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USAC Veterinaria - fundamentos científicos study materials
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USAC Veterinaria - fundamentos científicos practice questions

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Question 1Biología y razonamiento experimental

Which class of biological macromolecules serves as the primary structural component of animal cell plasma membranes, forming a stable amphipathic bilayer?

  1. Triglycerides
  2. Phospholipids Correct answer.
  3. Glycogen polymers
  4. Fibrous keratins

Explanation

Phospholipids are amphipathic molecules possessing hydrophilic phosphate heads and two hydrophobic fatty acid tails. In an aqueous biological environment, they spontaneously assemble into a stable lipid bilayer that constitutes the fundamental structural matrix of all animal cell membranes.

Question 2Biología y razonamiento experimental

In animal biochemistry, which monosaccharide is the primary metabolic fuel absorbed into systemic circulation from the small intestine to supply immediate energy via glycolysis?

  1. D-Fructose
  2. D-Galactose
  3. D-Glucose Correct answer.
  4. Ribose

Explanation

D-Glucose is the principal circulating aldohexose in animal blood plasma. It crosses cellular membranes via facilitated diffusion transporters (GLUT family) and enters the glycolytic pathway to generate ATP.

Question 3Biología y razonamiento experimental

Which level of protein organization is defined exclusively by the linear amino acid sequence linked by covalent peptide bonds?

  1. Primary structure Correct answer.
  2. Secondary structure
  3. Tertiary structure
  4. Quaternary structure

Explanation

The primary structure of a polypeptide is the unique, linear sequence of amino acids joined end-to-end by covalent peptide (amide) bonds, specified directly by the nucleotide sequence of the encoding gene.

Question 4Biología y razonamiento experimental

Which structural bond confers high tensile strength and chemical stability to tertiary protein architecture in animal keratin and immunoglobulins via covalent linkage between two cysteine residues?

  1. Hydrogen bond
  2. Hydrophobic interaction
  3. Ionic salt bridge
  4. Disulfide bridge Correct answer.

Explanation

A disulfide bridge is a covalent sulfur-sulfur bond formed by the oxidation of the sulfhydryl (-SH) thiol groups of two cysteine residues. Because it is a true covalent bond, it provides substantial structural stability to extracellular proteins like keratins and antibodies.

Question 5Biología y razonamiento experimental

Why are viruses strictly classified as obligate intracellular parasites rather than autonomous living cellular organisms?

  1. They contain both double-stranded DNA and double-stranded RNA simultaneously in all viral particles
  2. They lack their own metabolic machinery and ribosomes, requiring a host cell to replicate Correct answer.
  3. They perform chemiosmotic ATP synthesis only within extracellular physiological fluids
  4. They replicate through binary fission identical to prokaryotic eubacteria

Explanation

Viruses possess genetic material enclosed within a protein capsid (and sometimes a lipid envelope), but completely lack ribosomes, metabolic enzymes, and energy-generating machinery. Consequently, they cannot synthesize proteins or generate ATP independently and must hijack the host cell's translational apparatus to reproduce.

Question 6Biología y razonamiento experimental

Which eukaryotic organelle contains acid hydrolases and an internal acidic lumen (pH ~4.5–5.0) maintained by vacuolar H⁺-ATPases, functioning as the primary site for intracellular digestion and autophagocytosis?

  1. Peroxisome
  2. Lysosome Correct answer.
  3. Golgi apparatus
  4. Centrosome

Explanation

Lysosomes are membrane-bound digestive vesicles packed with over 50 different acid hydrolases. Their internal pH is maintained between 4.5 and 5.0 by membrane-bound proton pumps, providing the optimal environment to degrade engulfed debris, bacteria, and damaged cellular organelles.

Question 7Biología y razonamiento experimental

In eukaryotic cells, the rough endoplasmic reticulum (RER) is distinguished structurally and functionally from the smooth endoplasmic reticulum (SER) by the presence of:

  1. Membrane-bound 80S ribosomes synthesizing transmembrane and secretory proteins Correct answer.
  2. Double-membraned cristae synthesizing ATP via oxidative phosphorylation
  3. Enzymes specializing in steroid hormone biosynthesis and xenobiotic detoxification
  4. Hydrolytic enzymes that digest engulfed pathogenic microorganisms

Explanation

The rough endoplasmic reticulum is studded on its cytosolic face with ribosomes (80S) that translate polypeptides bearing N-terminal signal peptides, translocating them into the lumen for processing into membrane proteins, lysosomal enzymes, or secretory products.

Question 8Biología y razonamiento experimental

The sodium-potassium pump (Na⁺/K⁺-ATPase) in animal cell membranes hydrolyzes one molecule of ATP to perform which stoichiometric ion exchange?

  1. Import 3 Na⁺ ions and export 2 K⁺ ions along their concentration gradients
  2. Export 2 Na⁺ ions and import 3 K⁺ ions using passive facilitated diffusion
  3. Transport equal numbers (2 Na⁺ and 2 K⁺) across the membrane without membrane charge changes
  4. Export 3 Na⁺ ions and import 2 K⁺ ions against their electrochemical gradients Correct answer.

Explanation

The electrogenic Na⁺/K⁺-ATPase consumes one ATP molecule to actively pump 3 Na⁺ ions out of the cytoplasm and 2 K⁺ ions into the cell, each moving against their respective electrochemical gradients. This action maintains resting membrane potential and cellular volume.

Question 9Biología y razonamiento experimental

If canine erythrocytes are suspended in an aqueous solution of 0.2% NaCl (significantly lower osmolarity than physiological 0.9% saline), what cellular response will occur?

  1. Water leaves the erythrocytes rapidly, resulting in cell shrinkage and crenation
  2. Solutes diffuse outward until cellular volume and shape remain entirely unchanged
  3. Water rushes into the erythrocytes down its osmotic gradient, causing cell swelling and osmotic lysis (hemolysis) Correct answer.
  4. The cell wall prevents expansion and generates internal turgor pressure

Explanation

A 0.2% NaCl solution is hypotonic relative to mammalian erythrocyte cytoplasm (~0.9% NaCl equivalent). Water moves rapidly across the plasma membrane into the hyperosmotic cytoplasm by osmosis. Because animal cells lack rigid cell walls, hydrostatic swelling exceeds membrane tensile strength, leading to lysis (hemolysis).

Question 10Biología y razonamiento experimental

Under strenuous anaerobic exertion in equine or canine skeletal muscle fibers, what physiological reaction enables glycolysis to continue producing ATP despite the absence of mitochondrial oxygen?

  1. Pyruvate is decarboxylated to ethanol by pyruvate decarboxylase, regenerating FAD
  2. Pyruvate is reduced to lactate by lactate dehydrogenase, regenerating NAD⁺ from NADH Correct answer.
  3. Acetyl-CoA directly enters anaerobic beta-oxidation to yield GTP
  4. Lactate is oxidized into glycogen directly within the sarcoplasmic reticulum

Explanation

When cellular oxygen is depleted, the electron transport chain halts and mitochondrial NADH cannot be reoxidized. To prevent glycolysis from stalling due to NAD⁺ starvation, cytosolic lactate dehydrogenase reduces pyruvate to L-lactate, transferring electrons from NADH to yield NAD⁺, which sustains continued ATP generation through glycolysis.

Question 11Biología y razonamiento experimental

In eukaryotic cells, where are the multi-enzyme complexes of the electron transport chain (Complexes I–IV) and ATP synthase physically located?

  1. Mitochondrial outer membrane
  2. Mitochondrial matrix
  3. Cytosolic polysomes
  4. Inner mitochondrial membrane Correct answer.

Explanation

Complexes I, II, III, and IV of the electron transport chain and ATP synthase are embedded within the highly folded inner mitochondrial membrane (cristae). Protons are pumped from the matrix across this inner membrane into the intermembrane space, creating the proton motive force.

Question 12Biología y razonamiento experimental

What is the ultimate terminal electron acceptor in mammalian aerobic mitochondrial respiration, and what chemical product is formed upon its reduction?

  1. Inorganic phosphate (Pᵢ), forming ATP
  2. Carbon dioxide (CO₂), forming carbonic acid (H₂CO₃)
  3. Molecular oxygen (O₂), forming water (H₂O) Correct answer.
  4. Cytochrome c, forming reduced ubiquinone

Explanation

At Complex IV (cytochrome c oxidase), four electrons donated by cytochrome c combined with four protons from the matrix are transferred to molecular oxygen (O₂), reducing it to two molecules of water (H₂O). Oxygen's high electronegativity provides the driving thermodynamic potential for the entire respiratory chain.

Question 13Biología y razonamiento experimental

In veterinary histopathology, observing chromosomes aligned individually in a single plane along the equatorial plate of the mitotic spindle identifies which stage of mitosis?

  1. Metaphase Correct answer.
  2. Prophase
  3. Anaphase
  4. Telophase

Explanation

During metaphase, tension exerted by kinetochore microtubules from opposite spindle poles aligns the sister chromatid pairs along the equatorial plane (metaphase plate) midway between the centrosomes, immediately preceding anaphase separation.

Question 14Biología y razonamiento experimental

What is the critical evolutionary and genetic significance of crossing over (homologous genetic recombination) during Prophase I of meiosis?

  1. It doubles the chromosome number to ensure daughter gametes are diploid
  2. It permanently silences all paternal alleles in the developing zygote
  3. It exchanges reciprocal segments between non-sister chromatids of homologous chromosomes, generating novel allele combinations Correct answer.
  4. It prevents chromosome condensation so that transcription can continue during gametogenesis

Explanation

During pachynema of Prophase I, non-sister chromatids of synapsed homologous chromosome pairs undergo reciprocal breakage and crossing over at chiasmata. This physical exchange breaks linkage groups and generates recombinant chromatids bearing novel combinations of maternal and paternal alleles.

Question 15Biología y razonamiento experimental

In male domestic mammals (e.g., bulls, stallions, boars), how many functional, motile haploid spermatozoa are generated from the complete meiotic division of a single primary spermatocyte?

  1. One functional haploid ovum and three polar bodies
  2. Two diploid spermatozoa
  3. Eight haploid spermatids
  4. Four functional haploid spermatozoa Correct answer.

Explanation

In mammalian spermatogenesis, symmetrical meiotic divisions occur. One diploid primary spermatocyte completes Meiosis I to form two secondary spermatocytes, which then complete Meiosis II to yield four haploid spermatids. Each spermatid matures via spermiogenesis into a functional, motile spermatozoon.

Question 16Biología y razonamiento experimental

In mammalian female oogenesis, primary oocytes are arrested at which specific meiotic stage from embryonic development until ovulation occurs post-puberty?

  1. Prophase I (specifically the diplotene/dictyate stage) Correct answer.
  2. Metaphase II
  3. Anaphase I
  4. Telophase II

Explanation

In female mammals, oogonia enter meiosis during fetal gestation and arrest during Prophase I at the diplotene stage (also termed dictyate stage). They remain in this dormant meiotic arrest until luteinizing hormone (LH) surges at puberty stimulate individual follicles to resume meiosis.

Question 17Biología y razonamiento experimental

According to Woese's three-domain phylogenetic classification, which biological domains comprise cellular organisms with prokaryotic cellular organization lacking a membrane-bound nucleus?

  1. Bacteria and Eukarya
  2. Bacteria and Archaea Correct answer.
  3. Archaea and Protista
  4. Monera and Fungi

Explanation

Carl Woese established three evolutionary domains based on 16S/18S ribosomal RNA sequence divergence: Bacteria, Archaea, and Eukarya. Both Bacteria and Archaea possess prokaryotic cellular organization (no membrane-bound nucleus or membrane-bound organelles), though Archaea share key molecular features (RNA polymerases, histones) with Eukarya.

Question 18Biología y razonamiento experimental

Which molecular feature distinguishes the membrane lipids of organisms in the domain Archaea from those found in the domains Bacteria and Eukarya?

  1. Unbranched fatty acid chains joined to glycerol by ester linkages
  2. Absence of any polar head groups or phosphate moieties
  3. Branched isoprenoid hydrocarbons joined to glycerol by ether linkages Correct answer.
  4. Steroid hopanoids embedded within a triple-layered peptidoglycan coat

Explanation

Archaeal membrane lipids are uniquely composed of branched phytanyl (isoprenoid) chains connected to L-glycerol via ether linkages, forming either lipid bilayers or covalently linked monolayers. In contrast, Bacteria and Eukarya utilize unbranched fatty acids linked to D-glycerol by ester bonds.

Question 19Biología y razonamiento experimental

A cell biologist treats cultured feline fibroblasts with colchicine, a drug that specifically binds tubulin dimers and inhibits microtubule polymerisation. Which cellular event will be completely blocked?

  1. Semiconservative replication of genomic DNA in S phase
  2. Transcription of primary mRNA transcripts by RNA polymerase II
  3. Hydrolysis of ATP by the sarcoplasmic reticulum calcium pump (SERCA)
  4. Formation of the mitotic spindle and separation of sister chromatids during mitosis Correct answer.

Explanation

Colchicine depolymerizes microtubules by binding unpolymerized tubulin dimers. Because the mitotic spindle is composed of microtubules radiating from centrosomes to kinetochores, colchicine treatment arrests cells in metaphase by preventing spindle formation and subsequent chromosome segregation.

Question 20Biología y razonamiento experimental

In the ruminant digestive tract (e.g., cattle, sheep, goats), which of the four stomach compartments is the true glandular stomach that secretes hydrochloric acid and pepsinogen?

  1. Abomasum Correct answer.
  2. Rumen
  3. Reticulum
  4. Omasum

Explanation

The abomasum is the fourth compartment of the ruminant stomach and the only true glandular stomach. Its mucosa secretes gastric juice containing hydrochloric acid (HCl), pepsinogen, and lysozyme to digest incoming microbial biomass and unfermented feed bypass particles.

Question 21Biología y razonamiento experimental

What are the three principal Volatile Fatty Acids (VFAs) generated by anaerobic microbial fermentation of plant carbohydrates in the bovine rumen, which supply up to 70-80% of the animal's digestible energy?

  1. Lactate, pyruvate, and citrate
  2. Palmitate, stearate, and oleate
  3. Formate, succinate, and malate
  4. Acetate, propionate, and butyrate Correct answer.

Explanation

Ruminal microorganisms ferment cellulose, hemicellulose, and starches into short-chain volatile fatty acids: acetate (acetic acid, ~60-70%), propionate (propionic acid, ~15-20%), and butyrate (butyric acid, ~10-15%). Propionate is the primary gluconeogenic precursor in ruminant liver.

Question 22Biología y razonamiento experimental

In hindgut fermenters such as equids (horses) and lagomorphs (rabbits), in which anatomical region of the digestive tract does primary microbial cellulose fermentation occur?

  1. Simple glandular stomach
  2. Duodenum and jejunum
  3. Cecum and large colon Correct answer.
  4. Esophageal diverticulum

Explanation

Horses and rabbits are hindgut (post-gastric) fermenters. They possess a simple glandular stomach and small intestine for enzymatic digestion of non-structural feeds, while microbial fermentation of fiber occurs downstream in the expanded, highly vascularized cecum and large (ascending) colon.

Question 23Biología y razonamiento experimental

Which pancreatic zymogen is cleaved into its active proteolytic form by enterokinase (enteropeptidase) in the brush border of the feline duodenum, subsequently triggering activation of other digestive proteases?

  1. Trypsinogen (activated to trypsin) Correct answer.
  2. Pepsinogen (activated to pepsin)
  3. Procarboxypeptidase (activated to carboxypeptidase)
  4. Chymotrypsinogen (activated to chymotrypsin)

Explanation

Enterokinase (enteropeptidase), an enzyme anchored to the duodenal mucosa, cleaves an inhibitory hexapeptide from pancreatic trypsinogen to yield active trypsin. Active trypsin then autocatalytically activates more trypsinogen and initiates a cascade activating chymotrypsinogen, proelastase, and procarboxypeptidase.

Question 24Biología y razonamiento experimental

Trace the correct chronological path of deoxygenated blood returning from systemic tissues through the mammalian heart to the pulmonary circuit:

  1. Vena cava → Left atrium → Bicuspid valve → Left ventricle → Aortic semilunar valve → Pulmonary artery
  2. Vena cava → Right atrium → Tricuspid valve → Right ventricle → Pulmonary semilunar valve → Pulmonary artery Correct answer.
  3. Pulmonary vein → Right atrium → Bicuspid valve → Right ventricle → Pulmonary trunk
  4. Vena cava → Right ventricle → Tricuspid valve → Right atrium → Pulmonary vein

Explanation

Systemic venous blood enters the right atrium via the cranial and caudal venae cavae. During ventricular diastole, blood flows across the tricuspid (right atrioventricular) valve into the right ventricle. Upon ventricular systole, it is ejected through the pulmonary semilunar valve into the pulmonary trunk and arteries toward the lungs.

Question 25Biología y razonamiento experimental

Which specialized cardiac tissue serves as the primary intrinsic physiological pacemaker of the mammalian heart, exhibiting spontaneous phase 4 diastolic depolarization?

  1. Atrioventricular (AV) node
  2. Bundle of His
  3. Sinoatrial (SA) node Correct answer.
  4. Purkinje fiber network

Explanation

The sinoatrial (SA) node, located subepicardially at the junction of the cranial vena cava and right atrium, has the highest intrinsic rate of spontaneous diastolic depolarization (mediated by funny sodium channels, I_f). Consequently, it dictates the normal heart rate and acts as the primary pacemaker.

Question 26Biología y razonamiento experimental

What substance is synthesized and secreted by alveolar Type II pneumocytes in the mammalian lung to reduce alveolar surface tension and prevent atelectasis (alveolar collapse) at low lung volumes?

  1. Mucin glycoprotein
  2. Alpha-1 antitrypsin
  3. Bicarbonate-rich serous fluid
  4. Pulmonary surfactant (dipalmitoylphosphatidylcholine complex) Correct answer.

Explanation

Type II pneumocytes produce pulmonary surfactant, a lipoprotein mixture dominated by dipalmitoylphosphatidylcholine (DPPC). By inserting between water molecules lining the alveoli, surfactant lowers surface tension proportionately as alveoli shrink, preventing collapse (atelectasis) and reducing the work of breathing.

Question 27Biología y razonamiento experimental

In an exercising canine athlete, increased skeletal muscle metabolism leads to elevated PCO₂, increased [H⁺] (acidosis), and elevated temperature. How do these physiological changes affect the hemoglobin-oxygen dissociation curve (the Bohr effect)?

  1. The curve shifts to the right, decreasing hemoglobin's oxygen affinity and facilitating O₂ unloading to tissues Correct answer.
  2. The curve shifts to the left, increasing hemoglobin's oxygen affinity to prevent O₂ release
  3. The curve flattens completely into a hyperbolic shape independent of partial pressure
  4. Hemoglobin is irreversibly denatured into methemoglobin, stopping oxygen transport

Explanation

The Bohr effect describes how higher concentrations of carbon dioxide, hydrogen ions (lower pH), and elevated temperature stabilize the deoxygenated (Tense, T) conformation of hemoglobin. This shifts the oxygen-hemoglobin dissociation curve to the right, lowering oxygen affinity and promoting the release of O₂ into metabolically active tissues.

Question 28Biología y razonamiento experimental

In the mammalian kidney, which segment of the nephron is responsible for reabsorbing virtually 100% of filtered glucose and amino acids under normal physiological conditions via sodium-dependent cotransporters?

  1. Descending limb of the loop of Henle
  2. Distal convoluted tubule (DCT)
  3. Proximal convoluted tubule (PCT) Correct answer.
  4. Cortical collecting duct

Explanation

The proximal convoluted tubule (PCT) has extensive apical brush-border microvilli and abundant mitochondria. It reabsorbs ~65% of filtered water and sodium, and nearly 100% of filtered glucose (via SGLT2/SGLT1) and amino acids before tubular fluid leaves this initial segment.

Question 29Biología y razonamiento experimental

What physiological mechanism established in the mammalian renal medulla enables the formation of concentrated, hypertonic urine in response to dehydration?

  1. Passive bulk filtration of plasma proteins across the glomerular basement membrane
  2. The countercurrent multiplier system established by the hairpin loops of Henle and vasa recta Correct answer.
  3. Active excretion of large volumes of water by renal pelvic transitional epithelium
  4. Direct enzymatic hydrolysis of urea into ammonia in the proximal tubule

Explanation

The countercurrent multiplier system, generated by active NaCl transport in the thick ascending limb of the loop of Henle (impermeable to water) paired with the water-permeable descending limb and maintained by hairpin vasa recta capillaries, establishes an osmotic gradient from cortex to inner medulla. In the presence of ADH, water is drawn from collecting ducts into this hypertonic interstitium.

Question 30Biología y razonamiento experimental

Antidiuretic Hormone (ADH, or arginine vasopressin), released from the posterior pituitary, promotes renal water conservation by triggering which cellular action in principal cells of the renal collecting ducts?

  1. Inhibition of the Na⁺/K⁺-ATPase in the basolateral membrane
  2. Direct active transport of water molecules against their chemical gradient via ATP hydrolysis
  3. Stimulation of calcium oxalate crystal precipitation in the renal pelvis
  4. Exocytosis and apical insertion of Aquaporin-2 (AQP2) water channels Correct answer.

Explanation

ADH binds to basolateral V2 receptors on principal cells, activating adenylyl cyclase to increase intracellular cAMP. This stimulates protein kinase A, causing intracellular vesicles containing Aquaporin-2 (AQP2) water channels to fuse with the apical membrane, allowing water to flow passively along the medullary osmotic gradient into the interstitium.

Question 31Biología y razonamiento experimental

Aldosterone, a mineralocorticoid synthesized in the adrenal cortex (zona glomerulosa), acts on the late distal tubule and collecting ducts of domestic animals to produce which electrolyte effects?

  1. Increased Na⁺ and water reabsorption accompanied by increased K⁺ and H⁺ excretion Correct answer.
  2. Increased excretion of Na⁺ and water accompanied by retention of K⁺
  3. Complete inhibition of renal calcium and phosphate reclamation
  4. Selective reabsorption of bicarbonate with massive elimination of circulating sodium

Explanation

Aldosterone upregulates epithelial sodium channels (ENaC) on the apical membrane and Na⁺/K⁺-ATPase on the basolateral membrane of principal cells. This enhances sodium reabsorption (followed by water) while promoting potassium (K⁺) and hydrogen ion (H⁺ via intercalated cells) excretion into the tubular lumen.

Question 32Biología y razonamiento experimental

What is the physiological basis of the rapid depolarizing upstroke (phase 0) of an action potential in a mammalian peripheral motor neuron?

  1. Opening of voltage-gated K⁺ channels, allowing rapid K⁺ efflux into the extracellular space
  2. Closure of resting leak channels, allowing active ATP-driven pumping of calcium ions
  3. Rapid opening of voltage-gated Na⁺ channels, allowing massive Na⁺ influx down its electrochemical gradient Correct answer.
  4. Release of inhibitory neurotransmitter GABA into the intracellular axoplasm

Explanation

When the axon hillock membrane reaches its threshold potential (~ -55 mV), voltage-gated sodium channels rapidly open their activation gates. Driven by both concentration and electrical gradients, Na⁺ rushes into the axon, shifting the membrane potential towards +30 to +40 mV.

Question 33Biología y razonamiento experimental

How does the myelin sheath produced by Schwann cells (in the PNS) or oligodendrocytes (in the CNS) accelerate electrical impulse conduction along myelinated mammalian axons?

  1. It provides ATP directly to the axon core to pump sodium ions at supersonic velocity
  2. It acts as an electrical insulator, restricting action potential regeneration to the unmyelinated nodes of Ranvier (saltatory conduction) Correct answer.
  3. It completely eliminates the need for sodium or potassium channels along the entire nerve fiber
  4. It increases membrane capacitance so that impulses linger continuously at each centimeter

Explanation

Myelin has high electrical resistance and low capacitance. It insulates the internodal axon segments, preventing trans-membrane current loss. Voltage-gated Na⁺ channels are densely clustered at the unmyelinated nodes of Ranvier, allowing the action potential to leap from node to node (saltatory conduction), greatly increasing conduction velocity while conserving metabolic energy.

Question 34Biología y razonamiento experimental

Which neurotransmitter is released by all preganglionic autonomic fibers (both sympathetic and parasympathetic) as well as postganglionic parasympathetic nerve terminals?

  1. Norepinephrine (noradrenaline)
  2. Dopamine
  3. Serotonin (5-hydroxytryptamine)
  4. Acetylcholine (ACh) Correct answer.

Explanation

Acetylcholine is the universal neurotransmitter of all autonomic preganglionic neurons (acting on nicotinic cholinergic receptors), all parasympathetic postganglionic neurons (acting on muscarinic cholinergic receptors), and somatic motor neurons at the neuromuscular junction.

Question 35Biología y razonamiento experimental

In domestic mammals, which hormone produced by the beta cells of the endocrine pancreatic islets of Langerhans promotes glucose uptake into skeletal muscle and adipose tissue via GLUT4 translocation?

  1. Glucagon
  2. Somatostatin
  3. Insulin Correct answer.
  4. Pancreatic polypeptide

Explanation

Insulin is the primary anabolic hormone secreted by pancreatic beta cells during postprandial hyperglycemia. It binds tyrosine kinase receptors, triggering intracellular signaling cascades that translocate GLUT4 glucose transporter vesicles to the plasma membrane of striated muscle and adipocytes, facilitating cellular glucose uptake.

Question 36Biología y razonamiento experimental

In beef cattle, the polled (hornless) trait is governed by a dominant allele (P), while the horned condition is recessive (p). If a heterozygous polled bull (Pp) is mated with a horned cow (pp), what is the expected genotypic and phenotypic ratio among the offspring?

  1. All Pp (100% polled)
  2. 1 Pp : 1 pp (50% polled, 50% horned) Correct answer.
  3. 3 Pp : 1 pp (75% polled, 25% horned)
  4. 1 PP : 2 Pp : 1 pp (25% homozygous polled, 50% heterozygous polled, 25% horned)

Explanation

This is a classic Mendelian testcross. The heterozygous bull produces P and p gametes in equal proportions (1/2 each), while the horned cow produces only p gametes. Offspring genotypes will be 1/2 Pp (polled) and 1/2 pp (horned), resulting in a 1:1 ratio (50% polled, 50% horned).

Question 37Biología y razonamiento experimental

In Shorthorn cattle, coat color exhibits codominance governed by alleles Cᴿ (red) and Cᵂ (white). Heterozygous cattle (CᴿCᵂ) have a roan coat (an intermingled mixture of individual red and white hairs). If two roan individuals are mated, what phenotypic proportions are expected in their calves?

  1. 100% Roan
  2. 75% Red, 25% White
  3. 50% Red, 50% White
  4. 25% Red, 50% Roan, 25% White Correct answer.

Explanation

Because Cᴿ and Cᵂ are codominant, each genotype produces a distinct phenotype. A mating between two heterozygous roan parents (CᴿCᵂ × CᴿCᵂ) yields the genotypic ratio 1/4 CᴿCᴿ (Red) : 1/2 CᴿCᵂ (Roan) : 1/4 CᵂCᵂ (White), meaning the phenotypic ratio matches the 1:2:1 genotypic ratio.

Question 38Biología y razonamiento experimental

In domestic felines, the orange coat color gene is located on the X chromosome (X-linked). Allele Xᴼ produces orange fur, while Xᵒ produces black fur. Because random X-chromosome inactivation (lyonization) occurs in female somatic cells, what coat color pattern is observed almost exclusively in heterozygous female cats (XᴼXᵒ)?

  1. Solid uniform cream coat across the entire body
  2. Complete albinism due to melanin synthesis arrest
  3. Tortoiseshell or Calico (patches of orange and black/brown fur) Correct answer.
  4. Pointed Siamese coloration localized to the extremities

Explanation

Because the orange locus is X-linked, a heterozygous female (XᴼXᵒ) undergoes random X-inactivation early in embryonic development. In cells where the Xᴼ chromosome remains active, pheomelanin (orange) is produced, whereas in cells where the Xᵒ chromosome remains active, eumelanin (black) is produced. This clonal mosaicism yields the tortoiseshell or calico (with white spotting) pattern.

Question 39Biología y razonamiento experimental

In the Manx cat breed, a single copy of the dominant M allele causes taillessness (Mm), but homozygous dominant embryos (MM) fail to develop and die early in utero. If two tailless Manx cats (Mm × Mm) are mated, what phenotypic proportion is expected among the surviving live-born kittens?

  1. 3/4 Tailless Manx : 1/4 Normal-tailed
  2. 2/3 Tailless Manx : 1/3 Normal-tailed Correct answer.
  3. 1/2 Tailless Manx : 1/2 Normal-tailed
  4. 100% Tailless Manx

Explanation

The cross Mm × Mm yields 1 MM : 2 Mm : 1 mm at fertilization. Because MM individuals are embryonic lethals and perish before birth, the surviving live-born population consists only of 2 Mm (tailless Manx) and 1 mm (normal-tailed), producing a characteristic 2:1 (or 2/3 : 1/3) phenotypic ratio.

Question 40Biología y razonamiento experimental

DNA replication is described as semiconservative. What does that mean for the two daughter DNA molecules produced from one parental double helix?

  1. Both parental strands stay together in one daughter molecule, while the second daughter molecule is built entirely from new nucleotides
  2. The parental strands are cut into fragments that are randomly interspersed with newly made fragments in both daughter molecules
  3. Both daughter molecules are assembled entirely from new nucleotides, and the parental helix is degraded
  4. Each daughter molecule keeps one intact parental strand paired with one newly synthesised complementary strand Correct answer.

Explanation

Helicase unwinds the parental double helix and each separated strand serves as a template. DNA polymerase adds nucleotides that pair complementarily with the template (A with T, G with C), so every daughter molecule ends up with one old strand and one new strand. Meselson and Stahl confirmed this pattern experimentally in 1958 using nitrogen-isotope density labelling.

Question 41Biología y razonamiento experimental

During transcription in a mammalian cell, what does RNA polymerase produce from the DNA template, and which base appears in the product wherever the template carries adenine?

  1. A complementary DNA strand, with cytosine appearing opposite template adenine
  2. A finished polypeptide chain, with uracil appearing opposite template adenine
  3. A single-stranded messenger RNA molecule, with uracil appearing opposite template adenine Correct answer.
  4. A double-stranded RNA molecule, with thymine appearing opposite template adenine

Explanation

RNA polymerase reads the DNA template strand in the 3' → 5' direction and builds a single-stranded messenger RNA in the 5' → 3' direction. RNA uses uracil in place of thymine, so wherever the template strand carries adenine, the growing mRNA receives uracil; guanine and cytosine still pair with each other.

Question 42Biología y razonamiento experimental

A mature mRNA molecule reads 5'-AUG GGC UGU UAA-3'. During translation, how many amino acids does this message specify, and what is the role of the final codon?

  1. Three amino acids; UAA is a stop codon that terminates translation and releases the polypeptide Correct answer.
  2. Four amino acids; UAA specifies the amino acid tyrosine
  3. Two amino acids; AUG is only a ribosome-binding signal and is never translated
  4. Twelve amino acids, one for each nucleotide present in the sequence

Explanation

The genetic code is read in non-overlapping triplets. AUG is the initiation codon and is itself translated as methionine, GGC specifies glycine, and UGU specifies cysteine. UAA is one of the three stop codons (UAA, UAG, UGA); it binds a release factor rather than a transfer RNA, so translation ends and the three-residue peptide is released.

Question 43Química fundamental

In a controlled veterinary clinical trial evaluating a novel canine anthelmintic, what term identifies the variable deliberately manipulated by the investigator between experimental and control groups?

  1. Dependent variable
  2. Confounding variable
  3. Standard error variable
  4. Independent variable Correct answer.

Explanation

In scientific experimentation, the independent variable is the factor deliberately altered or manipulated by the researcher (e.g., administration of the anthelmintic drug versus placebo) to observe its effect on the dependent outcome variable (e.g., fecal parasite egg counts).

Question 44Química fundamental

What is the official base unit of thermodynamic temperature in the International System of Units (SI)?

  1. Degree Celsius (°C)
  2. Kelvin (K) Correct answer.
  3. Degree Fahrenheit (°F)
  4. Rankine (°R)

Explanation

The Kelvin (K) is the SI base unit of thermodynamic temperature, defined by fixing the numerical value of the Boltzmann constant. Degrees Celsius (°C) is an accepted derived unit, but Kelvin is the fundamental base unit.

Question 45Química fundamental

An environmental toxicologist tests a water supply for bovine livestock and determines a fluoride concentration of 2.5 parts per million (ppm, w/v). What mass of fluoride is present in a 400 L livestock watering trough?

  1. 250 mg (0.25 g)
  2. 100 mg (0.10 g)
  3. 1,000 mg (1.0 g) Correct answer.
  4. 10,000 mg (10 g)

Explanation

By definition, 1 ppm (w/v in dilute aqueous solution) equals 1 mg per liter (1 mg/L). Therefore, a concentration of 2.5 ppm corresponds to 2.5 mg/L. In a 400 L trough, the total mass of dissolved fluoride is: 2.5 mg/L × 400 L = 1,000 mg = 1.0 g.

Question 46Química fundamental

Which of the following processes represents a chemical change rather than a physical change?

  1. Metabolic oxidation of blood glucose to carbon dioxide and water Correct answer.
  2. Freezing of liquid water into crystalline ice in an animal waterer
  3. Grinding whole grain corn kernels into fine mash feed
  4. Evaporation of sweat from equine skin during exercise cooling

Explanation

The cellular oxidation of glucose (C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O) breaks and reforms covalent chemical bonds, producing entirely new chemical substances with distinct molecular identities-the hallmark of a chemical change. Freezing, grinding, and evaporating are physical changes where chemical composition remains unaltered.

Question 47Química fundamental

In veterinary pharmacology, how does a true homogeneous solution (such as 5% dextrose injection) differ fundamentally from a heterogeneous suspension (such as procaine penicillin G for injection)?

  1. A true solution exhibits strong light scattering (Tyndall effect), whereas suspensions are optically transparent
  2. A true solution can be separated by ordinary filter paper, whereas suspensions pass freely through any membrane
  3. Suspensions consist of single-phase uniform mixtures, whereas solutions always contain distinct liquid phases
  4. In a true solution, solute particles are molecular or ionic (<1 nm) and will never settle upon standing, whereas suspensions contain insoluble particles (>1,000 nm) that settle over time Correct answer.

Explanation

True solutions are homogeneous, single-phase mixtures where solute particles are dissolved as individual ions or small molecules (<1 nm) that do not scatter visible light and never settle out by gravity. Suspensions are heterogeneous mixtures containing solid particles large enough (>1,000 nm) to settle upon standing and require resuspension prior to administration.

Question 48Química fundamental

A neutral atom of iron has an atomic number Z = 26 and a mass number A = 56. What are the numbers of protons, neutrons, and electrons in this neutral atom?

  1. 26 protons, 56 neutrons, and 26 electrons
  2. 26 protons, 30 neutrons, and 26 electrons Correct answer.
  3. 30 protons, 26 neutrons, and 30 electrons
  4. 26 protons, 26 neutrons, and 30 electrons

Explanation

The atomic number (Z = 26) defines the number of nuclear protons. In a neutral atom, the number of electrons equals the number of protons (26). The mass number (A = 56) is the sum of protons and neutrons, so the number of neutrons is A - Z = 56 - 26 = 30 neutrons.

Question 49Química fundamental

Radioactive Iodine-131 (¹³¹I) is used in feline veterinary medicine to treat hyperthyroidism and has a half-life of 8.0 days. If a feline thyroid therapeutic dose initially contains 160 MBq of ¹³¹I, what activity remains after 24.0 days?

  1. 40 MBq
  2. 10 MBq
  3. 20 MBq Correct answer.
  4. 80 MBq

Explanation

The elapsed time is t = 24.0 days. The number of elapsed half-lives is n = 24.0 / 8.0 = 3 half-lives. The remaining activity is calculated as A = A₀ × (1/2)ⁿ = 160 MBq × (1/2)³ = 160 / 8 = 20 MBq.

Question 50Química fundamental

Which quantum rule dictates that when electrons occupy degenerate orbitals (such as the three 2p orbitals), they must occupy them singly with parallel spins before pairing?

  1. Hund's rule of maximum multiplicity Correct answer.
  2. Pauli exclusion principle
  3. Aufbau principle
  4. Heisenberg uncertainty principle

Explanation

Hund's rule states that for degenerate orbitals (orbitals of identical energy), the lowest energy configuration is attained when electrons occupy each orbital singly with their spins aligned in parallel before any orbital receives a second, spin-paired electron.