Examkin

Form and Function

256 questions

All types256Multiple choice131Matching71Ordering54
Question 161Question →

During nutrient transport and circulatory routing in mammals, blood absorbed from the small intestine must travel through specific vascular networks and heart chambers before reaching systemic organs. What is the correct physiological sequence of blood flow from the intestinal capillaries to the main systemic artery?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of blood flow from the small intestine to systemic arterial delivery is: (1) Intestinal villi capillaries to hepatic portal vein and liver sinusoids -> (2) Hepatic vein to inferior vena cava and right atrium -> (3) Right ventricle through pulmonary trunk to pulmonary capillaries -> (4) Pulmonary veins to left atrium and left ventricle -> (5) Ejection from left ventricle into the systemic aorta.
The sequence correctly traces blood through the mammalian cardiovascular system: intestinal capillaries feed into the hepatic portal system (liver sinusoids), exiting via hepatic veins into the inferior vena cava to enter the right atrium. Deoxygenated blood is then pumped by the right ventricle to the lungs via pulmonary arteries. Oxygenated blood returns through pulmonary veins into the left atrium, moves to the left ventricle, and is ejected into the aorta for systemic distribution.

Step-by-Step Solution

1
Trace hepatic portal movement
Blood carrying absorbed nutrients drains from intestinal capillaries into the hepatic portal vein to be processed in liver sinusoids.
Mammalian circulatory design directs blood from digestive capillaries directly to liver capillaries before systemic venous return.
2
Trace venous return to the heart
Blood leaves the liver via the hepatic vein, joins the inferior vena cava, and enters the right atrium.
Systemic venous return collects deoxygenated blood and returns it to the right atrium.
3
Trace pulmonary arterial delivery
Blood flows into the right ventricle and is pumped into the pulmonary trunk/arteries leading to alveolar capillaries.
The right ventricle supplies the low-pressure pulmonary circuit for oxygenation.
4
Trace pulmonary venous return to systemic heart
Oxygenated blood returns via pulmonary veins into the left atrium and moves into the left ventricle.
Double circulation routes pulmonary return exclusively to the left side of the heart.
5
Trace systemic arterial ejection
The left ventricle contracts, forcing blood into the systemic aorta.
High hydrostatic pressure generated by the muscular left ventricle distributes oxygenated blood across the systemic body tissues.

Key Concept

Integration of Hepatic Portal and Pulmonary-Systemic Circuits
Estimated Time:2m 0s
Question 162Question →

Match each plant excretory structure or product on the left with its corresponding physiological mechanism or mode of elimination on the right.

Click a left item, then click its matching right item

Items

Stomata and lenticels
Hydathodes
Old bark and heartwood
Calcium oxalate crystals (raphides)

Matches

Show answer & explanation

Answer

Stomata and lenticels match with the release of gaseous metabolic by-products via simple diffusion; Hydathodes match with exudation of liquid water containing dissolved salts through guttation; Old bark and heartwood match with deposition and long-term storage of tannins, resins, and gums in non-functional tissues; Calcium oxalate crystals (raphides) match with insoluble waste precipitation within vacuoles to prevent cellular toxicity.
Each plant excretory structure is correctly paired with its specific mechanism: stomata and lenticels eliminate gaseous by-products by diffusion; hydathodes eliminate liquid water droplets containing dissolved minerals through guttation; old bark and heartwood store secondary organic metabolites like tannins and gums; calcium oxalate crystals safely isolate metabolic oxalic acid in an insoluble crystalline form inside vacuoles.

Step-by-Step Solution

1
Identify the primary excretory pathway for plant gases.
Stomata (in epidermal tissues of leaves) and lenticels (in bark of woody stems) serve as diffusion channels for CO2CO_2 from respiration and O2O_2 from photosynthesis.
Gaseous waste elimination relies on direct kinetic movement of gas molecules down concentration gradients.
2
Determine the mechanism associated with hydathodes.
Hydathodes exude drops of liquid water and dissolved minerals during guttation when transpiration rates are low and root pressure is elevated.
Hydathodes are permanently open pores at leaf margins distinct from stomatal guard cells.
3
Analyze how plants store organic secondary metabolites.
Non-utilizable organic substances such as resins, gums, and tannins are stored in non-functional secondary xylem (heartwood) or bark prior to organ shedding.
Plants lack complex excretory organs and frequently use tissue sequestration followed by abscission.
4
Evaluate the role of calcium oxalate crystal formation.
Toxic oxalic acid is neutralized by binding with calcium ions to yield insoluble raphide crystals stored inertly inside vacuoles.
Precipitating waste as insoluble salt crystals prevents osmotic imbalance and cytoplasmic chemical toxicity.

Key Concept

Plant Excretory Mechanisms and Waste Storage
Question 163Question →

During a plant physiology experiment, a potted plant is supplied with carbon dioxide containing radioactively labeled oxygen-18 (C18O2\text{C}^{18}\text{O}_2) while exposed to bright sunlight. In which of the following photosynthetic products will the radioactive 18O^{18}\text{O} isotope primarily be detected?

Show answer & explanation

Answer: Glucose synthesized during the light-independent reactions in the stroma

Answer

The radioactively labeled oxygen (18O^{18}\text{O}) from carbon dioxide will primarily appear in glucose synthesized during the light-independent reactions (Calvin cycle).
Carbon dioxide enters the light-independent stage (Calvin cycle) where it combines with ribulose 1,5-bisphosphate (RuBP). The oxygen atoms present in CO2\text{CO}_2 form part of the backbone of 3-phosphoglycerate and are subsequently reduced into organic carbohydrates like glucose. Therefore, isotopic 18O^{18}\text{O} supplied as C18O2\text{C}^{18}\text{O}_2 ends up in glucose.

Step-by-Step Solution

1
Identify the origin of released oxygen gas (O2\text{O}_2) in photosynthesis.
Photolysis of water (H2O→2H++2e−+12O2\text{H}_2\text{O} \rightarrow 2\text{H}^+ + 2\text{e}^- + \frac{1}{2}\text{O}_2) during the light-dependent reaction produces atmospheric oxygen.
Water molecules are split in photosystem II inside the thylakoid lumen to supply electrons.
2
Trace the biochemical pathway of carbon dioxide (CO2\text{CO}_2) in photosynthesis.
Carbon dioxide is fixed by RuBisCO during the light-independent stage (Calvin cycle) and reduced to 3-phosphoglycerate and glyceraldehyde-3-phosphate (G3P).
The oxygen atoms attached to carbon in CO2\text{CO}_2 remain bound through the carbon fixation sequence.
3
Determine the final destination of 18O^{18}\text{O} from C18O2\text{C}^{18}\text{O}_2.
The labeled oxygen is incorporated directly into hexose sugars (glucose) and water formed during stromal dark reactions.
Since CO2\text{CO}_2 is used to build organic molecules, its isotopic oxygen tag ends up in carbohydrate molecules.

Key Concept

Origin of oxygen atoms in photosynthetic products (Photolysis vs. Calvin Cycle Carbon Fixation)
Estimated Time:1m 15s
Question 164Question →

An adult herbivorous mammal, such as a sheep, has a dental formula represented as i03,c01,pm33,m33i \frac{0}{3}, c \frac{0}{1}, pm \frac{3}{3}, m \frac{3}{3}. What is the total number of teeth present in the animal's mouth, and how does the structural modification of its upper jaw assist in feeding?

Show answer & explanation

Answer: 32 teeth; a fibrous horny pad on the upper jaw works against the lower incisors to crop and clip grass.

Answer

32 teeth; a fibrous horny pad on the upper jaw works against the lower incisors to crop and clip grass.
The dental formula i03,c01,pm33,m33i \frac{0}{3}, c \frac{0}{1}, pm \frac{3}{3}, m \frac{3}{3} represents one quadrant pair (half of the upper and lower jaws). Summing the upper teeth (0+0+3+3=60+0+3+3=6) and lower teeth (3+1+3+3=103+1+3+3=10) gives 16 teeth per side. Multiplying by 2 yields a total of 32 teeth. In herbivorous ruminants like sheep, upper incisors are absent and replaced by a tough dental pad that works against the lower incisors to pull and crop vegetation.

Step-by-Step Solution

1
Calculate the number of teeth on one side of the upper jaw.
Upper half-jaw = 0 incisors + 0 canines + 3 premolars + 3 molars = 6 teeth.
The top numbers in the dental formula indicate the teeth present in one half of the upper jaw.
2
Calculate the number of teeth on one side of the lower jaw.
Lower half-jaw = 3 incisors + 1 canine + 3 premolars + 3 molars = 10 teeth.
The bottom numbers in the dental formula indicate the teeth present in one half of the lower jaw.
3
Sum the upper and lower teeth for one side and multiply by 2 for the full mouth.
(6 + 10) × 2 = 16 × 2 = 32 total teeth.
Dental formulas represent only one half of the symmetrical skull, requiring multiplication by two to determine total dentition.
4
Analyze the functional adaptation of the upper jaw in ruminant herbivores.
Upper incisors and canines are absent and replaced by a firm, calloused dental (horny) pad.
During grazing, grass is held between the lower incisors and the horny pad and sheared off with a upward jerk of the head.

Key Concept

Mammalian dental formulas and specialized herbivore feeding adaptations
Question 165Question →

When a person rapidly shifts their visual gaze from a distant object on the horizon to read fine print in a book held close to the face, which of the following physiological responses occurs within the mammalian eye to bring the near object into sharp focus on the retina?

Show answer & explanation

Answer: The ciliary muscles contract, causing the suspensory ligaments to slacken, which allows the elastic lens to become thicker and more convex.

Answer

Near vision accommodation occurs when the ciliary muscles contract, tension on the suspensory ligaments decreases (slackens), and the crystalline lens becomes thicker and more convex to increase light refraction.
Accommodation for near vision requires the eye to increase its refractive power. When focusing on a near object, parasympathetic nerve signals cause the ciliary muscles to contract. Because the ciliary muscle forms a sphincter ring surrounding the lens, its contraction narrows the ring and reduces the outward pull on the suspensory ligaments (zonules of Zinn). With tension removed from the ligaments, the natural elasticity of the lens causes it to recoil into a thicker, more convex shape, sharply focusing divergent light rays onto the retina.

Step-by-Step Solution

1
Identify the visual requirement for focusing on a near object.
Light rays arriving from a near object are divergent and require stronger refraction (greater bending power) to converge accurately on the retina.
Near objects require a shorter focal length and higher optical refractive power.
2
Determine the required lens shape change for increased focal power.
The lens must become more spherical (thicker and more convex).
Increased curvature of the lens increases its refractive index and shortens the focal distance to focus light onto the retina.
3
Analyze the mechanical action of the ciliary muscles and suspensory ligaments.
Contraction of the sphincter-like ciliary body moves it closer to the lens, releasing tension on the suspensory ligaments and allowing lens bulge due to its natural elasticity.
Ciliary muscle contraction relaxes suspensory ligament strain, enabling maximum lens convexity.

Key Concept

Mechanism of visual accommodation in the mammalian eye
Question 166Question →

When an individual transitions from focusing on fine print in a brightly lit room to viewing a distant star in the night sky, which combination of ocular physiological changes occurs to adjust focal length and light entry?

Show answer & explanation

Answer: Ciliary muscles relax, suspensory ligaments become taut, and radial iris muscles contract.

Answer

Ciliary muscles relax, suspensory ligaments become taut, and radial iris muscles contract.
For distant vision, the ciliary muscles relax, causing the suspensory ligaments to become taut and pull the lens into a flatter shape. Simultaneously, under dim light conditions such as looking at the night sky, the radial iris muscles contract under sympathetic control to dilate the pupil and maximize light capture.

Step-by-Step Solution

1
Analyze accommodation for distant vision
For distant vision, the ciliary muscles relax, widening the ciliary body ring. This increases tension on the suspensory ligaments (making them taut), pulling the lens into a flatter, less convex shape to increase focal length.
Distant light rays require less refraction to focus accurately onto the retina.
2
Analyze pupillary reflex for low light conditions
In dim light (viewing a night sky), sympathetic nerve impulses stimulate the radial muscles of the iris to contract while circular muscles relax, dilating the pupil.
Pupillary dilation maximizes light entry into the eye under low ambient illumination.
3
Synthesize the combined physiological response
Combining distant accommodation and dim light adaptation yields relaxed ciliary muscles, taut suspensory ligaments, and contracted radial iris muscles.
Lens curvature adjustments and pupil aperture modifications occur concurrently when shifting focus and ambient lighting.

Key Concept

Visual accommodation and pupillary light reflex mechanisms in the human eye
Question 167Question →

Arrange the following physiological events involved in the perception of smell (olfaction) in a mammal in the correct sequential order from initial stimulus entry to brain perception.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequential order is: (1) Odorant molecules dissolve in the layer of mucus covering the olfactory epithelium, (2) Odorant molecules bind to specialized receptor proteins on the cilia of olfactory sensory neurons, (3) Nerve impulses travel along sensory nerve fibers through the cribriform plate to the olfactory bulb, and (4) Nerve impulses travel along the olfactory tract to the olfactory cortex of the brain for interpretation.
Olfaction begins when airborne odorants dissolve in nasal mucus. The dissolved chemicals bind to membrane receptors on olfactory cilia, generating nerve impulses. These impulses pass via sensory axons into the olfactory bulb and subsequently along the olfactory tract to the brain's olfactory cortex for processing.

Step-by-Step Solution

1
Identify the initial physical interaction of the stimulus
Inhaled odorant molecules dissolve in the fluid layer coating the nasal sensory epithelium.
Chemoreceptors require chemical substances to be in aqueous solution to interact with receptor sites.
2
Determine the signal transduction phase
Dissolved odorants bind to specific protein receptors on the sensory cilia, generating an action potential.
Receptor binding initiates depolarization in the olfactory neuron membrane.
3
Trace the initial neural pathway to the primary relay center
Action potentials pass through the cribriform plate into the olfactory bulb.
Olfactory nerve axons penetrate the ethmoid bone to synapse inside the olfactory bulb.
4
Follow the path to final sensory processing
Relay neurons carry the electrical signals along the olfactory tract to the olfactory cortex.
Conscious olfactory discrimination occurs in the higher brain center.

Key Concept

Olfactory transduction and neural pathway of smell perception.
Question 168Question →

In a ringing (girdling) experiment on a woody dicotyledonous stem, a outer ring of bark containing the phloem is removed while leaving the xylem intact. Which of the following observations will occur above the ring after several days?

Show answer & explanation

Answer: Swelling caused by the accumulation of manufactured organic nutrients

Answer

Swelling caused by the accumulation of manufactured organic nutrients
Removing a ring of bark removes the phloem, which translocates manufactured organic food downward from the leaves. Because the xylem is intact, water and minerals continue to travel upward, but sugars cannot move past the ring, resulting in food accumulation and swelling above the ringed area.

Step-by-Step Solution

1
Identify the tissue removed and its physiological function
Ringing removes phloem tissue from the stem, which is responsible for translocating organic solutes (photosynthates).
Phloem translocates organic food manufactured in leaves downward toward the roots.
2
Identify the tissue left intact and its physiological function
Xylem vessels remain intact, maintaining upward transport of water and mineral salts from the soil to the leaves.
Xylem is situated deeper within the vascular bundle and is not removed during superficial stem girdling.
3
Deduce the resulting accumulation point
Downwards movement of sugars is blocked at the ring, leading to nutrient buildup and cell expansion (swelling) immediately above the cut.
Accumulation of organic solutes above the blockage increases local osmotic concentration and cell division.

Key Concept

Phloem Translocation and Ringing Experiment
Question 169Question →

Pancreatic juice contains key digestive enzymes, including trypsin and pancreatic lipase, which act in the duodenum. Which of the following environmental conditions is required for these enzymes to function at their optimal rate?

Show answer & explanation

Answer: An alkaline medium (pH 7.5−8.5\text{pH } 7.5 - 8.5) at a body temperature of 37∘C37^\circ\text{C}

Answer

An alkaline medium (pH 7.5−8.5\text{pH } 7.5 - 8.5) at a body temperature of 37∘C37^\circ\text{C}
Pancreatic enzymes such as trypsin and lipase require an alkaline environment (pH 7.5−8.5\text{pH } 7.5 - 8.5) created by alkaline secretions in the duodenum, functioning at their maximum catalytic rate at mammalian body temperature (37∘C37^\circ\text{C}).

Step-by-Step Solution

1
Identify the site of action and origin of pancreatic enzymes
Trypsin and pancreatic lipase are secreted by the pancreas into the duodenum.
Understanding the physiological environment of the duodenum determines the required pH.
2
Determine the optimal pH for pancreatic enzyme function
The neutralization of acidic chyme by bile and sodium hydrogen carbonate creates an alkaline environment (pH 7.5−8.5\text{pH } 7.5 - 8.5).
Pancreatic enzymes require an alkaline pH to achieve their maximum catalytic rate.
3
Determine the optimal temperature condition for mammalian biological catalysis
Optimal enzyme action occurs at mammalian body temperature (37∘C37^\circ\text{C}).
Lower temperatures cause temporary inactivation due to low kinetic energy, while extreme high temperatures causes permanent denaturation.

Key Concept

Pancreatic enzyme activity dependence on alkaline pH and optimal body temperature
Estimated Time:1m 0s
Question 170Question →

In insects, gaseous exchange occurs through a network of internal air tubes called tracheae that terminate in fine, fluid-filled tracheoles in direct contact with active tissues. During intense muscular exertion, metabolic activity leads to lactic acid accumulation within muscle cells, significantly raising their internal solute concentration. Which of the following best describes the physiological effect of this osmotic shift on gas exchange at the tracheoles?

Show answer & explanation

Answer: Water moves by osmosis from the tracheoles into the muscle cells, drawing air deeper into the tracheoles and increasing the surface area available for gaseous diffusion.

Answer

Water moves by osmosis from the tracheoles into the muscle cells, drawing air deeper into the tracheoles and increasing the surface area available for gaseous diffusion.
During vigorous muscle contraction, anaerobic respiration generates lactic acid within the tissue cells. This increases the internal solute concentration of the muscle cells. Consequently, water moves by osmosis from the fluid-filled tracheole tips into the muscle cells. As the liquid recedes, air travels deeper into the tracheoles, bringing oxygen closer to the mitochondria and speeding up gaseous exchange because oxygen diffuses much faster in air than in liquid.

Step-by-Step Solution

1
Analyze the resting state of tracheoles
At rest, the terminal ends of tracheoles contain liquid, which limits the rate of oxygen diffusion because gases diffuse much slower through liquids than through air.
Understanding baseline physiological conditions is necessary to determine the direction of physical shifts.
2
Determine the osmotic change during intense exertion
Anaerobic respiration in active muscles produces lactic acid, increasing the solute concentration (hypertonicity) inside the muscle cells relative to the tracheole fluid.
Water moves down its water potential gradient toward areas of higher solute concentration.
3
Evaluate the movement of water and its effect on gas exchange
Water moves out of the tracheoles and into the muscle cells by osmosis. As the fluid level drops, air extends further down the tracheoles closer to the cell membranes, dramatically reducing diffusion distance and expanding the surface area for direct gas-phase diffusion.
Gaseous diffusion in air is thousands of times faster than dissolved gas diffusion in liquid.

Key Concept

Insect Tracheal Adaptation and Osmotic Control of Tracheole Fluid
Question 171Question →

Match each region of the mammalian brain in Column A with its primary physiological function in Column B.

Click a left item, then click its matching right item

Items

Cerebrum
Cerebellum
Medulla oblongata
Hypothalamus

Matches

Show answer & explanation

Answer

Cerebrum matches with controlling voluntary actions and conscious thought; Cerebellum matches with coordinating muscle movements and posture; Medulla oblongata matches with regulating autonomic involuntary processes like breathing and heartbeat; Hypothalamus matches with regulating body temperature and homeostatic balance.
Each brain region is matched to its physiological function: cerebrum governs voluntary actions and intelligence; cerebellum manages muscle posture and balance; medulla oblongata controls vital autonomic reflexes such as respiration and heartbeat; hypothalamus regulates homeostatic functions like thermoregulation.

Step-by-Step Solution

1
Identify the function of the cerebrum
It governs conscious sensory interpretation, reasoning, intelligence, and voluntary movement.
The cerebral cortex is the center for higher cognitive and voluntary motor functions.
2
Identify the function of the cerebellum
It integrates sensory inputs from muscles and joints to maintain balance and smooth motor coordination.
Precise muscle coordination and posture maintenance occur in the cerebellum.
3
Identify the function of the medulla oblongata
It controls life-sustaining involuntary actions including heart rate, vasomotion, and breathing rate.
Autonomic reflex centers for vital organ systems are located in the brainstem/medulla.
4
Identify the function of the hypothalamus
It acts as the primary integrator for autonomic homeostasis, thermoregulation, and osmoregulation.
Internal physiological equilibrium is monitored and adjusted by the hypothalamus.

Key Concept

Brain Region Functions in Central Nervous System Coordination
Estimated Time:1m 0s
Question 172Question →

In the alimentary canal of the cockroach (*Periplaneta americana*), specialized anatomical regions carry out distinct physiological processes during nutrition. Which of the following statements correctly describes the function of a named section of this digestive system?

Show answer & explanation

Answer: Hepatic caeca secrete digestive enzymes into the midgut to facilitate extracellular chemical digestion.

Answer

Hepatic caeca secrete digestive enzymes into the midgut to facilitate extracellular chemical digestion.
In the cockroach, six to eight finger-like projections called hepatic (gastric) caeca are present at the junction of the gizzard and the midgut. These caeca secrete digestive enzymes into the mesenteron (midgut), where the chemical hydrolysis of carbohydrates, proteins, and fats takes place extracellularly.

Step-by-Step Solution

1
Analyze the structural organization of the insect (cockroach) digestive tract.
The digestive tract is divided into foregut (stomodaeum), midgut (mesenteron), and hindgut (proctodaeum).
Understanding regional specialization is required to evaluate each functional claim.
2
Evaluate the physiological function of the hepatic (gastric) caeca.
Six to eight blind-ended hepatic caeca are situated at the junction of the foregut and midgut. They secrete digestive juice containing enzymes into the midgut.
This confirms that chemical digestion in the midgut relies on secretions from the hepatic caeca.
3
Differentiate digestive function from excretory function and enzyme temperature kinetics.
Malpighian tubules handle excretion, gizzards grind food mechanically inside the body, and low temperatures cause reversible inactivation rather than denaturation.
This rules out all incorrect alternatives based on biological principles.

Key Concept

Insect digestive system anatomy and physiological modifications in animal nutrition
Question 173Question →

Match each plant transport mechanism or structural feature listed on the left with its corresponding physiological description or function on the right.

Click a left item, then click its matching right item

Items

Apoplast pathway
Symplast pathway
Casparian strip
Transpiration pull

Matches

Show answer & explanation

Answer

The correct pairings are: Apoplast pathway matches water movement through non-living cell walls and intercellular spaces; Symplast pathway matches water movement through living cytoplasm via plasmodesmata; Casparian strip matches waxy suberin barrier in root endodermis; Transpiration pull matches tension generated by evaporation at leaf stomata.
Each plant transport term directly pairs with its anatomical definition or physiological role: apoplast with cell wall spaces, symplast with cytoplasm and plasmodesmata, Casparian strip with suberized endodermal barrier, and transpiration pull with evaporative tension.

Step-by-Step Solution

1
Differentiate extracellular and intracellular water transport routes in roots.
The apoplast route utilizes non-living cell walls, whereas the symplast route moves water through cytoplasm across plasmodesmata connections.
Apoplastic flow does not cross plasma membranes, while symplastic flow moves through the living protoplast continuum.
2
Identify the endodermal regulation mechanism for root water uptake.
The Casparian strip, composed of waterproof suberin, blocks apoplastic transport in the root endodermis.
This structural restriction forces water and dissolved ions to cross a selective plasma membrane into the symplast.
3
Identify the tension-generating force responsible for mass flow of water in xylem.
Transpiration pull creates negative pressure via evaporation at stomatal pores.
Evaporative water loss generates tension that pulls the continuous xylem water column upward.

Key Concept

Plant Transport Pathways and Driving Forces
Question 174Question →

Arrange the following anatomical structures in the correct chronological sequence through which a erythrocyte travels, starting from the systemic venous return via the vena cava until it is pumped into the systemic circulation through the aorta.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of blood flow from systemic return to systemic ejection is: Right atrium → Right ventricle → Pulmonary artery → Pulmonary vein → Left ventricle.
In mammals, double circulation ensures complete separation of oxygenated and deoxygenated blood. Deoxygenated blood returns via the vena cava into the right atrium, moves into the right ventricle, and is pumped through the pulmonary artery to the lungs for gaseous exchange. Oxygenated blood leaves the lungs through the pulmonary vein, enters the left atrium, moves to the left ventricle, and is subsequently propelled into the aorta for systemic distribution.

Step-by-Step Solution

1
Identify the entry chamber for systemic deoxygenated blood.
Deoxygenated blood from the body tissues enters the right atrium via the venae cavae.
The right atrium serves as the receiving chamber for systemic venous blood.
2
Trace the movement into the pulmonary pump chamber.
Blood passes through the tricuspid valve into the right ventricle.
The right ventricle is responsible for generating pressure to propel blood into the pulmonary circuit.
3
Determine the vessel leaving the right side of the heart.
The right ventricle pumps blood into the pulmonary artery.
Pulmonary arteries carry deoxygenated blood away from the heart to the lungs.
4
Trace the vessel returning oxygenated blood to the heart.
Blood is oxygenated in pulmonary capillaries and returns to the left atrium via the pulmonary vein.
Pulmonary veins carry oxygenated blood from the lungs back to the left side of the heart.
5
Identify the final muscular chamber prior to systemic distribution.
Blood moves from the left atrium into the left ventricle before being ejected into the aorta.
The thick-walled left ventricle generates high pressure to distribute oxygenated blood throughout systemic organs.

Key Concept

Mammalian Double Circulation and Pathway of Blood Flow
Question 175Question →

What is the primary physiological function of the myelin sheath surrounding axon fibers in the mammalian nervous system?

Show answer & explanation

Answer: To increase the speed of nerve impulse transmission along the axon

Answer

The primary physiological function of the myelin sheath is to increase the speed of nerve impulse transmission along the axon.
The myelin sheath provides electrical insulation along the axon, enabling action potentials to jump from one Node of Ranvier to the next in a process called saltatory conduction, which greatly increases impulse conduction speed.

Step-by-Step Solution

1
Identify the structure and composition of the myelin sheath.
The myelin sheath is a lipid-rich layer formed around nerve axons by Schwann cells in the peripheral nervous system and oligodendrocytes in the central nervous system.
Understanding its lipid composition clarifies its role as an electrical insulator.
2
Relate the insulating structure to nerve impulse propagation.
Depolarization occurs only at uninsulated gaps called Nodes of Ranvier, enabling action potentials to leap rapidly along the axon.
This process, known as saltatory conduction, significantly increases the velocity of nerve signal transmission.

Key Concept

Function of Myelin Sheath in Nerve Impulse Conduction
Question 176Question →

When an animal detects a sudden noxious chemical stimulus on its skin, a rapid spinal reflex action is initiated to withdraw the affected limb. Which sequence correctly describes the directional flow of nerve impulses along the neural pathway of this reflex arc?

Show answer & explanation

Answer: Sense receptor →\rightarrow afferent neuron →\rightarrow spinal interneuron →\rightarrow efferent neuron →\rightarrow muscle effector

Answer

Sense receptor →\rightarrow afferent neuron →\rightarrow spinal interneuron →\rightarrow efferent neuron →\rightarrow muscle effector
The correct response accurately traces the unidirectional impulse pathway: sensory receptors detect the stimulus, sensory (afferent) neurons carry the impulse to interneurons (relay neurons) in the central nervous system, and motor (efferent) neurons carry the impulse out to the effector muscle.

Step-by-Step Solution

1
Identify the initial reception of the stimulus
The noxious chemical stimulus activates specialized cutaneous sense receptors.
Receptors detect environmental changes and initiate electrical action potentials.
2
Trace sensory conduction to the central nervous system
Impulses travel along sensory (afferent) neurons into the dorsal horn of the spinal cord.
Afferent pathways transmit sensory information toward the central nervous system.
3
Identify central integration and motor transmission
The impulse passes across synapses via spinal interneurons to motor (efferent) neurons.
Interneurons process the signal in the spinal grey matter and relay it to efferent neurons exiting via the ventral root.
4
Trace motor conduction to the target organ
Efferent neurons carry impulses to skeletal muscle effectors, causing muscle contraction and limb withdrawal.
Effector organs carry out the physical response to remove the organism from the harmful stimulus.

Key Concept

Neuron sequence in a spinal reflex arc
Question 177Question →

During seed germination in cereal grains, gibberellin plays a vital role in mobilizing food reserves stored in the endosperm. Arrange the following steps of gibberellin-mediated seed germination in the correct physiological sequence from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct physiological sequence is: Water imbibition by the dry seed activates metabolic activity in the embryo -> The activated embryo synthesizes and secretes gibberellic acid -> Gibberellic acid diffuses across the seed tissue to the aleurone layer -> Target cells in the aleurone layer synthesize digestive enzymes, including alpha-amylase -> Alpha-amylase hydrolyzes stored insoluble starch in the endosperm into soluble sugars for seedling growth.
Seed germination begins with water absorption (imbibition), which stimulates the embryo to synthesize gibberellic acid. Gibberellin then diffuses to the aleurone layer, where it induces gene expression and synthesis of alpha-amylase. Alpha-amylase degrades insoluble endosperm starch into simple sugars that nourish the growing embryo.

Step-by-Step Solution

1
Identify the initial physical stimulus for germination.
Imbibition of water activates the embryo.
Water absorption hydrates seed tissues and initiates metabolic reactions.
2
Determine the initial endocrine signal produced by the embryo.
Embryo produces gibberellic acid.
Gibberellin is the primary plant growth regulator that triggers mobilization of reserve food.
3
Trace the pathway of hormone transport.
Gibberellin diffuses to the aleurone layer.
The aleurone layer consists of target tissue surrounding the endosperm.
4
Determine the response of the target aleurone cells.
Aleurone cells synthesize digestive enzymes like alpha-amylase.
Gibberellin stimulates the synthesis of hydrolytic enzymes needed for starch digestion.
5
Identify the final biochemical outcome of enzyme activity.
Starch is converted into soluble sugars to feed the growing seedling.
Soluble glucose and maltose supply energy for respiration and cell elongation in the developing shoot and root.

Key Concept

Gibberellin-Induced Mobilization of Endosperm Reserves
Question 178Question →

Match each nitrogenous metabolic waste product or excretory pigment listed on the left with its corresponding characteristic, primary organism group, or elimination pathway on the right.

Click a left item, then click its matching right item

Items

Ammonia
Urea
Uric acid
Bile pigments (Bilirubin & Biliverdin)

Matches

Show answer & explanation

Answer

Ammonia matches with highly toxic, water-soluble waste excreted by freshwater bony fishes; Urea matches with moderately toxic waste synthesized in the liver and excreted by mammals; Uric acid matches with insoluble, non-toxic paste excreted by birds and insects; Bile pigments match with waste products from hemoglobin breakdown eliminated in feces.
Each excretory product corresponds directly to the organism's evolutionary adaptation for water conservation and nitrogenous toxicity management. Ammonia is excreted by aquatic fishes, urea by mammals, uric acid by birds/insects, and bile pigments by the liver via the digestive tract.

Step-by-Step Solution

1
Identify the excretory toxicity and solubility characteristics of nitrogenous wastes.
Ammonia requires large volumes of water due to high toxicity; Urea is moderately soluble and moderately toxic; Uric acid precipitates easily and requires negligible water.
Organism excretory products adapt directly to habitat water availability.
2
Map each waste product to its representative organism class and physiological process.
Aquatic teleosts excrete Ammonia; Mammals synthesize Urea in the liver; Birds/insects precipitate Uric acid; Liver excretes Bile pigments from erythrocyte breakdown.
Metabolic pathways convert nitrogenous products according to water-conservation needs.

Key Concept

Nitrogenous Waste Elimination and Liver Excretory Functions
Estimated Time:1m 30s
Question 179Question →

Arrange the following structures of the mammalian nephron in the correct sequence through which fluid flows during the process of urine formation, starting from the site of ultrafiltration.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of fluid flow through the nephron is: Bowman's capsule → Proximal convoluted tubule → Loop of Henle → Distal convoluted tubule → Collecting duct.
During urine formation, ultrafiltration forces fluid out of the renal capillaries into Bowman's capsule. The filtrate then flows into the proximal convoluted tubule, travels down and up the loop of Henle, enters the distal convoluted tubule, and finally drains into the collecting duct.

Step-by-Step Solution

1
Identify the initial receiving structure for glomerular filtrate.
Ultrafiltration pushes fluid from the glomerulus directly into Bowman's capsule.
Bowman's capsule encapsulates the glomerulus and collects the fluid forced out under high hydrostatic pressure.
2
Trace the sequential pathway through the tubular regions of the nephron.
The filtrate travels from Bowman's capsule into the proximal convoluted tubule, down into the hairpin loop of Henle, and up into the distal convoluted tubule.
This anatomical order allows step-by-step selective reabsorption of glucose, amino acids, and ions followed by osmotic regulation.
3
Determine the final duct that collects urine from the nephron unit.
The processed fluid drains from the distal convoluted tubule into the collecting duct.
The collecting duct gathers urine from several distal convoluted tubules and routes it to the ureter.

Key Concept

Pathway of fluid flow through the functional unit (nephron) of the mammalian kidney during urine formation.
Question 180Question →

A flower collected from a tropical rainforest canopy possesses a deeply tubular corolla, secretes copious amounts of dilute nectar at the base of the floral tube, produces red unscented petals, and displays sturdy floral structures. Which of the following agents is primarily adapted to pollinate this flower?

Show answer & explanation

Answer: Sunbirds, which possess long slender beaks to reach nectar in deep corolla tubes and rely on visual cues rather than olfactory senses.

Answer

Sunbirds, which possess long slender beaks to reach nectar in deep corolla tubes and rely on visual cues rather than olfactory senses.
The combination of a deep tubular corolla, red coloration, unscented petals, and high production of dilute nectar is characteristic of bird pollination (ornithophily). Sunbirds have long slender beaks capable of probing deep floral tubes, excellent perception of red light, and high caloric needs satisfied by nectar, while requiring no floral scent due to their weak olfactory sense.

Step-by-Step Solution

1
Analyze the floral characteristics presented in the stem.
Identified key features: deeply tubular corolla, high volume of dilute nectar, red color, sturdy floral parts, and absence of fragrance.
Floral syndromes correspond directly to the sensory capabilities and morphological features of specific pollinator groups.
2
Evaluate pollinator sensory and physical match for the floral features.
Birds (such as sunbirds or hummingbirds) possess keen vision for red wavelengths, weak senses of smell (hence unscented flowers), long bills adapted to reach deep corollas, and require high volumes of dilute nectar to satisfy high metabolic demands.
Red tubular unscented flowers with copious dilute nectar are classic adaptations for ornithophily (bird pollination).
3
Differentiate from alternative pollination syndromes (anemophily, phalaenophily, melittophily).
Wind pollination lacks nectar and petals; moth pollination requires night-visible white petals and strong scents; bee pollination requires scent, landing platforms, and colors visible in the UV spectrum rather than deep unscented red tubes.
Matching structural adaptations eliminates insect and wind vectors.

Key Concept

Floral structural adaptations and pollination syndromes in angiosperms
Estimated Time:1m 30s
PreviousPage 9 / 13Next
Form and Function Practice Questions — JAMB UTME — Page 9 | Examkin