All practice questions

Question 7101Question →

In a municipal water purification plant, raw river water undergoes several sequential treatment processes to make it safe for public consumption. What is the correct chronological sequence of these major water treatment stages, from raw water intake to final distribution?

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Answer

The correct sequence of municipal water treatment is: Coagulation/flocculation → Sedimentation → Filtration → Disinfection.

The correct sequence follows the logical progression of municipal water purification: Coagulation chemically aggregates fine particles into flocs; Sedimentation allows these heavy flocs to settle to the bottom; Filtration passes the clarified liquid through sand/gravel to catch residual solids; Disinfection is performed last with chlorine to destroy pathogenic microorganisms in clear water.

Step-by-Step Solution

1

Identify the initial chemical coagulation step

Alum,

Al_2(SO_4)_3

, is added to raw water to neutralize colloidal charges and form floc particles.

Fine suspended particles will not settle naturally without first being coagulated into larger masses.

2

Determine the settling process

Water flows slowly through sedimentation tanks where heavy flocs sink to the bottom.

Gravity settling removes the bulk of solid impurities created during coagulation.

3

Identify the fine solid removal process

Clarified water passes through coarse sand, fine sand, and gravel beds.

Filtration removes any remaining microscopic suspended particles.

4

Determine the final biological purification step

Chlorine is injected into the clear filtered water.

Chlorination destroys disease-causing microbes and guarantees biological safety throughout the distribution network.

Key Concept

Stages of Municipal Water Treatment

Question 7102Question →

Complete the following statement regarding the qualitative test for hydrogen sulfide gas by filling in the missing physical characteristic of the precipitate formed.

Fill in the blanks below

When hydrogen sulfide gas (

H_2S

) is bubbled into an aqueous solution of lead(II) ethanoate, a insoluble precipitate of lead(II) sulfide (

PbS

) is observed.

Show answer & explanation

Answer

black

When hydrogen sulfide gas comes into contact with lead(II) ethanoate, lead(II) sulfide (

PbS

) is formed. Lead(II) sulfide is insoluble in water and has a characteristic black color, making 'black' the correct term to complete the sentence.

Step-by-Step Solution

1

Identify the chemical reaction between hydrogen sulfide gas (

H_2S

) and aqueous lead(II) ethanoate solution (

(CH_3COO)_2Pb

).

The double displacement reaction produces lead(II) sulfide precipitate (

PbS

) and ethanoic acid (

CH_3COOH

).

The equation is:

(CH_3COO)_2Pb_{(aq)} + H_2S_{(g)} \rightarrow PbS_{(s)} + 2CH_3COOH_{(aq)}

.

2

Determine the physical property (color) of the resulting lead(II) sulfide precipitate.

Lead(II) sulfide (

PbS

) is a distinct black solid.

This specific color change is the standard laboratory confirmatory test for identifying

H_2S

gas.

Key Concept

Qualitative test for hydrogen sulfide gas using lead(II) ethanoate paper or solution

Question 7103Question →

An aqueous solution of sodium fluoride ( $NaF$ ) is basic to litmus. Which species undergoes hydrolysis to cause this alkalinity?

Fluoride ion (

F^-

)

Sodium ion (

Na^+

)

Hydroxide ion (

OH^-

)

Hydrogen ion (

H^+

)

Show answer & explanation

Answer: Fluoride ion ( $F^-$ )

Answer

The fluoride ion (

F^-

) undergoes anion hydrolysis in water to produce hydroxide ions (

OH^-

), resulting in a basic solution.

Sodium fluoride (

NaF

) dissociates completely in water to yield

Na^+

and

F^-

. Because

HF

is a weak acid, its conjugate base (

F^-

) reacts with water molecules (anion hydrolysis) via

F^- + H_2O \rightleftharpoons HF + OH^-

, generating hydroxide ions which make the solution alkaline.

Step-by-Step Solution

1

Identify the parent acid and base of the salt.

NaF

is formed from a strong base (

NaOH

) and a weak acid (

HF

).

Salts of strong bases and weak acids produce basic solutions upon dissolving in water.

2

Determine which ion reacts with water (hydrolyzes).

The fluoride ion (

F^-

) reacts with water according to

F^- + H_2O \rightleftharpoons HF + OH^-

.

The conjugate base of a weak acid is strong enough to accept a proton from water.

3

Relate the generated ion to the acidity/alkalinity of the solution.

The production of extra

OH^-

ions increases the pH above 7, turning the solution basic.

An excess of hydroxide ions relative to hydronium ions causes basic/alkaline behavior.

Key Concept

Anion Hydrolysis of Weak Acid-Strong Base Salts

Question 7104Question →

During the thermal decomposition of dinitrogen pentoxide, $2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g)$ , the concentration of $\text{N}_2\text{O}_5$ decreases from $0.80\text{ mol dm}^{-3}$ to $0.20\text{ mol dm}^{-3}$ over a time interval of $5\text{ minutes}$ . What is the average rate of decomposition of $\text{N}_2\text{O}_5$ in $\text{mol dm}^{-3}\text{s}^{-1}$ ?

Show answer & explanation

Answer: 0.002

Answer

0.002 mol dm⁻³ s⁻¹

The average rate of decomposition of a reactant is calculated by dividing the decrease in its concentration by the elapsed time in seconds. Here, the concentration change is

\Delta C = 0.80 - 0.20 = 0.60\text{ mol dm}^{-3}

and the time in seconds is

5 \times 60 = 300\text{ s}

. Thus, the rate is

\frac{0.60}{300} = 0.002\text{ mol dm}^{-3}\text{s}^{-1}

.

Step-by-Step Solution

1

Calculate the change in concentration of the reactant

\Delta [\text{N}_2\text{O}_5] = 0.80\text{ mol dm}^{-3} - 0.20\text{ mol dm}^{-3} = 0.60\text{ mol dm}^{-3}

The rate depends on the amount of reactant consumed during the reaction period.

2

Convert time from minutes to seconds

\Delta t = 5\text{ min} \times 60\text{ s/min} = 300\text{ s}

The requested unit is per second (

\text{s}^{-1}

), so time must be converted to standard SI units.

3

Divide the concentration change by the total time in seconds

\text{Rate} = \frac{0.60\text{ mol dm}^{-3}}{300\text{ s}} = 0.002\text{ mol dm}^{-3}\text{s}^{-1}

Average rate of reaction is defined as the change in concentration per unit time.

Key Concept

Average Rate of Reaction from Concentration Change

Question 7105Question →

Match each mixture or suspension to the physical separation technique best suited for isolating its components.

Click a left item, then click its matching right item

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Answer

Ammonium chloride and sodium chloride are separated by sublimation; sulfur and iron filings are separated by magnetization; red blood cells in plasma are separated by centrifugation.

Each mixture matches its technique based on fundamental physical characteristics: ammonium chloride sublimes readily upon heating; iron is attracted to magnets; suspended blood cells require high-speed centrifugal acceleration to separate from plasma.

Step-by-Step Solution

1

Identify the key physical property difference in the solid mixture of ammonium chloride and sodium chloride.

Ammonium chloride is a sublimable solid, whereas sodium chloride is thermally stable and non-volatile at gentle heating temperatures.

Sublimation vaporizes ammonium chloride directly into gas, which condenses on a cool surface while sodium chloride remains behind.

2

Analyze the magnetic properties of sulfur and iron filings.

Iron is strongly magnetic (ferromagnetic) while sulfur is non-magnetic.

Applying a magnetic field pulls the iron particles out of the mixture cleanly.

3

Determine the appropriate method for separating solid blood cells from liquid blood plasma.

Blood cells exist as a fine suspension of particles with subtle density differences relative to plasma, which settle too slowly under normal gravity.

Centrifugation applies high rotational speed to force dense suspended cells to the bottom of the container rapidly.

Key Concept

Selecting separation techniques based on physical properties (sublimability, magnetism, and particle density in suspension)

Question 7106Question →

Match each ecological measuring instrument listed in Column A with the corresponding abiotic parameter it measures in Column B.

Click a left item, then click its matching right item

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Answer

Wind vane matches with direction of wind movement; Lux meter matches with intensity of light in a habitat; Rain gauge matches with amount of precipitation over a period; Maximum-minimum thermometer matches with daily temperature range and extremes.

Each instrument correctly corresponds to its specific environmental measurement: wind vanes indicate wind direction, lux meters measure light intensity, rain gauges record precipitation amount, and maximum-minimum thermometers capture daily temperature extremes.

Step-by-Step Solution

1

Identify the primary function of each ecological instrument.

Wind vane indicates wind direction; Lux meter quantifies light intensity; Rain gauge collects rainfall; Maximum-minimum thermometer measures extreme temperatures.

Abiotic environmental factors are quantified using specific instruments designed for physical measurements.

2

Pair each instrument from Column A to its matching abiotic factor in Column B.

All instruments are mapped correctly to their respective environmental measurement parameters.

Direct one-to-one mapping links each equipment item with its intended ecological variable.

Key Concept

Measurement of Abiotic Ecological Factors

Question 7107Question →

The solubility of a salt $Y$ (molar mass = $160\text{ g mol}^{-1}$ ) in water is $2.5\text{ mol dm}^{-3}$ at $70^\circ\text{C}$ and $1.2\text{ mol dm}^{-3}$ at $30^\circ\text{C}$ . If $300\text{ cm}^3$ of a saturated solution of $Y$ at $70^\circ\text{C}$ is cooled to $30^\circ\text{C}$ , what mass of salt $Y$ will crystallize out of solution?

62.4\text{ g}

120.0\text{ g}

208.0\text{ g}

0.39\text{ g}

Show answer & explanation

Answer: $62.4\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

62.4\text{ g}

.

The mass of salt precipitated is determined by taking the difference in molar solubility (

2.5 - 1.2 = 1.3\text{ mol dm}^{-3}

), multiplying by the volume fraction (

300/1000 = 0.3\text{ dm}^3

) to find the number of moles (

0.39\text{ mol}

), and then multiplying by the molar mass (

160\text{ g mol}^{-1}

) to obtain

62.4\text{ g}

.

Step-by-Step Solution

1

Determine the change in molar solubility upon cooling

\Delta S = 2.5\text{ mol dm}^{-3} - 1.2\text{ mol dm}^{-3} = 1.3\text{ mol dm}^{-3}

Solubility decreases as temperature drops, causing the excess solute to precipitate.

2

Calculate the amount in moles precipitated in

300\text{ cm}^3

of solution

n = 1.3\text{ mol dm}^{-3} \times \frac{300\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.39\text{ mol}

The solution volume is

300\text{ cm}^3

(

0.3\text{ dm}^3

), so the moles precipitated must be scaled from

1\text{ dm}^3

.

3

Convert the moles precipitated to mass in grams

\text{Mass} = 0.39\text{ mol} \times 160\text{ g mol}^{-1} = 62.4\text{ g}

Mass is found by multiplying the mole quantity by the given molar mass of the salt.

Key Concept

Solubility and Crystallization Calculations

Question 7108Question →

Arrange the following organic compounds, each having a relative molecular mass of approximately $58-60\text{ g/mol}$ , in order of INCREASING boiling point (from lowest boiling point to highest boiling point).

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Answer

The correct sequence in order of increasing boiling point is Butane (

\text{C}_4\text{H}_{10}

), followed by Methyl formate (

\text{HCOOCH}_3

), then Propan-1-ol (

\text{C}_3\text{H}_7\text{OH}

), and finally Ethanoic acid (

\text{CH}_3\text{COOH}

).

Boiling points depend on the relative strength of intermolecular forces when molecular masses are comparable (~60 g/mol). Butane is non-polar and exhibits only weak dispersion forces (lowest boiling point). Methyl formate is polar and exhibits dipole-dipole attractions. Propan-1-ol forms strong hydrogen bonds via its hydroxyl group. Ethanoic acid forms even stronger hydrogen bonds and stable cyclic dimers, giving it the highest boiling point.

Step-by-Step Solution

1

Identify the primary type of intermolecular force present in each compound of similar molar mass (

\approx 58-60\text{ g/mol}

).

Butane has London dispersion forces; Methyl formate has dipole-dipole forces; Propan-1-ol has hydrogen bonding; Ethanoic acid has extensive hydrogen bonding and dimer formation.

Boiling point increases as the strength of intermolecular forces holding the liquid molecules together increases.

2

Compare the compounds without hydrogen bonding capabilities (Butane vs. Methyl formate).

Butane is non-polar and exhibits only weak dispersion forces. Methyl formate has a polar carbonyl group (

\text{C=O}

) causing dipole-dipole attractions, making its boiling point higher than butane.

Permanent dipole-dipole attractions are stronger than instantaneous dispersion forces for molecules of similar size.

3

Compare the hydrogen-bonded compounds (Propan-1-ol vs. Ethanoic acid).

Propan-1-ol forms intermolecular hydrogen bonds through its single hydroxyl group. Ethanoic acid forms stronger hydrogen bonds using both its carbonyl oxygen and hydroxyl hydrogen to form stable cyclic dimers.

Dimerization in alkanoic acids effectively doubles the molecular interaction area, requiring significantly more thermal energy to break apart during vaporization.

4

Arrange the compounds in order of increasing boiling point.

Butane < Methyl formate < Propan-1-ol < Ethanoic acid.

The progression of intermolecular force strength directly dictates the trend in boiling points.

Key Concept

Intermolecular Forces and Boiling Point Trends in Alkanoic Acids, Esters, Alkanols, and Alkanes

Question 7109Question →

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

2

4

36

38

Show answer & explanation

Answer: $2$

Answer

The net yield of ATP molecules obtained during anaerobic respiration is

2

.

During anaerobic respiration, glucose undergoes glycolysis in the cytoplasm. Although glycolysis produces

4\text{ ATP}

molecules in total,

2\text{ ATP}

molecules are used during the phosphorylation steps, leaving a net gain of

2\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 (

4

) minus ATP invested (

2

) equals a net yield of

2\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

Question 7110Question →

A taxonomist documented four specimens during a biodiversity survey, recording their proposed scientific names alongside key structural characteristics:

* Specimen I: *amoeba proteus* — unicellular eukaryote with pseudopodia.
* Specimen II: *Rhizopus stolonifer* — multicellular eukaryotic heterotroph possessing a chitinous cell wall.
* Specimen III: *Bacillus subtilis* — unicellular prokaryote with a cell wall composed primarily of cellulose.
* Specimen IV: *Rabies virus* — acellular nucleoprotein entity classified under Kingdom Monera.

Which specimen entry strictly complies with the rules of binomial nomenclature while accurately stating its diagnostic kingdom-level characteristic?

Specimen I, because the species epithet and generic name are written in italics.

Specimen II, because the generic name is capitalized, the specific epithet is lowercase, and chitin is the major structural cell wall component of Fungi.

Specimen III, because the name is italicized and bacterial cell walls are constructed from cellulose.

Specimen IV, because viral nomenclature follows standard Linnaean binomial rules within Kingdom Monera.

Show answer & explanation

Answer: Specimen II, because the generic name is capitalized, the specific epithet is lowercase, and chitin is the major structural cell wall component of Fungi.

Answer

Specimen II is the only entry that strictly complies with binomial nomenclature formatting and features an accurate diagnostic kingdom-level trait.

The choice identifying Specimen II is correct because *Rhizopus stolonifer* adheres perfectly to Linnaean binomial rules (Genus capitalized, specific epithet lowercase, set in italics) and accurately reflects that fungi possess cell walls made of chitin.

Step-by-Step Solution

1

Analyze binomial nomenclature formatting rules across all four specimens.

Specimen I fails because the generic name *amoeba* is not capitalized. Specimen II (*Rhizopus stolonifer*) and Specimen III (*Bacillus subtilis*) follow binomial rules (capitalized genus, lowercase species epithet, italicized). Specimen IV uses a common English name format, as viruses do not follow formal Linnaean binomial nomenclature.

Binomial nomenclature requires a capitalized Genus name and a lowercase species epithet, both printed in italics or underlined when handwritten.

2

Evaluate the biological and kingdom-level diagnostic statements for the correctly formatted scientific names.

Specimen II correctly states that *Rhizopus stolonifer* (a fungus) has a chitinous cell wall. Specimen III incorrectly attributes a cellulose cell wall to a bacterium (*Bacillus subtilis*), whereas bacterial cell walls consist of peptidoglycan.

Taxonomic classification relies on fundamental cell wall composition and cellular organization to place organisms into appropriate biological kingdoms.

3

Synthesize results to identify the valid specimen entry.

Specimen II satisfies both nomenclature formatting and biological accuracy.

Only Specimen II meets all statutory rules of binomial designation and kingdom diagnostics.

Key Concept

Rules of Binomial Nomenclature and Diagnostic Characteristics of Biological Kingdoms

Question 7111Question →

In comparative vertebrate anatomy, which of the following groups possesses a circulatory system in which oxygenated blood from the lungs and deoxygenated blood from the body tissues enter two separate atria, but are subsequently pumped through a single, undivided ventricle?

Amphibians

Bony fishes

Birds

Mammals

Show answer & explanation

Answer: Amphibians

Answer

Amphibians possess a three-chambered heart made up of two distinct atria and one single ventricle.

Amphibians feature a three-chambered heart comprising two separate atria (left and right) and one undivided ventricle. Deoxygenated blood returns from systemic tissues into the right atrium, while oxygenated blood returns from the lungs and skin into the left atrium. Both streams flow into the common ventricle before being pumped out.

Step-by-Step Solution

1

Analyze the structural organization of vertebrate heart chambers across main taxonomic groups.

Fishes have 2 chambers (1 atrium, 1 ventricle); Amphibians have 3 chambers (2 atria, 1 ventricle); Reptiles have 3 chambers with an incomplete septum (except crocodilians); Birds and Mammals have 4 chambers (2 atria, 2 ventricles).

Identifying chamber counts establishes how blood flows through the heart in each class.

2

Determine the pathway of blood entering the heart in amphibians.

Deoxygenated blood from the body enters the right atrium, while oxygenated blood from the lungs and skin enters the left atrium. Both atria then empty into a single shared ventricle.

This two-atria, single-ventricle configuration results in incomplete separation of systemic and pulmonary blood flow.

Key Concept

Comparative Vertebrate Cardiac Anatomy and Blood Circulation Pathways

Question 7112Question →

In a constitutional presidential democracy, the doctrine of separation of powers demands complete structural and functional isolation among the executive, legislative, and judicial organs of government.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Separation of powers does not mean total functional isolation, as the complementary mechanism of checks and balances requires purposeful interaction and mutual oversight among organs of government.

The correct answer is False because Montesquieu's theory and modern constitutional practice combine the division of functions with checks and balances. Overlapping oversight mechanisms, such as judicial review of legislative acts or executive vetoes, explicitly contradict the idea of total isolation.

Step-by-Step Solution

1

Analyze the core premise of separation of powers.

Separation of powers assigns distinct functions—lawmaking, law execution, and adjudication—to separate government branches.

Dividing primary functions prevents any single branch from concentrating absolute governmental authority.

2

Examine how checks and balances intersect with separation of powers.

Checks and balances intentionally grant each organ specific powers to oversee, restrict, or modify actions of the other organs.

Complete isolation would make constitutional mechanisms like presidential vetoes, legislative impeachment, and judicial review impossible.

3

Evaluate the claim of absolute functional isolation.

Because checks and balances mandate inter-arm interaction, claiming that separation of powers requires complete isolation is incorrect.

Constitutional governance relies on partial checks between distinct bodies rather than total quarantine of powers.

Key Concept

Interdependence of Separation of Powers and Checks and Balances

Estimated Time:1m 30s

Question 7113Question →

Match each plant group on the left with its corresponding structural feature on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thallophytes correspond to an undifferentiated thallus lacking roots, stems, leaves, and vascular tissue; Bryophytes correspond to non-vascular plants anchored by rhizoids with gametophyte dominance; Pteridophytes correspond to seedless vascular plants with true roots, stems, leaves, and sporophyte dominance.

Thallophytes feature a simple thallus without specialized conducting tissue or organs. Bryophytes are non-vascular plants with rhizoids and gametophyte dominance. Pteridophytes are true vascular seedless plants with true roots, stems, leaves, and sporophyte dominance.

Step-by-Step Solution

1

Identify the characteristic features of Thallophytes.

Thallophytes represent the simplest algae/plant forms with an unspecialized thallus body devoid of vascular tissues.

They have not evolved vascular organs or tissue differentiation.

2

Identify the characteristic features of Bryophytes.

Bryophytes are non-vascular land plants (mosses, liverworts) anchored by root-like rhizoids where the gametophyte phase is dominant.

They lack lignified xylem and true roots.

3

Identify the characteristic features of Pteridophytes.

Pteridophytes are vascular seedless plants (ferns) possessing true vegetative structures (roots, stems, leaves) and a dominant sporophyte generation.

They are the first plant group to evolve conducting tissues (xylem and phloem).

Key Concept

Structural organization and vascular tissue presence across lower plant divisions

Question 7114Question →

Which of the following represents the correct chronological sequence of events during asexual reproduction by budding in the unicellular yeast *Saccharomyces*?

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Answer

The correct sequence begins with localized softening of the chitin cell wall to form a bud bulge, followed by mitotic nuclear division and migration into the bud, deposition of a chitinous septum across the bud neck, and finally enzymatic cleavage leading to daughter cell separation and bud scar formation.

In *Saccharomyces*, budding progresses in a strict order: localized enzymatic softening of the chitin cell wall permits bud emergence, after which mitotic nuclear division sends one daughter nucleus into the bud. Next, a chitinous septum is synthesized across the bud neck to seal both compartments, and finally, enzymatic cleavage breaks the septum to release the daughter cell and form a bud scar.

Step-by-Step Solution

1

Identify the initial cellular event initiating yeast budding.

Specific enzymes weaken the rigid cell wall at a designated bud site, allowing turgor pressure to push out a small cytoplasmic bulge.

Bud formation requires localized structural weakening of the cell wall before cellular contents can expand outward.

2

Trace the movement and division of genetic material.

The parent nucleus undergoes mitosis, elongating through the bud neck so that one daughter nucleus enters the growing bud while the other remains in the parent cell.

Asexual budding ensures genetic continuity through exact mitotic segregation.

3

Determine how the cytoplasm of the two cells is partitioned.

Chitin synthases deposit a primary chitin septum across the narrow bud neck between the mother and daughter cell membranes.

Septum synthesis creates a secure physical partition prior to actual detachment.

4

Identify the concluding event resulting in autonomous yeast cells.

Chitinases and glucanases hydrolyze the outer wall layers of the septum, freeing the daughter yeast cell and leaving a chitin-rich bud scar on the mother cell wall.

Enzymatic separation completes the division cycle while leaving a permanent structural marker on the parent cell.

Key Concept

Asexual Reproduction by Budding in Saccharomyces Yeast

Question 7115Question →

Which structural feature differentiates pteridophytes from bryophytes?

Presence of vascular tissues (xylem and phloem)

Presence of well-developed xylem and phloem in mosses

Xylem tissue that translocates manufactured food from leaves down to roots

Extracellular saprophytic digestion of dead organic matter

Show answer & explanation

Answer: Presence of vascular tissues (xylem and phloem)

Answer

Presence of vascular tissues (xylem and phloem)

Pteridophytes (such as ferns) are the first group of land plants to evolve vascular tissues (xylem and phloem) for the transport of water, minerals, and synthesized food, whereas bryophytes (mosses and liverworts) completely lack true vascular tissues.

Step-by-Step Solution

1

Analyze the structural characteristics of bryophytes.

Bryophytes (mosses and liverworts) are non-vascular plants lacking true roots, stems, leaves, and conducting tissues (xylem and phloem).

Identifying the baseline transport mechanism in bryophytes.

2

Analyze the structural characteristics of pteridophytes.

Pteridophytes (ferns) are vascular seedless plants possessing specialized conducting tissues (xylem and phloem).

Determining the primary anatomical advancement of pteridophytes.

3

Compare the two plant groups to find the key distinguishing feature.

The presence of vascular tissue (xylem and phloem) is unique to pteridophytes when compared to bryophytes.

Selecting the correct option based on plant division anatomy.

Key Concept

Vascular Organization in Lower Plants

Question 7116Question →

Which of the following human traits represents a physiological variation rather than a morphological variation?

Ability to taste phenylthiocarbamide (PTC)

Fingerprint ridge pattern

Shape of the human nose

Attached versus free earlobes

Show answer & explanation

Answer: Ability to taste phenylthiocarbamide (PTC)

Answer

Ability to taste phenylthiocarbamide (PTC)

The ability to taste phenylthiocarbamide (PTC) depends on functional chemical receptors on the tongue, which makes it a physiological variation.

Step-by-Step Solution

1

Distinguish between morphological and physiological variations in humans.

Morphological variations involve external appearance, anatomical structures, and physical form. Physiological variations involve body functions, biochemical processes, and cellular mechanisms.

Separating physical structure from biological function is necessary to classify human traits accurately.

2

Classify the given options based on structure versus function.

Fingerprint patterns, nose shape, and earlobe structure are physical body features (morphological). The ability to taste PTC is a sensory functional response based on receptor chemistry (physiological).

Identifying the trait driven by functional biochemistry pinpoints the physiological variation.

Key Concept

Classification of human variations into morphological (structural/physical form) and physiological (functional/biochemical process) categories.

Question 7117Question →

Which of the following air pollutants reduces the oxygen-carrying capacity of human blood by binding with hemoglobin?

Carbon monoxide

Sulfur dioxide

Chlorofluorocarbon

Carbon dioxide

Show answer & explanation

Answer: Carbon monoxide

Answer

Carbon monoxide

Carbon monoxide combines with blood hemoglobin to form carboxyhemoglobin, severely diminishing the capacity of red blood cells to transport oxygen.

Step-by-Step Solution

1

Identify the biological mechanism of carbon monoxide toxicity in humans.

Carbon monoxide (

CO

) has an affinity for hemoglobin over 200 times greater than oxygen, forming carboxyhemoglobin.

This stable complex reduces the available hemoglobin sites for oxygen binding and inhibits oxygen release to tissues.

Key Concept

Air Pollutants and Biological Effects of Carbon Monoxide

Estimated Time:45s

Question 7118Question →

A specific tax of $\text{₦}80$ per unit is imposed on a market commodity. The price elasticity of demand for the commodity is $0.6$ , while its price elasticity of supply is $1.4$ . What is the tax burden per unit borne by the consumer in Naira?

Show answer & explanation

Answer: 56

Answer

The tax burden per unit borne by the consumer is 56 Naira.

The incidence of tax on consumers depends on relative elasticity. The formula for the consumer's burden is

T \times \frac{E_s}{E_d + E_s}

. Substituting the given values gives

80 \times \frac{1.4}{0.6 + 1.4} = 80 \times 0.7 = 56

Naira.

Step-by-Step Solution

1

Identify given variables and elasticity values.

Tax per unit (

T

) =

\text{₦}80

, Price elasticity of demand (

E_d

) =

0.6

, Price elasticity of supply (

E_s

) =

1.4

.

These values determine the proportion of the tax burden shifted to buyers versus sellers.

2

Apply the tax incidence formula for the consumer's share.

\text{Consumer Share} = T \times \left( \frac{E_s}{E_d + E_s} \right)

Tax incidence on consumers is directly proportional to supply elasticity relative to the sum of demand and supply elasticities.

3

Compute the numerical value.

\text{Consumer Share} = 80 \times \left( \frac{1.4}{0.6 + 1.4} \right) = 80 \times 0.7 = 56

Multiplying the per-unit tax by the consumer incidence proportion yields the exact burden per unit in Naira.

Key Concept

Tax Incidence and Price Elasticity

Question 7119Question →

Match each plant growth regulator on the left with its primary physiological action on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Auxin pairs with cell elongation and apical dominance; Ethylene pairs with fruit ripening and leaf abscission; Abscisic acid pairs with stomatal closure during drought stress; Gibberellin pairs with seed germination and stem elongation.

Each plant growth regulator is matched directly to its physiological function: Auxin promotes cell elongation and apical dominance, Ethylene stimulates fruit ripening and abscission, Abscisic acid triggers stomatal closure during drought, and Gibberellin initiates seed germination.

Step-by-Step Solution

1

Identify the primary function of Auxin.

Auxin stimulates cell elongation and apical dominance.

Auxin is synthesized in apical meristems and controls longitudinal growth.

2

Identify the primary function of Ethylene.

Ethylene promotes fruit ripening and leaf abscission.

Ethylene acts as a gaseous growth regulator during organ maturation.

3

Identify the primary function of Abscisic acid.

Abscisic acid induces stomatal closure under water deficit.

Abscisic acid functions as an inhibitory stress hormone during drought conditions.

4

Identify the primary function of Gibberellin.

Gibberellin mobilizes food reserves to promote seed germination.

Gibberellins stimulate the synthesis of hydrolytic enzymes during seed germination.

Key Concept

Physiological Roles of Plant Hormones

Question 7120Question →

In a given financial year, an economy records a Net National Product ( $\text{NNP}$ ) of $\text{N}6,200\text{ million}$ and a Capital Consumption Allowance (Depreciation) of $\text{N}450\text{ million}$ . The factor income earned by domestic citizens from abroad is $\text{N}320\text{ million}$ , while the factor income paid to foreign residents within the economy is $\text{N}510\text{ million}$ . What is the value of the Gross Domestic Product ( $\text{GDP}$ ) of this economy in million Naira?

Show answer & explanation

Answer: 6840

Answer

6840 million Naira

To calculate Gross Domestic Product (GDP), first derive Gross National Product (GNP) by adding Capital Consumption Allowance (Depreciation) to Net National Product (NNP):

\text{GNP} = 6,200 + 450 = 6,650\text{ million Naira}

. Next, compute Net Factor Income from Abroad (NFIA) as factor income received from abroad minus factor income paid to foreigners:

\text{NFIA} = 320 - 510 = -190\text{ million Naira}

. Finally, apply the national accounting identity

\text{GNP} = \text{GDP} + \text{NFIA}

, rearranged as

\text{GDP} = \text{GNP} - \text{NFIA} = 6,650 - (-190) = 6,840\text{ million Naira}

.

Step-by-Step Solution

1

Calculate Gross National Product (GNP) from Net National Product (NNP) and Depreciation

GNP = 6,200 + 450 = 6,650 million Naira

Gross aggregates include depreciation, whereas net aggregates exclude it:

\text{GNP} = \text{NNP} + \text{Depreciation}

.

2

Determine Net Factor Income from Abroad (NFIA)

NFIA = 320 - 510 = -190 million Naira

NFIA is defined as factor income earned from abroad by citizens minus factor income paid to foreign residents domestically.

3

Determine Gross Domestic Product (GDP) using GNP and NFIA

GDP = 6,650 - (-190) = 6,840 million Naira

Because

\text{GNP} = \text{GDP} + \text{NFIA}

, rearranging gives

\text{GDP} = \text{GNP} - \text{NFIA}

. Subtracting a negative value is equivalent to adding its positive magnitude.

Key Concept

Relationship between basic national income aggregates (GDP, GNP, NNP, NDP), Depreciation, and Net Factor Income from Abroad (NFIA).

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