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Form and Function

256 questions

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Question 101Question →

Match each specialized reproductive structure in plants or animals with its corresponding physiological function.

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Items

Sertoli cells
Tapetum
Acrosome
Synergids

Matches

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Answer

Sertoli cells nourish developing spermatids during spermatogenesis; the tapetum nourishes microspores in the anther; the acrosome releases lysosomal enzymes to facilitate egg penetration; and synergids release chemical attractants to direct pollen tube entry.
Each listed structure carries out a distinct supportive or functional role in gametogenesis or fertilization: Sertoli cells nurse spermatids in mammalian testes; the tapetum nourishes microspores within angiosperm anthers; the acrosome provides digestive enzymes needed for sperm penetration during fertilization; and synergid cells produce chemotropic signals directing the pollen tube to the egg cell.

Step-by-Step Solution

1
Identify the cellular origin and function of animal male reproductive structures.
Sertoli cells function as nurse cells in the testes (matching with nourishment of developing germ cells), while the acrosome contains enzymes for egg penetration (matching with secretion of hydrolytic enzymes).
Both structures perform essential support roles in male gametes in animals.
2
Identify the cellular structures and functions involved in plant gametogenesis and fertilization.
The tapetum forms the nutritive inner layer of the microsporangium (matching with pollen nourishment), while synergid cells sit next to the egg cell in the ovule (matching with chemical guidance of the pollen tube).
Both structures are plant-specific adaptations for pollen development and double fertilization.

Key Concept

Structural and functional adaptations of specialized reproductive cells in plants and animals
Estimated Time:1m 30s
Question 102Question →

During photosynthesis, the photolysis of water takes place within the thylakoid lumen of chloroplasts. Which of the following correctly describes the immediate fate of the molecular oxygen gas generated from this reaction?

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Answer: It diffuses out of the chloroplast and leaves the leaf via the stomata or is consumed in cellular respiration.

Answer

The oxygen gas diffuses out of the chloroplast and exits the plant through stomata or is used internally for cellular respiration.
During photolysis of water in the light-dependent phase of photosynthesis, water molecules are split into hydrogen ions, electrons, and molecular oxygen gas. Since oxygen is a byproduct rather than a reactant for subsequent light-independent reactions, it diffuses out of the chloroplast into mesophyll air spaces and exits through stomatal pores, or is used by the cell's mitochondria during aerobic respiration.

Step-by-Step Solution

1
Identify the site and products of photolysis of water.
Photolysis occurs in the light-dependent stage within thylakoids, splitting water molecules (2H2O→4H++4e−+O22H_2O \rightarrow 4H^+ + 4e^- + O_2).
Water molecules supply replacement electrons for Photosystem II and generate protons for ATP synthesis.
2
Determine the metabolic destination of the released oxygen gas.
Oxygen gas is a byproduct of light reactions and is not used as a substrate in the Calvin cycle.
The light-independent reactions require CO2CO_2, ATP, and NADPH, but not oxygen gas.
3
Trace the physical exit path of oxygen from the photosynthetic cell.
Oxygen diffuses down its concentration gradient out of the chloroplast into mesophyll air spaces, exiting via stomata or entering mitochondria.
Gaseous exchange in leaves occurs primarily by simple diffusion through stomatal pores.

Key Concept

Photolysis of water and destination of photosynthetic byproducts
Question 103Question →

In certain hermaphroditic flowering plants, self-pollination is prevented because the anthers mature and release pollen before the stigma of the same flower becomes receptive. Which term describes this specific temporal adaptation?

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Answer: Protandry

Answer

Protandry is the temporal mechanism in which anthers mature before the stigma becomes receptive in the same flower.
Protandry is a form of dichogamy where male organs (androecium) mature before female organs (gynoecium). This temporal separation prevents self-fertilization in bisexual flowers by ensuring pollen is shed when the flower's own stigma is not yet receptive.

Step-by-Step Solution

1
Analyze the condition described in the stem
The male organs (anthers) mature earlier than the female organs (stigma) within a bisexual flower.
This is a temporal separation technique (dichogamy) designed to ensure cross-pollination.
2
Identify the biological terminology for male-first maturation
The prefix 'proto-' means first, and 'andry' relates to male organs (androecium). Thus, early male maturation is protandry.
Differentiating protandry from protogyny based on the etymology and function of androecium versus gynoecium.

Key Concept

Floral Adaptations for Cross-Pollination (Dichogamy)
Estimated Time:50s
Question 104Question →

Arrange the following physiological events occurring during the light-dependent stage of photosynthesis in their correct chronological sequence from start to finish.

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Answer

The correct sequence begins with the absorption of light photons by chlorophyll leading to photo-excitation of electrons, followed by the photolytic cleavage of water molecules into hydrogen ions, electrons, and oxygen, then the transfer of high-energy electrons through an electron transport chain to synthesize ATP, and culminates in the reduction of NADP+NADP^+ to NADPHNADPH by accepting electrons and hydrogen ions.
The light-dependent phase begins when light photons strike chlorophyll molecules, exciting electrons. To fill the electron hole left in chlorophyll, water undergoes photolysis to yield protons, electrons, and oxygen. As the energized electrons pass along carrier proteins in the thylakoid membrane, ADP is phosphorylated into ATP. Finally, the electrons and protons are accepted by NADP+ to form NADPH, completing the sequence.

Step-by-Step Solution

1
Identify the primary trigger of the light-dependent stage.
Photon absorption by chlorophyll photo-excites electrons.
Light absorption is required to initiate electron flow in photosystems.
2
Determine the source of replacement electrons.
Water photolysis occurs, producing H+H^+, electrons, and O2O_2.
Splitting water provides electrons to replace those lost by photo-excited chlorophyll.
3
Trace the movement of excited electrons and energy capture.
ATP is synthesized via photophosphorylation as electrons pass through membrane carriers.
Proton gradient creation drives ATP synthesis during electron transport.
4
Identify the final electron acceptor of the light reaction.
NADP+NADP^+ is reduced to NADPHNADPH.
NADP+NADP^+ accepts terminal electrons and hydrogen ions to store reducing power for carbon fixation.

Key Concept

Sequence of events in light-dependent reactions of photosynthesis
Question 105Question →

Match each seed germination structure or growth process on the left with its corresponding role during seedling development on the right.

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Items

Hypocotyl elongation in epigeal germination
Epicotyl elongation in hypogeal germination
Coleoptile emergence in monocot seedlings
Radicle emergence through the micropyle

Matches

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Answer

Hypocotyl elongation in epigeal germination pairs with pushing cotyledons above the soil surface. Epicotyl elongation in hypogeal germination pairs with keeping cotyledons below ground while pushing the shoot upward. Coleoptile emergence pairs with forming a protective sheath around the shoot tip during soil penetration. Radicle emergence pairs with anchoring the seedling and forming the primary root system.
Each plant developmental structure performs a distinct physiological role during germination: hypocotyl growth elevates cotyledons above ground in epigeal species; epicotyl growth keeps cotyledons underground in hypogeal species; the coleoptile protects the delicate young shoot tip in monocots; and the radicle emerges first to establish the primary root system.

Step-by-Step Solution

1
Differentiate between epigeal and hypogeal germination mechanisms.
Epigeal germination involves hypocotyl growth lifting cotyledons above ground, whereas hypogeal germination involves epicotyl growth leaving cotyledons below ground.
The region of rapid cellular division and elongation determines whether cotyledons are elevated or remain buried.
2
Identify the specialized protective role of the coleoptile in monocot development.
The coleoptile acts as a pointed sheath guarding fragile plumule tissues from abrasive soil particles during upward growth.
Monocot shoots require specialized structural protection during soil emergence.
3
Determine the sequence and role of embryonic root emergence.
The radicle exits first through the micropyle to secure anchorage and absorb water prior to shoot development.
Early root establishment is essential to supply hydration required for metabolic processes and growth.

Key Concept

Plant Seed Germination Types and Early Embryonic Growth Patterns
Question 106Question →

During photosynthesis, light energy absorbed by chlorophyll drives essential chemical conversions within the thylakoid membranes of chloroplasts. Which of the following biological processes occurs specifically during this light-dependent stage?

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Answer: Photolysis of water yielding oxygen gas, protons, and high-energy energy carriers

Answer

Photolysis of water yielding oxygen gas, protons, and high-energy energy carriers occurs specifically during the light-dependent stage.
The light-dependent stage of photosynthesis occurs in the thylakoid membranes of chloroplasts. Light energy absorbed by chlorophyll causes the photolysis of water (2H2O→4H++4e−+O22H_2O \rightarrow 4H^+ + 4e^- + O_2), releasing oxygen gas as a byproduct and generating ATP and reduced NADP (NADPH) required for the dark reaction.

Step-by-Step Solution

1
Identify the biological site and primary events of the light-dependent stage of photosynthesis.
The light-dependent reactions occur within the thylakoid membranes of chloroplasts, initiated by light absorption by chlorophyll.
Photon absorption excites electrons to drive water splitting and energy carrier generation.
2
Distinguish between light-dependent reaction products and light-independent Calvin cycle steps.
Photolysis of water (2H2O→4H++4e−+O22H_2O \rightarrow 4H^+ + 4e^- + O_2) produces oxygen gas, ATPATP, and NADPHNADPH in thylakoids, whereas carbon fixation, reduction, and sugar synthesis occur in the stroma.
Light energy is directly required to split water molecules, whereas carbon fixation relies on chemical products in the stroma.

Key Concept

Light-dependent stage of photosynthesis and photolysis of water
Question 107Question →

During insect development, the transition between developmental stages is regulated by the interaction of ecdysone (moulting hormone) and juvenile hormone. A high concentration of juvenile hormone paired with ecdysone promotes larval-to-larval moulting. Which of the following developmental outcomes occurs when the secretion of juvenile hormone declines significantly while ecdysone remains active?

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Answer: The larva undergoes metamorphosis to form a pupa

Answer

The larva undergoes metamorphosis to form a pupa
Juvenile hormone inhibits metamorphosis and keeps the insect in its immature larval state. When its concentration drops to a low level while ecdysone remains active, the insect undergoes metamorphosis from the larval stage into a pupa.

Step-by-Step Solution

1
Identify the primary hormones involved in insect growth and metamorphosis.
Ecdysone triggers moulting, while juvenile hormone maintains larval structural features.
Understanding hormone interaction is key to determining developmental pathways in arthropods.
2
Analyze the effect of changing hormone concentration ratios.
High juvenile hormone + Ecdysone = Larval moult; Low juvenile hormone + Ecdysone = Pupal metamorphosis; Absence of juvenile hormone + Ecdysone = Adult emergence.
Juvenile hormone acts as a suppressor of adult and pupal gene expression.
3
Deduce the outcome for a significant decrease in juvenile hormone.
The larva transitions into the pupal stage.
Lowering juvenile hormone level allows pupal genes to be expressed during the ecdysone-stimulated moult.

Key Concept

Hormonal regulation of metamorphosis in insects
Question 108Question →

Match each specialized digestive organ or structure found in animals with its primary physiological function in nutrition.

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Items

Rumen
Crop
Gizzard
Villi

Matches

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Answer

Rumen matches with cellulose fermentation by symbiotic microorganisms; Crop matches with temporary storage and softening of food; Gizzard matches with mechanical breakdown and grinding of food; Villi match with maximizing surface area for nutrient absorption.
Each specialized digestive organ corresponds directly to its anatomical role in comparative animal nutrition: the rumen houses microbes for fermenting plant fibers, the crop stores and moistens food, the gizzard mechanically grinds coarse materials, and intestinal villi expand the absorptive surface area.

Step-by-Step Solution

1
Identify the primary function of the rumen in ruminants.
The rumen acts as a fermentation vat containing symbiotic microflora to hydrolyze cellulose.
Ruminants lack endogenously produced cellulase enzymes and rely on microbial fermentation in the rumen.
2
Determine the role of the crop in birds and insects.
The crop temporarily stores swallowed food and lubricates it with secretions before it enters the stomach.
Birds lack teeth for mastication, requiring an anatomical site to hold and soften unchewed food items.
3
Determine the role of the gizzard.
The muscular gizzard performs physical triturating (grinding) of hard food seeds or grains.
Mechanical digestion must substitute for teeth in birds, earthworms, and certain invertebrates.
4
Match villi to their intestinal function.
Villi increase the total surface area available for diffusion and active transport of digested nutrients.
Epithelial foldings maximize contact area between chyme and intestinal capillaries/lacteals.

Key Concept

Comparative Animal Digestive Structures and Adaptations
Estimated Time:1m 0s
Question 109Question →

A student set up an experiment to observe hypogeal germination in a maize seed (*Zea mays*). In which chronological sequence do the following physiological and morphological events occur during this growth process?

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Answer

The correct developmental sequence for hypogeal germination is: (1) Imbibition of water through the micropyle, (2) Emergence of the radicle growing downwards, (3) Rapid elongation of the epicotyl pushing the plumule upward, and (4) Rupture of the protective coleoptile by expanding foliage leaves.
Germination starts with water imbibition via the micropyle, activating metabolic enzymes. The radicle emerges first to establish root anchorage and water uptake. Following this, epicotyl elongation pushes the protective coleoptile and plumule upward through the soil. Finally, exposure to light stimulates the first foliage leaves to rupture the coleoptile and unroll for photosynthesis.

Step-by-Step Solution

1
Identify the initial physical trigger required for seed metabolism.
Water imbibition occurs first, swelling the endosperm and activating digestive enzymes.
Metabolic processes and cell expansion cannot take place in dry seed tissues.
2
Determine the first embryonic organ to pierce the seed coat.
The radicle emerges downwards into the soil to form the primary root system.
Anchorage and water absorption are necessary before shoot growth commences.
3
Analyze shoot elongation specific to hypogeal germination.
The epicotyl elongates, carrying the sheath-enclosed plumule upward while the cotyledon stays below ground.
Hypocotyl growth is minimal in hypogeal germination, keeping the food storage organ underground.
4
Identify the final step in establishing autotrophic seedling growth.
Foliage leaves break out of the coleoptile upon reaching light and expand to photosynthesize.
This completes seedling development and transitions the plant to independent energy production.

Key Concept

Sequence of events in hypogeal seed germination
Estimated Time:1m 30s
Question 110Question →

In herbivorous mammals such as rabbits and sheep, a prominent toothless gap called a diastema separates the front incisors from the cheek teeth. What is the primary physiological function of the diastema during feeding?

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Answer: It provides a space for the tongue to manipulate, rotate, and hold food while grinding occurs at the molars.

Answer

The primary physiological function of the diastema is to provide a space for the tongue to manipulate and maneuver food during chewing.
The correct answer highlights the mechanical function of the diastema. In herbivores, this toothless gap gives the tongue sufficient space to manipulate plant material, pushing extra food into the cheek pockets or turning food onto the grinding surfaces of the premolars and molars.

Step-by-Step Solution

1
Identify the anatomical structure described in the stem.
The diastema is a toothless gap naturally located between the front cropping teeth (incisors) and back grinding teeth (premolars and molars) in many herbivores.
Understanding anatomical position helps determine its mechanical role.
2
Analyze the functional role of the mouth and tongue during herbivore mastication.
Herbivores need to gather vegetation with incisors and move it back to the molars without interfering with ongoing grinding.
The gap allows the tongue to manipulate, push, and turn unchewed plant material efficiently.

Key Concept

Mammalian Dentition and Feeding Adaptations
Estimated Time:1m 0s
Question 111Question →

When a growing dicotyledonous seedling stem is exposed to light coming from a single direction (unilateral illumination), the stem curves towards the light source. Which physiological mechanism explains this directional growth response?

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Answer: Auxin migrates away from the illuminated side to the shaded side, causing cells on the shaded side to elongate faster.

Answer

Auxin migrates away from the illuminated side to the shaded side, causing cells on the shaded side to elongate faster.
In stem tips, unilateral light induces the lateral transport of auxin to the shaded side. The higher concentration of auxin on the shaded side causes cells in that region to elongate more rapidly than those on the illuminated side, leading to differential growth that bends the stem toward the light.

Step-by-Step Solution

1
Identify the growth hormone responsible for shoot elongation in response to light stimuli.
Auxin (indole-3-acetic acid) regulates cell elongation in plant shoot tips.
Shoot phototropism is driven by asymmetric distribution of plant growth substances.
2
Analyze how unilateral light alters the spatial distribution of auxin in the shoot apex.
Unilateral light causes auxin to move laterally from the lit side to the shaded side.
Phototropin photoreceptors detect light direction and trigger lateral auxin translocation.
3
Determine the effect of higher auxin concentration on the shaded side cells.
Higher auxin concentration accelerates cell elongation on the shaded side relative to the illuminated side.
Differential elongation rates cause the stem to curve toward the light source.

Key Concept

Phototropism and Auxin Redistribution in Plant Growth
Question 112Question →

Organize the following processes in the sequential order in which food passes through and is processed in the alimentary canal of a domestic fowl (bird), starting from initial ingestion.

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Answer

The correct sequence of digestion in birds is: storage in the crop, chemical secretion in the proventriculus, mechanical grinding in the gizzard, enzymatic absorption in the small intestine, and egestion via the cloaca.
In the avian digestive tract, food follows a strict anteroposterior sequence: it is swallowed whole into the crop for softening, moves to the proventriculus for enzymatic chemical breakdown, enters the muscular gizzard for mechanical trituration with grit, proceeds through the small intestine for final digestion and absorption, and lastly exits via the cloaca.

Step-by-Step Solution

1
Identify the initial entry point after ingestion in birds.
Food travels down the oesophagus into the expanded crop for temporary storage and softening.
Birds lack teeth for mastication, so whole food is initially stored and moistened in the crop.
2
Trace food movement into the glandular stomach region.
Food enters the proventriculus where digestive juices containing acid and enzymes are secreted.
Chemical digestion begins in the proventriculus before physical pulverization.
3
Trace movement into the mechanical grinding organ.
Food is forced into the gizzard (ventriculus) where muscular contractions and grit grind the food particles.
The gizzard acts as the functional equivalent of teeth in birds.
4
Determine the site of primary nutrient absorption.
Finely ground food moves into the small intestine where digestion is completed and nutrients enter the bloodstream.
The small intestine contains intestinal enzymes and villi specialized for nutrient absorption.
5
Identify the terminal site of waste expulsion.
Undigested matter moves to the rectum and exits through the cloaca.
The cloaca is the common chamber for digestive waste and urinary excretion in birds.

Key Concept

Avian alimentary canal anatomy and sequence of digestive processing
Question 113Question →

A mixture of starch and protein was treated with pancreatic juice in a test tube maintained at 0∘C0^\circ\text{C} and optimal pH\text{pH} for two hours, resulting in no breakdown of either substrate. When the temperature of the mixture was subsequently raised to 37∘C37^\circ\text{C}, rapid digestion of both starch and protein was observed. Which of the following best explains why digestion occurred after the mixture was warmed?

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Answer: Low temperatures temporarily render enzymes inactive without destroying their structural integrity, allowing activity to resume upon warming.

Answer

Low temperatures temporarily render enzymes inactive without destroying their structural integrity, allowing activity to resume upon warming.
At 0∘C0^\circ\text{C}, enzymes like pancreatic amylase and trypsin lose kinetic energy, reducing the rate of effective collisions with substrate molecules. Because low temperature does not disrupt tertiary protein structure, the enzyme is merely inactivated, not denatured. Restoring the temperature to 37∘C37^\circ\text{C} restores kinetic motion and catalytic capability.

Step-by-Step Solution

1
Analyze the effect of low temperature (0∘C0^\circ\text{C}) on enzyme molecules.
Enzymes experience reduced kinetic energy, drastically lowering substrate collision frequency without altering their active site conformation.
Temperature affects molecular velocity, but low temperatures do not break the covalent or hydrogen bonds maintaining enzyme tertiary structure.
2
Evaluate the effect of raising the temperature to the optimal level (37∘C37^\circ\text{C}).
Kinetic energy increases, leading to frequent effective collisions between active sites and substrate molecules.
Since the enzymes remained intact during the cold phase, warming restores catalytic function instantly.

Key Concept

Effect of Temperature on Digestive Enzyme Activity and Reversibility of Low-Temperature Inactivation
Question 114Question →

Match each compartment of the ruminant stomach with its primary physiological function during digestion.

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Items

Rumen
Reticulum
Omasum
Abomasum

Matches

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Answer

Rumen matches with fermentation of plant cellulose and temporary storage; Reticulum matches with formation of cud boluses for regurgitation; Omasum matches with absorption of water and volatile fatty acids; Abomasum matches with secretion of hydrochloric acid and enzymes for protein digestion.
Ruminant digestion relies on four distinct stomach chambers: the rumen functions as the primary fermenter housing micro-organisms that break down cellulose; the reticulum collects dense particles and forms cud for remastication; the omasum extracts excess water and volatile fatty acids through its folded walls; and the abomasum operates as the true glandular stomach producing hydrochloric acid and pepsin.

Step-by-Step Solution

1
Identify the microbial fermentation chamber
The rumen harbors microbial symbionts that break down tough plant cell walls (cellulose).
Mammals lack cellulase enzymes and rely on rumen microorganisms.
2
Identify the chamber responsible for cud formation
The reticulum works closely with the rumen to form small food balls (cud) for regurgitation back to the mouth.
Rumination requires mechanical re-mastication to reduce particle size.
3
Identify the water-absorbing chamber
The omasum filters and reabsorbs water and electrolytes before material enters the true stomach.
Its highly folded internal surface area maximizes fluid reabsorption.
4
Identify the true enzymatic stomach
The abomasum secretes gastric juices containing hydrochloric acid and pepsin.
This is the only chamber with digestive glands producing endogenous enzymes in ruminants.

Key Concept

Ruminant Stomach Specialized Compartments
Question 115Question →

Bile salts are synthesized in the liver and stored in the gall bladder before being secreted into the duodenum. Although bile salts do not contain any digestive enzymes, they are essential for the digestion of lipids. What is the primary physiological role of bile salts during fat digestion?

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Answer: Emulsifying large fat droplets into tiny droplets to increase the surface area for pancreatic lipase activity

Answer

Emulsifying large fat droplets into tiny droplets to increase the surface area for pancreatic lipase activity
The correct answer highlights that bile salts perform emulsification, a mechanical process that disperses large fat globules into minute droplets. Because lipase is a water-soluble enzyme acting on water-insoluble fats, emulsification greatly expands the surface area available for pancreatic lipase to digest triglycerides into fatty acids and glycerol.

Step-by-Step Solution

1
Identify the nature of bile salts
Bile salts are non-enzymatic amphipathic molecules secreted by the liver into the duodenum.
Because they lack catalytic activity, they cannot break chemical bonds in nutrients directly.
2
Determine the physical action of bile salts on hydrophobic lipids
Lipids aggregate into large droplets in aqueous chyme. Bile salts lower surface tension, dispersing fats into microscopic droplets (emulsification).
Emulsification expands the lipid-water interface area.
3
Relate emulsification to enzyme action
Pancreatic lipase operates at the aqueous interface to hydrolyze fats into fatty acids and glycerol efficiently.
Increasing surface area accelerates the rate of enzymatic lipid digestion.

Key Concept

Role of bile in emulsification and physical breakdown of lipids
Estimated Time:1m 0s
Question 116Question →

Which of the following biological processes occurs specifically during the light-independent stage (Calvin cycle) of photosynthesis in green plants?

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Answer: Reduction of carbon dioxide to synthesize simple sugars

Answer

Reduction of carbon dioxide to synthesize simple sugars
During the light-independent stage (Calvin cycle) in the chloroplast stroma, carbon dioxide is chemically reduced to form triose phosphate and glucose using NADPH and ATP generated during the light stage.

Step-by-Step Solution

1
Identify the two main stages of photosynthesis and their cellular locations.
The light-dependent stage occurs in the thylakoids, while the light-independent stage (Calvin cycle) occurs in the stroma of chloroplasts.
Different enzymatic and chemical reactions are spatially compartmentalized within the chloroplast.
2
Determine the specific biochemical events of the light-independent stage.
Atmospheric carbon dioxide is fixed by RuBP and reduced using ATP and NADPH to form carbohydrates such as glucose.
This stage does not require direct light photons but relies on the chemical energy products formed during the light-dependent stage.

Key Concept

Light-Independent Reaction (Calvin Cycle)
Question 117Question →

Match each chloroplast component or biochemical agent involved in photosynthesis on the left with its correct physiological function or characteristic on the right.

Click a left item, then click its matching right item

Items

Thylakoid membrane
Stroma
NADP+\text{NADP}^+
Ribulose-1,5-bisphosphate (RuBP)

Matches

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Answer

Thylakoid membrane matches with the location of chlorophyll excitation, photolysis of water, and electron transport chain complexes; Stroma matches with the site of light-independent carbon dioxide fixation and soluble Calvin cycle enzyme activity; NADP+\text{NADP}^+ matches with the terminal electron and hydrogen acceptor reduced during light-dependent reactions; Ribulose-1,5-bisphosphate (RuBP) matches with the five-carbon organic compound that acts as the initial carbon dioxide acceptor.
Each structural and biochemical component is correctly matched to its specific role in photosynthesis: Thylakoid membranes harbor photosystems for light absorption and water photolysis; Stroma contains enzymes for carbon dioxide fixation in the Calvin cycle; NADP+\text{NADP}^+ acts as the terminal hydrogen/electron acceptor; and RuBP functions as the five-carbon CO2\text{CO}_2 acceptor molecule.

Step-by-Step Solution

1
Identify the structural site of light absorption and water splitting in the chloroplast.
Thylakoid membranes hold photosynthetic pigments and proteins responsible for photolysis and non-cyclic electron flow.
Light-dependent reactions rely on membrane-bound chlorophyll complexes and electron carriers.
2
Identify the fluid compartment where carbon fixation occurs.
The stroma surrounds the thylakoids and contains the enzymes needed for dark reactions.
The Calvin cycle takes place in the fluid matrix using ATP and NADPH synthesized at the thylakoid membrane.
3
Identify the primary coenzyme electron acceptor and the carbon dioxide acceptor molecule.
NADP+\text{NADP}^+ acts as the terminal electron acceptor forming NADPH, while RuBP is the 5-carbon5\text{-carbon} acceptor of CO2\text{CO}_2.
NADPH supplies reducing power for carbohydrate synthesis, and RuBP reacts with CO2\text{CO}_2 in the initial carboxylation step.

Key Concept

Chloroplast Functional Compartmentalization and Photosynthetic Reactions
Estimated Time:1m 30s
Question 118Question →

Match each specialized cell type or reproductive gland with its precise physiological role in plant or animal reproduction.

Click a left item, then click its matching right item

Items

Sertoli cells
Bulbourethral (Cowper's) glands
Tapetal layer (Tapetum)
Synergid cells

Matches

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Answer

Sertoli cells correspond to providing structural and metabolic support to spermatogenic cells; Bulbourethral glands correspond to secreting alkaline viscous fluid that neutralizes urethral acidity; Tapetal layer corresponds to nourishing developing microspores and contributing to pollen wall formation; Synergid cells correspond to secreting chemotropic signals guiding pollen tube entry through the micropyle.
Each pair correctly matches a specific cellular or glandular component of plant or animal reproductive anatomy with its precise physiological function: Sertoli cells support mammalian spermatogenesis, Bulbourethral glands neutralize male urethral acidity, the tapetum nourishes microspores in anthers, and synergid cells guide pollen tube growth toward the embryo sac.

Step-by-Step Solution

1
Analyze mammalian male reproductive cell structures
Identify Sertoli cells as somatic support cells in the seminiferous tubules (blood-testis barrier and germ cell nourishment) and Bulbourethral glands as accessory glands secreting neutralizing alkaline mucus into the male urethra.
Differentiates primary spermatogenic supportive tissue from accessory gland secretions.
2
Analyze angiosperm reproductive cellular layers and female gametophyte components
Identify the tapetum as the nutritive inner tissue of the microsporangium wall and synergid cells as specialized female gametophyte cells guiding double fertilization.
Establishes functional homologies and specialized roles in plant spore and gamete development.
3
Form correct matching pairs based on cellular physiology
Pair Sertoli cells with spermatogenic cell support/blood-testis barrier, Bulbourethral glands with urethral acidity neutralization, Tapetal layer with microspore nourishment, and Synergid cells with chemotropic pollen tube attraction.
Completes accurate cross-domain mapping of reproductive cellular functions.

Key Concept

Nutritive and regulatory cellular adaptations in plant and animal gametogenesis and fertilization.
Question 119Question →

A physiological examination of an aquatic invertebrate reveals that metabolic waste elimination and osmoregulation are performed by a pair of specialized excretory structures situated at the base of its antennae. Which organism possesses this excretory system?

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Answer: Freshwater prawn

Answer

The freshwater prawn is the organism that excretes metabolic waste through green glands located at the base of its antennae.
The option specifying the freshwater prawn is correct because crustaceans possess paired green glands (antennal glands) situated near the base of the antennae. These glands collect fluid, reabsorb essential nutrients, and excrete nitrogenous waste (mainly ammonia) directly to the exterior.

Step-by-Step Solution

1
Identify the anatomical location and type of excretory structure described in the question stem.
The structure described is located at the base of the antennae, which defines antennal glands or green glands.
Green glands are distinctive excretory organs found specifically at the base of antennae in certain arthropods.
2
Map the identified excretory structure to its corresponding animal group and representative organism.
Green glands are characteristic excretory organs of crustaceans, such as prawns, lobsters, and crayfish.
Different invertebrate phyla rely on distinct excretory structures tailored to their anatomy and habitat.

Key Concept

Excretory structures across invertebrate phyla
Estimated Time:1m 0s
Question 120Question →

What is the net yield of adenosine triphosphate (ATP) molecules obtained from the breakdown of one molecule of glucose during anaerobic respiration?

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Answer: 22

Answer

The net yield of ATP molecules obtained during anaerobic respiration is 22.
During anaerobic respiration, glucose undergoes glycolysis in the cytoplasm. Although glycolysis produces 4 ATP4\text{ ATP} molecules in total, 2 ATP2\text{ ATP} molecules are used during the phosphorylation steps, leaving a net gain of 2 ATP2\text{ ATP} molecules per glucose molecule.

Step-by-Step Solution

1
Identify the metabolic pathway operating under anaerobic conditions.
In the absence of oxygen, cellular respiration is restricted to glycolysis.
Pyruvate cannot enter the mitochondria to undergo the link reaction, Krebs cycle, or oxidative phosphorylation without oxygen.
2
Calculate the net ATP energy yield from glycolysis.
Gross ATP produced (44) minus ATP invested (22) equals a net yield of 2 ATP2\text{ ATP}.
Two ATP molecules are consumed during the activation steps converting glucose to fructose-1,6-bisphosphate.

Key Concept

Net ATP accounting in anaerobic respiration
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Form and Function Practice Questions — JAMB UTME — Page 6 | Examkin