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13931 questions

Question 4781Question

Match each mammalian reproductive cell or gland with its primary physiological function.

Click a left item, then click its matching right item

Items

Seminal vesicles
Cowper's glands
Sertoli cells
Leydig cells

Matches

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Answer

Seminal vesicles match with secreting fructose-rich alkaline fluid supplying energy for sperm motility; Cowper's glands match with secreting mucus fluid to lubricate the urethra and neutralize acidic urine traces; Sertoli cells match with nourishing developing spermatids and establishing the blood-testis barrier; Leydig cells match with synthesizing and secreting testosterone in response to luteinizing hormone.
Seminal vesicles produce fructose-rich fluid powering sperm movement. Cowper's glands secrete pre-ejaculatory alkaline mucus neutralizing urethral acidity. Sertoli cells provide physical and nutritional support to developing gametes while forming the blood-testis barrier. Leydig cells produce testosterone under LH control.

Step-by-Step Solution

1
Identify the main function of accessory seminal vesicles.
Seminal vesicles produce fructose, prostaglandins, and alkaline fluid which nourish sperm and aid motility.
Sperm cells require carbohydrate substrates (fructose) for aerobic respiration to power flagellar movement.
2
Analyze the protective role of Cowper's (bulbourethral) glands.
Cowper's glands release a pre-ejaculatory clear fluid into the urethra.
This fluid neutralizes residual acidic urine in the shared urogenital tract prior to sperm passage.
3
Distinguish between intratubular Sertoli cells and interstitial Leydig cells in the testis.
Sertoli cells act as nurse cells forming tight junctions for the blood-testis barrier, while Leydig cells reside outside the tubules to synthesize testosterone.
Germ cells need immune privilege provided by Sertoli cells, while systemic male secondary traits and spermatogenesis maintenance depend on Leydig cell testosterone secretion.

Key Concept

Structure and function of male mammalian reproductive glands and testicular cells
Question 4782Question

During selective reabsorption in the mammalian nephron, glucose and amino acids are completely reabsorbed back into the bloodstream from the glomerular filtrate. In which specific region of the nephron does this process primarily take place, and what structural feature enables it?

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Answer: Proximal convoluted tubule, supported by a dense brush border of microvilli and abundant mitochondria

Answer

Proximal convoluted tubule, supported by a dense brush border of microvilli and abundant mitochondria
Selective reabsorption of glucose, amino acids, and essential ions occurs predominantly in the proximal convoluted tubule (PCT). The epithelial cells lining the PCT are structurally adapted for intensive active transport: they possess a microvillar brush border that vastly increases the absorptive surface area and contain numerous mitochondria to generate the ATP required for active carrier-mediated transport back into the peritubular capillaries.

Step-by-Step Solution

1
Identify the primary site of selective reabsorption in the nephron
Selective reabsorption of essential nutrients like glucose and amino acids occurs in the proximal convoluted tubule (PCT).
Over 65-70% of filtrate volume, including 100% of glucose and amino acids under normal conditions, is reabsorbed immediately after leaving Bowman's capsule.
2
Identify the cellular adaptations required for active transport in the PCT
The epithelial cells of the PCT possess a brush border of microvilli to maximize surface area and high density of mitochondria.
Active transport requires cellular energy in the form of ATP generated by mitochondria, and microvilli dramatically increase the absorption surface area.

Key Concept

Nephron Physiology and Selective Reabsorption
Question 4783Question

A commercial farm in Kaduna operates with fixed total resources of 1010 hectares of arable land and 120120 worker-hours of labor per day. Producing 11 ton of maize requires 11 hectare of land and 1212 worker-hours of labor. Producing 11 ton of yams requires 0.50.5 hectares of land and 2020 worker-hours of labor. Initially, the farm devotes all its resources to maximize maize production, harvesting 1010 tons of maize per day. If the farm owner decides to reallocate resources to produce 33 tons of yams per day, what is the opportunity cost of this decision expressed in tons of maize foregone?

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Answer: 5

Answer

The opportunity cost of producing 3 tons of yams is 5 tons of maize.
Producing 3 tons of yams absorbs 60 worker-hours out of the total 120 worker-hours available. The remaining 60 worker-hours can only produce 5 tons of maize. Since initial maize production was 10 tons, the farm gives up 5 tons of maize (10 - 5 = 5 tons).

Step-by-Step Solution

1
Calculate resources consumed by the production of 3 tons of yams.
Land consumed = 3×0.5=1.53 \times 0.5 = 1.5 hectares. Labor consumed = 3×20=603 \times 20 = 60 worker-hours.
Opportunity cost depends on the amount of productive inputs diverted away from maize production.
2
Determine the remaining inputs available for maize production.
Remaining land = 101.5=8.510 - 1.5 = 8.5 hectares. Remaining labor = 12060=60120 - 60 = 60 worker-hours.
Maize can only be produced using the unallocated land and labor inputs.
3
Identify the binding constraint for maize output.
Labor restricts maize output to 6012=5\frac{60}{12} = 5 tons (since land would permit 8.58.5 tons). Thus, maximum feasible maize output is 55 tons.
Production is restricted by the scarcest resource (labor).
4
Subtract the new maximum maize output from the initial maize output to find the opportunity cost.
10 tons5 tons=5 tons of maize foregone10 \text{ tons} - 5 \text{ tons} = 5 \text{ tons of maize foregone}.
Opportunity cost is defined as the quantity of the alternative good given up.

Key Concept

Opportunity cost with constrained multi-resource allocation
Question 4784Question

A commercial flour milling enterprise located in an industrial estate in Kaduna increases all of its production inputs—both capital and labor—by 100%100\%. Due to internal managerial communication bottlenecks and operational delays, total output increases by only 65%65\%. Concurrently, the regional government constructs a shared high-capacity grain silo and transport infrastructure hub that reduces raw material storage and freight costs for all milling firms operating in that industrial estate. Which of the following statements correctly categorizes the firm's internal production relationship and the cost benefit derived from the regional infrastructure?

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Answer: The firm experiences internal decreasing returns to scale, while benefitting from external economies of scale.

Answer

The firm experiences internal decreasing returns to scale, while benefitting from external economies of scale.
The correct answer identifies that doubling all inputs (100%100\%) leads to a less-than-proportional increase in output (65%65\%), which by definition represents internal decreasing returns to scale. Furthermore, cost advantages arising from public infrastructure provided to an entire industrial cluster represent external economies of scale.

Step-by-Step Solution

1
Analyze the long-run input-output relationship of the firm
Inputs (capital and labor) increase by 100%100\%, whereas output increases by 65%65\%.
When all inputs change in equal proportion in the long run and the resulting percentage change in output is less than the percentage change in inputs (%ΔOutput<%ΔInputs\% \Delta \text{Output} < \% \Delta \text{Inputs}), the firm exhibits decreasing returns to scale (diseconomies of scale internally).
2
Analyze the cost advantage resulting from the shared grain silo and logistics hub
The cost reductions apply to all milling firms operating within the industrial estate due to industry localization and government infrastructure.
Cost reductions that accrue to an entire industry or cluster of firms as a result of external factors (such as regional infrastructure or localization of industry) are classified as external economies of scale.
3
Synthesize the internal and external cost dynamics
The firm exhibits internal decreasing returns to scale combined with external economies of scale.
Combining the firm-specific input-output ratio with the industry-wide external cost savings yields this exact classification.

Key Concept

Returns to Scale and Internal vs. External Economies of Scale
Question 4785Question

An uncatalyzed endothermic reaction has a forward activation energy of 120 kJ mol1120\text{ kJ mol}^{-1} and an overall enthalpy change (ΔH\Delta H) of +50 kJ mol1+50\text{ kJ mol}^{-1}. In the presence of a positive catalyst, the activation energy of the forward reaction is reduced by 40 kJ mol140\text{ kJ mol}^{-1}. What is the activation energy for the reverse reaction in the presence of the catalyst?

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Answer: 30 kJ mol130\text{ kJ mol}^{-1}

Answer

The activation energy for the catalyzed reverse reaction is 30 kJ mol130\text{ kJ mol}^{-1}.
For an endothermic reaction, the activation energy of the reverse reaction is equal to the forward activation energy minus the enthalpy change (Ea,reverse=Ea,forwardΔHE_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H). First, determine the forward activation energy under catalyzed conditions: 120 kJ mol140 kJ mol1=80 kJ mol1120\text{ kJ mol}^{-1} - 40\text{ kJ mol}^{-1} = 80\text{ kJ mol}^{-1}. Then, subtract the enthalpy change: 80 kJ mol150 kJ mol1=30 kJ mol180\text{ kJ mol}^{-1} - 50\text{ kJ mol}^{-1} = 30\text{ kJ mol}^{-1}.

Step-by-Step Solution

1
Calculate the forward activation energy of the catalyzed reaction.
Ea,forward, catalyzed=120 kJ mol140 kJ mol1=80 kJ mol1E_{a,\text{forward, catalyzed}} = 120\text{ kJ mol}^{-1} - 40\text{ kJ mol}^{-1} = 80\text{ kJ mol}^{-1}.
A positive catalyst lowers the activation energy barrier.
2
Relate forward activation energy, reverse activation energy, and enthalpy change.
Ea,reverse=Ea,forwardΔHE_{a,\text{reverse}} = E_{a,\text{forward}} - \Delta H.
For an endothermic reaction, the energy of products is higher than reactants by ΔH\Delta H.
3
Substitute the catalyzed forward activation energy and enthalpy change into the relation.
Ea,reverse, catalyzed=80 kJ mol150 kJ mol1=30 kJ mol1E_{a,\text{reverse, catalyzed}} = 80\text{ kJ mol}^{-1} - 50\text{ kJ mol}^{-1} = 30\text{ kJ mol}^{-1}.
Subtracting ΔH\Delta H from the catalyzed forward activation energy gives the required reverse activation energy.

Key Concept

Activation energy relationship for forward and reverse reactions with catalysts
Estimated Time:1m 15s
Question 4786Question

A gaseous hydrocarbon rapidly decolourizes bromine water in tetrachloromethane, but fails to form a precipitate when bubbled through an ammoniacal solution of silver nitrate. Which of the following hydrocarbons exhibits this chemical behavior?

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Answer: Propene

Answer

Propene
Propene is an alkene (CH3CH=CH2CH_3CH=CH_2). Its carbon-carbon double bond allows it to undergo electrophilic addition with bromine water, decolourizing the reddish-brown solution. However, because it is an alkene rather than a terminal alkyne, it lacks an acidic hydrogen atom bonded to a triply-bonded carbon atom, so it does not produce a precipitate when treated with ammoniacal silver nitrate.

Step-by-Step Solution

1
Analyze the reaction with bromine water.
Rapid decolourization of bromine water confirms the presence of unsaturation (a double or triple carbon-carbon bond). Saturated hydrocarbons like propane are eliminated.
Unsaturated hydrocarbons undergo addition reactions across their double or triple bonds to absorb bromine.
2
Analyze the outcome with ammoniacal silver nitrate solution.
The absence of a precipitate indicates that the hydrocarbon is NOT a terminal alkyne (RCCHR-\text{C}\equiv\text{C}-\text{H}).
Only terminal alkynes possess acidic acetylenic hydrogen atoms capable of being replaced by silver ions to form insoluble metallic acetylides.
3
Identify the hydrocarbon matching both criteria.
Propene (CH3CH=CH2CH_3CH=CH_2) is an alkene. It decolourizes bromine water due to unsaturation but yields a negative test with ammoniacal silver nitrate.
Alkenes contain carbon-carbon double bonds but lack acidic acetylenic hydrogen atoms.

Key Concept

Distinguishing alkenes from terminal alkynes using unsaturation and acetylenic hydrogen reagents
Estimated Time:1m 0s
Question 4787Question

A commercial farming enterprise in Oyo State uses its fixed land and labor resources to produce Cassava and Yam. The table below presents its monthly production possibility schedule:

CombinationCassava (bags)Yam (bags)
P1200
Q9045
R5080
S0100

What is the opportunity cost of increasing Yam production from 4545 bags to 8080 bags, expressed in bags of Cassava foregone?

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Answer: 40

Answer

The opportunity cost of increasing Yam production from 45 bags to 80 bags is 40 bags of Cassava.
Increasing Yam production from 4545 bags to 8080 bags requires moving from Combination Q to Combination R. At Combination Q, Cassava production is 9090 bags, whereas at Combination R, Cassava production decreases to 5050 bags. The opportunity cost is the foregone Cassava production, calculated as 9050=4090 - 50 = 40 bags of Cassava.

Step-by-Step Solution

1
Locate the initial production combination
At Combination Q, the firm produces 45 bags of Yam and 90 bags of Cassava.
Opportunity cost measures what must be sacrificed when shifting resources from one alternative to another.
2
Locate the target production combination
At Combination R, the firm produces 80 bags of Yam and 50 bags of Cassava.
Increasing Yam production from 45 to 80 bags requires moving along the schedule from Q to R.
3
Calculate the quantity of the sacrificed commodity (Cassava)
Opportunity Cost = 90 - 50 = 40 bags of Cassava.
The opportunity cost is the explicit quantity of Cassava sacrificed to gain the additional 35 bags of Yam.

Key Concept

Opportunity Cost from Production Schedule
Question 4788Question

An illuminated suspension of green algae undergoing steady-state photosynthesis was supplied with radioactively labeled carbon dioxide (14CO2^{14}CO_2). When the light source was abruptly turned off while maintaining the supply of 14CO2^{14}CO_2, analysis revealed an immediate accumulation of glycerate-3-phosphate (PGA) alongside a rapid decline in ribulose 1,5-bisphosphate (RuBP). Which of the following physiological mechanisms explains why glycerate-3-phosphate accumulates in the absence of light?

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Answer: The reduction of glycerate-3-phosphate to glyceraldehyde-3-phosphate requires ATP and NADPH from the light-dependent stage, whereas carbon dioxide fixation into glycerate-3-phosphate proceeds temporarily using remaining ribulose 1,5-bisphosphate.

Answer

The reduction of glycerate-3-phosphate to glyceraldehyde-3-phosphate requires ATP and NADPH synthesized during the light-dependent phase, while carbon dioxide fixation onto ribulose 1,5-bisphosphate continues until the pool of ribulose 1,5-bisphosphate is depleted.
In photosynthesis, the light-dependent reactions in the thylakoids yield ATP and NADPH. These assimilation products drive the reduction of glycerate-3-phosphate (PGA) to glyceraldehyde-3-phosphate (G3P/GALP) in the stroma. When light is removed, ATP and NADPH levels drop immediately. Carbon dioxide fixation onto ribulose 1,5-bisphosphate (RuBP) continues briefly using residual RuBP, forming PGA. However, because PGA cannot be reduced without ATP and NADPH, PGA accumulates while RuBP is consumed and cannot be regenerated.

Step-by-Step Solution

1
Identify the chemical requirements of the light-independent stage (Calvin cycle).
Carbon fixation converts carbon dioxide (CO2CO_2) and ribulose 1,5-bisphosphate (RuBP) into glycerate-3-phosphate (PGA) via RuBisCO.
Carbon fixation itself does not directly consume ATP or NADPH.
2
Analyze the effect of eliminating the light source.
Light-dependent reactions stop, cutting off the immediate production of ATP and reduced NADP (NADPH).
ATP and NADPH are photochemically generated in the thylakoid membranes.
3
Evaluate the metabolic bottleneck caused by the absence of light energy products.
The reduction step converting PGA to glyceraldehyde-3-phosphate (GALP/G3P) halts due to lack of ATP and NADPH, leading to PGA accumulation and RuBP depletion.
RuBP regeneration also requires ATP; hence, existing RuBP is converted to PGA but cannot be regenerated.

Key Concept

Interdependence of Photosynthetic Light and Dark Reactions
Question 4789Question

A commercial bakery operating in a small community decides against dividing its production process into specialized sub-tasks because the local population buys only a small quantity of bread daily. Which factor is primarily responsible for limiting the practice of division of labor in this bakery?

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Answer: The extent of the market

Answer

The extent of the market primarily limits the practice of division of labor.
The extent of the market determines the volume of goods a firm can sell. If demand is limited, breaking down production into highly specialized repetitive tasks leads to underutilization of workers and operational inefficiency.

Step-by-Step Solution

1
Identify the primary constraint presented in the scenario.
The bakery faces low daily demand for its product from the local community.
Specialization requires continuous mass production to justify dividing labor into discrete, narrow tasks.
2
Relate the constraint to economic principles of production.
Division of labor is limited by the extent of the market.
When total market demand is small, workers assigned to narrow tasks would spend most of their time idle.

Key Concept

Division of labor is limited by the extent of the market.
Estimated Time:45s
Question 4790Question

Match each type of gene mutation or chromosomal aberration on the left with its corresponding description on the right.

Click a left item, then click its matching right item

Items

Substitution mutation
Frameshift mutation
Aneuploidy
Polyploidy

Matches

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Answer

Substitution mutation matches replacement of a single nucleotide base; Frameshift mutation matches addition or deletion altering the reading frame; Aneuploidy matches gain or loss of specific individual chromosomes (2n+12n+1 or 2n12n-1); Polyploidy matches possession of extra complete chromosome sets (3n3n or 4n4n).
Substitution mutation corresponds to swapping a single nucleotide base for another. Frameshift mutation occurs when additions or deletions of nucleotides shift the triplet codon reading sequence. Aneuploidy describes the loss or gain of individual chromosomes, while polyploidy describes organisms having additional full sets of chromosomes.

Step-by-Step Solution

1
Distinguish between gene mutations and numerical chromosomal aberrations.
Substitution and frameshift are point gene mutations, while aneuploidy and polyploidy involve changes in chromosome number.
Gene mutations affect nucleotide sequences within a gene, whereas numerical aberrations alter chromosome counts.
2
Identify the mechanisms of point mutations.
Substitution replaces a single base, whereas frameshift adds or deletes bases shifting downstream codons.
Codons are read in triplets, so non-multiple-of-three insertions or deletions disrupt all subsequent amino acid translation.
3
Distinguish between aneuploidy and polyploidy.
Aneuploidy affects individual chromosome counts (2n±12n \pm 1), whereas polyploidy involves entire extra sets of chromosomes (3n,4n3n, 4n).
Non-disjunction of a single chromosome pair causes aneuploidy, whereas failure of spindle formation during total cell division leads to polyploidy.

Key Concept

Gene Mutations vs Chromosomal Numerical Aberrations
Question 4791Question

The table below presents the utility schedule for a consumer consuming successive units of commodity XX:

Units of Commodity XX (QQ)Total Utility (TUTU in utils)
118
232
342
448
548

What is the marginal utility (MUMU) derived from consuming the 4th4^{\text{th}} unit of commodity XX?

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Answer: 66 utils

Answer

The marginal utility derived from consuming the 4th unit of commodity X is 6 utils.
Marginal utility (MUMU) is defined as the addition to total utility derived from the consumption of an extra unit of a commodity (MU=ΔTUΔQMU = \frac{\Delta TU}{\Delta Q}). For the 4th unit, MU=TU4TU3=4842=6MU = TU_4 - TU_3 = 48 - 42 = 6 utils.

Step-by-Step Solution

1
Identify the formula for Marginal Utility (MU)
The formula is MUn=TUnTUn1MU_n = TU_n - TU_{n-1}, where nn is the unit number.
Marginal utility measures the extra satisfaction gained from consuming one additional unit of a good.
2
Extract Total Utility values from the table for Q=4Q = 4 and Q=3Q = 3
TU4=48TU_4 = 48 utils and TU3=42TU_3 = 42 utils.
To calculate the addition to total utility from the 4th unit, we compare total utility at 4 units with total utility at 3 units.
3
Compute the difference to find MU4MU_4
MU4=4842=6MU_4 = 48 - 42 = 6 utils.
Subtracting TU3TU_3 from TU4TU_4 yields the exact marginal contribution of the 4th unit.

Key Concept

Calculation and Interpretation of Marginal Utility
Estimated Time:1m 0s
Question 4792Question

In a socialist or command economy, key economic decisions regarding production, resource allocation, and distribution are primarily made by which of the following?

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Answer: The government through a central planning authority

Answer

The economic decisions in a command economy are made by the government through a central planning authority.
In a command or socialist economic system, the government owns the principal means of production and allocates resources through a central planning body to address basic economic problems.

Step-by-Step Solution

1
Identify the defining characteristic of a command economy.
In a command (socialist) economy, major factors of production and resources are public property controlled by the state.
State ownership eliminates decentralized market competition.
2
Determine how fundamental economic questions (what, how, and for whom to produce) are answered in this system.
The government utilizes a central planning board to allocate resources and schedule output according to state priorities.
Central planning replaces the price mechanism.

Key Concept

Centralized decision-making and state resource allocation in a command economy
Estimated Time:45s
Question 4793Question

A flour milling firm operating in an industrial zone in Ogun State increases all of its inputs of capital and labor by 100%100\%, leading to a 150%150\% increase in total output. However, as multiple neighboring factories expand simultaneously in the same area, severe traffic congestion and municipal power shortages cause the firm's long-run average cost (LRACLRAC) to increase. Which of the following correctly classifies the economic phenomena experienced by this firm?

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Answer: Increasing returns to scale internally and external diseconomies of scale.

Answer

The firm experiences increasing returns to scale internally, alongside external diseconomies of scale resulting from industry expansion.
Increasing inputs by 100%100\% while total output grows by 150%150\% signifies increasing returns to scale internally, as output expanded at a higher rate than inputs. Concurrently, rising long-run average costs resulting from surrounding factory expansion and shared infrastructure strain (traffic and power supply) constitute external diseconomies of scale.

Step-by-Step Solution

1
Analyze input-output proportion for returns to scale
Inputs increased by 100%100\%, while output increased by 150%150\%.
When output increases by a higher percentage than the proportional increase in all inputs, the firm experiences increasing returns to scale (economies of scale).
2
Identify the source of cost increases
Costs increase due to external industrial growth causing traffic congestion and power grid pressure.
Cost increases arising outside the firm's direct control due to the growth of the entire industry/region represent external diseconomies of scale.
3
Synthesize scale classifications
Internal increasing returns to scale combined with external diseconomies of scale.
The internal production function yields economies of scale, while the external operating environment imposes diseconomies.

Key Concept

Internal economies/returns to scale depend on a single firm's output-input expansion ratio, whereas external economies/diseconomies of scale arise from industry-wide expansion affecting all firms in a locality.
Question 4794Question

In monohybrid inheritance experiments involving guinea pig coat color, where the allele for black fur (BB) is completely dominant over the allele for white fur (bb), match each specified parental cross or test cross scenario on the left with its corresponding phenotypic or genotypic outcome in the offspring on the right.

Click a left item, then click its matching right item

Items

Cross between a heterozygous black guinea pig (BbBb) and a white guinea pig (bbbb)
Cross between two heterozygous black guinea pigs (Bb×BbBb \times Bb) yielding a total of 160 offspring
A test cross of a dominant black guinea pig that produces 100% black offspring across multiple litters
Cross between pure-breeding black (BBBB) and pure-breeding white (bbbb) parents to produce the F1F_1 generation

Matches

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Answer

The correct pairings match each monohybrid inheritance scenario to its precise Mendelian ratio or outcome: (1) Heterozygous black crossed with white matches the 1:1 genotypic and phenotypic ratio; (2) Two heterozygous black parents producing 160 offspring matches 120 black and 40 white offspring (3:1 ratio); (3) A test cross producing 100% dominant offspring confirms a homozygous dominant parent (BB); and (4) Crossing pure-breeding parents yields 100% uniform heterozygous F1 offspring.
Each cross directly demonstrates a fundamental aspect of Mendel's First Law. Heterozygote ×\times homozygous recessive gives a 1:11:1 ratio; two heterozygotes produce a 3:13:1 phenotypic ratio (120:40 out of 160); a test cross yielding zero recessive offspring confirms homozygous dominance; and contrasting pure lines produce uniform F1F_1 heterozygotes.

Step-by-Step Solution

1
Analyze the cross between BbBb and bbbb
Gametes: Parent 1 produces BB and bb in equal proportion (1:11:1), Parent 2 produces only bb. Offspring genotypes: 12Bb\frac{1}{2} Bb (black fur) and 12bb\frac{1}{2} bb (white fur).
Mendel's Law of Segregation dictates that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.
2
Calculate expected offspring numbers for Bb×BbBb \times Bb with total N=160N = 160
Monohybrid phenotypic ratio is 33 dominant : 11 recessive. Dominant count =34×160=120= \frac{3}{4} \times 160 = 120; Recessive count =14×160=40= \frac{1}{4} \times 160 = 40.
A monohybrid cross of two heterozygotes generates a genotypic ratio of 1BB:2Bb:1bb1 BB : 2 Bb : 1 bb, which simplifies to a 3:13:1 phenotypic ratio under complete dominance.
3
Evaluate the test cross of an unknown black parent with a recessive bbbb parent
If the parent were BbBb, white offspring would appear in approximately 50%50\% of cases. Since 100%100\% of offspring are black, the unknown parent must be homozygous dominant (BBBB).
A test cross uses a known homozygous recessive individual to reveal whether an organism expressing a dominant trait is homozygous or heterozygous.
4
Determine the outcome of crossing homozygous contrasting parents (BB×bbBB \times bb)
All F1F_1 offspring receive BB from the black parent and bb from the white parent, making 100%100\% of the F1F_1 generation heterozygous (BbBb).
This illustrates Mendel's principle of uniformity in the F1F_1 generation when crossing pure lines.

Key Concept

Mendel's Law of Segregation and Monohybrid Inheritance Ratios
Question 4795Question

In a perfectly competitive market, an individual firm is considered a price taker because its output is so small relative to total market supply that it cannot influence the market price.

Show answer & explanation

Answer: True

Answer

The statement is True.
The statement correctly identifies the fundamental assumption of price-taker behavior in perfect competition, which arises because each firm's market share is too small to affect price.

Step-by-Step Solution

1
Analyze the characteristic of firms in a perfectly competitive market structure.
A core assumption of perfect competition is that firms are price takers.
There are numerous small firms selling identical (homogeneous) products.
2
Evaluate the relationship between firm output and market price control.
An individual firm's output is an insignificant portion of aggregate supply, giving it zero control over price.
Market price is determined purely by the intersection of aggregate market demand and aggregate market supply.

Key Concept

Price-Taker Characteristic of Perfect Competition
Question 4796Question

A sample of salivary amylase was kept in an ice bath at 0C0^\circ\text{C} for two hours and then mixed with a starch solution maintained at body temperature (37C37^\circ\text{C}). Which of the following best describes the outcome of this reaction?

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Answer: Starch is hydrolyzed to maltose because low temperature causes temporary inactivation rather than denaturation.

Answer

Starch is hydrolyzed to maltose because low temperature causes temporary inactivation rather than denaturation.
Low temperatures render enzymes temporarily inactive due to reduced molecular collisions. When returned to an optimal temperature of 37C37^\circ\text{C}, salivary amylase regains catalytic activity and hydrolyzes starch into maltose.

Step-by-Step Solution

1
Analyze the effect of low temperature (0C0^\circ\text{C}) on enzyme structure and kinetic energy.
At 0C0^\circ\text{C}, salivary amylase molecules have very low kinetic energy, resulting in temporary inactivation without altering the active site shape.
Low temperature does not break the chemical bonds maintaining the enzyme's 3D structure.
2
Determine what happens when the enzyme is warmed to 37C37^\circ\text{C}.
The enzyme regains kinetic energy and normal catalytic activity at its optimum temperature of 37C37^\circ\text{C}.
Because the enzyme was not denatured, active sites can successfully bind starch substrates.
3
Identify the end product of starch breakdown by salivary amylase.
Salivary amylase breaks starch down into the disaccharide maltose.
Amylase specifically cleaves alpha-1,4-glycosidic bonds in starch.

Key Concept

Effect of Temperature on Digestive Enzymes
Estimated Time:45s
Question 4797Question

During a field study of an intertidal mangrove swamp in Cross River State, a student used a 0.5 m×0.5 m0.5\text{ m} \times 0.5\text{ m} quadrat frame to sample the periwinkle (*Tympanotonus fuscatus*) population. Across 1010 randomly placed quadrat throws, a total count of 180180 periwinkles was recorded. What is the estimated population density of *Tympanotonus fuscatus* per square metre in this habitat?

Show answer & explanation

Answer: 72 periwinkles/m272\text{ periwinkles/m}^2

Answer

The estimated population density is 72 periwinkles/m272\text{ periwinkles/m}^2.
The correct option correctly evaluates population density by calculating the total sampled area (10×0.25 m2=2.5 m210 \times 0.25\text{ m}^2 = 2.5\text{ m}^2) and dividing the total counted organisms (180180) by that area, yielding 72 periwinkles/m272\text{ periwinkles/m}^2.

Step-by-Step Solution

1
Calculate the area of a single quadrat frame
Area of 1 quadrat = 0.5 m×0.5 m=0.25 m20.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2.
Determines the surface area enclosed by one sampling unit.
2
Determine the total area sampled across all throws
Total sampled area = 10 throws×0.25 m2=2.5 m210 \text{ throws} \times 0.25\text{ m}^2 = 2.5\text{ m}^2.
Accounts for the cumulative ground area surveyed during the 10 quadrat throws.
3
Calculate the population density per square metre
Population density = Total number of organismsTotal sampled area=1802.5 m2=72 periwinkles/m2\frac{\text{Total number of organisms}}{\text{Total sampled area}} = \frac{180}{2.5\text{ m}^2} = 72\text{ periwinkles/m}^2.
Population density is expressed as the total number of individuals per unit area.

Key Concept

Quadrat Population Density Calculation
Question 4798Question

During gametogenesis, non-disjunction of chromosome 21 occurs specifically during Meiosis I. If all resulting gametes are subsequently fertilized by normal haploid gametes, what percentage of the produced zygotes will have trisomy 21?

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Answer: 50%

Answer

50% of the resulting zygotes will exhibit trisomy 21.
When non-disjunction of a chromosome pair takes place during Meiosis I, the pair fails to segregate. Consequently, one daughter cell receives both homologous chromosomes while the other receives none. Following normal chromatid separation in Meiosis II, half of the gametes carry an extra chromosome (n+1n+1) and half are deficient by one chromosome (n1n-1). Fertilization of the n+1n+1 gametes with normal haploid gametes (nn) produces trisomic (2n+12n+1) zygotes. Therefore, 50% of the zygotes will have trisomy 21.

Step-by-Step Solution

1
Analyze the meiotic stage where non-disjunction occurs
In Meiosis I, non-disjunction means homologous chromosomes fail to separate into distinct daughter cells.
Understanding the exact stage of meiotic failure determines the chromosome distribution in the daughter cells after Meiosis I.
2
Determine the chromosomal content of the resulting four gametes after Meiosis II
Meiosis I yields one cell with an extra chromosome (n+1n+1) and one cell missing that chromosome (n1n-1). Normal chromatid separation in Meiosis II creates two (n+1)(n+1) gametes and two (n1)(n-1) gametes.
All four gametes produced are genetically abnormal after a Meiosis I non-disjunction event.
3
Calculate the proportion of trisomic zygotes post-fertilization
Fertilization of the two (n+1)(n+1) gametes by normal haploid (nn) gametes gives (2n+1)(2n+1), which is trisomy 21. Two out of four total zygotes (50%) will be trisomic (and 50% will be monosomic).
Trisomy requires the combination of an (n+1)(n+1) gamete with a normal nn gamete.

Key Concept

Chromosomal Non-disjunction and Aneuploidy
Question 4799Question
For the gaseous reaction represented by the equation:
2A(g)+B(g)C(g)+3D(g)2\text{A}(g) + \text{B}(g) \rightarrow \text{C}(g) + 3\text{D}(g)
the quantity of substance A\text{A} present in a 2.0 dm32.0\text{ dm}^3 reaction vessel decreases from 0.80 mol0.80\text{ mol} to 0.32 mol0.32\text{ mol} in 40 s40\text{ s}. What is the average rate of formation of product D\text{D} in mol dm3 s1\text{mol dm}^{-3}\text{ s}^{-1}?
Show answer & explanation

Answer: 0.009

Answer

The average rate of formation of product D is 0.009 mol dm3 s10.009\text{ mol dm}^{-3}\text{ s}^{-1}.
The change in concentration of reactant A over 40 s40\text{ s} is 0.48 mol2.0 dm3=0.24 mol dm3\frac{0.48\text{ mol}}{2.0\text{ dm}^3} = 0.24\text{ mol dm}^{-3}. The rate of consumption of A is 0.2440=0.006 mol dm3 s1\frac{0.24}{40} = 0.006\text{ mol dm}^{-3}\text{ s}^{-1}. Because 2 moles2\text{ moles} of A produce 3 moles3\text{ moles} of D, the rate of formation of D is 32×0.006=0.009 mol dm3 s1\frac{3}{2} \times 0.006 = 0.009\text{ mol dm}^{-3}\text{ s}^{-1}.

Step-by-Step Solution

1
Calculate the change in concentration of reactant A during the time interval.
Δ[A]=0.80 mol0.32 mol2.0 dm3=0.24 mol dm3\Delta [A] = \frac{0.80\text{ mol} - 0.32\text{ mol}}{2.0\text{ dm}^3} = 0.24\text{ mol dm}^{-3}
Concentration is moles per unit volume.
2
Calculate the rate of consumption of reactant A per unit time.
RateA=Δ[A]Δt=0.24 mol dm340 s=0.006 mol dm3 s1\text{Rate}_A = -\frac{\Delta [A]}{\Delta t} = \frac{0.24\text{ mol dm}^{-3}}{40\text{ s}} = 0.006\text{ mol dm}^{-3}\text{ s}^{-1}
Reaction rate is defined as the change in concentration over elapsed time.
3
Use the stoichiometric coefficients from the balanced equation to calculate the rate of formation of D.
RateD=32×RateA=32×0.006=0.009 mol dm3 s1\text{Rate}_D = \frac{3}{2} \times \text{Rate}_A = \frac{3}{2} \times 0.006 = 0.009\text{ mol dm}^{-3}\text{ s}^{-1}
According to the balanced chemical equation, 22 moles of A are consumed for every 33 moles of D formed, so 13RateD=12RateA\frac{1}{3}\text{Rate}_D = \frac{1}{2}\text{Rate}_A.

Key Concept

Stoichiometric relationship between rates of consumption of reactants and rates of formation of products
Question 4800Question

In fruit flies (*Drosophila melanogaster*), grey body color (BB) is dominant over black body color (bb), and normal wing length (VV) is dominant over vestigial wings (vv). If two heterozygous grey-bodied, normal-winged flies (BbVvBbVv) are crossed and produce a total of 320 offspring, how many of these offspring are expected to exhibit a black body with normal wings?

Show answer & explanation

Answer: 60

Answer

60 offspring are expected to have a black body with normal wings.
In a dihybrid cross of two heterozygous individuals (BbVv×BbVvBbVv \times BbVv), the independent assortment of alleles yields an F2 phenotypic ratio of 9:3:3:19 : 3 : 3 : 1. The proportion of offspring displaying the recessive trait for the first gene (black body, bbbb) and the dominant trait for the second gene (normal wings, V_V\_) is 316\frac{3}{16}. Multiplying 316\frac{3}{16} by the total offspring count of 320320 gives exactly 6060.

Step-by-Step Solution

1
Determine parental genotypes and set up the dihybrid cross.
Parental cross is BbVv×BbVvBbVv \times BbVv, where both parents are heterozygous for both traits.
Mendel's Second Law states that alleles of different genes assort independently during gamete formation.
2
Identify the expected phenotypic ratio for the offspring.
The phenotypic ratio for a dihybrid cross between two double heterozygotes (BbVv×BbVvBbVv \times BbVv) is 9:3:3:19 : 3 : 3 : 1.
- Grey body, normal wings (B_V_B\_V\_) = 916\frac{9}{16}
- Grey body, vestigial wings (B_vvB\_vv) = 316\frac{3}{16}
- Black body, normal wings (bbV_bbV\_) = 316\frac{3}{16}
- Black body, vestigial wings (bbvvbbvv) = 116\frac{1}{16}
The probability of recessive body color (bb=14bb = \frac{1}{4}) combined with dominant wing length (V_=34V\_ = \frac{3}{4}) is calculated as 14×34=316\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}.
3
Calculate the expected number of offspring with black body and normal wings.
316×320=60\frac{3}{16} \times 320 = 60.
Multiplying the expected phenotypic fraction by the total offspring count yields the specific expected count.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
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