MCQ #1 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following hormones in the body of an animal is lipid-based in nature?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Hormones are classified chemically into peptides/proteins, amino acid derivatives, and steroids. Steroid hormones are lipid-soluble derivatives of cholesterol synthesized primarily by the adrenal cortex and gonads.
Formula / Rule / Reaction:Qualitative concept / Steroid hormone classification.Solution:- Aldosterone is a mineralocorticoid synthesized from cholesterol by the zona glomerulosa of the adrenal cortex.
- Because of its lipophilic nature, aldosterone readily diffuses across target cell plasma membranes to bind intracellular mineralocorticoid receptors.
Why other options are incorrect:- Option A: Insulin is a globular peptide hormone composed of 51 amino acids organized in two disulfide-linked chains.
- Option B: Follicle-stimulating hormone (FSH) is a dimeric glycoprotein gonadotropin secreted by the anterior pituitary gland.
- Option C: Oxytocin is a peptide nonapeptide synthesized in the hypothalamus and stored in the posterior pituitary.
MCQ #2 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following base pairings maintains the constant diameter of the DNA double helix as per Watson and Crick's model?
B
Pyrimidine - Pyrimidine
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The B-DNA double helix maintains a constant uniform diameter of 2.0 nm because each base pair spans the interior of the helix through complementary pairing between a double-ring purine and a single-ring pyrimidine.
Formula / Rule / Reaction:$$\text{Diameter of B-DNA} \approx 2.0\text{ nm (20 \AA)}$$
Solution:- Purines (adenine and guanine) possess a bicyclic planar ring structure measuring approximately 1.2 nm in length.
- Pyrimidines (thymine and cytosine) possess a monocyclic planar ring structure measuring approximately 0.8 nm in length.
- Pairing a purine with a pyrimidine always yields a total distance of 2.0 nm between the two deoxyribose-phosphate backbones across the entire helix axis.
Why other options are incorrect:- Option A: Purine - purine pairs would measure approximately 2.4 nm, producing an outward bulge in the helix.
- Option B: Pyrimidine - pyrimidine pairs would measure approximately 1.6 nm, producing an inward collapse of the helix.
- Option D: Sugar-phosphate units make up the covalent outer backbone and do not participate in complementary cross-helix base pairs.
MCQ #3 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which muscle type is under conscious control and is multinucleated?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Skeletal muscle tissue consists of elongated, multinucleated cylindrical cells (myofibers) displaying cross-striations that contract voluntarily under the control of the somatic nervous system.
Formula / Rule / Reaction:Qualitative concept / Muscle tissue histology.Solution:- During embryonic development, multiple mononucleated myoblasts fuse end-to-end to form a syncytium, resulting in multinucleated skeletal muscle fibers with peripherally located nuclei.
- Skeletal muscles receive somatic motor innervation, allowing conscious (voluntary) regulation of contraction.
Why other options are incorrect:- Option A: Smooth muscle consists of non-striated, spindle-shaped cells containing a single central nucleus under involuntary autonomic control.
- Option C: Cardiac muscle fibers are striated, branched, and typically uninucleated, contracting involuntarily under intrinsic pacemaker and autonomic control.
- Option D: Visceral muscle is synonymous with smooth muscle, which is involuntary and mononucleated.
MCQ #4 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following connective tissues is completely avascular and heals slowly?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Cartilage is a specialized supporting connective tissue whose extracellular matrix is completely devoid of internal blood vessels, lymphatics, and nerves, which severely limits its regenerative and repair capacity.
Formula / Rule / Reaction:Qualitative concept / Connective tissue vascularity.Solution:- Chondrocytes reside within isolated lacunae and rely entirely on passive diffusion of nutrients and oxygen through the dense extracellular matrix from blood vessels located in the surrounding perichondrium.
- Because diffusion distances are large and metabolic turnover of chondrocytes is low, damaged cartilage exhibits very poor and slow repair mechanisms.
Why other options are incorrect:- Option A: Bone is a vascularized tissue containing extensive Haversian and Volkmann canals, allowing rapid remodeling and fracture repair.
- Option C: Adipose tissue contains an extensive microvascular capillary network essential for lipid storage and mobilization.
- Option D: Areolar tissue is loose connective tissue rich in capillaries that support overlying epithelial layers.
MCQ #5 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Cardiac muscles are different from skeletal muscles because they:
B
Are voluntary and multinucleated
C
Are involuntary and multinucleated
D
Are striated but involuntary
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Cardiac muscle shares the sarcomeric arrangement of alternating actin and myosin filaments that produces striations with skeletal muscle, but operates involuntarily under autonomic and myogenic pacemaker control.
Formula / Rule / Reaction:Qualitative concept / Muscle tissue differentiation.Solution:- Both skeletal and cardiac muscle tissues display repeating sarcomeres resulting in microscopic cross-striations.
- While skeletal muscle is innervated by the voluntary somatic motor system, cardiac myocytes are excited by spontaneous pacemaker activity in the sinoatrial node and modulated involuntarily by the autonomic nervous system.
Why other options are incorrect:- Option A: Cardiac muscle displays striations; smooth muscle is the muscle type that lacks striations.
- Option B: Cardiac muscle is involuntary and typically possesses a single centrally located nucleus (occasionally binucleated), not multinucleated.
- Option C: Cardiac myocytes are typically mononucleated cells joined by intercalated discs, unlike the multinucleated syncytium of skeletal muscle.
MCQ #6 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which organelle gives rise to primary lysosomes through budding in eukaryotic cells?
A
Smooth endoplasmic reticulum (SER)
B
Rough endoplasmic reticulum (RER)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Primary lysosomes are acidic, enzyme-containing storage vesicles that assemble by terminal packaging and budding directly from the trans-Golgi network of Golgi cisternae.
Formula / Rule / Reaction:Qualitative concept / Endomembrane sorting pathway.Solution:- Acid hydrolase enzymes are synthesized by ribosomes bound to the rough endoplasmic reticulum and targeted to the Golgi apparatus via transport vesicles.
- Within the Golgi cisternae, these hydrolytic enzymes undergo mannose-6-phosphate tagging and sorting before budding off from the trans-Golgi cisternae as membrane-bound primary lysosomes.
Why other options are incorrect:- Option A: The smooth endoplasmic reticulum specializes in lipid biosynthesis, glycogen metabolism, and calcium sequestration.
- Option B: The rough endoplasmic reticulum synthesizes the polypeptide chains of lysosomal enzymes but does not bud off mature primary lysosomes.
- Option C: The plasma membrane forms endocytic vesicles through invagination rather than primary lysosomes.
MCQ #7 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A mutation prevents proper disulphide bond formation in a fibrous protein, causing brittle nails and slow hair growth. Which of the following proteinaceous substances would be affected?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Keratin is an insoluble fibrous structural protein rich in cysteine residues whose covalent disulfide cross-links provide high mechanical rigidity and resistance to hair, nails, and outer epidermal layers.
Formula / Rule / Reaction:$$\text{Cysteine-SH} + \text{HS-Cysteine} \xrightarrow{\text{Oxidation}} \text{Cysteine-S-S-Cysteine (Cystine)} + 2\text{H}^+ + 2e^-$$
Solution:- Alpha-keratin forms the primary structural constituent of nails and hair filaments.
- Extensive covalent disulfide bridges formed between adjacent cysteine side chains hold the protofibrils together; failure to establish these covalent cross-links leads directly to mechanical weakness, manifest as brittle nails and impaired hair growth.
Why other options are incorrect:- Option A: Elastin provides elasticity to lungs and large arteries, relying on unique desmosine cross-links rather than extensive disulfide-stabilized keratin fibrils.
- Option B: Actin is a globular cytoskeletal and contractile protein found in microfilaments.
- Option D: Collagen is stabilized by hydrogen bonds and covalent cross-links formed by hydroxylysine and hydroxyproline, primarily residing in bone, tendon, and dermis.
MCQ #8 of 180
Biology
SZABMU 2025
[SZABMU 2025]
In muscle fibers, T-tubules are extensions of the sarcolemma that penetrate into the cell interior. What is their main function during muscle contraction?
A
Unblock actin's binding sites
B
Conduct nerve impulses deep into the muscle cell
C
Synthesize glycogen in the sarcoplasm
D
Store calcium ions in the sarcoplasmic reticulum
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Transverse (T) tubules are deep tubular invaginations of the muscle sarcolemma that propagate action potentials rapidly from the cell surface into the deep interior of the myofiber.
Formula / Rule / Reaction:Qualitative concept / Excitation-contraction coupling.Solution:- Skeletal muscle fibers have large diameters, meaning surface electrical signals would propagate slowly to interior myofibrils if reliant solely on passive planar conduction.
- T-tubules rapidly transmit sarcolemmal depolarization inward to voltage-gated dihydropyridine (DHP) receptors, opening ryanodine channels in the adjacent sarcoplasmic reticulum terminal cisternae.
Why other options are incorrect:- Option A: Actin's binding sites are unblocked when released calcium ions bind troponin C, shifting the tropomyosin complex.
- Option C: Glycogen synthesis is an enzymatic metabolic process occurring in the sarcoplasm, unrelated to T-tubule electrophysiology.
- Option D: Calcium ions are sequestered and stored in the terminal cisternae of the sarcoplasmic reticulum, not inside the lumen of T-tubules.
MCQ #9 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Polysaccharides such as starch and glycogen are mainly used for:
A
Structural framework only
B
Short-term and long-term energy storage
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Starch and glycogen are high-molecular-weight polymers of alpha-glucose that serve as non-osmotic, compact storage depots of chemical energy in plants and animals, respectively.
Formula / Rule / Reaction:Qualitative concept / Storage polysaccharides.Solution:- Starch (comprising amylose and amylopectin) acts as the primary long-term carbohydrate reserve in plant tissues.
- Glycogen is a branched alpha-glucan stored in vertebrate liver and skeletal muscle, functioning as an intermediate and short-term reservoir that is mobilized between meals or during muscular exertion.
Why other options are incorrect:- Option A: Structural framework roles are performed by beta-linked polysaccharides like cellulose in plants and chitin in fungi/arthropods.
- Option C: Biological catalysis is carried out by proteinaceous enzymes and ribozymes, not storage carbohydrates.
- Option D: Free glucose and other monosaccharides act as the immediate ready source of cellular energy.
MCQ #10 of 180
Biology
SZABMU 2025
[SZABMU 2025]
In chromosomes, positively charged histone proteins are organized with negatively charged DNA. The positive charges of histones are due to an abundance of basic amino acids:
D
Phenylalanine and arginine
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Histone proteins contain high proportions of the basic amino acids lysine and arginine, whose positively charged side chains at physiological pH form electrostatic bonds with the negatively charged phosphate groups of DNA.
Formula / Rule / Reaction:$$\text{Lysine side chain: } -(\text{CH}_2)_4-\text{NH}_3^+, \quad \text{Arginine side chain: } -(\text{CH}_2)_3-\text{NH}-\text{C}(\text{NH}_2)=\text{NH}_2^+$$
Solution:- Histones H1, H2A, H2B, H3, and H4 are rich in basic residues (lysine and arginine), carrying net positive charges at physiological pH (\(\text{pH } \approx 7.4\)).
- These positive charges interact electrostatically with the negatively charged phosphodiester backbone of DNA, enabling tight winding into nucleosome octamers.
Why other options are incorrect:- Option A: Alanine has a nonpolar, uncharged aliphatic methyl side chain (\(-\text{CH}_3\)).
- Option C: Alanine does not carry a positive charge at physiological pH.
- Option D: Phenylalanine possesses an uncharged, nonpolar aromatic benzyl side chain.
MCQ #11 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The envelope in a virus is derived from:
D
Viral genome replication
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The viral envelope is a lipid bilayer acquired by certain animal viruses during maturation when the viral nucleocapsid buds through host cellular membranes.
Formula / Rule / Reaction:Qualitative concept / Viral envelope biogenesis.Solution:- During viral budding, the viral nucleocapsid extrudes through host membranes (such as the plasma membrane, nuclear envelope, or Golgi membranes).
- The resulting envelope consists of host-derived phospholipids and cholesterol, embedded with viral glycoprotein spikes.
Why other options are incorrect:- Option A: Host cell ribosomes synthesize polypeptides and do not possess or form lipid bilayer membranes.
- Option C: Capsid proteins form the internal geometric protein shell enclosing the viral genome, not the outer lipid envelope.
- Option D: Viral genome replication synthesizes nucleic acid copies, not lipid bilayers.
MCQ #12 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The myelin sheath acts as an:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The myelin sheath is a multilayered dielectric lipoprotein wrapping produced around axons that functions as an electrical insulator, preventing ion leakage across the internodal axolemma.
Formula / Rule / Reaction:Qualitative concept / Neurophysiology of myelin.Solution:- Myelin exhibits high electrical resistance and low electrical capacitance due to its high lipid content (approximately 80% lipid).
- This insulation prevents transmembrane ionic dissipation along internodes, restricting depolarization to the unmyelinated nodes of Ranvier and driving rapid saltatory conduction.
Why other options are incorrect:- Option A: Although myelin provides secondary physical shielding, its primary neurophysiological role is electrical insulation.
- Option C: Sensors are specialized sensory receptors or dendrites that transduce stimuli into electrical impulses.
- Option D: Mechanical support of nervous tissue is provided generally by neuroglial scaffolding and meninges, not the specific physiological role of myelin.
MCQ #13 of 180
Biology
SZABMU 2025
[SZABMU 2025]
DNA/RNA probes are most commonly used as:
A
Diagnostic tools of infectious diseases
C
Cementing material in somatic cell hybridization
D
Carrier or vector in gene therapy
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Nucleic acid probes are defined single-stranded fragments of DNA or RNA labeled with fluorescent, enzymatic, or radioactive markers that detect complementary target sequences through molecular hybridization.
Formula / Rule / Reaction:Qualitative concept / Molecular hybridization.Solution:- Probes are synthesized complementary to pathogen-specific unique nucleotide sequences.
- When hybridized with clinical patient samples (e.g., in Southern/Northern blots or in situ hybridization), they bind to complementary viral or bacterial genomes, serving as sensitive diagnostic tools for infectious diseases.
Why other options are incorrect:- Option B: Vectors such as plasmids, bacteriophages, and cosmids act as carriers in gene cloning.
- Option C: Polyethylene glycol (PEG) and inactivated Sendai virus serve as fusogenic agents in somatic cell hybridization.
- Option D: Viral vectors (e.g., retroviruses, adenoviruses) or liposomes act as carriers in gene therapy.
MCQ #14 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The event that occurs first during skeletal muscle contraction?
B
Release of calcium from sarcoplasmic reticulum
D
Myosin head binds to ATP
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In excitation-contraction coupling, depolarization of the T-tubular membrane must first cause calcium ion efflux from the terminal cisternae before actin active sites are unmasked for myofilament interaction.
Formula / Rule / Reaction:$$\text{Action Potential} \rightarrow \text{T-tubule depolarization} \rightarrow \text{Ca}^{2+} \text{ release from SR} \rightarrow \text{Troponin binding} \rightarrow \text{Cross-bridge cycle}$$
Solution:- Action potentials invading T-tubules activate DHP receptors, which mechanically open ryanodine receptor channels in the sarcoplasmic reticulum.
- This causes rapid efflux of stored \(\text{Ca}^{2+}\) into the sarcoplasm, initiating the downstream sequence of troponin binding and cross-bridge formation.
Why other options are incorrect:- Option A: Cross-bridge formation cannot occur until calcium binds troponin C to displace tropomyosin from actin.
- Option C: The power stroke occurs after cross-bridge attachment and the release of inorganic phosphate.
- Option D: Myosin binds a new ATP molecule at the end of a power stroke to detach from the actin filament.
MCQ #15 of 180
Biology
SZABMU 2025
[SZABMU 2025]
When genes are linked, they tend to:
C
Stay together during inheritance
D
Be inherited separately
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Gene linkage describes loci situated in proximity on the same chromosome that do not assort independently and tend to be co-transmitted into the same gamete during meiosis.
Formula / Rule / Reaction:Rule of Genetic Linkage (Morgan).Solution:- Syntenic alleles located closely along the same physical DNA molecule travel together during anaphase I of meiosis.
- Consequently, parental phenotypic combinations appear in the progeny at a frequency greater than the 50% predicted by Mendel's law of independent assortment.
Why other options are incorrect:- Option A: Random segregation and independent assortment occur for unlinked genes located on non-homologous chromosomes.
- Option B: Crossing over between linked genes occurs with a probability proportional to the physical distance separating them, not at every meiotic event.
- Option D: Separate inheritance is the defining characteristic of unlinked genes.
MCQ #16 of 180
Biology
SZABMU 2025
[SZABMU 2025]
According to Morgan's experiment, which of the following is essential for detecting gene linkage accurately?
A
Large number of progeny
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Detecting linkage and mapping genetic distances requires large sample sizes to statistically distinguish low-frequency crossover recombinant classes from sampling error.
Formula / Rule / Reaction:$$\text{Recombination Frequency (\%)} = \frac{\text{Number of Recombinants}}{\text{Total Progeny}} \times 100$$
Solution:- Recombination through crossing over between closely linked genes produces low numbers of recombinant offspring.
- A large number of progeny is statistically required to reliably calculate the recombination frequency and differentiate true linkage from independent assortment.
Why other options are incorrect:- Option B: Small sample sizes introduce severe sampling bias and mask small recombinant classes.
- Option C: Crossing over does not occur in male Drosophila melanogaster; using only males would prevent the observation of recombinant linkage mapping.
- Option D: Inducing mutations is a tool used to discover new genes, not a requirement to measure linkage between existing alleles.
MCQ #17 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The end of skeletal muscle attached with moveable bone is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Skeletal muscles span joints and attach to bones; the attachment site to the bone that is displaced during muscular contraction is termed the insertion.
Formula / Rule / Reaction:Qualitative concept / Muscle anatomical architecture.Solution:- The origin of a muscle is its attachment to the stationary, proximal bone.
- The insertion is its attachment to the moveable, distal bone that is pulled toward the origin when the muscle belly contracts.
Why other options are incorrect:- Option B: The origin anchors the muscle to the relatively fixed or immobile bone.
- Option C: A tendon is a cord of dense fibrous connective tissue linking muscle to bone, not the specific functional attachment site.
- Option D: The belly is the fleshy, contractile central portion of the muscle.
MCQ #18 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following is common between aldosterone, sex hormones and cortisol?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Aldosterone, cortisol, and sex steroids share a common chemical classification as steroid hormones derived enzymatically from cholesterol in the adrenal cortex and gonads.
Formula / Rule / Reaction:Core Structure: Cyclopentanoperhydrophenanthrene ring system (three cyclohexane rings, one cyclopentane ring).Solution:- Aldosterone (mineralocorticoid) and cortisol (glucocorticoid) are synthesized in the adrenal cortex, while sex hormones (estrogens, progesterone, androgens) are synthesized in gonads and adrenal tissue.
- All these hormones share the 17-carbon four-ring steroid nucleus derived from cholesterol and are lipophilic signaling molecules that act via intracellular nuclear receptors.
Why other options are incorrect:- Option A: Peptide hormones are polymers of amino acids, such as insulin and parathyroid hormone.
- Option B: Phospholipids are structural amphipathic membrane lipids containing glycerol, fatty acids, and a phosphate head.
- Option D: Glycoproteins are proteins with covalently attached carbohydrate groups, such as TSH, LH, and FSH.
MCQ #19 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which one of the following organism is uricotelic with reference to secretion of nitrogenous compound?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Uricotelic animals excrete nitrogenous waste primarily as relatively insoluble uric acid, an adaptation that conserves water in terrestrial and arid environments.
Formula / Rule / Reaction:$$\text{Uric acid excretion requires } \approx 1 - 3\text{ mL H}_2\text{O per gram of nitrogen}$$
Solution:- Reptiles (including snakes) and birds synthesize uric acid from ammonia to prevent excessive water loss.
- Because uric acid has very low water solubility, it precipitates out of solution and is eliminated as a thick white semi-solid paste through the cloaca.
Why other options are incorrect:- Option A: Humans are ureotelic, eliminating excess nitrogen primarily as water-soluble urea.
- Option B: Freshwater fish are ammonotelic, eliminating toxic ammonia directly into the surrounding water through their gills.
- Option D: Adult amphibians (frogs) are ureotelic, excreting urea.
MCQ #20 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Ureotelism (removal of nitrogenous waste in the form of urea) is an adaptation to the:
B
Less availability of water
C
Flooded environmental conditions
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Ureotelism is an evolutionary adaptation in semi-terrestrial and terrestrial animals that reduces water loss by converting highly toxic ammonia into less toxic, water-soluble urea.
Formula / Rule / Reaction:$$\text{Ammonotelism: } \approx 500\text{ mL H}_2\text{O/g N} \quad \text{vs.} \quad \text{Ureotelism: } \approx 50\text{ mL H}_2\text{O/g N}$$
Solution:- Excreting ammonia requires roughly \(500\text{ mL}\) of water per gram of nitrogen to prevent toxic cellular injury.
- Terrestrial vertebrates convert ammonia to urea via the ornithine cycle in the liver, which requires only about \(50\text{ mL}\) of water per gram of nitrogen, conserving body water under conditions of restricted water availability.
Why other options are incorrect:- Option A: Temperature variations do not determine the metabolic choice of nitrogenous excretion pathway.
- Option C: Flooded or open aquatic environments provide abundant water, favoring direct ammonotelism.
- Option D: Food shortages alter nitrogen balance but do not determine the biochemical excretion pathway.
MCQ #21 of 180
Biology
SZABMU 2025
[SZABMU 2025]
All of the following are the characteristics of limbic system EXCEPT:
A
Produce most basic and primitive emotions and drives
C
Consists of hypothalamus, amygdala, and hippocampus
D
Controls reflex movement of eyes
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The limbic system is a functional network of forebrain structures regulating emotions, basic drives, and memory consolidation; visual and eye movement reflexes are controlled by brainstem centers.
Formula / Rule / Reaction:Qualitative concept / Neuroanatomical functional localization.Solution:- Reflexive somatic movements of the eyes in response to visual and auditory stimuli are mediated by the superior colliculi of the midbrain and cranial nerve nuclei (III, IV, VI).
- The limbic system (incorporating the amygdala, hippocampus, cingulate gyrus, and hypothalamus) regulates primitive emotions, motivation, and memory.
Why other options are incorrect:- Option A: Primitive emotions (fear, rage, sexual drives) are primary functional outputs of the amygdala and limbic circuits.
- Option B: The hippocampus within the limbic system is required for the conversion of short-term memory into long-term memory.
- Option C: The hypothalamus, amygdala, and hippocampus are core anatomical components of the limbic system.
MCQ #22 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Sensory input from auditory and visual pathways, skin and within the body is received by _________ and distributed to _________
A
Hypothalamus; Cerebellum
C
Cerebrum; rest of the brain
D
Cerebellum; rest of the brain
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The thalamus serves as the primary sensory relay and processing station of the central nervous system, relaying ascending sensory information (except olfaction) to specific sensory cortices of the cerebrum.
Formula / Rule / Reaction:Sensory Pathway: Receptors $\rightarrow$ Second-Order Neurons $\rightarrow$ Thalamus $\rightarrow$ Cerebral Cortex.Solution:- Ascending sensory inputs from the optic tracts (lateral geniculate nucleus), auditory pathways (medial geniculate nucleus), and somatosensory tracts (ventral posterolateral/posteromedial nuclei) synapse within the thalamus.
- Third-order thalamocortical neurons project these signals to corresponding functional sensory areas of the cerebral cortex for conscious perception and interpretation.
Why other options are incorrect:- Option A: The hypothalamus is an autonomic and endocrine regulatory center, while the cerebellum coordinates motor activities.
- Option C: The cerebrum is the ultimate target of sensory projections, not the relay station that receives and redistributes them.
- Option D: The cerebellum receives proprioceptive inputs for balance and motor coordination, but does not relay general sensory pathways.
MCQ #23 of 180
Biology
SZABMU 2025
[SZABMU 2025]
According to the Lamarck theory of evolution, the organ that has not been used in several generations will:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Lamarck's evolutionary principle of use and disuse stated that organs frequently exercised develop and enlarge, whereas unexercised organs undergo progressive reduction, atrophy, and ultimate disappearance over generations.
Formula / Rule / Reaction:Lamarckian Postulate: Continuous disuse $\rightarrow$ Vestigialization $\rightarrow$ Eventual disappearance.Solution:- Lamarck proposed that environmental changes create new behavioral needs, leading an organism to cease using certain structures.
- According to this hypothesis, continuous disuse across successive generations causes an organ to progressively shrink and eventually disappear from the species.
Why other options are incorrect:- Option A: An organ becomes strong and enlarged through active, continuous use according to Lamarckism.
- Option B: Amputation is surgical or traumatic removal, not an evolutionary mechanism of morphological change.
- Option D: Development of an organ is the predicted outcome of repeated use, the exact opposite of disuse.
MCQ #24 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following statement is incorrect with reference to the enzyme action?
A
Substrate binds with the enzyme at its active site
B
Non-competitive inhibitor binds the enzyme at a site distinct from that of active site
C
Addition of a lot of succinate do not reverse the inhibition of succinic dehydrogenase by malonate
D
Malonate is a competitive inhibitor of succinic dehydrogenase
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Competitive inhibition is reversible; because the substrate and competitive inhibitor compete for the same active site, sufficiently increasing the substrate concentration outcompetes the inhibitor and restores \(V_{\max}\).
Formula / Rule / Reaction:$$\text{Competitive Inhibition: } [S] \rightarrow \infty \implies v = V_{\max}$$
Solution:- Malonate acts as a classical competitive inhibitor of succinate dehydrogenase due to its close structural resemblance to succinate.
- Adding an excess of succinate increases the probability of succinate binding to the active site over malonate, fully reversing the inhibition; therefore, the statement claiming it does not reverse the inhibition is incorrect.
Why other options are incorrect:- Option A: Substrates specifically bind to the catalytic active site of enzymes to form the enzyme-substrate complex.
- Option B: Non-competitive inhibitors bind to allosteric sites separate from the active site, altering the enzyme's catalytic efficiency.
- Option D: Malonate is a classic competitive inhibitor of succinate dehydrogenase.
MCQ #25 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The temperature of underground water remains constant due to:
D
Specific heat of vaporization
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Water's high specific heat capacity allows it to absorb or release large amounts of heat energy with minimal changes in temperature, providing thermal buffering in subterranean reservoirs.
Formula / Rule / Reaction:$$q = m c \Delta T \implies \Delta T = \frac{q}{m c}$$
Solution:- Water has a high specific heat capacity (\(c = 4.184\text{ J/g}\cdot^\circ\text{C}\)) resulting from extensive hydrogen bonding between water molecules.
- Because a large input of thermal energy produces only a small change in temperature (\(\Delta T\)), underground water insulated from surface atmospheric radiation maintains a stable temperature throughout the year.
Why other options are incorrect:- Option B: Polarity accounts for water's solvent capabilities and hydrogen bonding, but heat capacity is the specific thermodynamic property that stabilizes temperature.
- Option C: Heat of vaporization governs the phase change from liquid to gas at the boiling point.
- Option D: Specific heat of vaporization relates to phase transitions during evaporation rather than thermal buffering in liquid water.
MCQ #26 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The shape of adenovirus is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Adenoviruses are non-enveloped double-stranded DNA viruses with capsids displaying icosahedral symmetry, forming a 20-sided regular polyhedron.
Formula / Rule / Reaction:Qualitative concept / Viral capsid morphology (20 equilateral triangular faces, 12 vertices).Solution:- The adenovirus capsid is a polyhedral structure composed of 252 capsomeres (240 hexons on the triangular faces and 12 pentons at the vertices with radiating fibers).
- This regular polyhedral structure allows the virus to enclose its genome using identical protein subunits.
Why other options are incorrect:- Option A: Helical capsids are rod-shaped cylindrical structures, such as those of Tobacco mosaic virus and Rhabdoviruses.
- Option B: Tadpole morphology is characteristic of complex tailed bacteriophages (such as the T-even phages).
- Option D: Circular is a two-dimensional geometric term; viral capsids are three-dimensional structures.
MCQ #27 of 180
Biology
SZABMU 2025
[SZABMU 2025]
In human female normally ovulation occurs at _________ day of menstrual cycle.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:In a standard 28-day human ovarian cycle, peak estrogen levels trigger a surge of luteinizing hormone (LH), leading to ovulation at approximately day 14.
Formula / Rule / Reaction:$$\text{Day of Ovulation} = \text{Total Cycle Duration (28 days)} - \text{Fixed Luteal Phase (14 days)} = \text{Day 14}$$
Solution:- High levels of estradiol from the mature Graafian follicle trigger an LH surge from the anterior pituitary.
- Approximately 12 to 24 hours after the LH peak (and 14 days after the onset of menses), the mature follicle ruptures, releasing the secondary oocyte into the peritoneal cavity.
Why other options are incorrect:- Option A: Day 9 falls within the proliferative/follicular phase when follicles are still maturing under FSH stimulation.
- Option B: Day 11 is too early in a typical 28-day cycle; estrogen levels have not reached the threshold to trigger the LH surge.
- Option D: Day 15 corresponds to the early secretory/luteal phase following ovulation.
MCQ #28 of 180
Biology
SZABMU 2025
[SZABMU 2025]
HIV belongs to which class of virus with respect to genome type?
A
Single stranded DNA viruses
B
Double stranded RNA viruses
C
ssRNA; template for DNA synthesis
D
ssRNA; template for mRNA synthesis
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Human Immunodeficiency Virus (HIV) is a retrovirus (Baltimore Group VI) carrying two identical single-stranded positive-sense RNA molecules that serve as templates for reverse transcription into complementary DNA.
Formula / Rule / Reaction:$$\text{Viral (+)ssRNA} \xrightarrow{\text{Reverse Transcriptase}} \text{ssDNA} \rightarrow \text{dsDNA} \xrightarrow{\text{Integrase}} \text{Host Proviral DNA}$$
Solution:- HIV is an enveloped lentivirus containing two copies of positive-sense ssRNA.
- Upon infection, its virally encoded reverse transcriptase uses the ssRNA genome as a template to synthesize double-stranded viral DNA, which then integrates into the host chromosome.
Why other options are incorrect:- Option A: Single-stranded DNA genomes are characteristic of Parvoviruses (Baltimore Group II).
- Option B: Double-stranded RNA genomes are characteristic of Reoviruses and Rotaviruses (Baltimore Group III).
- Option D: ssRNA acting directly as a template for mRNA synthesis describes negative-sense ssRNA viruses (Baltimore Group V, like Influenza and Rhabdoviruses).
MCQ #29 of 180
Biology
SZABMU 2025
[SZABMU 2025]
How many ATPs are produced during oxidative phase of glycolysis?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The pay-off (oxidative) phase of glycolysis generates four ATP molecules per glucose molecule via substrate-level phosphorylation, yielding a net gain of two ATP after accounting for the two ATP invested in the preparatory phase.
Formula / Rule / Reaction:$$2\text{ (1,3-BPG)} + 2\text{ ADP} \rightarrow 2\text{ (3-PGA)} + 2\text{ ATP}$$
$$2\text{ (PEP)} + 2\text{ ADP} \rightarrow 2\text{ (Pyruvate)} + 2\text{ ATP}$$
$$\text{Total ATP produced in pay-off phase} = 2 + 2 = 4\text{ ATP}$$
Solution:- Glycolysis is divided into an energy-investment phase (which consumes 2 ATP) and an energy-yielding/oxidative phase.
- In the oxidative phase, two triose phosphate molecules undergo substrate-level phosphorylation at the phosphoglycerate kinase step (producing 2 ATP) and at the pyruvate kinase step (producing 2 ATP).
- The total number of ATP molecules synthesized during this phase is 4 ATP (yielding a net of \(4 - 2 = 2\text{ ATP}\) for the overall pathway).
Why other options are incorrect:- Option A: 2 is the net ATP yield of the overall glycolytic pathway after subtracting the 2 ATP invested in the preparatory phase.
- Option C: 34 represents the theoretical ATP yield generated across the complete respiratory chain via oxidative phosphorylation.
- Option D: 36 is the classical total theoretical net yield of ATP per glucose molecule from complete aerobic respiration.
MCQ #30 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following best describes the relationship between endoplasmic reticulum (ER) and Golgi apparatus?
A
Both ER and Golgi apparatus are completely separate organelles with no interaction
B
ER is precursor organelle to Golgi apparatus because material is transported from ER to Golgi
C
Golgi apparatus is precursor organelle to ER because material is transported from Golgi to ER
D
ER and Golgi apparatus have similar roles
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The endoplasmic reticulum acts as the biosynthetic and structural precursor to the Golgi apparatus within the secretory pathway, supplying proteins, lipids, and membrane vesicles to the cis-Golgi network.
Formula / Rule / Reaction:Biosynthetic-Secretory Pathway: RER / SER $\rightarrow$ COPII Transport Vesicles $\rightarrow$ cis-Golgi $\rightarrow$ trans-Golgi.Solution:- Polypeptides, glycoproteins, and lipids synthesized in the ER are packaged into COPII-coated transport vesicles.
- These vesicles travel to and fuse with the cis face of the Golgi cisternae, establishing the ER as the upstream precursor in membrane and material transport.
Why other options are incorrect:- Option A: The ER and Golgi apparatus interact continuously through anterograde and retrograde vesicular transport.
- Option C: The Golgi receives material from the ER; it does not serve as a precursor to the ER.
- Option D: The ER focuses on primary protein synthesis, initial folding, and lipid synthesis, whereas the Golgi specializes in oligosaccharide remodeling, sorting, and packaging.
MCQ #31 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Osmotic gradient of inner medulla of kidney is increased by:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The hyperosmotic medullary gradient required for urinary concentration is established by sodium chloride reabsorption in the loop of Henle and the medullary recycling of urea from the collecting ducts.
Formula / Rule / Reaction:$$\text{Total Medullary Osmolality} \approx [\text{NaCl}] + [\text{Urea}] \approx 1200 - 1400\text{ mOsm/L}$$
Solution:- In the presence of antidiuretic hormone (ADH), the inner medullary collecting duct becomes permeable to urea via UT-A1 and UT-A3 transporters.
- Urea diffuses into the medullary interstitium, contributing nearly half of the inner medullary osmotic gradient (up to 500 to 600 mOsm/L) to drive water reabsorption.
Why other options are incorrect:- Option B: Water reabsorption into the interstitium dilutes the solute concentration, dissipating the gradient if not drained by the vasa recta.
- Option C: Phosphate is primarily reabsorbed in the proximal tubule and functions as a urinary buffer rather than a medullary osmolyte.
- Option D: Interstitial potassium is maintained at low levels to prevent abnormal disruption of membrane potentials.
MCQ #32 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Decreased level of FSH hormone during menstrual cycle triggers release of:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:During the late follicular phase, rising estrogen and inhibin from the dominant follicle depress FSH secretion through negative feedback, while sustained high estrogen levels switch to positive feedback on the pituitary to trigger the LH surge.
Formula / Rule / Reaction:$$\uparrow \text{Estradiol} \xrightarrow{\text{Negative Feedback}} \downarrow \text{FSH}; \quad \uparrow\uparrow \text{Estradiol} \xrightarrow{\text{Positive Feedback}} \text{LH Surge}$$
Solution:- As the dominant follicle matures, it secretes high levels of estradiol, which lowers FSH release via negative feedback on the anterior pituitary.
- When circulating estradiol remains elevated above threshold for over 36 hours, it triggers a positive feedback surge of luteinizing hormone (LH), inducing ovulation.
Why other options are incorrect:- Option A: Human chorionic gonadotropin (hCG) is secreted by syncytiotrophoblasts of the blastocyst following implantation.
- Option C: Oxytocin is released from the posterior pituitary in response to uterine contraction during labor and suckling during lactation.
- Option D: Progesterone is secreted by the corpus luteum after the LH surge induces ovulation.
MCQ #33 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The digestive secretions from different parts of GIT have the ability to kill microbe present in the food. Which of the following best describe this role?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Antimicrobial agents are classified by their site of application; agents that destroy or inhibit microorganisms on living tissues or internal mucosal surfaces act as antiseptics.
Formula / Rule / Reaction:Qualitative concept / Antimicrobial classification.Solution:- Antiseptics are substances that prevent infection by destroying or inhibiting microorganisms on or within living tissues without causing tissue toxicity.
- Gastrointestinal secretions (including gastric hydrochloric acid and salivary lysozyme) kill ingested microorganisms along mucosal surfaces, functioning endogenously as natural antiseptics.
Why other options are incorrect:- Option B: Disinfectants are chemical agents applied to non-living, inanimate objects and are too toxic for living tissues.
- Option C: Chemotherapeutics are synthetic or biosynthesized drugs administered systemically to treat internal infections or cancers.
- Option D: Immunosuppressants dampen immune responses, increasing susceptibility to microbial infection.
MCQ #34 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A carrier female for hemophilia can pass the defective allele to:
C
Both sons and daughters
D
Neither sons nor daughters
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Hemophilia is an X-linked recessive disorder; a heterozygous carrier female has one mutant X chromosome (\(X^h\)) that she passes to 50% of her offspring regardless of sex.
Formula / Rule / Reaction:$$\text{Mother: } X^H X^h \xrightarrow{\text{Meiosis}} 50\% \text{ Gametes with } X^H, \quad 50\% \text{ Gametes with } X^h$$
Solution:- A carrier mother possesses the genotype \(X^H X^h\).
- During gametogenesis, her two X chromosomes segregate independently into ova.
- She contributes the \(X^h\) allele to 50% of her male offspring (who express hemophilia, \(X^h Y\)) and 50% of her female offspring (who become carriers, \(X^H X^h\)), meaning the defective allele can be inherited by both sons and daughters.
Why other options are incorrect:- Option A: Daughters also receive one X chromosome from their mother and can inherit the \(X^h\) allele to become carriers.
- Option B: Sons receive an X chromosome from their mother and can inherit the defective allele.
- Option D: A carrier mother produces gametes carrying the mutant allele in half of her ovulations.
MCQ #35 of 180
Biology
SZABMU 2025
[SZABMU 2025]
During ventricular systole, which pressure changes occur?
A
Ventricular pressure rises above atrial and arterial pressure
B
Atrial pressure rises above Ventricular pressure
C
Ventricular pressure falls below arterial pressures
D
Ventricular pressure falls below Atrial pressure
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:During ventricular systole, isovolumetric contraction increases intraventricular pressure above atrial pressure (closing the AV valves), and subsequent ejection raises it above arterial outflow pressure (opening the semilunar valves).
Formula / Rule / Reaction:$$\text{Ventricular Systole: } P_{\text{ventricle}} > P_{\text{atrium}} \quad \text{and} \quad P_{\text{ventricle}} > P_{\text{artery}}$$
Solution:- When the ventricles contract, intraventricular pressure rises rapidly above atrial pressure, snapping the atrioventricular (mitral and tricuspid) valves shut.
- As contraction continues, ventricular pressure exceeds the diastolic pressure in the aorta (\(> 80\text{ mmHg}\)) and pulmonary trunk (\(> 10\text{ mmHg}\)), opening the semilunar valves to eject blood.
Why other options are incorrect:- Option B: Atrial pressure exceeds ventricular pressure during ventricular diastole, driving ventricular filling.
- Option C: If ventricular pressure fell below arterial pressure, the semilunar valves would close and ejection would stop.
- Option D: Ventricular pressure falling below atrial pressure occurs during diastole, not systole.
MCQ #36 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which cell type is essential for stimulating cell division of both B-cells and T-cells?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Helper T-lymphocytes (CD4+) coordinate adaptive immunity by secreting cytokines that stimulate the proliferation and differentiation of both B-lymphocytes and cytotoxic T-lymphocytes.
Formula / Rule / Reaction:$$\text{Activated CD4}^+ \text{ T-cell} \xrightarrow{\text{IL-2, IL-4, IL-5}} \text{Clonal expansion of B-cells and T-cells}$$
Solution:- Upon antigen recognition on MHC Class II molecules, helper T-cells secrete interleukin-2 (IL-2).
- IL-2 acts as an autocrine and paracrine growth factor that stimulates clonal division of both cytotoxic T-cells (CD8+) and B-cells, driving adaptive immune responses.
Why other options are incorrect:- Option A: Plasma cells are terminally differentiated B-cells that produce antibodies and do not stimulate lymphocyte division.
- Option B: Memory cells persist to mediate accelerated secondary immune responses and do not direct primary lymphocyte proliferation.
- Option D: Neutrophils are innate phagocytic granulocytes that do not direct lymphocyte cell division.
MCQ #37 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The hormone, when overproduced can lead to hypercalcemia and stone formation is
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Parathyroid hormone (PTH) raises serum calcium by stimulating osteoclast-mediated bone resorption, renal tubular calcium reabsorption, and vitamin D activation; its overproduction produces hypercalcemia and nephrolithiasis.
Formula / Rule / Reaction:$$\uparrow \text{PTH} \rightarrow \uparrow \text{Bone Resorption} + \uparrow \text{Renal Ca}^{2+} \text{ Reabsorption} \rightarrow \text{Hypercalcemia} \rightarrow \text{Calcium Kidney Stones}$$
Solution:- Excessive PTH secretion (hyperparathyroidism) mobilizes large amounts of calcium from bone into the bloodstream.
- The filtered load of calcium exceeds the renal reabsorption capacity, leading to hypercalciuria; this excess urinary calcium precipitates as calcium oxalate or calcium phosphate kidney stones.
Why other options are incorrect:- Option A: Insulin regulates blood glucose levels and does not directly control calcium homeostasis.
- Option B: Thyroxine controls basal metabolic rate; excess levels do not typically cause primary hypercalcemia with kidney stones.
- Option D: Calcitonin lowers blood calcium by inhibiting osteoclasts, counteracting hypercalcemia and stone formation.
MCQ #38 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Fibrocartilage is found in:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Fibrocartilage contains dense, orderly bundles of type I collagen fibers that provide tensile strength and shock absorption in weight-bearing joints.
Formula / Rule / Reaction:Qualitative concept / Cartilage histological distribution.Solution:- Fibrocartilage forms the menisci of the knee joint, the pubic symphysis, and the intervertebral discs.
- In the knee joint, the fibrocartilaginous menisci act as shock absorbers that disperse compressive stresses during movement.
Why other options are incorrect:- Option A: The trachea is supported by C-shaped rings of hyaline cartilage.
- Option B: The structural framework of the external nose is made of hyaline cartilage.
- Option D: Ear flaps (auricles) are composed of flexible elastic cartilage.
MCQ #39 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Additional capillaries present in the juxtamedullary nephrons to form a loop of vessels are:
B
Peritubular capillaries
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Juxtamedullary nephrons possess long, hairpin-shaped capillary loops called the vasa recta that parallel the loops of Henle into the renal medulla to function as countercurrent exchangers.
Formula / Rule / Reaction:Qualitative concept / Renal microvasculature.Solution:- Efferent arterioles of juxtamedullary nephrons descend deep into the renal medulla to form hairpin loops called the vasa recta.
- These capillary loops remove reabsorbed water and solutes while maintaining the medullary hyperosmotic gradient through passive countercurrent exchange.
Why other options are incorrect:- Option A: Efferent arterioles carry blood out of the glomerulus and supply the capillary networks.
- Option B: Peritubular capillaries form an interconnected mesh surrounding the proximal and distal convoluted tubules in the renal cortex.
- Option D: Afferent arterioles deliver blood directly into the glomerular capillaries for filtration.
MCQ #40 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Y-linked inheritance refers to which mode of the inheritance?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Y-linked (holandric) inheritance describes the direct transmission of genes located on the non-homologous region of the Y chromosome exclusively from father to son.
Formula / Rule / Reaction:$$\text{Holandric Transmission: } \text{Father } (XY^*) \rightarrow 100\% \text{ Sons } (XY^*) \quad (0\% \text{ Daughters})$$
Solution:- Because the Y chromosome is present only in males, Y-linked traits pass along the direct male lineage across generations.
- This transmission is termed 'straight' (patrilineal) inheritance, contrasting with the crisscross pattern typical of X-linked recessive traits.
Why other options are incorrect:- Option A: Crisscross inheritance describes X-linked recessive traits passing from affected fathers to carrier daughters, then to grandsons.
- Option C: 'Loop' is not a recognized genetic inheritance pattern.
- Option D: 'Jumping' refers to transposable genetic elements, not chromosomal inheritance patterns.
MCQ #41 of 180
Biology
SZABMU 2025
[SZABMU 2025]
If a colorblind lady marries a normal man, their children will be:
A
Normal daughters and normal sons
B
Normal sons and carrier daughters
C
Colorblind sons and carrier daughters
D
Colorblind sons and colorblind daughters
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Red-green color blindness is an X-linked recessive trait; males inherit their single X chromosome from their mother, while daughters receive an X chromosome from each parent.
Formula / Rule / Reaction:$$\text{Cross: } X^c X^c \text{ (Colorblind Mother)} \times X^C Y \text{ (Normal Father)}$$
$$\text{Progeny: } 100\% \text{ Sons } X^c Y \text{ (Colorblind)}, \quad 100\% \text{ Daughters } X^C X^c \text{ (Carriers)}$$
Solution:- The colorblind mother contributes a recessive \(X^c\) allele to every child.
- Sons receive the father's Y chromosome and the mother's \(X^c\), making all sons colorblind (\(X^c Y\)).
- Daughters receive the father's dominant normal allele (\(X^C\)) and the mother's \(X^c\), making all daughters phenotypically normal carriers (\(X^C X^c\)).
Why other options are incorrect:- Option A: All sons are colorblind because they receive their only X chromosome from the affected mother.
- Option B: Sons cannot be normal because the mother contributes only the mutant \(X^c\) allele.
- Option D: Daughters are carriers rather than colorblind because they inherit a normal dominant \(X^C\) allele from the father.
MCQ #42 of 180
Biology
SZABMU 2025
[SZABMU 2025]
All of the following properties of water are associated with its capillary action. EXCEPT:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Capillary action is the spontaneous movement of water through narrow spaces driven by adhesive forces, cohesive intermolecular forces, and surface tension.
Formula / Rule / Reaction:$$h = \frac{2 \gamma \cos \theta}{\rho g r}$$
Solution:- Adhesion between polar water molecules and hydrophilic surfaces, combined with cohesive hydrogen bonding between water molecules, generates surface tension that pulls the liquid column upward.
- Although fluid density (\(\rho\)) determines the downward gravitational force acting on the column, it is not a driving force of capillary action.
Why other options are incorrect:- Option A: Adhesion provides the electrostatic attraction between water and tube surfaces that drives liquid ascent.
- Option C: Cohesion maintains the continuous column of water molecules via intermolecular hydrogen bonding.
- Option D: Surface tension acts at the meniscus to pull the water column upward against gravity.
MCQ #43 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which organelle serves as a packaging and a distribution center for molecules within the cells?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The Golgi apparatus chemically modifies, sorts, and packages proteins and lipids received from the endoplasmic reticulum into secretory or transport vesicles.
Formula / Rule / Reaction:Qualitative concept / Endomembrane transport.Solution:- Transport vesicles from the endoplasmic reticulum fuse with the cis-Golgi network.
- Within the Golgi cisternae, enzymes modify proteins and lipids, sorting and packaging them into transport vesicles at the trans face for delivery to lysosomes, the plasma membrane, or for secretion.
Why other options are incorrect:- Option B: Mitochondria generate cellular ATP via the Krebs cycle and oxidative phosphorylation.
- Option C: Ribosomes are ribonucleoprotein complexes responsible for translating mRNA into polypeptides.
- Option D: Vacuoles serve primarily for storage, turgor maintenance, and waste isolation.
MCQ #44 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following hormone is not produced in older women?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Menopause involves the depletion of ovarian follicles, resulting in the cessation of ovarian estrogen biosynthesis in older women.
Formula / Rule / Reaction:Qualitative concept / Postmenopausal endocrinology.Solution:- During reproductive years, the ovaries synthesize the vast majority of circulating estrogen (17-beta-estradiol).
- With menopause and follicular depletion, the ovaries cease estrogen production, causing a marked drop in systemic estrogen levels.
Why other options are incorrect:- Option B: Gastrin is secreted by gastric G-cells to stimulate gastric acid production, independent of age.
- Option C: Secretin is released by duodenal S-cells to stimulate pancreatic secretion, unaffected by menopause.
- Option D: Renin is produced by the juxtaglomerular cells of the kidney to regulate blood pressure throughout life.
MCQ #45 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which lymphatic structure absorbs dietary fats in the intestine?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Lacteals are specialized, blunt-ended lymphatic capillaries located within intestinal villi that absorb long-chain dietary fats packaged as chylomicrons.
Formula / Rule / Reaction:$$\text{Dietary Lipids} \rightarrow \text{Chylomicrons} \rightarrow \text{Central Lacteals} \rightarrow \text{Thoracic Duct} \rightarrow \text{Venous Circulation}$$
Solution:- Long-chain fatty acids absorbed by enterocytes are assembled into chylomicrons, which are too large to pass through continuous blood capillaries.
- The central lacteals of the intestinal villi possess wide endothelial fenestrations and discontinuous basal laminae that allow chylomicrons to enter lymphatic drainage.
Why other options are incorrect:- Option A: Peyer's patches are aggregates of lymphoid tissue in the ileum that mediate gut mucosal immunity.
- Option C: The pancreatic duct conveys digestive enzymes and bicarbonate from the pancreas to the duodenum.
- Option D: Lymph nodes filter lymph and house lymphocytes, but do not directly absorb lipids from the intestinal lumen.
MCQ #46 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Thinking, memory and voluntary actions are mainly controlled by the:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The cerebrum, particularly the cerebral cortex, is the primary control center for higher intellectual functions, conscious thought, reasoning, memory integration, and the initiation of voluntary motor activity.
Formula / Rule / Reaction:Qualitative concept / Functional neuroanatomy.Solution:- The cerebral hemispheres contain functional sensory, motor, and association areas responsible for perception, voluntary commands, speech, and intellect.
- While subcortical structures assist in memory processing, overall voluntary command and higher cognitive processes are directed by the cerebrum.
Why other options are incorrect:- Option B: The cerebellum coordinates muscular movements, posture, and balance, but does not initiate voluntary thought.
- Option C: The pons acts as a bridge connecting brain regions and houses pneumotaxic centers for respiratory rhythmicity.
- Option D: The hippocampus specializes in consolidating short-term memory into long-term memory, but does not govern voluntary actions or general thinking.
MCQ #47 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Ventricular diastole causes:
A
Ventricular contraction
B
Closure of atrioventricular valves
C
Opening of semilunar valves
D
Closure of semilunar valves
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Ventricular diastole is the relaxation phase of the ventricular myocardium during which intraventricular pressure drops below arterial pressure, causing the semilunar valves to snap shut.
Formula / Rule / Reaction:$$\text{Ventricular Diastole: } P_{\text{ventricle}} < P_{\text{artery}} \implies \text{Backflow closes aortic and pulmonary semilunar valves ($S_2$ sound)}$$
Solution:- As the ventricles relax, intraventricular pressure drops below the pressures in the aorta and pulmonary trunk.
- Elastic recoil of the great arteries forces blood back toward the ventricles, catching the valve cusps and causing closure of the semilunar valves to prevent regurgitation.
Why other options are incorrect:- Option A: Ventricular contraction defines ventricular systole, not diastole.
- Option B: Closure of atrioventricular (mitral and tricuspid) valves occurs at the onset of ventricular systole.
- Option C: Semilunar valves open during the ejection phase of ventricular systole when ventricular pressure exceeds arterial pressure.
MCQ #48 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following statements correctly describes the organelle that is not membrane bound?
B
Modification and packaging of proteins
C
Presence of digestive enzymes
D
Made of rRNA and protein
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Ribosomes are non-membrane-bound cellular organelles composed of ribosomal ribonucleic acid (rRNA) and structural proteins that carry out protein translation.
Formula / Rule / Reaction:$$\text{Eukaryotic Ribosome (80S)} = \text{Small Subunit (40S)} + \text{Large Subunit (60S)}$$
Solution:- Ribosomes lack a surrounding phospholipid bilayer membrane.
- Each ribosome consists of two ribonucleoprotein subunits assembled from rRNA molecules and specific ribosomal polypeptides.
Why other options are incorrect:- Option A: Cristae are inner membrane folds characteristic of mitochondria, which are double-membrane-bound organelles.
- Option B: Modification and packaging of proteins is the function of the Golgi apparatus, a membrane-bound organelle.
- Option C: Digestive hydrolytic enzymes are enclosed within lysosomes, which are single-membrane-bound vesicles.
MCQ #49 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A biological laboratory is developing gene therapy with a modified virus that infects birds only. Which of the following categories does this virus belong to?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Viruses are classified based on host tropism; viruses that specifically infect vertebrate and invertebrate animal hosts, including avian species (class Aves), are designated animal viruses.
Formula / Rule / Reaction:Qualitative concept / Viral host taxonomy.Solution:- Birds are members of the kingdom Animalia (phylum Chordata).
- A virus that infects avian host cells is classified as an animal virus.
Why other options are incorrect:- Option A: Algal viruses specifically infect photosynthetic eukaryotic algae.
- Option B: Fungal viruses (mycoviruses) specifically infect fungal organisms.
- Option D: Bacterial viruses (bacteriophages) infect bacterial prokaryotes.
MCQ #50 of 180
Biology
SZABMU 2025
[SZABMU 2025]
CFTR gene that encodes a protein which regulates:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The Cystic Fibrosis Transmembrane Conductance Regulator (CFTR) gene encodes an ABC transporter-class chloride ion channel that regulates salt and water balance across epithelial surfaces.
Formula / Rule / Reaction:$$\text{Active } \text{Cl}^- \text{ transport} \rightarrow \text{Passive } \text{Na}^+ \text{ flux} \rightarrow \text{Osmotic } \text{H}_2\text{O movement}$$
Solution:- CFTR regulates the secretion of chloride ions and the secondary reabsorption of sodium ions across mucosal membranes.
- Osmotic gradients generated by this salt transport drive the movement of water, maintaining proper hydration of mucus in the respiratory, digestive, and reproductive tracts.
Why other options are incorrect:- Option A: CFTR does not transport or regulate whole proteins across membranes.
- Option C: CFTR does not govern glucose or carbohydrate homeostasis.
- Option D: This option is an inaccurate and ambiguous description of electrolyte transport.
MCQ #51 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which organelle is usually referred as post office of the cell and why?
A
Golgi apparatus, because it receives, sorts and packages material for transport to other parts of cell
B
Rough endoplasmic reticulum, because it synthesizes protein used by the cell
C
Mitochondria, because it synthesizes and provides energy to other parts of the cells
D
Nucleus, because it controls other organelles of the cell
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The Golgi complex functions as the intracellular post office by receiving newly synthesized molecules from the endoplasmic reticulum, modifying them, attaching molecular destination tags, and packaging them into transport vesicles.
Formula / Rule / Reaction:Qualitative concept / Endomembrane transport.Solution:- Proteins and lipids arrive at the cis-Golgi network from the endoplasmic reticulum.
- Within the cisternae, they undergo glycosylation and phosphorylation, which serve as molecular address labels, before sorting and packaging at the trans face for delivery to lysosomes, the cell membrane, or secretory pathways.
Why other options are incorrect:- Option B: The rough endoplasmic reticulum serves as the site of translation for secretory and membrane proteins, not the sorting post office.
- Option C: Mitochondria are the powerhouses of the cell that generate ATP.
- Option D: The nucleus contains genetic material and serves as the cellular control center.
MCQ #52 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Components of reflex arc must contain neurons from:
A
Motor nerve and Mixed nerve
B
Sensory nerve and Intermediate nerve
C
Motor nerve and Sensory Nerve
D
Mixed nerve and Intermediate nerve
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:A reflex arc is the neural pathway that mediates a reflex action; minimally, it requires a sensory pathway to detect the stimulus and a motor pathway to stimulate the effector.
Formula / Rule / Reaction:$$\text{Reflex Arc: Receptor} \rightarrow \text{Sensory Neuron} \rightarrow [\text{Interneuron in CNS}] \rightarrow \text{Motor Neuron} \rightarrow \text{Effector}$$
Solution:- Every functional reflex arc must contain an afferent (sensory) neuron to transmit impulses to the central nervous system and an efferent (motor) neuron to convey response impulses to muscles or glands.
- Monosynaptic reflexes (such as the patellar stretch reflex) require only sensory and motor neurons, making them the essential components of the circuit.
Why other options are incorrect:- Option A: A mixed nerve is a composite anatomical trunk containing both sensory and motor axons, not a distinct functional class of neuron.
- Option B: Monosynaptic reflex arcs function without intermediate neurons (interneurons).
- Option D: This option lacks sensory neurons, which are required to detect the initial stimulus.
MCQ #53 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A confirmed COVID patient shows weak specific immunity. Which component is most likely deficient in this patient?
D
Suppressor T-Lymphocytes
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Specific (adaptive) immunity requires CD4+ helper T-lymphocytes to recognize viral antigens and release cytokines that activate B-cells for humoral immunity and cytotoxic T-cells for cell-mediated immunity.
Formula / Rule / Reaction:$$\text{CD4}^+ \text{ Helper T-cell} \xrightarrow{\text{Cytokines (IL-2, IFN-}\gamma\text{)}} \text{Activation of B-cells and CD8}^+ \text{ Cytotoxic T-cells}$$
Solution:- Helper T-lymphocytes coordinate adaptive immune responses against SARS-CoV-2.
- A deficiency or depletion of helper T-lymphocytes impairs both antibody production by B-cells and cytotoxic T-cell-mediated elimination of virally infected host cells.
Why other options are incorrect:- Option A: Red blood cells function in oxygen and carbon dioxide gas transport and do not direct specific immunity.
- Option C: Thrombocytes (platelets) are cell fragments involved in hemostasis and blood clotting.
- Option D: Suppressor (regulatory) T-lymphocytes downregulate immune activity; their deficiency causes autoimmunity rather than weak specific immunity.
MCQ #54 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following organ takes part in the T-cell maturation for both the lymphatic and endocrine systems?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The thymus is a primary lymphoid organ that provides the microenvironment for T-lymphocyte maturation and acts as an endocrine gland by secreting peptide hormones like thymosin.
Formula / Rule / Reaction:Qualitative concept / Dual lymphatic-endocrine physiology.Solution:- Immature pro-T cells migrate from the bone marrow to the thymus, where they undergo positive and negative selection to mature into immunocompetent T-lymphocytes.
- Thymic epithelial cells also synthesize and release hormones, such as thymosin and thymopoietin, qualifying the thymus as an endocrine gland.
Why other options are incorrect:- Option A: The spleen is a secondary lymphoid organ that filters blood, but does not support T-cell maturation or function as an endocrine gland.
- Option C: The liver is a metabolic organ and exocrine/endocrine gland, but does not mature T-lymphocytes.
- Option D: The kidney secretes erythropoietin and renin, but has no role in T-cell maturation.
MCQ #55 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The part of male reproductive system that maintains the temperature for sperm production is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Spermatogenesis requires an optimal temperature approximately \(2 - 3\,^\circ\text{C}\) below core body temperature, which is maintained by the external suspension and muscular activity of the scrotum.
Formula / Rule / Reaction:$$T_{\text{scrotal}} \approx 34 - 35\,^\circ\text{C} \quad (\text{Core body temperature } \approx 37\,^\circ\text{C})$$
Solution:- The scrotum is an extra-abdominal pouch that houses the testes outside the warmer pelvic cavity.
- Contraction and relaxation of the dartos and cremaster muscles adjust the distance of the testes from the body wall in response to ambient temperature changes.
Why other options are incorrect:- Option A: The testes are the gonads that produce spermatozoa and testosterone; they do not regulate their own external temperature.
- Option C: The vas deferens is the muscular duct that transports mature spermatozoa from the epididymis to the ejaculatory duct.
- Option D: The epididymis is the coiled tubular structure where spermatozoa mature and gain motility.
MCQ #56 of 180
Biology
SZABMU 2025
[SZABMU 2025]
After Liver transplant, doctors suppress the T-lymphocyte activity in patient. Why?
A
To increase the rate of antibody production
C
To avoid graft rejection by the immune system
D
To produce memory cells faster
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Acute allograft rejection is mediated by host T-lymphocytes that recognize foreign human leukocyte antigens (MHC molecules) on the donor organ, initiating cytotoxic destruction of the graft.
Formula / Rule / Reaction:$$\text{Host T-cells} + \text{Donor Foreign MHC} \rightarrow \text{Cell-mediated graft rejection}$$
Solution:- Recipient CD4+ and CD8+ T-cells recognize foreign major histocompatibility complex (MHC) proteins on the donor liver tissue.
- Immunosuppressive drugs (such as cyclosporine or tacrolimus) inhibit T-cell activation and cytokine release, preventing immune-mediated graft rejection.
Why other options are incorrect:- Option A: Suppressing T-lymphocyte activity decreases helper T-cell signals, thereby reducing antibody production.
- Option B: Suppressing T-cells impairs immune defenses, increasing vulnerability to opportunistic infections.
- Option D: Immunosuppression inhibits the formation and proliferation of memory cells.
MCQ #57 of 180
Biology
SZABMU 2025
[SZABMU 2025]
How do the kidneys contribute to maintain normal blood calcium level?
A
By storing excess calcium
B
By reabsorbing calcium and activating vitamin D
C
By secreting calcium directly into the blood
D
By converting calcium into urea for excretion
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The kidneys regulate extracellular calcium levels by modulating tubular calcium reabsorption and performing the final hydroxylation of vitamin D into active calcitriol.
Formula / Rule / Reaction:$$25\text{-(OH)D}_3 \xrightarrow{1\alpha\text{-hydroxylase (kidney)}} 1,25\text{-(OH)}_2\text{D}_3\text{ (Calcitriol)}$$
Solution:- Under the influence of parathyroid hormone (PTH), renal tubular epithelial cells increase active calcium reabsorption in the thick ascending limb and distal convoluted tubule.
- Kidney enzymes convert calcidiol into active calcitriol (1,25-dihydroxycholecalciferol), which stimulates intestinal absorption of dietary calcium.
Why other options are incorrect:- Option A: Kidneys do not serve as storage reservoirs for calcium; bones store calcium.
- Option C: Kidneys filter and reabsorb calcium rather than synthesizing and secreting it de novo into the blood.
- Option D: Calcium is a chemical element that cannot be converted into organic urea molecules.
MCQ #58 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Nodes of Ranvier along the length of axon fibers allow:
A
More diffusion of neurotransmitters
B
Saltatory conduction of nerve impulse
C
Formation of new synaptic connections
D
Synthesis of new neurons
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Nodes of Ranvier are unmyelinated gaps along a myelinated axon where high concentrations of voltage-gated sodium channels allow action potentials to jump from node to node via saltatory conduction.
Formula / Rule / Reaction:$$v_{\text{saltatory}} \gg v_{\text{continuous}} \quad (\text{Conduction velocity increases up to } 100 - 120\text{ m/s})$$
Solution:- The insulating myelin sheath prevents ion leakage along the internodes.
- Inward sodium currents generated during an action potential travel electrotonically along the axon to depolarize the next node of Ranvier, allowing rapid saltatory propagation.
Why other options are incorrect:- Option A: Neurotransmitter diffusion occurs exclusively across synaptic clefts at axon terminals, not along the axon shaft.
- Option C: Synaptogenesis occurs at axonal growth cones and dendritic spines, not along internodal shafts.
- Option D: Neurogenesis occurs in neurogenic niches of the central nervous system, unrelated to axonal nodes.
MCQ #59 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Arachidonic acid, which is the precursor molecule for prostaglandins is:
A
saturated fatty acid with 18 carbons
B
An unsaturated fatty acid with 18 carbons
C
A tri-unsaturated fatty acid with 20 carbons
D
A tetra-unsaturated fatty acid with 20 carbons
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Arachidonic acid is a 20-carbon polyunsaturated omega-6 fatty acid with four cis double bonds, serving as the direct biosynthetic precursor for prostaglandins, thromboxanes, and leukotrienes.
Formula / Rule / Reaction:$$\text{Arachidonic Acid: } \text{C}_{20}\text{H}_{32}\text{O}_2 \quad (20:4\,\Delta^{5,8,11,14})$$
Solution:- Arachidonic acid possesses a 20-carbon carboxylic acid chain containing four carbon-carbon double bonds (tetra-unsaturated).
- Cyclooxygenase enzymes (COX-1 and COX-2) convert liberated arachidonic acid into prostaglandin \(\text{H}_2\) (\(\text{PGH}_2\)), which is converted into active prostaglandins.
Why other options are incorrect:- Option A: Saturated 18-carbon fatty acid refers to stearic acid.
- Option B: Unsaturated 18-carbon fatty acids include oleic, linoleic, and alpha-linolenic acids, not arachidonic acid.
- Option C: Arachidonic acid has four double bonds, not three.
MCQ #60 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Atrial walls are thinner as compared to ventricles because:
A
Atria are small in size
B
Atria are present above the ventricles
C
Atria have to force blood into the ventricles which lie very close to them
D
Blood enters into the atria by osmosis
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The thickness of a cardiac chamber myocardium corresponds to the workload and resistance it encounters; atria pump against low resistance into adjacent ventricles, requiring less muscle mass.
Formula / Rule / Reaction:$$\text{Myocardial Wall Tension} \propto \text{Pressure} \times \text{Radius}$$
Solution:- The atria receive venous blood and pump it across low-resistance atrioventricular valves into the ventricles immediately below them.
- Because this requires minimal pressure, atrial walls possess a thin layer of pectinate muscle, whereas ventricles must generate high pressure to overcome systemic and pulmonary vascular resistance.
Why other options are incorrect:- Option A: Chamber volume does not dictate wall thickness; stroke volume in atria and ventricles is balanced.
- Option B: Anatomical position above the ventricles does not explain the physiological requirement for thin walls.
- Option D: Blood enters the atria via bulk flow driven by venous hydrostatic pressure gradients, not osmosis.
MCQ #61 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A patient with a genetic disorder undergoes treatment where a normal gene is inserted into their cells. This is called:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Gene therapy is a therapeutic technique that delivers functional copies of a gene into patient cells to correct, replace, or compensate for a defective or missing pathogenic allele.
Formula / Rule / Reaction:Qualitative concept / Recombinant biotechnology.Solution:- In gene therapy, viral vectors or non-viral delivery systems introduce functional therapeutic genes into target cells.
- Expression of the inserted gene produces functional proteins, alleviating the symptoms of the genetic disorder.
Why other options are incorrect:- Option B: Genetic testing involves analyzing DNA to identify mutations, not treating disorders.
- Option C: Protein engineering alters protein sequences and structures for industrial or pharmaceutical use.
- Option D: DNA fingerprinting identifies individuals using variable tandem repeats (VNTRs/STRs) for forensic analysis.
MCQ #62 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following is not a potential symptom of chronic renal failure?
A
Nephritis and bacterial infection
B
Bone pain and fractures
C
Impaired concentration and confusion
D
Muscle cramps due to electrolyte imbalance
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Chronic renal failure causes uremic syndrome, manifesting as bone disorders, neurological symptoms, and electrolyte abnormalities; nephritis and bacterial infections represent causes rather than resulting symptoms.
Formula / Rule / Reaction:Etiology vs. Clinical Manifestation in Renal Pathophysiology.Solution:- Chronic renal failure leads to secondary hyperparathyroidism causing bone pain/fractures, uremic encephalopathy causing confusion, and hypocalcemia/hyperkalemia causing muscle cramps.
- Nephritis and pyelonephritis are inflammatory diseases that can cause renal failure, but are not direct symptoms produced by the failing kidney.
Why other options are incorrect:- Option B: Renal osteodystrophy caused by deficient calcitriol synthesis leads directly to bone demineralization, pain, and pathological fractures.
- Option C: Accumulation of uremic neurotoxins causes uremic encephalopathy, characterized by impaired concentration and confusion.
- Option D: Electrolyte disturbances, such as hypocalcemia and hyperphosphatemia, produce muscle cramps and tetany.
MCQ #63 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Arthritis is a disease that primarily affects the:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Arthritis is a musculoskeletal disorder characterized by acute or chronic inflammation, erosion, and degenerative changes within articular cartilage and synovial joints.
Formula / Rule / Reaction:Qualitative concept / Pathology of the skeletal system.Solution:- Arthritis encompasses joint disorders such as osteoarthritis (degeneration of articular cartilage) and rheumatoid arthritis (autoimmune destruction of synovial membranes).
- The primary pathological manifestations are joint pain, swelling, stiffness, and reduced range of motion.
Why other options are incorrect:- Option A: Inflammatory diseases of the kidneys are classified as nephritis or nephrosis.
- Option B: Inflammatory diseases of the lungs include pneumonia, bronchitis, and pneumonitis.
- Option D: Muscle inflammation is termed myositis.
MCQ #64 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Osmoreceptors that detect the osmotic pressure of blood are primarily located in the:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Central osmoreceptors located in specialized circumventricular organs of the hypothalamus monitor plasma osmolality and coordinate water balance by regulating thirst and antidiuretic hormone (ADH) secretion.
Formula / Rule / Reaction:$$\uparrow \text{Plasma Osmolality } (> 295\text{ mOsm/kg}) \rightarrow \text{Hypothalamic Osmoreceptor Shrinkage} \rightarrow \uparrow \text{ADH Release}$$
Solution:- Osmoreceptive neurons reside in the anterior hypothalamus (including the organum vasculosum of the lamina terminalis and supraoptic nuclei).
- When blood osmotic pressure increases, these cells lose water via osmosis, shrink, and trigger neural signals that stimulate thirst and ADH release from the posterior pituitary.
Why other options are incorrect:- Option B: The cerebellum integrates vestibular and proprioceptive inputs to coordinate voluntary motor function and balance.
- Option C: The medulla oblongata houses autonomic centers controlling heart rate, vasomotor tone, and respiration.
- Option D: The cerebral cortex handles conscious sensory processing and reasoning, but does not house primary homeostatic osmoreceptors.
MCQ #65 of 180
Biology
SZABMU 2025
[SZABMU 2025]
A chromosome which contains the centromere at the center is called:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Chromosomes are categorized morphologically by centromere location; a metacentric chromosome features a centrally placed centromere that divides the chromosome into two arms of equal length.
Formula / Rule / Reaction:$$\text{Arm Ratio } (r = q/p) \approx 1.0 \implies \text{Metacentric (V-shaped during anaphase)}$$
Solution:- In a metacentric chromosome, the primary constriction (centromere) is positioned at the midpoint.
- This produces equal-length p and q arms, giving the chromosome a characteristic V-shape during anaphase movement toward spindle poles.
Why other options are incorrect:- Option A: Mesocentric is not a recognized cytogenetic classification term.
- Option C: Acrocentric chromosomes have the centromere positioned near one end, producing one very short arm (p) and one long arm (q).
- Option D: Telocentric chromosomes have the centromere located at the terminal end, leaving only a single arm.
MCQ #66 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Binding site for a substrate on enzyme is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The active site is the specific three-dimensional cleft on an enzyme where substrate molecules bind and undergo catalytic conversion into products.
Formula / Rule / Reaction:$$\text{E} + \text{S} \rightleftharpoons \text{ES complex} \rightarrow \text{E} + \text{P}$$
Solution:- The active site consists of a substrate-binding site that provides chemical specificity and a catalytic site with amino acid side chains that lower activation energy.
- Substrates bind to this site via non-covalent interactions (hydrogen bonds, ionic interactions, van der Waals forces).
Why other options are incorrect:- Option A: Allosteric sites are secondary binding sites separate from the active site that bind non-substrate regulatory molecules.
- Option B: Passive site is not a recognized biochemical term in enzymology.
- Option D: Regulatory sites bind activators or inhibitors to modulate enzyme conformation, rather than binding the primary reaction substrate.
MCQ #67 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The most abundant polysaccharide in plants, widely used in paper and as a source of dietary fibre for human is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Cellulose is a linear, unbranched homopolysaccharide of beta-D-glucose residues joined by beta-1,4-glycosidic bonds, forming the structural basis of plant cell walls and human dietary fiber.
Formula / Rule / Reaction:$$[\text{C}_6\text{H}_{10}\text{O}_5]_n \quad \text{with repeating } \beta(1\rightarrow 4)\text{-glycosidic linkages}$$
Solution:- Cellulose accounts for over 50% of all organic carbon in the plant kingdom, serving as the main structural material in plant cell walls and the raw material for paper production.
- Because humans lack the cellulase enzyme required to hydrolyze beta-1,4-glycosidic bonds, cellulose passes undigested through the gastrointestinal tract, functioning as dietary fiber.
Why other options are incorrect:- Option A: Chitin is a structural polysaccharide composed of N-acetylglucosamine found in fungal cell walls and arthropod exoskeletons.
- Option B: Glycogen is the branched storage polysaccharide of animals, not plants.
- Option C: Starch consists of alpha-glucose polymers (amylose and amylopectin) that are easily digested by human amylase, functioning as a nutrient source rather than dietary fiber.
MCQ #68 of 180
Biology
SZABMU 2025
[SZABMU 2025]
This given organelle is involved in the synthesis of oil, phospholipids and steroids:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The smooth endoplasmic reticulum (SER) contains membrane-bound enzymes that synthesize neutral lipids, fatty acids, phospholipids, and steroid hormones.
Formula / Rule / Reaction:Qualitative concept / Lipid metabolic pathways in endomembranes.Solution:- The smooth endoplasmic reticulum lacks ribosomes and specializes in lipid metabolism.
- It contains key enzymes that assemble phospholipids for cellular membranes and convert cholesterol into steroid hormones in endocrine tissues.
Why other options are incorrect:- Option A: Mitochondria generate ATP via oxidative phosphorylation and the Krebs cycle.
- Option B: The Golgi complex modifies, sorts, and packages macromolecules received from the ER, but does not carry out de novo lipid biosynthesis.
- Option D: Ribosomes translate messenger RNA into polypeptide chains and have no role in lipid synthesis.
MCQ #69 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which type of neurons stimulate the muscles to contract in a reflex arc?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Efferent (motor) neurons convey motor impulses away from the central nervous system to peripheral effector organs, such as skeletal muscle fibers, triggering contraction.
Formula / Rule / Reaction:$$\text{CNS Integration} \xrightarrow{\text{Action potential in efferent axon}} \text{Neuromuscular Junction} \rightarrow \text{Muscle Contraction}$$
Solution:- In a reflex arc, sensory information processed in the spinal cord triggers action potentials in efferent (motor) neurons.
- The efferent axons exit via the ventral roots and release acetylcholine at neuromuscular junctions, stimulating muscle contraction.
Why other options are incorrect:- Option B: Sensory neurons carry impulses from peripheral sensory receptors toward the central nervous system.
- Option C: Interneurons are local circuit neurons within the CNS that integrate information between sensory and motor pathways.
- Option D: Afferent neurons is another term for sensory neurons, which carry signals into the central nervous system.
MCQ #70 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Receptors that are located in the retina of eye are called:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Rod and cone cells are specialized neuroepithelial photoreceptors in the neural retina that transduce light photons into graded electrical potentials.
Formula / Rule / Reaction:$$\text{Photon Absorption by Rhodopsin/Photopsin} \rightarrow \text{11-cis to all-trans retinal} \rightarrow \text{Hyperpolarization}$$
Solution:- The retina contains two classes of photoreceptor cells: rods and cones.
- Rods contain rhodopsin and mediate dim-light (scotopic) monochromatic vision, while cones contain photopsins and mediate bright-light (photopic) high-acuity color vision.
Why other options are incorrect:- Option A: Merkel discs are slow-adapting mechanoreceptors located in the basal epidermis that detect light touch.
- Option B: Ruffini endings are slow-adapting mechanoreceptors in the dermis that detect skin stretch.
- Option C: Pacinian corpuscles are rapidly adapting mechanoreceptors in subcutaneous tissue that detect deep pressure and high-frequency vibration.
MCQ #71 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The bony roof of the oral cavity, which separates it from the nasal cavity, is formed by the:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The hard palate forms the rigid anterior bony septum separating the oral cavity from the nasal cavity, composed of the palatine processes of the maxillae and the horizontal plates of the palatine bones.
Formula / Rule / Reaction:$$\text{Hard Palate} = \text{Palatine Processes of Maxillae (anterior 2/3)} + \text{Horizontal Plates of Palatine Bones (posterior 1/3)}$$
Solution:- The superior surface of the hard palate forms the floor of the nasal cavity, while its inferior surface forms the bony roof of the oral cavity.
- This rigid plate provides a solid counter-surface for the tongue during mastication and deglutition.
Why other options are incorrect:- Option A: The soft palate is a posterior, flexible fibromuscular partition that lacks a bony internal framework.
- Option B: The mandible forms the lower jaw and the skeletal framework of the floor of the mouth.
- Option D: While the maxilla contributes its palatine processes to the hard palate, the anatomical structure separating the cavities is the hard palate itself, which also includes the palatine bones.
MCQ #72 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which structure in neuron is responsible for receiving information from other neurons?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Dendrites are branching cytoplasmic extensions of the neuron that receive synaptic inputs from other neurons and convey incoming graded potentials toward the soma.
Formula / Rule / Reaction:Information Flow in Neurons: Dendrites (reception) $\rightarrow$ Soma (integration) $\rightarrow$ Axon (conduction) $\rightarrow$ Terminal (transmission).Solution:- Dendritic branches possess dendritic spines equipped with ligand-gated ion channels and neurotransmitter receptors.
- Neurotransmitters released from presynaptic neurons bind these receptors, generating postsynaptic potentials that propagate toward the cell body.
Why other options are incorrect:- Option A: The axon conducts action potentials away from the soma toward target effectors.
- Option B: The soma houses the nucleus and organelles, integrating electrical potentials rather than serving as the primary receptive arbor.
- Option D: Axon terminals release neurotransmitters into the synaptic cleft to transmit signals to downstream cells.
MCQ #73 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Chemicals that stimulate the olfactory receptors enter the nasal cavity in the form of
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Olfactory chemoreceptors detect volatile airborne chemical compounds that enter the respiratory tract in the gaseous phase before dissolving in the mucosal layer of the olfactory epithelium.
Formula / Rule / Reaction:Qualitative concept / Olfactory sensory transduction.Solution:- Inhaled odorant molecules must be sufficiently volatile to travel as airborne gases into the superior nasal cavity.
- Upon reaching the olfactory epithelium, these gaseous molecules dissolve into the surface mucus layer and bind to G-protein coupled odorant receptors on olfactory cilia.
Why other options are incorrect:- Option A: Particulate solids cannot diffuse through the inhaled airstream to reach olfactory receptors.
- Option B: While odorants dissolve in mucus on mucosal surfaces, they enter the nasal cavity as airborne vapors/gases.
- Option D: Plasma is an ionized gas state found at high temperatures, not compatible with physiological respiration.
MCQ #74 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Neurosecretory cells are:
D
Neurons that have been adapted to secrete hormones
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Neurosecretory cells are specialized neurons that translate neural signals into chemical endocrine signals by releasing neurohormones directly into the bloodstream from their axon terminals.
Formula / Rule / Reaction:Neural Action Potential $\rightarrow$ Axon Terminal Depolarization $\rightarrow$ Neurohormone Exocytosis into Capillaries.Solution:- Neurosecretory cells (such as magnocellular neurons of the paraventricular and supraoptic hypothalamic nuclei) possess dendrites, a soma, and an axon.
- Instead of innervating another neuron or muscle at a synapse, their axon terminals terminate on capillary walls to release hormones (e.g., ADH and oxytocin) into circulation.
Why other options are incorrect:- Option A: Typical endocrine cells are non-neuronal epithelial cells that synthesize and release hormones.
- Option B: Exocrine cells discharge secretions through ducts onto epithelial surfaces, not into the bloodstream.
- Option C: Neuroglial cells provide metabolic and structural support in the nervous system and do not conduct impulses or secrete systemic neurohormones.
MCQ #75 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Lamarck theory of evolution was rejected experimentally by:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:August Weismann experimentally refuted Lamarck's hypothesis of the inheritance of acquired characteristics by demonstrating that somatic mutilations are not transmitted to offspring, establishing the Germ Plasm Theory.
Formula / Rule / Reaction:$$\text{Germ Plasm Theory: Somatoplasm changes } \not\to \text{ Germplasm (Heritable information)}$$
Solution:- Weismann cut off the tails of laboratory mice across 22 consecutive generations.
- In every generation, newborn mice developed tails of normal length, proving that acquired somatic changes do not modify the germline DNA inherited by offspring.
Why other options are incorrect:- Option A: Ernst Haeckel formulated the biogenetic law ('ontogeny recapitulates phylogeny') and supported evolutionary theory.
- Option B: Charles Darwin proposed natural selection as the mechanism of evolution, but did not experimentally disprove Lamarck.
- Option D: Thomas R. Malthus wrote an essay on population growth that influenced Darwin and Wallace, but did not conduct biological experiments on Lamarckism.
MCQ #76 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following characteristics is common among sucrose, lactose and maltose?
B
occurrence in living beings
C
chemical nature of their monosaccharide units
D
to act as reducing sugar
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Sucrose, lactose, and maltose are structural isomers belonging to the class of hexose disaccharides, all formed from two hexose units via condensation and sharing the same molecular formula.
Formula / Rule / Reaction:$$2\text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{Condensation}} \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O}$$
Solution:- Maltose (glucose + glucose), lactose (glucose + galactose), and sucrose (glucose + fructose) each contain 12 carbon, 22 hydrogen, and 11 oxygen atoms.
- All three disaccharides share the identical molecular formula \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\).
Why other options are incorrect:- Option B: Occurrence in living beings is a generic property of many biomolecules and does not uniquely define or unite these sugars.
- Option C: Their monosaccharide compositions differ: maltose contains two glucoses, lactose contains glucose and galactose, and sucrose contains glucose and fructose.
- Option D: Maltose and lactose are reducing sugars with free anomeric carbons, but sucrose is a non-reducing sugar because both anomeric carbons are tied up in the glycosidic bond.
MCQ #77 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Ribozymes are the only biocatalysts that are made up of:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Ribozymes are catalytically active ribonucleic acid (RNA) molecules capable of accelerating specific chemical reactions in the absence of proteinaceous enzymes.
Formula / Rule / Reaction:$$\text{Catalytic RNA (e.g., 23S rRNA in peptidyl transferase)} \rightarrow \text{Peptide bond synthesis}$$
Solution:- Discovered by Thomas Cech and Sidney Altman, ribozymes fold into defined tertiary structures using intramolecular base pairing.
- Examples include self-splicing group I/II introns, the RNA subunit of RNase P, and the peptidyl transferase center of ribosomal 23S/28S rRNA.
Why other options are incorrect:- Option A: DNA serves primarily as genetic storage and does not act as a natural biological catalyst in vivo.
- Option C: Conventional enzymes are composed of polypeptides (proteins), which ribozymes are contrasted against.
- Option D: Fatty acids are structural lipid units that do not possess intrinsic biocatalytic activity.
MCQ #78 of 180
Biology
SZABMU 2025
[SZABMU 2025]
The induced fit model differs from the Lock and Key model because the enzyme in this model:
A
Has a rigid and fixed active site
B
Changes shape to fit the substrate
C
Is non-specific in its action
D
Is denatured during reaction
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Daniel Koshland's induced fit model proposes that the enzyme active site is structurally flexible, undergoing a conformational change upon substrate binding to optimize catalytic interactions.
Formula / Rule / Reaction:$$\text{Enzyme (relaxed conformation)} + \text{Substrate} \rightarrow \text{Enzyme-Substrate Complex (induced complementary conformation)}$$
Solution:- Unlike Emil Fischer's rigid Lock and Key model, the induced fit model posits that the active site is not a static template.
- Substrate binding induces precise conformational changes in the enzyme active site, aligning catalytic residues to stabilize the transition state and lower the activation energy.
Why other options are incorrect:- Option A: A rigid, fixed active site describes the Lock and Key hypothesis.
- Option C: Enzymes operating by induced fit maintain substrate specificity based on complementary chemical interactions.
- Option D: Enzyme denaturation involves irreversible loss of tertiary structure, which does not occur during normal catalysis.
MCQ #79 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which competitive inhibitor blocks the enzyme responsible for bacterial cell wall synthesis?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Penicillin contains a beta-lactam ring that structurally mimics the D-alanyl-D-alanine terminus of bacterial peptidoglycan precursors, competitively and irreversibly inhibiting transpeptidase.
Formula / Rule / Reaction:$$\text{Penicillin} + \text{Transpeptidase (PBP)} \rightarrow \text{Inactivated Covalent Penicilloyl-Enzyme Complex}$$
Solution:- Bacterial cell wall synthesis requires transpeptidase enzymes to cross-link glycan chains with peptide bridges.
- Penicillin fits into the transpeptidase active site as a structural analogue, blocking cross-link formation and causing osmotic lysis of dividing bacteria.
Why other options are incorrect:- Option A: Sulfonamides competitively inhibit dihydropteroate synthase to block bacterial folic acid synthesis.
- Option B: Streptomycin is an aminoglycoside that binds the 30S ribosomal subunit to inhibit protein synthesis.
- Option C: Tetracycline blocks the binding of aminoacyl-tRNA to the A-site of the 30S ribosomal subunit.
MCQ #80 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Water commonly known as a universal solvent because:
A
It dissolves polar and ionic substances effectively
B
It supports chemical reactions without reacting itself
C
It has a high specific heat capacity
D
It dissolves many substances due to its small molecule size
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Water's bent geometry and electronegativity difference create a strong molecular dipole, allowing it to dissolve ionic compounds via hydration shells and polar substances via hydrogen bonding.
Formula / Rule / Reaction:$$\text{Dielectric constant of water } (\varepsilon_r \approx 80) \implies F = \frac{1}{4\pi\varepsilon_r\varepsilon_0}\frac{q_1 q_2}{r^2} \ll F_{\text{vacuum}}$$
Solution:- Water molecules possess a permanent electrical dipole with a partial negative charge on oxygen and partial positive charges on hydrogen.
- Water readily forms hydration shells around cations and anions, attenuating electrostatic attractions and dissolving a wide range of ionic and polar substances.
Why other options are incorrect:- Option B: Water frequently acts as a direct chemical reactant in hydrolysis and condensation reactions.
- Option C: High specific heat capacity provides thermal buffering, but does not explain solvent capacity.
- Option D: Solvation depends on polarity and dielectric properties, not simply small molecular volume.
MCQ #81 of 180
Biology
SZABMU 2025
[SZABMU 2025]
Which of the following structural change in water molecule leads to the low density of ice?
A
Formation of ionic bonds
B
Expansion due to hydrogen bonding
C
Increased kinetic energy
D
Loss of polarity in molecules
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:When water freezes, each molecule forms four hydrogen bonds arranged in a rigid, open hexagonal crystal lattice, causing the structure to expand and reducing its density relative to liquid water.
Formula / Rule / Reaction:$$\rho_{\text{ice}} (0\,^\circ\text{C}) \approx 0.917\text{ g/cm}^3 < \rho_{\text{water}} (4\,^\circ\text{C}) \approx 1.000\text{ g/cm}^3$$
Solution:- In liquid water, hydrogen bonds continuously break and reform, allowing molecules to pack closely together with maximum density at \(4\,^\circ\text{C}\).
- Upon freezing at \(0\,^\circ\text{C}\), water molecules are fixed into an open tetrahedral lattice held by four hydrogen bonds, increasing the total volume by roughly 9% and lowering the density of ice.
Why other options are incorrect:- Option A: Water molecules do not form ionic bonds; bonding in ice remains covalent with intermolecular hydrogen bonding.
- Option C: Kinetic energy decreases as liquid water cools and transitions into solid ice.
- Option D: Water molecules retain their polar dipole structure in the solid phase.
MCQ #82 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Reduction of carboxylic acid by using lithium aluminohydride produce
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Lithium aluminium hydride (\(\text{LiAlH}_4\)) is a strong reducing agent that reduces carboxylic acids to primary alcohols through nucleophilic acyl substitution followed by hydride addition.
Formula / Rule / Reaction:$$\text{R-COOH} \xrightarrow{1.\; \text{LiAlH}_4\text{, dry ether}} \xrightarrow{2.\; \text{H}_3\text{O}^+} \text{R-CH}_2\text{OH}$$
Solution:- Carboxylic acids are reduced by \(\text{LiAlH}_4\) via transfer of hydride ions (\(\text{H}^-\)) to the carbonyl carbon.
- The carboxylic acid is converted to an aldehyde intermediate, which undergoes immediate reduction to yield a primary alcohol as the final product.
Why other options are incorrect:- Option B: Acid halides are produced by reacting carboxylic acids with thionyl chloride (\(\text{SOCl}_2\)) or phosphorus halides (\(\text{PCl}_3, \text{PCl}_5\)).
- Option C: Esters are synthesized via Fischer esterification of carboxylic acids with alcohols in the presence of an acid catalyst.
- Option D: Reducing carboxylic acids directly to alkanes requires harsh conditions, such as red phosphorus with \(\text{HI}\).
MCQ #83 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
In separate vessel of volume 2000 cm³ and 1500cm³, 6g of hydrogen and 28g of nitrogen are mixed in each vessel at 25°C, ammonia formed is?
$$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$$
B
More in 1500 cm³ vessel
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:According to Le Chatelier's principle, increasing the pressure of a gaseous equilibrium system by decreasing the container volume shifts the equilibrium toward the side with fewer gas moles.
Formula / Rule / Reaction:$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad (\Delta n_g = 2 - (1 + 3) = -2)$$
Solution:- The forward synthesis of ammonia results in a decrease in gas moles from 4 moles of reactants to 2 moles of product.
- The 1500 cm³ vessel has a smaller volume than the 2000 cm³ (2 L) vessel, generating a higher total pressure for the same initial mole quantities.
- This higher pressure shifts the equilibrium toward the product side, producing a higher yield of ammonia in the 1500 cm³ vessel.
Why other options are incorrect:- Option A: The 2 L vessel has a larger volume and lower equilibrium pressure, shifting the reaction toward reactants.
- Option C: The yield is not equal because equilibrium position depends on system volume when \(\Delta n_g \neq 0\).
- Option D: The direction of the shift is predicted by Le Chatelier's principle.
MCQ #84 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The weakest intermolecular force fresent in liquid is?
C
Instantaneous dipole induced dipole force
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Instantaneous dipole-induced dipole forces (London dispersion forces) arise from temporary, transient fluctuations in electron density and represent the weakest class of intermolecular interactions.
Formula / Rule / Reaction:$$\text{Relative Strength: } \text{London Dispersion} < \text{Dipole-Dipole} < \text{Hydrogen Bonding} \ll \text{Covalent/Ionic}$$
Solution:- Temporary displacement of the electron cloud in a nonpolar or polar molecule creates a momentary instantaneous dipole.
- This temporary dipole induces a complementary dipole in an adjacent molecule, resulting in weak attractive forces that quickly dissipate.
Why other options are incorrect:- Option A: Dipole-dipole forces involve electrostatic attractions between permanent molecular dipoles, which are stronger than dispersion forces.
- Option B: Permanent dipole-induced dipole forces (Debye forces) are stronger than instantaneous dispersion interactions.
- Option D: Hydrogen bonding is a strong type of dipole-dipole attraction occurring between hydrogen and electronegative atoms (N, O, F).
MCQ #85 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Number of sigma bond in acetylene and ethylene are?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:A single covalent bond is a sigma bond; a double bond consists of one sigma and one pi bond; a triple bond consists of one sigma and two pi bonds.
Formula / Rule / Reaction:$$\text{Acetylene: } \text{H}-\text{C}\equiv\text{C}-\text{H} \implies 2\,(\text{C}-\text{H } \sigma) + 1\,(\text{C}-\text{C } \sigma) = 3\,\sigma\text{ bonds}$$
$$\text{Ethylene: } \text{H}_2\text{C}=\text{CH}_2 \implies 4\,(\text{C}-\text{H } \sigma) + 1\,(\text{C}-\text{C } \sigma) = 5\,\sigma\text{ bonds}$$
Solution:- In acetylene (ethyne, \(\text{C}_2\text{H}_2\)), the two C-H bonds are single sigma bonds and the central carbon-carbon triple bond contains one sigma and two pi bonds, totaling 3 sigma bonds.
- In ethylene (ethene, \(\text{C}_2\text{H}_4\)), the four C-H bonds are single sigma bonds and the carbon-carbon double bond contains one sigma and one pi bond, totaling 5 sigma bonds.
Why other options are incorrect:- Option B: Ethylene possesses 5 sigma bonds, not 3.
- Option C: Acetylene contains 3 sigma bonds, not 6.
- Option D: Reverses the order and miscounts sigma bonds in both hydrocarbons.
MCQ #86 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following has smallest bond angle?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Bond angles are determined by electron pair geometry and central atom hybridization; according to Drago's rule, hydrides of group 16 elements below oxygen utilize nearly pure p-orbitals, producing bond angles close to \(90^\circ\).
Formula / Rule / Reaction:$$\theta(\text{C}_2\text{H}_2) = 180^\circ, \quad \theta(\text{BeCl}_2) = 180^\circ, \quad \theta(\text{NH}_3) = 107.5^\circ, \quad \theta(\text{H}_2\text{S}) \approx 92.1^\circ$$
Solution:- Acetylene (\(\text{C}_2\text{H}_2\)) and beryllium chloride (\(\text{BeCl}_2\)) are linear molecules with \(\text{sp}\) hybridization and \(180^\circ\) bond angles.
- Ammonia (\(\text{NH}_3\)) exhibits \(\text{sp}^3\) hybridization with one lone pair, yielding a trigonal pyramidal angle of \(107.5^\circ\).
- Hydrogen sulfide (\(\text{H}_2\text{S}\)) involves a large sulfur atom with low electronegativity where bonding utilizes almost pure 3p orbitals, yielding an angle of approximately \(92.1^\circ\), the smallest among the choices.
Why other options are incorrect:- Option A: Acetylene has a linear bond angle of \(180^\circ\).
- Option B: Ammonia has a bond angle of \(107.5^\circ\), which is larger than that of \(\text{H}_2\text{S}\).
- Option D: Beryllium chloride has a linear bond angle of \(180^\circ\).
MCQ #87 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which one of the following has highest ionic character?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The percentage of ionic character in a chemical bond is directly proportional to the difference in electronegativity (\(\Delta \chi\)) between the bonded atoms, with values exceeding 1.7 typically denoting predominantly ionic bonds.
Formula / Rule / Reaction:$$\text{Pauling Electronegativities: } \chi(\text{Na}) = 0.93, \; \chi(\text{H}) = 2.20 \implies \Delta \chi = 1.27 \text{ (forming an ionic lattice of } \text{Na}^+ \text{ and } \text{H}^-)$$
Solution:- Sodium hydride (\(\text{NaH}\)) is an ionic crystalline salt containing discrete \(\text{Na}^+\) cations and \(\text{H}^-\) hydride anions with high melting and boiling points.
- By contrast, \(\text{BCl}_3\) and \(\text{PCl}_3\) are covalent molecular gases and liquids, while \(\text{AlCl}_3\) forms a covalent dimer (\(\text{Al}_2\text{Cl}_6\)) that readily sublimes, exhibiting strong covalent behavior due to the high charge density of \(\text{Al}^{3+}\).
Why other options are incorrect:- Option A: \(\text{AlCl}_3\) possesses significant covalent character due to polarization of chloride electron clouds by \(\text{Al}^{3+}\) (Fajans rules).
- Option B: \(\text{BCl}_3\) is a trigonal planar non-polar covalent molecule.
- Option C: \(\text{PCl}_3\) is a covalent molecular compound with polar covalent P-Cl bonds.
MCQ #88 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which one has highest bond energy?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Covalent bond energy increases with greater s-character of the overlapping hybrid orbital, because higher s-character brings the bonding electrons closer to the nucleus, yielding shorter and stronger bonds.
Formula / Rule / Reaction:$$\text{s-character: } \text{sp (50\%)} > \text{sp}^2\text{ (33.3\%)} > \text{sp}^3\text{ (25\%)}; \quad \text{Bond Energy } \propto \text{s-character}$$
Solution:- An \(\text{sp}\) hybrid orbital contains 50% s-character, making it smaller and more compact than \(\text{sp}^2\) (33.3%) or \(\text{sp}^3\) (25%) orbitals.
- When an \(\text{sp}\) orbital overlaps with an s-orbital, it achieves effective orbital overlap and a shorter internuclear distance, resulting in the highest bond dissociation energy.
Why other options are incorrect:- Option B: An \(\text{sp}^3\text{-s}\) bond has only 25% s-character, resulting in a longer, weaker bond.
- Option C: An \(\text{sp}^2\text{-s}\) bond contains 33.3% s-character, which produces a weaker bond than \(\text{sp-s}\).
- Option D: An \(\text{sp}^3\text{-p}\) bond involves directional p-orbitals with low s-character, producing a longer bond with lower dissociation energy.
MCQ #89 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The relationship of (+) tartaric acid and (-) Tartaric acid are:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Enantiomers are stereoisomers that are non-superimposable mirror images of each other, possessing identical physical properties except for the direction in which they rotate plane-polarized light.
Formula / Rule / Reaction:$$\text{Specific Rotation: } [\alpha]_D \text{ of (+) Tartaric acid} = -[\alpha]_D \text{ of (-) Tartaric acid}$$
Solution:- Tartaric acid possesses two chiral stereocenters (2R,3R and 2S,3S).
- The dextrorotatory (+) isomer and levorotatory (-) isomer represent non-superimposable mirror images that rotate plane-polarized light in equal and opposite directions, defining them as enantiomers.
Why other options are incorrect:- Option A: Geometric isomers (cis/trans) require restricted rotation, typically around double bonds or ring systems.
- Option C: Diastereomers are stereoisomers that are not mirror images of one another (such as meso-tartaric acid compared to (+) tartaric acid).
- Option D: Tartaric acid features tetrahedral \(\text{sp}^3\) chiral carbons and is a non-planar three-dimensional molecule.
MCQ #90 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The IUPAC name of compound C₆H₅COCH₃ is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Under IUPAC rules for aliphatic-aromatic ketones, the open carbon chain containing the principal carbonyl group serves as the parent alkane chain, with the aromatic ring designated as a substituent.
Formula / Rule / Reaction:$$\text{C}_6\text{H}_5-\text{C}(=\text{O})-\text{CH}_3 \implies \text{Parent: Ethane } (\text{C}_1-\text{C}_2) \rightarrow \text{Ethanone}; \quad \text{Substituent at } \text{C}_1: \text{Phenyl}$$
Solution:- The ketone carbonyl is part of a two-carbon aliphatic chain, establishing the parent stem as ethanone.
- The phenyl group (\(-\text{C}_6\text{H}_5\)) is attached to carbon-1 of this chain, giving the systematic IUPAC name 1-phenylethan-1-one (phenylethanone).
Why other options are incorrect:- Option A: Methyl phenyl ketone is the traditional common (radicofunctional) name, not the systematic IUPAC name.
- Option C: Octan-2-one is an unbranched eight-carbon aliphatic ketone.
- Option D: Benzylethanone implies an extra methylene unit (\(-\text{CH}_2-\text{C}_6\text{H}_5\)), representing 1-phenylpropan-2-one rather than acetophenone.
MCQ #91 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
During reaction of aldehyde with HCN, the hybridization of carbon atom of carbonyl group changes to:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Nucleophilic addition of hydrogen cyanide to a carbonyl group transforms the trigonal planar carbonyl carbon into a tetrahedral cyanohydrin carbon, altering its hybridization from sp² to sp³.
Formula / Rule / Reaction:$$\text{R-CH}=\text{O } (\text{sp}^2\text{, planar}) + \text{HCN} \xrightarrow{\text{OH}^-} \text{R-CH}(\text{OH})-\text{C}\equiv\text{N } (\text{sp}^3\text{, tetrahedral})$$
Solution:- In the parent aldehyde, the carbonyl carbon forms three sigma bonds and one pi bond, adopting trigonal planar geometry with sp² hybridization.
- Attack by the nucleophilic cyanide ion (\(\text{CN}^-\)) breaks the carbon-oxygen pi bond, generating a tetrahedral alkoxide intermediate with four sigma bonds and sp³ hybridization.
Why other options are incorrect:- Option A: The carbonyl carbon begins as sp², not linear sp.
- Option B: sp³ to sp² describes an elimination reaction producing a double bond, not nucleophilic addition.
- Option D: dsp² hybridization occurs in square planar transition metal complexes, not in main-group organic carbon atoms.
MCQ #92 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Acetone reacts with water in presence of catalyst produces:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Hydration of ketones involves reversible acid- or base-catalyzed nucleophilic addition of water across the carbonyl double bond, yielding a geminal diol (gem-diol).
Formula / Rule / Reaction:$$\text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 + \text{H}_2\text{O} \rightleftharpoons^+\text{ or OH}^-} \text{CH}_3-\text{C}(\text{OH})_2-\text{CH}_3\text{ (propane-2,2-diol)}$$
Solution:- Water acts as a weak nucleophile that adds to the electrophilic carbonyl carbon of propan-2-one (acetone).
- Proton transfer generates a geminal diol where two hydroxyl groups are attached to the same carbon (carbon-2), yielding propane-2,2-diol (acetone hydrate).
Why other options are incorrect:- Option A: Propan-2-ol is a secondary alcohol produced by catalytic hydrogenation or metal-hydride reduction of acetone, not hydration.
- Option B: Propan-1,2-diol is a vicinal diol produced by dihydroxylation of propene.
- Option D: Propanol (propan-1-ol) is a primary alcohol formed by reduction of propanal.
MCQ #93 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which one show positive test with Fehling solution:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Fehling solution is a mild oxidizing agent containing an alkaline bistartratocuprate(II) complex that oxidizes aliphatic aldehydes to carboxylic acid salts while precipitating brick-red copper(I) oxide.
Formula / Rule / Reaction:$$\text{CH}_3\text{CHO} + 2\text{Cu}^{2+} + 5\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{Cu}_2\text{O}\downarrow\text{ (brick-red)} + 3\text{H}_2\text{O}$$
Solution:- Acetaldehyde (ethanal) is an aliphatic aldehyde readily oxidized by Fehling solution, reducing copper(II) to insoluble red copper(I) oxide (\(\text{Cu}_2\text{O}\)).
- Ketones and aromatic aldehydes lack sufficient reducing power to reduce Fehling solution under standard alkaline test conditions.
Why other options are incorrect:- Option A: Acetophenone is an aromatic ketone that does not undergo oxidation with Fehling reagent.
- Option B: Metaformaldehyde (trioxane) is a cyclic trimer of formaldehyde lacking free oxidizable aldehyde functional groups.
- Option D: Benzaldehyde is an aromatic aldehyde; resonance stabilization of the formyl group by the benzene ring prevents reaction with Fehling solution.
MCQ #94 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Oxidation state of Tungsten in Na₂W₄O₁₃.H₂O is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The oxidation state of an element in a neutral coordination or polyoxometalate compound is calculated by equating the sum of the oxidation numbers of all constituent atoms to zero.
Formula / Rule / Reaction:$$\sum \text{Oxidation States} = 2(+1) + 4(x) + 13(-2) + 1(0) = 0$$
Solution:- In \(\text{Na}_2\text{W}_4\text{O}_{13}\cdot\text{H}_2\text{O}\), sodium has an oxidation state of +1, oxygen is -2, and neutral water of crystallization contributes 0.
- Setting up the algebraic equation: \(2(+1) + 4(x) + 13(-2) = 0 \implies 2 + 4x - 26 = 0\).
- Solving for \(x\): \(4x = 24 \implies x = +6\).
Why other options are incorrect:- Option A: An oxidation number of +5 would leave the polyoxometalate cluster with an uncompensated negative charge of -4.
- Option C: +8 exceeds the maximum possible valence shell oxidation state for tungsten (group 6 transition metal).
- Option D: Zero oxidation state is found only in elemental tungsten metal (\(\text{W}^0\)).
MCQ #95 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Orientation of reacting molecules is important in collision theory in the following cases EXCEPT:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:According to transition state and collision theory, the steric orientation factor \(p\) governs whether colliding species collide with their reactive centers aligned; at very low pressure, collision frequency rather than complex steric orientation dominates the rate-determining behavior.
Formula / Rule / Reaction:$$k = p \cdot Z \cdot e^{-E_a / RT} \quad (p = \text{steric or orientation factor})$$
Solution:- Steric orientation is critical for polyatomic and structurally complex molecules, as well as under high-density/high-pressure regimes where collisions are frequent and specific alignment dictates reaction success.
- Under low pressure conditions, the mean free path is large and the overall reaction rate is limited primarily by raw collision frequency rather than stereochemical orientation constraints.
Why other options are incorrect:- Option A: At high pressure, collision rates are high, making correct spatial orientation the primary discriminator for successful activation.
- Option B: Complex molecules have localized reactive sites, making steric orientation mandatory for product formation.
- Option C: Polyatomic molecules possess multiple non-reactive bonds, requiring collisions at precise orientations to initiate bond rearrangement.
MCQ #96 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Keeping temperature constant, if pressure is increased, density of gas will:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:From the ideal gas equation of state, the density of an ideal gas is directly proportional to its absolute pressure at constant temperature.
Formula / Rule / Reaction:$$P M = \rho R T \implies \rho = \frac{P M}{R T} \implies \rho \propto P \quad (\text{at constant } T)$$
Solution:- When pressure increases at constant temperature, gas molecules are forced closer together, decreasing the volume according to Boyle's law.
- Because mass remains constant, this reduction in volume increases the gas density (\(\rho = m / V\)).
Why other options are incorrect:- Option B: Density decreases when pressure is lowered or when temperature is raised at constant pressure.
- Option C: Density doubles only if pressure is doubled; for an unspecified pressure increase, stating it doubles is an unsupported assumption.
- Option D: Density remains constant only if mass and volume remain fixed in a rigid closed container.
MCQ #97 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
IUPAC name of chloroform is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Haloalkanes are named by prefixing the halogen substituent names onto the parent alkane chain; a one-carbon alkane substituted with three chlorine atoms is systematic trichloromethane.
Formula / Rule / Reaction:$$\text{CHCl}_3 \implies \text{Parent: Methane } (\text{CH}_4) \rightarrow \text{Substituents: Three chlorines} \rightarrow \text{Trichloromethane}$$
Solution:- Chloroform possesses the chemical formula \(\text{CHCl}_3\).
- The single carbon defines the parent alkane as methane, and the three chlorine atoms are designated by the prefix 'trichloro-', giving the IUPAC name trichloromethane.
Why other options are incorrect:- Option A: Methyl chloride is the common name for chloromethane (\(\text{CH}_3\text{Cl}\)).
- Option B: Methyltrichloride is a non-standard, incorrect hybrid term.
- Option D: Chloromethane is the IUPAC name for monochlorinated methane (\(\text{CH}_3\text{Cl}\)).
MCQ #98 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which among the following will react with ammonical cuprous chloride:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Terminal alkynes possess an acidic sp-hybridized terminal hydrogen that reacts with ammoniacal cuprous chloride to yield a characteristic red-brown copper acetylide precipitate.
Formula / Rule / Reaction:$$\text{CH}_3\text{CH}_2-\text{C}\equiv\text{CH} + [\text{Cu}(\text{NH}_3)_2]\text{Cl} \rightarrow \text{CH}_3\text{CH}_2-\text{C}\equiv\text{C-Cu}\downarrow\text{ (red-brown)} + \text{NH}_4\text{Cl} + \text{NH}_3$$
Solution:- 1-Butyne is a 1-alkyne (terminal alkyne) containing a weakly acidic terminal acetylenic hydrogen atom (\(\equiv\text{C}-\text{H}\)).
- In the presence of ammoniacal cuprous chloride (\([\text{Cu}(\text{NH}_3)_2]^+\)), the terminal proton is abstracted, precipitating cuprous 1-butynide as a red-brown solid.
Why other options are incorrect:- Option A: 1-Butene is an alkene and lacks acidic acetylenic protons, showing no reaction.
- Option C: 2-Butene is an alkene and does not react with ammoniacal copper(I) reagent.
- Option D: 2-Butyne is an internal alkyne (\(\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3\)) lacking terminal acetylenic hydrogens.
MCQ #99 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following is not a crystalline solid?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Crystalline solids exhibit long-range periodic order with sharp melting points, whereas amorphous solids lack regular three-dimensional atomic lattice symmetry.
Formula / Rule / Reaction:Qualitative concept / Solid-state structural classification.Solution:- Glass is an amorphous supercooled liquid consisting of an irregular, disordered network of silicate tetrahedra without long-range periodicity.
- It softens gradually over a broad temperature range rather than exhibiting a sharp, well-defined melting point.
Why other options are incorrect:- Option A: Potassium chloride (KCl) is an ionic crystalline solid with a face-centered cubic lattice.
- Option B: Iron metal (Fe) is a crystalline metallic solid adopting a body-centered cubic lattice at room temperature.
- Option D: Rhombic sulfur is a molecular crystalline allotrope composed of puckered \(\text{S}_8\) rings packed into an orthorhombic lattice.
MCQ #100 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Number of electron with l=2 in an atom having atomic number 23 is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The azimuthal quantum number \(l = 2\) designates d-subshell orbitals; the count of electrons with \(l = 2\) corresponds to the occupancy of d-orbitals in the ground-state electronic configuration.
Formula / Rule / Reaction:$$\text{Vanadium } (Z = 23): 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^3\, 4s^2 \quad (l=0\text{ for } s, \; l=1\text{ for } p, \; l=2\text{ for } d)$$
Solution:- An atom with atomic number 23 is vanadium (V).
- Following the Aufbau principle, its electronic configuration is \([\text{Ar}]\,3d^3\,4s^2\).
- Electrons characterized by \(l = 2\) are located exclusively in the 3d subshell, which contains exactly 3 electrons.
Why other options are incorrect:- Option A: 2 is the number of electrons in the 4s subshell (where \(l = 0\)).
- Option C: 4 electrons with \(l = 2\) would correspond to chromium's theoretical unpromoted configuration.
- Option D: 5 electrons with \(l = 2\) corresponds to manganese (\([\text{Ar}]\,3d^5\,4s^2\)).
MCQ #101 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Reaction having high activation energy are:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:According to the Arrhenius equation, the reaction rate constant depends exponentially on the activation energy; high activation energy barriers result in smaller rate constants and slower reaction velocities.
Formula / Rule / Reaction:$$k = A e^{-E_a / RT} \implies \uparrow E_a \implies \downarrow k \implies \text{Slow reaction}$$
Solution:- Activation energy (\(E_a\)) represents the minimum kinetic energy reactant molecules must possess to overcome the transition state barrier.
- A high activation energy means only a tiny fraction of colliding molecules possess sufficient energy, causing the overall reaction to proceed slowly at standard temperatures.
Why other options are incorrect:- Option A: Spontaneity is governed by Gibbs free energy change (\(\Delta G < 0\)), not kinetic activation energy.
- Option B: Fast reactions are characterized by low activation energies where many collisions are fruitful.
- Option D: High activation energies occur in both exothermic reactions (e.g., combustion of coal) and endothermic reactions.
MCQ #102 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following reagents is used to convert benzene to 2-chloro methyl benzene?
A
CH₃Cl/AlCl₃ followed by Cl₂ in the presence of anhydrous FeCl₃
B
CH₃Cl/AlCl₃ followed by Cl₂ in the presence of diffused sunlight
C
Cl₂/FeCl₃ followed by CH₃Cl in the presence of diffused sunlight
D
Cl₂/UV followed by CH₃Cl in the presence of anhydrous FeCl₃
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Synthesis of 2-chlorotoluene (2-chloro methyl benzene) requires introducing a methyl directing group via Friedel-Crafts alkylation, followed by ortho-directed electrophilic aromatic chlorination using a Lewis acid catalyst.
Formula / Rule / Reaction:$$\text{Step 1: } \text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{anhydrous AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3\text{ (Toluene)} + \text{HCl}$$
$$\text{Step 2: } \text{C}_6\text{H}_5\text{CH}_3 + \text{Cl}_2 \xrightarrow{\text{anhydrous FeCl}_3} \text{o-Chlorotoluene} + \text{p-Chlorotoluene} + \text{HCl}$$
Solution:- Alkylation of benzene with \(\text{CH}_3\text{Cl}/\text{AlCl}_3\) yields toluene.
- The methyl group is activating and ortho/para-directing; subsequent chlorination with \(\text{Cl}_2\) and \(\text{FeCl}_3\) directs chlorine to the ortho position to yield 2-chlorotoluene.
Why other options are incorrect:- Option B: \(\text{Cl}_2\) in diffused sunlight causes free-radical side-chain halogenation of toluene, producing benzyl chloride (\(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\)).
- Option C: Chlorinating benzene first gives chlorobenzene; chlorine is deactivating and makes subsequent Friedel-Crafts alkylation slow and inefficient.
- Option D: \(\text{Cl}_2/\text{UV}\) causes free-radical addition to benzene, producing benzene hexachloride.
MCQ #103 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
According to the Kinetic molecular theory, the kinetic energy of gas molecules increases when they are
A
Melted from solid to liquid state
B
Mixed with other molecules at lower temperature
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to Kinetic Molecular Theory, the average translational kinetic energy of molecules is directly proportional to absolute temperature; phase transitions that absorb heat energy increase molecular kinetic motion.
Formula / Rule / Reaction:$$\overline{KE} = \frac{3}{2} k_B T \implies \overline{KE} \propto T$$
Solution:- Melting is an endothermic phase transition where thermal energy is absorbed by the substance.
- This absorbed thermal energy disrupts the rigid crystal lattice and increases the translational and vibrational kinetic energy of the molecules.
Why other options are incorrect:- Option B: Mixing with molecules at lower temperatures decreases average kinetic energy via thermal equilibration.
- Option C: Freezing releases latent heat, decreasing molecular kinetic energy.
- Option D: Condensation is an exothermic phase change that lowers molecular kinetic energy.
MCQ #104 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
If the equilibrium expression of a reversible reaction is [C]²/[A][B]. The balanced chemical equation should be:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:According to the Law of Mass Action, the equilibrium constant expression \(K_c\) places product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of its stoichiometric coefficient.
Formula / Rule / Reaction:$$a\text{A} + b\text{B} \rightleftharpoons c\text{C} \implies K_c = \frac{[\text{C}]^c}{[\text{A}]^a [\text{B}]^b}$$
Solution:- The given expression is \(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]}\).
- The numerator \([\text{C}]^2\) indicates that 2 moles of product C are formed, while the denominator \([\text{A}][\text{B}]\) indicates that 1 mole of A and 1 mole of B are consumed.
- Matching stoichiometry yields the balanced equation: \(\text{A} + \text{B} \rightleftharpoons 2\text{C}\).
Why other options are incorrect:- Option A: For \(\text{C} \rightleftharpoons \text{A} + \text{B}\), the expression would be \(K_c = [\text{A}][\text{B}] / [\text{C}]\).
- Option B: For \(2\text{C} \rightleftharpoons \text{A} + \text{B}\), the expression would be \(K_c = [\text{A}][\text{B}] / [\text{C}]^2\).
- Option C: For \(\text{A} + \text{B} \rightleftharpoons \text{C}\), the expression would be \(K_c = [\text{C}] / ([\text{A}][\text{B}])\).
MCQ #105 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Common name of 1,3 benzene diol is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Benzenediols are aromatic dihydroxy compounds with three positional isomers: 1,2-benzenediol (catechol), 1,3-benzenediol (resorcinol), and 1,4-benzenediol (hydroquinone/quinol).
Formula / Rule / Reaction:$$\text{C}_6\text{H}_4(\text{OH})_2 \quad (1,3\text{-position}) \implies \text{Resorcinol}$$
Solution:- 1,3-Benzenediol has two hydroxyl groups attached meta to each other on the benzene ring.
- Its IUPAC-accepted common trivial name is resorcinol.
Why other options are incorrect:- Option A: Pyrogallol is 1,2,3-trihydroxybenzene, a benzenetriol.
- Option B: Cresol refers to methylphenols (hydroxytoluenes), containing one \(-\text{OH}\) and one \(-\text{CH}_3\) group.
- Option D: Catechol is the common name for 1,2-benzenediol (ortho isomer).
MCQ #106 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
How many sigma and pi bonds are present in maleic anhydride?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Maleic anhydride is a cyclic dicarboxylic acid anhydride (\(\text{C}_4\text{H}_2\text{O}_3\)); its total sigma bonds equal the sum of its single bonds plus one per double bond, and its pi bonds correspond to the double bonds.
Formula / Rule / Reaction:$$\text{Structure: Ring formed by } \text{O}-\text{C}(=\text{O})-\text{CH}=\text{CH}-\text{C}(=\text{O})$$
Solution:- Pi (\(\pi\)) bonds: One in the \(\text{C}=\text{C}\) bond and one in each of the two \(\text{C}=\text{O}\) carbonyl groups, giving \(1 + 2 = 3\,\pi\) bonds.
- Sigma (\(\sigma\)) bonds: Two \(\text{C}-\text{H}\) single bonds (2), two \(\text{C}-\text{C}\) single bonds (2), two \(\text{C}-\text{O}\) ring single bonds (2), one \(\sigma\) from \(\text{C}=\text{C}\) (1), and two \(\sigma\) from the two \(\text{C}=\text{O}\) groups (2), giving \(2 + 2 + 2 + 1 + 2 = 9\,\sigma\) bonds.
Why other options are incorrect:- Option A: Omits the sigma components of ring single bonds and \(\text{C}-\text{H}\) bonds.
- Option B: Neglects the sigma bonds in the carbonyl linkages.
- Option D: Overcounts total sigma bonds by treating pi bonds as additional sigma linkages.
MCQ #107 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Addition of three molecules of chlorine into benzene prove that benzene has
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Under ultraviolet radiation, benzene undergoes exhaustive free-radical addition of three halogen molecules, confirming the presence of three conjugated double bonds in its Kekule representation.
Formula / Rule / Reaction:$$\text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{h\nu \text{ / UV}} \text{C}_6\text{H}_6\text{Cl}_6\text{ (Benzene hexachloride / Gammaxene)}$$
Solution:- Each molecule of chlorine (\(\text{Cl}_2\)) adds across one carbon-carbon double bond.
- The consumption of exactly three chlorine molecules to form saturated 1,2,3,4,5,6-hexachlorocyclohexane proves that benzene possesses three pairs of shared pi electrons (three double bonds).
Why other options are incorrect:- Option B: While benzene is nonpolar due to symmetry, stoichiometry of chlorine addition does not establish polarity.
- Option C: Benzene is a nonpolar hydrocarbon.
- Option D: Benzene is a planar aromatic ring with sp² hybridization.
MCQ #108 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Propyne is allowed to react with hydrochloric acid two times then final product is
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electrophilic addition of hydrogen halides to unsymmetrical alkynes follows Markovnikov's rule in both sequential steps, directing both halogen atoms to the more substituted internal carbon.
Formula / Rule / Reaction:$$\text{CH}_3-\text{C}\equiv\text{CH} + \text{HCl} \rightarrow \text{CH}_3-\text{C}(\text{Cl})=\text{CH}_2\text{ (2-chloropropene)}$$
$$\text{CH}_3-\text{C}(\text{Cl})=\text{CH}_2 + \text{HCl} \rightarrow \text{CH}_3-\text{C}(\text{Cl})_2-\text{CH}_3\text{ (2,2-dichloropropane)}$$
Solution:- In step 1, the proton electrophile adds to carbon-1 to form the more stable secondary carbocation, followed by chloride attack to give 2-chloropropene.
- In step 2, Markovnikov addition repeats because carbocation intermediate stability is enhanced by resonance donation from the chlorine lone pair, directing the second chlorine to carbon-2 to yield 2,2-dichloropropane.
Why other options are incorrect:- Option A: 1,1-Dichloropropane requires anti-Markovnikov addition in both steps.
- Option B: 1,2-Dichloropropane requires addition of chlorine gas (\(\text{Cl}_2\)) followed by hydrogenation, not hydrochlorination.
- Option D: 1,2-Dichloropropene is an intermediate monochloroalkene that forms from halogenation, not the saturated product of two HCl additions.
MCQ #109 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Diethyl ether and Methyl n-propyl ether are
B
Functional group isomers
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Metamerism is constitutional isomerism exhibited by compounds having the same functional group and molecular formula but differing in the alkyl groups attached to the central polyvalent heteroatom.
Formula / Rule / Reaction:$$\text{Diethyl ether: } \text{C}_2\text{H}_5-\text{O}-\text{C}_2\text{H}_5 \quad \text{vs.} \quad \text{Methyl n-propyl ether: } \text{CH}_3-\text{O}-\text{CH}_2\text{CH}_2\text{CH}_3$$
Solution:- Both ethers share the identical molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) and the same ether functional group (\(-\text{O}-\)).
- They differ in the distribution of carbon atoms around the divalent oxygen atom (ethyl + ethyl vs. methyl + propyl), classifying them as metamers.
Why other options are incorrect:- Option A: Position isomers possess the same carbon skeleton but differ in the position of a substituent or functional group along the chain.
- Option B: Functional group isomers possess different functional groups (e.g., diethyl ether and butan-1-ol).
- Option D: Tautomers are constitutional isomers that interconvert rapidly through dynamic proton transfer.
MCQ #110 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
In the given reaction CH₃OH → HCOOH, oxidation state of carbon changes from:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The oxidation state of carbon in an organic compound is determined by assigning standard oxidation numbers to attached atoms (+1 for H, -2 for O) and balancing the total molecular charge to zero.
Formula / Rule / Reaction:$$\text{In } \text{CH}_3\text{OH}: x + 4(+1) + 1(-2) = 0 \implies x = -2$$
$$\text{In } \text{HCOOH}: x + 2(+1) + 2(-2) = 0 \implies x = +2$$
Solution:- In methanol (\(\text{CH}_3\text{OH}\)), carbon is bonded to three hydrogens, one oxygen, and shares a proton: \(x + 4 - 2 = 0 \implies x = -2\).
- In formic acid (\(\text{HCOOH}\)), carbon is bonded to one hydrogen, one hydroxyl group, and double-bonded to oxygen: \(x + 2 - 4 = 0 \implies x = +2\).
- The net change in carbon's oxidation state is from -2 to +2 (loss of 4 electrons).
Why other options are incorrect:- Option A: 0 is the oxidation state of carbon in formaldehyde (HCHO).
- Option C: -3 is the oxidation state of carbon in ethane (\(\text{C}_2\text{H}_6\)).
- Option D: Carbon in methanol has an oxidation state of -2, not 0.
MCQ #111 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
For endothermic reaction, enthalpy change is:
D
ΔHreactant = ΔHproduct = 0
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:An endothermic reaction absorbs heat from its surroundings, yielding products that possess a higher total enthalpy content than the reactants.
Formula / Rule / Reaction:$$\Delta H_{\text{rxn}} = H_{\text{products}} - H_{\text{reactants}} > 0 \implies H_{\text{reactants}} < H_{\text{products}}$$
Solution:- Because system enthalpy increases as thermal energy is converted into chemical potential energy, \(\Delta H\) is positive.
- This thermodynamic requirement means the enthalpy of reactants is less than the enthalpy of products (\(H_{\text{reactant}} < H_{\text{product}}\)).
Why other options are incorrect:- Option A: \(H_{\text{reactant}} > H_{\text{product}}\) defines an exothermic reaction where enthalpy is released.
- Option C: Equal reactant and product enthalpy means \(\Delta H = 0\), describing a thermoneutral process.
- Option D: Absolute zero enthalpy is not characteristic of chemical chemical systems.
MCQ #112 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The structure of four Chloroalkanes W, X, Y and Z are shown:
W: (CH₃)₂CHCH(Cl)CH₃
X: CH₂(Cl)CH₃
Y: (CH₃)₃CCH₂Cl
Z: (CH₃)₂CHC(CH₃)₂Cl
Select the Chloroalkane classified as tertiary:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A tertiary (3°) chloroalkane is one in which the halogen-bearing carbon atom is directly bonded to three other carbon atoms.
Formula / Rule / Reaction:$$\text{Tertiary Carbon: } \text{R}_3\text{C}-\text{Cl} \quad (\text{bonded to 3 alkyl carbons})$$
Solution:- In compound Z, \((\text{CH}_3)_2\text{CHC}(\text{CH}_3)_2\text{Cl}\), the chlorine-bearing carbon is directly bonded to two methyl groups and one isopropyl group (three carbon atoms), making it tertiary.
- W, \((\text{CH}_3)_2\text{CHCH(Cl)CH}_3\), has its chlorine attached to a secondary carbon bonded to two carbons.
- X (chloroethane) and Y (neopentyl chloride) have chlorines attached to primary carbons bonded to only one carbon.
Why other options are incorrect:- Option A: W is a secondary (2°) chloroalkane.
- Option B: X is a primary (1°) chloroalkane.
- Option C: Y is a primary (1°) chloroalkane with a bulky tert-butyl substituent.
MCQ #113 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The smallest repeating pattern from which the lattice is built in a crystalline solid is called
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The unit cell is the fundamental geometric structural unit of a crystalline solid that, when repeated translationally in three dimensions, generates the entire macroscopic crystal lattice.
Formula / Rule / Reaction:Qualitative concept / Crystallography.Solution:- A crystal lattice is an infinite three-dimensional array of points in space.
- The unit cell is the smallest repeating parallel-piped block displaying the full symmetry and stoichiometry of the chemical crystal.
Why other options are incorrect:- Option A: A crystallite is a microscopic domain or single-crystal grain in a polycrystalline material.
- Option B: An amorphous region lacks regular repeating structure.
- Option D: The crystal lattice is the complete three-dimensional network produced by repeating the unit cell.
MCQ #114 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following reaction will move backward by decreasing pressure?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to Le Chatelier's principle, decreasing pressure shifts a gaseous equilibrium toward the side with the greater number of gaseous moles.
Formula / Rule / Reaction:$$\Delta n_g = n_{\text{products}} - n_{\text{reactants}}; \quad \downarrow P \implies \text{Shift toward side with larger } n_g$$
Solution:- In reaction A (\(4\text{Q} \rightleftharpoons \text{T} + \text{S}\)), the reactant side has 4 moles of gas while the product side has 2 moles of gas.
- Decreasing pressure favors the side with more gaseous moles; therefore, the system shifts backward toward reactants (left) to relieve the decrease in pressure.
Why other options are incorrect:- Option B: \(2\text{A} \rightleftharpoons 3\text{C}\) has more moles on the product side (3 vs. 2), so decreasing pressure shifts it forward.
- Option C: \(\text{K} \rightleftharpoons \text{L}\) has equal moles on both sides (1 vs. 1), making it insensitive to pressure changes.
- Option D: \(\text{H} + 2\text{M} \rightleftharpoons 2\text{B} + 2\text{D}\) has 3 reactant moles and 4 product moles, shifting forward when pressure is decreased.
MCQ #115 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
According to Le Chatelier's principle, in Haber's process yield of ammonia can be increased by:
A
Decreasing pressure and increasing temperature
B
Increasing pressure and decreasing temperature
C
Increasing volume and decreasing temperature
D
Decreasing concentration of H₂
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The synthesis of ammonia via the Haber process is an exothermic gas reaction accompanied by a reduction in total gaseous moles; higher pressure and lower temperature shift equilibrium toward product formation.
Formula / Rule / Reaction:$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad (\Delta H = -92.4\text{ kJ/mol}, \; \Delta n_g = -2)$$
Solution:- Because 4 moles of reactants yield 2 moles of product, increasing total pressure shifts the system forward to minimize volume.
- Because the forward reaction is exothermic, lowering the operating temperature shifts equilibrium toward the right, increasing the equilibrium yield of \(\text{NH}_3\).
Why other options are incorrect:- Option A: Decreasing pressure shifts equilibrium backward, while increasing temperature favors the endothermic decomposition of ammonia.
- Option C: Increasing volume is equivalent to lowering pressure, which reduces ammonia yield.
- Option D: Decreasing reactant concentration (\(\text{H}_2\)) shifts the equilibrium backward toward reactants.
MCQ #116 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Hyperbola isotherm is obtained when graph is plotted between:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Boyle's law establishes that gas pressure is inversely proportional to volume at constant temperature, producing a rectangular hyperbola when pressure is plotted directly against volume.
Formula / Rule / Reaction:$$P \cdot V = k \implies P = \frac{k}{V} \quad (xy = c, \text{ standard equation of a rectangular hyperbola})$$
Solution:- An isotherm represents thermodynamic state variables measured at constant temperature.
- Plotting \(P\) along the vertical axis against \(V\) along the horizontal axis produces a smooth, asymptotic curved hyperbolic isotherm.
Why other options are incorrect:- Option A: Plotting \(P\) against \(1/V\) yields a straight line passing through the origin.
- Option B: Plotting \(PV\) against \(V\) yields a horizontal straight line parallel to the volume axis.
- Option D: Plotting \(V\) against \(1/P\) produces a linear relationship passing through the origin.
MCQ #117 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Highest energy for one quantum among radiation is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The energy of a single photon quantum is directly proportional to its electromagnetic frequency and inversely proportional to its wavelength.
Formula / Rule / Reaction:$$E = h \nu = \frac{h c}{\lambda} \implies E_{\text{UV}} > E_{\text{visible}} > E_{\text{IR}} > E_{\text{microwave}}$$
Solution:- Ultraviolet radiation has wavelengths ranging from 10 to 400 nm, corresponding to higher frequencies and energies per photon than visible, infrared, or microwave radiation.
- Because ultraviolet radiation has the shortest wavelength among the listed choices, each of its quanta carries the highest energy.
Why other options are incorrect:- Option A: Microwaves have long wavelengths (millimeters to centimeters) and very low quantum energies.
- Option B: Infrared radiation has lower frequencies and lower photon energies than visible and UV light.
- Option D: Visible light (400 to 700 nm) has longer wavelengths and lower photon energies than ultraviolet radiation.
MCQ #118 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following reagent is commonly used to prepare alkyne form vicinal dihalide?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Vicinal dihalides undergo sequential double dehydrohalogenation in the presence of strong bases to eliminate two equivalents of hydrogen halide, producing an alkyne.
Formula / Rule / Reaction:$$\text{R-CH(Br)-CH}_2\text{Br} \xrightarrow{\text{alcoholic KOH, } \Delta} \text{R-C(Br)}=\text{CH}_2 \xrightarrow{\text{NaNH}_2} \text{R-C}\equiv\text{CH} + 2\text{HBr}$$
Solution:- Initial dehydrohalogenation with hot alcoholic potassium hydroxide (KOH) eliminates one molecule of \(\text{HX}\) to yield a haloalkene.
- Because the haloalkene is less reactive toward elimination, the stronger base sodamide (\(\text{NaNH}_2\)) is employed to eliminate the second \(\text{HX}\) and generate the carbon-carbon triple bond.
Why other options are incorrect:- Option A: Hydrogen gas with a metal catalyst reduces alkynes to alkanes, rather than synthesizing alkynes.
- Option C: Potassium permanganate (\(\text{KMnO}_4\)) is an oxidizing agent that cleaves double and triple bonds.
- Option D: Lithium aluminium hydride reduces carbonyls and alkyl halides to alkanes.
MCQ #119 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
In a standard hydrogen electrode, the platinum electrode:
B
Provides a surface for oxidation and reduction
D
Provides voltage to the circuit
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In the Standard Hydrogen Electrode (SHE), platinum functions as an inert conducting catalyst that facilitates electron transfer between adsorbed hydrogen gas and hydrogen ions without reacting chemically.
Formula / Rule / Reaction:$$2\text{H}^+(aq, 1\text{ M}) + 2e^- \rightleftharpoons} \text{H}_2(g, 1\text{ atm}) \quad (E^\circ = 0.00\text{ V})$$
Solution:- Platinum foil coated with finely divided platinum black provides a large surface area for the adsorption of \(\text{H}_2\) molecules.
- It serves as an electron conductor facilitating oxidation (\(\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-\)) or reduction (\(2\text{H}^+ + 2e^- \rightarrow \text{H}_2\)), establishing dynamic redox equilibrium.
Why other options are incorrect:- Option A: Platinum is a noble, chemically inert metal that does not react with hydrogen gas.
- Option C: A salt bridge consists of an electrolyte gel (e.g., KCl in agar) that maintains electrical neutrality between half-cells.
- Option D: The standard hydrogen electrode is arbitrarily assigned a reference potential of 0.00 V and does not supply voltage on its own.
MCQ #120 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which of the following compound has lower Lattice Energy?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Lattice energy of an ionic crystal is directly proportional to ionic charge product and inversely proportional to internuclear separation distance between the ions.
Formula / Rule / Reaction:$$U \propto \frac{|z^+ \cdot z^-|}{r_c + r_a}$$
Solution:- Between LiCl and KCl, both compounds have monovalent ions (\(+1\) and \(-1\)), but the ionic radius of potassium (\(\text{K}^+\)) is significantly larger than that of lithium (\(\text{Li}^+\)).
- The larger internuclear distance in KCl reduces electrostatic attraction, giving KCl a lower lattice energy than LiCl.
- \(\text{CaCl}_2\) contains a divalent cation (\(\text{Ca}^{2+}\)), resulting in a much higher lattice energy.
Why other options are incorrect:- Option A: LiCl has a smaller cation radius than KCl, resulting in higher lattice energy.
- Option C: \(\text{CaCl}_2\) possesses a divalent \(\text{Ca}^{2+}\) cation, giving it a much higher lattice energy.
- Option D: Glucose is a covalent molecular solid held by intermolecular forces, not an ionic lattice.
MCQ #121 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Reaction of 2-Bromobutane with alcoholic KOH yields:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Reaction of a haloalkane with hot alcoholic potassium hydroxide induces dehydrohalogenation via an E2 mechanism, selectively producing the more substituted alkene as the major product (Saytzeff's rule).
Formula / Rule / Reaction:$$\text{CH}_3-\text{CH}_2-\text{CH(Br)}-\text{CH}_3 + \text{alcoholic KOH} \xrightarrow{\Delta} \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\text{ (2-butene, 80\%)} + \text{KBr} + \text{H}_2\text{O}$$
Solution:- Alcoholic KOH acts as a strong base that abstracts a beta-hydrogen from carbon-3 while bromide departs from carbon-2.
- According to Saytzeff's rule, beta-elimination yields the more substituted and thermodynamically stable alkene (but-2-ene) as the major product.
Why other options are incorrect:- Option A: 2-Butanol is formed via nucleophilic substitution (\(\text{S}_\text{N}2\)) using aqueous KOH, not alcoholic KOH.
- Option C: Formation of 2-butyne requires a dihaloalkane precursor undergoing double elimination.
- Option D: 1-Butanol is synthesized by nucleophilic substitution of 1-bromobutane.
MCQ #122 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which species contain an element with an oxidation number of +4?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The oxidation state of an atom in an oxoanion or neutral salt is determined using known oxidation rules (+1 for alkali metals, -2 for oxygen).
Formula / Rule / Reaction:$$\text{In } \text{Na}_2\text{CO}_3: 2(+1) + x + 3(-2) = 0 \implies 2 + x - 6 = 0 \implies x = +4$$
Solution:- In sodium carbonate (\(\text{Na}_2\text{CO}_3\)), sodium contributes +2 and three oxygens contribute -6.
- To maintain electrical neutrality, carbon must have an oxidation state of +4.
Why other options are incorrect:- Option A: In chromate (\(\text{CrO}_4^{2-}\)), chromium has an oxidation state of +6: \(x + 4(-2) = -2 \implies x = +6\).
- Option B: In manganate (\(\text{MnO}_4^{2-}\)), manganese has an oxidation state of +6: \(x + 4(-2) = -2 \implies x = +6\).
- Option C: In sulfuric acid (\(\text{H}_2\text{SO}_4\)), sulfur has an oxidation state of +6: \(2(+1) + x + 4(-2) = 0 \implies x = +6\).
MCQ #123 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The rate of reaction for a zero-order reaction:
A
Increases as the reaction proceeds
B
Decreases as the reaction proceeds
C
Remains the same as the reaction proceeds
D
May decrease or increase as the reaction proceeds
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:In a zero-order reaction, the reaction rate is independent of the concentrations of the reactants, remaining constant throughout the course of the reaction until the reactants are exhausted.
Formula / Rule / Reaction:$$\text{Rate} = k [A]^0 = k = \text{constant}$$
Solution:- The differential rate law for a zero-order reaction is \(-\frac{d[A]}{dt} = k\).
- Because the rate constant \(k\) depends only on temperature and catalyst presence, the velocity of the reaction remains unchanged over time as reactant is consumed.
Why other options are incorrect:- Option A: Reaction rate does not increase unless catalyzed autocatalytically or temperature is raised.
- Option B: A decreasing rate over time is characteristic of first-order, second-order, and higher-order reactions where rate depends on concentration.
- Option D: The rate of an isothermal zero-order reaction is constant, not variable.
MCQ #124 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
The order of chemical reaction is defined as:
A
The number of reactants involved in balanced equation
B
The number of products formed in balanced equation
C
The power to which the concentration of a reactant is raised in the rate equation
D
The rate constant of the reaction
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The order of a chemical reaction with respect to a given reactant is the exponent to which its concentration term is raised in the experimentally determined rate equation.
Formula / Rule / Reaction:$$\text{Rate} = k [A]^m [B]^n \implies m = \text{order with respect to } A; \quad (m+n) = \text{overall order}$$
Solution:- The rate equation expresses reaction velocity as a function of reactant concentrations raised to empirical powers.
- These powers (exponents) represent the reaction orders, which are experimentally determined and cannot be deduced simply from balanced stoichiometric coefficients.
Why other options are incorrect:- Option A: The stoichiometric count of reactants in a balanced chemical equation defines stoichiometric molecularity, not kinetic reaction order.
- Option B: The number of products formed does not determine reaction order.
- Option D: The rate constant (\(k\)) is the proportionality constant in the rate law, not the order.
MCQ #125 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Real gases show more positive deviation as pressure increases because of
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:At high pressures, gas molecules are forced close together, making intermolecular repulsive forces and finite molecular volume dominant, which drives the compressibility factor \(Z > 1\) (positive deviation).
Formula / Rule / Reaction:$$Z = \frac{PV}{nRT} > 1 \quad (\text{dominated by the excluded volume parameter } b \text{ and repulsive forces})$$
Solution:- At low to moderate pressures, attractive forces dominate, pulling molecules together and producing negative deviations (\(Z < 1\)).
- At high pressures, the physical volume occupied by molecules becomes significant relative to container volume, and electronic repulsion resists further compression, causing the molar volume to exceed ideal predictions (positive deviation).
Why other options are incorrect:- Option A: Real gases consist of neutral covalent atoms or molecules, not ionic species.
- Option C: Attractive forces cause negative deviations (\(Z < 1\)) at moderate pressures.
- Option D: Nonpolar gases experience dispersion forces, but polarity is not the primary cause of high-pressure positive deviation.
MCQ #126 of 180
Chemistry
SZABMU 2025
[SZABMU 2025]
Which statement applies to both ideal and real gases
A
Collision between molecules are elastic
B
Molecules are in constant random motion
C
Molecules attract each other
D
Molecules have zero size
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:A fundamental postulate of Kinetic Molecular Theory that applies to all gases is that gas molecules are in continuous, random, chaotic motion in all directions.
Formula / Rule / Reaction:Fundamental Postulate of Gas Kinetics: Continuous, random, Brownian translational motion.Solution:- Thermal energy ensures that particles of both ideal and real gases undergo continuous, rapid, straight-line random motion interrupted only by collisions.
- This property is universal across ideal models and real fluids.
Why other options are incorrect:- Option A: Perfectly elastic collisions without net kinetic energy loss are an idealized assumption; real molecular collisions involve temporary deformation and energy redistribution.
- Option C: Intermolecular attractive forces exist only in real gases, being neglected in ideal gases.
- Option D: Zero molecular volume is an idealized assumption; real gas particles occupy a finite volume.
MCQ #127 of 180
Physics
SZABMU 2025
[SZABMU 2025]
If voltage across a resistor is doubled then its resistance will become: (temperature and physical state of the conductor remains constant).
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Electrical resistance of an ohmic conductor is an intrinsic geometric and material property defined by resistivity, length, and cross-sectional area, remaining independent of applied voltage at constant temperature.
Formula / Rule / Reaction:$$R = \rho \frac{L}{A} \quad (\text{independent of } V \text{ and } I \text{ for an ohmic resistor})$$
Solution:- According to Ohm's law, \(V = I R\). When voltage \(V\) is doubled, the current \(I\) flowing through the conductor doubles proportionally.
- Because resistance depends on the physical dimensions and resistivity of the conductor, \(R\) remains constant as long as temperature and physical state are unchanged.
Why other options are incorrect:- Option A: Doubling voltage increases the current, not the resistance.
- Option B: Halving resistance would require changing the physical dimensions or material of the conductor.
- Option C: Resistance does not scale quadratically with voltage.
MCQ #128 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A particle is moving in a uniform circular path whose projection is executing simple harmonic motion on horizontal diameter. The ratio of instantaneous velocity to the maximum velocity of the projection while passing through the center is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The projection of uniform circular motion along a diameter executes simple harmonic motion, achieving its maximum instantaneous velocity when passing through the center (mean equilibrium position).
Formula / Rule / Reaction:$$v(x) = \omega \sqrt{A^2 - x^2}; \quad \text{At the center } (x = 0): v = \omega A = v_{\max}$$
Solution:- The instantaneous velocity of the projection is given by \(v = \omega \sqrt{A^2 - x^2}\).
- At the center of the diameter, the displacement is zero (\(x = 0\)), meaning the instantaneous velocity equals the maximum velocity: \(v_{\text{inst}} = \omega A = v_{\max}\).
- The ratio of instantaneous velocity to maximum velocity at this point is \(v_{\max} : v_{\max} = 1:1\).
Why other options are incorrect:- Option B: 1:2 occurs at a displacement of \(x = \frac{\sqrt{3}}{2} A\), not at the center.
- Option C: Instantaneous velocity can never exceed maximum velocity in simple harmonic motion.
- Option D: 1:4 corresponds to an instantaneous velocity near the extreme displacement points.
MCQ #129 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The option that shows the conditions used by Laplace for connecting the velocity of sound is options medium used thermodynamics process
1 solid adiabatic 2 liquids isobaric 3 gas adiabatic 4 gas isothermal
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Laplace corrected Newton's formula for the speed of sound by recognizing that rapid compressions and rarefactions in a gas occur adiabatically, without significant heat exchange with the surrounding medium.
Formula / Rule / Reaction:$$v_{\text{Laplace}} = \sqrt{\frac{\gamma P}{\rho}} \quad (\gamma = C_p / C_v, \; \text{adiabatic bulk modulus } E_s = \gamma P)$$
Solution:- Newton incorrectly assumed that acoustic compressions and rarefactions in a gas occurred isothermally.
- Laplace recognized that sound waves propagate so rapidly that thermal conduction is negligible, meaning acoustic pressure fluctuations in a gas take place under adiabatic conditions.
Why other options are incorrect:- Option A: Laplace's correction specifically addressed acoustic propagation in gases, not solids.
- Option B: Sound propagation does not occur under isobaric (constant pressure) conditions.
- Option D: An isothermal gas process was Newton's original assumption, which underestimated the measured speed of sound in air by approximately 16%.
MCQ #130 of 180
Physics
SZABMU 2025
[SZABMU 2025]
If different colored light beams have same total energy, which color beam will contain the smallest number of photons?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Total radiant energy is the product of photon count and single-photon energy; because photon energy is directly proportional to frequency, the color with the highest frequency requires the fewest photons to attain a given total energy.
Formula / Rule / Reaction:$$E_{\text{total}} = N \cdot E_{\text{photon}} = N \cdot \frac{h c}{\lambda} \implies N = \frac{E_{\text{total}} \cdot \lambda}{h c}$$
Solution:- Violet light has the shortest wavelength (\(\lambda \approx 400\text{ nm}\)) and the highest frequency among visible colors, giving each violet photon the highest individual energy.
- For a fixed total energy \(E_{\text{total}}\), the beam composed of higher-energy photons requires the smallest total number of photons \(N\).
Why other options are incorrect:- Option B: Blue light has a longer wavelength and lower photon energy than violet light, requiring more photons.
- Option C: Green light has intermediate photon energy and requires more photons than violet.
- Option D: Red light has the longest wavelength and lowest photon energy in the visible spectrum, requiring the largest number of photons.
MCQ #131 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A patient is injected with a radioactive isotope to trace blood flow and detect circulation issues. The isotope used is more likely to be:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Sodium-24 is a gamma-emitting radiotracer with a short half-life (15 hours) that behaves physiologically like natural sodium in blood plasma, making it suitable for diagnosing circulatory disorders.
Formula / Rule / Reaction:$$^{24}_{11}\text{Na} \xrightarrow{t_{1/2} = 15\text{ h}} \; ^{24}_{12}\text{Mg} + \beta^- + \gamma \quad (\text{used for hemodynamic tracing})$$
Solution:- Injected as a sterile isotonic saline solution (\(^{24}\text{NaCl}\)), sodium-24 distributes evenly throughout the vascular system.
- Detecting its emitted gamma radiation with external scintillation counters allows clinicians to trace blood flow velocity and identify site-specific circulatory obstructions.
Why other options are incorrect:- Option A: Iodine-131 concentrates in the thyroid gland to treat and diagnose thyroid disorders.
- Option B: Phosphorus-32 is a pure beta emitter used to treat polycythemia vera and label nucleic acids.
- Option C: Cobalt-60 is a high-energy gamma source used for external beam radiotherapy, not as an internal hemodynamic tracer.
MCQ #132 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Water flows through a pipe of 0.02 m2 with a speed of 3 m/s. The pipe narrows to 0.01 m2. The speed in narrower section is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:For an incompressible, steady fluid flow, the equation of continuity dictates that the product of cross-sectional area and fluid velocity remains constant along a streamline.
Formula / Rule / Reaction:$$A_1 v_1 = A_2 v_2 \implies v_2 = v_1 \left(\frac{A_1}{A_2}\right)$$
Solution:- Given values: \(A_1 = 0.02\text{ m}^2\), \(v_1 = 3\text{ m/s}\), and \(A_2 = 0.01\text{ m}^2\).
- Substitute into the continuity equation: \(0.02 \times 3 = 0.01 \times v_2\).
- Solve for \(v_2\): \(v_2 = \frac{0.06}{0.01} = 6\text{ m/s}\).
Why other options are incorrect:- Option A: 1.5 m/s would result if the pipe expanded to double its area.
- Option B: 3 m/s would mean no change in cross-sectional area occurred.
- Option D: 9 m/s would require the area to decrease to one-third of its original value.
MCQ #133 of 180
Physics
SZABMU 2025
[SZABMU 2025]
In the human circulatory system, turbulent blood flow is most likely to occur when the:
A
Vessel diameter is very small
B
Velocity of blood is very low
C
Blood flows through a clogged vessel
D
Blood viscosity increases
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Fluid turbulence is governed by the Reynolds number; when blood encounters an obstruction in a narrowed or clogged vessel, flow velocity increases locally and produces eddies and vortices.
Formula / Rule / Reaction:$$N_R = \frac{\rho v D}{\eta} \quad (N_R > 2000 \implies \text{Turbulent Flow})$$
Solution:- Atherosclerotic plaques or partial obstructions constrict the vessel lumen, forcing blood through the constriction at high localized velocities.
- The resulting flow separation and velocity shear exceed the critical Reynolds number, generating turbulent vortices heard clinically as bruits or murmurs.
Why other options are incorrect:- Option A: Very small vessel diameters (e.g., in capillaries) keep the Reynolds number low, maintaining laminar flow.
- Option B: Low velocity lowers the Reynolds number, promoting laminar flow.
- Option D: Increased blood viscosity increases the denominator of the Reynolds equation, stabilizing laminar flow.
MCQ #134 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A conducting ring is placed near a current-carrying coil. As the current in the coil increases, the induced current in the ring flows:
A
In the direction of magnetic field
B
To increases the magnetic flux
C
To oppose increasing magnetic flux
D
Randomly without a definite direction
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:According to Lenz's law and Faraday's law of induction, the direction of an induced current always opposes the change in magnetic flux that produces it.
Formula / Rule / Reaction:$$\mathcal{E} = -\frac{d\Phi_B}{dt}$$
Solution:- As current in the primary coil increases, the magnetic flux passing through the nearby conducting ring increases.
- By Lenz's law, an electromotive force is induced in the ring driving a current whose own magnetic field opposes this increase in magnetic flux.
Why other options are incorrect:- Option A: Induced current flows along the circular path of the conductor, perpendicular to the magnetic field vector.
- Option B: Flowing to increase flux would violate the principle of conservation of energy.
- Option D: The induced current has a deterministic direction dictated by electromagnetic laws.
MCQ #135 of 180
Physics
SZABMU 2025
[SZABMU 2025]
In progressive waves, energy is transferred from one point to another through:
A
Circular motion of particles
B
Oscillatory motion of particles
D
Translation of particles
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:A progressive mechanical wave transfers energy and momentum through a medium via localized simple harmonic oscillations of particles about their fixed equilibrium positions without bulk translation of matter.
Formula / Rule / Reaction:$$y(x,t) = A \sin(kx - \omega t) \quad (\text{Wave propagation with localized particle oscillation})$$
Solution:- As the wave propagates, medium particles oscillate back and forth about their mean equilibrium positions.
- Elastic forces between neighboring particles transmit kinetic and potential energy along the medium while the particles themselves remain in place on average.
Why other options are incorrect:- Option A: Circular particle motion is characteristic of surface water waves, not a general description of progressive waves.
- Option C: Rotational motion does not mediate longitudinal or transverse mechanical wave propagation.
- Option D: Medium particles do not undergo net translational displacement during wave transmission.
MCQ #136 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A capacitor is connected to an ac source. If the frequency of the AC source is doubled the current in a purely capacitive circuit will:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In a purely capacitive AC circuit, capacitive reactance is inversely proportional to frequency, meaning alternating current is directly proportional to frequency.
Formula / Rule / Reaction:$$X_C = \frac{1}{2\pi f C} \implies I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} (2\pi f C) \implies I \propto f$$
Solution:- Capacitive reactance represents the opposition offered by a capacitor to alternating current flow: \(X_C = 1 / (2\pi f C)\).
- When the AC frequency \(f\) is doubled to \(2f\), the reactance \(X_C\) is halved.
- Consequently, for a constant applied RMS voltage, the circuit current doubles proportionally.
Why other options are incorrect:- Option B: Current depends directly on alternating frequency in reactive capacitive circuits.
- Option C: Current halves in an inductive circuit (\(I \propto 1/f\)), not in a capacitive circuit.
- Option D: Current approaches zero as frequency approaches zero (DC limit), not when AC frequency is increased.
MCQ #137 of 180
Physics
SZABMU 2025
[SZABMU 2025]
When thermal energy is transferred from a hot object to a cold one, the change occurs in the hot object as:
A
Increases in internal energy
B
Decrease in internal energy
C
Increase in temperature
D
Increase in heat content
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:According to the First and Second Laws of Thermodynamics, when an isolated system transfers heat spontaneously to a colder body, the donor object loses thermal energy, reducing its internal energy.
Formula / Rule / Reaction:$$\Delta U = Q - W \quad (\text{Heat removed: } Q < 0 \implies \Delta U < 0)$$
Solution:- Thermal energy transfers spontaneously from a region of higher temperature to one of lower temperature until thermal equilibrium is established.
- Because energy is leaving the hotter body in the form of heat, its internal kinetic and potential microscopic energy decreases (\(\Delta U < 0\)).
Why other options are incorrect:- Option A: The hot body expels energy, making an increase in internal energy impossible.
- Option C: Temperature decreases proportionally with internal energy loss in the absence of phase changes.
- Option D: 'Heat content' is an obsolete caloric concept; thermodynamics defines heat strictly as energy in transit.
MCQ #138 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Two spectral lines in the Hydrogen atom spectrum are close together. This likely means:
A
The energy difference is zero
B
The energy difference is small
C
The light intensity is low
D
The atom becomes unstable
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The separation between spectral lines is proportional to the difference between their emission frequencies, which corresponds directly to the energy differences between the quantum transitions.
Formula / Rule / Reaction:$$\Delta E = h \nu = \frac{h c}{\lambda} \implies |\Delta E_1 - \Delta E_2| = h c \left|\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right|$$
Solution:- Spectral lines correspond to electronic transitions between discrete quantum energy levels.
- When two spectral lines lie very close together in wavelength space (\(\Delta \lambda \approx 0\)), the photons have nearly identical frequencies and energies, meaning the energy gap difference between the two transition pairs is small.
- Note on board erratum: Some provincial keys erroneously marked Option C; however, spectral line spacing is determined by quantum transition energy differences, not beam intensity.
Why other options are incorrect:- Option A: If the energy difference were zero, the two transitions would produce identical wavelengths and appear as a single line.
- Option C: Light intensity determines spectral line brightness, not the wavelength interval separating distinct transitions.
- Option D: Closely spaced spectral lines are standard features of atomic spectra (e.g., fine-structure doublet splitting) and do not imply nuclear or atomic instability.
MCQ #139 of 180
Physics
SZABMU 2025
[SZABMU 2025]
When two identical waves arrive in phase at same point in a region at the same time the resultant displacement is equal to:
A
Displacement of a single wave
B
Difference of their displacements
C
Sum of their displacements
D
Product of their displacements
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The principle of superposition states that when two or more propagating waves overlap at a spatial point, the net instantaneous displacement equals the algebraic sum of the individual wave displacements.
Formula / Rule / Reaction:$$y_{\text{net}} = y_1 + y_2 \quad (\text{For identical in-phase waves: } y_1 = y_2 = y \implies y_{\text{net}} = 2y)$$
Solution:- Constructive interference occurs when two identical waves meet with zero phase difference (in phase).
- The crests and troughs align simultaneously, causing the individual displacements to add constructively to yield twice the amplitude of a single wave.
Why other options are incorrect:- Option A: Displacement of a single wave represents an unperturbed solitary wave without interference.
- Option B: The difference of displacements occurs during completely destructive interference when waves meet \(180^\circ\) out of phase.
- Option D: Displacements add linearly in linear media; they do not multiply.
MCQ #140 of 180
Physics
SZABMU 2025
[SZABMU 2025]
High slope of Ohm's law graph (V vs I) means:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a standard voltage versus current (V-I) graph for an ohmic conductor, potential difference is plotted on the vertical axis and current on the horizontal axis, making the slope equal to resistance.
Formula / Rule / Reaction:$$\text{Slope} = \frac{\Delta V}{\Delta I} = R$$
Solution:- According to Ohm's law, \(V = I R\).
- In a V versus I coordinate system, the gradient of the straight-line plot is \(\Delta V / \Delta I\), which equals the electrical resistance \(R\).
- A steeper, higher slope indicates that a larger voltage is required to drive a unit of current, signifying high resistance.
Why other options are incorrect:- Option A: Low resistance produces a gentle, low slope on a V-I graph (or a high slope only if plotted as I vs V).
- Option C: An open circuit produces an infinite vertical line where current is zero regardless of voltage.
- Option D: Low current is a secondary circuit outcome, not the direct physical quantity defined by the slope.
MCQ #141 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The electric potential at a point due to a point charge is 100 J/C. If a test charge of magnitude 1 C placed at that point is replaced with a 2 C charge, then the physical quantity that will change is:
A
Electric Field at that point
B
Electric potential at that point
C
Electric Potential Energy of the system
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electric potential and electric field at a point in space are determined solely by the source charge and location, whereas electric potential energy depends directly on the magnitude of the placed test charge.
Formula / Rule / Reaction:$$V = \frac{k Q}{r} = \text{constant}; \quad U = q_0 V \implies U \propto q_0$$
Solution:- Electric potential \(V = 100\text{ J/C}\) is an intrinsic property of the electrostatic field established by the primary source charge.
- When the test charge \(q_0\) is increased from 1 C to 2 C, the electric potential energy increases from \(U_1 = (1\text{ C})(100\text{ J/C}) = 100\text{ J}\) to \(U_2 = (2\text{ C})(100\text{ J/C}) = 200\text{ J}\).
Why other options are incorrect:- Option A: The electric field \(E = k Q / r^2\) depends only on the source charge and coordinate distance, remaining unchanged.
- Option B: The electric potential is work per unit charge (\(100\text{ J/C}\)) and does not change with test charge magnitude.
- Option D: Electrostatic test charge substitutions do not alter ambient thermodynamic temperature.
MCQ #142 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The waves among the following which require a material medium for their propagation are:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Mechanical waves require an elastic material medium to propagate via particle oscillations, whereas electromagnetic waves propagate through vacuum via oscillating electric and magnetic fields.
Formula / Rule / Reaction:$$\text{Sound Speed: } v = \sqrt{\frac{E}{\rho}} \quad (\text{Medium required; cannot propagate in vacuum})$$
Solution:- Infrasonic waves are acoustic (sound) waves with frequencies below 20 Hz.
- Because sound is a longitudinal mechanical wave requiring particle-to-particle collisions to transmit compressions and rarefactions, it cannot propagate without a material medium.
Why other options are incorrect:- Option A: Gamma rays are high-energy electromagnetic waves that travel freely through the vacuum of space.
- Option B: X-rays are electromagnetic waves capable of propagating without a physical medium.
- Option C: Infrared waves are electromagnetic radiation and do not require matter to propagate.
MCQ #143 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Unequal-changes occurring in velocity of a body is called:
C
Instantaneous acceleration
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:When a moving body experiences unequal changes in velocity over equal intervals of time, its acceleration is non-constant, defining variable (non-uniform) acceleration.
Formula / Rule / Reaction:$$a(t) = \frac{dv}{dt} \neq \text{constant} \implies \text{Variable acceleration}$$
Solution:- Acceleration is defined as the rate of change of velocity with respect to time: \(a = \Delta v / \Delta t\).
- If \(\Delta v\) is not constant across successive identical time intervals \(\Delta t\), the acceleration varies with time, producing variable acceleration.
Why other options are incorrect:- Option A: Uniform acceleration requires equal changes in velocity in equal intervals of time.
- Option B: Uniform velocity means velocity remains constant with zero acceleration.
- Option C: Instantaneous acceleration is the acceleration evaluated at a specific moment in time (\(\lim_{\Delta t \to 0} \Delta v / \Delta t\)).
MCQ #144 of 180
Physics
SZABMU 2025
[SZABMU 2025]
What will potential difference, if a wire has resistance of 10 ohms and current 2A flow through it:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:According to Ohm's law, the potential difference across a conductor is the product of the electric current flowing through it and its electrical resistance.
Formula / Rule / Reaction:$$V = I \cdot R$$
Solution:- Given: Resistance \(R = 10\,\Omega\) and current \(I = 2\text{ A}\).
- Substitute the values into Ohm's law: \(V = 2\text{ A} \times 10\,\Omega = 20\text{ V}\).
- Note on board erratum: Some unofficial answer keys mistakenly marked Option B (10 V); Option C (20 V) is the mathematically sound solution.
Why other options are incorrect:- Option A: 5 V results from dividing resistance by current (\(10 / 2\)), violating Ohm's law.
- Option B: 10 V represents a calculation error where current is assumed to be 1 A.
- Option D: 40 V assumes a current of 4 A.
MCQ #145 of 180
Physics
SZABMU 2025
[SZABMU 2025]
According to the particle model of light, a photon is:
A
Particle with mass and charge
B
A quantum of energy with zero rest mass and zero charge
C
A continuous energy wave
D
A particle that travels slower than light
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In quantum electrodynamics and Einstein's particle model, light consists of localized discrete quanta called photons, each carrying energy \(E = h\nu\), zero rest mass, and zero electric charge.
Formula / Rule / Reaction:$$E = h \nu, \quad p = \frac{h}{\lambda}, \quad m_0 = 0, \quad q = 0, \quad v = c$$
Solution:- Photons are fundamental gauge bosons mediating the electromagnetic force.
- They possess zero invariant rest mass (\(m_0 = 0\)), carry no electric charge (\(q = 0\)), and travel at the speed of light in vacuum (\(c \approx 3 \times 10^8\text{ m/s}\)).
Why other options are incorrect:- Option A: Photons carry neither rest mass nor electric charge.
- Option C: A continuous wave describes classical Maxwellian wave theory, contrasting with the discrete packet model of quantum theory.
- Option D: Photons travel at the speed of light in vacuum; they cannot travel slower than \(c\) in free space.
MCQ #146 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The current in forward biased PN junction is mainly due to:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Applying forward bias reduces the potential barrier across a PN junction, allowing majority charge carriers to diffuse across the depletion region.
Formula / Rule / Reaction:$$I = I_0 \left(e^{e V / k_B T} - 1\right) \quad (\text{Forward current dominated by majority carrier diffusion})$$
Solution:- Under forward bias, the positive terminal connects to the p-type material and the negative terminal to the n-type material, opposing the built-in potential.
- This lowered barrier allows electrons in the n-region and holes in the p-region (majority carriers) to diffuse across the junction, generating large forward current.
Why other options are incorrect:- Option B: Minority carriers dominate the reverse saturation current under reverse bias conditions.
- Option C: Leakage current is the tiny reverse current caused by thermally generated minority carriers.
- Option D: Thermionic emission is the thermal ejection of electrons from a heated cathode surface into vacuum, not junction diffusion.
MCQ #147 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Stationary waves are formed in a stretched string of 2m length, such that two vibrating loops are formed. The distance between consecutive nodes formed is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a standing wave on a string fixed at both ends, the length of one vibrating loop equals half a wavelength, which represents the distance between two consecutive nodes.
Formula / Rule / Reaction:$$L = n \left(\frac{\lambda}{2}\right); \quad d_{\text{node-node}} = \frac{\lambda}{2} = \frac{L}{n}$$
Solution:- Given: String length \(L = 2\text{ m}\) and number of vibrating loops \(n = 2\).
- The wavelength is \(\lambda = L = 2\text{ m}\).
- The distance between two consecutive nodes equals the length of one loop: \(d = \lambda / 2 = 2 / 2 = 1\text{ m}\).
Why other options are incorrect:- Option A: 0.5 m is the distance between adjacent nodes in a four-loop mode (\(n = 4\)).
- Option C: 2 m is the total string length and the full wavelength of the second harmonic.
- Option D: 3 m exceeds the length of the string.
MCQ #148 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A football is kicked with a speed of 20 m/s at an angle of 30° With the horizontal, the maximum height it attains is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The maximum vertical height reached by a projectile depends on the square of the vertical component of its initial velocity and the acceleration due to gravity.
Formula / Rule / Reaction:$$H_{\max} = \frac{v_0^2 \sin^2\theta}{2g}$$
Solution:- Given: Initial speed \(v_0 = 20\text{ m/s}\), launch angle \(\theta = 30^\circ\), and \(g \approx 10\text{ m/s}^2\).
- Vertical velocity component: \(v_{0y} = v_0 \sin(30^\circ) = 20 \times 0.5 = 10\text{ m/s}\).
- Maximum height: \(H_{\max} = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5\text{ m}\).
Why other options are incorrect:- Option B: 10 m would be attained if the ball were launched vertically at 30° without squaring the sine factor.
- Option C: 15 m is an overestimation that fails to resolve initial velocity into its orthogonal components.
- Option D: 20 m would require launching the projectile straight up at \(90^\circ\).
MCQ #149 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A high internal resistance battery is not suitable for heavy loads due to:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:When a heavy load draws large electric current from a battery, internal resistance causes an internal potential drop, reducing the terminal voltage delivered to the load.
Formula / Rule / Reaction:$$V_t = E - I r \quad (\text{where } I r = \text{internal voltage drop})$$
Solution:- A heavy electrical load has low load resistance (\(R_L\)) and demands a large current \(I\).
- If the internal resistance \(r\) of the battery is high, the internal voltage drop \(I r\) becomes large, leaving insufficient terminal potential difference \(V_t\) to operate the load.
Why other options are incorrect:- Option B: Terminal voltage drops significantly under heavy load, rather than remaining high.
- Option C: Electromotive force (EMF) is a finite chemical property of the cell, never infinite.
- Option D: Current varies inversely with total circuit resistance and does not remain constant.
MCQ #150 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Two equal charges experience a certain force when placed in vacuum. When the charges are placed in a medium, the force becomes 1/4 th. The dielectric constant of the medium is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The electrostatic Coulomb force between two charges is reduced in a dielectric medium by a factor equal to the relative permittivity (dielectric constant) of the medium.
Formula / Rule / Reaction:$$F_{\text{med}} = \frac{F_{\text{vac}}}{\varepsilon_r} \implies \varepsilon_r = \frac{F_{\text{vac}}}{F_{\text{med}}}$$
Solution:- Given: \(F_{\text{med}} = \frac{1}{4} F_{\text{vac}}\).
- Rearranging for the dielectric constant: \(\varepsilon_r = \frac{F_{\text{vac}}}{F_{\text{vac}} / 4} = 4\).
- The dielectric constant of the medium is 4.
Why other options are incorrect:- Option A: 0.25 is the reciprocal of the dielectric constant, representing \(1 / \varepsilon_r\).
- Option B: A dielectric constant of 2 would reduce the force to half, not one-fourth.
- Option D: 16 squares the dielectric constant value.
MCQ #151 of 180
Physics
SZABMU 2025
[SZABMU 2025]
If a test charge of magnitude 2C experience forces of 100N 1 and 200N at two points A and B in an electric field respectively. The ratio of electric fields at A to that of B is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The electric field intensity at any point is defined as the electrostatic force experienced per unit positive test charge placed at that location.
Formula / Rule / Reaction:$$E = \frac{F}{q} \implies \frac{E_A}{E_B} = \frac{F_A / q}{F_B / q} = \frac{F_A}{F_B}$$
Solution:- Because the identical test charge (\(q = 2\text{ C}\)) is used at both locations, the electric field strength is directly proportional to the measured force.
- The ratio of field intensities is: \(E_A : E_B = 100\text{ N} : 200\text{ N} = 1:2\).
Why other options are incorrect:- Option A: 1:1 implies equal forces were measured at points A and B.
- Option C: 1:4 incorrectly squares the force ratio.
- Option D: 4:1 inverts and squares the ratio of electric field strengths.
MCQ #152 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A body of mass 2 kg moving with velocity 3 m/s collides with a body of mass 1 kg at rest. If they stick together, their common velocity after collision is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a perfectly inelastic collision where interacting bodies stick together, total linear momentum is conserved in the absence of external net forces.
Formula / Rule / Reaction:$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \implies v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$$
Solution:- Given: \(m_1 = 2\text{ kg}\), \(v_1 = 3\text{ m/s}\), \(m_2 = 1\text{ kg}\), and \(v_2 = 0\text{ m/s}\).
- Total initial momentum: \(P_i = (2\text{ kg})(3\text{ m/s}) + (1\text{ kg})(0) = 6\text{ kg}\cdot\text{m/s}\).
- Combined mass after collision: \(m_1 + m_2 = 2 + 1 = 3\text{ kg}\).
- Final common velocity: \(v_f = \frac{6\text{ kg}\cdot\text{m/s}}{3\text{ kg}} = 2\text{ m/s}\).
Why other options are incorrect:- Option A: 1 m/s results from dividing initial momentum by an incorrect mass of 6 kg.
- Option C: 3 m/s would mean the moving mass transferred no momentum to the stationary body.
- Option D: 4 m/s violates the principle of conservation of linear momentum.
MCQ #153 of 180
Physics
SZABMU 2025
[SZABMU 2025]
Two balls thrown with equal speeds but at different angles cover equal horizontal distance. If one is thrown at an angle of 40°, then the angle of projection of the other is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:For a projectile launched across level ground at a given initial speed, complementary angles of projection (angles that sum to 90°) produce identical horizontal ranges.
Formula / Rule / Reaction:$$R = \frac{v_0^2 \sin(2\theta)}{g}; \quad \sin(2(90^\circ - \theta)) = \sin(180^\circ - 2\theta) = \sin(2\theta) \implies \theta_2 = 90^\circ - \theta_1$$
Solution:- Given launch angle: \(\theta_1 = 40^\circ\).
- The complementary angle yielding the exact same range is: \(\theta_2 = 90^\circ - 40^\circ = 50^\circ\).
Why other options are incorrect:- Option A: 20° has a complement of 70°, yielding a shorter range than at 40°.
- Option B: 30° is complementary to 60°, not 40°.
- Option C: 45° is the unique angle that maximizes horizontal projectile range.
MCQ #154 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The work done on a body is stored in it in the form of:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:By the work-energy theorem, mechanical work performed by net external forces on a system results in an equivalent change and storage of mechanical energy.
Formula / Rule / Reaction:$$W = \Delta E = \Delta K + \Delta U$$
Solution:- When work is done against conservative forces (such as gravitational or elastic forces), it is stored as potential energy.
- When work accelerates an object, it is converted into kinetic energy, confirming that work done is stored in the form of energy.
Why other options are incorrect:- Option A: Power is the time rate of doing work (\(P = dW/dt\)), not a stored physical quantity.
- Option C: Momentum is the product of mass and velocity (\(p = mv\)), not equivalent to accumulated work.
- Option D: Impulse is the integral of force over time (\(J = \int F dt\)), producing a change in linear momentum.
MCQ #155 of 180
Physics
SZABMU 2025
[SZABMU 2025]
If 20 waves pass a point in 2 seconds and with a speed of 5m/s, then the wavelength of wave is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Wavelength is the ratio of wave propagation speed to frequency, where frequency is the number of complete wave cycles passing a fixed point per unit time.
Formula / Rule / Reaction:$$f = \frac{N}{t}; \quad v = f \lambda \implies \lambda = \frac{v}{f}$$
Solution:- Calculate wave frequency: \(f = \frac{20\text{ waves}}{2\text{ seconds}} = 10\text{ Hz}\).
- Given wave speed \(v = 5\text{ m/s}\), solve for wavelength: \(\lambda = \frac{5\text{ m/s}}{10\text{ Hz}} = 0.5\text{ m}\).
Why other options are incorrect:- Option B: 1 m would result if the wave speed were 10 m/s.
- Option C: 1.5 m does not satisfy the wave speed equation \(v = f \lambda\).
- Option D: 2 m inverts the frequency calculation (\(10 / 5\)).
MCQ #156 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The unit of RC in case of charging a capacitor is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The product of electrical resistance \(R\) and capacitance \(C\) defines the capacitive time constant \(\tau\) of a circuit, possessing the physical dimension of time.
Formula / Rule / Reaction:$$[\tau] = [R][C] = \left(\frac{\text{Volt}}{\text{Ampere}}\right) \times \left(\frac{\text{Coulomb}}{\text{Volt}}\right) = \frac{\text{Coulomb}}{\text{Coulomb/Second}} = \text{Seconds}$$
Solution:- In transient RC charging circuits, the voltage across the capacitor rises according to \(V(t) = V_0 (1 - e^{-t / RC})\).
- Because the exponent \(-t/RC\) must be dimensionless, the product \(RC\) has the unit of seconds.
Why other options are incorrect:- Option A: The Farad is the unit of capacitance \(C\).
- Option C: The Ohm is the unit of resistance \(R\).
- Option D: The Volt is the unit of electric potential difference.
MCQ #157 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A longitudinal wave has a frequency of 500 Hz and wavelength of 0.6 m, its speed is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The propagation velocity of any wave is equal to the product of its temporal frequency and spatial wavelength.
Formula / Rule / Reaction:$$v = f \cdot \lambda$$
Solution:- Given: Frequency \(f = 500\text{ Hz}\) and wavelength \(\lambda = 0.6\text{ m}\).
- Calculate speed: \(v = 500\text{ s}^{-1} \times 0.6\text{ m} = 300\text{ m/s}\).
Why other options are incorrect:- Option A: 30 m/s misplaces the decimal point during multiplication.
- Option B: 83 m/s results from dividing frequency by wavelength.
- Option D: 1200 m/s results from multiplying by an incorrect scalar factor.
MCQ #158 of 180
Physics
SZABMU 2025
[SZABMU 2025]
The true statement about angular displacement is:
A
It always increases with time
B
It is treated as vector for small rotations
C
It is always a scalar quantity
D
It is measured in meters
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Finite angular displacements do not obey the commutative law of vector addition, whereas infinitesimal (small) angular rotations commute and are treated as true axial vectors.
Formula / Rule / Reaction:$$d\vec{\theta}_1 + d\vec{\theta}_2 = d\vec{\theta}_2 + d\vec{\theta}_1 \quad (\text{Commutative for infinitesimal rotations})$$
Solution:- Large, finite angular rotations cannot be represented as vectors because the final orientation depends on the order of rotation (\(\theta_1 + \theta_2 \neq \theta_2 + \theta_1\)).
- Infinitesimal angular displacements commute under addition, allowing them to be treated as vectors whose direction is given by the right-hand rule.
Why other options are incorrect:- Option A: Angular displacement can decrease, oscillate, or reverse sign depending on rotation direction.
- Option C: Small angular displacements behave as axial vectors, not scalars.
- Option D: Angular displacement is measured in dimensionless radians or degrees, not meters.
MCQ #159 of 180
Physics
SZABMU 2025
[SZABMU 2025]
π radians are equivalent to:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A full circular revolution traverses an angle of \(360^\circ\), which corresponds to \(2\pi\) radians; therefore, \(\pi\) radians is equal to \(180^\circ\).
Formula / Rule / Reaction:$$\theta_{\text{deg}} = \theta_{\text{rad}} \times \left(\frac{180^\circ}{\pi}\right) \implies \pi \times \left(\frac{180^\circ}{\pi}\right) = 180^\circ$$
Solution:- By definition, one complete rotation around a circle subtends \(2\pi\) radians at the center, equivalent to \(360^\circ\).
- Dividing both sides by 2 yields \(\pi\text{ radians} = 180^\circ\).
Why other options are incorrect:- Option A: 30° corresponds to \(\pi / 6\) radians.
- Option B: 270° corresponds to \(3\pi / 2\) radians.
- Option C: 90° corresponds to \(\pi / 2\) radians.
MCQ #160 of 180
Physics
SZABMU 2025
[SZABMU 2025]
A person moves 600 m towards north then 300 m towards south. Its displacement to distance ratio is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Distance is the total scalar path length traversed, whereas displacement is the net vector change in position from initial to final point.
Formula / Rule / Reaction:$$\text{Distance} = d_1 + d_2; \quad \text{Displacement} = |\vec{d}_1 + \vec{d}_2|$$
Solution:- Total path distance: \(600\text{ m} + 300\text{ m} = 900\text{ m}\).
- Net displacement: Taking north as positive, \(\Delta x = +600\text{ m} - 300\text{ m} = +300\text{ m}\text{ (north)}\).
- Displacement to distance ratio: \(300\text{ m} : 900\text{ m} = 1:3\).
- Note on board erratum: Some preliminary keys incorrectly printed 1:2; the mathematically sound ratio is 1:3.
Why other options are incorrect:- Option A: 1:2 is a ratio error that miscalculates net displacement or distance.
- Option C: 2:1 incorrectly states that displacement exceeds path distance.
- Option D: 3:1 inverts the ratio, representing distance to displacement.
MCQ #161 of 180
Physics
SZABMU 2025
[SZABMU 2025]
In a stationary wave the point that undergoes zero acceleration is:
C
Midpoint between node and antinode
D
Every point along the wave
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In a standing wave, a node is a point of permanent destructive interference where displacement remains zero at all times; consequently, its velocity and acceleration are identically zero.
Formula / Rule / Reaction:$$y(x,t) = 2A \sin(kx) \cos(\omega t); \quad a(x,t) = -\omega^2 y(x,t) \implies \text{At a node } (y = 0): a = 0$$
Solution:- In simple harmonic motion, particle acceleration is directly proportional to displacement: \(a = -\omega^2 y\).
- At nodal points, destructive interference ensures that the medium displacement \(y\) is zero at all times, meaning the acceleration is permanently zero.
Why other options are incorrect:- Option B: Antinodes experience maximum displacement and therefore undergo maximum acceleration.
- Option C: Points between nodes and antinodes oscillate with intermediate amplitudes and experience non-zero acceleration.
- Option D: Most points along a standing wave oscillate with non-zero acceleration, except nodes.
MCQ #162 of 180
Physics
SZABMU 2025
[SZABMU 2025]
If the speed of sound is measured at sea level and at the top of a mountain, both at the same temperature. It will be:
B
Greater at the mountain top
C
The same at both places
D
Greater where the air is denser
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The speed of sound in an ideal gas depends solely on the adiabatic index, the gas constant, and the absolute temperature, remaining independent of pressure and density changes at constant temperature.
Formula / Rule / Reaction:$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}$$
Solution:- According to Boyle's law, at constant temperature, pressure is directly proportional to density (\(P / \rho = \text{constant}\)).
- Although air pressure and density are lower at the top of a mountain, their ratio \(P / \rho\) remains constant at equal temperatures.
- Therefore, the speed of sound depends only on temperature and is identical at both locations.
Why other options are incorrect:- Option A: Sound speed does not increase with higher pressure if temperature is held constant.
- Option B: Lower air density at mountain altitudes does not increase sound velocity because pressure decreases proportionally.
- Option D: Denser air does not increase sound speed at constant temperature because the increased inertia offsets the pressure change.
MCQ #163 of 180
English
SZABMU 2025
[SZABMU 2025]
The book was written by Rashid Khan. Identify the correct usage of voice.
A
Rashid Khan writes the book.
B
Rashid Khan wrote the book.
C
Rashid Khan written the book.
D
Rashid Khan has write the book.
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In converting a passive voice sentence in the simple past tense ('was written') to active voice, the auxiliary verb is omitted and the main verb takes its simple past form ('wrote').
Formula / Rule / Reaction:$$\text{Passive: Object} + \text{was/were} + \text{V}_3 + \text{by Subject} \implies \text{Active: Subject} + \text{V}_2\text{ (past simple)} + \text{Object}$$
Solution:- The sentence 'The book was written by Rashid Khan' uses the simple past passive form ('was written').
- In active voice, the agent ('Rashid Khan') becomes the grammatical subject, taking the simple past tense verb 'wrote' followed by the direct object 'the book'.
Why other options are incorrect:- Option A: 'Writes' is simple present tense, changing the temporal meaning of the sentence.
- Option C: 'Written' is a past participle that cannot function as a standalone finite verb without an auxiliary.
- Option D: 'Has write' is grammatically incorrect; the present perfect requires the past participle 'written'.
MCQ #164 of 180
English
SZABMU 2025
[SZABMU 2025]
Identify the example of complex sentence.
A
I both thanked him and rewarded him.
B
Life is what we make it.
C
He owed his success to his father.
D
Jumping up, he ran away.
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:A complex sentence consists of exactly one independent clause and at least one dependent (subordinate) clause joined by a subordinating conjunction or relative pronoun.
Formula / Rule / Reaction:$$\text{Complex Sentence} = \text{Independent Clause} + \text{Dependent Subordinate Clause}$$
Solution:- In 'Life is what we make it', the main independent clause is 'Life is [predicate noun]' and the dependent clause is the noun clause 'what we make it', functioning as the predicative complement.
- This structure satisfies the grammatical definition of a complex sentence.
Why other options are incorrect:- Option A: 'I both thanked him and rewarded him' is a simple sentence containing a compound predicate.
- Option C: 'He owed his success to his father' is a simple sentence containing one independent clause.
- Option D: 'Jumping up, he ran away' is a simple sentence modified by a participial phrase ('Jumping up').
MCQ #165 of 180
English
SZABMU 2025
[SZABMU 2025]
Pick out the sentence which illustrates the use of infinitive.
A
He refused to obey the orders.
B
He refused obeying the orders.
C
He refused obey the orders.
D
He refused obeyed the orders.
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:An infinitive is the base form of a verb typically preceded by the particle 'to' ('to + verb'), functioning as a noun, adjective, or adverb.
Formula / Rule / Reaction:$$\text{Full Infinitive} = \text{'to'} + \text{base verb (V}_1\text{)}$$
Solution:- The verb 'refuse' takes a full infinitive complement.
- In 'He refused to obey the orders', 'to obey' functions as the full infinitive acting as the direct object of 'refused'.
Why other options are incorrect:- Option B: 'Obeying' is a gerund; the verb 'refuse' does not take a gerund complement in standard English.
- Option C: Omitting 'to' creates an ungrammatical bare infinitive after 'refuse'.
- Option D: 'Obeyed' is a past tense form that cannot follow the finite verb 'refused'.
MCQ #166 of 180
English
SZABMU 2025
[SZABMU 2025]
What is the effect of using different shades of meaning in satire?
A
To create a serious tone
B
To highlight the absurdity or irony of a situation
D
To add complexity to the language
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Satire utilizes nuanced diction, irony, and lexical connotations to expose societal follies, vices, and incongruities through deliberate double meanings.
Formula / Rule / Reaction:Rhetorical function of satire: Juxtaposition of connotation $\rightarrow$ Exposure of absurdity and irony.Solution:- Satirists select words with subtle nuances and distinct shades of meaning to contrast superficial appearance with reality.
- This lexical selection accentuates the irony and absurdity of the subject being critiqued.
Why other options are incorrect:- Option A: Satire uses humor, irony, and ridicule rather than attempting to maintain a solemn, serious tone.
- Option C: Effective satire clarifies critique through wit rather than creating pointless reader confusion.
- Option D: Adding linguistic complexity is merely stylistic and not the primary communicative purpose of satire.
MCQ #167 of 180
English
SZABMU 2025
[SZABMU 2025]
People approach the situation differently depending on who they are. The word 'approach' in this sentence means:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In context, the verb 'approach' means to deal with, address, or relate to a task, problem, or situation from a particular perspective.
Formula / Rule / Reaction:Contextual Lexical Semantics: 'Approach a situation' $\equiv$ deal with / address / relate to.Solution:- The sentence refers to how diverse individuals engage with and respond to circumstances based on their personal traits.
- Among the choices, 'relate' (to interact with or respond to) best captures the contextual meaning of approaching a situation.
Why other options are incorrect:- Option B: 'Coordinate' means to harmonize or organize disparate elements into a unified whole.
- Option C: 'Comprehend' means to mentally understand or grasp, omitting the active behavioral response implied by 'approach'.
- Option D: 'Verify' means to authenticate, confirm, or prove true.
MCQ #168 of 180
English
SZABMU 2025
[SZABMU 2025]
The team won the match, ............ they broke the school record.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Additive transitions introduce additional supporting information that reinforces or builds upon an established positive clause.
Formula / Rule / Reaction:$$\text{Clause 1 (Achievement)} + \text{Additive Conjunctive Adverb (Furthermore)} + \text{Clause 2 (Additional Achievement)}$$
Solution:- The sentence connects two complementary achievements: winning the match and breaking the school record.
- 'Furthermore' is an additive transitional adverb that appropriately links and amplifies the second achievement.
Why other options are incorrect:- Option A: 'Similarly' denotes comparison between two analogous situations, rather than adding a supplementary accomplishment.
- Option B: 'Nevertheless' introduces an adversative concession or unexpected contrast.
- Option D: 'Instead' indicates replacement or mutual exclusion between alternatives.
MCQ #169 of 180
English
SZABMU 2025
[SZABMU 2025]
Identify the sentence with no spelling errors.
A
Rabia completes homework assignments well in time
B
Rabia complete homwork assignments well in time.
C
Rabia completes homework asignments well in time.
D
Rabia completes homework assigments well in time.
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Standard English orthography requires precise letter patterns: 'homework' (compound noun) and 'assignments' (double 's', single 'g', 'n-m-e-n-t').
Formula / Rule / Reaction:$$\text{Correct orthography: } \text{'homework'} + \text{'assignments'}$$
Solution:- Option A correctly spells 'completes', 'homework', and 'assignments', while maintaining subject-verb agreement.
- All other options contain typographical or grammatical errors.
Why other options are incorrect:- Option B: Contains the misspelling 'homwork' and lacks singular agreement on 'complete'.
- Option C: Contains the misspelling 'asignments' (missing second 's').
- Option D: Contains the misspelling 'assigments' (missing 'n').
MCQ #170 of 180
English
SZABMU 2025
[SZABMU 2025]
Traveling in hot, dusty terrain gives me no pleasure. The underlined part of the sentence is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:A gerund phrase consists of a gerund and its associated modifiers and objects, functioning collectively as a noun phrase to serve as the subject of a clause.
Formula / Rule / Reaction:$$\text{[Traveling in hot, dusty terrain]} = \text{Subject (Noun Phrase / Gerund Phrase)}$$
Solution:- The phrase 'Traveling in hot, dusty terrain' occupies the subject position of the finite verb 'gives'.
- Because it functions syntactically as the subject of the sentence, it is classified as a noun phrase.
Why other options are incorrect:- Option A: Adverb phrases modify verbs, adjectives, or other adverbs, rather than serving as grammatical subjects.
- Option B: While 'in hot, dusty terrain' is a prepositional phrase, the full subject phrase is headed by the gerund 'traveling'.
- Option D: An infinitive phrase must be headed by 'to + base verb', which is absent here.
MCQ #171 of 180
English
SZABMU 2025
[SZABMU 2025]
What does "I've told you a million times" means?
A
To provide an exact count
B
To emphasize frustration or impatience
C
To confuse the listener
D
To make the character sound foolish
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Hyperbole is a figure of speech that uses intentional exaggeration not meant to be taken literally, employed here to convey strong emotion such as annoyance or frustration.
Formula / Rule / Reaction:Figurative Language: Hyperbolic exaggeration $\rightarrow$ Emotional emphasis of impatience.Solution:- The expression 'a million times' is a common hyperbolic idiom.
- It emphasizes that the speaker has repeated an instruction many times, conveying frustration or impatience rather than stating a literal numerical tally.
Why other options are incorrect:- Option A: The phrase is figurative hyperbole, not an exact mathematical count.
- Option C: The idiom is used to reinforce a point clearly, not to confuse the listener.
- Option D: The expression reflects standard conversational emphasis, not foolishness.
MCQ #172 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
An outbreak of a corona disease is reported in the city. What should be the first course of action?
A
Develop a vaccination plan.
B
Conduct research on the disease.
C
Implement quarantine measures.
D
Inform the public by using all means.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:In public health crisis intervention and logical decision-making, the immediate containment of a contagious viral pathogen requires rapid containment and quarantine measures to break transmission chains.
Formula / Rule / Reaction:Epidemic Containment Sequence: Immediate Quarantine and Isolation $\rightarrow$ Public Advisories $\rightarrow$ Vaccination and Research.Solution:- An infectious respiratory outbreak presents an immediate risk of geometric spread through contact.
- Immediate enforcement of quarantine and isolation measures halts the transmission vector, preventing the collapse of regional medical infrastructure while secondary measures are organized.
Why other options are incorrect:- Option A: Developing and manufacturing a vaccine requires months or years, offering no immediate containment.
- Option B: Academic research is a long-term strategy that does not stop active viral transmission.
- Option D: Public communication is essential, but must accompany direct containment measures to halt spread.
MCQ #173 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
Find the next number: 5, 11, 23, 47, 95, ____?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a recursive arithmetic sequence, each successive term is generated by multiplying the preceding term by a constant and adding a fixed integer, or by adding successive powers of 2 multiplied by an initial difference.
Formula / Rule / Reaction:$$T_n = 2 \cdot T_{n-1} + 1$$
Solution:- \(5 \times 2 + 1 = 11\)
- \(11 \times 2 + 1 = 23\)
- \(23 \times 2 + 1 = 47\)
- \(47 \times 2 + 1 = 95\)
- Next term: \(95 \times 2 + 1 = 190 + 1 = 191\).
Why other options are incorrect:- Option A: 190 doubles 95 without adding the required constant 1.
- Option C: 192 adds 2 instead of 1.
- Option D: 193 adds 3 instead of 1.
MCQ #174 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
Which number comes in the missing place? 1, 3, 7, 15, 31, ____.
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In this numerical sequence, each term is obtained by doubling the preceding term and adding 1, representing numbers of the form \(2^n - 1\).
Formula / Rule / Reaction:$$T_n = 2^n - 1 \quad \text{or} \quad T_n = 2 \cdot T_{n-1} + 1$$
Solution:- \(T_1 = 2^1 - 1 = 1\)
- \(T_2 = 2^2 - 1 = 3\)
- \(T_3 = 2^3 - 1 = 7\)
- \(T_4 = 2^4 - 1 = 15\)
- \(T_5 = 2^5 - 1 = 31\)
- Next term: \(T_6 = 2^6 - 1 = 64 - 1 = 63\).
Why other options are incorrect:- Option A: 61 fails to follow the \(2^n - 1\) progression.
- Option C: 65 corresponds to \(2^6 + 1\), adding 1 instead of subtracting 1.
- Option D: 67 deviates from the recursive rule.
MCQ #175 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
Every alphabet in the word "SURGEON" represent a fixed numerical value and the numbers when added, sum up to 99. Based on this, what will the sum of the word "EORGNSU".
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The commutative property of addition states that the sum of a finite set of real numbers remains unchanged regardless of the order in which they are added.
Formula / Rule / Reaction:$$\sum_{i=1}^n x_i = x_{\pi(1)} + x_{\pi(2)} + \dots + x_{\pi(n)}$$
Solution:- The word 'SURGEON' consists of the distinct letters: S, U, R, G, E, O, N.
- The word 'EORGNSU' is an exact anagram containing the identical seven letters: E, O, R, G, N, S, U.
- Because letter values are fixed and addition is commutative, the sum of their numerical values remains 99.
Why other options are incorrect:- Option A: 102 incorrectly adds values to an identical set of letters.
- Option B: 78 assumes letters were omitted from the set.
- Option D: 56 incorrectly reduces the sum of the original letters.
MCQ #176 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
What is the primary difference between hypothesis and a theory?
A
hypothesis is a proven fact, while a theory is an unproven data.
B
A hypothesis is a specific prediction, while a theory is a broad explanation.
C
A hypothesis is a qualitative statement, while a theory is quantitative statement.
D
A hypothesis is a tentative explanation, while a theory is a well-established explanation.
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A hypothesis is a preliminary, testable proposition formulated to explain an observation, whereas a scientific theory is an empirically validated framework supported by experimental evidence.
Formula / Rule / Reaction:$$\text{Observation} \rightarrow \text{Hypothesis (tentative)} \xrightarrow{\text{Rigorous testing}} \text{Theory (well-substantiated)}$$
Solution:- A hypothesis is an initial, tentative explanation that can be tested through experimentation and falsification.
- A theory is a broad, well-substantiated explanation supported by extensive empirical data and independent experimental confirmation.
Why other options are incorrect:- Option A: A hypothesis is never a proven fact, and a scientific theory is not unproven conjecture.
- Option B: A hypothesis is an explanatory proposal rather than merely a prediction.
- Option C: Both hypotheses and theories incorporate qualitative and quantitative components.
MCQ #177 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
ZX, WU, TR, QO.. What comes next?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In alphabetical sequence puzzles, examine the step pattern of the first and second letters independently using their numerical positions in the alphabet.
Formula / Rule / Reaction:$$\text{Position Series: } L_1: 26 \xrightarrow{-3} 23 \xrightarrow{-3} 20 \xrightarrow{-3} 17 \xrightarrow{-3} 14; \quad L_2: 24 \xrightarrow{-3} 21 \xrightarrow{-3} 18 \xrightarrow{-3} 15 \xrightarrow{-3} 12$$
Solution:- First letters: Z (26), W (23), T (20), Q (17). Subtracting 3 gives position 14, which corresponds to N.
- Second letters: X (24), U (21), R (18), O (15). Subtracting 3 gives position 12, which corresponds to L.
- Combining both letters gives the next pair: NL.
Why other options are incorrect:- Option B: LN inverts the letter order.
- Option C: NM subtracts 2 instead of 3 for the second letter.
- Option D: KH subtracts 6 from Q and 7 from O, skipping a step.
MCQ #178 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
In a row of students, a student is at 13th position from the left end and at 20th position from the right end. How many students are there in the row?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:When calculating total elements in a single ordered line from the positions of a single person counted from both ends, subtract one to avoid double-counting that person.
Formula / Rule / Reaction:$$\text{Total} = \text{Position}_{\text{left}} + \text{Position}_{\text{right}} - 1$$
Solution:- The student is 13th from the left and 20th from the right.
- Summing the two ranks: \(13 + 20 = 33\).
- Because the student has been counted twice, subtract 1: \(33 - 1 = 32\) total students.
Why other options are incorrect:- Option A: 30 subtracts 3 students arbitrarily.
- Option B: 31 subtracts the student twice.
- Option D: 33 fails to subtract the overlapping count of the reference student.
MCQ #179 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
A is older than B but younger than F. C is younger than B but older than D. Who is the youngest of them?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Linear order problems are solved by translating comparative statements into transitive inequality relationships.
Formula / Rule / Reaction:$$\text{Premise 1: } F > A > B; \quad \text{Premise 2: } B > C > D \implies F > A > B > C > D$$
Solution:- From 'A is older than B but younger than F': \(F > A > B\).
- From 'C is younger than B but older than D': \(B > C > D\).
- Combining the two inequalities gives the descending age order: \(F > A > B > C > D\).
- D is the youngest member of the group.
Why other options are incorrect:- Option A: A is older than B, C, and D.
- Option B: B is older than C and D.
- Option C: C is older than D.
MCQ #180 of 180
Logical Reasoning
SZABMU 2025
[SZABMU 2025]
A hiker climbed to a higher altitude, where he felt the air pressure was lower as compared to the lower altitude. What is the likely effect?
A
The hiker felt more energetic.
B
The hiker felt no difficulty in breathing.
C
The hiker might experience altitude sickness.
D
The hiker felt healthier and fresh.
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:As barometric pressure falls at high altitudes, the partial pressure of oxygen decreases, causing arterial hypoxemia and the onset of acute altitude sickness.
Formula / Rule / Reaction:$$\downarrow P_{\text{atm}} \implies \downarrow P_{\text{O}_2} = 0.21 \times P_{\text{atm}} \implies \text{Hypoxemia} \rightarrow \text{Acute Mountain Sickness}$$
Solution:- Ascending to high altitude exposes the body to lower atmospheric pressure and reduced partial pressure of oxygen.
- Without prior acclimatization, the resulting drop in arterial oxygen saturation leads to altitude sickness, characterized by headaches, fatigue, dizziness, and dyspnea.
Why other options are incorrect:- Option A: Hypoxia impairs aerobic ATP synthesis, causing fatigue rather than increased energy.
- Option B: Breathing difficulty (hyperventilation and dyspnea) is an immediate physiological response to high-altitude hypoxia.
- Option D: Acute drop in oxygen availability causes physical distress rather than enhanced freshness.