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Question 8881Question

A population of crop-damaging insects was repeatedly sprayed with a synthetic chemical insecticide over several years. Initially, the chemical killed almost all insects, but after consecutive generations, the population became predominantly resistant to the chemical. According to modern evolutionary theory, which of the following best explains how this resistance developed within the gene pool?

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Answer: Pre-existing random gene mutations provided resistance in a few individuals, and natural selection increased the frequency of these resistant alleles over generations.

Answer

Pre-existing random gene mutations provided resistance in a few individuals, and natural selection increased the frequency of these resistant alleles over generations.
According to modern evolutionary theory, random gene mutations introduce variation into a gene pool independently of environmental needs. When an environmental stressor such as an insecticide is introduced, it acts as a selective pressure. Individuals already possessing the resistant allele survive and reproduce, passing the genetic trait to their offspring. Over generations, the frequency of the resistant allele increases within the population.

Step-by-Step Solution

1
Identify the source of variation in modern evolutionary theory (Neo-Darwinism).
Gene mutations occur randomly in the gene pool before any environmental change occurs.
Evolutionary changes rely on pre-existing genetic variation in germ cells rather than post-exposure adaptations.
2
Analyze the role of the environmental factor (the insecticide).
The insecticide acts as a selective agent, eliminating susceptible individuals while resistant individuals survive.
Natural selection shifts allele frequencies in favor of advantageous traits already present.
3
Evaluate population gene pool shifts across generations.
Surviving resistant individuals reproduce and pass resistant alleles to offspring, increasing the allele frequency.
Differential reproductive success leads to microevolutionary change in the population.

Key Concept

Modern Evolutionary Theory (Neo-Darwinism) and Natural Selection on Genetic Variation
Question 8882Question

Match each characteristic of viruses in Column A with its corresponding biological implication in Column B.

Click a left item, then click its matching right item

Items

Absence of metabolic machinery and organelles
Ability to form crystals outside host cells
Presence of nucleic acid (DNA or RNA)
Specific surface glycoprotein spikes

Matches

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Answer

Absence of metabolic machinery and organelles matches with Requires obligate intracellular parasitism for replication; Ability to form crystals outside host cells matches with Demonstrates non-living chemical property when isolated from host; Presence of nucleic acid matches with Provides genetic instructions for viral protein synthesis; Specific surface glycoprotein spikes match with Enables recognition and attachment to host membrane receptors.
The correct matches align viral structural characteristics with their exact biological implications: lack of cellular organelles enforces obligate intracellular parasitism, crystallization demonstrates non-living chemical behavior outside a host, nucleic acids provide genetic hereditary information, and glycoprotein spikes facilitate specific binding to host cell surface receptors.

Step-by-Step Solution

1
Analyze the biological implication of lacking metabolic machinery
Lacking organelles means viruses cannot generate ATP or synthesize proteins independently, making them obligate intracellular parasites.
Viruses rely entirely on the host cell's metabolic infrastructure.
2
Analyze the implication of viral crystallization
Forming crystals in non-living environments behaves like chemical compounds rather than living cells.
Crystallization reflects metabolic inertness outside a living host.
3
Analyze the function of viral nucleic acid
Nucleic acid (either DNA or RNA) carries genetic code for replication.
The genome directs host ribosomes to manufacture viral components.
4
Analyze the role of surface glycoprotein spikes
Spikes mediate host specificity by anchoring to host cell surface receptors.
Viral attachment depends on specific molecular fitting between spikes and host receptors.

Key Concept

Viral Characteristics and Structure
Question 8883Question

In a computerized accounting system, the physical electronic equipment used to input, process, store, and display financial transactions is categorized as hardware.

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Answer: True

Answer

The statement is true because the physical equipment of a computer system used in processing accounting data is defined as hardware.
The statement accurately identifies that all physical electronic devices used in financial data processing constitute the hardware component of a computerized accounting system.

Step-by-Step Solution

1
Identify the basic components of a computerized accounting system.
A computerized accounting system comprises five core elements: hardware, software, data, procedures, and peopleware.
Understanding component definitions helps distinguish physical items from operational programs.
2
Evaluate the definition of physical electronic equipment.
Tangible machinery such as monitors, central processing units, keyboards, and printers fall specifically under hardware.
Hardware is defined by its physical, tangible nature in contrast to software applications.

Key Concept

Hardware component of a computerized accounting system
Question 8884Question

Match each fungal structure or specialized cell type with its corresponding diagnostic anatomical or reproductive characteristic.

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Items

Rhizoid (*Rhizopus*)
Basidium (*Agaricus*)
Zygospore (*Rhizopus*)
Pseudohypha (*Saccharomyces*)

Matches

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Answer

Rhizoid (*Rhizopus*) matches the specialized root-like hyphal branch that penetrates substrate for extracellular digestion; Basidium (*Agaricus*) matches the microscopic club-shaped structure lining mushroom gills where meiosis yields sexual spores; Zygospore (*Rhizopus*) matches the thick-walled, resistant zygotic structure from gametangial fusion; Pseudohypha (*Saccharomyces*) matches the short chain of unseparated daughter cells formed during budding.
Each fungal archetype possesses unique structural and reproductive adaptations: *Rhizopus* utilizes subterranean-like rhizoids for substrate penetration and enzyme secretion, as well as zygospores for sexual survival; *Agaricus* develops basidia on gill surfaces for sexual basidiospore production; and *Saccharomyces* forms pseudohyphae when budding daughter cells remain attached in chains.

Step-by-Step Solution

1
Analyze the nutrition and vegetative structures of moulds
Identify that rhizoids in *Rhizopus* serve an anchoring and absorptive function by penetrating the growth medium and secreting extracellular enzymes.
Rhizoids are specialized subterranean-like hyphae distinct from aerial sporangiophores and horizontal stolons.
2
Evaluate the reproductive structures of macrofungi (mushrooms)
Identify the basidium as the diagnostic spore-bearing club cell of *Agaricus* situated on the gill surfaces.
Karyogamy followed by meiosis occurs inside the basidium, extruding four haploid basidiospores externally.
3
Examine sexual reproduction in Zygomycota
Connect the zygospore to gametangial conjugation in moulds like *Rhizopus*.
When opposite mating strains meet, multinucleate gametangia fuse to create a highly resistant, dark, thick-walled zygospore.
4
Analyze asexual growth patterns in unicellular yeasts
Associate pseudohyphae with budding in *Saccharomyces*.
When budded cells fail to detach during rapid mitotic growth, they form elongated, linked chains known as pseudohyphae.

Key Concept

Structural and reproductive differentiation among Kingdom Fungi archetypes (moulds, yeasts, and mushrooms)
Question 8885Question

Match each sex-linked inheritance pattern or sex determination concept on the left with its correct biological description on the right.

Click a left item, then click its matching right item

Items

Y-linked gene transmission
X-linked recessive carrier mother
Homogametic male sex determination
Heterogametic female sex determination

Matches

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Answer

Y-linked gene transmission matches with direct father-to-son inheritance; X-linked recessive carrier mother matches with 50% allele transmission probability; Homogametic male sex determination matches with ZZ males in birds; Heterogametic female sex determination matches with ZW females in birds and reptiles.
Each genetic concept correctly aligns with its characteristic mode of chromosome transmission or gametic constitution: Y-linked traits are holandric (father-to-son), X-linked recessive carrier mothers have a 50% chance per child of passing the mutated allele, ZZ represents homogametic males in birds, and ZW represents heterogametic females in birds and reptiles.

Step-by-Step Solution

1
Identify the inheritance mode of Y-linked genes.
Since only males possess the Y chromosome, Y-linked traits pass exclusively from fathers to sons.
Y chromosomes are inherited solely through the paternal line.
2
Determine the transmission probability for a heterozygous carrier mother.
A carrier mother (XHXhX^H X^h) passes the XhX^h allele to 50% of her offspring.
Segregation during meiosis distributes each X chromosome into half of the gametes.
3
Differentiate between homogametic and heterogametic sex determination systems.
In birds (ZZ-ZW system), males have two identical sex chromosomes (ZZ, homogametic) while females have two distinct sex chromosomes (ZW, heterogametic).
Homogametic individuals produce only one type of sex gamete, whereas heterogametic individuals produce two different types.

Key Concept

Sex Determination Mechanisms and Sex-Linked Inheritance
Question 8886Question

In maize plants (*Zea mays*), the allele for purple kernels (PP) is completely dominant to the allele for yellow kernels (pp). A true-breeding purple-kernel plant is crossed with a true-breeding yellow-kernel plant. If the F1F_1 plants are self-pollinated, what is the expected genotypic ratio in the F2F_2 generation?

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Answer: 1:2:11 : 2 : 1

Answer

The expected genotypic ratio in the F2F_2 generation is 1:2:11 : 2 : 1 (1 PP:2 Pp:1 pp1\ PP : 2\ Pp : 1\ pp).
Crossing two heterozygous F1F_1 plants (Pp×PpPp \times Pp) yields genotypes PPPP, PpPp, and pppp in the proportions 1/4 PP1/4\ PP, 1/2 Pp1/2\ Pp, and 1/4 pp1/4\ pp. This corresponds to a genotypic ratio of 1:2:11 : 2 : 1.

Step-by-Step Solution

1
Determine parental genotypes and F1F_1 genotype
True-breeding parents are PPPP and pppp; all F1F_1 offspring are PpPp.
Homozygous dominant (PPPP) crossed with homozygous recessive (pppp) produces offspring that all inherit one PP and one pp allele.
2
Perform the F1F_1 self-cross (Pp×PpPp \times Pp)
Gametes PP and pp combine to produce PPPP, PpPp, pPpP, and pppp.
According to Mendel's Law of Segregation, alleles segregate into gametes with equal probability.
3
Calculate the genotypic ratio of the F2F_2 offspring
Genotypic ratio is 1 PP:2 Pp:1 pp1\ PP : 2\ Pp : 1\ pp, or 1:2:11 : 2 : 1.
Grouping offspring by genetic constitution yields 1 homozygous dominant (PPPP), 2 heterozygous (PpPp), and 1 homozygous recessive (pppp).

Key Concept

Monohybrid Cross Genotypic Ratio under Mendel's First Law
Estimated Time:1m 0s
Question 8887Question

In a comparative anatomical study of invertebrates, four different specimens were analyzed to identify their primary excretory organs: a crayfish (crustacean), a planarian (flatworm), a housefly (insect), and an earthworm (annelid). Which of the following correctly pairs each organism with its characteristic excretory structure?

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Answer: Crayfish: Green gland; Planarian: Flame cell; Housefly: Malpighian tubule; Earthworm: Nephridium

Answer

Crayfish possess green glands (antennal glands), planarians use specialized flame cells (protonephridia), houseflies rely on Malpighian tubules, and earthworms excrete metabolic waste using nephridia (metanephridia).
Different invertebrate groups possess specialized excretory structures adapted to their body plans and habitats. Crustaceans such as the crayfish excrete via green glands (antennal glands) located at the base of the antennae. Platyhelminthes (flatworms like planarians) utilize protonephridia equipped with cilia-propelled flame cells. Insects like the housefly use Malpighian tubules extending into the hemolymph to discharge waste into the alimentary canal. Annelids such as earthworms possess segmentally arranged nephridia.

Step-by-Step Solution

1
Identify the taxonomic class and corresponding excretory structure for each organism.
Crayfish (Crustacea) -> Green gland; Planarian (Platyhelminthes) -> Flame cell; Housefly (Insecta/Arthropoda) -> Malpighian tubule; Earthworm (Annelida) -> Nephridium.
Different invertebrate phyla have evolved distinct physiological organs to eliminate nitrogenous waste and maintain osmoregulation.
2
Evaluate the option choices to find the accurate set of pairings.
The pairing matching Crayfish to Green gland, Planarian to Flame cell, Housefly to Malpighian tubule, and Earthworm to Nephridium is correct.
All four organism-to-organ mappings in this combination align precisely with invertebrate excretory system anatomy.

Key Concept

Invertebrate Excretory Organs and Phylum Mapping
Question 8888Question

In diploid eukaryotic organisms, genetic information is organized into chromosomes containing specific functional sequences. Which of the following statements accurately describes the biological distinction between alleles and gene loci during meiotic division?

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Answer: A gene locus defines the fixed physical position of a gene on a chromosome, whereas alleles are alternative sequence variants of that gene located at identical loci on homologous chromosomes.

Answer

A gene locus defines the fixed physical position of a gene on a chromosome, whereas alleles are alternative sequence variants of that gene located at identical loci on homologous chromosomes.
The correct response accurately highlights that a gene locus is the precise structural coordinate on a chromosome, whereas alleles represent the specific molecular sequence variations of that gene present at corresponding loci on homologous pairs.

Step-by-Step Solution

1
Define the term 'gene locus'.
A gene locus (plural: loci) refers to the specific, fixed location of a particular gene along the length of a chromosome.
Establishing physical chromosomal position is essential before analyzing variant forms.
2
Define the term 'allele' in relation to homologous chromosomes.
Alleles are different functional variants of the same gene that occupy identical corresponding loci on homologous chromosome pairs.
Diploid organisms carry two alleles for each autosomal gene, one inherited from each parent.
3
Compare definitions to evaluate statement validity.
The option stating that a locus is the fixed position while alleles are sequence variants at identical loci on homologous chromosomes is scientifically correct.
This maintains precise distinction between physical locus coordinates and allele variants.

Key Concept

Gene Locus vs. Allele in Homologous Chromosomes
Estimated Time:1m 30s
Question 8889Question

In paleontological studies of evolutionary history, sedimentary rock strata preserve a chronological record of major biological transitions. Arrange the following key fossilized organisms in order of their first appearance in the global geological record, starting from the oldest (earliest geological period) to the most recent (youngest geological period).

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Answer

The correct chronological order from oldest to most recent geological appearance is: Ediacaran fauna (Precambrian) → Tiktaalik roseae (Devonian) → Seymouria (Permian) → Archaeopteryx lithographica (Jurassic) → Eohippus (Eocene).
The geological fossil record documents a clear temporal progression of life forms. Ediacaran fauna represent Precambrian multicellular organisms (~550 Ma). Tiktaalik roseae represents the Devonian transition of aquatic vertebrates to land (~375 Ma). Seymouria marks the Permian transition from primitive amphibians to early amniote/reptilian forms (~280 Ma). Archaeopteryx lithographica represents the Jurassic divergence of birds from theropod reptiles (~150 Ma). Eohippus represents Cenozoic mammalian radiation in the Eocene (~50 Ma). Ordering these from oldest to youngest gives: Ediacaran fauna → Tiktaalik roseae → Seymouria → Archaeopteryx lithographica → Eohippus.

Step-by-Step Solution

1
Determine the geological time period associated with the first appearance of each fossilized taxon in the fossil record.
Ediacaran fauna (~550 Ma, Precambrian), Tiktaalik roseae (~375 Ma, Devonian), Seymouria (~280 Ma, Permian), Archaeopteryx lithographica (~150 Ma, Jurassic), and Eohippus (~50 Ma, Eocene).
Paleontological age determination relies on relative stratigraphy and radiometric dating of surrounding rock strata.
2
Order the corresponding geological eras and periods chronologically from earliest to most recent according to the principle of superposition.
Precambrian → Devonian → Permian → Jurassic → Eocene.
Lower, undisturbed sedimentary layers correspond to older geological time spans compared to upper, younger strata.
3
Map each fossil organism to its respective position on the established geological scale.
Ediacaran fauna → Tiktaalik roseae → Seymouria → Archaeopteryx lithographica → Eohippus.
This establishes the verified evolutionary sequence of major vertebrate and pre-vertebrate milestones.

Key Concept

Stratigraphic succession and chronological timeline of transitional fossils in paleontology
Question 8890Question

During mammalian aerobic respiration, inhaled oxygen in the pulmonary alveoli must travel through multiple anatomical compartments and fluid media to serve as a terminal electron acceptor in tissue cellular respiration. What is the correct sequential path taken by an oxygen molecule from alveolar air space to its final metabolic reduction inside a muscle cell mitochondrion?

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Answer

The correct sequential order begins with oxygen diffusing across the alveolar and capillary endothelial walls into blood plasma, followed by binding to hemoglobin in red blood cells, circulatory transport through pulmonary veins to the left heart and systemic arterial system, dissociation and diffusion across tissue capillaries into cell cytoplasm, and ultimately its reduction to water at the inner mitochondrial membrane.
Inhaled oxygen moves from the alveolar air space across the thin alveolar epithelial membrane and pulmonary capillary endothelium into blood plasma. It then passes into red blood cells where it binds reversibly to the heme iron of hemoglobin. This oxygenated blood travels via pulmonary veins into the left side of the heart, which pumps it into systemic arterial circulation. Upon reaching systemic capillaries in active tissues, the low partial pressure of oxygen induces dissociation from hemoglobin, allowing oxygen to diffuse into tissue interstitial fluid and cell cytoplasm. Finally, oxygen diffuses into the mitochondrial matrix and inner mitochondrial membrane, acting as the terminal electron acceptor in oxidative phosphorylation to produce water.

Step-by-Step Solution

1
Identify the primary site of external gas exchange across the respiratory membrane.
Oxygen moves out of the alveolar lumen, passing sequentially through the alveolar epithelial cell layer, basement membrane, and endothelial cell layer into blood plasma.
Diffusion occurs passively from a region of high partial pressure of oxygen (PO2104 mmHgPO_2 \approx 104\text{ mmHg}) in the alveoli to lower partial pressure in deoxygenated capillary blood.
2
Determine how oxygen is bound for bulk transport in blood.
Dissolved oxygen in plasma moves across erythrocyte cell membranes and binds reversibly to the ferrous iron (Fe2+Fe^{2+}) center of hemoglobin.
Hemoglobin binding allows the blood to transport significantly higher volumes of oxygen than dissolved plasma alone.
3
Trace the macro-circulatory movement of oxygenated blood.
Oxygenated blood flows from pulmonary capillaries into pulmonary veins, entering the left atrium, passing to the left ventricle, and being propelled into the systemic arterial tree.
Pulmonary veins carry oxygen-rich blood back to the heart to provide hydraulic pressure for systemic tissue distribution.
4
Analyze the mechanism of oxygen delivery to metabolizing tissue cells.
In systemic capillaries, low tissue PO2PO_2 promotes oxygen dissociation from hemoglobin; free oxygen diffuses across the capillary wall, through interstitial fluid, and across the plasma membrane into cytosol.
Active tissue metabolism continuously consumes oxygen, creating a steep concentration gradient favoring unloading.
5
Identify the final intracellular biochemical sink for oxygen.
Oxygen diffuses into mitochondria, reaching the inner mitochondrial membrane where it accepts electrons from Complex IV (cytochrome c oxidase) and combines with protons (H+H^+) to yield water (H2OH_2O).
Oxygen acts as the ultimate electron acceptor in oxidative phosphorylation, enabling the continued flow of electrons along the electron transport chain.

Key Concept

Respiratory gas exchange pathway and cellular oxygen delivery in mammalian physiological respiration
Question 8891Question

An ecological survey conducted along a microclimatic gradient from a high-altitude montane biome (such as the Jos Plateau) down into the surrounding Guinea Savanna recorded distinct shifts in plant morphology, ambient temperature, relative humidity, and soil characteristics. Which combination of environmental factors and plant structural adaptations correctly distinguishes the high-altitude montane zone from the lower-elevation Guinea Savanna?

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Answer: Lower mean annual temperature, higher relative humidity, and prevalence of dwarf or stunted trees laden with epiphytic mosses and lichens

Answer

Lower mean annual temperature, higher relative humidity, and prevalence of dwarf or stunted trees laden with epiphytic mosses and lichens
The correct answer accurately identifies that montane biomes (such as the Jos Plateau in Nigeria) experience lower mean annual temperatures due to altitude, high relative humidity from cloud/mist interception, and support specialized vegetation including stunted trees and rich epiphytic growths (mosses and lichens).

Step-by-Step Solution

1
Analyze atmospheric and climatic shifts along an altitudinal gradient in terrestrial biomes
As altitude increases, atmospheric pressure and temperature decrease (environmental lapse rate), while moisture condensation produces frequent mist/cloud cover, increasing relative humidity.
Temperature and humidity gradients strongly dictate the vegetative profile of montane environments compared to lowland savannas.
2
Evaluate morphological adaptations of flora in high-altitude montane biomes versus lowland savannas
Strong winds, cooler temperatures, and high moisture levels in montane zones result in stunted or dwarf tree growth (elfin forest characteristics) covered with epiphytic bryophytes and lichens.
Epiphytic mosses and lichens thrive in cool, persistently humid air, whereas lowland trees adapt to seasonal drought and bushfires.

Key Concept

Montane Biome Abiotic Factors and Flora Adaptations
Estimated Time:2m 0s
Question 8892Question

In a mammalian spinal reflex arc, which neural component carries nerve impulses away from the central nervous system directly to an effector organ?

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Answer: Motor neuron

Answer

Motor neuron
The motor neuron (efferent neuron) is structurally and functionally adapted to conduct action potentials away from the gray matter of the spinal cord directly to muscle fibers or glands, triggering a reflex action.

Step-by-Step Solution

1
Trace the directional pathway of nerve impulse transmission in a spinal reflex arc.
Sensory receptor → Sensory neuron → Central Nervous System (Relay neuron) → Motor neuron → Effector organ.
Reflex arcs operate along a strict unidirectional path from stimulus detection to physiological response.
2
Identify which neuron functions efferently by connecting the central integration center to the response tissue.
The motor neuron conducts impulses outgoing from the spinal cord to muscle fibers or secretory glands.
Effector organs require motor innervation to carry out involuntary responses.

Key Concept

Directionality and functional classification of neurons in a spinal reflex arc
Question 8893Question

During the light-dependent stage of photosynthesis in the unicellular green alga *Chlamydomonas*, water molecules undergo photolysis inside the chloroplast. Which molecule is released as a direct byproduct of this light reaction?

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Answer: Oxygen

Answer

Oxygen gas is the direct byproduct released when water molecules undergo photolysis during the light-dependent stage of photosynthesis in unicellular green algae.
During photolysis in the light-dependent stage of photosynthesis in unicellular green algae, light energy absorbed by chlorophyll drives the splitting of water molecules, yielding hydrogen ions, electrons, and oxygen gas as a byproduct.

Step-by-Step Solution

1
Identify the specific photosynthetic process occurring in the chloroplast of the unicellular alga.
The process is photolysis of water during the light-dependent stage of photosynthesis.
Photolysis splits water molecules using light energy absorbed by photosynthetic pigments.
2
Determine the chemical products resulting from the photolysis reaction.
Water (2H2O2H_2O) breaks down into hydrogen ions (4H+4H^+), electrons (4e4e^-), and oxygen gas (O2O_2).
Water serves as the primary electron donor to replace electrons excited from chlorophyll.
3
Identify which of the resulting substances is released as a gas into the environment.
Oxygen gas (O2O_2) diffuses out of the organelle and cell as a byproduct.
Hydrogen ions and electrons are retained for ATP and NADPH synthesis, whereas molecular oxygen is released.

Key Concept

Photolysis of water in unicellular algal photosynthesis
Question 8894Question

During rapid transpiration in a tall dicotyledonous tree, water moves continuously from the soil through the plant body to the atmosphere along a water potential (Ψ\Psi) gradient. Which of the following statements correctly describes the physical state of water and the primary driving mechanism within the xylem vessels during this process?

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Answer: Water is maintained under negative hydrostatic pressure (tension) and pulled upward due to transpirational evaporation at the leaf surface.

Answer

Water is maintained under negative hydrostatic pressure (tension) and pulled upward due to transpirational evaporation at the leaf surface.
According to the cohesion-tension theory, transpiration causes evaporation of water from leaf mesophyll cells into sub-stomatal air spaces. This lowers the water potential of the leaves and creates a continuous suction force that puts the water inside dead xylem vessels under tension (negative pressure). The cohesive properties of water molecules maintain an unbroken column pulled upward from the roots.

Step-by-Step Solution

1
Analyze the water potential gradient along the transpiration stream.
Water moves spontaneously from regions of higher (less negative) water potential in the soil and roots toward lower (more negative) water potential in the leaf air spaces and atmosphere.
Evaporation of water vapor through stomata creates a strong suction force at the top of the plant.
2
Evaluate the mechanical state of water in xylem vessels according to the cohesion-tension theory.
The continuous evaporation generates tension (negative hydrostatic pressure) in the xylem sap, pulling the water column upward intact due to hydrogen bonding (cohesion between water molecules and adhesion to vessel walls).
Passive transpirational pull under tension is the primary driving mechanism for long-distance xylem transport in tall plants during high transpiration rates.

Key Concept

Cohesion-Tension Theory and Water Potential Gradient in Xylem Transport
Estimated Time:2m 0s
Question 8895Question

In maize (*Zea mays*), the allele for starchy endosperm (SuSu) is completely dominant over the allele for sugary endosperm (susu). Based on Mendel's First Law (Law of Segregation), which expected progeny outcome on the right correctly matches each monohybrid parental cross scenario on the left?

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Items

Cross between two heterozygous starchy plants (Susu×SusuSu\,su \times Su\,su), evaluated for the genotypic ratio among starchy progeny only
Test cross of a heterozygous starchy plant (Susu×susuSu\,su \times su\,su), evaluated for the ratio of dominant allele (SuSu) carriers to non-carriers
Self-pollination of an F1F_1 plant (SusuSu\,su), evaluated for the total percentage of pure-breeding (homozygous) progeny
Cross between a heterozygous starchy plant (SusuSu\,su) and a homozygous starchy plant (SuSuSu\,Su), evaluated for the percentage of sugary (sususu\,su) progeny

Matches

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Answer

Cross between two heterozygous starchy plants matches 1 SuSu:2 Susu1\ Su\,Su : 2\ Su\,su; Test cross of a heterozygous starchy plant matches 1:11 : 1 ratio (50%50\% carriers to 50%50\% non-carriers); Self-pollination of an F1F_1 plant matches 50%50\% of total progeny; Cross between a heterozygous starchy plant and homozygous starchy plant matches 0%0\% of total progeny.
Each monohybrid parental cross is correctly matched by calculating the specific genotypic or phenotypic probabilities based on allele segregation during meiosis according to Mendel's First Law.

Step-by-Step Solution

1
Analyze the cross between two heterozygous starchy plants (Susu×SusuSu\,su \times Su\,su).
Genotypes produced are 1/4 SuSu1/4\ Su\,Su, 1/2 Susu1/2\ Su\,su, and 1/4 susu1/4\ su\,su.
Mendel's Law of Segregation states alleles segregate into gametes with equal probability (1/2 Su1/2\ Su, 1/2 su1/2\ su). Starchy plants comprise SuSuSu\,Su and SusuSu\,su. Among starchy progeny, the genotypic ratio is 1 SuSu:2 Susu1\ Su\,Su : 2\ Su\,su.
2
Analyze the test cross of a heterozygous starchy plant (Susu×susuSu\,su \times su\,su).
Genotypes produced are 1/2 Susu1/2\ Su\,su and 1/2 susu1/2\ su\,su.
Carriers of the dominant allele SuSu are SusuSu\,su (50%50\%), while non-carriers are sususu\,su (50%50\%), yielding a ratio of 1:11 : 1.
3
Analyze self-pollination of an F1F_1 plant (Susu×SusuSu\,su \times Su\,su) for pure-breeding individuals.
Pure-breeding individuals (SuSuSu\,Su and sususu\,su) constitute 50%50\% of offspring.
Pure-breeding offspring are homozygous. 25% SuSu+25% susu=50%25\%\ Su\,Su + 25\%\ su\,su = 50\% of total progeny.
4
Analyze the cross Susu×SuSuSu\,su \times Su\,Su for sugary (sususu\,su) progeny.
Proportion of sususu\,su offspring is 0%0\%.
The homozygous dominant parent contributes a dominant SuSu allele to all offspring, preventing the expression of the recessive sususu\,su genotype.

Key Concept

Mendel's First Law and Monohybrid Cross Proportions
Estimated Time:2m 30s
Question 8896Question

Human physiological traits such as blood pressure and resting pulse rate exhibit discontinuous variation because individuals strictly belong to distinct, non-overlapping phenotypic categories controlled solely by single-gene inheritance.

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Answer: False

Answer

False
The statement is false because blood pressure and resting pulse rate exhibit continuous physiological variation, characterized by a smooth quantitative gradient across a population under polygenic control and environmental modification.

Step-by-Step Solution

1
Analyze the pattern of variation exhibited by physiological traits like blood pressure and resting pulse rate.
Blood pressure and pulse rate display a continuous spectrum of phenotypic values across a population spectrum without distinct breaks.
Continuous variation presents a gradual quantitative range of phenotypic expressions rather than distinct categories.
2
Examine the underlying genetic basis and environmental interaction for these physiological traits.
These traits are polygenic (controlled by multiple genes working additively) and are substantially modified by environmental factors.
Polygenic inheritance combined with environmental modifications produces a smooth, bell-shaped normal distribution curve.
3
Contrast continuous physiological traits with discontinuous variations.
Discontinuous variations (such as ABO blood groups or tongue rolling) consist of clear-cut non-overlapping phenotypic categories controlled by single genes and are generally unaffected by environmental conditions.
Mislabelling continuous physiological traits as discontinuous fails to account for polygenic control and environmental susceptibility.

Key Concept

Continuous vs. Discontinuous Variation in Human Physiological Traits
Question 8897Question

In Nigeria, a botanical survey records a vegetation zone characterized by an extensive ground cover of tall grasses (1.53 m1.5 - 3\text{ m}) interspersed with scattered, broad-leaved trees possessing thick, corky bark adapted to frequent seasonal bushfires. Which of the following ecological zones does this vegetation represent?

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Answer: Southern Guinea Savanna

Answer

Southern Guinea Savanna is characterized by tall grasses interspersed with broad-leaved, fire-resistant trees.
The Southern Guinea Savanna is the largest vegetation belt in Nigeria. It receives moderate rainfall (10001500 mm1000 - 1500\text{ mm} annually) and is typified by tall perennial grasses and scattered broad-leaved deciduous trees with thick, fire-resistant bark capable of surviving annual dry-season fires.

Step-by-Step Solution

1
Analyze the structural vegetation features provided in the scenario
Identified tall grasses (1.53 m1.5 - 3\text{ m}) and scattered broad-leaved trees with fire-resistant corky bark.
Vegetation structure directly reflects the moisture availability and fire frequency of specific biomes.
2
Compare the identified characteristics against Nigerian vegetation zones
Southern Guinea Savanna is the largest vegetation zone in Nigeria, transitional between the rainforest and northern drier savannas, possessing tall grasses and fire-adapted broad-leaved trees.
Shorter grasses and thorny leaves characterize drier northern zones (Sudan/Sahel), while dense canopies with minimal grasses characterize rainforests.

Key Concept

Nigerian Biomes and Vegetation Characteristics
Question 8898Question

According to the Harrod-Domar growth model, if a nation has a savings rate (ss) of 15%15\% and an incremental capital-output ratio (kk) of 33, what is its expected annual economic growth rate (gg)?

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Answer: 5%5\%

Answer

The expected annual economic growth rate is 5%5\%.
In the Harrod-Domar growth model, the economic growth rate (gg) is directly proportional to the savings rate (ss) and inversely proportional to the incremental capital-output ratio (kk), expressed as g=skg = \frac{s}{k}. Substituting s=15%s = 15\% and k=3k = 3 gives g=15%3=5%g = \frac{15\%}{3} = 5\%.

Step-by-Step Solution

1
Identify the given variables and the Harrod-Domar growth model formula
Savings rate s=15%s = 15\%, Incremental Capital-Output Ratio k=3k = 3. The Harrod-Domar formula is g=skg = \frac{s}{k}.
The model states that national economic growth depends directly on savings rate and inversely on capital productivity requirements.
2
Substitute the values into the formula to compute growth rate gg
g=15%3=5%g = \frac{15\%}{3} = 5\%.
Dividing the savings rate by the capital-output ratio yields the rate of GDP expansion.

Key Concept

Harrod-Domar Growth Model
Question 8899Question

A botanist analyzed three cryptogamic plant specimens (XX, YY, and ZZ) collected from different microhabitats during an ecological survey. Specimen XX consists of an simple, undifferentiated thallus without vascular tissues or organ differentiation; Specimen YY possesses multicellular rhizoids and a dominant gametophytic generation lacking lignified conducting elements; Specimen ZZ exhibits a dominant sporophytic generation with true roots, fronds, and lignified xylem and phloem. Which of the following correctly categorizes specimens XX, YY, and ZZ into their respective taxonomic divisions?

Show answer & explanation

Answer: Specimen X is a Thallophyte, Specimen Y is a Bryophyte, and Specimen Z is a Pteridophyte.

Answer

Specimen X is a Thallophyte, Specimen Y is a Bryophyte, and Specimen Z is a Pteridophyte.
The classification is derived from key evolutionary and anatomical distinctions among cryptogams: Thallophytes (Specimen X) have an unsegmented thallus with no vascular system; Bryophytes (Specimen Y) possess multicellular rhizoids and an avascular, gametophyte-dominant body; Pteridophytes (Specimen Z) are vascular cryptogams characterized by true vegetative organs (roots, stems, leaves) and a dominant sporophyte phase.

Step-by-Step Solution

1
Analyze the structural characteristics of Specimen X
Specimen X has an undifferentiated thallus body without root, stem, or leaf differentiation and lacks specialized conducting tissue, placing it in Division Thallophyta (Algae/Fungi).
Thallophytes represent the simplest plant organisation characterized by a non-vascular thallus body.
2
Evaluate the reproductive and anatomical features of Specimen Y
Specimen Y possesses rhizoids and displays a dominant gametophytic phase (nn) while lacking true vascular tissues (xylem and phloem), identifying it as a Bryophyte (mosses/liverworts).
Bryophytes are non-vascular land plants with gametophyte-dominant alternation of generations.
3
Assess the vascularization and alternation of generations in Specimen Z
Specimen Z features true roots, leaves, lignified vascular tissues (xylem and phloem), and a dominant sporophytic phase (2n2n), placing it in Division Pteridophyta (ferns).
Pteridophytes are seedless vascular plants exhibiting a dominant, independent sporophyte generation.

Key Concept

Taxonomic differentiation of lower plant groups based on body organisation, vascularization, and dominant life cycle generation
Question 8900Question

A culture of bacteria in a nutrient-rich broth undergoes rapid population expansion under optimal environmental conditions. Which mechanism is primarily responsible for the asexual reproduction and multiplication of these prokaryotic organisms?

Show answer & explanation

Answer: Binary fission, resulting in two genetically identical daughter cells

Answer

Binary fission, resulting in two genetically identical daughter cells
In Kingdom Monera, bacteria reproduce asexually under favorable conditions through binary fission. The single circular chromosome replicates, attaches to the plasma membrane, and the cell elongates before invaginating to yield two genetically identical cells.

Step-by-Step Solution

1
Identify the biological group and kingdom of bacteria
Bacteria belong to Kingdom Monera and are prokaryotic organisms.
Prokaryotic cells lack membrane-bound organelles, a true nucleus, and mitotic structures.
2
Determine the standard mode of cellular reproduction in prokaryotes under favorable conditions
Prokaryotes replicate their single circular chromosome and divide by binary fission.
Binary fission allows rapid exponential growth by splitting one parent cell into two identical daughter cells without mitosis.

Key Concept

Bacterial Reproduction via Binary Fission
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