Form and Function

256 questions

Question 61Question

During the light-dependent stage of photosynthesis, oxygen gas is released into the atmosphere as a byproduct. Which of the following processes is directly responsible for the release of this oxygen?

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Answer: The photolysis of water molecules

Answer

The photolysis of water molecules is directly responsible for releasing oxygen gas during photosynthesis.
Photolysis of water molecules occurs when light energy absorbed by chlorophyll splits water into hydrogen ions, electrons, and oxygen gas (O2O_2). This reaction takes place inside the thylakoid membranes during the light-dependent stage of photosynthesis.

Step-by-Step Solution

1
Identify the biological origin of oxygen evolved during photosynthesis.
Oxygen gas originates exclusively from the splitting of water molecules during the light-dependent phase.
Absorbed light energy drives the photolysis of water (2H2O4H++4e+O22H_2O \rightarrow 4H^+ + 4e^- + O_2) inside the thylakoid lumen.
2
Distinguish light-dependent reactions from light-independent carbon fixation.
Photolysis generates oxygen gas in the thylakoids, whereas carbon dioxide assimilation forms sugar in the stroma.
Carbon dioxide contributes to the synthesis of glucose, not the free molecular oxygen released into the atmosphere.

Key Concept

Photolysis of Water in Photosynthesis
Question 62Question

During strenuous physical exertion, human skeletal muscle cells experience localized oxygen deficiency and temporarily undergo lactic acid fermentation. What is the net yield of ATP molecules produced per molecule of glucose metabolized in this anaerobic pathway?

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Answer: 2 ATP molecules

Answer

2 ATP molecules
During anaerobic respiration (lactic acid fermentation), glucose undergoes partial oxidation through glycolysis in the cytoplasm. Because 2 ATP molecules are consumed to phosphorylate glucose and 4 ATP molecules are subsequently synthesized, the net gain is precisely 2 ATP molecules per glucose molecule.

Step-by-Step Solution

1
Identify the metabolic pathway described in the scenario.
The scenario describes lactic acid fermentation, which is an anaerobic pathway.
In the absence of sufficient oxygen, cells cannot utilize the electron transport chain or Krebs cycle in mitochondria and rely solely on glycolysis.
2
Calculate the net energy yield of glycolysis during anaerobic respiration.
Glycolysis consumes 2 ATP2\text{ ATP} during phosphorylation and generates 4 ATP4\text{ ATP} via substrate-level phosphorylation, giving a net yield of 42=2 ATP4 - 2 = 2\text{ ATP}.
Pyruvate is reduced to lactate to regenerate NAD+\text{NAD}^+ for continuing glycolysis, without yielding any further ATP molecules.

Key Concept

Anaerobic Respiration Net ATP Yield
Question 63Question

Arrange the following physiological events and anatomical stages of human gaseous exchange in the correct sequential order, starting from atmospheric inhalation to oxygen binding in pulmonary blood.

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Answer

The correct sequence of human gaseous exchange starts with air entering the nasal cavity, proceeding through the trachea and main bronchi, conducting through smaller bronchioles to the alveoli, diffusing across the alveolar-capillary membrane, and finally binding to hemoglobin inside red blood cells.
The correct sequence follows the anatomical path of inhalation (nasal cavity → trachea/bronchi → bronchioles → alveoli) followed by physiological diffusion across the respiratory surface into blood plasma and final binding to hemoglobin.

Step-by-Step Solution

1
Identify the entry point of atmospheric air into the respiratory system.
Air intake begins at the nasal cavity for conditioning (filtering, warming, and moistening).
This is the initial anatomical barrier air encounters during inhalation.
2
Trace the passage through major conducting airways.
Air moves past the larynx and down the trachea into the primary mainstem bronchi.
The trachea serves as the trunk conducting air into the right and left lungs.
3
Follow the air deeper into the pulmonary branch network.
Air passes through small bronchioles into the alveolar clusters.
Bronchioles lead directly into the microscopic alveolar sacs where exchange takes place.
4
Determine the physical process of gas transfer.
Oxygen diffuses across the thin alveolar epithelium and capillary endothelium.
Passive diffusion down a partial pressure gradient is responsible for gas transfer into blood.
5
Identify the final chemical step of oxygen transport.
Oxygen binds to hemoglobin inside red blood cells to form oxyhemoglobin.
Binding to hemoglobin allows efficient transport of oxygen throughout the circulatory system.

Key Concept

Path of inhalation and alveolar gaseous exchange in human physiology
Question 64Question

A botanist observes an angiosperm flower species in which the anthers mature and release pollen several days before the stigma of the same flower becomes receptive. The flower also features large, scented petals with nectaries at its base. Which reproductive phenomenon is demonstrated by this flower, and what is its primary biological advantage?

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Answer: Protandry, which enforces cross-pollination by preventing self-fertilization within the flower.

Answer

Protandry, which enforces cross-pollination by preventing self-fertilization within the flower.
Protandry is a temporal form of dichogamy where anthers release pollen prior to the stigma becoming receptive in the same flower. This structural and physiological adaptation prevents autogamy (self-pollination) and ensures outcrossing (cross-pollination) via insect vectors attracted by the bright, scented petals and nectar.

Step-by-Step Solution

1
Analyze the temporal sequence of maturation of the reproductive organs.
The anthers (male organs) mature and shed pollen before the stigma (female organ) is receptive.
Temporal separation of male and female organ maturation within the same flower is termed dichogamy. Specifically, male-first maturation is known as protandry.
2
Determine the functional significance of protandry combined with floral features (scented petals, nectaries).
Scented petals and nectaries attract insect pollinators (entomophily), while protandry prevents pollen of the flower from fertilizing its own ovules.
Preventing self-fertilization forces pollen transfer between different individual plants, promoting outcrossing and genetic variation.

Key Concept

Dichogamy and mechanisms promoting cross-pollination in angiosperms
Question 65Question

Match each plant nutrient ion or chloroplast structure listed on the left with its precise physiological function or biochemical reaction site during plant nutrition and photosynthesis listed on the right.

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Items

Magnesium ions (Mg2+Mg^{2+})
Manganese ions (Mn2+Mn^{2+})
Stroma of the chloroplast
Thylakoid membrane

Matches

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Answer

Magnesium ions pair with the central component of chlorophyll's porphyrin ring; Manganese ions pair with the water-splitting complex for photolysis; the Stroma pairs with carbon dioxide fixation and Calvin cycle reactions; the Thylakoid membrane pairs with proton gradient creation and photophosphorylation.
Each structural item and mineral ion is matched according to its precise biochemical function in plant nutrition and photosynthesis: Magnesium forms the central metal atom in chlorophyll's porphyrin ring; Manganese acts as a vital cofactor for the water-splitting enzyme complex; the stroma provides the fluid enzymatic environment for carbon dioxide reduction in the Calvin cycle; and the thylakoid membrane houses the electron transport assemblies responsible for photophosphorylation.

Step-by-Step Solution

1
Analyze the biochemical role of mineral nutrients in chlorophyll synthesis and photosynthetic chemistry.
Magnesium (Mg2+Mg^{2+}) is the structural centerpiece of the chlorophyll porphyrin ring, whereas Manganese (Mn2+Mn^{2+}) is required catalytically to split water molecules (2H2OO2+4H++4e2H_2O \rightarrow O_2 + 4H^+ + 4e^-).
Differentiating macro- and micro-nutrients by their exact molecular function separates structural elements from catalytic trace cofactors.
2
Map chloroplast compartmentalization to the light-dependent and light-independent phases of photosynthesis.
Thylakoid membranes embed Photosystems I and II for light absorption and ATP generation, while the fluid stroma holds enzymes for carbon fixation.
Structural compartmentalization isolates high proton concentrations inside the thylakoid lumen while biochemical synthesis occurs in the surrounding stroma.
3
Correlate each left item precisely with its unique right partner.
Magnesium maps to porphyrin ring structure, Manganese maps to photolysis catalysis, Stroma maps to RuBisCO carbon fixation, and Thylakoid membrane maps to photophosphorylation.
Ensures all structural and ionic roles are accurately assigned without biochemical overlap.

Key Concept

Compartmentalization of photosynthesis stages and specific mineral nutrient functions in autotrophic nutrition.
Question 66Question

A student performed an experiment to test for the presence of starch in a green leaf exposed to sunlight. What is the correct sequence of steps for carrying out this procedure from start to finish?

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Answer

The correct sequence of steps for testing a leaf for starch is: boiling the leaf in water, boiling the leaf in ethanol, rinsing the leaf in warm water, and adding iodine solution.
The leaf starch test follows a specific functional sequence: boiling in water kills cells and makes membranes permeable, boiling in ethanol extracts green chlorophyll, rinsing in warm water softens the stiffened leaf, and applying iodine solution confirms starch presence via a blue-black color change.

Step-by-Step Solution

1
Determine the initial step to stop biological activity in the leaf.
The leaf is boiled in water.
Boiling kills the cell protoplasm and breaks cell membranes, making them permeable to testing reagents.
2
Determine the step required to remove leaf pigments.
The leaf is boiled in ethanol using a water bath.
Chlorophyll must be extracted so its green color does not mask the blue-black reaction with iodine.
3
Determine the step required to restore leaf flexibility.
The leaf is rinsed in warm water.
Alcohol dehydrates the leaf and makes it stiff; warm water restores its softness.
4
Determine the final step to test for starch.
Iodine solution is added to the leaf.
Iodine solution reacts with starch produced during photosynthesis to yield a characteristic blue-black color.

Key Concept

Experimental procedure for leaf starch testing as evidence of photosynthetic activity
Question 67Question

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

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Items

Apical meristem
Lateral meristem
Intercalary meristem

Matches

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Answer

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.

Step-by-Step Solution

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.

Key Concept

Types of plant meristems and their specific developmental functions
Question 68Question

Match each mammalian reproductive cell or gland with its primary physiological function.

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Items

Seminal vesicles
Cowper's glands
Sertoli cells
Leydig cells

Matches

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Answer

Seminal vesicles match with secreting fructose-rich alkaline fluid supplying energy for sperm motility; Cowper's glands match with secreting mucus fluid to lubricate the urethra and neutralize acidic urine traces; Sertoli cells match with nourishing developing spermatids and establishing the blood-testis barrier; Leydig cells match with synthesizing and secreting testosterone in response to luteinizing hormone.
Seminal vesicles produce fructose-rich fluid powering sperm movement. Cowper's glands secrete pre-ejaculatory alkaline mucus neutralizing urethral acidity. Sertoli cells provide physical and nutritional support to developing gametes while forming the blood-testis barrier. Leydig cells produce testosterone under LH control.

Step-by-Step Solution

1
Identify the main function of accessory seminal vesicles.
Seminal vesicles produce fructose, prostaglandins, and alkaline fluid which nourish sperm and aid motility.
Sperm cells require carbohydrate substrates (fructose) for aerobic respiration to power flagellar movement.
2
Analyze the protective role of Cowper's (bulbourethral) glands.
Cowper's glands release a pre-ejaculatory clear fluid into the urethra.
This fluid neutralizes residual acidic urine in the shared urogenital tract prior to sperm passage.
3
Distinguish between intratubular Sertoli cells and interstitial Leydig cells in the testis.
Sertoli cells act as nurse cells forming tight junctions for the blood-testis barrier, while Leydig cells reside outside the tubules to synthesize testosterone.
Germ cells need immune privilege provided by Sertoli cells, while systemic male secondary traits and spermatogenesis maintenance depend on Leydig cell testosterone secretion.

Key Concept

Structure and function of male mammalian reproductive glands and testicular cells
Question 69Question

During selective reabsorption in the mammalian nephron, glucose and amino acids are completely reabsorbed back into the bloodstream from the glomerular filtrate. In which specific region of the nephron does this process primarily take place, and what structural feature enables it?

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Answer: Proximal convoluted tubule, supported by a dense brush border of microvilli and abundant mitochondria

Answer

Proximal convoluted tubule, supported by a dense brush border of microvilli and abundant mitochondria
Selective reabsorption of glucose, amino acids, and essential ions occurs predominantly in the proximal convoluted tubule (PCT). The epithelial cells lining the PCT are structurally adapted for intensive active transport: they possess a microvillar brush border that vastly increases the absorptive surface area and contain numerous mitochondria to generate the ATP required for active carrier-mediated transport back into the peritubular capillaries.

Step-by-Step Solution

1
Identify the primary site of selective reabsorption in the nephron
Selective reabsorption of essential nutrients like glucose and amino acids occurs in the proximal convoluted tubule (PCT).
Over 65-70% of filtrate volume, including 100% of glucose and amino acids under normal conditions, is reabsorbed immediately after leaving Bowman's capsule.
2
Identify the cellular adaptations required for active transport in the PCT
The epithelial cells of the PCT possess a brush border of microvilli to maximize surface area and high density of mitochondria.
Active transport requires cellular energy in the form of ATP generated by mitochondria, and microvilli dramatically increase the absorption surface area.

Key Concept

Nephron Physiology and Selective Reabsorption
Question 70Question

An illuminated suspension of green algae undergoing steady-state photosynthesis was supplied with radioactively labeled carbon dioxide (14CO2^{14}CO_2). When the light source was abruptly turned off while maintaining the supply of 14CO2^{14}CO_2, analysis revealed an immediate accumulation of glycerate-3-phosphate (PGA) alongside a rapid decline in ribulose 1,5-bisphosphate (RuBP). Which of the following physiological mechanisms explains why glycerate-3-phosphate accumulates in the absence of light?

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Answer: The reduction of glycerate-3-phosphate to glyceraldehyde-3-phosphate requires ATP and NADPH from the light-dependent stage, whereas carbon dioxide fixation into glycerate-3-phosphate proceeds temporarily using remaining ribulose 1,5-bisphosphate.

Answer

The reduction of glycerate-3-phosphate to glyceraldehyde-3-phosphate requires ATP and NADPH synthesized during the light-dependent phase, while carbon dioxide fixation onto ribulose 1,5-bisphosphate continues until the pool of ribulose 1,5-bisphosphate is depleted.
In photosynthesis, the light-dependent reactions in the thylakoids yield ATP and NADPH. These assimilation products drive the reduction of glycerate-3-phosphate (PGA) to glyceraldehyde-3-phosphate (G3P/GALP) in the stroma. When light is removed, ATP and NADPH levels drop immediately. Carbon dioxide fixation onto ribulose 1,5-bisphosphate (RuBP) continues briefly using residual RuBP, forming PGA. However, because PGA cannot be reduced without ATP and NADPH, PGA accumulates while RuBP is consumed and cannot be regenerated.

Step-by-Step Solution

1
Identify the chemical requirements of the light-independent stage (Calvin cycle).
Carbon fixation converts carbon dioxide (CO2CO_2) and ribulose 1,5-bisphosphate (RuBP) into glycerate-3-phosphate (PGA) via RuBisCO.
Carbon fixation itself does not directly consume ATP or NADPH.
2
Analyze the effect of eliminating the light source.
Light-dependent reactions stop, cutting off the immediate production of ATP and reduced NADP (NADPH).
ATP and NADPH are photochemically generated in the thylakoid membranes.
3
Evaluate the metabolic bottleneck caused by the absence of light energy products.
The reduction step converting PGA to glyceraldehyde-3-phosphate (GALP/G3P) halts due to lack of ATP and NADPH, leading to PGA accumulation and RuBP depletion.
RuBP regeneration also requires ATP; hence, existing RuBP is converted to PGA but cannot be regenerated.

Key Concept

Interdependence of Photosynthetic Light and Dark Reactions
Question 71Question

A sample of salivary amylase was kept in an ice bath at 0C0^\circ\text{C} for two hours and then mixed with a starch solution maintained at body temperature (37C37^\circ\text{C}). Which of the following best describes the outcome of this reaction?

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Answer: Starch is hydrolyzed to maltose because low temperature causes temporary inactivation rather than denaturation.

Answer

Starch is hydrolyzed to maltose because low temperature causes temporary inactivation rather than denaturation.
Low temperatures render enzymes temporarily inactive due to reduced molecular collisions. When returned to an optimal temperature of 37C37^\circ\text{C}, salivary amylase regains catalytic activity and hydrolyzes starch into maltose.

Step-by-Step Solution

1
Analyze the effect of low temperature (0C0^\circ\text{C}) on enzyme structure and kinetic energy.
At 0C0^\circ\text{C}, salivary amylase molecules have very low kinetic energy, resulting in temporary inactivation without altering the active site shape.
Low temperature does not break the chemical bonds maintaining the enzyme's 3D structure.
2
Determine what happens when the enzyme is warmed to 37C37^\circ\text{C}.
The enzyme regains kinetic energy and normal catalytic activity at its optimum temperature of 37C37^\circ\text{C}.
Because the enzyme was not denatured, active sites can successfully bind starch substrates.
3
Identify the end product of starch breakdown by salivary amylase.
Salivary amylase breaks starch down into the disaccharide maltose.
Amylase specifically cleaves alpha-1,4-glycosidic bonds in starch.

Key Concept

Effect of Temperature on Digestive Enzymes
Estimated Time:45s
Question 72Question

Organisms living in arid terrestrial environments often conserve water by excreting nitrogenous waste in the form of insoluble uric acid. Which of the following correctly pairs the specialized excretory organ with an organism that utilizes this adaptation?

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Answer: Malpighian tubules in the desert locust

Answer

Malpighian tubules in the desert locust
The correct option identifies the desert locust, a terrestrial insect that utilizes Malpighian tubules to excrete insoluble uric acid. This structural and physiological combination allows maximum water reabsorption prior to defecation, serving as a critical adaptation for survival in dry terrestrial habitats.

Step-by-Step Solution

1
Identify the nitrogenous waste adaptation described in the stem.
Uric acid excretion (uricotelism) is an adaptation for extreme water conservation in terrestrial habitats.
Uric acid is insoluble and non-toxic, requiring minimal water for excretion.
2
Evaluate the organisms listed to determine which group is uricotelic and uses the correct excretory organ.
Insects are uricotelic organisms that possess Malpighian tubules.
Malpighian tubules lie in the hemolymph and discharge nitrogenous waste into the digestive tract for water reabsorption.
3
Cross-reference each organism with its anatomical excretory organ.
The desert locust (an insect) uses Malpighian tubules.
Other pairings misassign flatworm flame cells to earthworms, annelid nephridia to planarians, and crustacean green glands to insects.

Key Concept

Excretory structures and physiological adaptations to nitrogenous waste disposal in invertebrates
Estimated Time:1m 0s
Question 73Question

During urine formation in the mammalian kidney, blood plasma is filtered and processed along the nephron. What is the correct anatomical sequence of structures through which renal filtrate flows, starting from the site of ultrafiltration to urine collection?

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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.
In the mammalian kidney, urine formation follows a precise anatomical pathway: blood plasma undergoes ultrafiltration at the glomerulus into Bowman's capsule, proceeds through the proximal convoluted tubule (selective reabsorption), down and up the loop of Henle (concentration gradient), through the distal convoluted tubule (tubular secretion), and finally gathers in the collecting duct to be excreted.

Step-by-Step Solution

1
Identify the site of initial filtration
High hydrostatic pressure in the glomerulus forces water and small solutes into Bowman's capsule.
Bowman's capsule receives the initial glomerular filtrate, making it the first structure in the pathway.
2
Trace filtrate movement into the first tubular segment
Filtrate flows from Bowman's capsule into the proximal convoluted tubule.
The proximal convoluted tubule directly connects to Bowman's capsule and is specialized for bulk reabsorption.
3
Follow filtrate down into the renal medulla
Filtrate moves from the proximal convoluted tubule into the U-shaped loop of Henle.
The loop of Henle dips into the medulla to regulate water retention and salt concentration gradients.
4
Trace filtrate back up into the distal tubule segment
Filtrate exits the ascending limb of Henle's loop and enters the distal convoluted tubule.
The distal convoluted tubule sits downstream of the loop of Henle in the cortex to handle fine ion adjustments.
5
Determine the final collection segment
Filtrate empties into the collecting duct, which leads to the ureter.
Multiple nephrons drain into a shared collecting duct, which carries concentrated urine toward the renal pelvis.

Key Concept

Nephron Fluid Pathway and Urine Formation
Question 74Question

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

Answer

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).

Step-by-Step Solution

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.

Key Concept

Discontinuous growth and ecdysis in arthropods
Question 75Question

Match each organ of the mammalian digestive tract with its primary physiological function in nutrition.

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Items

Stomach
Small intestine
Large intestine

Matches

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Answer

Stomach matches with Secretion of hydrochloric acid and initiation of protein digestion; Small intestine matches with Completion of chemical digestion and absorption of digested nutrients; Large intestine matches with Reabsorption of water and mineral salts from indigestible waste.
Each digestive organ is correctly matched to its specific physiological role: the stomach initiates protein digestion in an acidic environment, the small intestine completes enzymatic breakdown and absorbs nutrient monomers, and the large intestine reabsorbs excess water and mineral salts.

Step-by-Step Solution

1
Identify the primary function of the stomach in mammalian digestion.
The stomach secretes gastric acid (HCl) and pepsinogen to begin chemical hydrolysis of proteins.
Gastric parietal cells produce HCl which lowers pH to activate pepsin.
2
Identify the primary function of the small intestine.
The small intestine is the chief site where enzyme digestion finishes and absorption of nutrient monomers occurs.
Villi and microvilli provide a broad surface area for active and passive uptake of nutrients.
3
Identify the primary function of the large intestine.
The large intestine absorbs remaining water and mineral salts to form solid feces.
Reabsorbing water maintains fluid balance in the organism.

Key Concept

Functions of mammalian alimentary canal organs
Question 76Question

In C4 plants, atmospheric carbon dioxide is initially fixed in mesophyll cells before being transferred to bundle sheath cells to minimize photorespiration. What is the correct chronological sequence of biochemical events in the C4 (Hatch-Slack) pathway from initial carbon fixation to its entry into the Calvin cycle?

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Answer

The correct sequence of the Hatch-Slack (C4) pathway is: (1) Carboxylation of PEP to oxaloacetate in mesophyll cells -> (2) Reduction of oxaloacetate to malate -> (3) Translocation of malate into bundle sheath cells -> (4) Decarboxylation of malate releasing CO2 and pyruvate -> (5) Fixation of released CO2 by RuBisCO in the Calvin cycle.
The correct order follows the spatial and biochemical sequence of the C4 pathway: atmospheric carbon dioxide is first fixed by PEP carboxylase in mesophyll cells to form oxaloacetate, which is reduced to malate. Malate is translocated through plasmodesmata into bundle sheath cells, where it undergoes decarboxylation to yield pyruvate and release carbon dioxide. Finally, RuBisCO fixes this released carbon dioxide in the Calvin cycle.

Step-by-Step Solution

1
Identify the initial primary carbon fixation reaction in mesophyll cells
Atmospheric CO2CO_2 reacts with phosphoenolpyruvate (PEP) catalyzed by PEP carboxylase to yield oxaloacetate.
PEP carboxylase has a high affinity for CO2CO_2 and lacks oxygenase activity, preventing photorespiration at the entry point.
2
Trace the organic acid reduction step in mesophyll chloroplasts
Oxaloacetate is reduced to malate.
Malate serves as the mobile 4-carbon transport intermediate carrying fixed carbon to the bundle sheath.
3
Identify the intercellular translocation step across Kranz anatomy
Malate moves from mesophyll cells into bundle sheath cells through plasmodesmata.
Spatial separation requires physical movement of the organic acid into bundle sheath cells.
4
Trace the carbon-release step inside bundle sheath chloroplasts
Malate undergoes decarboxylation to produce pyruvate and free CO2CO_2.
Decarboxylation elevates local CO2CO_2 concentration around RuBisCO to outcompete oxygen binding.
5
Connect the released CO2CO_2 to the Calvin cycle entry step
RuBisCO fixes the concentrated CO2CO_2 into 3-phosphoglycerate (PGA).
This completes the transfer of carbon into the standard light-independent reactions of photosynthesis.

Key Concept

Spatial carbon fixation and intercellular shuttle in C4 photosynthesis (Hatch-Slack pathway)
Question 77Question

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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Items

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

Matches

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Answer

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.

Step-by-Step Solution

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.

Key Concept

Plant Germination Dynamics, Seed Mobilization Biochemistry, and Hormonal Regulation of Ecdysis
Question 78Question

An experiment was conducted to investigate the rate of protein digestion by incubating boiled egg white with pancreatic juice under different physiological conditions. In tube 1, sodium hydrogen carbonate was added at 37C37^\circ\text{C}. In tube 2, dilute hydrochloric acid was added at 37C37^\circ\text{C}. In tube 3, sodium hydrogen carbonate was added after boiling the pancreatic juice at 100C100^\circ\text{C}. Which of the following observations and physiological explanations correctly accounts for the results?

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Answer: Rapid digestion occurs only in tube 1 because trypsin and chymotrypsin require an alkaline medium provided by bicarbonate ions at body temperature.

Answer

Rapid digestion occurs only in tube 1 because pancreatic proteolytic enzymes (such as trypsin) operate optimally in an alkaline environment provided by sodium hydrogen carbonate at optimum body temperature (37°C).
Pancreatic juice contains proteolytic enzymes such as trypsinogen and chymotrypsinogen. In the duodenum, sodium hydrogen carbonate secreted in pancreatic juice neutralizes gastric acid to establish an alkaline pH (pH 7.58.0\text{pH } 7.5 - 8.0), which is optimal for pancreatic enzyme activity at body temperature (37C37^\circ\text{C}). Tube 1 replicates these optimal physiological conditions, leading to rapid protein hydrolysis.

Step-by-Step Solution

1
Analyze the components of pancreatic juice and their optimal pH environment.
Pancreatic juice contains digestive enzymes like trypsin and chymotrypsin which hydrolyze proteins in the alkaline environment of the duodenum.
Sodium hydrogen carbonate neutralizes acidic chyme from the stomach and provides the alkaline pH needed for pancreatic enzymes.
2
Evaluate the effect of acid in tube 2.
Hydrochloric acid lowers the pH, creating an acidic environment in which trypsin is inactive.
Pancreatic enzymes require alkaline pH, unlike gastric pepsin which functions in acidic pH.
3
Evaluate the effect of thermal denaturation in tube 3.
Boiling pancreatic juice at 100°C permanently denatures the active sites of its enzymes.
High temperatures break hydrogen and disulfide bonds in enzyme tertiary structure, rendering them inactive even if optimal pH is supplied.

Key Concept

Enzyme specificity, pH optima, and thermal denaturation in mammalian digestive tracts
Question 79Question

Match each mammalian nephron region on the left with its primary physiological function during urine formation on the right.

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Items

Glomerulus
Proximal convoluted tubule
Loop of Henle
Distal convoluted tubule

Matches

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Answer

Glomerulus matches with Ultrafiltration under high hydrostatic pressure; Proximal convoluted tubule matches with Selective reabsorption of glucose, amino acids, and bulk filtrate water; Loop of Henle matches with Establishment of an osmotic medullary gradient; Distal convoluted tubule matches with Hormone-regulated reabsorption of sodium and selective tubular secretion.
Each nephron section performs a distinct step in urine formation: the glomerulus filters blood plasma under pressure; the proximal convoluted tubule reabsorbs vital solutes like glucose; the Loop of Henle generates the medullary osmotic gradient; and the distal convoluted tubule carries out selective secretion and hormone-controlled electrolyte tuning.

Step-by-Step Solution

1
Identify the primary function of the renal corpuscle/glomerulus
High blood pressure forces water and dissolved solutes out of capillaries into Bowman's capsule (ultrafiltration).
Afferent arteriole diameter is larger than efferent arteriole diameter, creating hydrostatic filtration pressure.
2
Determine the main process occurring in the proximal convoluted tubule
Obligatory reabsorption of 100% glucose, amino acids, and ~65-70% of salts and water occurs here.
Epithelial lining has a microvillar brush border that maximizes surface area for active transport.
3
Identify the role of the hair-pin shaped Loop of Henle
Creates a hyperosmolal gradient in the renal medulla via countercurrent multiplication.
Descending limb is permeable to water, while ascending limb actively pumps out sodium and chloride ions.
4
Determine the functional role of the distal convoluted tubule
Fine-tunes electrolyte balance via active secretion of ions (K+K^+, H+H^+) and aldosterone-stimulated Na+Na^+ absorption.
Target region for hormonal regulation to maintain systemic pH and blood volume balance.

Key Concept

Nephron functional anatomy and urine formation processes
Estimated Time:1m 15s
Question 80Question

Match each digestive secretion or enzyme of the mammalian alimentary canal with its primary physiological function or mode of action in animal nutrition.

Click a left item, then click its matching right item

Items

Bile salts
Enterokinase (Enteropeptidase)
Ptyalin (Salivary amylase)
Rennin (Chymosin)

Matches

Show answer & explanation

Answer

Bile salts pair with lipid emulsification; Enterokinase pairs with trypsinogen activation; Ptyalin pairs with starch hydrolysis to maltose; Rennin pairs with caseinogen coagulation.
Each secretion or enzyme fulfills a distinct biochemical function: bile salts physically emulsify fats to assist lipase, enterokinase converts inactive trypsinogen into active trypsin in the small intestine, ptyalin hydrolyzes starch to maltose in the buccal cavity, and rennin precipitates milk protein in juvenile stomachs.

Step-by-Step Solution

1
Identify the non-enzymatic role of liver secretions in fat digestion.
Bile salts reduce lipid surface tension without chemical cleavage, performing emulsification.
Increasing the lipid surface area enhances subsequent chemical digestion by lipase.
2
Determine the activation cascade of pancreatic proteolytic enzymes.
Enterokinase, secreted by the succus entericus, activates trypsinogen to trypsin.
Proteolytic enzymes are produced as inactive zymogens to prevent autolysis of gland tissues.
3
Analyze carbohydrate hydrolysis in the anterior alimentary canal.
Ptyalin (salivary amylase) breaks down cooked starch to maltose.
Amylase acts specifically on alpha-1,4 glycosidic bonds in cooked carbohydrates.
4
Examine specialized gastric protein digestion in young mammals.
Rennin converts soluble caseinogen to insoluble casein in the presence of calcium ions.
Precipitating milk protein slows gut transit time so pepsin can effectively digest it.

Key Concept

Digestive Secretions, Enzymes, and Functional Specificity
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