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13931 questions

Question 4721Question

Match each biological trait on the left with its corresponding category of variation and genetic control on the right.

Click a left item, then click its matching right item

Items

Human height
ABO blood group
Fingerprint pattern

Matches

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Answer

Human height matches continuous variation controlled by polygenes and environmental factors; ABO blood group matches discontinuous variation controlled by a single gene locus with multiple alleles; Fingerprint pattern matches discontinuous variation exhibiting distinct qualitative categories.
Human height is a quantitative feature showing continuous variation influenced by polygenes and the environment. In contrast, ABO blood groups and fingerprint patterns are qualitative traits exhibiting discontinuous variation with distinct, clear-cut phenotypic categories.

Step-by-Step Solution

1
Analyze human height variation.
Height displays a continuous range of phenotypes with no clear separation between types.
It is a quantitative trait controlled by polygenes and influenced by environmental conditions.
2
Analyze ABO blood group variation.
Blood groups exist in distinct classes (A, B, AB, O) without intermediate phenotypes.
It is a monogenic trait controlled by multiple alleles at a single locus.
3
Analyze fingerprint pattern variation.
Fingerprint patterns fall into distinct categories such as arches, loops, and whorls.
It exhibits clear-cut discrete categories characteristic of discontinuous variation.

Key Concept

Continuous variation presents an unbroken spectrum of intermediate phenotypes regulated by polygenes, whereas discontinuous variation exhibits distinct, non-overlapping phenotypic categories controlled by one or a few genes.
Question 4722Question

Ada has ₦5,000 and must choose between buying a set of past examination questions or a new school bag, both priced at ₦5,000. If she decides to purchase the past examination questions, what is the opportunity cost of her decision?

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Answer: The school bag foregone

Answer

The school bag foregone
Opportunity cost (or real cost) refers to the next best alternative foregone when a choice is made under conditions of scarcity. In this scenario, purchasing the past examination questions means Ada must give up the school bag. Therefore, the school bag represents the opportunity cost of her decision.

Step-by-Step Solution

1
Identify the options available and the decision taken
Ada chose the set of past examination questions over the school bag.
Determining opportunity cost requires evaluating the selected choice against the alternative available.
2
Determine the alternative that was sacrificed
The sacrificed alternative is the school bag.
Opportunity cost is defined as the real commodity or opportunity foregone when a choice is made.

Key Concept

Opportunity Cost
Estimated Time:45s
Question 4723Question

An autotrophic plant possesses lignified vascular tissues and a prominent sporophyte generation, yet relies on environmental moisture for flagellated sperm to reach the archegonium. Which plant division is characterized by these features?

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Answer: Pteridophytes

Answer

Pteridophytes
Pteridophytes (such as ferns and horsetails) are the earliest group of vascular plants with a dominant sporophyte phase. Although they have developed true xylem and phloem for internal conduction, they retain the primitive reproductive requirement of needing free water so that motile sperm can swim to fertilize the egg in the archegonium.

Step-by-Step Solution

1
Analyze anatomical characteristics provided in the stem
The presence of lignified vascular tissues (xylem and phloem) and a dominant sporophyte rules out non-vascular divisions.
Thallophytes and bryophytes lack true vascular transport structures.
2
Evaluate the reproductive constraint regarding fertilization
The dependence on environmental water for swimming flagellated sperm identifies the group as seedless vascular plants.
Spermatophytes transfer male gametes via wind or pollinators using pollen tubes without needing a film of water.

Key Concept

Structural traits and reproductive dependencies of Pteridophytes
Estimated Time:50s
Question 4724Question

A genetic counselor assesses a couple planning to have children. Both parents are asymptomatic carriers of the sickle-cell allele (HbAHbSHb^A Hb^S). The father has blood group ABAB (IAIBI^A I^B) and the mother has blood group OO (iiii). What is the probability that their first child will be a carrier of the sickle-cell trait and also have a blood group that can safely receive red blood cells from a type AA (IAiI^A i) donor without transfusion agglutination?

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Answer: 14\frac{1}{4} (or 25%25\%)

Answer

The probability that the child is both a sickle-cell carrier and compatible with type A donor blood is 14\frac{1}{4} (or 25%25\%).
Crossing two sickle-cell carriers (HbAHbS×HbAHbSHb^A Hb^S \times Hb^A Hb^S) gives a 12\frac{1}{2} probability of producing a carrier child (HbAHbSHb^A Hb^S). Crossing a parent of blood type ABAB (IAIBI^A I^B) with a parent of type OO (iiii) yields offspring with blood types AA (IAiI^A i) or BB (IBiI^B i), each with a probability of 12\frac{1}{2}. For red blood cell transfusion from a type AA donor, only type AA offspring can receive the blood safely because type BB offspring possess anti-A antibodies that would cause agglutination. Since the two gene loci assort independently, the overall combined probability is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} (or 25%25\%).

Step-by-Step Solution

1
Determine the probability of inheriting the sickle-cell carrier genotype (HbAHbSHb^A Hb^S).
Crossing two carrier parents (HbAHbS×HbAHbSHb^A Hb^S \times Hb^A Hb^S) produces genotypes 1HbAHbA:2HbAHbS:1HbSHbS1\, Hb^A Hb^A : 2\, Hb^A Hb^S : 1\, Hb^S Hb^S. The probability of a carrier offspring (HbAHbSHb^A Hb^S) is 24=12\frac{2}{4} = \frac{1}{2}.
Monohybrid inheritance of autosomal recessive trait yields a 1:2:11:2:1 genotypic ratio.
2
Determine the blood group genotypes of the offspring and identify donor compatibility.
Crossing father ABAB (IAIBI^A I^B) with mother OO (iiii) yields 12\frac{1}{2} blood group AA (IAiI^A i) and 12\frac{1}{2} blood group BB (IBiI^B i). Only blood group AA recipients can safely receive type AA red blood cells without antibody-antigen agglutination.
Blood group BB recipients possess anti-A antibodies in their blood plasma, causing hemolysis/agglutination if transfused with type AA cells.
3
Calculate the combined probability of both independent genetic events.
P(Carrier AND Type A)=P(Carrier)×P(Type A)=12×12=14P(\text{Carrier AND Type A}) = P(\text{Carrier}) \times P(\text{Type A}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} (or 25%25\%).
The hemoglobin locus and ABO locus reside on different chromosomes and assort independently according to Mendel's Second Law.

Key Concept

Application of Mendel's laws of independent assortment to human medical genetic counseling involving disease carriers and ABO blood transfusion compatibility.
Question 4725Question

What is the correct equilibrium constant expression, KcK_c, for the reaction 2NO2(g)N2O4(g)2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)?

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Answer: Kc=[N2O4][NO2]2K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}

Answer

The correct equilibrium constant expression is Kc=[N2O4][NO2]2K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}.
According to the Law of Mass Action, the equilibrium constant expression KcK_c for a general reversible reaction aA+bBcC+dDaA + bB \rightleftharpoons cC + dD is given by Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}. For the reaction 2NO2(g)N2O4(g)2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g), the concentration of the product N2O4\text{N}_2\text{O}_4 is raised to the power of 1 in the numerator, and the concentration of the reactant NO2\text{NO}_2 is raised to the power of 2 in the denominator, yielding Kc=[N2O4][NO2]2K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}.

Step-by-Step Solution

1
Identify the products and reactants along with their stoichiometric coefficients from the balanced equation.
Reactant: NO2\text{NO}_2 with coefficient 2; Product: N2O4\text{N}_2\text{O}_4 with coefficient 1.
Equilibrium constant expressions require product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of its stoichiometric coefficient.
2
Formulate the equilibrium constant ratio Kc=[Products]coefficients[Reactants]coefficientsK_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}.
Kc=[N2O4]1[NO2]2K_c = \frac{[\text{N}_2\text{O}_4]^1}{[\text{NO}_2]^2}.
Placing [N2O4][\text{N}_2\text{O}_4] in the numerator and [NO2]2[\text{NO}_2]^2 in the denominator satisfies the law of mass action.

Key Concept

Equilibrium Constant Expression (KcK_c)
Question 4726Question

A biological survey of poikilothermic vertebrates identifies three distinct species based on their anatomical and physiological traits:

- Species X exhibits a two-chambered heart consisting of one atrium and one ventricle, excretes ammonia as its main nitrogenous waste, and breathes via gills.
- Species Y possesses a three-chambered heart with an incomplete ventricular septum, excretes uric acid, and lays shelled (cleidoic) eggs.
- Species Z undergoes metamorphosis during development, possesses a simple three-chambered heart without any ventricular division, and utilizes moist skin for cutaneous gas exchange.

Which of the following lists the correct scientific names corresponding to Species X, Species Y, and Species Z, respectively?

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Answer: Tilapia zillii, Agama agama, and Bufo regularis

Answer

Species X is the fish Tilapia zillii, Species Y is the reptile Agama agama, and Species Z is the amphibian Bufo regularis.
The combination of Tilapia zillii (bony fish), Agama agama (reptile), and Bufo regularis (amphibian) accurately aligns with the physiological profiles of Species X, Y, and Z respectively. Fish possess a 2-chambered heart and gill respiration; reptiles possess a 3-chambered heart with an incomplete septum, uricotelic excretion, and cleidoic eggs; and amphibians possess a 3-chambered heart, undergo metamorphosis, and exhibit cutaneous respiration.

Step-by-Step Solution

1
Analyze the circulatory, excretory, and respiratory features of Species X
Species X has a 2-chambered heart (one atrium, one ventricle), excretes ammonia, and breathes using gills, which uniquely identifies it as belonging to class Pisces (e.g., Tilapia zillii).
Fish possess a single-circuit circulatory system with a two-chambered heart.
2
Analyze the anatomical and reproductive traits of Species Y
Species Y has a 3-chambered heart with an incomplete ventricular septum, excretes uric acid, and lays cleidoic (shelled) eggs, which are characteristic features of class Reptilia (e.g., Agama agama).
Reptiles have adapted to fully terrestrial life via uricotelic excretion, cleidoic eggs, and partial ventricular separation.
3
Analyze the developmental and respiratory features of Species Z
Species Z undergoes metamorphosis, has a 3-chambered heart without a septum, and uses skin for respiration, which characterizes class Amphibia (e.g., Bufo regularis).
Amphibians undergo dual-stage life cycles and rely heavily on cutaneous gas exchange.
4
Verify scientific binomial nomenclature formatting
The genus must begin with a capital letter and the specific epithet with a lowercase letter (e.g., Tilapia zillii, Agama agama, Bufo regularis).
Standard botanical and zoological taxonomy strictly enforces binomial nomenclature capitalization.

Key Concept

Comparative anatomy and physiological adaptations across poikilothermic vertebrate classes (Pisces, Amphibia, Reptilia)
Estimated Time:1m 30s
Question 4727Question

Which of the following terms describes an organism that possesses two identical alleles for a specific gene?

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Answer: Homozygous

Answer

Homozygous
Homozygous is the genetic term for an organism carrying two identical alleles for a given gene locus.

Step-by-Step Solution

1
Define genetic zygosity terms.
Alleles are alternative forms of a gene located at the same locus on homologous chromosomes.
Comparing maternal and paternal alleles determines the zygosity of the gene locus.
2
Identify the term for matching alleles.
When an individual inherits identical alleles from both parents for a gene, the genetic condition is homozygous.
The prefix 'homo-' means same, indicating identical alleles.

Key Concept

Basic Genetics Terminology (Homozygous vs. Heterozygous)
Question 4728Question

Complete the following statement on the scope and methodology of economic analysis.

Fill in the blanks below

Economic analysis that focuses on objective explanation and verifiable facts about 'what is' is known as economics, whereas analysis based on value judgments and subjective opinions about 'what ought to be' is known as economics.
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Answer

The first blank is 'positive' and the second blank is 'normative'.
Positive economics is concerned with objective, testable statements about economic reality ('what is'). In contrast, normative economics deals with value judgments, fairness, and ethical views regarding what policy goals or conditions 'ought to be'.

Step-by-Step Solution

1
Analyze the branch of economic enquiry dealing with empirical facts, testable hypotheses, and objective observations.
This defines positive economics, which describes economic phenomena as they exist without personal bias.
Positive economics focuses strictly on cause-and-effect relationships and factual statements.
2
Analyze the branch of economic enquiry dealing with ethics, policy recommendations, and subjective statements.
This defines normative economics, which discusses what policy measures or outcomes should be pursued.
Normative economics relies on value judgments that cannot be proven or disproven by data alone.

Key Concept

Positive and Normative Economics
Estimated Time:1m 0s
Question 4729Question

After a severe bushfire destroys a tropical forest ecosystem, fast-growing herbaceous plants and grasses quickly colonize the area by utilizing ash nutrients and seeds preserved within the intact soil. Which of the following best explains why this ecological sequence is classified as secondary succession?

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Answer: The process originates on a pre-existing soil substrate that already contains organic matter and dormant propagules.

Answer

Secondary succession is defined by ecological recolonization that takes place on pre-existing soil containing organic material and dormant seeds following a disturbance.
Secondary succession occurs when a disturbance disrupts an existing ecosystem without eliminating the soil. Because soil, nutrients, and dormant seeds (propagules) remain intact, recolonization proceeds rapidly through pioneer herbs and grasses.

Step-by-Step Solution

1
Identify the key environmental conditions present after the bushfire disturbance.
The fire destroyed vegetation but left intact soil, nutrients in ash, and dormant seeds.
Determining whether a substrate has pre-existing soil is the fundamental criterion for differentiating succession types.
2
Distinguish between primary and secondary succession based on substrate characteristics.
Primary succession starts on bare substrate (e.g., bare rock, volcanic lava, sand dunes) with no soil. Secondary succession starts on established soil after partial destruction.
Existing soil allows faster plant establishment and seed germination.
3
Select the statement that correctly accounts for secondary succession.
The presence of pre-existing soil containing organic matter and propagules defines secondary succession.
This directly aligns with ecological succession definitions.

Key Concept

Distinction between primary and secondary ecological succession based on pre-existing substrate and soil presence.
Estimated Time:1m 0s
Question 4730Question

In an experiment to measure the rate of a chemical reaction, 0.50 g0.50\text{ g} of calcium carbonate reacts completely with excess dilute hydrochloric acid in 25 seconds25\text{ seconds}. What is the average rate of reaction with respect to the loss of mass of calcium carbonate in g s1\text{g s}^{-1}?

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Answer: 0.02

Answer

The average rate of reaction with respect to the mass of calcium carbonate consumed is 0.02 g s10.02\text{ g s}^{-1}.
The average rate of a reaction is calculated as the ratio of the change in amount of reactant or product to the time taken. Substituting the given values gives Rate=0.50 g25 s=0.02 g s1\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}.

Step-by-Step Solution

1
Extract the given values from the problem statement.
Mass of CaCO3=0.50 g\text{CaCO}_3 = 0.50\text{ g}, Time =25 s= 25\text{ s}.
These parameters define the total change in quantity and the time interval for the reaction.
2
Calculate the average rate of reaction by dividing the change in mass by the time elapsed.
Rate=0.50 g25 s=0.02 g s1\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}.
The rate of a chemical reaction measures how rapidly a reactant is consumed per unit time.

Key Concept

Rate of Reaction Calculation
Estimated Time:1m 0s
Question 4731Question

Match each stage of cellular respiration listed on the left with its precise subcellular location and characteristic biochemical process on the right.

Click a left item, then click its matching right item

Items

Glycolysis
Link Reaction (Pyruvate Oxidation)
Krebs Cycle
Electron Transport Chain and Chemiosmosis

Matches

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Answer

Glycolysis corresponds to the cytosolic pathway yielding pyruvate, net 2 ATP2\text{ ATP}, and 2 NADH2\text{ NADH}. The Link Reaction corresponds to matrix oxidative decarboxylation forming Acetyl-CoA and CO2\text{CO}_2. The Krebs Cycle corresponds to matrix oxidation of acetyl groups generating CO2\text{CO}_2, NADH\text{NADH}, FADH2\text{FADH}_2, and ATP\text{ATP}. Electron Transport and Chemiosmosis correspond to cristae-bound electron transfer coupled to proton-gradient driven ATP synthesis.
Each stage of respiration occurs at a distinct cellular location optimized for its pathway: Glycolysis in the cytosol, the Link Reaction and Krebs Cycle within the mitochondrial matrix, and the Electron Transport Chain across the inner mitochondrial membrane (cristae).

Step-by-Step Solution

1
Identify the site and products of Glycolysis
Glycolysis is an anaerobic process taking place in the cytoplasm/cytosol, converting glucose to pyruvate with a net generation of 2 ATP2\text{ ATP} and 2 NADH2\text{ NADH}.
Enzymes for glycolysis are soluble in the cytosol, not membrane-bound in mitochondria.
2
Identify the site and products of the Link Reaction
Pyruvate enters the mitochondrial matrix where it undergoes oxidative decarboxylation to produce Acetyl-CoA, CO2\text{CO}_2, and NADH\text{NADH}.
This bridges cytosolic glycolysis to the matrix-localized Krebs cycle.
3
Identify the site and products of the Krebs Cycle
The cyclic breakdown of acetyl groups occurs in the mitochondrial matrix, producing CO2\text{CO}_2, reduced coenzymes (NADH\text{NADH}, FADH2\text{FADH}_2), and ATP\text{ATP}.
Enzymes of the citric acid cycle are dissolved in the fluid matrix of the mitochondrion.
4
Identify the site and mechanisms of Electron Transport and Chemiosmosis
Electron carriers and ATP synthase complexes are located on the inner mitochondrial membrane (cristae), generating the vast majority of ATP via oxidative phosphorylation.
The folded cristae maximize surface area for respiratory electron carrier complexes.

Key Concept

Subcellular Localization and Pathways of Cellular Respiration
Question 4732Question

A genetic counselor evaluates a married couple who are both heterozygous carriers of the sickle-cell gene (HbAHbSHb^A Hb^S). What is the probability that their first child will be an asymptomatic carrier of the sickle-cell trait?

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Answer: 50%

Answer

The probability that the couple's first child will be an asymptomatic carrier of the sickle-cell trait is 50%.
When two heterozygous individuals (HbAHbSHb^A Hb^S) produce offspring, Mendelian segregation yields a genotypic ratio of 1 HbAHbAHb^A Hb^A (normal, 25%) : 2 HbAHbSHb^A Hb^S (carrier trait, 50%) : 1 HbSHbSHb^S Hb^S (sickle-cell anemia, 25%). Therefore, there is a 50% chance that any child born to them will be an asymptomatic carrier.

Step-by-Step Solution

1
Determine parental genotypes
Both parents are heterozygous carriers with the genotype HbAHbSHb^A Hb^S.
Sickle-cell trait is an autosomal codominant condition where carriers possess one normal hemoglobin allele (HbAHb^A) and one sickle hemoglobin allele (HbSHb^S).
2
Perform a monohybrid cross between the parents (HbAHbS×HbAHbSHb^A Hb^S \times Hb^A Hb^S)
The possible offspring genotypes are 1 HbAHbAHb^A Hb^A : 2 HbAHbSHb^A Hb^S : 1 HbSHbSHb^S Hb^S.
Gametes from each parent separate independently during meiosis, giving equal likelihood for inheriting either allele.
3
Calculate the proportion corresponding to asymptomatic carriers (HbAHbSHb^A Hb^S)
Heterozygous carrier probability = 24=50%\frac{2}{4} = 50\%.
Two out of the four possible genotypic outcomes result in the carrier state (HbAHbSHb^A Hb^S).

Key Concept

Application of genetic inheritance in medical counseling for sickle-cell trait
Estimated Time:1m 0s
Question 4733Question

Match each male mammalian reproductive organ with its primary function.

Click a left item, then click its matching right item

Items

Testis
Epididymis
Vas deferens
Prostate gland

Matches

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Answer

Testis matches with production of sperm and testosterone; Epididymis matches with temporary storage and maturation of sperm; Vas deferens matches with conduction of sperm from epididymis to the urethra; Prostate gland matches with secretion of alkaline fluid to nourish sperm.
Each structure performs a distinct organ-level role in male reproduction: gametogenesis in the testis, maturation in the epididymis, transport in the vas deferens, and fluid production in the prostate gland.

Step-by-Step Solution

1
Identify the primary male gonad.
The testis produces sperm cells and testosterone.
It is responsible for gamete formation and primary androgen secretion.
2
Identify the site of sperm maturation.
The epididymis stores and matures sperm.
Sperm acquire motility within the long coiled tube of the epididymis.
3
Identify the sperm transport canal.
The vas deferens conducts sperm away from the storage site.
It acts as a passage tube connecting the epididymis to the urethra.
4
Identify the accessory gland.
The prostate gland secretes protective alkaline seminal fluid.
Accessory glands supply fluids necessary for sperm activation and viability.

Key Concept

Structure and Function of Mammalian Male Reproductive System
Estimated Time:1m 0s
Question 4734Question

Match each essential plant nutrient or chloroplast structural feature on the left with its corresponding biological role or reaction site on the right.

Click a left item, then click its matching right item

Items

Magnesium
Nitrogen
Stroma
Thylakoid membrane

Matches

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Answer

Magnesium matches with forming the central atom of chlorophyll; Nitrogen matches with being an essential component of proteins and nucleic acids; Stroma matches with the site of carbon dioxide fixation; Thylakoid membrane matches with the site of light absorption and photolysis of water.
Each term is accurately matched with its primary function or location: Magnesium constitutes the central ion of chlorophyll, Nitrogen is required for amino acid and protein synthesis, the stroma hosts carbon dioxide fixation, and thylakoid membranes carry out light absorption and water photolysis.

Step-by-Step Solution

1
Identify the primary structural role of Magnesium in photosynthesis.
Magnesium forms the core metallic atom in chlorophyll pigments.
Magnesium deficiency directly leads to chlorosis due to lack of chlorophyll synthesis.
2
Identify the physiological function of Nitrogen in plant nutrition.
Nitrogen forms the base of amino acids, proteins, and DNA/RNA.
Nitrogen is crucial for general plant growth and cellular structure.
3
Determine the specific chloroplast region where dark reactions occur.
The stroma hosts carbon fixation enzymes.
The light-independent Calvin cycle occurs in the fluid stroma surrounding thylakoids.
4
Determine the chloroplast region where light reactions occur.
Thylakoid membranes host light absorption and photolysis.
Photosystems embedded in the thylakoid membrane capture light and split water molecules.

Key Concept

Plant mineral functions and structural sites of photosynthetic reactions
Question 4735Question

Match each basic economic problem of society on the left with its corresponding fundamental economic decision on the right.

Click a left item, then click its matching right item

Items

What to produce
How to produce
For whom to produce

Matches

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Answer

'What to produce' matches selecting the types and quantities of goods and services to create; 'How to produce' matches choosing the combination of factors of production and technique to adopt; 'For whom to produce' matches determining the distribution and allocation of finished goods and services among members of society.
Each economic problem directly aligns with its standard economic definition: 'What to produce' determines resource allocation toward specific goods, 'How to produce' determines the technical method of production, and 'For whom to produce' governs distribution among consumers.

Step-by-Step Solution

1
Identify the meaning of 'What to produce'.
It addresses the choice of output types and quantities given resource constraints.
Society cannot produce everything, so it must prioritize which commodities to produce.
2
Identify the meaning of 'How to produce'.
It addresses the production technique and resource efficiency.
Producers must choose between labor-intensive and capital-intensive methods to minimize cost and maximize output.
3
Identify the meaning of 'For whom to produce'.
It addresses the distribution of the final output.
It determines how the total national output is shared among consumers based on purchasing power or social policy.

Key Concept

Basic Economic Problems of Society
Estimated Time:45s
Question 4736Question

An estuarine wetland ecosystem in a coastal industrial corridor experiences severe eutrophication due to untreated organic sewage and industrial runoff, leading to a critical decline in dissolved oxygen and widespread fish kills. Which integrated environmental management approach provides the most ecologically sustainable solution to restore the water quality and preserve biodiversity in the wetland?

Show answer & explanation

Answer: Implementation of biological wastewater treatment and bioremediation using micro-organisms, combined with enforcing statutory effluent limits on industrial discharge.

Answer

Implementation of biological wastewater treatment and bioremediation using micro-organisms, combined with enforcing statutory effluent limits on industrial discharge.
Combining biological wastewater treatment and bioremediation directly neutralizes excess organic nutrients and pollutants. Enforcing effluent discharge limits prevents continuous influx of contaminants, offering a sustainable, long-term solution for wetland restoration.

Step-by-Step Solution

1
Identify the primary environmental problem
Eutrophication caused by excess organic nutrients in untreated sewage and industrial effluent.
Excess nutrients trigger algal blooms whose subsequent decomposition consumes dissolved oxygen, leading to hypoxia.
2
Evaluate sustainable restoration methods
Bioremediation and biological wastewater treatment degrade organic pollutants into harmless inorganic substances.
Biological treatment utilizes decomposers naturally without introducing toxic chemical residues.
3
Integrate regulatory management controls
Enforcing statutory effluent standards prevents ongoing nutrient input.
Sustainable conservation requires combining biological remediation techniques with legal and institutional enforcement.

Key Concept

Water pollution control and biological environmental management
Question 4737Question

Match each piece of biochemical or biogeographical evolutionary evidence on the left with its corresponding evolutionary implication or mechanism on the right.

Click a left item, then click its matching right item

Items

Cytochrome c amino acid sequence homologies across diverse taxa
Disjunct global distribution of flightless ratite birds
Quantitative serological precipitation testing of serum proteins
Adaptive radiation of Galápagos finch species

Matches

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Answer

Cytochrome c sequence homologies match with indicating conservation of essential metabolic proteins from a common ancestor. Disjunct ratite distribution matches with demonstrating vicariance resulting from continental drift. Serological precipitation testing matches with measuring antigenic cross-reactivity for phylogenetic proximity. Adaptive radiation of Galápagos finches matches with illustrating speciation driven by ecological niche diversification.
Each evidence type correctly pairs with its established evolutionary conclusion: Cytochrome c sequence conservation reveals universal metabolic heritage; ratite distribution reflects tectonic vicariance; serological precipitation quantifies serum protein homology; and island finch diversity demonstrates adaptive radiation.

Step-by-Step Solution

1
Evaluate biochemical evidence from Cytochrome c
Cytochrome c is involved in electron transport; its highly conserved amino acid sequence across unicellular and multicellular organisms provides direct evidence of deep evolutionary homology.
Universal cellular enzymes reflect shared genetic ancestry.
2
Analyze biogeographical distribution of ratites (rheas, ostriches, emus, cassowaries)
Flightless birds could not have migrated across modern oceans; their presence on South America, Africa, and Australia is attributed to the Mesozoic breakup of Gondwana.
Vicariance through continental drift explains wide spatial separation of closely related terrestrial taxa.
3
Analyze immunological serological testing
Antisera produced against human serum proteins react most strongly (forming dense precipitate) with chimpanzee serum and progressively weaker with more distantly related mammals.
Precipitate quantity directly correlates with structural homology of serum albumins and globulins.
4
Examine adaptive radiation in island archipelagos
A single ancestral seed-eating finch colonizing the islands diversified into species specialized for seeds, insects, and nectar.
Geographical isolation combined with natural selection drives morphological divergence into vacant ecological niches.

Key Concept

Evidence for Evolution: Comparative Biochemistry and Biogeography
Question 4738Question

Calculate the quantity of electricity, in Coulombs, transferred when a steady electric current of 5.0 A5.0\text{ A} is passed through an electrolytic cell for 20 minutes20\text{ minutes}.

Show answer & explanation

Answer: 6000

Answer

The quantity of electricity transferred is 6000 C6000\text{ C}.
According to Faraday's laws of electrolysis, the total charge QQ passed through an electrolyte is calculated by Q=I×tQ = I \times t. Converting 20 minutes20\text{ minutes} to seconds gives 20×60=1200 s20 \times 60 = 1200\text{ s}. Multiplying by the current 5.0 A5.0\text{ A} gives Q=5.0×1200=6000 CQ = 5.0 \times 1200 = 6000\text{ C}.

Step-by-Step Solution

1
Convert time from minutes to seconds
t=1200 st = 1200\text{ s}
Electric current in Amperes measures charge per second, so time must be converted to seconds.
2
Calculate electric charge using Q=I×tQ = I \times t
Q=6000 CQ = 6000\text{ C}
The quantity of electricity (QQ) in Coulombs equals current (II) in Amperes multiplied by time (tt) in seconds.

Key Concept

Calculation of Quantity of Electricity (Q=I×tQ = I \times t)
Question 4739Question

A student with a limited budget of N12,000\text{N}12,000 constructs a scale of preference listing four items in descending order of priority: Textbooks (N7,000\text{N}7,000), School Shoes (N5,000\text{N}5,000), Scientific Calculator (N4,000\text{N}4,000), and School Bag (N5,000\text{N}5,000). Match each aspect of the student's economic decision on the left with its corresponding concept or interpretation on the right.

Click a left item, then click its matching right item

Items

Selecting Textbooks and School Shoes for purchase
Forgoing the Scientific Calculator
Ranking Textbooks higher than School Shoes
Inability to purchase all items on the list

Matches

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Answer

The correct pairings are: 'Selecting Textbooks and School Shoes for purchase' matches 'Satisfaction of effective demand within the income constraint'; 'Forgoing the Scientific Calculator' matches 'Opportunity cost (real cost) of the selected items'; 'Ranking Textbooks higher than School Shoes' matches 'Evaluation of relative marginal utility'; and 'Inability to purchase all items on the list' matches 'Direct consequence of scarcity of economic resources'.
A scale of preference is a list of unsatisfied wants arranged in order of relative importance. It enables a rational consumer to maximize utility under income constraints. Here, purchasing Textbooks and Shoes fulfills effective demand within the N12,000\text{N}12,000 limit. The highest-ranked unfulfilled want, the Scientific Calculator, is the opportunity cost. The ordering of items reflects relative marginal utility, and the overall shortfall of funds illustrates scarcity.

Step-by-Step Solution

1
Calculate total expenditure from available income
Income = N12,000\text{N}12,000. Purchasing Textbooks (N7,000\text{N}7,000) and School Shoes (N5,000\text{N}5,000) exhausts the budget total (N12,000\text{N}12,000).
Determines which wants become effective demand.
2
Identify the opportunity cost of the choice
The next highest priority item that remains unfulfilled is the Scientific Calculator (N4,000\text{N}4,000).
Opportunity cost is defined as the next best alternative forgone, not the total monetary outlay or all unfulfilled wants combined.
3
Analyze the priority order on the scale of preference
Textbooks are ranked above School Shoes because they yield higher relative marginal utility to the student.
A scale of preference ranks wants according to their urgency and utility to the individual.
4
Relate unfulfilled wants to underlying economic principles
The total cost of all desired items is N21,000\text{N}21,000, exceeding the N12,000\text{N}12,000 income.
Demonstrates economic scarcity, forcing the individual to make choices based on a preference schedule.

Key Concept

Scale of Preference, Scarcity, and Opportunity Cost
Question 4740Question

Arrange the following ecological stages of primary succession on bare rock (xerosere) in correct sequential order from the pioneer stage to the climax community. Which sequence accurately reflects this ecological progression?

Drag items to arrange them in the correct order

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Answer

The correct succession sequence begins with pioneer crustose lichens, followed by foliose lichens and mosses, then herbaceous grasses, followed by perennial shrubs, and culminates in a climax forest community.
Primary ecological succession on bare rock (xerosere) follows a predictable sequence of seral stages: crustose lichens pioneer soil formation, followed by foliose lichens and mosses deepening soil, herbaceous grasses colonizing, perennial shrubs establishing, and finally mature climax trees reaching ecological equilibrium.

Step-by-Step Solution

1
Identify the pioneer stage on bare substrate.
Crustose lichens colonize bare rock first due to their extreme xerophytic tolerance and ability to weather rock chemically.
Primary succession requires pioneer organisms capable of initiating soil formation on subaerial rock surfaces.
2
Determine the early seral invaders following initial weathering.
Foliose lichens and mosses invade the thin layer of weathered rock particles and organic dust.
Mosses require small amounts of accumulated moisture and organic debris to anchor their rhizoids.
3
Sequence the emergence of vascular herbaceous species.
Annual herbs and grasses establish as soil depth and humus content increase.
Vascular root systems need sufficient soil volume, which accumulates through the decay of mosses and foliose lichens.
4
Identify the transition to woody vegetation.
Perennial shrubs displace grasses due to superior light competition and deeper root structures.
Enriched soil supports larger perennial roots, allowing taller shrub canopy growth.
5
Determine the final equilibrium community.
A mature forest climax community establishes.
Climax trees represent the maximum biomass and biodiversity sustainable under the prevailing climate.

Key Concept

Sequential seral progression in primary lithosere/xerosere ecological succession
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