Page 4 of 4
AN72.1-AN73.3 | Integumentary System — Gate Quiz
Click any question card to reveal the correct answer.
The epidermal layer that contains keratohyalin granules and lamellar bodies (lipid-rich structures that create the waterproof barrier) is the:
Correct! The stratum granulosum (granular layer) is characterised by two types of cytoplasmic granules: (1) keratohyalin granules (contain profilaggrin and loricrin — cross-links keratin filaments), and (2) lamellar bodies (Odland bodies) that exocytose lipids into the intercellular space, forming the waterproof barrier.
The waterproof barrier is formed at the interface of the granulosum and corneum. Lamellar body contents (ceramides, free fatty acids, cholesterol esters) create a "brick and mortar" structure with corneocytes as bricks and lipids as mortar. Disruption of this barrier occurs in eczema/atopic dermatitis (filaggrin mutations).
Incorrect. The stratum basale has stem cells; the spinosum has desmosomes and Langerhans cells; the corneum has dead, anucleate corneocytes. The granular layer (granulosum) contains keratohyalin granules and lamellar bodies.
Click to reveal answer
A histology slide of thick skin from the palm shows coiled secretory tubules deep in the dermis and extending into the hypodermis. The secretory cells are arranged in two layers: inner secretory cells and outer myoepithelial cells. The ducts open directly onto the skin surface. These structures are:
Correct! Eccrine (merocrine) sweat glands are coiled tubular glands present in all skin including the palm (thick skin, which lacks hair and sebaceous glands). They have inner secretory cells (clear and dark cell types) + outer myoepithelial cells, and their ducts spiral through the epidermis to open directly on the skin surface. They produce watery sweat for thermoregulation.
Eccrine glands: (1) All body surfaces including palm/sole; (2) Duct opens directly to surface; (3) Merocrine/eccrine secretion (exocytosis); (4) Thermoregulation (can produce up to 10 L/day in extreme heat). Absent in thick skin: hair follicles and sebaceous glands. Present in thick skin: eccrine sweat glands only.
Incorrect. Apocrine glands are larger, open into hair follicles (not directly to surface), and are confined to axilla/groin/areola — not found in palmar thick skin. Sebaceous glands have holocrine secretion, acinar structure, and are absent in thick skin. Mucous glands are not a skin appendage type.
Click to reveal answer
A 25-year-old man has a small pigmented mole (naevus) on his arm. Biopsy shows increased numbers of melanin-producing cells. Melanocytes, which produce melanin and transfer it to surrounding keratinocytes, are normally found in which epidermal layer?
Correct! Melanocytes are dendritic cells derived from neural crest cells that migrate to the epidermis. They reside in the stratum basale (one melanocyte per ~36 basal keratinocytes in light skin). Melanocytes synthesise melanin in melanosomes and transfer them via their dendrites to surrounding keratinocytes (1 melanocyte : ~36 keratinocytes = the epidermal melanin unit).
Cell types in epidermis by layer: Basale — keratinocyte stem cells + melanocytes + Merkel cells. Spinosum — keratinocytes + Langerhans cells (immune surveillance, dendritic cells). Granulosum — maturing keratinocytes. Corneum — dead corneocytes only.
Incorrect. Melanocytes are located in the basal layer (stratum basale), not in upper layers. Langerhans cells are in the stratum spinosum. No living cells in the stratum corneum (all anucleate).
Click to reveal answer
Acrocentric chromosomes have a centromere positioned very near one end, resulting in a very short p arm. Which of the following groups consists entirely of acrocentric chromosomes?
Correct! Chromosomes 13, 14, 15 (Group D) and 21, 22 (Group G) are all acrocentric. Their very short p arms bear satellites and nucleolar organiser regions (NOR) containing ribosomal RNA genes. These chromosomes are particularly important because their p arms can participate in Robertsonian translocations (e.g., the most common cause of familial Down syndrome involves a 14;21 Robertsonian translocation).
Denver classification: Acrocentric chromosomes (centromere near tip) = D group (13, 14, 15) + G group (21, 22). Also the Y chromosome is acrocentric. These 5 pairs + Y bear satellites with rRNA genes (NOR). Robertsonian translocations between these chromosomes (especially 14;21) cause familial Down syndrome.
Incorrect. Chromosomes 1, 3 are metacentric; 2 is submetacentric. Chromosomes 6–12 are submetacentric (Group C). Chromosomes 16 is metacentric; 17, 18 are submetacentric (Group E). Acrocentric = 13, 14, 15, 21, 22 (+ Y).
Click to reveal answer
In the preparation of a karyotype, cells are arrested at metaphase using colchicine. If colchicine were omitted from the protocol, what would happen to the chromosomes?
Correct! Without colchicine, cells would complete mitosis normally — chromosomes would separate and decondense into interphase nuclei where they cannot be individually visualised. Colchicine specifically poisons the spindle by binding tubulin dimers, freezing cells at metaphase (the only stage where individual chromosomes are condensed and countable).
Metaphase is chosen for karyotyping because chromosomes are maximally condensed (individual chromosomes visible), spindle is forming (arrested here by colchicine), and each chromosome consists of two sister chromatids joined at the centromere (giving the characteristic X-shape).
Incorrect. Colchicine does not affect staining properties, G2 arrest, or apoptosis — it specifically binds tubulin and blocks spindle polymerisation, trapping cells in metaphase.
Click to reveal answer
A 28-year-old woman from Puducherry is referred to a genetics clinic after two consecutive miscarriages and the birth of a child with Down syndrome. Her karyotype shows 45 chromosomes with a chromosome that appears as a large metacentric chromosome formed by fusion of the long arms of chromosomes 14 and 21. This finding represents:
Correct! A Robertsonian translocation between chromosomes 14 and 21 creates a large derivative chromosome (rob(14;21)) made by fusion of the long arms. The carrier has 45 chromosomes but is phenotypically normal (because no genetic material is lost — the short arms of 14 and 21 carry only rRNA genes which are present on other acrocentric chromosomes). However, at meiosis, the carrier can produce gametes with an extra copy of chromosome 21 → Down syndrome offspring. This is the molecular basis of familial (not sporadic) Down syndrome.
Translocation Down syndrome: 45 chromosomes (not 47). The extra chromosome 21 material is attached to chromosome 14. Carrier parent is phenotypically normal but has 25–30% risk of Down syndrome offspring. Critical for genetic counselling: recurrence risk is much higher than sporadic Trisomy 21 (maternal age-dependent, ~1%).
Incorrect. Standard Trisomy 21 (nondisjunction) shows 47 chromosomes with three distinct chromosomes 21. Mosaicism shows a mixture of 46 and 47-chromosome cells. Isochromosome 21 is a different structural abnormality.
Click to reveal answer
A patient with Klinefelter syndrome has a karyotype of 47,XXY. How many Barr bodies would be expected in a buccal smear from this patient?
Correct! The number of Barr bodies = total X chromosomes minus 1. A 47,XXY individual has two X chromosomes, so one X is inactivated → 1 Barr body. Normal females (46,XX) also have 1 Barr body. Normal males (46,XY) have 0 Barr bodies. Triple X females (47,XXX) have 2 Barr bodies.
Barr body formula: n(Barr bodies) = n(X chromosomes) − 1. Applications: Normal female = 1, Normal male = 0, Turner (45,X) = 0, Klinefelter (47,XXY) = 1, Triple X (47,XXX) = 2, XXXY = 2. Historically used in buccal smears for sex determination (now replaced by molecular methods).
Incorrect. Rule: Barr bodies = X chromosomes − 1. 47,XXY has 2 X chromosomes → 2 − 1 = 1 Barr body.
Click to reveal answer
According to Lyon's hypothesis, X-inactivation in female mammals is brought about by coating of the inactive X chromosome by a non-coding RNA. This RNA is produced by which gene?
Correct! The XIST (X-inactive specific transcript) gene is located on the X chromosome and produces a long non-coding RNA (lncRNA) that coats the X chromosome in cis (on the same chromosome). XIST RNA recruits polycomb proteins, leading to histone modifications (H3K27me3) and DNA methylation that silence the inactive X, converting it into the Barr body.
XIST gene: On the X chromosome, in the X-inactivation centre (XIC). Expressed exclusively from the inactive X. Produces a lncRNA that coats and silences the chromosome. The active X expresses TSIX (antisense to XIST), which represses XIST on the active X. XIST is one of the best-characterised non-coding RNA genes in mammalian biology.
Incorrect. SRY (Y chromosome) is the sex-determining gene. BRCA1 is a tumour suppressor on chromosome 17. MECP2 is an X-linked gene that reads methylated DNA; mutations cause Rett syndrome. XIST is the master regulator of X-inactivation.
Click to reveal answer
A 20-year-old woman is a known carrier of haemophilia A (X-linked recessive; factor VIII deficiency). Her mother is a carrier and her maternal uncle has haemophilia. She herself has normal bleeding times. Lyon's hypothesis explains her normal phenotype because:
Correct! As a carrier female (X^H X^h — one normal X, one haemophilia X), inactivation is random. Approximately 50% of her hepatocytes (which produce factor VIII) express the normal X and produce functional factor VIII. This is sufficient to maintain normal or near-normal clotting factor levels. Only if inactivation is heavily skewed toward the normal X being inactivated would she show symptoms.
Lyon's hypothesis explains carrier females: random inactivation creates a mosaic of cells — ~50% express normal gene, ~50% express mutant gene. For haemophilia (factor VIII produced by liver), ~50% of hepatocytes producing functional factor VIII is sufficient for normal haemostasis. This mosaic protection does not apply to all X-linked genes (e.g., Duchenne dystrophy carriers may have patchy muscle weakness).
Incorrect. Carrier females are XX (no Y). Both X chromosomes are present but one is inactivated per cell. The normal X is not necessarily inactivated. X-inactivation is random, not preferential for the mutant allele.
Click to reveal answer
A 50-year-old woman from Chennai presents with multiple flaccid blisters on the skin and oral mucosa. Skin biopsy with immunofluorescence shows IgG antibodies directed against desmogleins (desmosomal proteins) in the intercellular spaces of the epidermis. The blisters form by separation of cells in which epidermal layer?
Correct! Pemphigus vulgaris is caused by IgG autoantibodies against desmoglein 3 (and in severe cases, desmoglein 1). Desmosomes are most abundant in the stratum spinosum ("prickle cell" layer — the spines ARE desmosomes). Anti-desmosomal antibodies disrupt cell-cell adhesion in the spinous layer → acantholysis (loss of cell cohesion) → suprabasal intraepidermal blisters. The cleavage is within the epidermis (intraepidermal), above the basal layer.
Blistering disease levels: (1) Pemphigus vulgaris — suprabasal intraepidermal (spinous layer) — anti-desmoglein 3. (2) Pemphigus foliaceus — subcorneal (superficial) — anti-desmoglein 1. (3) Bullous pemphigoid — subepidermal (basement membrane) — anti-hemidesmosome (BP180, BP230). Blister level = clinical clue: flaccid (intraepidermal) vs tense (subepidermal).
Incorrect. Subepidermal blistering (at the basement membrane) occurs in bullous pemphigoid (anti-hemidesmosome antibodies). Stratum corneum separation occurs in superficial impetigo or pemphigus foliaceus (anti-desmoglein 1, superficial layer). Pemphigus vulgaris targets the spinous layer (desmoglein 3 concentrated there).
Click to reveal answer