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پستهای کانال
The correct answer is: E. Müllerian agenesis (Mayer-Rokitansky-Küster-Hauser [MRKH] syndrome).
Why?
This patient has the classic features of MRKH syndrome:
✅ Primary amenorrhea (never had a menstrual period)
✅ Dyspareunia (painful intercourse)
✅ Normal secondary sexual characteristics (Tanner stage 5 breasts and pubic hair)
✅ Absent or very short vaginal canal on examination
In MRKH syndrome:
The Müllerian (paramesonephric) ducts fail to develop.
This results in:
Absent uterus
Absent cervix
Upper two-thirds of the vagina absent
The ovaries are normal (they develop from the genital ridge, not the Müllerian ducts), so estrogen production is normal, leading to normal breast development and pubic hair.
Key associations
46,XX karyotype
Normal female external genitalia
Normal ovarian function
Often associated with renal anomalies (e.g., unilateral renal agenesis) and skeletal abnormalities.
Why not the others?
A. Hyperprolactinemia
Causes amenorrhea but does not cause absence of the vagina or uterus.
Often associated with galactorrhea.
B. Exposure to DES in utero
Causes uterine abnormalities (e.g., T-shaped uterus) and infertility.
Does not cause complete absence of the vagina or uterus.
C. Turner syndrome (45,X)
Presents with primary amenorrhea due to ovarian failure.
Patients have poor secondary sexual development (minimal breast development), short stature, webbed neck, etc.
D. PCOS
Causes oligomenorrhea or secondary amenorrhea, not primary amenorrhea with an absent vaginal canal.
Patients have a normal uterus.
High-yield comparison
Feature
MRKH
Androgen Insensitivity Syndrome
Karyotype
46,XX
46,XY
Uterus
Absent
Absent
Vagina
Short/absent upper part
Short blind-ending
Ovaries/Testes
Normal ovaries
Undescended testes
Breasts
Normal
Normal
Pubic/Axillary hair
Normal
Sparse or absent
Exam pearl:
Primary amenorrhea + normal breasts + absent uterus/vagina = Think MRKH first (46,XX). If pubic hair is absent, think androgen insensitivity syndrome (46,XY) instead.
| 2 |  بدون متن... | 7 |
| 3 | The correct answer is: ✅ C. Interleukin-5 (IL-5)
Diagnosis
This patient most likely has neurocysticercosis caused by Taenia solium.
Clues:
Recent travel to rural Mexico
Ate undercooked pork (classically associated in exam questions)
Multiple cystic brain lesions on CT
Seizures, headache, altered mental status
CSF eosinophilia (15%) → suggests a helminthic (parasitic) infection
Why IL-5?
Helminth infections stimulate a Th2 immune response.
The key Th2 cytokines are:
IL-4 → IgE class switching
IL-5 → Eosinophil growth, activation, and recruitment
IL-13 → Mucus production and alternative macrophage activation
Thus, IL-5 is the cytokine most directly responsible for the eosinophilic response in the CSF.
Why the other options are incorrect
A. Interferon-γ (IFN-γ) → Activates macrophages; important in Th1 responses against intracellular pathogens (e.g., TB).
B. TNF-α → Mediates inflammation, fever, and septic shock; not specific for eosinophils.
D. IL-2 → Promotes T-cell proliferation.
E. IL-12 → Drives differentiation into Th1 cells and stimulates IFN-γ production.
NEET PG / INI-CET High-Yield Table
Cytokine
Main Function
IL-4
IgE class switching, Th2 differentiation
IL-5
Eosinophil activation and growth ✅
IL-13
Mucus secretion, airway hyperreactivity
IL-2
T-cell proliferation
IL-12
Th1 differentiation
IFN-γ
Macrophage activation
TNF-α
Acute inflammation, fever, septic shock
Exam Pearl
Helminths → Th2 response → IL-5 → Eosinophilia.
Answer: ✅ C. Interleukin-5. | 12 |
| 4 |  بدون متن... | 12 |
| 5 |  بدون متن... | 21 |
| 6 |  بدون متن... | 18 |
| 7 | The correct answer is C. Togavirus with envelope containing E1 and E2 glycoproteins ✅
Why?
Eastern equine encephalitis virus (EEEV) is:
Family: Togaviridae
Genus: Alphavirus
Genome: Single-stranded, positive-sense RNA (+ssRNA)
Capsid: Icosahedral
Envelope: Enveloped virus with E1 and E2 glycoproteins
Transmission: Mosquito vector (classically Culiseta melanura)
Reservoir: Birds
Humans: Dead-end hosts
Clinical: Severe encephalitis with high fever, headache, altered mental status, seizures, and focal neurologic deficits
Mnemonic:
🦟 Alphavirus = Togavirus → E1 + E2 envelope glycoproteins.
Answer: C. Togavirus with envelope containing E1 and E2 glycoproteins. | 16 |
| 8 | The correct answer is C. Sympathetic cholinergic fibers via muscarinic receptors with acetylcholine as the neurotransmitter.
🧠 Key concept: Sweat glands are the exception!
Most postganglionic sympathetic neurons release norepinephrine (NE), but eccrine sweat glands are innervated by sympathetic cholinergic fibers that release acetylcholine (ACh).
Pathway:
Hypothalamus → Sympathetic preganglionic neuron → ACh → Nicotinic receptor (Nn) in autonomic ganglion → Sympathetic postganglionic cholinergic neuron → ACh → Muscarinic receptor (M3) on eccrine sweat gland → Sweating
So:
Preganglionic: ACh → Nicotinic
Postganglionic to eccrine sweat gland: ACh → Muscarinic (M3)
Why the others are wrong?
A. Parasympathetic ❌ — Sweating is controlled by the sympathetic nervous system.
B. Sympathetic cholinergic + nicotinic + norepinephrine ❌ — Neurotransmitter is ACh, and the sweat gland receptor is muscarinic, not nicotinic.
C. Sympathetic cholinergic + muscarinic + ACh ✅ Correct
D. Sympathetic adrenergic + alpha + NE ❌ — This is the typical sympathetic pathway, but eccrine sweat glands are the classic exception.
E. Somatic motor ❌ — Sweating is an autonomic function, not somatic motor activity.
📌 High-yield mnemonic
"Sympathetic sweating is cholinergic."
Primary hyperhidrosis = excessive activity of sympathetic cholinergic innervation of eccrine sweat glands.
Exam trap: Don't confuse the nicotinic receptor in the autonomic ganglion with the muscarinic receptor on the sweat gland. | 22 |
| 9 |  بدون متن... | 17 |
| 10 |  بدون متن... | 26 |
| 11 | Correct answer: C. Eustachian tube (pharyngotympanic/auditory tube) ✅
Why?
The Eustachian tube connects the middle ear to the nasopharynx. It:
Equalizes air pressure between the middle ear and atmosphere.
Provides a route through which upper respiratory tract infections can spread to the middle ear, causing acute otitis media.
Key clue: A 4-year-old with recent URI + fever, ear pain, bulging and immobile tympanic membrane → Acute otitis media.
Quick anatomy:
Round window → allows fluid movement in cochlea.
Semicircular canals → balance and rotational acceleration.
Oval window → transmits vibrations from stapes to inner ear.
External auditory meatus → conducts sound to tympanic membrane.
Eustachian tube → middle ear ↔ nasopharynx; pressure equalization + infection pathway ⭐ | 24 |
| 12 | The correct answer is:
✅ D. KOH preparation
Diagnosis:
The patient has Tinea corporis (ringworm), suggested by:
Itchy rash
Erythematous, scaly plaque
Central clearing (classic "ring-shaped" lesion)
Wrestler (close skin-to-skin contact is a common risk factor; also called tinea corporis gladiatorum)
Best diagnostic test:
KOH (potassium hydroxide) preparation
Skin scrapings from the edge of the lesion are placed in KOH.
KOH dissolves keratin, allowing visualization of branching, septate fungal hyphae under the microscope.
It is the fastest and most commonly used bedside test.
Why not the others?
A. Culture on Sabouraud agar – Can grow dermatophytes but is not the initial test because it takes days to weeks.
B. Culture on Eaton agar – Used for Mycoplasma pneumoniae.
C. Culture on Thayer-Martin agar – Used for Neisseria gonorrhoeae and N. meningitidis.
E. Wood's lamp examination – Most dermatophytes causing tinea corporis do not fluoresce (only some species like Microsporum do), so it is less useful.
High-yield point for exams:
Annular, scaly lesion with central clearing → Think Tinea corporis → Confirm with KOH preparation showing septate branching hyphae.
✅ Correct answer: D. KOH preparation | 27 |
| 13 |  بدون متن... | 22 |
| 14 | The correct answer is: C. Secondary active transporters fail to completely reabsorb glucose in the renal tubules. ✅
Why?
This patient has poorly controlled type 2 diabetes mellitus, evidenced by:
Polyuria and polydipsia
HbA1c = 9%
Obesity
Diabetic retinopathy (microaneurysms)
Peripheral neuropathy (loss of vibration sense)
Proteinuria (early diabetic nephropathy)
Glycosuria without ketones
Mechanism of glycosuria
Normally:
Glucose is freely filtered at the glomerulus.
Nearly 100% is reabsorbed in the proximal convoluted tubule by SGLT2 (and SGLT1).
These are secondary active transporters that use the sodium gradient created by the Na⁺/K⁺-ATPase (primary active transport).
In uncontrolled diabetes:
Blood glucose exceeds the renal threshold (~180 mg/dL).
The transport maximum (Tm) of SGLT transporters is exceeded.
The transporters are normal, but they become saturated.
Excess glucose remains in the urine → glycosuria.
Thus, secondary active transporters cannot completely reabsorb the filtered glucose, making Option C correct.
Why the other options are wrong
A. ❌ Yeast does not produce urinary glucose. It grows because glucose is already present in the urine.
B. ❌ Primary active transport (Na⁺/K⁺-ATPase) is not disrupted. The sodium gradient is intact.
D. ❌ Glucose reabsorption is not passive; it occurs via secondary active transport.
E. ❌ Increased GFR alone does not cause glycosuria. Glycosuria occurs when the filtered glucose load exceeds the transport maximum of SGLT transporters.
High-yield exam point (MBBS/NEET PG)
Glucose filtration: Passive at the glomerulus.
Glucose reabsorption: Secondary active transport via SGLT2 (early PCT) and SGLT1 (late PCT).
Energy source: Na⁺ gradient maintained by Na⁺/K⁺-ATPase (primary active transport).
Renal threshold for glucose: ~180 mg/dL.
Transport maximum (Tm): ~375 mg/min.
Answer: ✅ C. Secondary active transporters fail to completely reabsorb glucose in the renal tubules (because they become saturated). | 26 |
| 15 | A 61-year-old man presents to the clinic with complaints of excessive thirst, frequent urination, and partial vision loss in both eyes. His family history is significant for type 2 diabetes mellitus in his mother and cousin. His weight is 112 kg (246.9 lb), height is 187 cm (6 ft 1 in), blood pressure is 150/90 mm Hg, pulse is 89/min, respiratory rate is 14/min, and temperature is 36.7°C (98.4°F). The physical examination is significant for dry skin, a pustular rash over the shoulders and back, an accentuated second heart sound (S2) best heard in the second intercostal space at the right sternal border, and distal loss of vibration sensitivity in both feet. A fundoscopic examination shows small red dots in the superficial retinal layers suggestive of microaneurysms. His HbA1c is 9% and his urinalysis shows the following:
Color Pale yellow (light/pale-to-dark/deep amber)
Clarity Cloudy
pH 6.6
Specific gravity 1.010
Glucose 199 mg/dL
Ketones None
Nitrites Negative
Leukocyte esterase Negative
Bilirubin Negative
Urinary bilirubin Traces
Red blood cells 3 RBCs
Protein 120 mg/d
RBCs ≤ 2/hpf
WBCs 22/hpf
Epithelial cells 27/hpf
Casts 5/lpf
Crystals Occasional
Bacteria None
Yeast Present
Which of the following statements best describes the cause of this man’s glycosuria?
A.
Yeast converts glucogenic urinary amino acids into glucose
7%
B.
There is a disruption of the primary active transport of glucose in the proximal renal tubules
13%
C.
Secondary active transporters fail to completely reabsorb glucose in the renal tubules
64%
D.
There is a disruption of the passive transport of the glucose in the proximal renal tubules
8%
E.
Glucosuria results from an increased glomerular filtration rate
8%
Result
Incorrect | 21 |
| 16 | The correct answer is: E. Oxygen free radical formation causing direct cellular damage and inflammatory response in lung tissue.
Diagnosis: Bronchopulmonary Dysplasia (BPD)
This 28-week premature infant has classic features of bronchopulmonary dysplasia, a chronic lung disease of prematurity.
Clues in the question:
Premature infant (28 weeks)
Severe RDS initially treated with surfactant
Mechanical ventilation + 80% oxygen for 10 days
Worsening respiratory status after initial improvement
Chest X-ray: "Bubbly" appearance (alternating hyperinflation and atelectasis)
Histology:
Thickened alveolar walls
Hyaline membranes
Fibroblast proliferation (fibrosis)
Type I pneumocyte injury
Type II pneumocyte hyperplasia
These findings are characteristic of BPD.
Pathogenesis
The major mechanism is:
Prolonged exposure to high concentrations of oxygen (hyperoxia) and positive-pressure ventilation
→ Generation of reactive oxygen species (oxygen free radicals)
→ Injury to alveolar epithelial and endothelial cells
→ Inflammatory cytokine release
→ Fibrosis, impaired alveolar development, and chronic lung disease
Therefore, Option E is correct.
Why the other options are wrong
A. Surfactant deficiency
Causes Neonatal Respiratory Distress Syndrome (NRDS) immediately after birth.
This infant initially had NRDS but later developed a different complication (BPD) due to oxygen therapy and ventilation.
B. Bacterial pneumonia
Would show neutrophilic infiltrates and infection, not the classic bubbly CXR and fibrosis.
C. Alpha-1 antitrypsin deficiency
Causes emphysema in adults, not neonatal chronic lung disease.
D. Persistent pulmonary hypertension
Causes hypoxemia due to fetal circulation but does not produce fibrosis or bubbly lung changes.
High-yield MBBS/NEET PG point
Disease
Cause
Histology
Neonatal RDS
Surfactant deficiency
Hyaline membranes
Bronchopulmonary dysplasia
Oxygen toxicity + mechanical ventilation → oxygen free radicals
Fibrosis, type II pneumocyte hyperplasia, impaired alveolar development
Exam pearl:
Prematurity + prolonged oxygen therapy/ventilation + bubbly chest X-ray = Bronchopulmonary Dysplasia (BPD) → Oxygen free radical injury (Option E). | 41 |
| 17 |  بدون متن... | 30 |
| 18 |  بدون متن... | 26 |
| 19 | The correct answer is:
D. Linear band of immunoglobulin G (IgG) on the epidermal side of the basement membrane
Diagnosis: Bullous Pemphigoid
Clues in the question:
35-year-old man (although bullous pemphigoid is more common in the elderly, the pathology is classic)
Large, tense blisters
Flexor surfaces and trunk
Subepidermal blisters
Eosinophil-rich infiltrate
These findings are characteristic of bullous pemphigoid.
Pathophysiology
Autoantibodies (IgG) are directed against hemidesmosomal proteins BP180 (BPAG2) and BP230 (BPAG1) in the basement membrane.
This causes separation of the epidermis from the dermis, producing subepidermal, tense bullae.
Direct immunofluorescence (DIF)
Shows a linear band of IgG and C3 along the basement membrane, specifically on the epidermal side of salt-split skin.
✅ Answer: D
Why the other options are wrong
A. Autoantibodies to desmoglein 1 → Pemphigus foliaceus (superficial intraepidermal blisters)
B. Autoantibodies to desmoglein 3 → Pemphigus vulgaris (flaccid bullae, mucosal involvement, suprabasal acantholysis)
C. Granular IgA deposits in dermal papillae → Dermatitis herpetiformis
E. Linear IgA in basement membrane → Linear IgA bullous dermatosis
NEET PG High-Yield Table
Disease
Antibody
Level of blister
DIF pattern
Bullous pemphigoid
BP180, BP230 (hemidesmosomes)
Subepidermal
Linear IgG & C3
Pemphigus vulgaris
Desmoglein 3 (±1)
Suprabasal
Fish-net IgG
Pemphigus foliaceus
Desmoglein 1
Subcorneal
Fish-net IgG
Dermatitis herpetiformis
IgA to transglutaminase
Subepidermal
Granular IgA
Linear IgA disease
IgA
Subepidermal
Linear IgA
NEET PG mnemonic:
PemphiGOID = GOes to basement membrane → Linear IgG.
PemphiGUS = Gives "fish-net" IgG between cells. | 36 |
| 20 |  بدون متن... | 25 |
