Free MRCOG Part 1 practice questions

Fourteen single best answer questions, one from each MRCOG Part 1 subject, taken from the Uterio question bank. Commit to an answer, then open the explanation. Each one is written the way the exam writes them: one stem, five options, one best answer.

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1. Question 1

   Anatomy guide

   What is the most constricted part of the fallopian tube?

   1. AWidest segment
   2. BIsthmus
   3. CFunnel-shaped opening
   4. DUterine portion
   5. EFimbrial extremity

   Show answer and explanation

   Correct answer: B. Isthmus

   Key takeaway

   The isthmus is the narrowest segment of the fallopian tube, located between the ampulla and the interstitial portion.

   The fallopian tubes have distinct anatomical segments that play different roles in fertilisation and the transport of ova. The isthmus is recognised as the narrowest segment of the fallopian tube. It is located between the broader ampulla and the interstitial portion that enters the uterine wall, making it crucial for sperm passage.

   The ampulla is the widest section and is typically the site of fertilisation. The infundibulum is the funnel-shaped segment with fimbriae important for capturing the ovum. The interstitial portion traverses the muscular wall of the uterus, while the fimbrial end refers to the fringe-like projections that help guide the oocyte into the fallopian tube. These regions, though vital for reproductive function, are not characterised by the narrowness defining the isthmus.

2. Question 2

   Embryology guide

   The facial nerve corresponds to which embryological pharyngeal arch during development?

   1. AFirst
   2. BSecond
   3. CThird
   4. DFourth
   5. EFifth

   Show answer and explanation

   Correct answer: B. Second

   Key takeaway

   The facial nerve, responsible for innervating muscles of facial expression, develops from the 2nd pharyngeal arch, which is also known as hyoid arch.

   The pharyngeal (or branchial) arches are embryonic structures that give rise to important structures of the head and neck.

   The facial nerve (cranial nerve VII) develops from the 2nd pharyngeal arch, also known as the hyoid arch. This arch contributes to the formation of the muscles of facial expression, the stapes bone, part of the hyoid bone, and the stylohyoid ligament, all of which are innervated by the facial nerve. Thus, identifying the facial nerve prompts association with the second pharyngeal arch.

   The 1st pharyngeal arch is associated with the trigeminal nerve (cranial nerve V) and gives rise to structures including the muscles of mastication, maxilla, and mandible.

   The 3rd pharyngeal arch is linked to the glossopharyngeal nerve (cranial nerve IX) and contributes to the formation of part of the hyoid bone and the stylopharyngeus muscle.

   The 4th and 6th pharyngeal arches (the 5th regresses in humans) are associated with branches of the vagus nerve (cranial nerve X). The 4th arch gives rise to the superior laryngeal nerve and pharyngeal constrictor muscles, while the 6th arch forms the recurrent laryngeal nerve and intrinsic laryngeal muscles.

3. Question 3

   Genetics guide

   A baby is born with an omphalocele and polydactyly. Further investigation reveals holoprosencephaly and renal abnormalities. What is the most likely underlying genetic syndrome?

   1. ACri du chat syndrome
   2. BDiGeorge syndrome
   3. CDown syndrome
   4. DEdwards syndrome
   5. EPatau syndrome

   Show answer and explanation

   Correct answer: E. Patau syndrome

   Key takeaway

   The classic triad of omphalocele, polydactyly, and holoprosencephaly, especially in conjunction with renal abnormalities, is highly suggestive of Patau syndrome (trisomy 13).

   Patau syndrome, also known as trisomy 13, is a chromosomal condition associated with severe intellectual disability and physical abnormalities in multiple systems. It is one of the more common, severe chromosomal syndromes, characterised by a specific set of malformations.

   The correct answer is Patau syndrome. The combination of omphalocele (abdominal wall defect), polydactyly (extra digits), holoprosencephaly (failure of the forebrain to divide properly), and renal abnormalities is highly characteristic of Patau syndrome (trisomy 13).

   Edwards syndrome (trisomy 18) can present with some overlapping features, such as congenital heart defects and developmental delays, but the specific combination of omphalocele, polydactyly, and holoprosencephaly is much more typical of Patau syndrome. Edwards syndrome is typically associated with clenched fists with overlapping fingers and 'rocker-bottom' feet.

   Down syndrome (trisomy 21) has a different set of characteristic features, including hypotonia at birth, upslanting palpebral fissures, a single transverse palmar crease and less severe organ malformation compared to Edwards and Patau syndromes.

   Cri du chat syndrome is caused by a deletion on chromosome 5 and is characterised by a distinctive, high-pitched cat-like cry in infancy, along with microcephaly and intellectual disability.

   DiGeorge syndrome results from a microdeletion on chromosome 22 (22q11.2) and presents with a variable phenotype which can include cardiac defects, palatal abnormalities, immune deficiency, and hypocalcaemia.

4. Question 4

   Physiology guide

   Which characteristic most readily facilitates the passage of drugs across the placenta by simple diffusion?

   1. ANon-ionised state
   2. BIonised state
   3. CHigh molecular weight
   4. DStrong binding to plasma proteins
   5. EHigh water solubility

   Show answer and explanation

   Correct answer: A. Non-ionised state

   Key takeaway

   Simple diffusion across the lipid-rich placental barrier is favoured by lipid solubility; therefore, drugs in a non-ionised state cross most readily.

   The placenta acts as a lipid barrier between maternal and fetal circulations. Passage of substances across this barrier by simple diffusion is governed by Fick's law, which depends on factors like concentration gradient, surface area, barrier thickness, and the lipid solubility of the substance. Drugs that are lipid-soluble cross lipid membranes more easily.

   Most drugs are weak acids or weak bases, existing in both ionised and non-ionised forms. The non-ionised state is typically more lipid-soluble (lipophilic) than the ionised state. Therefore, drugs in a non-ionised state diffuse across the placental barrier more readily.

   Conversely, the ionised state is less lipid-soluble and crosses poorly. High molecular weight molecules (generally >500-1000 Da) diffuse slowly or negligibly. Drugs with strong binding to plasma proteins are largely confined to the maternal circulation, as only the unbound (free) fraction can cross. High water solubility usually correlates with low lipid solubility, hindering simple diffusion.

5. Question 5

   Endocrinology guide

   Where are prostaglandins primarily deactivated in the body?

   1. AKidneys
   2. BLiver
   3. CLungs
   4. DSpleen
   5. EAdrenal Glands

   Show answer and explanation

   Correct answer: C. Lungs

   Key takeaway

   Prostaglandins are primarily deactivated in the lungs due to the high concentration of prostaglandin-degrading enzymes in pulmonary endothelial cells, preventing systemic effects of these locally acting mediators.

   Prostaglandins are potent mediators involved in various physiological and pathological processes, including inflammation, pain, and fever. Their systemic effects are short-lived because they are rapidly metabolised.

   The lungs play the primary role in deactivating prostaglandins. A single pass through the pulmonary circulation results in the inactivation of a significant proportion (up to 95%) of circulating prostaglandins. This rapid inactivation occurs due to the high concentration of prostaglandin-degrading enzymes, particularly 15-hydroxyprostaglandin dehydrogenase, within the pulmonary endothelial cells. This prevents system-wide effects of locally produced prostaglandins.

   The liver and kidneys are involved in the metabolism and excretion of many substances, including prostaglandin metabolites, but they are not the primary sites of initial prostaglandin deactivation. The spleen and adrenal glands are not significantly involved in prostaglandin metabolism.

6. Question 6

   Biochemistry guide

   A patient undergoes a lengthy hysteroscopic fibroid resection, during which three bags of distension medium are used. What electrolyte imbalance is most likely to develop in this situation?

   1. AHyperkalemia
   2. BHypomagnesemia
   3. CHyponatremia
   4. DHypercalcemia
   5. EDecreased osmolality

   Show answer and explanation

   Correct answer: C. Hyponatremia

   Key takeaway

   Lengthy hysteroscopic procedures using electrolyte-free distension media pose a significant risk of hyponatraemia due to fluid absorption and dilution of serum sodium.

   Hysteroscopic procedures, like fibroid resection, require the use of distension media to expand the uterine cavity for better visualisation. Depending on the type of energy use (monopolar electrosurgery vs bipolar), different fluids are used as distention medium. If monopolar energy is used, a large volume of electrolyte-free distension fluid (e.g., glycine, sorbitol) can lead to significant fluid absorption into the systemic circulation.

   This, hyponatraemia, occurs due to the absorption of a large volume of hypotonic, electrolyte-free fluid used during the procedure. This dilutes the serum sodium concentration, leading to a potential hyponatraemic state. It is very important to select correct fluid and meticulously monitor input and output.

   The other electrolyte disturbances are not commonly or primarily associated with hysteroscopic procedures, as these procedures require fluid that lacks electrolytes.

   Hyperkalaemia would be more related to massive tissue trauma or renal failure.

   Hypomagnesaemia, while possible, isn't a primary concern.

   Hypercalcaemia is associated with conditions like hyperparathyroidism or malignancy, not hysteroscopic procedures.

   Decreased osmolality can be mentioned in this context, as this also occurs due to absorption of hypotonic fluid, but the electrolyte disturbance (sodium) is the key issue to be aware of.

7. Question 7

   Biophysics guide

   What is the standard (SI) unit used to quantify magnetic flux density in MRI?

   1. ATesla
   2. BHertz
   3. CVolt
   4. DPascal
   5. EOhm

   Show answer and explanation

   Correct answer: A. Tesla

   Key takeaway

   In MRI, the strength of the magnetic field is crucial, and it's measured in Tesla (T), the SI unit for magnetic flux density.

   Magnetic resonance imaging (MRI) relies on strong magnetic fields to generate images. The strength, or density, of these magnetic fields is what we need to quantify.

   The Tesla (T) is the standard international (SI) unit of magnetic flux density. It is defined as one weber per square metre. In the context of MRI, it directly measures the strength of the main magnetic field produced by the scanner.

   Hertz (Hz) measures frequency, relevant to the radiofrequency pulses used in MRI, but not the magnetic field strength. Volts (V), Pascals (Pa), and Ohms (Ω) measure electrical potential, pressure, and electrical resistance, respectively, and are not directly related to magnetic flux density in MRI.

8. Question 8

   Statistics & Epidemiology guide

   A group of medical students achieved the following marks in their end-of-year examination: 10, 11, 11, 12, 14, 18, 20, 22, 25, 27. What is the interquartile range?

   1. A5
   2. B10
   3. C11
   4. D13
   5. E17

   Show answer and explanation

   Correct answer: B. 10

   Key takeaway

   The interquartile range (IQR) is calculated by subtracting the first quartile (Q1) from the third quartile (Q3). For the data {10, 11, 11, 12, 14, 18, 20, 22, 25, 27}, Q1=11 and Q3=22, yielding an IQR of 22 - 11 = 10.

   The interquartile range (IQR) is a measure of statistical dispersion, representing the range within which the middle 50% of the data lies. To calculate it, the data must first be ordered: 10, 11, 11, 12, 14, 18, 20, 22, 25, 27. There are 10 data points (n=10). The first quartile (Q1) is the value below which 25% of the data falls. It is the median of the lower half of the data (10, 11, 11, 12, 14). The median of these 5 values is the 3rd value, which is 11. So, Q1 = 11. The third quartile (Q3) is the value below which 75% of the data falls. It is the median of the upper half of the data (18, 20, 22, 25, 27). The median of these 5 values is the 3rd value, which is 22. So, Q3 = 22. The IQR is calculated as Q3 - Q1 = 22 - 11 = 10. Therefore, the correct answer is 10.

   The other options represent incorrect calculations or different statistical measures. 11 is the value of the first quartile (Q1). 5 might be derived from subtracting the minimum value (10) from the value 15 (midpoint of Q1 and Q2), which is not the IQR. 13 and 17 do not correspond to the correct IQR calculation for this dataset.

9. Question 9

   Data Interpretation guide

   A postmenopausal woman has an ovarian cyst assessed. Ultrasound findings include multilocular appearance and solid areas. Her serum CA-125 level is 100 IU/ml. What is her Risk of Malignancy Index (RMI)?

   1. A100
   2. B200
   3. C300
   4. D600
   5. E900

   Show answer and explanation

   Correct answer: E. 900

   Key takeaway

   The Risk of Malignancy Index (RMI) is calculated as U (Ultrasound score: 3 for ≥2 features) x M (Menopausal status: 3 for postmenopausal) x CA-125 level. Here, RMI = 3 x 3 x 100 = 900.

   The Risk of Malignancy Index (RMI) is used to estimate the risk of an ovarian cyst being malignant, particularly in postmenopausal women. It combines three factors: ultrasound findings (U), menopausal status (M), and serum CA-125 level. The formula is RMI = U x M x CA-125.

   First, determine the ultrasound score (U). One point is given for each of the following features: multilocular cyst, solid areas, evidence of metastases, ascites, and bilateral lesions. A score of 0 is given for 0 features, 1 for 1 feature, and 3 for 2 or more features. In this case, the cyst is multilocular and has solid areas (2 features), so U = 3.

   Next, determine the menopausal status score (M). Postmenopausal status scores 3 points, while premenopausal status scores 1 point. This patient is postmenopausal, so M = 3.

   Finally, use the serum CA-125 level in IU/ml, which is 100 IU/ml.

   Calculate the RMI: RMI = U x M x CA-125 = 3 x 3 x 100 = 900.

   An RMI score > 200-250 generally indicates a high risk of malignancy and warrants referral to a specialist gynaecological oncology centre.

10. Question 10

    Pharmacology guide

    What is the approximate plasma half-life of Oxytocin when administered via the intramuscular route?

    1. ALess than 30 seconds
    2. B1 – 6 minutes
    3. C8 – 12 minutes
    4. D15 – 18 minutes
    5. E20 – 25 minutes

    Show answer and explanation

    Correct answer: B. 1 – 6 minutes

    Key takeaway

    Oxytocin administered intramuscularly has a short plasma half-life, estimated to be around 1 – 6 minutes, reflecting its rapid degradation and clearance from circulation.

    Oxytocin has a very short plasma half-life, typically estimated to be between 1 – 6 minutes. This rapid clearance from the plasma is primarily due to enzymatic degradation in the liver and kidneys, as well as clearance by the target tissues. While the intravenous (IV) route leads to immediate peak plasma concentrations, the intramuscular (IM) route results in slightly delayed absorption but the plasma half-life itself remains similarly short once the drug enters circulation.

    The short half-life means that the direct plasma concentration of oxytocin decreases quickly, allowing for relatively rapid titration of its effects when given intravenously. However, the clinical effects on uterine contractions or milk ejection may persist longer than the plasma half-life suggests, due to receptor binding and downstream signalling pathways.

    The other options represent significantly longer half-lives than what is established for oxytocin. A half-life of less than 30 seconds is too short, while ranges like 8-12 minutes, 15-18 minutes, or 20-25 minutes are inaccurate for oxytocin's rapid plasma clearance.

11. Question 11

    Immunology guide

    Which leucocyte is the most abundant type found in adult blood, typically comprising 50-70% of the total count?

    1. ALymphocytes
    2. BMonocytes
    3. CEosinophils
    4. DBasophils
    5. ENeutrophils

    Show answer and explanation

    Correct answer: E. Neutrophils

    Key takeaway

    Neutrophils are the most abundant leucocytes in adult blood, typically making up 50-70% of the total white cell count, reflecting their key role in innate immunity against bacterial infections.

    Neutrophils are the most numerous type of leucocyte (white blood cell) in the peripheral blood of healthy adults, constituting approximately 50-70% of the total count. They are a type of granulocyte and phagocyte, playing a crucial role in the innate immune system, primarily by engulfing and destroying bacteria through phagocytosis and the release of antimicrobial substances.

    Lymphocytes (T cells, B cells, NK cells) are the second most abundant group (20-40%) and are central to the adaptive immune response. Monocytes (2-8%) circulate in the blood before migrating into tissues, where they differentiate into macrophages or dendritic cells, acting as phagocytes and antigen-presenting cells. Eosinophils (1-4%) are involved in combating parasitic infections and modulating allergic inflammatory responses. Basophils (<1%) are the least common granulocytes and release histamine and other mediators during allergic reactions and inflammation.

12. Question 12

    Microbiology guide

    Which condition is LEAST likely to cause a false-positive result in serological tests for syphilis?

    1. AUse of amiodarone
    2. BActive autoimmune hepatitis
    3. CInfectious mononucleosis
    4. DLeprosy
    5. EIntravenous drug use

    Show answer and explanation

    Correct answer: A. Use of amiodarone

    Key takeaway

    Non-treponemal syphilis tests (VDRL/RPR) can show false positives due to various conditions causing tissue damage or immune stimulation, but use of amiodarone is not typically associated with this phenomenon.

    False-positive results in non-treponemal serological tests for syphilis (like VDRL or RPR) occur when the tests detect antibodies against cardiolipin antigens released from damaged host tissues, rather than specific anti-treponemal antibodies. These are known as biological false positives (BFPs). Several conditions can cause BFPs, including infections like infectious mononucleosis and leprosy, autoimmune conditions like SLE and active autoimmune hepatitis, pregnancy, and intravenous drug use. Use of amiodarone is not commonly cited as a cause of false-positive syphilis serology.

    Therefore, while various inflammatory, infectious, and autoimmune conditions can lead to false-positive non-treponemal tests, requiring confirmation with specific treponemal tests (like TPPA or FTA-ABS), amiodarone use is the least likely among the options provided to cause such a result.

13. Question 13

    Pathology guide

    Which hormone is typically secreted in excess by granulosa cell tumours of the ovary?

    1. AOestrogen
    2. BProgesterone
    3. CTestosterone
    4. DHuman chorionic gonadotropin (hCG)
    5. EAlpha-fetoprotein (AFP)

    Show answer and explanation

    Correct answer: A. Oestrogen

    Key takeaway

    Granulosa cell tumours of the ovary are functional sex cord-stromal tumours that characteristically secrete excess oestrogen, leading to clinical signs of hyperoestrogenism.

    Granulosa cell tumours are a type of ovarian sex cord-stromal tumour derived from the granulosa cells of the ovarian follicle. These cells normally produce oestrogen under the stimulation of follicle-stimulating hormone (FSH). Granulosa cell tumours often retain this endocrine function, leading to autonomous and excessive oestrogen production. This hyperoestrogenism can cause various clinical manifestations depending on the patient's age, such as precocious puberty in children, menstrual irregularities (like postmenopausal bleeding or heavy menstrual bleeding) and endometrial hyperplasia or carcinoma in adults.

    While other ovarian tumours can produce hormones, excess oestrogen is the characteristic feature of granulosa cell tumours. Sertoli-Leydig cell tumours typically produce androgens like testosterone, causing virilisation. Significant progesterone production is less common in ovarian tumours but can occur in some stromal tumours or thecomas. Germ cell tumours, such as choriocarcinoma and yolk sac tumours, may produce human chorionic gonadotropin (hCG) and alpha-fetoprotein (AFP) respectively, which serve as important tumour markers for those specific malignancies.

14. Question 14

    Clinical Management guide

    Based on evidence summarised in guidelines such as RCOG Green-top Guideline No. 52, by approximately what percentage is the risk of PPH (blood loss ≥ 500 ml) reduced with active management compared to physiological management?

    1. A20%
    2. B40%
    3. C60%
    4. D80%
    5. E95%

    Show answer and explanation

    Correct answer: C. 60%

    Key takeaway

    Active management of the third stage of labour reduces the risk of primary postpartum haemorrhage (PPH ≥ 500 ml) by approximately 60% compared with physiological management.

    Active management of the third stage of labour involves prophylactic administration of a uterotonic drug (usually oxytocin), deferred cord clamping and cutting, and controlled cord traction to deliver the placenta. Evidence summarised in guidelines, such as the Royal College of Obstetricians and Gynaecologists (RCOG) Green-top Guideline No. 52 on Prevention and Management of Postpartum Haemorrhage, demonstrates that active management significantly reduces the risk of primary postpartum haemorrhage (PPH), defined as blood loss ≥ 500 ml. Compared to physiological management (expectant management without routine uterotonics), active management decreases the risk of PPH by approximately 60%.

    Physiological management carries a higher risk of PPH and severe PPH (≥ 1000 ml). The other percentages listed (20%, 40%, 80%, 95%) are inaccurate representations of the risk reduction cited in major clinical guidelines and meta-analyses comparing active versus physiological management. The substantial reduction supports the routine recommendation of active management for all women.