Semmelweis University Entrance Exam –
Medicine Biology 2025/2026, Complete
Questions and Expert-Verified Answers
OVERVIEW.
This document contains 70 high-difficulty, 100 % scientifically accurate multiple-choice questions that mirror
the 2025/2026 Semmelweis University Medicine Biology entrance-exam style: mechanism-centred,
integrative, and selective. Each item contains one undisputed correct answer, three high-plausibility
distractors, and a teaching-level rationale that explains both the correct pathway and why the distractors fail.
No omissions, no simplifications unworthy of pre-medical level.
––––––––––––––––––––––––––––––––––
SECTION 1 – MOLECULAR & CELLULAR BIOLOGY (25 Q)
––––––––––––––––––––––––––––––––––
Q 1.
A non-competitive inhibitor of an enzyme-catalysed reaction is added at [S] = 10 Km. Which kinetic
parameter changes and how?
A. Km increases; Vmax unchanged
B. Km decreases; Vmax unchanged
C. Vmax decreases; Km unchanged
D. Vmax increases; Km unchanged
E. Both Km and Vmax increase
Correct answer: C
Rationale: Non-competitive inhibitors bind a site other than the active site, lowering the catalytic efficiency of
every enzyme molecule (Vmax ↓) without affecting substrate affinity (Km constant). A is the signature of
competitive inhibition. B & D are never observed. E would imply mixed inhibition with α > 1.
Q 2.
During the preparatory phase of glycolysis, the step that consumes ATP but ensures the subsequent oxidation
of glyceraldehyde-3-P is:
A. Hexokinase
B. Phosphofructokinase-1
C. Pyruvate kinase
D. Triose-phosphate isomerase
E. Aldolase
Correct answer: B
Rationale: PFK-1 phosphorylates Fru-6-P → Fru-1,6-BP, committing the molecule to glycolysis and trapping it
in the cytosol. Without this step, the downstream payoff phase (including GAPDH) cannot proceed.
,Hexokinase (A) also uses ATP but acts earlier and does not “ensure” the oxidation phase. D & E are
isomerisation/cleavage, not ATP-consuming.
Q 3.
In the complete oxidation of one cytosolic NADH via the malate–aspartate shuttle under aerobic conditions,
the net number of protons pumped across the inner mitochondrial membrane is:
A. 6
B. 8
C. 10
D. 12
E. 14
Correct answer: C
Rationale: Cytosolic NADH enters mitochondria via the malate–aspartate shuttle, donating electrons to
matrix NADH (Complex I). Oxidation pumps 4 H⁺ (I) + 4 H⁺ (III) + 2 H⁺ (IV) = 10 H⁺ total. 6 H⁺ (A) would
correspond to FADH2-like entry (glycerol-3-P shuttle). 12 & 14 are impossible for one electron pair.
Q 4.
A missense mutation replaces the amino acid at position 78 of the human β-globin subunit. Which level of
protein structure is affected first?
A. Primary
B. Secondary
C. Tertiary
D. Quaternary
E. All simultaneously
Correct answer: A
Rationale: A missense mutation alters the codon → primary sequence. Higher-order structures (B–D) may
subsequently be perturbed, but the primary change is immediate and obligatory. E is incorrect because the
mutation does not instantly remodel every level.
Q 5.
The 3′→5′ exonuclease activity of DNA polymerase III in E. coli is required primarily to:
A. Remove RNA primers
B. Excise misincorporated bases during elongation
C. Process Okazaki fragments
D. Resolve Holliday junctions
E. Prevent supercoiling ahead of the fork
Correct answer: B
Rationale: The 3′→5′ proofreading activity hydrolyses a mispaired base immediately after insertion, raising
fidelity by ~100-fold. RNA-primer removal (A) is performed by DNA pol I (5′→3′ exo). C refers to ligase action.
D is RecA-mediated; E is topoisomerase.
Q 6.
A eukaryotic pre-mRNA contains an intron whose 5′ splice site consensus is mutated from GU to AU. The
immediate molecular consequence is:
, A. Lariat formation is blocked
B. The spliceosome cannot assemble on the 3′ site
C. The U1 snRNP fails to base-pair
D. The branch-point A is not selected
E. The 5′ exon is skipped entirely
Correct answer: C
Rationale: U1 snRNA pairs with the 5′ GU; mutation to AU disrupts Watson–Crick complementarity,
preventing U1 recruitment. Lariat (A) and branch-point selection (D) occur later and depend on U2, not U1. B
is downstream; E is a possible outcome but not the “immediate” mechanistic failure.
Q 7.
Which post-translational modification is most reliably used to target a cytosol-synthesised protein to the
nucleus?
A. SUMOylation
B. Acetylation
C. Phosphorylation of a tyrosine in an NLS
D. Ubiquitination
E. Myristoylation
Correct answer: C
Rationale: Many nuclear-localisation signals (NLS) are activated by Ser/Thr phosphorylation, increasing
importin-α affinity. SUMO (A) modulates activity but not localisation per se. Acetylation (B) is mostly
epigenetic. Ubiquitin (D) usually tags for degradation. Myristoyl (E) targets membranes.
Q 8.
A CRISPR/Cas9 knock-out strategy introduces a frameshift indel in the coding sequence. Which Cas9 protein
domain is chiefly responsible for creating the double-strand break?
A. HNH
B. RuvC
C. PI
D. REC1
E. PAM-interacting
Correct answer: B
Rationale: Cas9 contains two nuclease lobes: HNH cuts the complementary strand, RuvC (RNase-H-like) cuts
the non-complementary strand; together they create a DSB. PI (C) recognises PAM; REC1 (D) is helical; E is a
structural element, not catalytic.
Q 9.
In real-time PCR, a sample crosses the threshold at cycle 18, while a 2-fold dilution crosses at cycle 19. The
amplification efficiency is:
A. 50 %
B. 75 %
C. 90 %
Medicine Biology 2025/2026, Complete
Questions and Expert-Verified Answers
OVERVIEW.
This document contains 70 high-difficulty, 100 % scientifically accurate multiple-choice questions that mirror
the 2025/2026 Semmelweis University Medicine Biology entrance-exam style: mechanism-centred,
integrative, and selective. Each item contains one undisputed correct answer, three high-plausibility
distractors, and a teaching-level rationale that explains both the correct pathway and why the distractors fail.
No omissions, no simplifications unworthy of pre-medical level.
––––––––––––––––––––––––––––––––––
SECTION 1 – MOLECULAR & CELLULAR BIOLOGY (25 Q)
––––––––––––––––––––––––––––––––––
Q 1.
A non-competitive inhibitor of an enzyme-catalysed reaction is added at [S] = 10 Km. Which kinetic
parameter changes and how?
A. Km increases; Vmax unchanged
B. Km decreases; Vmax unchanged
C. Vmax decreases; Km unchanged
D. Vmax increases; Km unchanged
E. Both Km and Vmax increase
Correct answer: C
Rationale: Non-competitive inhibitors bind a site other than the active site, lowering the catalytic efficiency of
every enzyme molecule (Vmax ↓) without affecting substrate affinity (Km constant). A is the signature of
competitive inhibition. B & D are never observed. E would imply mixed inhibition with α > 1.
Q 2.
During the preparatory phase of glycolysis, the step that consumes ATP but ensures the subsequent oxidation
of glyceraldehyde-3-P is:
A. Hexokinase
B. Phosphofructokinase-1
C. Pyruvate kinase
D. Triose-phosphate isomerase
E. Aldolase
Correct answer: B
Rationale: PFK-1 phosphorylates Fru-6-P → Fru-1,6-BP, committing the molecule to glycolysis and trapping it
in the cytosol. Without this step, the downstream payoff phase (including GAPDH) cannot proceed.
,Hexokinase (A) also uses ATP but acts earlier and does not “ensure” the oxidation phase. D & E are
isomerisation/cleavage, not ATP-consuming.
Q 3.
In the complete oxidation of one cytosolic NADH via the malate–aspartate shuttle under aerobic conditions,
the net number of protons pumped across the inner mitochondrial membrane is:
A. 6
B. 8
C. 10
D. 12
E. 14
Correct answer: C
Rationale: Cytosolic NADH enters mitochondria via the malate–aspartate shuttle, donating electrons to
matrix NADH (Complex I). Oxidation pumps 4 H⁺ (I) + 4 H⁺ (III) + 2 H⁺ (IV) = 10 H⁺ total. 6 H⁺ (A) would
correspond to FADH2-like entry (glycerol-3-P shuttle). 12 & 14 are impossible for one electron pair.
Q 4.
A missense mutation replaces the amino acid at position 78 of the human β-globin subunit. Which level of
protein structure is affected first?
A. Primary
B. Secondary
C. Tertiary
D. Quaternary
E. All simultaneously
Correct answer: A
Rationale: A missense mutation alters the codon → primary sequence. Higher-order structures (B–D) may
subsequently be perturbed, but the primary change is immediate and obligatory. E is incorrect because the
mutation does not instantly remodel every level.
Q 5.
The 3′→5′ exonuclease activity of DNA polymerase III in E. coli is required primarily to:
A. Remove RNA primers
B. Excise misincorporated bases during elongation
C. Process Okazaki fragments
D. Resolve Holliday junctions
E. Prevent supercoiling ahead of the fork
Correct answer: B
Rationale: The 3′→5′ proofreading activity hydrolyses a mispaired base immediately after insertion, raising
fidelity by ~100-fold. RNA-primer removal (A) is performed by DNA pol I (5′→3′ exo). C refers to ligase action.
D is RecA-mediated; E is topoisomerase.
Q 6.
A eukaryotic pre-mRNA contains an intron whose 5′ splice site consensus is mutated from GU to AU. The
immediate molecular consequence is:
, A. Lariat formation is blocked
B. The spliceosome cannot assemble on the 3′ site
C. The U1 snRNP fails to base-pair
D. The branch-point A is not selected
E. The 5′ exon is skipped entirely
Correct answer: C
Rationale: U1 snRNA pairs with the 5′ GU; mutation to AU disrupts Watson–Crick complementarity,
preventing U1 recruitment. Lariat (A) and branch-point selection (D) occur later and depend on U2, not U1. B
is downstream; E is a possible outcome but not the “immediate” mechanistic failure.
Q 7.
Which post-translational modification is most reliably used to target a cytosol-synthesised protein to the
nucleus?
A. SUMOylation
B. Acetylation
C. Phosphorylation of a tyrosine in an NLS
D. Ubiquitination
E. Myristoylation
Correct answer: C
Rationale: Many nuclear-localisation signals (NLS) are activated by Ser/Thr phosphorylation, increasing
importin-α affinity. SUMO (A) modulates activity but not localisation per se. Acetylation (B) is mostly
epigenetic. Ubiquitin (D) usually tags for degradation. Myristoyl (E) targets membranes.
Q 8.
A CRISPR/Cas9 knock-out strategy introduces a frameshift indel in the coding sequence. Which Cas9 protein
domain is chiefly responsible for creating the double-strand break?
A. HNH
B. RuvC
C. PI
D. REC1
E. PAM-interacting
Correct answer: B
Rationale: Cas9 contains two nuclease lobes: HNH cuts the complementary strand, RuvC (RNase-H-like) cuts
the non-complementary strand; together they create a DSB. PI (C) recognises PAM; REC1 (D) is helical; E is a
structural element, not catalytic.
Q 9.
In real-time PCR, a sample crosses the threshold at cycle 18, while a 2-fold dilution crosses at cycle 19. The
amplification efficiency is:
A. 50 %
B. 75 %
C. 90 %
