Growth, Development, and Metamorphosis

23 soru

Soru 1Soru

In insects such as the housefly, growth and development occur through complete metamorphosis. Which of the following sequences correctly represents the sequential stages of complete metamorphosis?

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Cevap: Egg → Larva → Pupa → Imago

Cevap

Egg → Larva → Pupa → Imago
Complete metamorphosis (holometabolous growth) proceeds sequentially through four distinct developmental forms: egg, larva, pupa, and adult (imago). The larva is specialized for feeding and growth, while the pupa undergoes internal cellular restructuring to form adult structures.

Adım Adım Çözüm

1
Identify the type of metamorphosis requested.
The question specifies complete (holometabolous) metamorphosis.
Complete metamorphosis involves four distinct morphological stages.
2
Trace the developmental stages in chronological order.
The embryo hatches from the egg into an active feeding larva (maggot/caterpillar), transforms into a non-feeding pupa, and finally emerges as a sexually mature adult (imago).
The larval stage builds energy reserves, while the pupal stage undergoes tissue reorganization.

Anahtar Kavram

Complete metamorphosis (Holometabolous development)
Soru 2Soru

Match each developmental signal or growth mechanism on the left with its corresponding physiological outcome or developmental process on the right.

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Öğeler

Ecdysone secretion in the presence of high juvenile hormone concentration
Ecdysone secretion following the degeneration of the corpus allatum (low juvenile hormone)
High ratio of auxin to cytokinin maintained at the shoot apex
Rapid elongation of the hypocotyl during seed germination

Eşleşmeler

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Cevap

Ecdysone with high juvenile hormone pairs with larval-to-larval molting; ecdysone with low juvenile hormone pairs with metamorphosis; high auxin to cytokinin ratio pairs with apical dominance suppression of lateral buds; hypocotyl elongation pairs with epigeal germination elevating cotyledons above soil.
Growth and development in insects and plants are tightly regulated by hormonal concentrations and axial tissue elongation. Ecdysone induces molting; high juvenile hormone maintains larval traits, whereas its decline leads to metamorphosis. In plants, high apical auxin maintains apical dominance over axillary buds. In seed germination, hypocotyl elongation carries cotyledons above the surface in epigeal germination.

Adım Adım Çözüm

1
Analyze insect endocrine regulation of ecdysis and metamorphosis.
Ecdysone triggers cuticle shedding. High juvenile hormone preserves larval stage (larval-to-larval molt). Absence/low juvenile hormone allows ecdysone to induce pupation and metamorphosis.
Juvenile hormone acts as a status-quo hormone modifying the action of ecdysone.
2
Analyze plant hormonal control of meristematic activity.
Apical dominance is maintained when auxin concentrations from the apical bud are significantly higher relative to cytokinins.
Auxin inhibits lateral (axillary) bud growth directly or indirectly through hormonal signaling pathways.
3
Differentiate seedling germination biomechanics.
Hypocotyl growth below cotyledons raises them above the soil line (epigeal), whereas epicotyl growth leaves cotyledons below ground (hypogeal).
The site of maximal cellular elongation determines whether cotyledons are pushed upward or remain buried.

Anahtar Kavram

Hormonal control of growth, apical dominance, germination patterns, and insect metamorphosis
Soru 3Soru

Match each type of plant meristematic tissue on the left with its correct primary growth role on the right.

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Öğeler

Apical meristem
Lateral meristem
Intercalary meristem

Eşleşmeler

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Cevap

Apical meristem matches elongation at root and shoot apexes; Lateral meristem matches increase in stem girth; Intercalary meristem matches internodal extension at nodes.
Apical meristems produce length extension at growing tips, lateral meristems increase stem girth through secondary thickening, and intercalary meristems enable growth at internodes in monocots.

Adım Adım Çözüm

1
Identify the function of apical meristem
Apical meristems drive growth in length at shoot and root tips.
Active cell division at the apex produces primary plant tissues and extends plant height.
2
Identify the function of lateral meristem
Lateral meristems drive growth in thickness or diameter.
Cambium tissue layers divide laterally to form secondary xylem, secondary phloem, and cork.
3
Identify the function of intercalary meristem
Intercalary meristems drive growth at internodes and leaf bases.
Found predominantly in grasses, these meristems allow rapid stem elongation even after grazing.

Anahtar Kavram

Types of plant meristems and their specific developmental functions
Soru 4Soru

A student monitored the growth of a cockroach nymph over a six-week period by measuring its linear body length at regular intervals. The resulting plot showed a distinct staircase (step-like) growth curve with flat horizontal plateaus interrupted by vertical increases, rather than a continuous smooth curve. Which of the following biological processes accounts for the rapid increase in body length between the horizontal plateaus?

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Cevap: Periodic shedding of the rigid chitinous exoskeleton during ecdysis, allowing tissue expansion before the new cuticle hardens

Cevap

Periodic shedding of the rigid chitinous exoskeleton during ecdysis, allowing tissue expansion before the new cuticle hardens
Arthropods possess an inelastic chitinous exoskeleton that prevents continuous expansion in linear dimensions. During the intermoult period (represented by the horizontal plateaus), tissue mass increases internally. However, visible increases in length occur abruptly during ecdysis (molting), when the old cuticle is shed and the soft, new cuticle expands rapidly prior to sclerotization (hardening).

Adım Adım Çözüm

1
Analyze the nature of growth curves in arthropods versus non-arthropod organisms.
Identify that arthropods exhibit discontinuous (step-like) linear growth curves due to their non-expandable chitinous exoskeleton.
The rigid cuticle restricts continuous increase in external dimensions such as body length.
2
Evaluate the physiological events occurring during the horizontal plateau phase.
Recognize that cell division and accumulation of dry mass occur during intermoult (instar) periods without changes in external linear measurements.
Internal tissue growth compresses within the fixed exoskeleton.
3
Identify the cause of the rapid vertical jump in the staircase curve.
Conclude that ecdysis (molting) allows rapid intake of air or water to expand body volume before the new cuticle hardens.
Linear growth occurs in short bursts immediately following the shedding of the old exoskeleton.

Anahtar Kavram

Discontinuous growth and ecdysis in arthropods
Soru 5Soru

Match each plant growth phenomenon or developmental regulator on the left with its corresponding physiological function or developmental outcome on the right.

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Öğeler

Hypocotyl rapid elongation
Epicotyl rapid elongation
Gibberellin synthesis upon seed imbibition
Ecdysone secretion by prothoracic glands

Eşleşmeler

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Cevap

Hypocotyl rapid elongation matches with pulling cotyledons above soil in epigeal germination; Epicotyl rapid elongation matches with pushing the plumule upward while cotyledons stay underground in hypogeal germination; Gibberellin synthesis upon imbibition matches with inducing the aleurone layer to produce amylase; Ecdysone secretion matches with triggering apolysis and cuticle synthesis during ecdysis.
Hypocotyl elongation brings cotyledons above ground (epigeal germination). Epicotyl elongation elevates the plumule while keeping cotyledons below ground (hypogeal germination). Gibberellin signals the aleurone layer to produce α\alpha-amylase during germination. Ecdysone promotes apolysis and cuticle formation during arthropod ecdysis.

Adım Adım Çözüm

1
Analyze seedling germination types
Hypocotyl elongation lifts cotyledons above ground (epigeal), whereas epicotyl elongation leaves cotyledons underground (hypogeal).
The site of stem elongation relative to the cotyledons determines whether germination is epigeal or hypogeal.
2
Evaluate seed dormancy breakdown biochemistry
Imbibition activates gibberellin release from the embryo, targeting the aleurone layer.
Gibberellins stimulate gene transcription of hydrolytic enzymes like α\alpha-amylase to break down stored endosperm starch into soluble glucose.
3
Analyze arthropod growth and hormonal control
Ecdysone directly controls epidermal cell division and cuticle synthesis.
Ecdysone is the primary moulting hormone in insects that coordinates shedding of the old exoskeleton.

Anahtar Kavram

Plant Germination Dynamics, Seed Mobilization Biochemistry, and Hormonal Regulation of Ecdysis
Soru 6Soru

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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Öğeler

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

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Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Plant Seed Germination Types and Early Embryonic Growth Patterns
Soru 7Soru

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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Cevap: The larva undergoes metamorphosis to form a pupa

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Hormonal regulation of metamorphosis in insects
Soru 8Soru

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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Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Sequence of events in hypogeal seed germination
Tahmini Süre:1m 30s
Soru 9Soru

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Phototropism and Auxin Redistribution in Plant Growth
Soru 10Soru

Match each type of plant meristematic tissue on the left with its correct growth function on the right.

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Öğeler

Apical meristem
Lateral meristem
Intercalary meristem

Eşleşmeler

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Cevap

Apical meristem pairs with responsible for primary elongation at the apex of stems and roots; Lateral meristem pairs with facilitates increase in girth or secondary growth in stems and roots; Intercalary meristem pairs with promotes internodal elongation located at the base of leaf blades or nodes.
Apical meristems cause lengthening at shoot and root apices. Lateral meristems increase stem and root diameter. Intercalary meristems enable internodal lengthening in monocot stems.

Adım Adım Çözüm

1
Analyze the primary role of apical meristems in plant growth.
Apical meristems produce primary growth leading to increased length at tips.
These meristems exist at root and shoot apices where active cell division extends the plant bodies longitudinally.
2
Analyze the role of lateral meristems.
Lateral meristems produce secondary growth increasing thickness.
Located parallel to the long axis, lateral meristems add vascular and cork layers outward and inward.
3
Analyze the function of intercalary meristems.
Intercalary meristems drive internodal extension in monocots.
They remain active at leaf bases and stem nodes, allowing stems to elongate quickly.

Anahtar Kavram

Plant Meristematic Tissues and Primary vs Secondary Growth
Tahmini Süre:45s
Soru 11Soru

Arthropods exhibit a discontinuous, step-like growth curve because their outer body covering cannot expand continuously as internal tissue accumulates. Which structural feature of arthropods necessitates this periodic moulting (ecdysis) process?

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Cevap: A rigid, non-living chitinous exoskeleton that restricts expansion

Cevap

A rigid, non-living chitinous exoskeleton that restricts expansion
Arthropods are covered by a hard, non-living exoskeleton made of chitin and proteins. Because this outer layer cannot grow or stretch, the animal must shed its old exoskeleton (ecdysis) at regular developmental intervals to allow its body size to enlarge, producing a characteristic step-like growth curve.

Adım Adım Çözüm

1
Identify the primary physical barrier to continuous outward body expansion in arthropods.
The exoskeleton (cuticle) made of chitin is rigid, non-expandable, and non-living.
Because the cuticle cannot stretch once fully hardened, the organism must periodically shed it (ecdysis) to increase in size.
2
Evaluate alternative structural features regarding their role in growth pattern regulation.
Segmentation, jointed appendages, and circulatory organization are anatomical traits, not physical barriers to tissue expansion.
Only the impermeable and inflexible exoskeleton directly causes the intermittent step-like increase in body size.

Anahtar Kavram

Discontinuous growth and ecdysis in arthropods due to a rigid exoskeleton
Tahmini Süre:45s
Soru 12Soru

Match each developmental stage or chemical regulator of insect metamorphosis on the left with its corresponding biological role or characteristic on the right.

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Öğeler

Nymph
Pupa
Ecdysone
Juvenile Hormone

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Cevap

Nymph matches with the immature form in incomplete metamorphosis; Pupa matches with the non-feeding stage in complete metamorphosis where reorganization occurs; Ecdysone matches with the steroid hormone stimulating moulting; Juvenile Hormone matches with the hormone preserving larval traits.
Each concept is matched accurately: Nymph corresponds to the immature form in incomplete metamorphosis; Pupa corresponds to the non-feeding reorganization stage of complete metamorphosis; Ecdysone corresponds to the steroid hormone triggering moulting; and Juvenile Hormone corresponds to the hormone preserving larval features.

Adım Adım Çözüm

1
Identify the characteristic developmental stages of hemimetabolous vs holometabolous insects.
Nymphs belong to incomplete metamorphosis and resemble adults, while pupae belong to complete metamorphosis as a transitional reorganization stage.
Distinguishing between complete and incomplete metamorphosis depends on identifying their unique developmental stages.
2
Analyze the physiological functions of insect developmental hormones.
Ecdysone promotes shedding of the cuticle and metamorphosis, whereas juvenile hormone inhibits metamorphosis to preserve larval features.
Insect metamorphosis is regulated by the physiological balance between ecdysone and juvenile hormone.

Anahtar Kavram

Insect Metamorphosis and Endocrine Control
Soru 13Soru

During root development in vascular plants, tissue regions are structurally organized from the growing apex upward. What is the correct sequence of these regions starting from the extreme root tip and moving upward toward the main stem?

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Cevap

The correct sequence of root regions from the root tip upward is: Root cap, Zone of cell division (Apical meristem), Zone of cell elongation, and Zone of cell maturation (Differentiation zone).
The root apex grows sequentially starting with the protective root cap at the tip, followed by the zone of cell division where new cells are generated, then the zone of cell elongation where cells increase in length, and finally the zone of cell maturation where cells differentiate into specialized functional tissues.

Adım Adım Çözüm

1
Identify the protective terminal structure at the absolute tip of the root.
The root cap occupies the lowest position at the apex to shield delicate underlying tissues from friction against soil particles.
Terminal protection is required as the root apex advances through the soil.
2
Identify the region directly behind the protective cap.
The zone of cell division (apical meristem) lies immediately superior to the root cap.
Mitotic cell division produces new cells continuously at the root apex.
3
Determine where primary root extension occurs.
Cells produced by division move into the zone of elongation, expanding lengthwise to drive root penetration.
Cell elongation immediately follows cellular production before structural specialization.
4
Identify the final mature region furthest from the tip.
The zone of cell maturation lies above the elongation zone, featuring differentiated tissues like root hairs, xylem, and phloem.
Cells complete differentiation and acquire functional specialization after elongation stops.

Anahtar Kavram

Regions of Apical Root Growth
Soru 14Soru

In organisms such as butterflies and houseflies, the life cycle consists of four distinct developmental stages: egg, larva, pupa, and adult. Which type of metamorphosis is illustrated by this life cycle?

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Cevap: Complete metamorphosis

Cevap

Complete metamorphosis
Complete metamorphosis involves four distinct life stages: egg, larva, pupa, and adult. The presence of the pupal stage, in which profound structural transformation takes place, defines complete metamorphosis.

Adım Adım Çözüm

1
Identify the developmental stages listed in the question.
The four stages are egg, larva, pupa, and adult.
The presence of a distinct, non-feeding pupal stage is the primary distinguishing feature between developmental types.
2
Classify the life cycle based on the number and structural characteristics of the stages.
Complete metamorphosis (holometabolous development).
Only complete metamorphosis incorporates a pupal stage during which larval tissues are dismantled and adult structures develop.

Anahtar Kavram

Metamorphosis in Insects
Soru 15Soru

During an investigation into seedling growth, a student recorded a significant increase in the fresh weight of plants following heavy irrigation, but observed no change in their dry mass. Which of the following best explains why dry mass is considered a more reliable parameter for measuring biological growth than fresh weight?

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Cevap: Dry mass measures the irreversible accumulation of synthesized organic material and cellular structures, excluding temporary water fluctuations.

Cevap

Dry mass measures the irreversible accumulation of synthesized organic material and cellular structures, excluding temporary water fluctuations.
Growth is defined as a permanent and irreversible increase in size and dry mass resulting from cell division and the synthesis of new organic cellular material. Fresh weight varies considerably depending on water uptake, humidity, and transpiration rates, whereas dry mass measures the actual organic content synthesized by the plant.

Adım Adım Çözüm

1
Define biological growth in living organisms
Biological growth is defined as an irreversible, permanent increase in size, dry mass, and cell number.
Temporary changes in shape or volume due to water uptake do not constitute true biological growth.
2
Compare fresh weight and dry mass parameters
Fresh weight includes total plant mass, composed largely of water subject to transpiration and absorption changes. Dry mass measures constant organic matter after water evaporation.
Water content varies rapidly with humidity, irrigation, and physiological state, making fresh weight an unreliable indicator of true cellular synthesis.
3
Identify the correct explanation
The option stating that dry mass measures the irreversible accumulation of synthesized organic material excluding water fluctuations is correct.
It accurately highlights why dry mass reflects true organic matter synthesis.

Anahtar Kavram

Measurement of growth (dry mass versus fresh weight)
Soru 16Soru

A biological study recorded the growth parameters of germinating bean seeds (*Phaseolus vulgaris*) kept in total darkness over a ten-day period. Which statement correctly describes the trajectory of the seedling's dry mass and the underlying physiological process responsible for this outcome?

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Cevap: Dry mass decreases because stored organic reserves in the cotyledons are catabolized during cellular respiration to supply metabolic energy.

Cevap

Dry mass decreases because stored organic reserves in the cotyledons are catabolized during cellular respiration to supply metabolic energy.
The correct answer highlights that dry mass measures organic content exclusive of water. When a germinating seedling is kept in total darkness, photosynthesis cannot take place to fix carbon. The embryo relies on stored nutrient reserves within the cotyledons, breaking them down via cellular respiration into carbon dioxide gas and water. The escape of carbon dioxide leads to a measurable net decrease in total dry mass.

Adım Adım Çözüm

1
Define dry mass versus wet (fresh) mass in biological growth measurement.
Dry mass represents the mass of organic matter remaining after all water is removed by drying at low heat.
Water content fluctuates with environmental hydration, making dry mass the standard for measuring true metabolic growth.
2
Analyze environmental constraints during germination in complete darkness.
In total darkness, the light-dependent reactions of photosynthesis cannot take place, preventing carbon fixation.
Without photosynthetic carbon fixation, no new organic molecules can be synthesized from atmospheric carbon dioxide.
3
Evaluate the metabolic source of energy for seedling development before light exposure.
The seedling oxidizes stored carbohydrates, lipids, and proteins in the cotyledons through cellular respiration to generate ATP, releasing carbon dioxide gas into the atmosphere.
The loss of carbon as released carbon dioxide causes a continuous net decline in total seedling dry mass until photosynthetic tissue becomes functional in light.

Anahtar Kavram

Dry Mass Measurement and Metabolic Cost during Seed Germination
Tahmini Süre:1m 50s
Soru 17Soru

Arrange the following physiological and biochemical events during seed germination in the correct chronological order from the onset of germination to the protrusion of the embryonic axis.

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Cevap

The correct chronological sequence is: Imbibition of water resulting in hydration -> Synthesis and release of gibberellins -> Transcription and synthesis of hydrolytic enzymes -> Enzymatic hydrolysis of stored starch -> Cell elongation and emergence of the radicle.
Seed germination begins physically with water imbibition. Hydration triggers the embryo to synthesize gibberellin hormones, which diffuse to the aleurone layer. The aleurone layer then synthesizes hydrolytic enzymes (such as alpha-amylase) that breakdown insoluble endosperm starch into soluble glucose. Finally, the embryo utilizes this glucose for respiration and growth, causing radicle elongation and emergence through the seed coat.

Adım Adım Çözüm

1
Identify the initial physical trigger of germination.
Water imbibition hydrates the seed coat and embryonic tissues.
Dormant seeds have low water potential and must absorb water to reactivate metabolic functions.
2
Trace the hormone signalling pathway initiated by hydration.
The activated embryo synthesizes and secretes gibberellins.
Gibberellins act as the biochemical signal instructing storage tissues to mobilize nutrients.
3
Determine the site of action for gibberellins.
Gibberellins bind to aleurone layer cells to induce production of hydrolytic enzymes like alpha-amylase.
Hydrolytic enzymes are synthesized de novo in response to gibberellin signals.
4
Identify the enzymatic digestion stage.
Insoluble starch in the endosperm is converted into soluble glucose.
Enzymes break down complex macromolecules into transportable molecules.
5
Identify the structural outgrowth stage resulting from nutrient utilization.
The radicle elongates and ruptures the seed coat.
Soluble sugars provide energy and building blocks for cell expansion at the radicle tip.

Anahtar Kavram

Physiological and Biochemical Sequence of Seed Germination
Tahmini Süre:2m 0s
Soru 18Soru

During an ecological survey of a grassland habitat, a student collected an immature arthropod featuring three distinct body regions (head, thorax, and abdomen) and three pairs of jointed walking legs attached strictly to the thoracic segment. Microscopic examination revealed external wing pads on the immature stages, which reached maturity through a succession of nymphal instars without passing through a quiescent pupal stage. Which of the following correctly classifies the organism and its mode of development?

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Cevap: An insect undergoing incomplete (hemimetabolous) metamorphosis

Cevap

An insect undergoing incomplete (hemimetabolous) metamorphosis
The correct answer identifies the organism as an insect undergoing incomplete metamorphosis because it possesses the diagnostic anatomical features of Class Insecta (three body regions and three pairs of thoracic legs) and exhibits hemimetabolous development characterized by progressive nymphal stages, external wing buds, and the absence of a pupal stage.

Adım Adım Çözüm

1
Analyze anatomical features to identify the arthropod class.
The presence of three distinct body divisions (head, thorax, abdomen) and three pairs of legs attached to the thorax places the organism in Class Insecta (insects).
Arachnids have two body regions and four pairs of legs, while crustaceans typically have a cephalothorax and five or more pairs of legs.
2
Examine the developmental pattern to determine the type of metamorphosis.
Development involving nymphal instars with external wing pads and lacking a pupal stage indicates incomplete (hemimetabolous) metamorphosis.
Complete (holometabolous) metamorphosis requires a distinct larval stage, internal wing disc development, and a non-feeding pupal stage.

Anahtar Kavram

Insect Metamorphosis and Arthropod Classification
Tahmini Süre:1m 30s
Soru 19Soru

Tadpoles of the African bullfrog (*Pyxicephalus adspersus*) were raised in two separate aquatic environments. Environment 1 contained natural pond water, while Environment 2 contained mineral-free water deficient in dissolved iodine salts. The tadpoles in Environment 1 developed limbs, absorbed their tails, and successfully metamorphosed into juvenile frogs. In contrast, the tadpoles in Environment 2 increased in body size as giant tadpoles but failed to develop limbs or reabsorb their tails. Which hormone deficiency accounts for the halted metamorphosis in Environment 2, and which gland synthesizes this hormone?

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Cevap: Thyroxin deficiency, synthesized by the thyroid gland

Cevap

Thyroxin deficiency, synthesized by the thyroid gland
Amphibian metamorphosis is under strict endocrine control mediated by thyroxin, a hormone secreted by the thyroid gland. Thyroxin synthesis requires dietary or environmental iodine. When iodine is absent, thyroxin levels remain insufficient to trigger metamorphic events such as limb development, lung formation, and tail reabsorption, resulting in neotenic or oversized larval tadpoles.

Adım Adım Çözüm

1
Analyze the experimental observation
Tadpoles deprived of iodine failed to undergo structural transformation (metamorphosis) into frogs despite growing larger in size.
Iodine is an indispensable elemental component required for the biochemical synthesis of thyroid hormones.
2
Identify the specific hormone and endocrine gland involved
The thyroid gland requires iodine to produce thyroxin, which regulates gene expression for limb bud growth, lung maturation, and tail resorption.
Without iodine, thyroxin cannot be produced, halting amphibian metamorphosis at the larval stage.
3
Differentiate amphibian hormones from insect metamorphic hormones
Juvenile hormone and ecdysone govern arthropod/insect metamorphosis, whereas thyroxin governs amphibian metamorphosis.
Conflating arthropod endocrine systems with vertebrate amphibian systems is a common conceptual misconception.

Anahtar Kavram

Hormonal Regulation of Amphibian Metamorphosis
Soru 20Soru

Arrange the following physiological and anatomical events of ecdysis (moulting) in arthropods in the correct chronological sequence from initiation to final hardening of the new exoskeleton.

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Cevap

The correct chronological sequence of ecdysis events is: (1) Detachment of the epidermis from the old cuticle and secretion of inactive fluid, (2) Enzymatic digestion of the old endocuticle and synthesis of a new soft cuticle, (3) Internal pressure buildup to crack the old exoskeleton along suture lines, (4) Emergence from the exuviae and expansion of the soft body, and (5) Sclerotization to permanently harden the new cuticle.
Ecdysis begins with apolysis, where the living epidermal layer detaches from the old cuticle and releases inactive fluid. Once activated, enzymes break down the old endocuticle so materials can be reabsorbed while a new cuticle forms beneath. The arthropod then inflates its internal pressure to crack the old exuviae along ecdysial lines. After emerging, it expands its body volume to stretch the soft new cuticle. Finally, sclerotization chemically hardens the stretched cuticle to complete the process.

Adım Adım Çözüm

1
Identify the cellular initiation event of ecdysis.
Epidermal cells separate from the old cuticle (apolysis) and secrete inactive moulting enzymes into the resulting gap.
Moulting begins when living epidermal tissue detaches from the non-living outer layer.
2
Determine how the old cuticle is processed and recycled.
Enzymes in the fluid become active, digesting chitin and proteins in the endocuticle while a new procuticle forms underneath.
Recycling the endocuticle conserves nutrients and thins out the old shell for easier shedding.
3
Identify the mechanism for breaking open the weakened old shell.
The arthropod swallows air or water to expand hemolymph volume and exert pressure along weak ecdysial suture lines.
Mechanical force is required to split the remaining outer epicuticle.
4
Determine the step where body growth actually occurs.
The organism crawls out of the old skin (exuviae) and inflates itself to stretch the newly exposed, soft procuticle.
Increase in body size can only take place while the newly exposed outer layer remains stretchable.
5
Identify the final stabilizing phase.
Sclerotization (tanning) occurs, cross-linking cuticular proteins to darken and harden the new exoskeleton.
Hardening secures the larger body size and restores structural protection for muscle attachment.

Anahtar Kavram

Chronological Sequence of Arthropod Ecdysis
Sayfa 1 / 2Sonraki
Growth, Development, and Metamorphosis Alıştırma Soruları — JAMB UTME | Examkin