All Practice Exams
Free Practice Questions for Concours APESA
Exam-style questions and explanations by OpenExamPrep.
✓ No registration✓ No credit card
100+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More Moroccan Grandes Écoles Entrance Examinations
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
Sample Concours APESA Practice Questions
Try these sample questions to review concepts for the Concours APESA exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1In plant and animal cellular bioenergetics, ATP synthesis during oxidative phosphorylation and photophosphorylation is coupled to transmembrane proton gradients. Which statement accurately describes the orientation of proton pumping and the catalytic domain of ATP synthase in mitochondria?
A.Protons are pumped from the matrix into the intermembrane space, and the F1 catalytic headpiece faces the mitochondrial matrix
B.Protons are pumped from the intermembrane space into the matrix, and the F1 catalytic headpiece faces the intermembrane space
C.Protons are pumped from the matrix into the cytoplasm, and ATP is synthesized exclusively within the outer membrane pore
D.Protons accumulate in the mitochondrial matrix creating an acidic matrix, and the F0 channel pumps protons outward into the cristae
Explanation: During mitochondrial electron transport, respiratory complexes I, III, and IV translocate protons across the inner mitochondrial membrane from the matrix into the intermembrane space. This creates an electrochemical proton gradient (proton-motive force) with a lower pH in the intermembrane space. Protons flow back into the matrix through the F0 rotor channel of ATP synthase, driving rotational conformational changes in the F1 catalytic headpiece, which projects into the matrix where ADP and Pi are condensed into ATP.
2During semiconservative DNA replication in eukaryotic chromosomes, discontinuous synthesis on the lagging strand requires multiple coordinated enzymes. What is the precise sequential action required to process Okazaki fragments into a continuous DNA strand?
A.DNA ligase synthesizes an RNA primer, DNA helicase extends the DNA strand, and DNA topoisomerase seals the remaining nicks
B.RNA primase lays down an RNA primer, DNA polymerase extends the fragment, RNA primers are excised, DNA polymerase fills the gap, and DNA ligase catalyzes phosphodiester bond formation
C.DNA polymerase alpha synthesizes the entire Okazaki fragment continuously, followed by exonuclease cleavage of the 3' poly-A tail
D.Reverse transcriptase synthesizes a complementary cDNA primer, which is elongated by RNA polymerase and spliced by snRNPs
Explanation: On the lagging template strand (running 3' to 5' in the direction of fork opening), synthesis must proceed discontinuously away from the fork. RNA primase (in complex with DNA pol alpha) synthesizes a short RNA primer, providing a free 3'-OH group. DNA polymerase delta then elongates the Okazaki fragment until it reaches the preceding primer. The RNA primer is excised (by RNase H / FEN1), the resulting gap is filled with deoxynucleotides by DNA polymerase, and DNA ligase seals the nick by forming a covalent 3'-5' phosphodiester bond using ATP.
3In molecular genetics, the universal canonical genetic code is described as degenerate, non-overlapping, and unambiguous. What does 'degeneracy' (or redundancy) of the genetic code mean?
A.Adjacent codons share overlapping nucleotides during translation along the mRNA strand
B.A single triplet codon can code for multiple distinct amino acids depending on cellular conditions
C.Most amino acids are encoded by more than one triplet codon
D.Codons can be read interchangeably in either the 5'-to-3' or 3'-to-5' direction
Explanation: Degeneracy of the genetic code refers to the biological fact that with 64 triplet codons and only 20 standard proteinogenic amino acids, several amino acids are specified by two, three, four, or six synonymous codons (for example, leucine and arginine are each encoded by six distinct codons). Conversely, 'unambiguous' means that any given codon specifies exactly one amino acid. 'Non-overlapping' means adjacent codons are read sequentially three bases at a time without sharing nucleotides.
4A mature eukaryotic mRNA transcript contains an uninterrupted open reading frame (ORF) of exactly 1,200 nucleotides, beginning with the initiation codon AUG and ending with the stop codon UAA. Assuming no post-translational cleavage and an average amino acid molecular mass of 110 Daltons (Da), what is the number of amino acids in the translated polypeptide and its approximate molecular mass?
A.398 amino acids, with an approximate molecular mass of 43.78 kDa
B.400 amino acids, with an approximate molecular mass of 44.00 kDa
C.1,197 amino acids, with an approximate molecular mass of 131.67 kDa
D.399 amino acids, with an approximate molecular mass of 43.89 kDa
Explanation: The total number of triplet codons in the open reading frame is 1,200 / 3 = 400 codons. The final codon (UAA) is a nonsense (stop) codon that binds a release factor rather than a tRNA carrying an amino acid, so it does not encode an amino acid. Thus, the translated polypeptide comprises 400 - 1 = 399 amino acids. With an average mass of 110 Da per residue, the total molecular mass is 399 * 110 Da = 43,890 Da, or approximately 43.89 kDa.
5In enzyme kinetics governed by the Michaelis-Menten model, how does a reversible competitive inhibitor alter the apparent Michaelis constant (Km) and the maximum reaction velocity (Vmax)?
A.Apparent Km decreases while Vmax decreases
B.Apparent Km increases while Vmax remains unchanged
C.Apparent Km remains unchanged while Vmax decreases
D.Both apparent Km and Vmax increase proportionally
Explanation: A competitive inhibitor structurally resembles the natural substrate and competes directly for binding at the enzyme's catalytic active site. At saturating substrate concentrations ([S] >> Km), substrate molecules outcompete the inhibitor, allowing the enzyme to achieve its full catalytic capacity, so Vmax is unchanged. However, achieving half of this maximum velocity requires a higher concentration of substrate in the presence of the inhibitor, resulting in an increased apparent Km (lower apparent affinity).
6An agricultural soil urease extracted from rhizospheric bacteria exhibits Michaelis-Menten kinetics with a maximum velocity Vmax = 120 micromol/(min*mg protein) and a Michaelis constant Km = 15 mM. When urea is added to a soil assay at a substrate concentration [S] = 30 mM, what is the initial enzymatic reaction velocity (v0)?
A.90 micromol/(min*mg protein)
B.60 micromol/(min*mg protein)
C.80 micromol/(min*mg protein)
D.40 micromol/(min*mg protein)
Explanation: Using the Michaelis-Menten equation: v0 = (Vmax * [S]) / (Km + [S]). Substituting the provided values: v0 = (120 * 30) / (15 + 30) = 3,600 / 45 = 80 micromol/(min*mg protein). Notice that when [S] = 2 * Km, the reaction velocity must equal (2/3) * Vmax = (2/3) * 120 = 80 micromol/(min*mg protein).
7Glycolysis is the primordial anaerobic catabolic pathway occurring in the cytosol of eukaryotic cells. What is the net chemical equation for the conversion of one molecule of glucose during glycolysis?
A.1 Glucose + 2 NAD+ + 32 ADP + 32 Pi -> 6 CO2 + 2 NADH + 32 ATP + 6 H2O
B.1 Glucose + 4 NAD+ + 4 ADP + 4 Pi -> 2 Pyruvate + 4 NADH + 4 H+ + 4 ATP + 2 H2O
C.1 Glucose + 2 FAD + 2 ADP + 2 Pi -> 2 Lactate + 2 FADH2 + 2 ATP
D.1 Glucose + 2 NAD+ + 2 ADP + 2 Pi -> 2 Pyruvate + 2 NADH + 2 H+ + 2 ATP + 2 H2O
Explanation: During glycolysis, the energy investment phase consumes 2 ATP molecules (catalyzed by hexokinase and phosphofructokinase-1) to phosphorylate glucose into fructose-1,6-bisphosphate. In the subsequent energy payoff phase, 4 ATP molecules are produced by substrate-level phosphorylation alongside 2 NADH and 2 pyruvate molecules. Subtracting the 2 consumed ATP yields a net gain of 2 ATP, 2 NADH, 2 H+, and 2 H2O per glucose molecule.
8Using modern biochemical stoichiometry with mechanistic P/O ratios (2.5 ATP produced per matrix NADH oxidized and 1.5 ATP per FADH2 oxidized), what is the total theoretical net yield of ATP synthesized from the complete aerobic oxidation of one mole of glucose in a cell utilizing the malate-aspartate shuttle?
A.38 ATP
B.32 ATP
C.30 ATP
D.26 ATP
Explanation: The breakdown per glucose molecule is as follows: Glycolysis produces 2 net ATP (substrate-level) + 2 cytosolic NADH. Using the malate-aspartate shuttle, these 2 cytosolic NADH transfer electrons to matrix NAD+, yielding 2 * 2.5 = 5 ATP. The transition step (pyruvate dehydrogenase) yields 2 matrix NADH = 2 * 2.5 = 5 ATP. The Krebs cycle (2 turns) yields 2 GTP/ATP (substrate-level) + 6 matrix NADH (6 * 2.5 = 15 ATP) + 2 FADH2 (2 * 1.5 = 3 ATP) = 20 ATP. Summing these values: 2 + 5 + 5 + 20 = 32 ATP.
9In the mitochondrial electron transport chain, how does the metabolic poison cyanide (CN-) inhibit cellular respiration and what is its immediate effect on mitochondrial oxygen consumption?
A.Cyanide uncouples the inner membrane by acting as a protonophore, stimulating maximum oxygen consumption without ATP synthesis
B.Cyanide competitively inhibits succinate dehydrogenase (Complex II), shifting electrons directly to Complex IV without affecting oxygen use
C.Cyanide binds with high affinity at cytochrome c oxidase (Complex IV), blocking electron transfer to oxygen
D.Cyanide blocks the F0 subunit of ATP synthase, allowing electron transport and oxygen reduction to continue unhindered
Explanation: Cyanide binds with very high affinity at the heme a3-CuB center of cytochrome c oxidase (Complex IV). This blocks terminal electron transfer to molecular oxygen, sharply suppressing electron flow, oxygen consumption, proton pumping, and oxidative phosphorylation. The biochemical effect is severe without requiring the binding to be described as universally irreversible.
10During the light-dependent reactions of oxygenic photosynthesis in higher plants, what is the immediate biochemical source of the molecular oxygen (O2) released into the atmosphere?
A.Decarboxylation of 3-phosphoglycerate during the reduction phase of the Calvin cycle
B.Splitting of carbon dioxide by RuBisCO in the chloroplast stroma
C.Oxidation of plastoquinone by the cytochrome b6f complex
D.Photolysis of water at the oxygen-evolving manganese cluster of Photosystem II
Explanation: In Photosystem II (PSII), light energy absorbed by antenna pigments excites the reaction center chlorophyll P680 to P680*. Upon donating an electron to pheophytin, P680 becomes the powerful oxidant P680+, which extracts electrons from water via the oxygen-evolving complex (Mn4CaO5 cluster) on the lumenal face of the thylakoid membrane. The oxidation (photolysis) of two water molecules yields 4 electrons to reduce P680+, 4 protons released into the lumen, and one molecule of molecular oxygen (O2).
About the Concours APESA Exam
The official 2026–2027 page schedules the written APESA test for 19 July 2026, limits candidates to those under 23 on 1 September, and lists the eligible Moroccan Baccalauréat streams. It does not publish the tested subjects, item format, duration, count, pass mark, or assessment language. Accordingly, this independent English-language MCQ bank draws its science clusters from the eligible scientific backgrounds and the Institute's published APESA curriculum; those clusters are study choices, not a claimed official entrance-test blueprint or format simulation.
Exam sponsor: Institut Agronomique et Vétérinaire Hassan II. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.
Assessment
Baccalauréat-based preselection followed by a written admission test.
Time Limit
Not published in the reviewed official 2026 page.
Passing Score
Competitive ranking; no fixed pass mark published.
Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.
Our practice resources: topics covered
We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.
30 questions in this practice bank; not an official test weight
Life and Earth Sciences
Biology, physiology, genetics, ecology, geology, and agricultural foundations.
25 questions in this practice bank; not an official test weight
Mathematics
Algebra, functions, probability, and quantitative analysis.
30 questions in this practice bank; not an official test weight
Physics-Chemistry
Mechanics, electricity, thermodynamics, and chemistry.
15 questions in this practice bank; not an official test weight
Scientific Reasoning
Data interpretation, experimental reasoning, and applied quantitative judgment.
Preparing for the Concours APESA Exam
What You Need to Know
- Passing score: Competitive ranking; no fixed pass mark published.
- Assessment: Baccalauréat-based preselection followed by a written admission test.
- Time limit: Not published in the reviewed official 2026 page.
- Exam / certification fees: 150 MAD non-refundable dossier-processing fee Official sources
Using Our Practice Resources
- Work through all 100 available questions
- Review every answer and explanation
- Track weak areas and revisit them
- Use our AI tutor for tough concepts
Concours APESA: Suggested Study Strategy
1Use this bank for foundational review, not format prediction.
2Consult the annual official competition page for eligibility and instructions.
3Practice written scientific reasoning in addition to MCQs.
Frequently Asked Questions
Does IAV publish a 2026 subject blueprint on the reviewed competition page?
No. The practice clusters are based on eligible backgrounds and the published APESA curriculum, not a claimed official test blueprint.
Is the official test confirmed as MCQ?
No. The MCQ format here is only a study adaptation.