I85PREP
⌘ K
Bac Sciences Expérimentales Algeria - independent subject practice study materials
Bac Sciences Expérimentales Algeria - independent subject practice study resources

Bac Sciences Expérimentales Algeria - independent subject practice practice questions

One question at a time. Build your Bac Sciences Expérimentales Algeria - independent subject practice understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Cells, biomolecules, heredity and metabolism

During transcription of genetic information in eukaryotic cells, which enzyme is directly responsible for synthesizing pre-mRNA by polymerizing ribonucleotides complementary to the template DNA strand?

  1. RNA polymerase II Correct answer.
  2. DNA polymerase III
  3. DNA ligase
  4. Reverse transcriptase

Explanation

RNA polymerase II is the primary eukaryotic enzyme responsible for transcribing protein-coding genes into precursor messenger RNA (pre-mRNA). It reads the DNA template strand in the 3' to 5' direction and synthesizes the RNA transcript in the 5' to 3' direction using ribonucleoside triphosphates.

Question 2Cells, biomolecules, heredity and metabolism

Which sequence of molecular modifications converts eukaryotic precursor mRNA (pre-mRNA) into mature mRNA capable of being exported from the nucleus to the cytoplasm?

  1. Removal of exons, addition of a 3' inverted guanosine cap, and circularization of the transcript
  2. Addition of a 5' 7-methylguanosine cap, addition of a 3' poly-A tail, and splicing out of non-coding introns Correct answer.
  3. Cleavage of the promoter region, phosphorylation of adenine residues, and methylation of all uracil bases
  4. Binding to the small ribosomal subunit, translation of leader sequences, and exon ligation

Explanation

In eukaryotic post-transcriptional processing, primary pre-mRNA undergoes 5' capping with 7-methylguanosine (protecting the 5' end and facilitating translation initiation), 3' polyadenylation with a poly-A tail (providing nuclear export and stability), and spliceosome-mediated splicing to excise non-coding introns and join coding exons.

Question 3Cells, biomolecules, heredity and metabolism

An mRNA codon changes from GAA to GAG, but the translated amino acid remains glutamate. Which property of the genetic code explains this result?

  1. A single codon can specify multiple distinct amino acids depending on cellular conditions
  2. Codons overlap along the mRNA strand so that adjacent nucleotides participate in two codons
  3. Multiple distinct codons can code for the same amino acid Correct answer.
  4. Stop codons mutate spontaneously into initiation codons during stressful conditions

Explanation

Both GAA and GAG specify glutamate. The existence of multiple codons for the same amino acid is called degeneracy or redundancy. The code remains unambiguous because each individual codon specifies only one amino acid.

Question 4Cells, biomolecules, heredity and metabolism

During amino acid activation (aminoacylation) prior to translation, what specific biochemical event occurs to couple the amino acid to its cognate tRNA?

  1. GTP hydrolysis forms a peptide bond between the amino acid carboxyl group and the tRNA anticodon loop
  2. The amino acid is phosphorylated at its amino group and spontaneously binds the 5' triphosphate end of tRNA
  3. Peptidyl transferase creates an amide linkage between the tRNA D-loop and the amino acid side chain
  4. Aminoacyl-tRNA synthetase uses ATP to form an ester bond between the amino acid carboxyl group and the 3'-CCA hydroxyl end of tRNA Correct answer.

Explanation

Aminoacyl-tRNA synthetase enzymes ensure fidelity by activating each amino acid using ATP to form an aminoacyl-AMP intermediate, then transferring the amino acid to the 2' or 3' hydroxyl group of the adenosine nucleotide at the conserved 3'-CCA terminus of the cognate tRNA, producing an energetic ester bond.

Question 5Cells, biomolecules, heredity and metabolism

Which event marks the successful initiation phase of translation in eukaryotic protein synthesis?

  1. Release of the completed polypeptide chain by eukaryotic release factors (eRF)
  2. Binding of the small ribosomal subunit and initiator tRNA to the 5' AUG codon, followed by recruitment of the large ribosomal subunit Correct answer.
  3. Formation of a covalent disulfide bond between the first two incoming amino acids in the E site
  4. Translocation of the uncharged tRNA directly from the A site to the extracellular matrix

Explanation

Translation initiation begins when the small (40S) ribosomal subunit, preloaded with initiator Met-tRNA, binds to the 5' cap of mRNA, scans for the start codon (AUG), and forms base pairs. The large (60S) subunit then joins, positioning the initiator tRNA in the P site and creating a functional 80S initiation complex.

Question 6Cells, biomolecules, heredity and metabolism

During the elongation phase of translation, which ribosomal component catalyzes the formation of the peptide bond between the amino acid in the A site and the growing polypeptide chain in the P site?

  1. DNA helicase situated in the ribosomal exit channel
  2. Aminoacyl-tRNA synthetase associated with the 40S subunit
  3. Peptidyl transferase, an RNA enzyme (ribozyme) catalytic center of the large ribosomal subunit Correct answer.
  4. Elongation factor G acting exclusively as a proteolytic cleavage enzyme

Explanation

Peptidyl transferase activity is intrinsic to the 28S ribosomal RNA (in eukaryotes) of the large ribosomal subunit, demonstrating that the ribosome is a ribozyme. It catalyzes the nucleophilic attack of the amino group of the aminoacyl-tRNA in the A site on the ester carbon of the peptidyl-tRNA in the P site.

Question 7Cells, biomolecules, heredity and metabolism

Under electron microscopy, a polyribosome (polysome) appears as an mRNA strand associated with several ribosomes simultaneously. What is the functional significance of this structure?

  1. It enables rapid and simultaneous synthesis of multiple identical copies of the same polypeptide from a single mRNA molecule Correct answer.
  2. It allows each attached ribosome to produce a completely different protein with unrelated amino acid sequences
  3. It prevents mRNA degradation by completely halting translation until environmental stress passes
  4. It facilitates the reverse transcription of mRNA back into nuclear genomic DNA

Explanation

As soon as the first ribosome moves past the start codon, another ribosome can assemble and initiate translation on the same mRNA. A polysome allows a cell to amplify translation efficiency, synthesizing many identical polypeptide chains concurrently from a single transcript before it is degraded.

Question 8Cells, biomolecules, heredity and metabolism

A single nucleotide substitution in an exon changes the codon UAC (tyrosine) into UAA. What type of mutation is this, and what is its expected phenotypic consequence?

  1. Silent mutation; the protein retains full length and wild-type enzymatic activity
  2. Frameshift mutation; the reading frame shifts downstream, altering all subsequent amino acids
  3. Missense mutation; a single nonpolar amino acid replaces tyrosine without truncating the chain
  4. Nonsense mutation; UAA acts as a stop codon, causing premature termination and yielding a truncated, usually nonfunctional protein Correct answer.

Explanation

UAA is one of three termination codons (along with UAG and UGA). A point mutation that converts an amino acid codon into a stop codon is a nonsense mutation. Translation halts prematurely at this position, producing a truncated polypeptide that typically lacks functional activity and is rapidly degraded.

Question 9Cells, biomolecules, heredity and metabolism

The biological activity of a globular enzyme depends on its specific tertiary conformation. Which types of chemical interactions are primarily responsible for stabilizing this three-dimensional folding?

  1. Continuous covalent peptide bonds linking adjacent alpha-carbons along the backbone only
  2. Phosphodiester linkages and glycosidic bonds formed between lipid side chains
  3. Hydrophobic interactions, hydrogen bonds, ionic bonds, and covalent disulfide bridges between cysteine R-groups Correct answer.
  4. Magnetic interactions between iron ions and nuclear histone complexes

Explanation

Tertiary structure arises from interactions between the variable side chains (R-groups) of amino acids. Non-covalent forces (hydrophobic collapse in the protein core, hydrogen bonding, and electrostatic salt bridges) alongside strong covalent disulfide bonds (-S-S-) between oxidized cysteine residues stabilize the functional native conformation.

Question 10Cells, biomolecules, heredity and metabolism

An amino acid with an uncharged, non-ionizable side chain (such as alanine) is placed in an aqueous solution where the pH equals its isoelectric point (pI). What is the predominant ionic state and net charge of this amino acid?

  1. Zwitterion (dipolar ion) with a protonated amino group (-NH3+) and a deprotonated carboxyl group (-COO-), giving a net charge of 0 Correct answer.
  2. Cation with a protonated amino group (-NH3+) and a neutral carboxyl group (-COOH), giving a net charge of +1
  3. Anion with a neutral amino group (-NH2) and a deprotonated carboxyl group (-COO-), giving a net charge of -1
  4. Fully uncharged molecule with neutral -NH2 and neutral -COOH groups, giving a net charge of 0

Explanation

At its isoelectric point (pI), an amino acid exists predominantly as a zwitterion (dipolar ion). The alpha-carboxyl group is deprotonated (-COO-, pKa ~ 2.3) while the alpha-amino group is protonated (-NH3+, pKa ~ 9.7). The equal and opposite charges cancel out, resulting in a net electrical charge of zero and zero electrophoretic mobility.

Question 11Cells, biomolecules, heredity and metabolism

Enzymes are characterized by high biological specificity. What constitutes the 'double specificity' of an enzyme?

  1. Specificity for the cell nucleus and specificity for the extracellular matrix
  2. Specificity for cold temperatures and specificity for high atmospheric pressures
  3. Specificity for hydrophilic substrates and specificity for hydrophobic solvents
  4. Substrate specificity (acting on a specific reactant) and reaction specificity (catalyzing a specific chemical transformation) Correct answer.

Explanation

Enzymatic action exhibits double specificity: (1) substrate specificity, meaning the enzyme binds only to a particular substrate or family of substrates possessing complementary spatial and chemical properties; and (2) reaction (action) specificity, meaning it catalyzes only one specific biochemical transformation (such as hydrolysis, oxidation, or decarboxylation).

Question 12Cells, biomolecules, heredity and metabolism

Within an enzyme's active site, what distinguishes the 'binding site' residues from the 'catalytic site' residues?

  1. Binding residues cleave covalent bonds, whereas catalytic residues form the outer lipid membrane
  2. Binding residues recognize and orient the substrate via weak non-covalent interactions, while catalytic residues directly participate in bond making or breaking Correct answer.
  3. Binding residues operate exclusively at 100 °C, while catalytic residues function at 0 °C
  4. Binding residues are located in the nucleus, while catalytic residues are located in the cytoplasm

Explanation

The active site consists of two functional subgroups of amino acids: binding residues (which possess steric and chemical complementarity to position the substrate firmly through hydrogen bonds, ionic bonds, or van der Waals forces) and catalytic residues (whose reactive side chains donate/accept protons or electrons to accelerate the chemical transformation).

Question 13Cells, biomolecules, heredity and metabolism

In an enzymatic reaction with a fixed enzyme concentration, why does the initial reaction velocity (V0) plateau at high substrate concentrations [S]?

  1. All enzyme active sites are fully saturated with substrate, making the rate independent of further substrate increases (zero-order kinetics) Correct answer.
  2. The substrate molecules begin to denature irreversibly due to high kinetic friction in the solution
  3. The activation energy of the reaction increases exponentially, halting the catalytic cycle
  4. ATP molecules required for non-covalent binding are completely exhausted in the medium

Explanation

At high substrate concentrations, every enzyme molecule's active site is occupied by substrate in the enzyme-substrate (ES) complex. Under this saturation state, the reaction velocity reaches its maximum (Vmax) and transitions from first-order kinetics to zero-order kinetics, where increasing [S] cannot increase the rate.

Question 14Cells, biomolecules, heredity and metabolism

In Michaelis-Menten enzyme kinetics, how is the Michaelis constant (Km) defined and what biological property does it reflect?

  1. Km is the maximum velocity divided by the substrate concentration; it measures the total enzyme quantity
  2. Km is the time required to complete the reaction; it reflects the activation energy barrier
  3. Km is the substrate concentration at which the reaction velocity equals half of Vmax; it is an inverse measure of the enzyme's affinity for its substrate Correct answer.
  4. Km is the optimal temperature of the reaction; it reflects thermal stability of the peptide backbone

Explanation

By definition, the Michaelis constant Km is the substrate concentration [S] at which the initial reaction velocity V0 equals Vmax / 2. A smaller Km value indicates that the enzyme reaches half-maximal velocity at a lower substrate concentration, representing a higher affinity between the enzyme active site and the substrate.

Question 15Cells, biomolecules, heredity and metabolism

When an enzyme solution is heated well beyond its optimum temperature (e.g., above 65 °C), enzymatic activity drops to zero. What causes this loss of catalytic activity?

  1. The primary covalent peptide bonds between all adjacent amino acids are completely hydrolyzed
  2. Thermal agitation disrupts weak tertiary and secondary bonds, irreversibly denaturing the active site conformation Correct answer.
  3. The substrate molecules change their optical chirality from D-form to L-form
  4. The enzyme becomes allosterically inhibited by dissolved atmospheric nitrogen

Explanation

Elevated thermal kinetic energy breaks weak non-covalent bonds (hydrogen bonds, ionic interactions, and hydrophobic associations) that maintain the enzyme's secondary and tertiary structure. This thermal denaturation alters the spatial geometry of the active site so that it can no longer bind the substrate, causing irreversible inactivation.

Question 16Cells, biomolecules, heredity and metabolism

How does a significant departure from an enzyme's optimum pH impair its catalytic ability?

  1. It causes spontaneous dimerization of all substrate molecules into inert crystals
  2. It converts the enzyme's ribonucleotide backbone into single-stranded DNA
  3. It permanently degrades the ribosome responsible for producing the enzyme
  4. It alters the ionization states of amino acid R-groups in the active site and substrate, disrupting essential electrostatic and hydrogen bonds Correct answer.

Explanation

Variations in [H+] alter the protonation state of ionizable acidic and basic amino acid side chains (such as aspartate, glutamate, lysine, and histidine) in the active site. This disrupts salt bridges and electrostatic interactions vital for maintaining active site conformation and substrate binding, reducing or abolishing catalysis.

Question 17Cells, biomolecules, heredity and metabolism

In the presence of a competitive inhibitor, how do the kinetic parameters Vmax and apparent Km of an enzyme change?

  1. Vmax decreases significantly, while apparent Km remains strictly unchanged
  2. Both Vmax and apparent Km decrease in exact proportion to the inhibitor concentration
  3. Apparent Km increases (affinity appears lower), while Vmax remains unchanged because high substrate concentration overcomes the competition Correct answer.
  4. Both Vmax and apparent Km increase simultaneously by a factor of ten

Explanation

A competitive inhibitor structurally mimics the substrate and competes directly for the active site. Because the inhibitor and substrate compete for the same binding locus, a higher substrate concentration can outcompete the inhibitor and eventually achieve the same maximum velocity (Vmax unchanged), resulting in an increased apparent Km (Km' > Km).

Question 18Cells, biomolecules, heredity and metabolism

In the ideal case of pure non-competitive inhibition, how are the Michaelis-Menten parameters affected?

  1. It decreases Vmax without altering Km, because it binds to a distinct regulatory site and reduces catalytic efficiency regardless of substrate concentration Correct answer.
  2. It increases Km without altering Vmax, because it competes directly with the substrate for the active site
  3. It increases both Vmax and Km by facilitating faster release of products from the active site
  4. It converts the hyperbolic Michaelis-Menten curve into an irreversible linear vertical line

Explanation

In pure non-competitive inhibition, the inhibitor binds free enzyme and the enzyme-substrate complex equally well. The apparent Km therefore remains unchanged, while the amount of catalytically effective enzyme falls and Vmax decreases. The effect is not necessarily irreversible.

Question 19Cells, biomolecules, heredity and metabolism

According to Daniel Koshland's 'induced-fit' model of enzyme action, what happens when a substrate encounters the enzyme's active site?

  1. The active site remains completely rigid and unaltered, functioning strictly like a solid lock into which the substrate key fits
  2. The substrate permanently breaks down the enzyme into short dipeptides prior to catalysis
  3. The enzyme molecule unwinds into an unstructured polypeptide chain and diffuses out of the cell
  4. Binding of the substrate induces a reciprocal conformational adjustment in the active site, aligning catalytic residues into an optimal arrangement Correct answer.

Explanation

Unlike the rigid lock-and-key model proposed by Emil Fischer, Koshland's induced-fit model posits that the active site is flexible. Initial non-covalent interactions between substrate and binding residues induce conformational changes in the enzyme that position the catalytic groups with precise geometric orientation around the substrate bonds.

Question 20Cells, biomolecules, heredity and metabolism

On a Lineweaver-Burk double-reciprocal plot (1/V0 plotted against 1/[S]), which features correspond to the y-intercept and the x-intercept, respectively?

  1. The y-intercept represents -Km, and the x-intercept represents Vmax
  2. The y-intercept represents 1/Vmax, and the x-intercept represents -1/Km Correct answer.
  3. The y-intercept represents Vmax / 2, and the x-intercept represents [S] / 2
  4. The y-intercept represents 1/Km, and the x-intercept represents 1/Vmax

Explanation

Taking the reciprocal of the Michaelis-Menten equation yields 1/V0 = (Km / Vmax)(1/[S]) + (1/Vmax). In this linear y = mx + b form, the y-intercept (when 1/[S] = 0) is 1/Vmax, and the x-intercept (when 1/V0 = 0) is -1/Km, providing a clear graphical determination of both kinetic parameters.

Question 21Cells, biomolecules, heredity and metabolism

During glycolysis occurring in the eukaryotic cytoplasm, what is the net yield of energy-rich molecules and end products per mole of oxidized glucose?

  1. 2 moles of pyruvate, 2 moles of ATP (net), and 2 moles of NADH + H+ Correct answer.
  2. 1 mole of acetyl-CoA, 4 moles of ATP (net), and 2 moles of FADH2
  3. 2 moles of lactate, 36 moles of ATP (net), and 6 moles of CO2
  4. 6 moles of CO2, 6 moles of H2O, and 38 moles of ATP

Explanation

Glycolysis breaks down one 6-carbon glucose molecule into two 3-carbon pyruvate molecules. The pathway consumes 2 ATP molecules in the preparatory phase and generates 4 ATP by substrate-level phosphorylation, yielding a net gain of 2 ATP along with 2 reduced coenzymes (NADH + H+).

Question 22Cells, biomolecules, heredity and metabolism

Under aerobic conditions, what biochemical transition must pyruvate undergo before its carbon atoms can enter the mitochondrial Krebs cycle?

  1. Reduction into ethanol and carbon dioxide catalyzed by alcohol dehydrogenase
  2. Oxidative decarboxylation in the mitochondrial matrix by pyruvate dehydrogenase, forming acetyl-CoA, NADH, and CO2 Correct answer.
  3. Direct condensation with oxaloacetate in the intermembrane space without coenzyme A
  4. Phosphorylation into glucose-6-phosphate via reverse gluconeogenesis

Explanation

Pyruvate is transported across the mitochondrial membranes into the matrix, where the multienzyme pyruvate dehydrogenase complex carries out oxidative decarboxylation: pyruvate (3C) loses a carboxyl group as CO2, while the remaining 2C acetyl unit is transferred to coenzyme A to form acetyl-CoA, reducing NAD+ to NADH.

Question 23Cells, biomolecules, heredity and metabolism

For each complete turn of the citric acid (Krebs) cycle processing one mole of acetyl-CoA, what are the stoichiometric yields of reduced electron carriers and high-energy phosphates?

  1. 1 NADH, 1 FADH2, and 4 ATP molecules
  2. 4 NADH, 0 FADH2, and 2 ATP molecules
  3. 3 NADH + H+, 1 FADH2, and 1 ATP (or GTP) via substrate-level phosphorylation Correct answer.
  4. 6 NADH, 2 FADH2, and 36 ATP molecules

Explanation

Each acetyl-CoA entering the Krebs cycle combines with oxaloacetate (4C) to form citrate (6C). Through two successive oxidative decarboxylations and further dehydrogenations, each turn yields 3 NADH, 1 FADH2, 1 ATP (or GTP) by substrate-level phosphorylation, and 2 CO2, regenerating oxaloacetate.

Question 24Cells, biomolecules, heredity and metabolism

In the mitochondrial electron transport chain of cellular respiration, what is the ultimate electron acceptor, and what product is formed upon its reduction?

  1. NAD+, which is reduced to form NADH in the outer membrane
  2. Pyruvate, which is reduced to form lactic acid in the cristae
  3. Cytochrome c, which transfers electrons directly to ATP synthase
  4. Molecular oxygen (O2), which accepts electrons and protons at complex IV to form water (H2O) Correct answer.

Explanation

At complex IV (cytochrome c oxidase), four electrons from reduced cytochrome c are transferred to molecular oxygen (O2), which combines with matrix protons to produce two molecules of water (1/2 O2 + 2 H+ + 2 e- -> H2O). Oxygen is thus the terminal electron acceptor of aerobic respiration.

Question 25Cells, biomolecules, heredity and metabolism

According to Peter Mitchell's chemiosmotic hypothesis, what immediate driving force powers the phosphorylation of ADP to ATP by mitochondrial ATP synthase (F0-F1 complex)?

  1. A proton electrochemical gradient (proton-motive force) established by proton pumping across the inner mitochondrial membrane into the intermembrane space Correct answer.
  2. Direct mechanical friction generated by the active transport of glucose across the outer mitochondrial membrane
  3. A sodium ion (Na+) concentration gradient maintained exclusively across the nuclear envelope
  4. Spontaneous cleavage of covalent carbon-carbon bonds within the ATP synthase catalytic head

Explanation

Electron transport through complexes I, III, and IV pumps protons from the mitochondrial matrix into the intermembrane space, creating both a concentration difference (delta pH) and an electrical potential (delta psi). This proton-motive force drives protons back into the matrix through the F0 channel of ATP synthase, driving rotational catalysis in the F1 head to synthesize ATP.

Question 26Cells, biomolecules, heredity and metabolism

During the light-dependent reactions of photosynthesis, what is the biochemical role and source of electrons for Photosystem II (P680)?

  1. Carbon dioxide molecules are oxidized to yield electrons and oxygen gas
  2. Water molecules undergo photolysis in the thylakoid lumen, producing oxygen, protons, and electrons to reduce P680+ Correct answer.
  3. Glucose is broken down to release electrons directly to ferredoxin
  4. ATP is hydrolyzed to pump electrons across the outer chloroplast envelope

Explanation

When the reaction center P680 of Photosystem II absorbs photon energy, it gets photo-oxidized to P680+. The oxygen-evolving complex (OEC) in the thylakoid lumen catalyzes the photolysis of water (2 H2O -> O2 + 4 H+ + 4 e-), extracting electrons to reduce P680+ back to its ground state while releasing O2 as a byproduct.

Question 27Cells, biomolecules, heredity and metabolism

What is the non-cyclic pathway of electron transport through the photosynthetic thylakoid membrane complexes?

  1. PSI -> Plastoquinone -> Cytochrome b6f -> PSII -> NADPH
  2. Ferredoxin -> PSII -> ATP synthase -> Plastocyanin -> O2
  3. PSII -> Plastoquinone -> Cytochrome b6f complex -> Plastocyanin -> PSI -> Ferredoxin -> NADP+ reductase Correct answer.
  4. Cytochrome b6f -> Calvin cycle -> RuBisCO -> PSI -> PSII

Explanation

In non-cyclic photophosphorylation (the Z-scheme), electrons excited in PSII flow through plastoquinone (PQ), the cytochrome b6f complex, and plastocyanin (PC) to reach photo-oxidized PSI (P700+). From PSI, photo-excited electrons pass through ferredoxin (Fd) to ferredoxin-NADP+ reductase (FNR), which reduces NADP+ to NADPH in the stroma.

Question 28Cells, biomolecules, heredity and metabolism

In chloroplasts, how does the proton gradient that drives photophosphorylation across the thylakoid membrane compare in orientation to that of mitochondria?

  1. Protons accumulate in the stroma and flow into the thylakoid lumen through ATP synthase
  2. Protons accumulate exclusively in the intermembrane space of the chloroplast double membrane
  3. Protons are consumed by chlorophyll pigments to synthesize carotenoids directly
  4. Protons accumulate inside the thylakoid lumen (lowering lumen pH) and flow outward into the stroma through CF0-CF1 ATP synthase Correct answer.

Explanation

During photosynthetic electron transport, water photolysis and plastoquinone/cytochrome b6f proton pumping translocate protons into the thylakoid lumen, acidifying the lumen (pH ~ 5) relative to the alkaline stroma (pH ~ 8). Protons flow down this gradient from the lumen into the stroma through CF0-CF1 ATP synthase, driving ATP generation in the stroma where the Calvin cycle occurs.

Question 29Cells, biomolecules, heredity and metabolism

Which key carboxylating enzyme catalyzes the initial fixation of atmospheric CO2 onto ribulose-1,5-bisphosphate (RuBP) during the Calvin-Benson cycle?

  1. Ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO) Correct answer.
  2. Pyruvate decarboxylase
  3. Phosphofructokinase-1
  4. DNA ligase

Explanation

RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase) is the primary carbon-fixing enzyme located in the chloroplast stroma. It catalyzes the carboxylation of the 5-carbon sugar acceptor RuBP with CO2 to form an unstable 6-carbon intermediate that immediately cleaves into two molecules of 3-phosphoglycerate (3-PGA).

Question 30Cells, biomolecules, heredity and metabolism

During the Calvin cycle, what are the roles of ATP and NADPH produced during the light-dependent reactions?

  1. They are exported directly to the roots to power mineral nutrient absorption
  2. They provide energy and reducing equivalents to reduce 3-phosphoglycerate (3-PGA) into glyceraldehyde-3-phosphate (G3P) and regenerate RuBP Correct answer.
  3. They cleave starch granules in the stroma into individual sucrose molecules
  4. They act as structural lipids incorporated into the chloroplast thylakoid bilayer

Explanation

In the reduction phase of the Calvin cycle, ATP phosphorylates 3-PGA to 1,3-bisphosphoglycerate, which is then reduced by NADPH to form the triose phosphate glyceraldehyde-3-phosphate (G3P). Additional ATP is consumed in the regeneration phase to convert five G3P molecules back into three RuBP acceptors.

Question 31Physiology, immunity, ecology and evolution

What ionic mechanism is primarily responsible for establishing and maintaining the resting membrane potential (approximately -70 mV) of an unexcited neuron?

  1. Complete impermeability of the plasma membrane to all inorganic ions
  2. Rapid continuous influx of calcium ions through ligand-gated channels
  3. High resting membrane permeability to K+ via leak channels, combined with active transport by the electrogenic Na+/K+ ATPase pump (3 Na+ out / 2 K+ in) Correct answer.
  4. Massive active extrusion of chloride ions by vesicular exocytosis

Explanation

At rest, the neuronal membrane is 50 to 100 times more permeable to K+ than to Na+ due to open non-gated K+ leak channels. K+ diffuses outward down its concentration gradient until electrical attraction opposes further efflux. The electrogenic Na+/K+ ATPase pump continuously expels 3 Na+ for every 2 K+ imported, maintaining the concentration gradients and steady negative resting potential.

Question 32Physiology, immunity, ecology and evolution

During the generation of a neuronal action potential, what event is directly responsible for the rapid depolarization phase from the threshold potential to +30 mV?

  1. Sudden closure of all sodium channels and opening of chloride pores
  2. Active extrusion of potassium ions by the sodium-potassium pump
  3. Passive diffusion of proteins out of the axoplasm into the extracellular fluid
  4. Rapid opening of voltage-gated Na+ channels causing a massive influx of Na+ down its electrochemical gradient Correct answer.

Explanation

When membrane depolarization reaches threshold (around -50 mV), voltage-gated Na+ activation gates rapidly open. Driven by both a high concentration gradient and electrical attraction toward the negative interior, Na+ rushes into the axoplasm, creating a regenerative influx that drives membrane potential toward the sodium equilibrium potential (+30 to +40 mV).

Question 33Physiology, immunity, ecology and evolution

What ionic event causes the repolarization and subsequent transient hyperpolarization phase of a neuronal action potential?

  1. Inactivation of voltage-gated Na+ channels and sustained opening of voltage-gated K+ channels leading to K+ efflux Correct answer.
  2. Simultaneous opening of voltage-gated Na+ and Ca2+ channels
  3. Reversal of the concentration gradient of sodium by mitochondrial respiration
  4. Massive entry of bicarbonate anions through the myelin sheath

Explanation

At peak depolarization (+30 mV), voltage-gated Na+ channel inactivation gates close, halting Na+ entry. Simultaneously, slower voltage-gated K+ channels open fully, allowing K+ to exit rapidly down its electrochemical gradient. The delayed closing of these K+ channels causes the membrane potential to temporarily dip below resting potential (hyperpolarization).

Question 34Physiology, immunity, ecology and evolution

What is the physiological cause and functional importance of the absolute refractory period in an axon?

  1. Exhaustion of all cytoplasmic ATP; it forces the neuron to undergo cell division
  2. Inactivation of voltage-gated Na+ channels; it ensures unidirectional action potential propagation and limits maximum firing frequency Correct answer.
  3. Depletion of myelin lipid molecules; it allows neurotransmitters to escape into the blood
  4. Continuous opening of calcium channels; it converts electrical signals into magnetic waves

Explanation

During the absolute refractory period, the inactivation gates of voltage-gated Na+ channels remain closed and cannot reopen regardless of stimulus strength. This prevents backwards (retrograde) propagation of the action potential toward recently activated axonal regions and establishes an upper limit on neuronal firing frequency.

Question 35Physiology, immunity, ecology and evolution

How does the presence of a myelin sheath increase the conduction velocity of action potentials along an axon (saltatory conduction)?

  1. It heats the axoplasm to 45 °C, speeding up cytoplasmic enzymatic reactions
  2. It allows sodium ions to cross the membrane uniformly along the entire length of the internode
  3. It acts as an electrical insulator, forcing depolarization to regenerate exclusively at the unmyelinated nodes of Ranvier where voltage-gated channels are concentrated Correct answer.
  4. It eliminates the need for any ion movement across the axonal plasma membrane

Explanation

Myelin provides high electrical resistance and low membrane capacitance along internodal segments. Local currents spread passively and rapidly by electrotonic conduction through the axoplasm, regenerating action potentials exclusively at the unmyelinated nodes of Ranvier where voltage-gated Na+ channels are densely clustered, achieving velocities up to 120 m/s.

Question 36Physiology, immunity, ecology and evolution

When an action potential invades the presynaptic terminal of a chemical synapse, which immediate ionic event triggers the exocytosis of neurotransmitter molecules into the synaptic cleft?

  1. Efflux of potassium ions through leak channels
  2. Active uptake of glucose from the synaptic cleft into vesicles
  3. Depolarization-induced extrusion of magnesium ions into the extracellular space
  4. Opening of voltage-gated Ca2+ channels leading to an influx of Ca2+ into the presynaptic cytoplasm Correct answer.

Explanation

Depolarization of the presynaptic terminal triggers the opening of voltage-gated Ca2+ channels. Because extracellular [Ca2+] (~1-2 mM) is orders of magnitude higher than resting cytosolic [Ca2+] (~100 nM), Ca2+ rushes into the terminal and binds to synaptotagmin, triggering SNARE-mediated fusion of synaptic vesicles with the presynaptic membrane and neurotransmitter exocytosis.

Question 37Physiology, immunity, ecology and evolution

What causes an Excitatory Postsynaptic Potential (EPSP) on the postsynaptic membrane following neurotransmitter binding?

  1. Binding of an excitatory neurotransmitter (such as acetylcholine) opens ligand-gated cation channels, producing a localized net inward Na+ current that depolarizes the membrane Correct answer.
  2. Binding of glycine opens ligand-gated chloride channels, causing membrane hyperpolarization
  3. Activation of the Na+/K+ ATPase pump producing rapid outward movement of positive charge
  4. Direct diffusion of the neurotransmitter across the postsynaptic membrane into the cell nucleus

Explanation

Excitatory neurotransmitters (e.g., acetylcholine at nicotinic receptors or glutamate at AMPA receptors) bind to ligand-gated ion channels, opening pores permeable to monovalent cations. Because the electrochemical gradient favors Na+ influx over K+ efflux, a net inward positive current occurs, producing a graded, depolarizing postsynaptic potential (EPSP).

Question 38Physiology, immunity, ecology and evolution

An Inhibitory Postsynaptic Potential (IPSP) reduces neuronal excitability because:

  1. It completely destroys the postsynaptic membrane lipid bilayer
  2. The opening of ligand-gated Cl- or K+ channels causes hyperpolarization, moving the membrane potential further away from the firing threshold Correct answer.
  3. It accelerates the synthesis of myelin along dendrites
  4. It causes massive influx of positively charged calcium ions

Explanation

Inhibitory neurotransmitters (such as GABA and glycine) bind to ionotropic receptors that are selectively permeable to chloride (Cl-) or potassium (K+). Cl- influx or K+ efflux increases membrane negativity (hyperpolarization, e.g., to -80 mV), moving the membrane potential further away from the threshold required to fire an action potential.

Question 39Physiology, immunity, ecology and evolution

How does the axon hillock (initial segment) integrate multiple synaptic inputs to determine whether an action potential is triggered?

  1. It ignores all inhibitory inputs and triggers an action potential whenever any single EPSP occurs
  2. It multiplies the amplitudes of all inputs regardless of their spatial location or arrival time
  3. It performs spatial and temporal algebraic summation of EPSPs and IPSPs; if net depolarization reaches the threshold potential, an action potential is initiated Correct answer.
  4. It stores incoming charges in mitochondrial vacuoles for release several days later

Explanation

The axon hillock has the lowest threshold in the neuron due to an extremely high density of voltage-gated Na+ channels. It integrates inputs via spatial summation (adding potentials arriving simultaneously at different synaptic locations) and temporal summation (adding potentials arriving in rapid succession from the same synapse). If the net algebraic sum meets or exceeds threshold, an all-or-none action potential is fired.

Question 40Physiology, immunity, ecology and evolution

In the human stretch (myotatic) reflex arc (such as the patellar reflex), what is the neuronal circuit responsible for the simultaneous contraction of the stretched muscle and relaxation of the antagonist muscle?

  1. Polysynaptic activation of both muscles by cortical motor neurons in the cerebrum
  2. Direct mechanical linkage between tendons without any spinal cord involvement
  3. Release of adrenaline from sympathetic nerve endings directly onto the quadriceps muscle
  4. Monosynaptic excitation of the homonymous motor neuron by sensory Ia afferents, accompanied by polysynaptic reciprocal inhibition of the antagonist motor neuron via an inhibitory interneuron Correct answer.

Explanation

The myotatic reflex is a spinal reflex. Stretching muscle spindles activates sensory Ia afferent fibers that enter the dorsal horn of the spinal cord. They make a direct monosynaptic excitatory synapse with alpha motor neurons of the agonist muscle (causing contraction), while collateral branches synapse on inhibitory interneurons (Ia interneurons) that inhibit alpha motor neurons of the antagonist muscle (reciprocal inhibition).

Question 41Physiology, immunity, ecology and evolution

What is the primary difference in cellular distribution and function between Class I and Class II Major Histocompatibility Complex (MHC / HLA) molecules?

  1. MHC-I is expressed on all nucleated cells presenting endogenous antigens to CD8+ cytotoxic T cells, whereas MHC-II is expressed on professional antigen-presenting cells presenting exogenous antigens to CD4+ helper T cells Correct answer.
  2. MHC-I is found only on red blood cells, while MHC-II is found only on circulating blood platelets
  3. MHC-I binds soluble free antibodies in serum, while MHC-II synthesizes histamine in mast cells
  4. MHC-I presents viral RNA directly to B cell receptors without requiring peptide fragments

Explanation

MHC Class I molecules are expressed on virtually all nucleated somatic cells, presenting peptides derived from intracellular (endogenous) proteins to CD8+ cytotoxic T lymphocytes. MHC Class II molecules are restricted to professional antigen-presenting cells (dendritic cells, macrophages, B cells) and present endocytosed (exogenous) peptides to CD4+ helper T lymphocytes.

Question 42Physiology, immunity, ecology and evolution

Following specific recognition of an antigen by its membrane immunoglobulin (BCR) and stimulation by helper T cell cytokines, what is the fate of an activated B lymphocyte?

  1. It differentiates exclusively into natural killer cells that engulf foreign bacteria
  2. It undergoes clonal selection and proliferation, differentiating into antibody-secreting plasma cells and long-lived memory B cells Correct answer.
  3. It loses its nucleus and transforms into a non-nucleated erythrocyte
  4. It migrates directly to the thymus to undergo negative selection

Explanation

Antigen binding to the BCR provides signal 1, and interaction with CD4+ helper T cells (via CD40-CD40L and interleukins like IL-4) provides signal 2. This drives clonal expansion of the antigen-specific B cell clone into plasma cells (effector cells packed with rough ER that secrete high volumes of specific antibodies) and memory B cells (ensuring rapid secondary immune response).

Question 43Physiology, immunity, ecology and evolution

An immunoglobulin G (IgG) antibody molecule possesses a Y-shaped quaternary structure. What functional roles are assigned to the Fab and Fc regions, respectively?

  1. The Fab region binds complement proteins, while the Fc region contains the antigen-binding hypervariable loops
  2. The Fab region is hydrophobic and anchors into the nuclear membrane, while the Fc region produces cytokines
  3. The Fab region (variable domains) binds specifically to the antigen epitope, while the Fc region (constant domains) mediates effector functions such as opsonization and complement activation Correct answer.
  4. The Fab region catalyzes the replication of viral RNA, while the Fc region neutralizes bacterial toxins

Explanation

The two Fab (fragment antigen-binding) arms contain the variable domains of light and heavy chains (VL and VH) whose complementary determining regions (CDRs) recognize and bind specific antigen epitopes. The Fc (fragment crystallizable) stem consists of constant heavy chain domains that bind Fc receptors on phagocytes (promoting opsonization) and initiate the classical complement pathway via C1q.

Question 44Physiology, immunity, ecology and evolution

How do circulating antibodies eliminate foreign antigens during a humoral immune response?

  1. They directly digest foreign bacteria by secreting hydrochloric acid into the bloodstream
  2. They penetrate host cell membranes and destroy mitochondrial ribosomes
  3. They convert foreign bacterial proteins into human albumin molecules
  4. They neutralize toxins/viruses, agglutinate particulate antigens, and opsonize pathogens to facilitate phagocytosis and trigger the complement system Correct answer.

Explanation

Antibodies do not directly destroy pathogens themselves; they act as molecular tags. They neutralize pathogens by blocking binding epitopes, cross-link multiple antigens into aggregates (agglutination and precipitation), coat pathogens to enhance phagocytic engulfment via Fc receptors (opsonization), and trigger complement-mediated lysis.

Question 45Physiology, immunity, ecology and evolution

How does a CD8+ cytotoxic T lymphocyte (CTL) destroy a virus-infected or cancerous target cell displaying foreign peptide-MHC-I complexes?

  1. By releasing perforin to form pores in the target membrane and granzymes that enter to activate caspase cascades, triggering apoptotic cell death Correct answer.
  2. By engulfing the target cell intact through macropinocytosis and storing it in a lysosome
  3. By secreting high concentrations of circulating IgG antibodies that coat the cell surface
  4. By synthesizing a thick peptidoglycan wall around the target cell to starve it of nutrients

Explanation

Upon TCR and CD8 coreceptor recognition of the foreign peptide presented on MHC-I, the CTL reorients its Golgi apparatus and releases cytolytic granules toward the immunological synapse. Perforin polymerizes in the target membrane to form hydrophilic pores, allowing serine proteases (granzymes) to enter the target cytoplasm, cleaving procaspases and inducing controlled apoptotic suicide.

Question 46Physiology, immunity, ecology and evolution

Why are CD4+ T helper (Th) lymphocytes considered the central orchestrators (pivotal cells) of the adaptive immune system?

  1. They differentiate into phagocytes that clear cellular debris in the spleen
  2. Upon recognizing antigen-MHC-II complexes, they secrete key interleukins (such as IL-2) that stimulate the proliferation and differentiation of both B cells and cytotoxic T cells Correct answer.
  3. They produce all circulating serum complement proteins in the bone marrow
  4. They form the primary epithelial barrier in skin and mucous membranes

Explanation

CD4+ T helper cells act as the central communication hub of adaptive immunity. Upon activation by antigen-presenting cells displaying antigen-MHC-II, Th cells proliferate and differentiate into Th1 and Th2 subsets, secreting cytokines such as interleukin-2 (IL-2), IL-4, and interferon-gamma that are required to activate B cell antibody production and CTL cytotoxicity.

Question 47Physiology, immunity, ecology and evolution

During T lymphocyte maturation in the thymus, what is the essential difference between positive selection and negative selection?

  1. Positive selection occurs in bone marrow for B cells, while negative selection occurs in the spleen for red blood cells
  2. Positive selection destroys all cells possessing a T cell receptor, while negative selection stimulates plasma cells
  3. Positive selection in the cortex preserves thymocytes whose TCR recognizes self-MHC molecules, whereas negative selection in the medulla eliminates autoreactive thymocytes that bind self-antigens with high affinity Correct answer.
  4. Positive selection stimulates cell division with bacterial lipopolysaccharide, while negative selection tests viral resistance

Explanation

Thymic education ensures both MHC restriction and self-tolerance. In the thymic cortex, positive selection spares immature thymocytes whose TCR binds self-MHC with moderate affinity (preventing apoptosis by death by neglect). In the medulla, negative selection induces apoptosis in thymocytes that recognize self-antigen-MHC complexes with high affinity, eliminating potential autoreactive clones.

Question 48Physiology, immunity, ecology and evolution

What is the primary cellular mechanism by which the Human Immunodeficiency Virus (HIV) causes progressive collapse of the human immune system?

  1. HIV destroys epidermal keratinocytes, causing bacterial skin infections
  2. HIV selectively infects and lyses red blood cells, causing acute hemolytic anemia
  3. HIV paralyzes the spinal cord motor neurons, blocking reflex responses
  4. HIV viral glycoprotein gp120 targets CD4 receptors on helper T cells, replicating inside them and leading to severe CD4+ T cell depletion Correct answer.

Explanation

The envelope glycoprotein gp120 of HIV binds with high affinity to the CD4 molecule present on T helper lymphocytes (and macrophages), using chemokine coreceptors (CCR5 or CXCR4) for entry. Viral replication, syncytia formation, and cytotoxic elimination lead to a catastrophic decline in CD4+ T cell counts (below 200 cells/uL), paralyzing adaptive immunity and leading to AIDS.

Question 49Physical chemistry, amounts, reactions and equilibria

A projectile is launched from the origin (0, 0) with initial velocity v0 at an angle alpha above the horizontal in a uniform gravitational field g (neglecting air resistance). What is the Cartesian equation of its trajectory y(x)?

  1. y(x) = (g / (2 * v0^2)) * x^2 + cos(alpha) * x
  2. y(x) = -g * x + v0 * sin(alpha)
  3. y(x) = - (g / (2 * v0^2 * cos^2(alpha))) * x^2 + tan(alpha) * x Correct answer.
  4. y(x) = sqrt(v0^2 - 2 * g * x) * sin(alpha)

Explanation

The parametric equations of motion are x(t) = v0 * cos(alpha) * t and y(t) = -(1/2) * g * t^2 + v0 * sin(alpha) * t. Substituting t = x / (v0 * cos(alpha)) into y(t) eliminates the time parameter, yielding the standard parabolic trajectory equation y(x) = - (g / (2 * v0^2 * cos^2(alpha))) * x^2 + tan(alpha) * x.

Question 50Physical chemistry, amounts, reactions and equilibria

Which combination of characteristics defines a geostationary satellite relative to an observer on the surface of the Earth?

  1. Polar orbit, period of 90 minutes, moving from north to south at 500 km altitude
  2. Elliptical orbit inclined at 45° to the equator with an orbital period of 12 hours
  3. Retrograde orbit moving from east to west in any orbital plane at 10,000 km altitude
  4. Circular orbit located strictly in the equatorial plane, orbiting west-to-east with an orbital period equal to the Earth's sidereal day (approximately 24 hours, at ~35,800 km altitude) Correct answer.

Explanation

To remain fixed above the same point on Earth's equator, a satellite must fulfill three strict conditions: (1) its orbit must be circular and lie directly in Earth's equatorial plane; (2) its orbital period T must equal Earth's rotational period (sidereal day ~ 86,164 s ~ 24 h); and (3) it must revolve in the same direction as Earth's rotation (west to east), which places it at an altitude of approximately 35,800 km.