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Sample ENADE Ciências Biológicas Practice Questions
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1According to the fluid mosaic model, plasma membranes consist of a dynamic lipid bilayer embedded with proteins. How does cholesterol regulate plasma membrane fluidity across varying physiological temperatures in eukaryotic cells?
A.It acts as a bidirectional fluidity buffer, restraining phospholipid movement at high temperatures and preventing close packing/crystallization of acyl chains at low temperatures.
B.It permanently solidifies the lipid bilayer at elevated temperatures by covalently cross-linking adjacent sphingolipids and unsaturated fatty acids.
C.It promotes constant passive ion leakage at low temperatures by forming transmembrane aqueous pore complexes across the hydrophobic core.
D.It exclusively increases membrane fluidity at high temperatures by accelerating lateral phospholipid exchange and decreasing transition temperature.
Explanation: Cholesterol functions as a bidirectional fluidity buffer in animal cell membranes. At relatively high temperatures (such as 37°C in warm-blooded vertebrates), cholesterol's rigid steroid ring structure hinders excessive lateral movement of phospholipids, thereby decreasing fluidity and permeability. At lower temperatures, cholesterol disrupts the tight regular packing of phospholipid hydrocarbon tails, preventing membrane crystallization/phase transition to a rigid gel state.
2In animal epithelial cells, the apical uptake of D-glucose against its concentration gradient is coupled with the influx of sodium ions via the SGLT1 symporter. What mechanism provides the fundamental driving force that sustains this secondary active transport?
A.Direct enzymatic hydrolysis of cytosolic ATP by the catalytic nucleotide-binding domain of the SGLT1 symporter protein itself.
B.The electrochemical Na+ gradient maintained across the basolateral membrane by primary active transport via the Na+/K+ ATPase pump.
C.Passive facilitated diffusion of K+ out of the cell through apical voltage-gated potassium channels down its chemical gradient.
D.Generation of an alkaline intracellular pH gradient driven by mitochondrial vacuolar-type H+-ATPases.
Explanation: Secondary active transport via SGLT1 (sodium-glucose linked transporter 1) utilizes the electrochemical potential difference of Na+ across the plasma membrane. This Na+ gradient is continuously established and energized by the primary active transport of the basolateral Na+/K+ ATPase, which pumps 3 Na+ out of the cell and 2 K+ into the cell per hydrolyzed ATP molecule. As Na+ flows down its steep inward electrochemical gradient, glucose is cotransported against its uphill chemical concentration gradient.
3During the synthesis and sorting of secretory and lysosomal proteins in eukaryotic cells, which molecular sequence of events directs the nascent polypeptide to the lumen of the rough endoplasmic reticulum (RER)?
A.A C-terminal KDEL retrieval signal binds to cytoplasmic importin-alpha receptors, directing the ribosome directly to nuclear pore complexes.
B.A core oligosaccharide is transferred to cytosolic lysine residues by chaperone BiP, triggering passive diffusion through peroxisomal pores.
C.An N-terminal hydrophobic signal sequence emerges from the ribosome, is recognized by the Signal Recognition Particle (SRP), halts translation, and docks onto the SRP receptor at the Sec61 translocon.
D.Clathrin triskelions assemble around the ribosome on the cytosolic face, driving endocytic engulfment into cis-Golgi cisternae.
Explanation: Targeting of nascent secretory and endomembrane proteins to the rough ER begins when an N-terminal hydrophobic signal peptide emerges from the ribosome exit tunnel. The Signal Recognition Particle (SRP), a ribonucleoprotein complex, binds the signal peptide and temporarily pauses translation elongation. The ribosome-SRP complex then docks with the SRP receptor located on the cytosolic surface of the ER membrane, transferring the nascent chain to the Sec61 aqueous translocon channel for co-translational translocation into the ER lumen.
4In human cellular pathology, I-cell disease (mucolipidosis type II) is a severe autosomal recessive disorder characterized by the absence of multiple hydrolytic enzymes in lysosomes and their aberrant hypersecretion into the extracellular fluid. What is the underlying biochemical defect causing this cellular phenotype?
A.Constitutive activation of dynamin GTPase, resulting in premature fusion of late endosomes with the mitochondrial outer membrane.
B.Defect in the Sec61 translocon channel, preventing the entry of newly synthesized acid hydrolases into the lumen of the rough endoplasmic reticulum.
C.Total absence of the vacuolar H+-ATPase (V-ATPase) proton pump in the lysosomal membrane, leading to an alkaline lysosomal lumen.
D.Mutation in the gene encoding UDP-N-acetylglucosamine:lysosomal-enzyme N-acetylglucosaminyl-1-phosphotransferase, preventing the generation of mannose-6-phosphate (M6P) targeting signals in the cis-Golgi.
Explanation: Soluble lysosomal enzymes are tagged in the cis-Golgi apparatus with mannose-6-phosphate (M6P) residues via a two-step enzymatic reaction initiated by GlcNAc-phosphotransferase (UDP-N-acetylglucosamine:lysosomal-enzyme N-acetylglucosaminyl-1-phosphotransferase). Trans-Golgi M6P receptors recognize this modification and sort the hydrolases into clathrin-coated vesicles destined for endosomes/lysosomes. In I-cell disease, deficiency in GlcNAc-phosphotransferase causes lysosomal hydrolases to lack the M6P tag; consequently, they follow the default constitutive secretory pathway and are secreted into the extracellular space, leaving lysosomes empty and full of undegraded inclusion bodies (I-cells).
5The eukaryotic cell cycle is tightly controlled by cyclin-dependent kinases (CDKs) and their regulatory cyclin subunits. What molecular event is required for a mammalian cell to pass the G1/S restriction point (R-point) and commit to DNA replication?
A.Hyperphosphorylation of the Retinoblastoma protein (pRb) by Cyclin D-CDK4/6 and Cyclin E-CDK2 complexes, releasing active E2F to transcribe S-phase genes.
B.Dephosphorylation of the Retinoblastoma protein (pRb) by protein phosphatase 1, which locks the E2F transcription factor in a repressive complex.
C.Proteasomal degradation of the Anaphase-Promoting Complex/Cyclosome (APC/C) triggered by high levels of Cyclin B1.
D.Direct ubiquitin-mediated destruction of DNA polymerase delta and single-stranded DNA binding protein RPA by the p53 tumor suppressor.
Explanation: In early G1 phase, the unphosphorylated/hypophosphorylated Retinoblastoma protein (pRb) binds and inhibits the E2F family of transcription factors. In response to mitogenic signaling, Cyclin D levels rise, forming active Cyclin D-CDK4/6 complexes that initiate pRb phosphorylation. Subsequent activation of Cyclin E-CDK2 causes pRb hyperphosphorylation, triggering a conformational change that dissociates pRb from E2F. Free E2F then activates the transcription of essential S-phase genes (including Cyclin E, Cyclin A, and enzymes for dNTP synthesis and DNA replication), driving the cell past the restriction point into S phase.
6During prophase I of meiosis, the physical exchange of genetic material between non-sister chromatids of homologous chromosomes occurs at chiasmata. What protein structure mediates the precise point-for-point pairing (synapsis) of homologous chromosomes prior to crossing-over?
A.The Kinetochore-microtubule bridge complex.
B.The Synaptonemal complex.
C.The Nuclear lamina lattice.
D.The Centrosomal pericentriolar matrix.
Explanation: The synaptonemal complex is a tripartite proteinaceous lattice consisting of two lateral elements (derived from axial cores of homologous chromosomes) and a central element linked by transverse filaments. Assembling during the zygotene stage of prophase I, it stabilizes homologous chromosome alignment (synapsis) at an exact ~100 nm spacing, facilitating double-strand break repair and reciprocal homologous recombination (crossing-over) during pachytene.
7In eukaryotic nuclear DNA replication, how are the discontinuous Okazaki fragments on the lagging strand synthesized, processed, and covalently sealed into an uninterrupted strand?
A.DNA Pol gamma synthesizes both leading and lagging strands continuously from RNA primers without requiring endonuclease processing.
B.Reverse transcriptase removes the RNA primer and inserts ribonucleotides, which are then cross-linked by topoisomerase II alpha.
C.DNA Pol alpha/primase synthesizes an RNA-DNA primer, DNA Pol delta extends the fragment, flap endonuclease 1 (FEN1) and Dna2 remove the displaced primer flap, and DNA ligase I seals the phosphodiester nick.
D.DNA Pol beta removes RNA primers via 5'->3' exonuclease activity and joins DNA fragments using pyrophosphate energy.
Explanation: On the eukaryotic lagging strand, DNA polymerase alpha-primase complex synthesizes an initial ~10 nt RNA primer followed by ~20-30 nt of initiator DNA. The sliding clamp PCNA and replicative DNA polymerase delta (Pol delta) take over processive elongation. When Pol delta encounters the 5' end of the downstream Okazaki fragment, it displaces the RNA/DNA primer into a 5' flap. Flap endonuclease 1 (FEN1) (assisted by Dna2 for long flaps) cleaves the flap, and DNA ligase I catalyzes ATP-dependent phosphodiester bond formation to seal the remaining nick.
8Linear eukaryotic chromosomes encounter the 'end-replication problem' due to the inability of conventional DNA polymerases to replicate the extreme 5' ends of lagging strands. How does the enzyme telomerase overcome this limitation in germline and stem cells?
A.It acts as a DNA-dependent RNA polymerase that transcribes circular plasmids into double-stranded telomeric repeats.
B.It recruits DNA topoisomerase I to unwind the double helix and synthesize poly-adenine tails on chromosome termini.
C.It degrades the 3' single-stranded overhang and ligates blunt chromosome ends directly to the nuclear envelope matrix.
D.It uses an intrinsic internal RNA template within its TERC subunit and a reverse transcriptase catalytic subunit (TERT) to extend the single-stranded 3' G-rich overhang of chromosomal ends.
Explanation: Telomerase is a specialized ribonucleoprotein complex containing a catalytic reverse transcriptase subunit (TERT in humans) and an integral non-coding RNA component (TERC/hTR). TERC provides an internal RNA template complementary to telomeric repeats (5'-TTAGGG-3' in vertebrates). Telomerase binds the 3' protruding single-stranded G-rich overhang of the chromosome and uses its internal RNA template to synthesize tandem telomeric DNA repeats via reverse transcription. Subsequently, standard primase and DNA polymerase alpha/delta synthesize the complementary C-rich strand.
9Eukaryotic cells utilize three distinct nuclear RNA polymerases. Which RNA polymerase is specifically responsible for transcribing all protein-coding messenger RNAs (mRNAs) and most small nuclear RNAs (snRNAs)?
A.RNA Polymerase II
B.RNA Polymerase I
C.RNA Polymerase III
D.RNA Polymerase IV
Explanation: RNA Polymerase II (RNA Pol II) is the eukaryotic nuclear enzyme responsible for synthesizing all protein-coding precursor mRNAs, as well as most snRNAs (such as U1, U2, U4, U5 involved in splicing) and microRNAs (miRNAs). In contrast, RNA Polymerase I synthesizes the large ribosomal RNA precursor (28S, 18S, and 5.8S rRNAs), while RNA Polymerase III transcribes transfer RNAs (tRNAs), 5S rRNA, and the U6 snRNA.
10During pre-mRNA splicing in eukaryotic nuclei, which biochemical intermediate and catalytic mechanism characterize the excision of introns by the spliceosome?
A.Direct hydrolytic cleavage of phosphodiester bonds at the 5' and 3' splice junctions by topoisomerase I without intermediate formation.
B.Two successive transesterification reactions: the 2'-OH of an invariant branch-point adenosine attacks the 5' splice site, forming a lariat intermediate, followed by 3'-OH attack on the 3' splice site.
C.Phosphorylation of exon termini by polynucleotide kinase, followed by endonucleolytic removal of linear double-stranded intron hairpins.
D.S-adenosylmethionine (SAM)-dependent transmethylation that converts intron GU dinucleotides into poly-U tracts for exonuclease degradation.
Explanation: The spliceosome (composed of U1, U2, U4/U6, and U5 snRNPs) catalyzes pre-mRNA splicing through two sequential transesterification reactions that conserve phosphodiester energy. First, the 2'-hydroxyl group of a specific conserved branch-point adenosine nucleotide makes a nucleophilic attack on the phosphodiester bond at the 5' splice site (GU donor). This cleaves the 5' exon-intron junction and forms a 2'-5' phosphodiester bond, creating a branched 'lariat' intron-exon intermediate. In the second step, the free 3'-hydroxyl of the upstream exon attacks the phosphodiester bond at the 3' splice site (AG acceptor), ligating the two exons together and releasing the excised intron in lariat form.
About the ENADE Ciências Biológicas Exam
Ciências Biológicas has distinct 2026 ENADE routes. Bachelor's students take the four-hour bachelor assessment, while licensure students use the PND for theory and complete a separate practical teaching evaluation. Both official routes are in Portuguese. This English-language MCQ study adaptation supports shared biology knowledge but is not an official translation or a substitute for discursive and practical work.
Exam sponsor: INEP / Ministério da Educação (MEC). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.
Assessment
Bacharelado: Formação Geral and specific theory. Licenciatura: PND general/specific theory and separate supervised-teaching evaluation.
Time Limit
4 hours for Bacharelado; 5 hours 30 minutes for PND, with practical evaluation scheduled separately
Passing Score
No individual ENADE pass mark; regularity requirements and institutional scoring apply
Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.
Official sources
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.
Not published
Cell, molecular, biochemical, and physiological biology
Cell structure, genetics, metabolism, signaling, and organismal function.
Not published
Evolution, ecology, and biodiversity
Evolutionary processes, phylogeny, ecosystems, conservation, and Brazilian biodiversity.
Not published
Microbiology and biology education
Microorganisms, parasites, public and environmental health, inquiry, curriculum, and teaching practice.
Preparing for the ENADE Ciências Biológicas Exam
What You Need to Know
- Passing score: No individual ENADE pass mark; regularity requirements and institutional scoring apply
- Assessment: Bacharelado: Formação Geral and specific theory. Licenciatura: PND general/specific theory and separate supervised-teaching evaluation.
- Time limit: 4 hours for Bacharelado; 5 hours 30 minutes for PND, with practical evaluation scheduled separately
- Exam / certification fees: No separate fee for enrolled ENADE students Official sources
Using Our Practice Resources
- Work through all 100 available questions
- Review every answer and explanation
- Track weak areas and revisit them
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ENADE Ciências Biológicas: Suggested Study Strategy
1Confirm whether your institution registered you in the bachelor or licensure route.
2Use the applicable 2026 INEP matrix and practice Portuguese discursive or teaching-performance tasks separately.
Frequently Asked Questions
Do bachelor and licensure students take the same 2026 test?
No. Bachelor students use the bachelor ENADE format; licensure students take the PND theoretical stage and complete a separate practical teaching evaluation.
Is this an official English version?
No. It is an English-language four-option MCQ study adaptation of official Portuguese mixed-format assessments.