unlikely to be powered enough to detect what may
be a rare toxic effect. Remember the drug is established so there have been many patients taking it already and only
lately a letter is published in a medical journal. A "meta-analysis" would look at combining previous randomized
controlled trials and there would have at least been some of the trials that looked at photosensitivity for it to be of any
use in this case; it therefore seems to be excluded by simple logic. A "dose ranging study" is really for another purpose
- to decide the correct dose in early clinical trials so is hardly going to be of any use here. A "sequential" trial would be
comparing one therapy to another sequentially (usually with wash out periods in between). Again there are unlikely to
be enough subjects in the trial for this small risk. This leaves the "case control study" which seems the logical choice.
This would look at cases of photosensitivity (perhaps in subjects taking any antidepressant medication) and compare
them to age matched (or other criteria matched) control subjects to see if they were more / less / equally likely to be on
the antidepressant in question. This is by far the most rapid (since adverse drug reactions will have already been
collected) and efficient means of answering the question.
A 55 year old male presents with anorexia and weight loss of 12 months duration. Over this year he has had two
deep vein thromboses and had the last whilst his INR was 2. He remains on long-term warfarin therapy with an
INR above 2.6. Examination reveals that he is pigmented and has a postural drop in his blood pressure of 15
Investigations are as follows:
sodium concentration 131 mmol/l
potassium 5.0 mmol/l
INR 3.0
A Short synacthen test reveals a baseline cortisol concentration at time 0 of 120 nmol/l which rises to 155 nmol/l
after 30 minutes (Normal response >550 nmol/l).
Which single diagnosis would explain this patient's illness

1)Addison's disease
2)Anti-phospholipid syndrome
3)Autoimmune Polyendocrine Syndrome (Schmidt's disease)
4)Protein S deficiency
5)Pituitary infarction
Answer 2)

With a history of recurrent DVT and confirmed hypoadrenalism this patient is likely to have the antiphospholipid
syndrome. Antiphospholipid syndrome is a primary diagnosis or may co-exist with SLE. Anti-Cardiolipin antibodies or
Lupus anticoagulant may be present. It is associated with arterial and venousthrombosis and has a predilection for the
adrenal veins causing adrenal infarction with consequent hypoadrenalism. Addison's disease is an autoimmune
phenomenon and is not associated with DVT. The pigmentation (due to increased ACTH in hypoadrenalism) would
exclude pituitary infarction as the cause of the hypoadrenalism. Hypoadrenalism is not associated with protein S
deficiency. Autoimmune Polyendocrine syndrome is associated with hypothyroidism, type 1 diabetes, addison's
A new antihypertensive drug needs to be investigated to establish its relative potency.
Which of the following techniques is most appropriate for this purpose

2)case-control study
3)double-blind, randomized, placebo controlled study
4)postmarketing surveillance
5)sequential trial
Answer 1)

Biological assays are designed to measure the relative potency of different preparations. Blood pressure is highly
variable and is subject to variability because of the patient's level of anxiety and the method used by the observer to
measure it. In a test of EFFICACY of an antihypertensive drug, a double-blind, randomized design would be 0
favourable. A sequential trial (a trial in which the data are analysed after each participant's results become available,
and the trial continues until a clear benefit is seen in one of the comparison groups) could also be used to assess
efficacy, but there would have to be a large expected difference from placebo
Which of the following conditions may be detectable by growth monitoring

4)XYY Syndrome
5)Insulin dependent diabetes mellitus
Answer 2)
Benefits of growth monitoring include:
Early detection of conditions such as:
* hypothyroidism.
MRCP Question Bank, 2003
* growth hormone insufficiency.
* syndromes: Turners, Russell-Silver, Noonan's, skeletal dysplasias.
* growth impairment e.g. coeliac disease, inflammatory bowel disease or chronic renal failure.
* intracranial tumours.
* short normal children.
* children with short stature.
* Health promotion: impaired growth may be associated with child abuse or neglect for example.
* Focus of interest for parents.

Public health aspects:
* secular trend of increasing growth.
* linking growth patterns in fetal life and early infancy with adult patterns of disease.
* link between height and social circumstances.
A clinical trial assessing a new lipid lowering therapy for stroke allocates 1000 patients to active treatment and
another 1000 patients to placebo. Results demonstrate that number needed to treat (NNT) is 20 for the
prevention of the primary end-point. Which of the following best describes the results
1)20 patients in the treatment group were protected from stroke.
2)20 extra patients in the placebo group had a stroke
3)For 1000 patients treated with active therapy, there would be 20 fewer strokes
4)For 1000 patients treated with active therapy, there would be 50 fewer strokes.
5)For every 1000 patients treated with active therapy there would be 100 fewer strokes

Answer 4)

This prevention study for stroke reveals that 20 patients need to be treated to prevent one event. Thus if you treat a
1000 patients then you will expect to have 50 fewer strokes.
A 55 year old man presents with ataxia and bilateral gynaecomastia. Which of the following is the most likely diagnosis

1)Kleinfelters Syndrome
2)Long term treatment with cyclophosphamide for Wegener's Granulomatosis
3)Long term treatment with oral steroids for chronic asthma
4)Bronchial Carcinoma
5) Hypereosinophilic Syndrome
Answer 4)
Gynaecomastia is a non metastatic paraneoplastic syndrome usually due to Squamous cell lung cancer. It can be painful
and may be associated with testicular atrophy. Ataxia can occur as a result of cerebellar degeneration associated with
the malignancy.
Which of the following stimulate the generation of cyclic AMP as the second messenger

1)Nitric Oxide
3)Tissue Necrosis Factor (TNF) alpha
4)Cholera toxin
5)Growth hormone
Answer 4)

Nitric oxide generates cGMP as the second message and rosiglitazone acts through agonism of PPAR gamma.
Calcitonin Cholera toxin binds to the Ganglioside receptors and causes excessive production of cAMP which leads to
the activation of luminal sodium pumps and the secretory diarrhoea.. GH like TNF alpha acts on the GH/cytokine
superfamily of receptor which function via the JAK-STAT pathway.
GH act through which second messenger
 Cytokine kinase
A 60-year-old woman diagnosed with giant cell arteritis was commenced on high dose prednisolone therapy.
What is the most appropriate treatment for the prevention of steroid-induced osteoporosis

1)Bisphosphonate therapy
2)Calcium and vitamin D
3)Hormone replacement therapy
5)Salmon Calcitonin
Answer 1)

The most appropriate therapy advocated by the National Osteoporosis Society for the prevention of steroid-induced
Osteoporosis would be bisphosphonate therapy such as Didronel or alendronate. These are the only class of drug shown
to offer osteoprotection with steroid therapy. Patients taking 7.5 mg or more of prednisolone daily for 3 months or
longer should be offered osteoprotection. HRT would not really be appropriate for this subject who is ten years past the
menopause and likely to be free of all menopausal symptoms.
Bone densitometry performed on a 48-year-old woman demonstrates bone mass decreased more than 2 standard
deviations below the mean for her age in her left femoral head, wrist, and lumbar vertebral region. Six months
later, the amount of bone loss is seen to be increased by repeat densitometry examination.
These findings are most likely to be associated with with which of the following serum laboratory test

1)Intact parathormone of 5 pmol/L (1.2 - 5.8)
2)Cortisol of 2060 mmol/L (110 - 607)
3)Total serum globulin of 35 g/L
4)Uric acid of 930 micromol/L (149 - 446)
5)Total cholesterol of 10 mmol/L (< 5.17)
Answer 2)

She has osteoporosis with decreased bone mass. Most cases do not have a specific etiology, but Cushing's syndrome
with hypercortisolism can promote osteoporosis. Her age should make you suspicious. Hypoparathyroidism is not
going to accelerate bone loss. The bone resorption that accompanies hyperparathyroidism can cause osteoporosis. Over
95% of cases of osteoporosis are 'primary' with unknown cause. Elevated serum globulin should make you suspect a
monoclonal gammopathy, but myeloma leads to focal bone lytic lesions. Hyperuricemia can be associated with gout
that can cause focal bone destrution near affected joints, the bone mass overall is not decreased.
Statistical independence may be assumed in which of the following circumstances

1)Successive measures taken on the same individual.
2)Stratified sampling from a target population.
3)Two matched individuals in a case control study.
4)Response to antibiotics of children with otitis media who have a CRP of above 100.
5)Diagnosis of pyloric stenosis by ultrasound scan in patients attending a tertiary referral centre.
Answer 3)

Independent events do not affect each other. Thus, the chance of event A occurring is completely unaffected by the
chance of B occurring. Two measures on the same individual are clearly dependent, but the same also applies to 2
matched individuals such as in a case control study. The outcome of patients attending a tertiary referral centre is bound
to depend on the patients referred, and the facilities and expertise available at that centre. In patients with elevated
CRP, one might suspect that there is a higher risk of bacterial otitis media, and, therefore, one might expect a greater
response to antibiotics. The response to antibiotics is, therefore, dependent to some extent on the raised CRP.
Which of the following statements is correct regarding standard error of the mean (SEM) and standard
deviation (SD)

1)Standard error of mean is calculated by taking the square root of the standard deviation of the sample means
2)Standard deviation invariably falls with increasing sample size
3)Standard error of mean increases with sample size
4)if standard deviation is greater than the mean the distribution is negative
5)Student's t test is a non-parametric test
Answer 1)

The Standard error of the Mean = SD/sq root n. SD does not necessarily fall with sample size as the distribution of
values may increase and hence SD increase. SEM would decrease with sample size as can be seen in the above
calculation. The SD would only be greater than the mean if the sample was Negatively Distributed - i.e. the data was
negative. This is not the same as 'negatively skewed' where the distribution of data about the mean tails off to the left
with the majority of points being greater (the median and the mode are greater than the mean).
This question is a trick ... and quite a trivial one at that. A negative distribution's mean would be, of course, negative
but the standard deviation would still be a positive number!
Student's T test is a parametric test comparing normally distributed data.