All practice questions

Question 8701Question →

In domestic fowl (*Gallus gallus*), rose comb ( $R$ ) is dominant over single comb ( $r$ ), and feathered legs ( $F$ ) are dominant over clean legs ( $f$ ). A rooster heterozygous for both traits ( $RrFf$ ) is test-crossed with a hen possessing a single comb and clean legs ( $rrff$ ). If a total of $400$ chicks are hatched from this cross, how many are expected to exhibit a rose comb and clean legs?

75

100

200

225

Show answer & explanation

Answer: $100$

Answer

The expected number of offspring displaying rose comb and clean legs is

100

.

A cross between a double heterozygote (

RrFf

) and a double recessive organism (

rrff

) is a dihybrid test cross. According to Mendel's Law of Independent Assortment, the heterozygous parent forms four types of gametes (

RF

,

Rf

,

rF

,

rf

) in equal

1:1:1:1

frequencies, while the recessive parent yields only

rf

gametes. This produces four phenotypic classes in equal proportions (

\frac{1}{4}

each). Out of

400

total offspring, the expected count for rose comb and clean legs (

Rrff

) is

\frac{1}{4} \times 400 = 100

.

Step-by-Step Solution

1

Determine the parental genotypes and gamete types.

The heterozygous rooster (

RrFf

) produces four gamete types in equal proportions:

RF

,

Rf

,

rF

, and

rf

. The homozygous recessive hen (

rrff

) produces only one gamete type:

rf

.

Mendel's Law of Independent Assortment states that alleles of different genes segregate independently into gametes during meiosis.

2

Derive the offspring genotypes and phenotypic proportions.

Combining gametes yields four distinct offspring genotypes:

RrFf

(rose comb, feathered legs),

Rrff

(rose comb, clean legs),

rrFf

(single comb, feathered legs), and

rrff

(single comb, clean legs) in a

1:1:1:1

ratio.

A dihybrid test cross always generates a 1:1:1:1 phenotypic frequency among offspring.

3

Calculate the expected count for the target phenotype.

Target phenotype proportion (

Rrff

) =

\frac{1}{4} = 0.25

. Expected count =

0.25 \times 400 = 100

.

Multiplying the phenotypic probability by the total sample size gives the expected frequency.

Key Concept

Dihybrid Test Cross Ratio

Estimated Time:2m 0s

Question 8702Question →

Arrange the following hydrogen halides in order of increasing boiling point, starting from the compound with the lowest boiling point to the compound with the highest boiling point.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing boiling point is Hydrogen chloride (

HCl

), Hydrogen bromide (

HBr

), Hydrogen iodide (

HI

), and Hydrogen fluoride (

HF

).

The correct sequence ranks the Group 17 hydrides by increasing boiling point:

HCl < HBr < HI < HF

. For

HCl

,

HBr

, and

HI

, boiling points rise systematically with increasing atomic size and molar mass because larger electron clouds enhance polarizability and London dispersion forces.

HF

breaks this trend and has the highest boiling point in the group due to the presence of strong intermolecular hydrogen bonding.

Step-by-Step Solution

1

Identify the primary intermolecular forces present in each hydrogen halide.

HCl

,

HBr

, and

HI

interact mainly via London dispersion forces and dipole-dipole interactions, while

HF

forms strong intermolecular hydrogen bonds.

Fluorine is extremely small and electronegative, fulfilling the requirements for hydrogen bonding.

2

Compare van der Waals forces among

HCl

,

HBr

, and

HI

.

Boiling point increases progressively from

HCl

to

HBr

to

HI

.

As molecular size and electron cloud volume increase down Group 17, polarizability increases, leading to stronger temporary dipoles and stronger London dispersion forces.

3

Determine the position of

HF

within the series.

HF

has an anomalously high boiling point compared to the rest of the group.

Intermolecular hydrogen bonding is considerably stronger than van der Waals dispersion forces, making

HF

the least volatile of the four hydrogen halides.

Key Concept

Boiling point trends in hydrides are governed by the interplay of London dispersion forces (which scale with molar mass and polarizability) and hydrogen bonding.

Estimated Time:1m 0s

Question 8703Question →

Match each ecological structural term on the left with its corresponding definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ecotone matches the transitional boundary zone between distinct biological communities; Microhabitat matches the localized small-scale physical area with distinct microclimate; Saprotroph matches the heterotrophic organism absorbing nutrients from dead organic substrate; Ecological Guild matches the group of species exploiting the same resources in a similar manner.

Each ecological term correctly pairs with its definitive structural role or spatial unit in ecosystem organization: Ecotone is the transition zone, Microhabitat is the localized physical area, Saprotroph is the decomposer organism absorbing dissolved decay products, and Ecological Guild describes species sharing a resource utilization strategy.

Step-by-Step Solution

1

Analyze the term 'Ecotone'

Identify that 'ecotone' refers to boundary zones where two ecosystems meet and overlap.

The term describes transitional gradient areas between distinct plant and animal communities.

2

Analyze the term 'Microhabitat'

Connect 'microhabitat' to small-scale physical locations with specific microclimatic features.

Sub-environments like crevices in rocks or leaf litter are classic examples of microhabitats.

3

Analyze the term 'Saprotroph'

Associate saprotrophic nutrition with extracellular digestion of decaying organic matter.

Saprotrophs break down dead organic matter and absorb soluble nutrients directly.

4

Analyze the term 'Ecological Guild'

Match 'guild' with species occupying overlapping functional roles or resource utilization tactics.

Guild members share functional niches within a community regardless of phylogenetic distance.

Key Concept

Ecosystem Structure and Ecological Terminology

Question 8704Question →

In Jean-Baptiste Lamarck's theory of evolution, a changing environment causes direct physical alterations to an organism's body without requiring changes in the organism's behavioral habits or organ usage.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because Lamarck specified that environmental change acts indirectly. A changing environment creates new needs, which compel changes in an organism's behavioral habits. These changed habits determine which organs undergo increased use or disuse, thereby altering their physical structure prior to inheritance.

Step-by-Step Solution

1

Analyze Lamarck's foundational sequence for evolutionary adaptation.

Lamarck postulated that changes in the environment alter an organism's needs (besoins), which in turn induce new habits and actions.

Establishing the sequence of cause-and-effect in Lamarckism is necessary to differentiate indirect environmental influence from direct physical induction.

2

Evaluate the mediating role of behavioral habits and organ use or disuse.

Anatomical modifications develop as a direct consequence of altered organ usage resulting from changed habits, rather than from passive environmental exposure.

Lamarck stressed that behavioral changes mediate between environmental demands and anatomical modifications.

3

Determine the validity of the statement.

The statement asserts direct structural modification without habit or usage changes, which contradicts Lamarck's explicit mechanism.

Because Lamarckian evolution relies strictly on behavioral habits and differential use/disuse as intermediate steps, the statement is false.

Key Concept

Lamarckian Evolutionary Sequence: Environmental Changes → Modified Needs → Changed Habits → Organ Use/Disuse → Acquired Structural Modifications

Estimated Time:1m 30s

Question 8705Question →

Match each ecological pyramid concept on the left with its correct biological or structural feature on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pyramid of Energy matches with 'Always remains upright across all natural ecosystems because energy dissipates as heat at each successive trophic step.' Aquatic Pyramid of Biomass matches with 'Features an inverted structure where the producer level has a smaller standing crop than primary consumers due to rapid turnover.' Parasitic Pyramid of Numbers matches with 'Exhibits an inverted shape starting from a single host supporting numerous individuals at progressively higher trophic levels.' Ten Percent Law matches with 'Quantifies the average proportion of energy converted into biomass and transferred to the next higher trophic level.'

Each ecological concept correctly matches its underlying biological rule: the pyramid of energy is universally upright due to metabolic heat loss; the aquatic biomass pyramid can invert due to rapid producer turnover; parasitic numerical pyramids invert due to host-parasite population ratios; and the ten percent law defines ecological energy transfer efficiency.

Step-by-Step Solution

1

Analyze the thermodynamic properties of energy flow.

Energy transfer is unidirectional and governed by thermodynamic loss, meaning a pyramid of energy can never be inverted and is always upright.

Identify the fundamental physical law governing energy flow.

2

Examine ecosystem-specific biomass dynamics.

Open-water aquatic systems exhibit inverted biomass pyramids due to high photosynthetic turnover rates of microscopic producers.

Distinguish standing crop biomass from energy productivity.

3

Evaluate trophic structure in parasitic food chains.

A single host organism harbouring hundreds of ecto- or endoparasites creates an inverted pyramid of numbers.

Recognize numerical population distributions across specialized trophic roles.

4

Associate numerical transfer efficiency rules with ecological principles.

The ten percent law specifically defines ecological efficiency between trophic tiers.

Match numerical energetic transfer definitions with their scientific names.

Key Concept

Trophic dynamics, energetic decay, and structural variations in ecological pyramids

Question 8706Question →

Potassium tetraoxomanganate(VII),

\text{KMnO}_4

, reacts with excess concentrated hydrochloric acid according to the redox equation:

2\text{KMnO}_4(s) + 16\text{HCl}(aq) \rightarrow 2\text{KCl}(aq) + 2\text{MnCl}_2(aq) + 8\text{H}_2\text{O}(l) + 5\text{Cl}_2(g)

Calculate the volume of chlorine gas (in

\text{dm}^3

) produced at s.t.p. when

15.8\text{ g}

of

\text{KMnO}_4

reacts completely. [Molar mass of

\text{KMnO}_4 = 158\text{ g mol}^{-1}

, molar volume of gas at s.t.p. =

22.4\text{ dm}^3\text{mol}^{-1}

]

Show answer & explanation

Answer: 5.6

Answer

The volume of chlorine gas liberated at s.t.p. is

5.6\text{ dm}^3

.

Converting

15.8\text{ g}

of

\text{KMnO}_4

gives

0.10\text{ mol}

. According to the balanced equation,

2\text{ mol}

of

\text{KMnO}_4

yields

5\text{ mol}

of

\text{Cl}_2

, giving

0.25\text{ mol}

of

\text{Cl}_2

. Multiplying

0.25\text{ mol}

by the molar volume at s.t.p. (

22.4\text{ dm}^3\text{mol}^{-1}

) yields

5.6\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the moles of potassium tetraoxomanganate(VII),

\text{KMnO}_4

\text{Moles of KMnO}_4 = \frac{15.8\text{ g}}{158\text{ g mol}^{-1}} = 0.10\text{ mol}

Converting the given mass of reactant to moles allows stoichiometric comparison.

2

Use the balanced redox equation to determine the mole ratio between

\text{KMnO}_4

and

\text{Cl}_2

\text{Moles of Cl}_2 = 0.10\text{ mol} \times \frac{5}{2} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

of

\text{KMnO}_4

produce

5\text{ moles}

of

\text{Cl}_2

gas.

3

Calculate the volume of

\text{Cl}_2

gas produced at s.t.p.

\text{Volume} = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 5.6\text{ dm}^3

At s.t.p.,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

.

Key Concept

Stoichiometric calculations in the laboratory preparation of chlorine from oxidation of hydrochloric acid

Question 8707Question →

At extremely high pressures, the compressibility factor ( $Z = \frac{PV}{nRT}$ ) of a real gas is observed to be greater than $1.0$ . Which assumption of the kinetic molecular theory breaks down to cause this positive deviation?

The volume occupied by individual gas molecules is negligible compared to the total volume of the container.

There are no attractive or repulsive intermolecular forces acting between the gas molecules.

Collisions between gas molecules and the container walls are perfectly elastic.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature.

Show answer & explanation

Answer: The volume occupied by individual gas molecules is negligible compared to the total volume of the container.

Answer

The assumption that the volume of individual gas molecules is negligible breaks down at extremely high pressures, causing the compressibility factor to exceed 1.0.

At extremely high pressures, gas particles are forced close together so that the actual volume occupied by the gas molecules becomes significant relative to the container volume. This breaks the ideal gas postulate of zero molecular volume, making the actual molar volume larger than ideal and resulting in

Z > 1.0

.

Step-by-Step Solution

1

Analyze the compressibility factor equation

Z = \frac{PV}{nRT}

.

For an ideal gas,

Z = 1.0

. When

Z > 1.0

, the real volume occupied by the gas is larger than predicted by the ideal gas law (

V_{\text{real}} > V_{\text{ideal}}

).

Understanding

Z

allows identification of whether volume exclusion or attractive forces dominate.

2

Identify the cause of positive deviation (

Z > 1.0

) at very high pressure.

At very high pressures, molecules are compressed into a small container volume. The finite volume of the gas molecules themselves (

b

in van der Waals equation) can no longer be ignored.

The Kinetic Molecular Theory postulate stating gas particles have negligible volume breaks down under high pressure.

Key Concept

Compressibility Factor and Molecular Volume Exclusion in Real Gases

Question 8708Question →

Continuous application of ammonium-based nitrogen fertilizers increases soil acidity, which mobilizes toxic heavy metal cations by displacing them from soil colloid exchange sites into the soil solution.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. Biological nitrification of ammonium fertilizers generates hydrogen ions (

H^+

), lowering soil pH. The increased acidity causes

H^+

ions to displace bound heavy metal cations (

Cd^{2+}

,

Pb^{2+}

,

Al^{3+}

) from soil colloids into the soil solution, enhancing their mobility and bioavailability.

The statement accurately describes environmental soil chemistry. Soil nitrification of ammonium produces

H^+

ions, decreasing pH. The elevated

H^+

concentration displaces toxic heavy metal cations from negatively charged clay-humus exchange complexes, bringing them into the soil solution.

Step-by-Step Solution

1

Analyze the chemical effect of ammonium-based fertilizers on soil pH

Nitrifying bacteria oxidize

NH_4^+

to

NO_3^-

, releasing

H^+

ions according to the reaction:

NH_4^+ + 2O_2 \rightarrow NO_3^- + H_2O + 2H^+

. This increases soil acidity.

Establishing the source of increased soil acidity is essential for evaluating soil chemical dynamics.

2

Examine the interaction between hydrogen ions and heavy metal cations on soil colloids

Soil colloids carry negative surface charges that adsorb metal cations (

Ca^{2+}

,

Mg^{2+}

,

Cd^{2+}

,

Pb^{2+}

,

Al^{3+}

). High concentrations of

H^+

ions displace these bound cations into the soil solution through cation exchange.

Understanding ion-exchange equilibria on soil particles explains how acidity affects metal solubility.

3

Determine the environmental mobility and toxicity outcome

Displaced heavy metal cations enter the soil water solution, significantly increasing their mobility, leaching potential, and root uptake toxicity.

Confirms that increased soil acidity leads to heavy metal mobilization.

Key Concept

Soil Acidification and Heavy Metal Mobilization

Question 8709Question →

During a paper chromatography analysis of a mixture of amino acids, the solvent front travels a distance of $16.0\text{ cm}$ from the baseline. A spot corresponding to a particular amino acid is observed to travel a distance of $6.4\text{ cm}$ from the baseline. What is the retardation factor ( $R_f$ ) of this amino acid?

0.40

2.50

0.67

9.60

Show answer & explanation

Answer: $0.40$

Answer

The retardation factor (

R_f

) of the amino acid is

0.40

.

The retardation factor (

R_f

) is determined by dividing the distance moved by the solute spot (

6.4\text{ cm}

) by the distance moved by the solvent front (

16.0\text{ cm}

), giving

\frac{6.4}{16.0} = 0.40

.

Step-by-Step Solution

1

Identify the formula for the retardation factor (

R_f

)

R_f = \frac{\text{Distance travelled by solute spot}}{\text{Distance travelled by solvent front}}

By definition in paper chromatography,

R_f

is the ratio of solute displacement to solvent displacement relative to the origin line.

2

Substitute the given values into the equation

R_f = \frac{6.4\text{ cm}}{16.0\text{ cm}}

The solute displacement is

6.4\text{ cm}

and the solvent front displacement is

16.0\text{ cm}

.

3

Calculate the dimensionless

R_f

value

R_f = 0.40

Performing the division gives

0.40

, which is less than

1.0

as required for valid

R_f

values.

Key Concept

Calculation of Retardation Factor (Rf) in Paper Chromatography

Estimated Time:1m 0s

Question 8710Question →

In the industrial Chlor-Alkali process, concentrated sodium chloride solution (brine) is electrolyzed using inert carbon electrodes. Which substance is liberated at the cathode, and what is the reason for its discharge?

Hydrogen gas, because

H^+

ions are lower than

Na^+

ions in the electrochemical series and require less energy to gain electrons.

Sodium metal, because

Na^+

ions are present in a significantly higher concentration in brine than

H^+

ions.

Chlorine gas, because

Cl^-

ions migrate to the cathode to undergo reduction.

Oxygen gas, because

OH^-

ions are preferentially discharged over

Cl^-

ions at the cathode.

Show answer & explanation

Answer: Hydrogen gas, because $H^+$ ions are lower than $Na^+$ ions in the electrochemical series and require less energy to gain electrons.

Answer

Hydrogen gas is liberated at the cathode because hydrogen ions (

H^+

) lie lower in the electrochemical series than sodium ions (

Na^+

) and are thus preferentially reduced.

In the electrolysis of concentrated aqueous sodium chloride (brine), water provides

H^+

and

OH^-

ions alongside

Na^+

and

Cl^-

ions. At the cathode (the negative electrode), cations

Na^+

and

H^+

compete for discharge. Hydrogen (

H^+

) is much lower than sodium (

Na^+

) in the electrochemical series, meaning it accepts electrons far more easily. Therefore, hydrogen ions are preferentially reduced to yield hydrogen gas (

H_2

).

Step-by-Step Solution

1

Identify the ions present in the electrolyte

In aqueous

NaCl

(brine), the ions present are

Na^+

and

Cl^-

from sodium chloride, and

H^+

and

OH^-

from the auto-ionization of water.

Electrolysis of aqueous solutions involves ions from both the solute and the solvent.

2

Determine which ions migrate to the cathode

Cations (

Na^+

and

H^+

) migrate to the negatively charged cathode.

Oppositely charged ions are attracted to the electrodes.

3

Apply the principles of preferential discharge at the cathode

H^+

is placed much lower than

Na^+

in the reactivity/electrochemical series, so

2H^+ + 2e^- \rightarrow H_2(g)

occurs.

Ions lower in the electrochemical series gain electrons (are reduced) more readily than ions higher up.

Key Concept

Preferential discharge of cations during electrolysis of aqueous brine in the Chlor-Alkali industry

Question 8711Question →

An offshore oil platform made of steel is submerged in seawater. To protect the submerged steel structure from corrosion, an impressed current cathodic protection system is installed using an external direct current source and an inert anode. Which of the following best describes the electrochemical role and behavior of the steel structure in this system?

It acts as the cathode, receiving electrons from the external power source to prevent iron from oxidizing.

It acts as the anode, releasing electrons into the DC power source to corrode preferentially.

It acts as the cathode, undergoing continuous oxidation to form a protective metal oxide layer.

It acts as the anode, attracting hydroxide ions to passivate the steel surface.

Show answer & explanation

Answer: It acts as the cathode, receiving electrons from the external power source to prevent iron from oxidizing.

Answer

The steel structure acts as the cathode, receiving electrons from the external DC power source to prevent iron from oxidizing.

In impressed current cathodic protection (ICCP), an external DC power source supplies electrons to the steel structure. By forcing electrons into the steel, the structure becomes the cathode of the electrochemical cell, preventing the oxidation of iron (

Fe \rightarrow Fe^{2+} + 2e^-

) and effectively stopping corrosion.

Step-by-Step Solution

1

Identify the electrochemical mechanism of Cathodic Protection via Impressed Current (ICCP).

An external direct current (DC) power source forces electrons onto the metal structure to be protected.

Corrosion of iron occurs via an anodic oxidation process (

Fe \rightarrow Fe^{2+} + 2e^-

). Supplying electrons reverses/prevents this oxidation.

2

Determine the electrode polarities in the cathodic protection circuit.

The negative terminal of the DC power source is connected to the steel structure, making it the cathode. The positive terminal is connected to an inert material, making it the anode.

Reduction or electron supply occurs at the cathode, ensuring the protected metal remains in its unoxidized elemental state.

Key Concept

Impressed Current Cathodic Protection (ICCP)

Question 8712Question →

A unicellular autotrophic organism isolated from a pond sample performs oxygenic photosynthesis despite lacking chloroplasts or a membrane-bound nucleus. Chemical testing confirms that its cell wall structural matrix consists of peptidoglycan (murein). Which of the following statements accurately classifies this organism and describes its structural characteristics within Kingdom Monera?

The organism is a cyanobacterium whose cell wall is composed of cellulose, matching the cell wall composition of green plants.

The organism is a cyanobacterium possessing a cell wall made of peptidoglycan, distinguishing it from eukaryotic plant cells.

The organism is an acellular photosynthetic virus surrounded by a protein capsid rather than a peptidoglycan cell wall.

The organism is a nitrifying bacterium that manufactures glucose by oxidising ammonia into nitrates inside chloroplasts.

Show answer & explanation

Answer: The organism is a cyanobacterium possessing a cell wall made of peptidoglycan, distinguishing it from eukaryotic plant cells.

Answer

The organism is a cyanobacterium possessing a cell wall made of peptidoglycan, distinguishing it from eukaryotic plant cells.

The organism described exhibits prokaryotic organization (absence of chloroplasts and membrane-bound nucleus) paired with oxygenic photosynthesis, which defines cyanobacteria within Kingdom Monera. All Moneran cell walls feature a peptidoglycan (murein) matrix rather than the cellulose found in plant cells.

Step-by-Step Solution

1

Analyze cellular organization from the problem description.

The organism lacks a nucleus and membrane-bound organelles, placing it in Kingdom Monera (prokaryotes).

Monerans are prokaryotic organisms without membrane-bound organelles such as chloroplasts or nuclei.

2

Determine the metabolic and structural group within Kingdom Monera.

Unicellular prokaryotes that perform oxygenic photosynthesis are classified as cyanobacteria (blue-green algae).

Cyanobacteria utilize thylakoid membranes in the cytoplasm to perform photosynthesis.

3

Identify cell wall composition.

Prokaryotic cell walls are composed of peptidoglycan (murein), whereas eukaryotic plant cell walls are made of cellulose.

Peptidoglycan is the signature structural polymer of bacterial and cyanobacterial cell walls.

Key Concept

Structural features of Kingdom Monera (peptidoglycan cell wall, prokaryotic organization, thylakoids in cyanobacteria)

Estimated Time:1m 30s

Question 8713Question →

Match each specialized prokaryotic cell inclusion or biochemical component of Kingdom Monera on the left with its corresponding biological role or structural property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Gas vesicles match with proteinaceous hollow structures providing buoyancy; Calcium-dipicolinate complex matches with core constituent conferring thermal and chemical resistance to bacterial endospores; Magnetosomes match with membrane-enclosed iron oxide inclusions directing movement along geomagnetic field lines; Cyanophycin granules match with non-ribosomal polypeptide inclusions serving as nitrogen storage reserves.

Each Moneran cellular inclusion body serves a distinct ecological and physiological adaptation: gas vesicles adjust aquatic buoyancy, calcium-dipicolinate protects endospore genetic material from heat, magnetosomes guide navigation along Earth's magnetic fields, and cyanophycin granules store organic nitrogen.

Step-by-Step Solution

1

Analyze the physical function of Gas vesicles in aquatic prokaryotes.

Gas vesicles trap gas within rigid protein shells to control vertical positioning in the water column.

This allows cyanobacteria to stay in the photic zone for optimal photosynthesis.

2

Identify the biochemical agent responsible for endospore extreme heat resistance.

The calcium-dipicolinate complex accumulates in the endospore core, promoting severe dehydration.

Dehydration protects core enzymes and nucleic acids from denaturation under extreme environmental stress.

3

Determine the role of Magnetosomes in bacterial navigation.

Magnetosomes house magnetic mineral crystals enclosed in invaginated plasma membranes.

This structural alignment guides magnetotactic bacteria toward favorable low-oxygen aquatic strata.

4

Examine nitrogen accumulation inclusions in cyanobacteria.

Cyanophycin granules store multi-L-arginyl-poly-L-aspartic acid polymers.

Cyanobacteria accumulate this reserve during non-growing phases when fixed nitrogen is available.

Key Concept

Prokaryotic Cellular Inclusions and Biochemical Adaptations in Kingdom Monera

Question 8714Question →

Complete the statements regarding the artificial nuclear transmutation of aluminium-27 by filling in the blanks with the correct numerical values.

Fill in the blanks below

When aluminium-27 (

^{27}_{13}\text{Al}

) is bombarded with an alpha particle (

^{4}_{2}\alpha

), it produces phosphorus-30 (

^{30}_{15}\text{P}

) and a neutron (

^{A}_{Z}\text{n}

). According to the law of conservation of mass number and atomic number, the mass number

A

of the emitted neutron is and its atomic number

Z

is .

Show answer & explanation

Answer

The mass number A is 1 and the atomic number Z is 0.

In any balanced nuclear reaction, the sum of the superscripts (mass numbers) and the sum of the subscripts (atomic numbers) must be equal on both sides of the equation. For the reactants (

^{27}_{13}\text{Al} + ^{4}_{2}\alpha

), the total mass number is

27 + 4 = 31

and the total atomic number is

13 + 2 = 15

. For the products (

^{30}_{15}\text{P} + ^{A}_{Z}\text{n}

), setting

30 + A = 31

yields

A = 1

, and setting

15 + Z = 15

yields

Z = 0

. Thus, the emitted particle is a neutron (

^{1}_{0}\text{n}

).

Step-by-Step Solution

1

Balance the total mass numbers on both sides of the nuclear equation.

Total mass number on reactants side = 27 + 4 = 31. Mass number of phosphorus-30 = 30. Therefore, mass number of the neutron A = 31 - 30 = 1.

The total sum of mass numbers must remain equal before and after a nuclear reaction.

2

Balance the total atomic numbers (nuclear charge) on both sides of the nuclear equation.

Total atomic number on reactants side = 13 + 2 = 15. Atomic number of phosphorus-30 = 15. Therefore, atomic number of the neutron Z = 15 - 15 = 0.

The total sum of atomic numbers (positive charges) must be conserved in nuclear reactions.

Key Concept

Conservation of mass number and atomic number in nuclear transmutation equations.

Estimated Time:1m 0s

Question 8715Question →

Match each chemical reaction involving a carbonyl compound in Column A with its corresponding chemical product or visual observation in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming ethanal with Fehling's solution matches the formation of a brick-red precipitate of copper(I) oxide. Treating propanone with aqueous iodine and sodium hydroxide matches the formation of a pale yellow precipitate of triiodomethane. Reducing butan-2-one with lithium tetrahydridoaluminate(III) matches the production of a secondary alcohol, butan-2-ol. Oxidizing propanal with acidified potassium dichromate(VI) matches the production of propanoic acid with an orange to green color change.

Each carbonyl compound reacts according to its specific structural features: alkanals (ethanal, propanal) are easily oxidized by mild and strong oxidizing agents like Fehling's solution and acidified dichromate, respectively. Methyl ketones (propanone) uniquely yield yellow iodoform upon treatment with alkaline iodine solution. Ketones (butan-2-one) reduce under hydride transfer (

LiAlH_4

) to form secondary alcohols.

Step-by-Step Solution

1

Analyze the distinction test for ethanal (alkanal) using Fehling's solution

Alkanals reduce Fehling's solution containing copper(II) tartrate complex to insoluble red copper(I) oxide (

Cu_2O

).

Alkanals are easily oxidized to alkanoic acids due to the presence of the carbonyl hydrogen atom.

2

Analyze the triiodomethane (iodoform) reaction of propanone

Propanone contains the methyl carbonyl structure (

CH_3-CO-

), which reacts with

I_2/OH^-

to precipitate yellow

CHI_3

.

The iodoform test specifically identifies compounds containing a methyl group attached directly to a carbonyl carbon.

3

Determine the reduction product of the alkanone (butan-2-one)

Reduction of a ketone yields a secondary alcohol, turning

C=O

into

CH-OH

. Thus, butan-2-one gives butan-2-ol.

The carbonyl group of an alkanone has two alkyl substituents, forming a secondary alcohol carbon upon addition of hydrogen.

4

Determine the oxidation product of propanal

Oxidation of propanal adds oxygen across the C-H bond to yield propanoic acid, while reducing

Cr_2O_7^{2-}

(orange) to

Cr^{3+}

(green).

Acidified

K_2Cr_2O_7

acts as a strong oxidizing agent towards alkanals.

Key Concept

Chemical tests and redox behavior of alkanals vs. alkanones

Question 8716Question →

An environmental biologist isolated a microscopic organism from soil and documented its cellular structure: it is unicellular, lacks a nuclear membrane, possesses a rigid cell wall composed of peptidoglycan, and reproduces by binary fission. In a preliminary research paper, four different proposals were made for its binomial designation and kingdom classification:

1. *bacillus Subtilis* (Kingdom Monera)
2. *Bacillus subtilis* (Kingdom Protista)
3. *Bacillus Subtilis* (Kingdom Monera)
4. *Bacillus subtilis* (Kingdom Monera)

Which proposal correctly adheres to the established rules of binomial nomenclature while accurately placing the organism into its correct kingdom?

Proposal 4, because the generic name begins with a capital letter, the specific epithet is entirely in lowercase, both are italicized, and prokaryotes with peptidoglycan cell walls belong to Kingdom Monera.

Proposal 3, because both the genus name and specific epithet must begin with capital letters to indicate formal biological status within Kingdom Monera.

Proposal 2, because unicellular organisms possessing a cell wall and carrying out independent cellular functions are classified under Kingdom Protista.

Proposal 1, because generic names are written in lowercase while specific epithets are capitalized to emphasize species identity within Monera.

Show answer & explanation

Answer: Proposal 4, because the generic name begins with a capital letter, the specific epithet is entirely in lowercase, both are italicized, and prokaryotes with peptidoglycan cell walls belong to Kingdom Monera.

Answer

Proposal 4 is correct because the generic name begins with a capital letter, the specific epithet is entirely in lowercase, both are italicized, and prokaryotes with peptidoglycan cell walls belong to Kingdom Monera.

The correct answer identifies Proposal 4. According to Carl Linnaeus's rules of binomial nomenclature, a scientific name consists of two parts: the genus name, which must always begin with an uppercase letter, and the specific epithet, which must be entirely in lowercase. When printed, both names must be italicized. Furthermore, the described organism lacks a nuclear membrane (prokaryotic) and possesses a cell wall with peptidoglycan, which are diagnostic traits of Kingdom Monera.

Step-by-Step Solution

1

Evaluate the cellular characteristics to determine the correct Kingdom.

The organism is unicellular, lacks a nuclear membrane (prokaryotic), and contains peptidoglycan in its cell wall. These features definitively define Kingdom Monera (Bacteria).

Kingdom Protista contains eukaryotic unicellular organisms, whereas Monera comprises prokaryotes with peptidoglycan cell walls.

2

Apply the standard formatting rules of Linnaean binomial nomenclature.

The genus name must start with a capital letter ('Bacillus'), the specific epithet must be entirely in lowercase ('subtilis'), and both terms must be italicized (or underlined when handwritten).

Binomial nomenclature requires Genus (capitalized) + species epithet (lowercase) in italics.

3

Combine kingdom diagnostic traits with formatting rules to select the correct proposal.

Proposal 4 (*Bacillus subtilis* in Kingdom Monera) satisfies both taxonomic classification principles and Linnaean formatting rules.

Only Proposal 4 gets both the formatting conventions and kingdom assignment right.

Key Concept

Linnaean Binomial Nomenclature Rules and Monera Kingdom Diagnostic Features

Question 8717Question →

Match each organism with its characteristic anatomical structure or cellular mechanism used in nutrition and digestion.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Amoeba matches with pseudopodia for intracellular ingestion; Cockroach matches with chitinous teeth in the proventriculus for crushing food; Domestic Fowl matches with a muscular gizzard containing grit for grinding seeds; Rabbit matches with an enlarged caecum for microbial cellulose fermentation.

Each animal demonstrates structural adaptations tailored to its mode of feeding: Amoeba utilizes pseudopodia for phagocytosis; insects like cockroaches use chitinous teeth in the proventriculus to crush solid particles; birds utilize a muscular gizzard with ingested grit to grind grains; and non-ruminant herbivores like rabbits depend on an enlarged caecum containing symbiotic microbes to break down plant cellulose.

Step-by-Step Solution

1

Identify the unicellular mode of nutrition

Amoeba ingests microscopic food via pseudopodia forming food vacuoles.

Single-celled protists lack organs and rely on cellular engulfment.

2

Identify mechanical digestive structures in insects and birds

Cockroaches possess chitinous proventricular teeth, whereas birds possess a muscular gizzard with ingested stones.

Both organisms need mechanical breakdown mechanisms to substitute for oral chewing.

3

Identify herbivorous intestinal adaptations

Rabbits possess a specialized enlarged caecum for hindgut microbial fermentation.

Cellulose breakdown in non-ruminant mammals occurs via symbiotic bacteria in the caecum.

Key Concept

Comparative Digestive Structures and Adaptations in Animals

Question 8718Question →

Two plant specimens, $X$ and $Y$ , were cataloged during a botanical field investigation. Specimen $X$ was recorded as *Zea mays* L., while Specimen $Y$ was recorded as *Zea luxurians*. Based on the principles of Linnaean binomial nomenclature and taxonomic hierarchy, which of the following statements correctly interprets their taxonomic relationship and scientific designation?

Both specimens belong to the same genus but represent different species, with 'L.' designating the abbreviated authority name of the taxonomist who first published the species name.

Specimen

X

is written using trinomial nomenclature because it contains three terms, placing it in a lower taxonomic rank than Specimen

Y

.

The scientific designation of Specimen

Y

is invalid because binomial nomenclature requires both the generic name and the species epithet to begin with upper-case initial letters.

Specimen

X

and Specimen

Y

belong to Kingdom Fungi because organisms with italicized scientific names obtain nutrients exclusively through saprophytic extracellular digestion.

Show answer & explanation

Answer: Both specimens belong to the same genus but represent different species, with 'L.' designating the abbreviated authority name of the taxonomist who first published the species name.

Answer

Both specimens belong to the same genus (*Zea*) but represent distinct species, and 'L.' denotes the authority (Carl Linnaeus) who first described the species.

The correct answer accurately identifies that both specimens share the same genus (*Zea*) but are distinct species (*mays* and *luxurians*), and that 'L.' is the standard author citation representing Carl Linnaeus.

Step-by-Step Solution

1

Identify the generic name of both specimens

Specimen

X

(*Zea mays* L.) and Specimen

Y

(*Zea luxurians*) share the first word *Zea*.

In binomial nomenclature, the first capitalized word indicates the genus to which the organism belongs.

2

Analyze the specific epithets and additional notation

*mays* and *luxurians* are distinct specific epithets. The letter 'L.' stands for Linnaeus.

Different second words indicate distinct species within the same genus. An abbreviated author name following the species epithet denotes authority, not a third name component.

3

Verify capitalization and formatting conventions

The genus name is capitalized, the specific epithet is lowercase, and both are italicized.

Linnaean rules mandate capitalizing the genus while keeping the species epithet in lowercase.

Key Concept

Linnaean binomial nomenclature rules and authority citations

Estimated Time:1m 0s

Question 8719Question →

The solubility of lead(II) trioxonitrate(V), $\text{Pb(NO}_3\text{)}_2$ , in water is $3.5\text{ mol dm}^{-3}$ at $80^\circ\text{C}$ and $1.5\text{ mol dm}^{-3}$ at $30^\circ\text{C}$ . What mass of $\text{Pb(NO}_3\text{)}_2$ will crystallize out when $250\text{ cm}^3$ of its saturated solution is cooled from $80^\circ\text{C}$ to $30^\circ\text{C}$ ? [Molar mass of $\text{Pb(NO}_3\text{)}_2 = 331\text{ g mol}^{-1}$ ]

165.5\text{ g}

0.50\text{ g}

662.0\text{ g}

289.6\text{ g}

Show answer & explanation

Answer: $165.5\text{ g}$

Answer

The mass of

\text{Pb(NO}_3\text{)}_2

that will crystallize out is

165.5\text{ g}

.

Subtracting the molar solubility at

30^\circ\text{C}

from that at

80^\circ\text{C}

gives a precipitation rate of

2.0\text{ mol dm}^{-3}

. Multiplying by the molar mass (

331\text{ g mol}^{-1}

) gives

662.0\text{ g}

per

\text{dm}^3

. Scaling this value for

250\text{ cm}^3

(

0.25\text{ dm}^3

) yields

165.5\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in molar solubility between

80^\circ\text{C}

and

30^\circ\text{C}

.

\text{Molar solubility difference} = 3.5\text{ mol dm}^{-3} - 1.5\text{ mol dm}^{-3} = 2.0\text{ mol dm}^{-3}

This determines the amount of solute in moles that precipitates out per

\text{dm}^3

of solvent.

2

Calculate the mass of solute that crystallizes out from

1\text{ dm}^3

(

1000\text{ cm}^3

).

\text{Mass crystallized in } 1000\text{ cm}^3 = 2.0\text{ mol dm}^{-3} \times 331\text{ g mol}^{-1} = 662.0\text{ g dm}^{-3}

Converting the solubility difference from moles to grams using the given molar mass.

3

Scale the mass to the given solution volume of

250\text{ cm}^3

.

\text{Mass crystallized} = 662.0\text{ g} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3} = 165.5\text{ g}

The volume provided (

250\text{ cm}^3

) is one-quarter of

1000\text{ cm}^3

(

1\text{ dm}^3

).

Key Concept

Mass of solute precipitated upon cooling a saturated solution

Question 8720Question →

Arrange the standard experimental steps for testing a green leaf for the presence of starch in the correct chronological sequence from start to finish.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps to test a green leaf for starch is: (1) Boil the green leaf in water, (2) Boil the leaf in ethanol using a water bath, (3) Rinse the decolorized leaf in warm water, and (4) Spread the leaf on a white tile and add a few drops of iodine solution.

The correct experimental sequence ensures that plant cell membranes are permeable, green pigments that mask color changes are extracted safely in a water bath, the leaf is rehydrated to soften it, and iodine solution can react clearly with any stored starch.

Step-by-Step Solution

1

Boil the green leaf in water

Cell membranes become permeable and enzymatic activity ceases.

High temperature kills the cells and prevents further biochemical reactions.

2

Extract chlorophyll pigment using ethanol in a water bath

The leaf turns pale white/yellowish as chlorophyll dissolves in ethanol.

Decolorization is necessary because green chlorophyll masks the blue-black color of positive starch reaction.

3

Rinse the leaf in warm water

The leaf becomes soft and pliable.

Alcohol dehydration makes the leaf stiff and brittle; water restores flexibility.

4

Add iodine solution

A blue-black color develops if starch is present.

Iodine reacts specifically with starch molecules to form a characteristic blue-black complex.

Key Concept

Starch test procedure as evidence of photosynthesis in green plants

Estimated Time:45s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept