hllj/quesmed huggingface dataset:
A 32-year-old woman presents with 12 months of infertility despite regular unprotected intercourse. She reports regular periods, though increasingly heavy in recent years. She had normal childhood development and menarche at age 11, with up-to-date negative cervical smears for HPV, and has no past medical history. She is in a monogamous relationship with a negative sexually transmitted infection screen 6 months ago.


On examination, she has a firm, palpable painless 10cm smooth mass in the suprapubic region. Abdomen is otherwise soft non-tender with no shifting dullness or fluid thrill.


Routine bloods reveal microcytic iron deficiency anaemia but are otherwise unremarkable.


Urine dip and beta HCG are negative.


What is the most appropriate next investigation?

Options:
explanation: These would be reasonable suggestions to investigate subfertility/infertility if the patient did not present with an abdominal mass and ultrasound scans were unremarkable
name: Mid luteal phase progesterone
label: c
answer: no
explanation: MRI Pelvis provides excellent anatomic delineation of gynaecological structures. It is the most _accurate_ modality for detecting, localising, and characterising uterine fibroids. It is, however, not generally required for diagnosis and would not be the _first_ investigation of choice
name: MRI Pelvis
label: e
answer: no
explanation: This may help in confirming the aetiology of the anaemia, although the clinical information provides enough information to reasonably assume the anaemia was caused either by menorrhagia or from malignancy. The patient has presented with an abdominal mass which must be investigated first
name: Low vaginal swab for chlamydia and gonorrhoea
label: d
answer: no
explanation: This patient most likely has uterine fibroids which has resulted in her menorrhagia, subfertility, and anaemia. She has a smooth benign-feeling suprapubic mass; however this still needs to be formally assessed sonographically to be characterised and assessed for malignancy
name: Transvaginal ultrasound scan
label: a
answer: yes
explanation: This may help in confirming the aetiology of the anaemia, although the clinical information provides enough information to reasonably assume the anaemia was caused either by menorrhagia or from malignancy. The patient has presented with an abdominal mass which must be investigated first
name: Blood film
label: b
answer: no

Correct answer: a

Transvaginal ultrasound scan

Learning point: A transvaginal ultrasound scan is a diagnostic imaging technique used to visualize uterine fibroids, providing detailed images of their size, number, and location within the uterus, aiding in diagnosis and treatment planning.
A 6-year-old boy presents with his father to the emergency department with two days of progressive shortness of breath on exertion and new visible jaundice. He is normally fit and well and has reached all his developmental milestones. On questioning regarding the patient's diet, his father reports the child has been eating fava beans with every meal for the past week.


On examination, he appears well but is jaundiced. His chest is clear, and he has a smooth, palpable mass in the left upper quadrant; the abdomen is soft and non-tender.  

Blood tests:

|Serum|Result|Reference Range|
|-------------------|------|------------|
|Haemoglobin|68|130 - 170 mg/dL|
|White Cell Count|5.1|3.0 - 10.0 x 10⁹ /L|
|Platelets|162|150 - 400 x 10⁹/L|
|Reticulocytes|220|25 - 100 x 10⁹/L|
|Haptoglobin|19|50-220 mg/dL|
|Lactate Dehydrogenase|636|70 - 250 IU/L|
|Creatinine|95|60 - 120 µmol/L|
|Bilirubin|85|< 17 µmol/L|
|Alanine Aminotransferase|35|10 - 40 IU/L|
|Alkaline Phosphatase|89|25 - 115 IU/L|
|Gamma glutamyl transferase|21|9 - 40 U/L|


Peripheral blood film shows polychromasia and bite cells.


Which of the following is the next best investigation?

Options:
explanation: This patient likely has acute haemolytic anaemia from glucose-6-phosphate dehydrogenase deficiency following ingestion of fava beans. Direct antiglobulin test (direct Coombs test) is used to determine whether the patient has immune-mediated haemolytic anaemia
name: Direct antiglobulin test
label: a
answer: yes
explanation: This child is experiencing acute haemolytic anaemia, likely triggered ingestion of fava beans with underlying glucose-6-phosphate dehydrogenase deficiency. Bone marrow biopsy would not further aid in the diagnosis of the definitive cause of haemolytic anaemia
name: Bone marrow biopsy
label: c
answer: no
explanation: G6PD enzyme assay is used to diagnose glucose-6-phosphate dehydrogenase deficiency. It is checked approximately 3 months following an episode of acute haemolysis, and so it would not be the next investigation
name: G6PD enzyme assay
label: b
answer: no
explanation: G6PD enzyme assay is used to diagnose glucose-6-phosphate dehydrogenase deficiency. It is checked approximately 3 months following an episode of acute haemolysis, and so it would not be the next investigation
name: Screen for red blood cell pyruvate kinase activity
label: d
answer: no
explanation: Acute hepatitis would present with grossly deranged liver enzymes and raised bilirubin. A raised bilirubin with anaemia and normal liver enzymes is more typical for haemolytic anaemia
name: Hepatitis panel
label: e
answer: no

Correct answer: a

Direct antiglobulin test

Learning point: Direct antiglobulin test (direct Coombs test) is used to determine whether the patient has immune-mediated haemolytic anaemia
A 72-year-old man presents to the GP with 3 months of general fatigue, malaise, and night sweats.


Bloods reveal the following:


||||
|--------------|:-------:|---------------|
|Haemoglobin|78  g/L|(M) 130 - 170, (F) 115 - 155|
|White Cell Count|16.3x10<sup>9</sup>/L|3.0 - 10.0|
|Neutrophils|3.1x10<sup>9</sup>/L|2.0 - 7.5|
|Lymphocytes|8.2x10<sup>9</sup>/L|1.5 - 4.0|


Peripheral blood film shows smudge cells and small mature lymphocytes with round nuclei and clumped chromatin.


Given the likely diagnosis, which of the following is the next best investigation?

Options:
explanation: Serum electrophoresis is commonly used to investigate multiple myeloma or monoclonal gammopathy of uncertain significance (MGUS). While chronic lymphocytic leukaemia may produce IgM paraproteinaemia, it is not routinely used in its diagnosis
name: Serum electrophoresis
label: d
answer: no
explanation: This patient most likely has chronic lymphocytic leukaemia. Immunophenotyping detects tumour markers in peripheral blood by labelling the white blood cells with antibodies directed against cell surface proteins. This allows for differentiation of leukaemic cells to show which specific type of leukaemia the patient has and so which treatment would be most appropriate
name: Immunophenotyping
label: a
answer: yes
explanation: Chronic lymphocytic leukaemia can be subcategorised using immunophenotyping and flow cytometry from peripheral blood. Therefore, evaluation of bone marrow biopsy is not usually necessary unless there is diagnostic doubt from the former tests
name: Bone marrow biopsy
label: c
answer: no
explanation: Chronic lymphocytic leukaemia can be subcategorised using immunophenotyping and flow cytometry from peripheral blood. Evaluation of lymph node biopsy is not usually necessary unless there is diagnostic doubt from the former tests
name: Lymph node biopsy
label: b
answer: no
explanation: Iron studies are commonly used to investigate the cause of anaemia and will likely be done in this patient as they are anaemic, but it is not the next best investigation for their leukaemia
name: Iron studies
label: e
answer: no

Correct answer: a


Learning point: Immunophenotyping in chronic lymphocytic leukemia (CLL) is a diagnostic process that uses flow cytometry to identify specific cell surface markers, such as CD5, CD19, CD20, and CD23, to confirm the presence of clonal B-cell populations and distinguish CLL from other lymphoid disorders.
A 32-year-old woman has presented to the GP for advice about fertility. She has had four miscarriages prior to 20 weeks gestation, following which tests revealed no anatomical, hormonal, or chromosomal precipitating causes. She is otherwise well. Tests are requested to screen for thrombophilia with the results shown below:

| Serum                           | Result   | Result at 12 weeks |
| ------------------------------- | -------- | ------------------ |
| Lupus anticoagulant (LA)        | positive | positive           |
| Anticardiolipin (aCL) antibody  | positive | positive           |
| Anti-b2-glycoprotein I antibody | negative | negative           |

Which of the following is the most likely diagnosis?

Options:
explanation: While aCL antibodies can be found in Behçet's syndrome, there would likely also be a history of recurrent aphthous ulcers and uveitis. With positive LA antibodies, this points to antiphospholipid syndrome
name: Behçet's syndrome
label: b
answer: no
explanation: The name lupus anticoagulant is a misnomer and has only a weak association with SLE. It is a prothrombotic antibody where most patients with positive antibodies do not have SLE, and only a small proportion proceeds to develop the disease. The patient is otherwise well with no other systemic features and so makes the diagnosis of SLE unlikely
name: Systemic lupus erythematosus (SLE)
label: d
answer: no
explanation: Protein C deficiency is a rare cause of thrombosis, which can present with recurrent miscarriages. It is not, however, associated with LA, aCL, or anti-b2-glycoprotein I antibodies
name: Protein C deficiency
label: e
answer: no
explanation: All three of the antibodies above can be present in antiphospholipid syndrome, resulting in thrombophilia. This can often present as either unexplained recurrent miscarriages or vascular thrombosis
name: Antiphospholipid syndrome
label: a
answer: yes
explanation: Factor V Leiden is a common cause of thrombosis, which can present with recurrent miscarriages. It is not, however, associated with LA, aCL, or anti-b2-glycoprotein I antibodies
name: Factor V Leiden
label: c
answer: no

Correct answer: a


Learning point: Lupus anticoagulant, anticardiolipin antibody, and anti-β2-glycoprotein I antibody are autoantibodies linked to antiphospholipid syndrome, which increases the risk of blood clots and pregnancy complications, indicating a higher risk of thrombotic events.
A novel screening test has been developed to detect breast cancer early in high-risk individuals. It is tested on 1000 individuals with risk factors for breast cancer:

|                   | Breast cancer present | Breast cancer absent |
| ----------------- | --------------------- | -------------------- |
| **Test positive** | 280                   | 70                   |
| **Test negative** | 10                    | 640                  |

What is the specificity of the new test?

Options:
explanation: This value corresponds to the negative predictive value - the chance that the patient does not have the condition if the test is negative. This is calculated as the true negative/(true negative + false negative) = 640/(640+10) = 98.5%
name: 98.5%
label: d
answer: no
explanation: This value is the positive predictive value. Positive predictive value is the chance that the patient has the condition if the test is positive. This corresponds to true positive/(true positive + false positive) = 280/(280+70) = 80%
name: 80%
label: c
answer: no
explanation: The specificity is the proportion of patients without the condition with a negative test result. This is calculated as the true negative/(true negative + false positive) = 640/(640+70) = 90.1%
name: 69.6%
label: e
answer: no
explanation: Splitting the table up into true positive (280), false positive (70), false negative (10), and true negative (640) can make answering these tables easier to understand. The specificity is the proportion of patients without the condition with a negative test result. This is calculated as the true negative/(true negative + false positive) = 640/(640+70) = 90.1%
name: 90.1%
label: a
answer: yes
explanation: This value is the sensitivity. The sensitivity is the proportion of patients with the condition who have a positive test result. This is calculated as the true positive/(true positive + false negative) = 280/(280+10) = 96.6%
name: 96.6%
label: b
answer: no

Correct answer: a


Learning point: Specificity is a measure of a diagnostic test's ability to correctly identify those without the condition (true negatives), calculated as the ratio of true negatives to the total number of individuals without the condition: Specificity = (True Negatives / (True Negatives + False Positives)) × 100.
A new treatment has been developed for ulcerative colitis. In a large randomised controlled trial, 20% of participants in the placebo arm had an acute flare within the next year. In the treatment arm, 12% of participants developed an acute flare within the next year.

What is the relative risk reduction of the new treatment?

Options:
explanation: Relative risk reduction shows how much the treatment has reduced the risk of bad outcomes relative to the control. The risk reduction is (20-12%)/20% = 0.4 or 40%, i.e. a patient is 40% less likely to develop a colitic flair within the next year by taking the drug vs taking a placebo
name: 40%
label: a
answer: yes
explanation: If the treatment causes an increased risk of bad outcome, then it is termed the "relative risk increase". 66.7% would be the relative risk increase if the treatment arm resulted in 20% of colitic flares within 1 year and the placebo arm resulted in 12%
name: 66.7%
label: c
answer: no
explanation: 8% corresponds to the absolute relative risk, calculated as 20%-12%. This means that if you treated 100 people with this drug, 8 would be prevented from having a colitic flare in the next year. The number needed to treat (NNT) is calculated from this figure as 100/8 = 12.5, i.e. 12.5 people would need to be treated with this drug to prevent one episode
name: 8%
label: b
answer: no
explanation: 60% is the value of the relative risk (in percentage form). Relative risk is a proportional measure to determine the size of effect of a treatment compared to other interventions or a control. It is the proportion of bad outcomes in the intervention group divided by the proportion of bad outcomes in the control group. In this case 12%/20% = 0.6
name: 60%
label: d
answer: no
explanation: 140% relative risk reduction cannot exist as the maximum reduction of any bad outcome is 100%
name: 140%
label: e
answer: no

Correct answer: a


Learning point: Relative risk reduction (RRR) is a measure of how much the risk of an event is reduced in the treatment group compared to the control group, calculated as 1 minus the relative risk (RR) and expressed as a percentage: RRR = (1 - RR) × 100, where RR is the risk in the treatment group divided by the risk in the control group.
The following diagram is a forest plot from a systematic review by Crowley et al. showing the results of seven studies looking at the effect of corticosteroids on hastening neonatal lung development when given to women about to give birth prematurely.

[lightgallery]

Which of the following statements is correct?

Options:
explanation: The forest plot will include all studies investigated, regardless of their outcomes, analysed to determine the standardised mean difference, represented as the diamond in the bottom right of the image
name: Only the Auckland study showed significance, and so the other studies' results can be discounted
label: e
answer: no
explanation: The 95% confidence intervals, represented as the horizontal edges of the summary measure diamond, do not overlap with the odds ratio of 1 and therefore, there is a significant effect of corticosteroids on neonatal mortality (mean odds ratio 0.53)
name: There is no significant effect of corticosteroids on neonatal mortality
label: b
answer: no
explanation: The heterogeneity, or I squared, is often used in forest plots, but it is not shown in the data set here. Heterogeneity is the degree to which data from multiple studies displaying the same effect overlap each other. The more heterogeneous the studies are within the met-analysis, the less conclusive the data is
name: The heterogeneity of the study is 0.53
label: d
answer: no
explanation: The summary measure of the meta-analysis is portrayed as a diamond, with the vertical points lying on the mean effect and the horizontal points measuring the confidence intervals
name: The width of the diamond corresponds to the 95% confidence interval of the summary measure
label: a
answer: yes
explanation: The area of the squares correspond to the proportional study's weight in the meta-analysis and is unrelated to the p-value
name: The area of the square corresponds inversely to the study's p-value
label: c
answer: no

Correct answer: a


Learning point: The summary measure of the meta-analysis is portrayed as a diamond, with the vertical points lying on the mean effect and the horizontal points measuring the confidence intervals
A new blood test to help diagnose acute myocardial infarction (MI) has been developed. It is tested on 1000 individuals with suspected myocardial infarction in the emergency department :

|                   | MI present | MI absent |
| ----------------- | ---------- | --------- |
| **Test positive** | 600        | 60        |
| **Test negative** | 50         | 290       |

What is the positive predictive value of the new test?

Options:
explanation: Positive predictive value is the chance that the patient has the condition if the test is positive. This corresponds to true positive/(true positive + false positive) = 600/(600+60) = 90.9%
name: 67.4%
label: e
answer: no
explanation: This value corresponds to sensitivity - the proportion of patients with the condition who have a positive test result. This is calculated as the true positive/(true positive + false negative) = 600/(600+50) = 92.3%
name: 92.3%
label: b
answer: no
explanation: This value corresponds to the specificity - the proportion of patients without the condition with a negative test result. This is calculated as the true negative/(true negative + false positive) = 290 / (290+60) =82.9%
name: 82.9%
label: c
answer: no
explanation: Splitting the table up into true positive (600), false positive (60), false negative (50), and true negative (290) can make answering these tables easier to understand. Positive predictive value is the chance that the patient has the condition if the test is positive. This corresponds to true positive/(true positive + false positive) = 600/(600+60) = 90.9%
name: 90.9%
label: a
answer: yes
explanation: This value corresponds to the negative predictive value - the chance that the patient does not have the condition if the test is negative. This is calculated as the true negative/(true negative + false negative) = 290/(290+50) = 85.3%
name: 85.3%
label: d
answer: no

Correct answer: a


Learning point: Positive predictive value (PPV) is the probability that individuals with a positive test result truly have the condition, calculated as the ratio of true positives to the total number of positive test results (true positives + false positives).
A 23-year-old woman with antiphospholipid syndrome is 14 weeks pregnant and develops lower abdominal pain and subsequent vaginal bleeding. A transvaginal ultrasound scan confirms an inevitable miscarriage. She is known to be rhesus negative.

Which of the following is the best next step in the management of this patient with regards to anti-D?

Options:
explanation: An indirect Coombs test can be used to determine whether there are antibodies to the Rhesus factor in the mother's blood and therefore whether she has been "sensitised". A positive Coombs test indicates whether a Rh-positive foetus has the possibility of being harmed from rhesus D alloimmunisation. In this case, a sensitisation event has occurred in a miscarriage >12 weeks, and so anti-D must be given regardless
name: Perform indirect Coombs test before administration of anti-D
label: e
answer: no
explanation: A Kleihauer test measures the amount of foetal haemoglobin transferred from a foetus to a mother's bloodstream following exposure to a sensitising event from a RhD positive baby. It is conducted on all RhD negative women who have delivered a RhD positive baby, in pregnant women who have had a potentially sensitising event at >20 weeks gestation, and on all pregnant women who have had an invasive procedure. The test helps with dosing Anti-D immunoglobulin. In this case, however, a sensitisation event has occurred in a miscarriage >12 weeks, and so anti-D must be given regardless
name: Perform a Kleihauer test before administration of anti-D
label: d
answer: no
explanation: Anti-D is given regardless of the biological father's rhesus status
name: Anti-D only required if the biological father is rhesus positive
label: c
answer: no
explanation: In pregnancies between 12 and 20 weeks gestation, anti-D immunoglobulin should be given in any sensitising event such as miscarriage, even if managed medically. No tests are required prior to administration
name: Give anti-D immunoglobulin within 72hrs
label: a
answer: yes
explanation: Miscarriage in pregnancies >12 weeks require anti-D given to prevent rhesus D alloimmunisation and haemolytic disease of the newborn. It is also required in a number of conditions, including termination of pregnancy, ectopic pregnancy, or abdominal trauma
name: No need for anti-D immunoglobulin
label: b
answer: no

Correct answer: a


Learning point: Administer anti-D immunoglobulin within 72 hours after a miscarriage in Rh-negative women to prevent sensitisation.
A 29-year-old primigravida woman has delivered a healthy baby via vaginal delivery. The placenta was intact on delivery, and there were no initial complications. Unfortunately, the mother continues to bleed following delivery and has now lost 1.1L. Bimanual uterine compression is conducted and the mother is catheterised to empty the bladder. Despite this, she continues to haemorrhage and you begin transfusing blood.

Which of the following is the best next step in the management of this patient?

Options:
explanation: Following basic resuscitation measures, IV or IM ergometrine is given 2nd line for postpartum haemorrhage if IV syntocinon is unsuccessful or contraindicated
name: IV ergometrine
label: b
answer: no
explanation: Recombinant factor VIIa is increasingly used in the treatment and control of acute bleeding. While it can be a useful adjunct in achieving haemostasis, simple measures first with IV syntocinon must be given first
name: Give recombinant factor VIIa
label: c
answer: no
explanation: This woman is experiencing significant postpartum haemorrhage, most likely due to uterine atony. After standard resuscitation measures and bimanual uterine compression, the next step is to give IV syntocinon (oxytocin) to induce uterine contractions
name: IV syntocinon
label: a
answer: yes
explanation: Selective arterial embolisation is reserved for uncontrolled postpartum haemorrhage despite physical and pharmacological measures. Embolising the artery under radiological guidance reduces the blood supply to the uterus and, therefore capacity to bleed. Conservative measures such as IV syntocinon should be imitated first in this case
name: Selective arterial embolisation
label: e
answer: no
explanation: Misoprostol is a synthetic prostaglandin E1 analogue and induces uterine contractions to combat postpartum haemorrhage from uterine atony. NICE guidance currently opts for intravenous syntocinon first, although PR misoprostol may be helpful, particularly in low resource settings as it does not need refrigeration or infusion
name: PR misoprostol
label: d
answer: no

Correct answer: a


Learning point: IV syntocinon (oxytocin) is commonly used in the management of postpartum hemorrhage to stimulate uterine contractions, helping to reduce bleeding by promoting uterine tone.