Tüm alıştırma soruları

Soru 1361Soru →

Match each organic nitrogen compound on the left with its corresponding relative basicity or acid-base structural property on the right.

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Cevap

Phenylamine matches with being weaker than ammonia due to aromatic delocalization; Methylamine matches with being stronger than ammonia due to +I inductive effect; Ethanamide matches with being neutral due to carbonyl resonance; Dimethylamine matches with being stronger than methylamine in aqueous solution.

Phenylamine is less basic than ammonia because its lone pair is delocalized across the aromatic ring. Methylamine is more basic than ammonia due to inductive electron donation by the methyl group. Ethanamide is neutral because its lone pair participates in resonance with the carbonyl group. Dimethylamine is more basic than methylamine in water due to two electron-donating methyl groups.

Adım Adım Çözüm

1

Analyze how structural features influence nitrogen lone pair availability.

Electron-donating alkyl groups (+I effect) enhance lone pair availability (increasing basicity), while electron-withdrawing groups or resonance delocalization decrease lone pair availability (decreasing basicity).

Lewis/Brønsted-Lowry basicity of nitrogen compounds depends directly on lone pair availability to accept a proton.

2

Evaluate phenylamine and ethanamide.

Phenylamine delocalizes its lone pair into the benzene ring, making it weaker than

NH_3

. Ethanamide delocalizes its lone pair into the

C=O

double bond, making it neutral in aqueous solution.

Resonance delocalization significantly stabilizes the unprotonated state and lowers basicity.

3

Compare methylamine and dimethylamine.

Methylamine has one alkyl group increasing basicity over

NH_3

. Dimethylamine has two alkyl groups supplying greater electron density, making it more basic than methylamine in aqueous solution.

Inductive electron donation by methyl groups stabilizes the positive conjugate ammonium ion.

Anahtar Kavram

Relative basicity of amines and amides based on inductive and resonance effects

Soru 1362Soru →

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

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Cevap

Helium matches with weather balloons and diving gas mixtures; Neon matches with advertising signs; Argon matches with arc welding inert atmosphere; Radon matches with cancer radiotherapy.

Each noble gas is matched to its unique application based on its physical and chemical properties: Helium for low density and low blood solubility, Neon for glowing light discharge, Argon for unreactive shielding during welding, and Radon for cancer radiotherapy.

Adım Adım Çözüm

1

Identify the primary application of Helium.

Helium's low density and minimal blood solubility pair it with weather balloons and deep-sea diving mixtures.

Prevents decompression sickness in divers and provides buoyancy in balloons.

2

Identify the primary application of Neon.

Neon emits a distinct reddish-orange light in electrical discharge tubes used for advertising signs.

Excited neon gas emits characteristic light when electrical discharge occurs.

3

Identify the primary application of Argon.

Argon serves as an inert shielding gas in electric arc welding.

Prevents atmospheric oxygen and nitrogen from reacting with hot metals being welded.

4

Identify the primary application of Radon.

Radon is radioactive and used in cancer radiotherapy.

Radiation emitted during radioactive decay destroys targeted cancer cells.

Anahtar Kavram

Industrial applications and chemical inertness of noble gases

Soru 1363Soru →

Match each industrial separation task on the left with the appropriate physical separation method used in the chemical industry on the right.

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Cevap

Removal of iron impurities from cassiterite matches Magnetic separation; Separation of crude oil into petrol, kerosene, and diesel matches Fractional distillation; Recovery of vegetable oil using hexane matches Solvent extraction; Obtaining pure sugar crystals matches Fractional crystallization.

Each industrial process isolates target substances based on distinct physical properties: magnetic properties allow removal of magnetic iron from non-magnetic tin ore; boiling point differences allow separation of petroleum fractions via fractional distillation; organic solubility allows solvent extraction of vegetable oil; and solubility differences upon cooling allow recovery of pure sugar by fractional crystallization.

Adım Adım Çözüm

1

Identify the physical properties exploited in each industrial scenario.

Magnetic susceptibility, differences in boiling points, selective solubility in liquid solvents, and temperature-dependent crystallization/solubility.

Industrial separation techniques rely directly on differences in physical properties between components of raw materials or mixtures.

2

Match each industrial scenario to its corresponding separation technique.

Iron impurities in tin ore are removed magnetically; crude oil fractions are separated by boiling point in a fractionating column; vegetable oil is dissolved out of seeds using solvent extraction; sugar is purified by crystallization from solution.

Connecting physical property differences to their correct industrial unit operations.

Anahtar Kavram

Industrial applications of physical separation methods based on unique physical properties.

Soru 1364Soru →

Match each industrial separation requirement on the left with its corresponding separation technique applied in chemical manufacturing on the right.

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Cevap

The extraction of heat-sensitive essential oils matches steam distillation; the concentration of low-grade copper sulfide ore matches froth flotation; the recovery of residual vegetable oil from oilseed cake matches solvent extraction; and the harvesting of sodium chloride from seawater brine matches solar evaporative crystallization.

Each industrial process matches a specific physical property of the mixture: steam distillation isolates thermolabile oils without heat damage; froth flotation separates sulfide ores from rock gangue based on differential surface wettability; solvent extraction uses non-polar liquids like hexane to dissolve residual oils from crushed seeds; and solar evaporative crystallization deposits sodium chloride salts as water evaporates naturally from brine.

Adım Adım Çözüm

1

Identify the separation technique suitable for thermolabile volatile compounds.

Essential oils decompose when heated directly to high temperatures, making steam distillation the ideal process because steam lowers the effective boiling temperature of immiscible volatile liquids.

Reduces thermal degradation of delicate plant extracts.

2

Determine the ore beneficiation method based on surface wettability.

Copper sulfide minerals are hydrophobic and readily adhere to air bubbles generated in a froth flotation cell, while hydrophilic rock gangue sinks to the bottom.

Exploits differences in surface tension and hydrophobic properties.

3

Analyze how residual non-polar compounds are extracted from solid agricultural residue.

Solvent extraction using non-polar organic solvents (e.g., hexane) efficiently dissolves lipids from oilseed cakes where mechanical pressing is insufficient.

Utilizes selective solubility in organic solvents.

4

Select the industrial technique for salt recovery from saline water.

Solar evaporative crystallization uses solar radiation to continuously remove water from seawater until sodium chloride exceeds its solubility product and precipitates out.

Cost-effective, large-scale crystallization driven by solar evaporation.

Anahtar Kavram

Industrial applications of separation methods based on physical properties such as thermal stability, surface wettability, solubility, and volatility.

Soru 1365Soru →

Match each nitrogen oxide or nitrogen cycle component in Column I with its correct physical property or biological role in Column II.

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Cevap

Dinitrogen monoxide (

N_2O

) matches with 'A sweet-smelling, neutral gas that relights a glowing splint'. Nitrogen dioxide (

NO_2

) matches with 'A reddish-brown, acidic gas that dissolves in water to form a mixture of two acids'. Nitrosomonas bacteria matches with 'Converts soil ammonium ions (

NH_4^+

) into trioxonitrate(III) ions (

NO_2^-

)'. Nitrobacter bacteria matches with 'Converts soil trioxonitrate(III) ions (

NO_2^-

) into trioxonitrate(V) ions (

NO_3^-

)'.

Each nitrogen oxide and nitrifying bacterium is correctly associated with its characteristic chemical behavior or distinct biochemical pathway in the nitrogen cycle.

Adım Adım Çözüm

1

Differentiate between the physical and chemical properties of the nitrogen oxides.

N_2O

is neutral and sweet-smelling while supporting combustion.

NO_2

is acidic, reddish-brown, and forms

HNO_2

and

HNO_3

upon reaction with water.

Oxides of nitrogen vary in color, acidity, and combustion-supporting capabilities depending on the oxidation state of nitrogen.

2

Differentiate the roles of nitrifying bacteria in the nitrogen cycle.

Nitrosomonas oxidizes ammonium to nitrite (

NO_2^-

), whereas Nitrobacter oxidizes nitrite (

NO_2^-

) to nitrate (

NO_3^-

).

Nitrification proceeds in two distinct enzymatic steps mediated by specialized microbial species.

Anahtar Kavram

Properties of Nitrogen Oxides and Biological Nitrification Stages

Soru 1366Soru →

Pair each basic genetic term listed on the left with its correct biological description on the right.

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Cevap

Heterozygous matches with possessing two non-identical alleles; Genotype matches with the specific allele composition; Dominant allele matches with an allele that masks its contrasting form; Phenotype matches with the observable physical traits.

Each genetic term matches its corresponding definition precisely: Heterozygous indicates possessing two non-identical alleles at a locus; Genotype represents the organism's specific allele composition; Dominant allele describes an allele that masks the expression of its contrasting alternative form; and Phenotype refers to the observable physical or physiological traits.

Adım Adım Çözüm

1

Define Heterozygous

Heterozygous refers to having different alleles at a given gene locus.

The prefix 'hetero-' means different, indicating non-identical alleles.

2

Define Genotype

Genotype represents the internal genetic code or allele constitution.

Genotype specifies the genetic information rather than the outward appearance.

3

Define Dominant allele

A dominant allele masks the phenotypic effect of its alternative allele.

Dominance means the trait is expressed whenever at least one copy of the allele is present.

4

Define Phenotype

Phenotype represents the observable attributes and physical characteristics.

Phenotype is the external expression resulting from genotypic and environmental factors.

Anahtar Kavram

Basic Genetics Terminology and Concepts

Tahmini Süre:1m 0s

Soru 1367Soru →

Match each organism or ecological role on the left with its corresponding trophic level or function in energy flow on the right.

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Cevap

Green plants and phytoplankton match with Primary producers (Trophic Level 1); Herbivores such as grasshoppers match with Primary consumers (Trophic Level 2); Carnivores such as frogs match with Secondary consumers (Trophic Level 3); Saprophytic fungi and bacteria match with Decomposers recycling organic matter.

Organisms are categorized into trophic levels based on their source of energy: photosynthetic autotrophs are primary producers (Level 1), plant-eaters are primary consumers (Level 2), animal-eaters preying on herbivores are secondary consumers (Level 3), and decay organisms recycle organic matter as decomposers.

Adım Adım Çözüm

1

Identify the energy-capturing organisms in an ecosystem.

Green plants and phytoplankton capture solar energy to synthesize food, placing them at Trophic Level 1 as primary producers.

Primary producers form the foundational trophic level in all food chains.

2

Identify organisms that directly feed on producers.

Herbivores such as grasshoppers consume plant matter directly, placing them at Trophic Level 2 as primary consumers.

Direct consumers of autotrophs occupy the second trophic position.

3

Identify organisms that prey on primary consumers.

Carnivores such as frogs feed on primary consumers (herbivores), placing them at Trophic Level 3 as secondary consumers.

Predators of herbivores occupy the third trophic position in energy transfer.

4

Identify organisms responsible for breaking down dead organic waste.

Saprophytic fungi and bacteria digest non-living organic matter, functioning as decomposers.

Decomposers facilitate nutrient recycling back into abiotic ecosystem pools.

Anahtar Kavram

Trophic level classification and functional roles in food chains

Soru 1368Soru →

Match each salt in aqueous solution to the correct chemical description of its hydrolysis behavior and resulting pH at $25^\circ\text{C}$ .

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Cevap

Ammonium sulfate pairs with the acidic cation-hydrolyzing description; Sodium propanoate pairs with the alkaline anion-hydrolyzing description; Potassium nitrate pairs with the neutral non-hydrolyzing description; Ammonium cyanide pairs with the alkaline dual-hydrolyzing description where

K_b > K_a

.

The solution pH resulting from salt hydrolysis depends directly on the relative strengths of the parent acid and base. Salts derived from weak bases and strong acids yield acidic solutions via cation hydrolysis. Salts derived from strong bases and weak acids yield alkaline solutions via anion hydrolysis. Salts of strong acids and strong bases do not undergo net hydrolysis, remaining neutral. For salts derived from both a weak acid and a weak base, both ions undergo hydrolysis, and the solution acidity or alkalinity is determined by comparing the

K_b

of the weak base to the

K_a

of the weak acid.

Adım Adım Çözüm

1

Identify the parent acid and parent base for each salt.

(NH_4)_2SO_4

comes from

NH_3

(weak base) and

H_2SO_4

(strong acid).

CH_3CH_2COONa

comes from

NaOH

(strong base) and

CH_3CH_2COOH

(weak acid).

KNO_3

comes from

KOH

(strong base) and

HNO_3

(strong acid).

NH_4CN

comes from

NH_3

(weak base) and

HCN

(weak acid).

The strength of parent acids and bases determines which ions undergo hydrolysis in water.

2

Determine which ion hydrolyzes for single-weak component salts.

In

(NH_4)_2SO_4

,

NH_4^+

hydrolyzes to produce

H_3O^+

(acidic,

pH < 7

). In

CH_3CH_2COONa

,

CH_3CH_2COO^-

hydrolyzes to produce

OH^-

(alkaline,

pH > 7

). In

KNO_3

, neither ion hydrolyzes (neutral,

pH = 7

).

Conjugate ions of weak species react with water, whereas conjugate ions of strong species do not hydrolyze.

3

Compare ionization constants for the weak acid-weak base salt.

For

NH_4CN

, compare

K_b(NH_3) = 1.8 \times 10^{-5}

with

K_a(HCN) = 6.2 \times 10^{-10}

. Since

K_b > K_a

,

CN^-

anion hydrolysis produces more

OH^-

than

NH_4^+

cation hydrolysis produces

H_3O^+

, resulting in an alkaline solution (

pH > 7

).

When both ions hydrolyze, the relative magnitudes of

K_a

and

K_b

govern whether

OH^-

or

H_3O^+

is in excess.

Anahtar Kavram

Salt Hydrolysis and Solution Acidity/Alkalinity

Tahmini Süre:1m 30s

Soru 1369Soru →

In fruit flies (*Drosophila melanogaster*), the allele for normal wings ( $V$ ) is completely dominant over the allele for vestigial wings ( $v$ ). Match each parental cross listed on the left with its corresponding expected offspring phenotypic or genotypic ratio on the right.

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Cevap

Cross

Vv \times Vv

matches 3:1 phenotypic ratio (normal:vestigial). Test cross

Vv \times vv

matches 1:1 phenotypic ratio. Cross

VV \times vv

matches 100% normal wings (all

Vv

). Cross

VV \times Vv

matches 100% normal wings (1:1 genotypic ratio

VV:Vv

).

Each monohybrid cross yields specific offspring ratios according to Mendel's Law of Segregation. Crossing two heterozygotes (

Vv \times Vv

) yields a 3:1 phenotypic ratio. A test cross (

Vv \times vv

) yields a 1:1 phenotypic ratio. Crossing homozygous dominant with homozygous recessive (

VV \times vv

) produces 100% heterozygous offspring (

Vv

). Crossing homozygous dominant with a heterozygote (

VV \times Vv

) yields 100% dominant phenotype with a 1:1 genotypic ratio of

VV

to

Vv

.

Adım Adım Çözüm

1

Analyze the cross

Vv \times Vv

Produces genotypes

1\ VV : 2\ Vv : 1\ vv

. Since

V

is dominant, 3 parts show normal wings and 1 part shows vestigial wings (3:1 phenotypic ratio).

Mendel's Law of Segregation states that two alleles of a gene separate during gamete formation.

2

Analyze the test cross

Vv \times vv

Gametes from

Vv

are

V

and

v

; gametes from

vv

are

v

. Offspring are 50%

Vv

and 50%

vv

(1:1 phenotypic ratio).

A monohybrid test cross pairs a heterozygous individual with a homozygous recessive individual.

3

Analyze the cross

VV \times vv

All offspring inherit

V

from the dominant parent and

v

from the recessive parent, resulting in 100%

Vv

(100% normal wings).

Homozygous dominant crossed with homozygous recessive produces uniformly heterozygous F1 offspring.

4

Analyze the cross

VV \times Vv

Offspring genotypes are 50%

VV

and 50%

Vv

. Because all possess at least one dominant allele

V

, 100% display normal wings.

The dominant allele masks the recessive allele in heterozygous individuals.

Anahtar Kavram

Mendel's First Law and Monohybrid Inheritance Ratios

Soru 1370Soru →

Match each alloy listed on the left with its characteristic elemental composition and primary application on the right.

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Cevap

Duralumin pairs with

\text{Al}

,

\text{Cu}

,

\text{Mg}

,

\text{Mn}

(aircraft construction); Brass pairs with

\text{Cu}

,

\text{Zn}

(musical instruments/fittings); Stainless Steel pairs with

\text{Fe}

,

\text{Cr}

,

\text{Ni}

,

\text{C}

(cutlery/surgical tools); Soft Solder pairs with

\text{Pb}

,

\text{Sn}

(joining electrical connections).

Each alloy is correctly matched according to its primary constituent elements and application: Duralumin (

\text{Al}

,

\text{Cu}

,

\text{Mg}

,

\text{Mn}

) for aircraft bodywork; Brass (

\text{Cu}

,

\text{Zn}

) for musical instruments and fittings; Stainless Steel (

\text{Fe}

,

\text{Cr}

,

\text{Ni}

,

\text{C}

) for corrosion-resistant cutlery and medical tools; Soft Solder (

\text{Pb}

,

\text{Sn}

) for low-temperature electrical joint soldering.

Adım Adım Çözüm

1

Identify the base metals and secondary additives for light-engineering alloys.

Duralumin is an aluminium-based alloy with

\text{Cu}

,

\text{Mg}

, and

\text{Mn}

engineered for aerospace applications due to low density and high mechanical strength.

Aluminium provides low mass while added metals induce lattice distortion to increase hardness.

2

Distinguish between copper-zinc and copper-tin alloys.

Brass is made of copper and zinc, which differs from Bronze (copper and tin). Brass is widely used for decorative fittings and musical instruments.

Zinc substitution in the copper matrix enhances workability and corrosion resistance.

3

Identify steel variations based on anti-corrosion alloying elements.

Stainless steel contains iron, chromium, nickel, and carbon. Chromium imparts a self-healing passive oxide coating.

Chromium content (typically >10.5%) resists oxidative rusting in moist atmospheric conditions.

4

Recall low-melting-point joining alloys.

Soft solder is an alloy of lead and tin engineered to melt at relatively low temperatures (

< 250^\circ\text{C}

).

The eutectic composition of lead and tin depresses the melting point below that of either constituent element.

Anahtar Kavram

Elemental compositions, structural properties, and functional uses of key industrial alloys

Soru 1371Soru →

Match each chemical reaction or process involving alkanoic acid derivatives on the left with its corresponding principal product on the right.

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Cevap

The reaction of ethanoic acid and ethanol (esterification) pairs with Ethyl ethanoate and water. The alkaline hydrolysis of oil (saponification) pairs with Sodium salt of fatty acid (soap) and glycerol. The addition of hydrogen to liquid oil (catalytic hydrogenation) pairs with Solid saturated fat (margarine).

Esterification of ethanoic acid and ethanol yields ethyl ethanoate and water. Saponification of vegetable oils with aqueous sodium hydroxide yields fatty acid sodium salts (soap) and glycerol. Catalytic hydrogenation of unsaturated vegetable oils yields solid saturated fats (margarine).

Adım Adım Çözüm

1

Identify the products of esterification

Ethanoic acid reacts with ethanol in the presence of concentrated tetraoxosulfate(VI) acid to produce ethyl ethanoate and water.

The hydroxyl group of the acid combines with hydrogen from the alkanol to form water, linking the remaining fragments into an ester.

2

Identify the products of saponification

Triglycerides react with boiling aqueous sodium hydroxide to yield sodium alkanoates (soap) and propane-1,2,3-triol (glycerol).

Alkaline cleavage of ester linkages in fats yields carboxylate salts and frees the triol backbone.

3

Identify the products of catalytic hydrogenation

Unsaturated fatty acid chains in liquid vegetable oils undergo addition of hydrogen to form saturated chains, hardening the oil into solid fat.

Reducing double bonds increases the melting point, transforming liquid oils into margarine.

Anahtar Kavram

Reactions and Industrial Products of Alkanoic Acids, Esters, Fats, and Oils

Soru 1372Soru →

Match each structural component of a virus on the left with its corresponding chemical composition or biological function on the right.

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Cevap

Capsid matches with the protein coat enclosing the nucleic acid; Genetic core matches with DNA or RNA carrying hereditary instructions; Envelope matches with lipid membrane derived from host cell; Tail fibers match with protein appendages for host receptor attachment.

Each viral component serves a specific structural or biochemical role: Capsid protects genetic material via a protein coat; Genetic core contains DNA or RNA genome; Envelope is host-derived lipid bilayer; Tail fibers facilitate target cell attachment.

Adım Adım Çözüm

1

Identify the primary protective layer of a virus

The capsid is composed of capsomeres (proteins) and surrounds the nucleic acid core.

Proteins form the outer structural coat of all non-enveloped and enveloped viruses.

2

Identify the nucleic acid component

The genetic core consists of either single-stranded or double-stranded DNA or RNA.

Viruses possess a single type of nucleic acid carrying their genetic code.

3

Identify the lipid-containing structure

The viral envelope consists of lipids acquired from host cellular membranes.

Enveloped viruses exit host cells by budding, picking up host membrane lipids.

4

Identify the host attachment apparatus

Tail fibers function in host cell recognition and anchoring.

Bacteriophage tail fibers bind specifically to bacterial cell wall receptors.

Anahtar Kavram

Structural organization and chemical composition of viral components

Soru 1373Soru →

Match each industrial electrolytic process on the left with its corresponding chemical characteristic or operating condition on the right.

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Cevap

Hall-Héroult Process matches with 'Electrolysis of molten alumina dissolved in molten cryolite using carbon electrodes'. Electrorefining of Copper matches with 'Impure metal serves as the dissolving anode while pure metal deposits at the cathode'. Electroplating of Iron with Silver matches with 'The article to be coated serves as the cathode in an electrolyte containing silver ions'. Chlor-Alkali Process matches with 'Electrolysis of concentrated brine yielding chlorine gas, hydrogen gas, and sodium hydroxide'.

Each industrial process relies on specific redox reactions and cell configurations: Hall-Héroult process uses molten cryolite as a solvent for alumina; copper refining uses impure copper as the dissolving anode; electroplating places the article to be coated at the cathode; chlor-alkali electrolyzes brine to yield chlorine, hydrogen, and sodium hydroxide.

Adım Adım Çözüm

1

Identify the key components and conditions of aluminium extraction.

Aluminium is extracted by electrolyzing molten alumina (

Al_2O_3

) dissolved in cryolite (

Na_3AlF_6

) to lower its melting point.

This corresponds to the Hall-Héroult process operating condition.

2

Identify the electrode roles in metal purification.

Impure metal dissolves at the anode and deposits as pure metal at the cathode.

This defines electrorefining of copper.

3

Identify the arrangement for electroplating.

The target object to be coated forms the cathode where metal cations are reduced.

This describes the electroplating of iron with silver.

4

Identify the reactants and products of brine electrolysis.

Concentrated

NaCl

solution electrolyzed gives

Cl_2

,

H_2

, and

NaOH

.

This defines the chlor-alkali industrial process.

Anahtar Kavram

Industrial applications of electrolysis including metal extraction, refining, electroplating, and chlor-alkali synthesis

Soru 1374Soru →

Match each Lamarckian evolutionary postulate or related historical concept on the left with its corresponding descriptive statement or experimental critique on the right.

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Cevap

Principle of Use and Disuse matches the statement regarding organ hypertrophy through exertion and atrophy through disuse. Inheritance of Acquired Characteristics matches the statement regarding somatic modifications being passed to progeny. Environmental Need (Besoin) matches the statement about environmental shifts creating new demands that drive behavioral and structural adaptations. Germplasm Theory Critique matches the experimental proof that somatic changes do not affect germ cells.

Each concept correctly aligns with its historical definition or scientific critique. The principle of use and disuse concerns organ development through exertion or atrophy through neglect. Inheritance of acquired characteristics describes the transfer of somatic traits to progeny. Environmental need explains how changing surroundings prompt adaptive responses. Germplasm theory refutes Lamarckism by demonstrating the separation of somatic and reproductive cells.

Adım Adım Çözüm

1

Analyze the core postulates of Lamarckism and the primary historical counter-evidence.

Identify the definitions of use and disuse, acquired traits, environmental needs, and Weismann's germplasm theory.

Clear differentiation between Lamarck's mechanisms and their experimental refutation is necessary to pair each concept accurately.

2

Match 'Principle of Use and Disuse' to its mechanism.

Pairs with the description detailing organ enlargement from frequent use and degeneration from disuse.

Lamarck proposed that physical exertion directly alters organ structure within an individual's lifetime.

3

Match 'Inheritance of Acquired Characteristics' to its definition.

Pairs with the statement that lifespan modifications are transmitted to subsequent generations.

This postulate erroneously assumed somatic changes could be inherited.

4

Match 'Environmental Need (Besoin)' and 'Germplasm Theory Critique' to their remaining descriptions.

Environmental Need pairs with environmental shifts creating new demands, and Germplasm Theory Critique pairs with the distinction between somatic and germ cells.

Lamarck emphasized environmental stimulus for adaptation, while Weismann proved germ cells are isolated from somatic changes.

Anahtar Kavram

Lamarckian Evolutionary Principles and Germplasm Counter-Evidence

Soru 1375Soru →

Match each organic reaction involving an amine or amide on the left with its corresponding principal product on the right.

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Cevap

Reduction of propanamide with

LiAlH_4

pairs with propylamine; Reaction of propanamide with

Br_2/KOH

pairs with ethylamine; Alkaline hydrolysis of propanamide pairs with sodium propanoate and ammonia; Acylation of methylamine with ethanoyl chloride pairs with

N

-methylethanamide.

Each reaction pair is determined by its specific mechanistic pathway:

LiAlH_4

reduces the carbonyl group to methylene (retaining carbon count to form propylamine);

Br_2/KOH

undergoes Hofmann degradation to lose the carbonyl carbon (forming ethylamine); basic hydrolysis cleaves the

C-N

bond (yielding sodium propanoate and ammonia); and acylation of methylamine with ethanoyl chloride produces the substituted amide (

N

-methylethanamide).

Adım Adım Çözüm

1

Analyze the reduction reaction of primary amides.

Reducing

CH_3CH_2CONH_2

with

LiAlH_4

reduces the

C=O

bond to a

-CH_2-

group without altering the total carbon count, yielding

CH_3CH_2CH_2NH_2

(propylamine).

Amide reduction retains the full carbon skeleton.

2

Identify the reaction of primary amides with

Br_2

and

KOH

.

This is Hofmann degradation, which removes the carbonyl carbon (

C=O

) as carbonate, reducing the carbon length by 1. Propanamide (3 carbons) yields ethylamine (2 carbons).

Hofmann degradation shortens the carbon chain by one atom.

3

Examine the basic hydrolysis of amides.

Nucleophilic attack of

OH^-

on the carbonyl carbon of propanamide cleaves the amide bond to generate propanoate anion (forming sodium propanoate with

Na^+

) and ammonia gas.

Base hydrolysis of amides yields a carboxylate salt and ammonia.

4

Examine the nucleophilic substitution between methylamine and ethanoyl chloride.

The nitrogen lone pair of methylamine attacks ethanoyl chloride, releasing

HCl

to form a secondary amide,

N

-methylethanamide (

CH_3CONHCH_3

).

Primary amines undergo acylation to form secondary amides.

Anahtar Kavram

Chemical Reactions and Interconversions of Amines and Amides

Tahmini Süre:1m 30s

Soru 1376Soru →

Match each viral structural component on the left with its corresponding biochemical nature or function on the right.

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Cevap

Capsid matches with the protective protein shell surrounding and shielding the viral genome; Capsomere matches with individual protein sub-units forming the viral coat; Envelope matches with the outer lipid bilayer membrane derived from host cell membranes; Nucleic acid core matches with the central hereditary genetic material containing either DNA or RNA, but never both.

Viruses possess an acellular structure made of a protein capsid (composed of capsomeres) encapsulating a genome of either DNA or RNA. Enveloped viruses possess an outer lipid membrane acquired from host membranes.

Adım Adım Çözüm

1

Analyze the protective protein architecture of viruses.

The capsid forms the overall protective protein shell surrounding viral genetic material, constructed from smaller protein subunits termed capsomeres.

Viruses rely on structural protein coats to shield genetic material from environmental damage.

2

Distinguish between viral membrane modifications and genetic material properties.

Envelopes consist of host-derived lipids, whereas the nucleic acid core contains exclusively a single type of nucleic acid (either DNA or RNA).

Viruses are acellular and do not contain both nucleic acid types simultaneously.

Anahtar Kavram

Structural components and biochemical makeup of viruses

Soru 1377Soru →

Match each characteristic property of transition metals on the left with its fundamental atomic or electronic explanation on the right.

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Cevap

Formation of colored ions matches excitation of electrons between split d-orbital energy levels; Variable oxidation states matches small energy difference between (n-1)d and ns subshells; Paramagnetism matches presence of unpaired d-electrons; High catalytic efficiency matches ability to adopt multiple oxidation states and provide active surface sites.

Transition elements owe their distinct properties to incompletely filled

d

-subshells. Colored compounds are created by

d

-

d

electron transitions when visible light is absorbed. Variable oxidation states arise because

3d

and

4s

energy levels are very close, so electrons from both subshells participate in reaction pathways. Paramagnetism originates from unpaired

d

-electrons, while catalytic behavior is driven by vacant/partially filled

d

-orbitals that adsorb reactants and facilitate intermediate oxidation states.

Adım Adım Çözüm

1

Analyze the cause of color in transition metal complexes.

Ligands split the degenerate

d

-orbitals into different energy levels. Absorption of visible light promotes an electron (

d

-

d

transition), imparting color.

Relates the macroscopic color property to internal crystal field splitting.

2

Analyze why transition metals exhibit multiple oxidation states.

The energy gap between

(n-1)d

and

ns

subshells (such as

3d

and

4s

) is minimal, enabling electrons from both subshells to participate in bonding.

Explains why metals like Iron can exist as

Fe^{2+}

and

Fe^{3+}

.

3

Determine the electronic basis of paramagnetism.

Unpaired electrons possess a net magnetic spin moment, causing the ion or atom to be attracted into an external magnetic field.

Distinguishes paramagnetism (unpaired electrons) from diamagnetism (all paired electrons).

4

Examine how transition elements act as catalysts.

Partially filled

d

-orbitals adsorb reactants onto active sites, and variable oxidation states allow the metal to lower activation energy by forming intermediate species.

Connects surface adsorption and redox cycles to catalytic mechanism.

Anahtar Kavram

Electronic Configurations and Characteristic Properties of Transition Elements

Soru 1378Soru →

In cattle, coat color is determined by a single gene exhibiting complete dominance, where the allele for black coat ( $B$ ) is dominant over the allele for red coat ( $b$ ). Match each monohybrid cross scenario with its correct genotypic or phenotypic outcome outcome/deduction.

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Cevap

The monohybrid cross scenarios match their expected outcomes as follows: The cross between a heterozygous black bull and a red cow yields a phenotypic ratio of 1 black : 1 red; the cross between two heterozygous black cattle yields a genotypic ratio of 1 BB : 2 Bb : 1 bb; the cross between a homozygous black bull and a heterozygous black cow produces a genotypic ratio of 1 BB : 1 Bb (100% black); and a test cross producing 100% black offspring leads to the deduction that the tested parent is homozygous dominant (BB).

Each monohybrid inheritance scenario follows Mendel's First Law (Law of Segregation). Gametes segregate cleanly during meiosis, recombining in predictable frequencies: a heterozygous backcross (

Bb \times bb

) yields equal proportions of black and red progeny (

1:1

); two heterozygotes (

Bb \times Bb

) yield the classic

1\ BB : 2\ Bb : 1\ bb

genotypic ratio (

3:1

phenotype); a dominant homozygote crossed with a heterozygote (

BB \times Bb

) yields equal proportions of

BB

and

Bb

(

100\%

black); and a test cross yielding zero recessive phenotypes confirms homozygosity of the dominant parent.

Adım Adım Çözüm

1

Analyze the cross

Bb \times bb

Gametes from

Bb

are

B

(

50\%

) and

b

(

50\%

). Gametes from

bb

are all

b

(

100\%

). Offspring genotypes are

50\%\ Bb

(black coat) and

50\%\ bb

(red coat).

Demonstrates a standard testcross ratio of

1 : 1

for phenotypes.

2

Analyze the cross

Bb \times Bb

Punnett square analysis gives

25\%\ BB

,

50\%\ Bb

, and

25\%\ bb

.

This confirms Mendel's classic monohybrid F2 genotypic ratio of

1\ BB : 2\ Bb : 1\ bb

and phenotypic ratio of

3 : 1

.

3

Analyze the cross

BB \times Bb

The homozygous dominant parent provides only

B

alleles. Offspring genotypes are

50\%\ BB

and

50\%\ Bb

. All offspring display the black coat phenotype.

This matches the genotypic ratio

1\ BB : 1\ Bb

with

100\%

dominant phenotype.

4

Analyze the test cross principle for an unknown dominant phenotype

Crossing

B\_

with

bb

gives

bb

offspring only if the parent carries a hidden

b

allele. Receiving

100\%

dominant offspring confirms the parent is

BB

.

Confirms the diagnostic utility of test crosses for establishing zygosity.

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Genotypic/Phenotypic Ratios

Tahmini Süre:2m 0s

Soru 1379Soru →

Match each copper-containing ore with its corresponding chemical formula.

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Cevap

Malachite matches

\text{CuCO}_3\cdot\text{Cu(OH)}_2

, Copper pyrites matches

\text{CuFeS}_2

, Cuprite matches

\text{Cu}_2\text{O}

, and Chalcocite matches

\text{Cu}_2\text{S}

.

Each copper ore correctly maps to its characteristic chemical formula: Malachite is basic copper carbonate

\text{CuCO}_3\cdot\text{Cu(OH)}_2

; Copper pyrites is copper iron disulfide

\text{CuFeS}_2

; Cuprite is copper(I) oxide

\text{Cu}_2\text{O}

; and Chalcocite is copper(I) sulfide

\text{Cu}_2\text{S}

.

Adım Adım Çözüm

1

Identify the chemical nature of Malachite

Malachite is a basic carbonate mineral of copper with the formula

\text{CuCO}_3\cdot\text{Cu(OH)}_2

.

Basic copper carbonate consists of copper carbonate and copper hydroxide in a 1:1 mole ratio.

2

Identify the chemical nature of Copper pyrites

Copper pyrites (chalcopyrite) has the formula

\text{CuFeS}_2

.

It is the chief ore of copper containing both iron and copper sulfides.

3

Identify the chemical nature of Cuprite

Cuprite is copper(I) oxide,

\text{Cu}_2\text{O}

.

Cuprite is a red oxide ore containing copper in the +1 oxidation state.

4

Identify the chemical nature of Chalcocite

Chalcocite is copper(I) sulfide,

\text{Cu}_2\text{S}

.

Chalcocite is a dark sulfide ore of copper.

Anahtar Kavram

Chemical composition and formulas of primary copper ores

Tahmini Süre:1m 0s

Soru 1380Soru →

Match each core concept of Charles Darwin's theory of natural selection on the left with its corresponding description on the right.

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Overproduction matches with organisms producing more offspring than the environment can support; Struggle for existence matches with individuals competing for limited environmental resources; Survival of the fittest matches with organisms with advantageous traits being more likely to survive and reproduce; Inheritance of variation matches with favorable inherited traits being transmitted to subsequent generations.

Each core concept of natural selection directly aligns with a key evolutionary process: overproduction creates resource pressure; scarce resources lead to competition (struggle for existence); advantageous traits enable differential reproductive success (survival of the fittest); and passing those favorable traits to offspring alters population characteristics over time.

Adım Adım Çözüm

1

Define overproduction within Darwin's theoretical framework.

Species have a natural tendency to produce more offspring than their habitat can sustain.

High reproductive potential creates population pressure against finite natural resources.

2

Identify the primary consequence of overproduction and limited resources.

Organisms must engage in a struggle for existence.

Scarcity of essential resources like food, light, and shelter leads to competition.

3

Relate individual differences to survival outcomes.

Organisms with advantageous variations undergo differential survival (survival of the fittest).

Traits that better adapt an individual to its environment increase its probability of surviving and reproducing.

4

Explain how natural selection affects future generations.

Advantageous variations are inherited by offspring.

Heritable favorable traits become increasingly common in subsequent generations.

Anahtar Kavram

Core Concepts and Principles of Natural Selection

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