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Question 8841Question

During asexual reproduction in the unicellular fungus Saccharomyces (yeast), a precise sequence of cellular events leads to the formation and independent release of a daughter cell. What is the correct chronological sequence of these events during the budding process?

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Answer

The correct chronological sequence of budding in Saccharomyces begins with localized cell wall weakening and protrusion under turgor pressure, followed by nuclear mitosis and migration into the bud, then chitinous septum synthesis at the cell neck, and concludes with cell separation leaving a bud scar.
The budding process in yeast begins when wall-modifying enzymes locally weaken the cell wall, allowing hydrostatic turgor pressure to force out a small protrusion. As this bud grows, the parent nucleus undergoes mitotic division, and one daughter nucleus migrates through the neck into the bud. Following nuclear transfer, chitin synthesizers lay down a primary septum across the neck to seal off both cellular compartments. Finally, chitinase enzymes cleave the connecting wall layers, releasing the independent daughter cell while leaving a prominent chitinous bud scar on the parent.

Step-by-Step Solution

1
Identify the initial mechanical trigger for bud emergence.
Enzymatic softening of glucan/chitin wall bonds allows turgor pressure to push out a daughter bud.
Cell expansion cannot occur without localized relaxation of the rigid fungal wall.
2
Trace the movement of genetic material into the growing daughter structure.
Mitotic division occurs, and motor proteins transport one daughter nucleus into the bud.
Nuclear inheritance must precede physical isolation of the daughter cytoplasm.
3
Identify the structural partitioning step between parent and offspring.
A chitinous primary septum is synthesized at the neck junction.
Septum formation seals both cells prior to final physical detachment.
4
Identify the final separation and scar-marking event.
Enzymes digest the glucan layer joining the cells, detaching the daughter cell and leaving a permanent bud scar.
This completes cytokinesis and restores independence to both organisms.

Key Concept

Mechanism of Budding and Cytokinesis in Saccharomyces (Yeast)
Question 8842Question

During a biology practical session, students recorded various inherited characteristics in their peer group. Which of the following traits demonstrates discontinuous variation?

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Answer: ABO blood group

Answer

ABO blood group
ABO blood group exhibits discontinuous variation because individuals fall into clear-cut, distinct phenotypic categories (A, B, AB, and O) without intermediate phenotypes, and the trait is strictly inherited without environmental modification.

Step-by-Step Solution

1
Identify the distinguishing characteristic of discontinuous variation.
Discontinuous variation produces distinct, non-overlapping phenotypic categories without intermediate forms, usually under monogenic control.
Qualitative traits are determined by major alleles at one or very few gene loci.
2
Evaluate the phenotypic distribution of each listed trait.
Height, body weight, and skin color display continuous gradations across a population spectrum. In contrast, ABO blood group classifies individuals strictly into discrete groups (A, B, AB, or O).
Blood groups show clear-cut phenotypic distinction unaffected by environmental factors.

Key Concept

Discontinuous phenotypic variation in human traits
Estimated Time:45s
Question 8843Question

A coastal sand dune that was completely devoid of organic matter is slowly colonized over time, eventually developing into a stable woodland community. Which of the following factors primarily distinguishes this ecological process from the recovery of an abandoned farmland?

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Answer: The absence of a pre-existing soil substrate and seed bank prior to pioneer colonization

Answer

The absence of a pre-existing soil substrate and seed bank prior to pioneer colonization
Primary ecological succession occurs on substrates completely devoid of pre-existing organic soil and seed banks, such as sand dunes or exposed volcanic rock. Conversely, secondary succession occurs in areas where a previous community was disturbed but fertile topsoil and dormant seeds remain present.

Step-by-Step Solution

1
Identify the type of succession described in the scenario.
Colonization of a bare sand dune lacking organic soil is an example of primary succession.
Primary succession occurs on newly exposed or uncolonized land where no previous community or organic soil exists.
2
Compare primary succession with the recovery of abandoned farmland.
Abandoned farmland undergoes secondary succession because organic topsoil, seed banks, and micro-organisms remain intact after disturbance.
Secondary succession starts on existing soil, making the succession process significantly faster than primary succession.
3
Identify the key distinguishing feature between the two ecological processes.
The absence of pre-existing soil and seed banks in the sand dune environment.
Soil formation must take place first during primary succession before complex plant species can be supported.

Key Concept

Distinction between primary and secondary ecological succession substrates
Estimated Time:1m 0s
Question 8844Question

Match each organism and its physiological state with the corresponding primary structure and mechanism utilized for gaseous exchange.

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Items

Adult African toad (*Sclerophrys regularis*) dormant during estivation
Freshwater bony fish (*Tilapia zillii*) actively swimming
Grasshopper (*Locusta migratoria*) during vigorous flight
Dicotyledonous leaf (*Hibiscus*) during peak daylight photosynthesis

Matches

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Answer

The correct pairings are: Adult African toad during estivation matches cutaneous diffusion across moist vascularized skin; Freshwater bony fish matches countercurrent exchange across gill lamellae; Grasshopper during flight matches abdominal contractions forcing air into spiracles and tracheoles; and Dicotyledonous leaf during daylight matches inward CO2 diffusion through guard cell-regulated stomata.
Each organism utilizes specialized respiratory surfaces matched to its environment and metabolic activity: dormant adult amphibians rely on cutaneous skin diffusion; bony fish employ countercurrent flow across gill lamellae; terrestrial insects use abdominal pumping into tracheoles; and green leaves regulate stomatal diffusion via guard cell turgidity.

Step-by-Step Solution

1
Analyze the metabolic demands and structural adaptations of the estivating adult toad.
Estivation lowers metabolism and suppresses lung expansion, making cutaneous respiration across moist skin the main mode of exchange.
Amphibians switch respiratory surface reliance depending on environment and metabolic state.
2
Determine the gaseous exchange mechanism of active bony fish.
Water flowing over gill lamellae opposite to blood flow creates a countercurrent gradient ensuring efficient oxygen uptake.
Water has lower dissolved oxygen content than air, requiring a countercurrent mechanism to maximize uptake.
3
Evaluate gaseous transport in flying insects.
Insects lack hemoglobin for gas transport; active flight relies on abdominal ventilation pushing air directly through spiracles into tracheoles.
The tracheal system delivers gases directly to tissue cells without involving the circulatory fluid.
4
Identify leaf gas exchange dynamics during daylight.
High photosynthetic rate creates a CO2 concentration gradient, causing net CO2 entry through open stomata governed by guard cell turgor pressure.
Stomatal aperture changes based on osmotic water uptake by guard cells.

Key Concept

Respiratory Surface Adaptations across Diverse Taxa
Estimated Time:2m 0s
Question 8845Question

Which of the following represents the correct sequential order of plant community stages during primary ecological succession on a bare rock surface (xerosere), from initial colonizers to the mature climax community?

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Answer

The correct sequence of primary succession on bare rock is: Crustose lichens → Foliose lichens and mosses → Herbaceous grasses and weeds → Climax forest.
Primary succession on bare rock begins with crustose lichens breaking down rock surfaces. Decomposing lichen material forms a thin soil layer that allows mosses and foliose lichens to colonize. As soil accumulates further, herbaceous grasses establish, eventually yielding to a stable climax forest community.

Step-by-Step Solution

1
Identify the pioneer community
Crustose lichens are the first organisms capable of colonizing bare rock substrate where soil is entirely absent.
Pioneer species in primary succession must tolerate extreme exposure and contribute to initial weathering of substrate.
2
Arrange the intermediate seral stages by soil depth requirement
Mosses and foliose lichens establish next, followed by herbaceous grasses as organic matter builds up.
Bryophytes require minimal soil pockets, whereas herbaceous plants need shallow organic topsoil to anchor root systems.
3
Identify the stable terminal community
The progression culminates in a mature climax forest.
The climax community represents the final, self-perpetuating stage in equilibrium with the prevailing climate.

Key Concept

Sequential Seral Stages of Primary Ecological Succession (Xerosere)
Question 8846Question

According to modern evolutionary theory, which statement correctly describes the primary contribution of sexual reproduction to genetic variation within a population?

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Answer: It reshuffles existing alleles into unique combinations through crossing over and independent assortment.

Answer

Sexual reproduction reshuffles existing alleles into unique combinations through crossing over and independent assortment.
In modern evolutionary theory (Neo-Darwinism), sexual reproduction contributes to genetic diversity mainly by shuffling existing alleles during meiosis (crossing over and independent assortment) and random fertilization. This creates new genotypic combinations without generating brand-new genetic alleles.

Step-by-Step Solution

1
Distinguish between sources of new alleles and sources of genetic recombination.
Gene mutations are the ultimate source of novel alleles, whereas sexual reproduction rearranges existing alleles.
Modern synthesis clearly separates mutation (creation of new genetic material) from recombination (shuffling of existing alleles).
2
Identify the key mechanisms of genetic variation during sexual reproduction.
Processes such as crossing over during meiosis, independent assortment of chromosomes, and random fertilization generate new genotype combinations.
These mechanisms create unique offspring phenotypes without altering the fundamental genetic code of individual alleles.

Key Concept

Role of Sexual Reproduction and Recombination in Modern Evolutionary Theory
Question 8847Question

Unlike angiosperms, which enclose their ovules within an ovary that matures into a fruit and feature a reduced female gametophyte (embryo sac) lacking archegonia, gymnosperms produce exposed ovules on megasporophylls and develop distinct archegonia within their female gametophytes.

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Answer: True

Answer

The statement is TRUE.
The statement accurately highlights key evolutionary and anatomical distinctions: angiosperms enclose ovules in ovaries that form fruits and have an 8-nucleate female gametophyte (embryo sac) without archegonia. Gymnosperms bear naked ovules on megasporophylls and form archegonia within their female gametophytes.

Step-by-Step Solution

1
Analyze ovule and seed enclosure in Gymnosperms versus Angiosperms.
Gymnosperms bear exposed (naked) ovules on megasporophylls or cones, whereas angiosperms enclose ovules within carpels/ovaries that mature into protective fruits.
Ovary enclosure of ovules is a fundamental anatomical distinction of angiosperms.
2
Examine female gametophyte structure and archegonia presence.
Gymnosperms develop multicellular female gametophytes containing archegonia (female sex organs), whereas the angiosperm female gametophyte is reduced to an 8-nucleate embryo sac completely devoid of archegonia.
Evolutionary reduction of gametophytes in angiosperms eliminated discrete archegonia.
3
Evaluate the complete statement against biological facts.
Both clauses accurately describe the reproductive morphology and evolutionary differences distinguishing angiosperms from gymnosperms.
No part of the statement contains factual errors or misleading claims.

Key Concept

Reproductive morphology and gametophyte evolution distinguishing Gymnosperms and Angiosperms
Estimated Time:1m 15s
Question 8848Question

An ecology student investigated the population of spear grass (*Imperata cylindrica*) on a 400 m2400\text{ m}^2 plot in Jos, Plateau State. A rectangular quadrat measuring 1.0 m×0.5 m1.0\text{ m} \times 0.5\text{ m} was thrown randomly 16 times across the plot, yielding a total count of 120 spear grass plants. What is the estimated population density of spear grass in plants/m2\text{plants/m}^2?

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Answer: 15.0 plants/m215.0\text{ plants/m}^2

Answer

15.0 plants/m215.0\text{ plants/m}^2
The population density is defined as the total number of individuals of a species per unit area sampled. The area of one quadrat is 1.0 m×0.5 m=0.5 m21.0\text{ m} \times 0.5\text{ m} = 0.5\text{ m}^2. Sampling 16 times gives a total sampled area of 16×0.5 m2=8.0 m216 \times 0.5\text{ m}^2 = 8.0\text{ m}^2. Dividing the total count of 120 plants by 8.0 m28.0\text{ m}^2 yields 15.0 plants/m215.0\text{ plants/m}^2.

Step-by-Step Solution

1
Calculate the area of a single quadrat frame
Area of 1 quadrat=1.0 m×0.5 m=0.5 m2\text{Area of 1 quadrat} = 1.0\text{ m} \times 0.5\text{ m} = 0.5\text{ m}^2
Determines the sampling surface area covered by one throw.
2
Calculate the total area sampled across all quadrat throws
Total sampled area=16 throws×0.5 m2=8.0 m2\text{Total sampled area} = 16 \text{ throws} \times 0.5\text{ m}^2 = 8.0\text{ m}^2
Finds the total ground space inspected during the field survey.
3
Calculate the population density of spear grass per square metre
Population Density=Total plant countTotal sampled area=120 plants8.0 m2=15.0 plants/m2\text{Population Density} = \frac{\text{Total plant count}}{\text{Total sampled area}} = \frac{120\text{ plants}}{8.0\text{ m}^2} = 15.0\text{ plants/m}^2
Determines average number of organisms per unit area.

Key Concept

Quadrat Population Density Calculation
Estimated Time:1m 30s
Question 8849Question

Match each sex determination mechanism or sex-linked inheritance phenomenon on the left with its corresponding biological characteristic or inheritance pattern on the right.

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Items

XX-XO sex-determination system in grasshoppers (*Melanoplus* species)
ZZ-ZW sex-determination system in birds (*Gallus gallus*)
X-linked recessive phenotypic expression in Turner syndrome females (45,X45, X)
Holandric (Y-linked) trait inheritance in humans

Matches

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Answer

XX-XO in grasshoppers matches males being heterogametic (X0X0) and females homogametic (XXXX); ZZ-ZW in birds matches females being heterogametic (ZWZW) and males homogametic (ZZZZ); X-linked recessive expression in Turner syndrome females matches hemizygosity due to monosomy X; Holandric inheritance matches exclusive father-to-son transmission without female carriers.
Each mechanism accurately corresponds to its defining chromosomal configuration or inheritance pattern: XX-XO grasshoppers have X0X0 heterogametic males; ZZ-ZW birds have ZWZW heterogametic females; Turner syndrome females are hemizygous (45,X45, X) expressing X-linked recessives directly; holandric Y-linked traits transmit exclusively from fathers to sons.

Step-by-Step Solution

1
Analyze the sex determination system in grasshoppers (XX-XO).
Identify that grasshopper females are XXXX (homogametic) and males are X0X0 (heterogametic).
The absence of a Y chromosome means males have 23 chromosomes (22+X022 + X0) while females have 24 (22+XX22 + XX).
2
Analyze the sex determination system in birds (ZZ-ZW).
Identify that female birds are ZWZW (heterogametic) and male birds are ZZZZ (homogametic).
This reverses the male heterogametic pattern seen in mammals.
3
Evaluate the genetic condition of Turner syndrome females (45,X45, X) regarding X-linked traits.
Determine that monosomy X creates a hemizygous state in females.
Without a second X chromosome to mask a recessive allele, a single X-linked recessive allele is expressed phenotypically.
4
Evaluate holandric (Y-linked) inheritance in humans.
Determine that Y-linked genes pass strictly from male parent to male offspring.
Females do not inherit a Y chromosome and therefore cannot carry or pass on holandric traits.

Key Concept

Chromosomal mechanisms of sex determination (XX-XY, XX-XO, ZZ-ZW) and hemizygous expression of sex-linked genes.
Estimated Time:2m 0s
Question 8850Question

Match each ecological measuring instrument in the left column with its corresponding abiotic factor, operating principle, and measurement unit in the right column.

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Items

Hair hygrometer
Anemometer
Soil tensiometer
Salinometer

Matches

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Answer

Hair hygrometer matches atmospheric relative humidity measured via dimensional changes in organic fibers; Anemometer matches wind velocity measured in meters per second; Soil tensiometer matches soil matric potential (moisture tension); Salinometer matches total dissolved salt concentration in parts per thousand.
Each instrument is accurately linked to its specific ecological factor and measurement principle: the hair hygrometer monitors relative humidity through fiber length variation; the anemometer records wind speed in meters per second; the soil tensiometer quantifies soil matric potential (suction force); and the salinometer determines aquatic salinity in parts per thousand.

Step-by-Step Solution

1
Analyze atmospheric moisture measurement devices.
The hair hygrometer operates on the physical principle of fiber length modification due to atmospheric moisture, measuring relative humidity.
Transpiration rates and atmospheric moisture levels depend heavily on relative humidity.
2
Analyze atmospheric air movement instruments.
The anemometer uses rotating cups or propellers driven by wind movement to quantify speed in m/s\text{m/s}.
Wind velocity directly impacts plant pollination, evaporation, and animal behavior.
3
Evaluate edaphic moisture tension measurement devices.
The soil tensiometer measures the matric suction pressure exerted by soil particles on capillary water.
Root absorption efficiency depends on overcoming soil matric suction pressure.
4
Evaluate aquatic ionic strength and dissolved solid measurement tools.
The salinometer determines salt concentration in aquatic environments, expressed in parts per thousand (ppt\text{ppt}).
Salinity dictates osmoregulatory dynamics in marine and brackish organisms.

Key Concept

Ecological Factors and Their Measurement Instruments
Question 8851Question

Match each population ecology term on the left with its correct definition on the right.

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Items

Carrying capacity
Environmental resistance
Biotic potential
Natality

Matches

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Answer

Carrying capacity matches the maximum population size sustained indefinitely by a habitat; Environmental resistance matches the sum of factors restricting growth; Biotic potential matches the maximum growth rate under ideal conditions; Natality matches the rate of adding new individuals through birth.
Each ecological term is matched to its precise definition: Carrying capacity represents the maximum population size an ecosystem can sustain; Environmental resistance includes all physical and biological limiting factors; Biotic potential is the maximum reproductive capability under ideal conditions; Natality is the birth rate of a population.

Step-by-Step Solution

1
Define carrying capacity
Carrying capacity matches the description of maximum sustainable population size.
Ecosystem resources like food and space place an upper bound on population size.
2
Define environmental resistance
Environmental resistance matches the sum of all factors restricting population growth.
Biotic and abiotic factors work together to curb unlimited population expansion.
3
Define biotic potential
Biotic potential matches the maximum theoretical reproductive rate under ideal conditions.
It represents physiological capability to reproduce without environmental constraints.
4
Define natality
Natality matches the birth rate of individuals added per unit time.
Natality specifically concerns reproductive additions to the population.

Key Concept

Population Dynamics Parameters
Estimated Time:1m 0s
Question 8852Question

In human ABO blood group inheritance, an individual carrying both the IAI^A and IBI^B alleles expresses both A and B antigens on the surface of their red blood cells, resulting in blood type AB. Which genetic phenomenon is directly demonstrated by the simultaneous, full expression of both alleles in the heterozygous state?

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Answer: Codominance

Answer

Codominance
The correct answer is codominance because both the IAI^A and IBI^B alleles contribute fully and independently to the phenotype. Heterozygous individuals (IAIBI^A I^B) produce both A and B functional agglutinogens (antigens) on their red blood cell surfaces without blending or masking.

Step-by-Step Solution

1
Analyze the expression of alleles IAI^A and IBI^B in a heterozygous individual (IAIBI^A I^B).
Both antigen A and antigen B are independently produced and present on the erythrocyte membrane.
Neither allele masks the other, nor do they blend to form an intermediate antigen structure.
2
Relate this joint phenotypic expression to standard non-Mendelian genetic definitions.
The simultaneous full expression of two different alleles at a locus is defined as codominance.
This contrasts with complete dominance (where one allele masks another) and incomplete dominance (where a intermediate phenotype is formed).

Key Concept

Codominance in human blood groups
Question 8853Question
In the industrial synthesis of methanol, carbon(II) oxide gas reacts with hydrogen gas according to the equation:
CO(g)+2H2(g)CH3OH(g)\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)

At a given temperature, an equilibrium mixture in a 2.0 dm32.0\text{ dm}^3 sealed container contains 0.40 mol0.40\text{ mol} of CO(g)\text{CO}(g), 0.40 mol0.40\text{ mol} of H2(g)\text{H}_2(g), and 0.16 mol0.16\text{ mol} of CH3OH(g)\text{CH}_3\text{OH}(g). What is the numerical value of the equilibrium constant, KcK_c, for this reaction?

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Answer: 10.0 dm6 mol210.0\text{ dm}^6\text{ mol}^{-2}

Answer

10.0 dm6 mol210.0\text{ dm}^6\text{ mol}^{-2}
To find KcK_c, first divide the moles of each gas at equilibrium by the volume of the vessel (2.0 dm32.0\text{ dm}^3) to obtain their equilibrium concentrations: [CO]=0.20 mol dm3[\text{CO}] = 0.20\text{ mol dm}^{-3}, [H2]=0.20 mol dm3[\text{H}_2] = 0.20\text{ mol dm}^{-3}, and [CH3OH]=0.08 mol dm3[\text{CH}_3\text{OH}] = 0.08\text{ mol dm}^{-3}. Then substitute these values into the equilibrium expression Kc=[CH3OH][CO][H2]2K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}, giving 0.080.20×(0.20)2=10.0 dm6 mol2\frac{0.08}{0.20 \times (0.20)^2} = 10.0\text{ dm}^6\text{ mol}^{-2}.

Step-by-Step Solution

1
Calculate the equilibrium concentration of each species
[CO]=0.40 mol2.0 dm3=0.20 mol dm3[\text{CO}] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}, [H2]=0.40 mol2.0 dm3=0.20 mol dm3[\text{H}_2] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3}, [CH3OH]=0.16 mol2.0 dm3=0.08 mol dm3[\text{CH}_3\text{OH}] = \frac{0.16\text{ mol}}{2.0\text{ dm}^3} = 0.08\text{ mol dm}^{-3}
Equilibrium constant KcK_c requires molar concentrations in mol dm3\text{mol dm}^{-3}, calculated using C=nVC = \frac{n}{V}.
2
Write the equilibrium constant expression
Kc=[CH3OH][CO][H2]2K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}
Products are placed in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient.
3
Substitute concentrations into the expression and solve
Kc=0.08(0.20)×(0.20)2=0.080.20×0.04=0.080.008=10.0 dm6 mol2K_c = \frac{0.08}{(0.20) \times (0.20)^2} = \frac{0.08}{0.20 \times 0.04} = \frac{0.08}{0.008} = 10.0\text{ dm}^6\text{ mol}^{-2}
Accurate substitution and exponent evaluation.

Key Concept

Calculation of equilibrium constant (KcK_c) from equilibrium amounts and container volume
Estimated Time:1m 30s
Question 8854Question

During the infection cycle of the parasitic protozoan responsible for malaria, motile sporozoites are introduced into the human bloodstream via a mosquito bite. Before invading erythrocytes to cause clinical symptoms, which specific primary target organ cells must these sporozoites first invade and multiply within during the exo-erythrocytic schizogony phase?

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Answer: Hepatocytes of the liver

Answer

Hepatocytes of the liver are the primary target cells invaded by Plasmodium sporozoites during exo-erythrocytic schizogony.
Sporozoites injected into the human host by an infected female Anopheles mosquito travel through the bloodstream to the liver. There, they invade hepatocytes and undergo exo-erythrocytic schizogony, producing thousands of merozoites that subsequently rupture from liver cells to infect erythrocytes.

Step-by-Step Solution

1
Identify the infective stage of Plasmodium entering the human host.
Infective sporozoites are injected into human blood capillaries by an infected female Anopheles mosquito.
Sporozoites represent the motile stage produced in the mosquito salivary glands.
2
Trace the initial migration path of sporozoites within human tissues.
Sporozoites quickly leave the vascular circulation and specifically target the liver parenchymal cells (hepatocytes).
Surface circumsporozoite proteins bind specifically to receptors on hepatocytes.
3
Analyze the intracellular developmental stage prior to red blood cell invasion.
Asexual multiplication (exo-erythrocytic schizogony) occurs inside hepatocytes, generating thousands of merozoites.
Merozoites are the specific life-cycle stage adapted to invade human red blood cells.

Key Concept

Plasmodium life cycle and tissue tropism during exo-erythrocytic schizogony
Estimated Time:1m 30s
Question 8855Question

Match each microevolutionary genetic phenomenon on the left with its corresponding population genetics mechanism or outcome on the right.

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Items

Persistent retention of a deleterious or lethal recessive allele in a gene pool despite strong negative selection against homozygotes
Stochastic shifts in allele frequencies resulting from a severe, non-selective reduction in population size
Bimodal distribution of phenotypic traits caused by simultaneous selection against intermediate heterozygous or median phenotypes
Alteration of allele frequencies and reduction of genetic divergence between sub-populations caused by inter-population movement of fertile individuals

Matches

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Answer

The phenomenon of retaining a deleterious recessive allele corresponds to balancing selection via heterozygote advantage. Stochastic allele frequency shifts after population reduction correspond to genetic drift via the bottleneck effect. Bimodal distribution of traits corresponds to disruptive selection. The movement of individuals reducing divergence between populations corresponds to gene flow via migration.
The retention of deleterious alleles occurs through balancing selection because heterozygous carriers gain a survival advantage (e.g., malaria resistance in sickle-cell heterozygotes). Sudden random allele changes following a drastic population drop constitute genetic drift operating through the bottleneck effect. Selection favoring both extreme phenotypes while penalizing intermediate forms defines disruptive selection. The exchange of genetic material between distinct populations through migration defines gene flow.

Step-by-Step Solution

1
Analyze the retention of lethal recessive alleles in a population.
Identify that when heterozygous individuals possess higher fitness than either homozygous form, the recessive allele is preserved in the population (balancing selection / heterozygote superiority).
This explains why negative selection against homozygotes fails to eliminate harmful recessive alleles.
2
Examine random frequency changes due to catastrophic size reduction.
Map non-selective population drop to sampling error and random allele loss/fixation known as genetic drift (bottleneck effect).
When population size drops precipitously, survival is stochastic rather than fitness-based.
3
Evaluate the mechanism generating a bimodal phenotypic distribution.
Pair selective pressure against intermediate phenotypes with disruptive selection.
Disruptive selection acts against average phenotypes, favoring both extreme tail traits.
4
Determine the effect of inter-population movement of individuals.
Connect movement of gametes or breeding individuals between populations to gene flow.
Gene flow introduces new alleles and homogenizes gene pools across geographic regions.

Key Concept

Mechanisms of Microevolution and Population Genetics
Estimated Time:2m 0s
Question 8856Question

Match each specified plant genotype with the number of genetically distinct gamete types it can produce during meiosis according to Mendel's Law of Independent Assortment.

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Items

YyRrYyRr (Heterozygous at two gene loci)
YYRrYYRr (Homozygous dominant at one locus and heterozygous at another)
yyrryyrr (Homozygous recessive at both gene loci)
YyRrSsYyRrSs (Heterozygous at three independently assorting gene loci)

Matches

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Answer

The correct matches are: YyRrYyRr matches with 4 distinct gamete types; YYRrYYRr matches with 2 distinct gamete types; yyrryyrr matches with 1 distinct gamete type; and YyRrSsYyRrSs matches with 8 distinct gamete types.
According to Mendel's Second Law, alleles of unlinked genes assort independently during meiosis. The number of genetically distinct gametes produced by an organism is given by 2n2^n, where nn is the number of heterozygous gene pairs. Thus, YyRrYyRr has 2 heterozygous pairs giving 4 gametes (222^2), YYRrYYRr has 1 heterozygous pair giving 2 gametes (212^1), yyrryyrr has 0 heterozygous pairs giving 1 gamete (202^0), and YyRrSsYyRrSs has 3 heterozygous pairs giving 8 gametes (232^3).

Step-by-Step Solution

1
Identify the formula for determining the number of distinct gametes.
The formula is 2n2^n, where nn represents the number of heterozygous gene pairs.
Mendel's Law of Independent Assortment states that alleles for different traits segregate independently during gamete formation.
2
Calculate gamete numbers for each given genotype.
For YyRrYyRr, n=2    22=4n=2 \implies 2^2 = 4; for YYRrYYRr, n=1    21=2n=1 \implies 2^1 = 2; for yyrryyrr, n=0    20=1n=0 \implies 2^0 = 1; for YyRrSsYyRrSs, n=3    23=8n=3 \implies 2^3 = 8.
Counting the number of heterozygous pairs (nn) directly determines the variety of gametes produced.

Key Concept

Gamete Genotype Determination and Mendel's Law of Independent Assortment
Estimated Time:45s
Question 8857Question

During the bending (flexion) of the human forearm at the elbow joint, which of the following muscular actions occurs?

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Answer: The biceps muscle contracts while the triceps muscle relaxes

Answer

The biceps muscle contracts while the triceps muscle relaxes.
Skeletal movement at joints is produced by antagonistic muscle pairs. During flexion (bending) of the forearm at the elbow joint, the biceps muscle (flexor) contracts to pull the bone forward, while the triceps muscle (extensor) relaxes to allow the movement to take place.

Step-by-Step Solution

1
Identify the movement described in the stem
The movement is forearm flexion (bending the elbow joint).
Flexion decreases the angle between the upper arm and forearm.
2
Determine the role of the antagonistic muscle pair involved
The biceps acts as the flexor (agonist) and the triceps acts as the extensor (antagonist).
Muscles can only pull when contracting; movement requires paired opposing actions.
3
Match muscle states required for flexion
Contraction of the biceps produces the upward pull, while relaxation of the triceps permits motion.
Antagonistic coordination ensures smooth directional movement across synovial joints.

Key Concept

Antagonistic Muscle Action in Locomotion
Question 8858Question

In comparative biochemistry, the degree of evolutionary relationship between different organisms can be determined by comparing the amino acid sequences of conserved proteins such as Cytochrome c. If human Cytochrome c differs by 00 amino acids from chimpanzees, 11 amino acid from rhesus monkeys, 1212 amino acids from horses, and 4545 amino acids from baker's yeast, which of the following conclusions is most valid?

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Answer: Humans share the most recent common ancestor with chimpanzees and the most distant common ancestor with baker's yeast.

Answer

Humans share the most recent common ancestor with chimpanzees and the most distant common ancestor with baker's yeast.
The correct answer correctly interprets comparative biochemical evidence. The fewer the differences in the amino acid sequence of a universal, conserved protein such as Cytochrome c between two species, the more recently those species diverged from a common ancestor. Since human and chimpanzee Cytochrome c sequences are identical, they share the most recent common ancestor, while the large difference of 45 amino acids between humans and yeast indicates a very ancient divergence.

Step-by-Step Solution

1
Analyze the quantitative biochemical data provided in the stem.
Sequence difference count from humans: Chimpanzee = 0, Rhesus monkey = 1, Horse = 12, Baker's yeast = 45.
Fewer amino acid differences indicate less time elapsed since divergence from a shared ancestor.
2
Relate sequence similarity to evolutionary distance.
Humans are most closely related to chimpanzees (0 differences) and most distantly related to baker's yeast (45 differences).
Homologous proteins mutate at a relatively constant rate over geological time; lower divergence reflects a more recent common ancestor.

Key Concept

Comparative Biochemistry as Evidence for Evolution
Estimated Time:1m 0s
Question 8859Question

In guinea pigs (*Cavia porcellus*), black coat color (BB) is dominant over white coat color (bb), and short hair (SS) is dominant over long hair (ss). If a heterozygous black, short-haired guinea pig (BbSsBbSs) is mated with a white, long-haired guinea pig (bbssbbss) and they produce a total of 640 offspring, how many of the offspring are expected to display a black coat and long hair?

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Answer: 160

Answer

160 offspring are expected to have a black coat and long hair.
In a dihybrid testcross involving a double heterozygote (BbSsBbSs) and a homozygous recessive individual (bbssbbss), the offspring phenotypes appear in equal ratios of 1:1:1:1 (25% for each phenotypic class). The black coat, long hair phenotype (BbssBbss) corresponds to 1/4 of the total offspring. Multiplying 1/4 by 640 yields exactly 160 expected offspring.

Step-by-Step Solution

1
Determine the type of genetic cross and parental genotypes.
The cross is a dihybrid testcross between BbSsBbSs and bbssbbss.
One parent is heterozygous for both independently assorting traits (BbSsBbSs), and the other parent is homozygous recessive (bbssbbss).
2
Determine the proportion of offspring expected to have the phenotype black coat and long hair (BbssBbss).
The proportion of BbssBbss offspring is 14\frac{1}{4} (or 25%25\%).
The BbSsBbSs parent produces four types of gametes (BSBS, BsBs, bSbS, bsbs) in equal proportions (14\frac{1}{4} each). Combining BsBs with bsbs yields BbssBbss.
3
Calculate the expected count out of 640 total offspring.
14×640=160\frac{1}{4} \times 640 = 160.
Multiplying the expected phenotypic fraction by the total offspring count yields the absolute expected count.

Key Concept

Dihybrid testcross ratio and probability calculation
Question 8860Question

Following a volcanic eruption, a newly formed island of solid basalt rock is completely devoid of soil and organic matter. Which of the following organisms is most likely to act as a pioneer species to initiate primary ecological succession on this substrate?

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Answer: Lichens

Answer

Lichens act as the pioneer species on bare volcanic rock during primary succession.
Primary succession begins on newly exposed, completely sterile surfaces where no soil layer exists. Lichens serve as pioneer colonizers because their symbiotic algal and fungal components allow them to photosynthesize, withstand drought, and dissolve rock surfaces chemically to create the first layers of soil.

Step-by-Step Solution

1
Identify the initial ecosystem conditions described.
The habitat is bare volcanic rock devoid of topsoil, organic matter, and pre-existing seed banks.
Primary succession takes place on previously uncolonized, soil-free substrates.
2
Determine which organism type can colonize bare substrate without soil.
Lichens are specialized pioneer organisms that tolerate harsh, nutrient-poor conditions and weather rock faces into soil.
Pioneer species must initiate soil formation to allow subsequent plant communities to establish.

Key Concept

Pioneer species in primary ecological succession
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