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Question 8801Question

Match each noble gas listed on the left with its corresponding primary industrial application or physical property on the right.

Click a left item, then click its matching right item

Items

Helium (HeHe)
Neon (NeNe)
Argon (ArAr)
Krypton (KrKr)

Matches

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Answer

Helium matches deep-sea diving breathing mixtures (heliox); Neon matches orange-red advertising discharge tubes; Argon matches inert shielding atmosphere in arc welding; Krypton matches high-efficiency airport runway lighting.
Each noble gas possesses distinct physical properties leading to specific applications: Helium's low blood solubility makes it essential for diving mixtures; Neon's electrical excitation spectrum yields orange-red sign lighting; Argon's abundance and chemical inertness provide a protective shield during welding; Krypton's high atomic mass improves filament life in specialized high-intensity lighting.

Step-by-Step Solution

1
Identify the low solubility property of Helium
Helium replaces nitrogen in deep-sea breathing gas (heliox).
Prevents painful decompression sickness because helium is significantly less soluble in human blood under high pressure.
2
Determine the atomic spectrum property of Neon
Neon produces a characteristic orange-red glow in gas discharge lamps.
Electron transitions in excited neon gas release photons with wavelengths corresponding to reddish-orange light.
3
Analyze the industrial application of Argon in metallurgy
Argon serves as an inert protective blanket in electric arc welding.
Being non-reactive and atmospheric abundant, it displaces atmospheric oxygen and nitrogen during metal joining.
4
Relate Krypton's atomic mass to incandescent lighting efficiency
Krypton is used in high-intensity airport runway bulbs.
Heavy noble gas atoms retard the thermal evaporation of tungsten filaments.

Key Concept

Specific industrial applications and unique physical/chemical characteristics of noble gases (Group 18).
Estimated Time:1m 30s
Question 8802Question

In humans, Duchenne muscular dystrophy is inherited as an X-linked recessive disorder (XdX^d), while the normal allele is dominant (XDX^D). A phenotypically normal woman seeks genetic counseling. Her maternal grandfather had Duchenne muscular dystrophy, whereas her maternal grandmother was homozygous normal. Her father is phenotypically normal. If this woman marries a phenotypically normal man, what is the probability (expressed as a percentage) that their first male child will be affected by the disorder?

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Answer: 25

Answer

The probability that their first male child will be affected by Duchenne muscular dystrophy is 25%.
The maternal grandfather (XdYX^d Y) passes his XdX^d chromosome to his daughter (the woman's mother), making her an obligate carrier (XDXdX^D X^d). When this carrier mother has a daughter with a normal male (XDYX^D Y), the daughter has a 50%50\% (0.50.5) chance of being a carrier (XDXdX^D X^d). If the woman is a carrier, any male child she has has a 50%50\% (0.50.5) chance of receiving the XdX^d allele and being affected. Multiplying these independent probabilities (0.5×0.50.5 \times 0.5) yields 0.250.25, or 25%25\%.

Step-by-Step Solution

1
Determine the genotype of the woman's mother from her maternal grandparents.
The woman's mother inherited XdX^d from her father (XdYX^d Y) and XDX^D from her mother (XDXDX^D X^D), making her an obligate carrier (XDXdX^D X^d).
Fathers always pass their single X chromosome to their daughters.
2
Calculate the probability that the woman inherited the recessive allele from her mother.
Probability that the woman is a carrier (XDXdX^D X^d) is 0.50.5 (or 50%50\%).
A carrier mother (XDXdX^D X^d) and normal father (XDYX^D Y) have a 50%50\% chance of producing a carrier daughter.
3
Calculate the probability that a male child of a carrier woman receives the recessive X-linked allele.
If the woman is a carrier, the probability of an affected son (XdYX^d Y) is 0.50.5 (or 50%50\%).
A male child receives his only X chromosome from his mother.
4
Multiply the independent probabilities to find the overall risk for the first male child.
P(Affected male child)=0.5 (mother is carrier)×0.5 (son inherits Xd)=0.25=25%P(\text{Affected male child}) = 0.5 \text{ (mother is carrier)} \times 0.5 \text{ (son inherits } X^d) = 0.25 = 25\%.
Both independent events (mother being a carrier and son inheriting the mutated allele) must occur.

Key Concept

Sex-Linked Recessive Inheritance and Pedigree Carrier Probability
Question 8803Question

In an ecosystem, energy is lost as heat at each progressive trophic level according to the laws of thermodynamics. Which of the following ecological pyramids is ALWAYS upright in shape across all natural ecosystems?

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Answer: Pyramid of energy

Answer

The pyramid of energy is the only ecological pyramid that is always upright in shape across all functional ecosystems.
The correct answer is the pyramid of energy because energy transfer between trophic levels is inefficient due to metabolic loss (heat, respiration, non-consumed materials). Consequently, each successive trophic level inevitably contains less usable energy than the level below it, keeping the pyramid perpetually upright.

Step-by-Step Solution

1
Analyze energy flow thermodynamics
Energy enters primary producers via photosynthesis and is transferred through consumer levels, with approximately 90% lost as metabolic heat and waste at each step.
The Second Law of Thermodynamics dictates that energy transformations are inefficient, ensuring higher trophic levels always receive less energy than lower levels.
2
Evaluate pyramid representations
Because energy flow is strictly unidirectional and decreases continuously, a pyramid of energy can never be inverted or spindle-shaped.
Pyramids of numbers and biomass measure static quantities at one instant and can be inverted, whereas energy pyramids represent rate of energy flow over time.

Key Concept

Thermodynamic energy loss in trophic pyramids
Estimated Time:45s
Question 8804Question

Immunological serological tests demonstrate high precipitin cross-reactivity when human antibodies are mixed with chimpanzee serum proteins, indicating close structural homology resulting from a recent common evolutionary ancestor. Is this statement True or False?

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Answer: True

Answer

True
The statement is true because comparative serology utilizes antigen-antibody reactions to quantify biochemical similarity. High precipitin formation between human anti-serum and chimpanzee serum proteins demonstrates that their plasma proteins share extremely similar amino acid sequences, providing strong molecular evidence for a recent common ancestor.

Step-by-Step Solution

1
Analyze the principle of comparative serology in evolutionary biochemistry.
Serum proteins (such as albumin and globulins) evolve through gene mutations over long periods. Closely related species share highly similar protein structures.
Fewer structural protein differences accumulate when organisms share a recent common ancestor.
2
Evaluate the mechanism of the precipitin test in cross-reactivity.
Antibodies produced against human serum proteins bind specifically to homologous protein epitopes in chimpanzee serum, causing high precipitation.
Antigen-antibody binding affinity increases with protein structural similarity.
3
Determine the validity of the statement based on biochemical evidence.
High cross-reactivity confirms a high degree of biochemical homology and recent shared ancestry.
The statement accurately describes the biochemical evidence for primate evolution.

Key Concept

Comparative serology and immunological cross-reactivity as evidence for evolutionary relationships
Question 8805Question

Which of the following characteristics is a defining feature of discontinuous variation in a biological population?

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Answer: The presence of clear-cut, distinct phenotypic categories with no intermediate forms

Answer

The presence of clear-cut, distinct phenotypic categories with no intermediate forms
Discontinuous variation produces non-overlapping phenotypic classes with distinct differences and no intermediate stages between them. Examples include human ABO blood groups, ability to roll the tongue, and presence or absence of horns in cattle.

Step-by-Step Solution

1
Define discontinuous variation in genetic terms.
Discontinuous variation refers to phenotypic traits that are split into distinct, non-overlapping classes.
Understanding the core definition differentiates it from continuous variation.
2
Evaluate the option choices against key features of discontinuous variation.
Discontinuous traits are typically monogenic (controlled by one gene), unaffected by environment, and show discrete phenotypic categories without intermediate states.
Traits like blood groups or sex determination are either present or absent in fixed categories.

Key Concept

Discontinuous Variation Features
Question 8806Question

In maize (*Zea mays*), purple endosperm color (PP) is dominant over yellow endosperm color (pp), and starchy kernel texture (SS) is dominant over waxy kernel texture (ss). A plant heterozygous for both traits (PpSsPpSs) is allowed to self-pollinate, producing 480 kernels in the F2F_2 generation. How many of these kernels are expected to display the purple endosperm and waxy kernel texture phenotype?

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Answer: 90

Answer

90 kernels are expected to show the purple endosperm and waxy kernel texture phenotype.
In a dihybrid cross between two heterozygous individuals (PpSs×PpSsPpSs \times PpSs), the offspring phenotypes segregate in a classical 9:3:3:1 ratio. The phenotype representing one dominant trait and one recessive trait (purple endosperm and waxy texture, P_ssP\_ss) occurs with a frequency of 316\frac{3}{16}. Out of 480 kernels, 316×480=90\frac{3}{16} \times 480 = 90 kernels will manifest this specific phenotype.

Step-by-Step Solution

1
Determine parental cross and gametes
The self-pollinated parent is PpSsPpSs. Gametes formed are PSPS, PsPs, pSpS, and psps in equal proportions.
Mendel's Law of Independent Assortment states that alleles of different genes segregate independently during gamete formation.
2
Calculate the expected phenotypic fraction for purple and waxy kernels
Probability of purple (P_P\_) = 34\frac{3}{4}; Probability of waxy (ssss) = 14\frac{1}{4}. Combined probability = 34×14=316\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}.
Because the two genes assort independently, the joint probability is the product of their individual probabilities.
3
Calculate expected count from total offspring
Expected count = 316×480=90\frac{3}{16} \times 480 = 90.
Multiply the expected phenotypic frequency by the total kernel population size.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid F2F_2 Phenotypic Calculations
Question 8807Question

Match each respiratory process with its corresponding characteristic end-products.

Click a left item, then click its matching right item

Items

Glycolysis
Krebs cycle
Electron transport chain
Alcoholic fermentation

Matches

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Answer

Glycolysis matches with Pyruvate, ATP, and NADH; Krebs cycle matches with Carbon dioxide, ATP, NADH, and FADH2; Electron transport chain matches with Water and a high yield of ATP; Alcoholic fermentation matches with Ethanol, carbon dioxide, and ATP.
Each respiratory metabolic pathway yields distinct chemical end-products. Glycolysis yields pyruvate, ATP, and NADH. The Krebs cycle produces carbon dioxide, ATP, NADH, and FADH2. The electron transport chain synthesizes water and a high yield of ATP. Alcoholic fermentation produces ethanol, carbon dioxide, and ATP.

Step-by-Step Solution

1
Identify the primary end-products of glycolysis.
Glycolysis splits one glucose molecule into two pyruvate molecules while producing net 2 ATP2\text{ ATP} and 2 NADH2\text{ NADH}.
This represents the initial cytoplasm-based stage of glucose degradation.
2
Identify the main products of the Krebs cycle.
The breakdown of acetyl-CoA in the mitochondrial matrix releases CO2\text{CO}_2 along with reduced electron carriers (NADH and FADH2) and ATP.
This accounts for the complete decarboxylation and oxidation of carbon intermediates.
3
Determine the output of the electron transport chain.
Electrons passed to oxygen form H2O\text{H}_2\text{O}, driving oxidative phosphorylation to generate the bulk of ATP.
Oxygen serves as the final electron acceptor in aerobic respiration.
4
Match anaerobic alcoholic fermentation with its characteristic products.
In yeast, anaerobic pathway breakdown yields ethyl alcohol (ethanol), carbon dioxide gas, and ATP.
Fermentation regenerates NAD+ necessary to keep glycolysis operational without oxygen.

Key Concept

Cellular Respiration Pathways and End-Products
Question 8808Question

Match each ecological measuring instrument in the left column with the corresponding abiotic factor it measures in the right column.

Click a left item, then click its matching right item

Items

Hygrometer
Anemometer
Secchi disc
Barometer

Matches

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Answer

Hygrometer pairs with Relative humidity; Anemometer pairs with Wind speed; Secchi disc pairs with Water turbidity and light penetration; Barometer pairs with Atmospheric pressure.
Each measuring instrument is correctly matched to its specific environmental variable: Hygrometer to relative humidity, Anemometer to wind speed, Secchi disc to aquatic turbidity and transparency, and Barometer to atmospheric pressure.

Step-by-Step Solution

1
Determine the parameter measured by a hygrometer.
A hygrometer quantifies moisture levels in the atmosphere.
Atmospheric moisture level is referred to as relative humidity.
2
Determine the parameter measured by an anemometer.
An anemometer quantifies the rate of airflow in terrestrial environments.
Airflow velocity is defined as wind speed.
3
Determine the parameter measured by a Secchi disc.
A Secchi disc measures clarity in aquatic environments based on visual disappearance depth.
Clarity in aquatic environments corresponds to water transparency or turbidity.
4
Determine the parameter measured by a barometer.
A barometer measures force exerted per unit area by the weight of air above.
This force per unit area is atmospheric pressure.

Key Concept

Measurement of Abiotic Ecological Factors
Estimated Time:1m 0s
Question 8809Question

In humans, red-green color blindness is an X-linked recessive trait (XcX^c), whereas normal vision is controlled by the dominant allele (XCX^C). Albinism is an autosomal recessive disorder (aa), whereas normal skin pigmentation is controlled by the dominant allele (AA). A woman with normal vision and normal skin pigmentation, whose father was both color-blind and albino, marries a man with normal vision who is a carrier for albinism. If this couple produces a male child (son), what is the percentage probability that the son will be both color-blind and albino?

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Answer: 12.5

Answer

The percentage probability that a son born to this couple will be both color-blind and albino is 12.5%.
The mother's father was albino (aaaa) and color-blind (XcYX^c Y), meaning she inherited aa and XcX^c from him. Given her normal phenotype, her genotype is AaXCXcAa X^C X^c. The father is AaXCYAa X^C Y. When determining traits for a son, the son receives the YY chromosome from the father, so his vision phenotype depends entirely on which XX chromosome he receives from his mother (50% chance of XcX^c). The probability of being albino from two carrier parents (Aa×AaAa \times Aa) is 25% (14\frac{1}{4}). Multiplying these independent probabilities yields 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}, which equals 12.5%.

Step-by-Step Solution

1
Determine parental genotypes from the pedigree information provided.
Mother's genotype: AaXCXcAa X^C X^c; Father's genotype: AaXCYAa X^C Y.
The mother received recessive alleles aa and XcX^c from her affected father (aaXcYaa X^c Y). The father is stated to have normal vision (XCYX^C Y) and to be a carrier for albinism (AaAa).
2
Calculate the probability of the male child inheriting the X-linked color blindness trait.
Probability of color-blind son = 12\frac{1}{2} (50%).
For male offspring, sex is fixed by inheriting the YY chromosome from the father. The mother has a 50% chance of passing her XcX^c allele.
3
Calculate the probability of the child inheriting autosomal albinism.
Probability of albino phenotype (aaaa) = 14\frac{1}{4} (25%).
Crossing two heterozygous carriers (Aa×AaAa \times Aa) yields a 1 in 4 chance of an autosomal recessive aaaa offspring.
4
Apply the product rule for independent genetic events.
Combined probability = 12×14=18=12.5%\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} = 12.5\%.
Autosomal inheritance and X-linked inheritance are independent genetic events, so their probabilities are multiplied.

Key Concept

Independent assortment of an autosomal recessive trait and an X-linked recessive trait in human pedigree analysis.
Question 8810Question

A study conducted on a human population recorded two phenotypic traits: resting systolic blood pressure (Trait I) and the presence or absence of the Rhesus D antigen on red blood cells (Trait II). Trait I produced a smooth, bell-shaped frequency distribution curve across a continuous gradient of values, whereas Trait II resulted in two distinct, non-overlapping categories with no intermediate forms. Which of the following correctly explains the genetic mechanism and environmental susceptibility responsible for these observed patterns of variation?

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Answer: Trait I is polygenic and modified by environmental factors, whereas Trait II is controlled by a single gene locus and unaffected by environmental conditions.

Answer

Trait I represents continuous variation, which is polygenic (controlled by multiple genes) and influenced by environmental factors, whereas Trait II represents discontinuous variation, which is monogenic (controlled by one gene pair or single locus) and independent of environmental changes.
The option stating that Trait I is polygenic and modified by environmental factors, whereas Trait II is controlled by a single gene locus and unaffected by environment is correct. Continuous traits like blood pressure are polygenic (controlled by multiple genes whose effects combine additively) and show significant variation due to external environmental factors like diet and stress, resulting in a continuous bell-shaped curve. In contrast, discontinuous traits like the Rhesus factor are monogenic (determined by alleles at a single locus), resulting in distinct, clear-cut phenotypic classes that environment cannot alter.

Step-by-Step Solution

1
Analyze the phenotypic distribution of Trait I (systolic blood pressure).
The smooth, bell-shaped normal distribution curve indicates continuous variation with quantitative grading between extremes.
Continuous variation occurs when traits are polygenic (governed by multiple additive genes) and sensitive to environmental influences.
2
Analyze the phenotypic distribution of Trait II (Rhesus antigen presence/absence).
Two distinct, non-overlapping categories with no intermediate forms indicate discontinuous variation.
Discontinuous variation is driven by monogenic inheritance (a single gene pair or major locus) and is largely unaffected by environmental conditions.
3
Synthesize the genetic mechanisms and environmental impacts for both traits.
Trait I is polygenic and environmentally modified, while Trait II is monogenic and environmentally stable.
Matching phenotypic distribution patterns (bell-shaped vs discrete categories) directly to their genetic architecture resolves the question.

Key Concept

Polygenic inheritance causing continuous variation versus monogenic inheritance causing discontinuous variation
Question 8811Question

Match each core Lamarckian evolutionary postulate or related historical criticism on the left with its accurate biological mechanism or empirical evaluation on the right.

Click a left item, then click its matching right item

Items

Principle of Use and Disuse
Inheritance of Acquired Characteristics
Internal Vital Impulse
Weismann's Germplasm Barrier

Matches

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Answer

Principle of Use and Disuse pairs with organ hypertrophy or atrophy from functional demand; Inheritance of Acquired Characteristics pairs with direct transmission of somatic changes to offspring; Internal Vital Impulse pairs with the proposed innate drive for complexity from non-living matter; Weismann's Germplasm Barrier pairs with the experimental refutation showing somatic changes do not affect germ line genetics.
Each concept correctly aligns with its historical and biological definition: Use and Disuse describes organ changes within a lifespan; Inheritance of Acquired Characteristics describes the proposed transfer of those changes to offspring; Internal Vital Impulse accounts for the drive toward complexity; and Weismann's Germplasm Barrier provides the classical empirical refutation distinguishing germline inheritance from somatic changes.

Step-by-Step Solution

1
Analyze the primary tenets of Lamarckism
Identify Use and Disuse as somatic modification during life, Inheritance of Acquired Traits as intergenerational transfer of those modifications, and Vital Impulse as the inherent drive for complexity.
Lamarck's theory relies on these distinct physiological and evolutionary mechanisms.
2
Evaluate historical scientific critiques of Lamarckian mechanisms
Recognize Weismann's experiment as establishing the barrier between germline (hereditary) and soma (body) cells.
Modern genetics refutes Lamarckism because somatic adaptations do not alter gametic DNA.
3
Match each postulate and critique to its corresponding definition
Connect left item 1 to right item 3, left item 2 to right item 4, left item 3 to right item 1, and left item 4 to right item 2.
Ensures precise conceptual mapping based on evolutionary biological definitions.

Key Concept

Lamarckian Postulates and Historical Refutation
Question 8812Question

A single virion contains both DNA and RNA concurrently enclosed within its protein capsid to enable independent protein synthesis outside a living host cell.

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Answer: False

Answer

The statement is False. A virus contains either DNA or RNA as its genetic material, never both concurrently, and lacks metabolic machinery for independent protein synthesis.
The statement is false because virions carry either DNA or RNA as their genetic material, never both within the same capsid, and possess no cellular machinery to perform independent protein synthesis.

Step-by-Step Solution

1
Examine the genomic architecture of viruses.
Viruses possess a core of genetic material composed of either single-stranded or double-stranded DNA, or single-stranded or double-stranded RNA, but never both nucleic acids within the same virion.
This single-type nucleic acid genome is a defining biochemical feature separating viruses from cellular organisms.
2
Assess the metabolic and cellular capabilities of viruses outside host cells.
Viruses lack cytoplasm, cellular organelles (such as ribosomes and mitochondria), and autonomous metabolic pathways.
Because they lack translational machinery, viruses are obligate intracellular parasites that cannot synthesize proteins independently.

Key Concept

Viral Genome Composition and Acellular Inertness
Question 8813Question

A waterlogged agricultural field experiences prolonged anaerobic conditions following excessive irrigation. Laboratory analysis of the soil reveals a marked decline in soil nitrate (NO3NO_3^-) concentration accompanied by a corresponding release of dinitrogen gas (N2N_2) into the atmosphere. Which microbial process and bacterial genus are primarily responsible for this nitrogen transformation?

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Answer: Denitrification by Pseudomonas species

Answer

Denitrification by Pseudomonas species
Denitrification is the anaerobic microbial reduction of soil nitrates (NO3NO_3^-) to atmospheric dinitrogen gas (N2N_2). Species of the bacterial genus Pseudomonas thrive in oxygen-depleted environments such as waterlogged soils, utilizing nitrate during anaerobic respiration and consequently reducing soil nitrogen levels.

Step-by-Step Solution

1
Analyze environmental conditions and chemical change
Waterlogging creates anoxic (anaerobic) soil conditions, leading to the reduction of soil nitrate (NO3NO_3^-) into gaseous dinitrogen (N2N_2).
When oxygen is depleted, facultative anaerobic microbes utilize nitrate as an alternative terminal electron acceptor in cellular respiration.
2
Identify the specific metabolic pathway
The conversion of nitrate (NO3NO_3^-) to gaseous nitrogen (N2N_2) is denitrification.
Denitrification returns fixed soil nitrogen back to the atmospheric reservoir as dinitrogen gas.
3
Match the pathway with the correct microbial agent
Pseudomonas species (and Thiobacillus denitrificans) are classic denitrifying bacteria.
Nitrosomonas performs nitrification (ammonia oxidation), Azotobacter fixes N2N_2 gas, and ammonification decomposes organic residues into ammonium.

Key Concept

Denitrification in Anaerobic Soil Microenvironments
Question 8814Question

During the electrolysis of concentrated hydrochloric acid (HCl(aq)HCl_{(aq)}) using inert graphite electrodes, a gaseous product XX is evolved at the anode while gas YY is evolved at the cathode. Which option correctly identifies gas XX, gas YY, and the primary factor responsible for the preferential discharge that yields gas XX?

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Answer: Gas XX is Cl2Cl_2, Gas YY is H2H_2, driven by the high concentration of chloride ions.

Answer

Gas XX is Cl2Cl_2, Gas YY is H2H_2, driven by the high concentration of chloride ions.
In concentrated hydrochloric acid, the ions migrating to the anode are ClCl^- and OHOH^-. Because the solution is concentrated, the concentration of ClCl^- ions is far higher than that of OHOH^-. Consequently, the concentration factor predominates over position in the electrochemical series, resulting in ClCl^- being preferentially oxidized to chlorine gas (Cl2Cl_2). At the cathode, H+H^+ ions gain electrons to form hydrogen gas (H2H_2).

Step-by-Step Solution

1
Identify the ions present in concentrated hydrochloric acid (HCl(aq)HCl_{(aq)}).
The cations present are H+H^+ (from HClHCl and H2OH_2O) and the anions present are ClCl^- and OHOH^-.
Hydrochloric acid ionizes completely in water into H+H^+ and ClCl^-, alongside minor autoionization of water.
2
Determine the reaction taking place at the anode (positive electrode).
Chloride ions (ClCl^-) are discharged preferentially to form chlorine gas (Cl2Cl_2).
Although OHOH^- is higher in the electrochemical series than ClCl^-, the significantly higher concentration of ClCl^- in concentrated HClHCl overrides position in the series.
3
Determine the reaction taking place at the cathode (negative electrode).
Hydrogen ions (H+H^+) are discharged to form hydrogen gas (H2H_2).
H+H^+ is the only cation present in solution, gaining electrons at the cathode.

Key Concept

Concentration effect on preferential discharge of anions during electrolysis
Question 8815Question

In classic Mendelian genetics, a testcross is carried out by mating an organism exhibiting a dominant phenotype with an individual that is homozygous dominant for the trait in question.

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Answer: False

Answer

The statement is False because a testcross requires mating with a homozygous recessive individual, not a homozygous dominant one.
The statement is false because a testcross relies on a homozygous recessive tester (aaaa) so that the offspring phenotypes directly reveal the gametic contribution and genotype of the dominant parent.

Step-by-Step Solution

1
Identify the biological purpose of a testcross.
A testcross determines whether an organism expressing a dominant trait is homozygous dominant (AAAA) or heterozygous (AaAa).
Organisms with genotypes AAAA and AaAa are phenotypically identical.
2
Determine the required genotype of the tester parent.
The tester individual must be homozygous recessive (aaaa).
A homozygous recessive tester produces only recessive gametes (aa), allowing hidden recessive alleles from the tested parent to be expressed in the offspring's phenotype.
3
Evaluate the effect of using a homozygous dominant tester (AAAA).
Crossing either AAAA or AaAa with AAAA yields 100% dominant phenotype offspring.
The dominant allele from the AAAA tester masks any recessive allele contributed by a heterozygous parent.

Key Concept

Definition and methodology of a genetic testcross
Estimated Time:1m 0s
Question 8816Question

Helium is used in deep-sea diving gas mixtures (heliox) to prevent decompression sickness and in meteorological balloons to provide lift. Which set of properties makes helium suitable for both of these applications?

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Answer: Low solubility in blood under pressure, low density, and non-flammability

Answer

Low solubility in blood under pressure, low density, and non-flammability
Helium's 1s21s^2 stable electronic duplet makes it non-flammable and chemically inert. Its small molar mass (4 g/mol4\text{ g/mol}) makes it less dense than air, providing lift for weather balloons. Furthermore, its exceptionally low solubility in blood under pressure prevents decompression sickness ('the bends') in deep-sea divers.

Step-by-Step Solution

1
Analyze the requirements for deep-sea diving gas mixtures (heliox).
The gas mixed with oxygen must have low solubility in blood at high underwater pressures to prevent nitrogen narcosis and painful bubble formation upon decompression.
Helium replaces nitrogen in heliox because of its minimal blood solubility under pressure.
2
Analyze the requirements for meteorological (weather) balloons.
The gas must be less dense than air to provide upward buoyant force and non-flammable to prevent explosion hazards.
Helium has a molar mass of 4 g/mol (much lower than air's average of 29 g/mol) and a complete 1s21s^2 valence shell, making it non-flammable and safe compared to hydrogen.
3
Synthesize the properties into a single matching choice.
The combination of low blood solubility under pressure, low density, and chemical non-flammability correctly describes helium's behavior.
This set of physical and chemical properties uniquely qualifies helium for both applications.

Key Concept

Properties and Applications of Helium
Estimated Time:1m 0s
Question 8817Question

Unicellular organisms such as *Chlamydomonas* require light to synthesize organic compounds. Which specialized subcellular organelle functions as a light-sensitive region that enables the organism to perceive light and exhibit positive phototaxis?

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Answer: Stigma

Answer

The stigma (also known as the eyespot) is the organelle that perceives light and enables directional movement toward light sources.
The stigma, also referred to as the eyespot, is a pigmented organelle found in unicellular photosynthetic protists like *Chlamydomonas* and *Euglena*. It filters light and allows the cell to perceive illumination intensity and direction, driving positive phototaxis.

Step-by-Step Solution

1
Identify the primary physiological requirement mentioned in the stem.
The organism needs to sense light direction to perform phototaxis toward light for photosynthesis.
Phototaxis relies on a specialized photoreceptor structure.
2
Evaluate the organelle associated with light perception in unicellular algae like *Chlamydomonas*.
The stigma contains carotenoid pigments that filter light and help the cell orient its flagellar swimming toward illumination.
This differentiates the light-perceiving organelle from metabolic or osmoregulatory structures.

Key Concept

Function of the Stigma (Eyespot) in Phototaxis
Question 8818Question

A comparative biological study analyzes three distinct fungal organisms: unicellular yeast (*Saccharomyces*), filamentous bread mould (*Rhizopus*), and a macrofungal mushroom (*Agaricus*). Which of the following statements correctly identifies a fundamental structural and metabolic feature common to all three organisms that distinguishes Kingdom Fungi from plants and bacteria?

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Answer: They possess cell walls composed primarily of chitin and absorb soluble nutrients following extracellular enzymatic digestion.

Answer

They possess cell walls composed primarily of chitin and absorb soluble nutrients following extracellular enzymatic digestion.
The correct answer highlights the two unifying traits of Kingdom Fungi across diverse morphological forms (yeasts, moulds, and mushrooms): a cell wall constructed of chitin and an absorptive mode of nutrition mediated by extracellular enzyme secretion.

Step-by-Step Solution

1
Analyze cell wall composition across kingdoms
Bacterial cell walls contain peptidoglycan, plant cell walls contain cellulose, and fungal cell walls (in yeasts, moulds, and mushrooms) contain chitin.
Chitin is a defining structural marker of Kingdom Fungi.
2
Evaluate nutritional modes of fungal archetypes
Saccharomyces, Rhizopus, and Agaricus all exhibit saprophytic (absorptive) heterotrophy.
They secrete exoenzymes into their surroundings to break down complex polymers into simple soluble compounds, which are then absorbed across their cell membranes.
3
Synthesize structural and physiological criteria to select the correct statement
The presence of chitinous cell walls combined with extracellular saprophytic digestion correctly characterizes all three fungal representatives.
This differentiates fungi from autotrophic plants and peptidoglycan-walled bacteria.

Key Concept

Structural and Nutritional Characteristics of Kingdom Fungi
Estimated Time:1m 30s
Question 8819Question

Analogous structures, such as the camera-type eye of an octopus and the eye of a mammal, share a common embryonic origin and basic structural arrangement inherited from a shared recent ancestor.

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Answer: False

Answer

False. Analogous structures perform similar functions due to convergent evolution but do not share a common embryonic origin or ancestral structure.
The correct answer is False because analogous structures evolve independently to perform similar functions in response to similar environmental demands (convergent evolution), without sharing a common embryonic origin or recent ancestral structure.

Step-by-Step Solution

1
Define the biological concepts of homology and analogy in comparative anatomy.
Homologous structures share a common ancestry and embryonic development regardless of function, whereas analogous structures serve similar functions in different organisms due to environmental pressures but differ in embryonic origin.
Establishing accurate definitions is necessary to evaluate statements regarding evolutionary evidence.
2
Examine the specific anatomical structures presented (octopus eye vs. mammal eye).
Although both eyes perform the function of forming focused images, they develop from distinct embryonic layers (molluscan skin invagination vs. vertebrate neural tube outgrowth) and display key structural differences, such as the orientation of retinal photoreceptors.
Demonstrates that functional similarity in these organs is the result of convergent evolution rather than shared lineage.
3
Determine the truth value of the stem statement.
The statement incorrectly attributes shared embryonic origin and common ancestry to analogous structures.
Because analogous structures do not share a common embryonic origin or ancestral structure, the statement is false.

Key Concept

Distinction between homologous and analogous structures in comparative anatomy
Question 8820Question

An ecologist compared two disturbed ecosystems: Site X, an abandoned agricultural field where topsoil and organic seed banks remained intact, and Site Y, a newly exposed volcanic basalt ledge devoid of organic soil. Which of the following observations correctly contrasts the ecological succession dynamics between Site X and Site Y?

Show answer & explanation

Answer: Site X undergoes secondary succession characterized by rapid colonization from surviving seeds and soil microbes, whereas Site Y undergoes primary succession requiring crustose lichens to initiate substrate weathering.

Answer

Site X undergoes secondary succession characterized by rapid colonization from surviving seeds and soil microbes, whereas Site Y undergoes primary succession requiring crustose lichens to initiate substrate weathering.
The correct answer accurately identifies that Site X undergoes secondary succession because topsoil and organic propagules are already present, allowing for rapid vegetation recovery. Conversely, Site Y undergoes primary succession on bare volcanic rock, where pioneer species like crustose lichens must first colonize, secrete organic acids to weather substrate, and accumulate organic matter before vascular plants can establish.

Step-by-Step Solution

1
Analyze the starting substrate conditions of Site X and Site Y.
Site X has pre-existing topsoil, organic matter, and seed banks (abandoned agricultural field). Site Y has bare volcanic basalt with no pre-existing soil.
The presence or absence of pre-existing soil determines whether succession is primary or secondary.
2
Determine the type of ecological succession and pioneer organisms for each site.
Site X undergoes secondary succession (rapid, starts with grasses/herbaceous plants). Site Y undergoes primary succession (slow, starts with lichens/mosses to form soil).
Secondary succession builds upon established soil, whereas primary succession must build soil from uncolonized substrate.
3
Evaluate the option statements against established ecological principles.
The statement identifying Site X as secondary succession and Site Y as primary succession initiated by lichen weathering is correct.
Lichens secrete organic acids that weather bare rock into soil, enabling subsequent plant establishment during primary succession.

Key Concept

Distinction between primary and secondary ecological succession based on initial soil availability and pioneer community traits
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