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Exam Review

Key Facts: FY13CE Biology Exam

3 hours + 10 min

Official exam duration including reading time

Ministry of Education Fiji

28 / 16 / 36

Marks for Strands 1, 2 and 3 of the FY13CE Biology paper

MoE Fiji FY13CE 2025 Biology paper

20 marks

Two compulsory essay questions on the FY13CE Biology paper

MoE Fiji FY13CE 2025 Biology paper

$5 / $25 FJD

Ministry recount and remark fee per subject

MoE Fiji Annual Report 2022-2023

100

Multiple-choice practice questions in this bank

OpenExamPrep

Fiji FY13CE Biology is a single external 3-hour written paper marked out of 100 across three strands plus two compulsory essays, set by the Ministry of Education Examinations and Assessment Unit. This free practice bank provides 100 English-language MCQ revision questions on cellular processes, ecology and genetics as a study aid, not a format simulation.

Sample FY13CE Biology Practice Questions

Try these sample questions to review concepts for the FY13CE Biology exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which compartment of the chloroplast contains the enzymes responsible for the carbon-fixing reactions of the Calvin cycle?
A.Thylakoid lumen
B.Stroma
C.Outer envelope membrane
D.Granum intermembrane space
Explanation: The stroma is the fluid-filled space surrounding the thylakoid membranes in chloroplasts. It contains soluble enzymes, notably RuBisCO, required for carbon fixation and the subsequent steps of the Calvin cycle. The thylakoid membranes, in contrast, house the pigments and electron transport chains of the light-dependent reactions.
2What primary structural feature of the mitochondrial inner membrane increases its capacity for ATP generation?
A.Extensive folding into cristae to maximize surface area
B.A dense layer of peptidoglycan for structural rigidity
C.Large aqueous pores formed by porin proteins
D.A high concentration of cholesterol that immobilizes phospholipids
Explanation: The inner mitochondrial membrane is folded into numerous convolutions termed cristae. This folding significantly increases the available surface area for embedding complexes of the electron transport chain and ATP synthase molecules. As a result, the rate of oxidative phosphorylation and chemiosmotic ATP synthesis is greatly amplified.
3How do eukaryotic cytoplasmic ribosomes differ from the ribosomes found within the matrix of mitochondria and the stroma of chloroplasts?
A.Cytoplasmic ribosomes are 70S complexes, whereas organellar ribosomes are 80S complexes
B.Cytoplasmic ribosomes are 80S complexes, whereas organellar ribosomes are 70S complexes
C.Cytoplasmic ribosomes lack ribosomal RNA, whereas organellar ribosomes contain rRNA
D.Cytoplasmic ribosomes consist of three subunits, whereas organellar ribosomes consist of a single subunit
Explanation: Eukaryotic cytosolic ribosomes are 80S particles composed of 60S and 40S subunits. In contrast, mitochondria and chloroplasts possess smaller 70S ribosomes (consisting of 50S and 30S subunits) that closely resemble bacterial ribosomes, providing strong evidence for the endosymbiotic theory.
4Which cellular activity is predominantly carried out by the smooth endoplasmic reticulum rather than the rough endoplasmic reticulum?
A.Synthesis of phospholipids, steroid hormones, and detoxification of drugs
B.Translation of secretory polypeptide chains on bound ribosomes
C.Folding and initial N-linked glycosylation of membrane proteins
D.Packaging of hydrolytic enzymes destined for primary lysosomes
Explanation: The smooth endoplasmic reticulum lacks ribosomes and specializes in lipid biosynthesis (including phospholipids and steroid hormones), carbohydrate metabolism, calcium ion storage, and enzymatic detoxification of toxic metabolic products or xenobiotics. In contrast, the rough endoplasmic reticulum is studded with ribosomes and specializes in synthesizing secretory and membrane proteins.
5In the secretory pathway of a eukaryotic cell, what is the functional role of the trans-Golgi network?
A.Receiving newly synthesized transport vesicles directly from the rough endoplasmic reticulum
B.Sorting, tagging, and packaging modified proteins into specific vesicles for targeted delivery
C.Synthesizing ribosomal RNA precursors within an unsegmented matrix
D.Executing oxidative phosphorylation to supply ATP directly to the cisternae
Explanation: The trans-Golgi network is the exit face of the Golgi apparatus. It performs the final sorting, biochemical tagging (such as phosphorylation or sulfation), and packaging of modified proteins and lipids into distinct vesicle populations directed to lysosomes, the plasma membrane, or secretory granules.
6According to the fluid mosaic model, why is the biological plasma membrane described as 'amphipathic'?
A.It consists solely of hydrophilic carbohydrate residues facing both extracellular and intracellular fluids
B.Phospholipids possess hydrophilic polar phosphate heads and hydrophobic nonpolar fatty acid tails
C.It allows free, unrestricted passage of both polar ions and nonpolar macromolecules
D.Integral membrane proteins are entirely water-insoluble along their entire polypeptide chain
Explanation: Phospholipids are amphipathic molecules because they contain both a hydrophilic (water-loving) polar phosphate head group and two hydrophobic (water-fearing) nonpolar hydrocarbon fatty acid tails. In an aqueous environment, they spontaneously assemble into a stable bilayer with hydrophobic tails sequestered in the interior and hydrophilic heads oriented toward the aqueous cytoplasm and extracellular fluid.
7Which characteristic distinguishes primary active transport from facilitated diffusion across a cell membrane?
A.Primary active transport moves solutes down their electrochemical gradient without requiring energy
B.Primary active transport directly hydrolyzes ATP to move solutes against their electrochemical gradient
C.Facilitated diffusion requires direct metabolic ATP hydrolysis to open channel proteins
D.Facilitated diffusion can transport solutes from a region of low concentration to high concentration
Explanation: Primary active transport uses transmembrane carrier proteins (pumps) that directly couple the hydrolysis of ATP to the transport of ions or molecules against their electrochemical or concentration gradients. Facilitated diffusion, by contrast, is a passive process that utilizes channel or carrier proteins to allow solutes to diffuse down their concentration gradient without metabolic energy expenditure.
8Which statement accurately describes the mechanism of receptor-mediated endocytosis in eukaryotic cells?
A.Non-specific engulfment of extracellular fluid via invagination of smooth plasma membrane
B.Specific binding of extracellular ligands to cell-surface receptors triggering clathrin-coated vesicle formation
C.Continuous exocytosis of waste products via fusion of secretory vesicles with the outer membrane
D.Direct translocation of naked DNA plasmids across nuclear pore complexes into the cytosol
Explanation: Receptor-mediated endocytosis is a highly selective form of endocytosis in which specific extracellular macromolecules (ligands) bind to complementary transmembrane receptor proteins. This binding triggers the recruitment of adaptor proteins and clathrin to form coated pits, which invaginate and pinch off as clathrin-coated endocytic vesicles.
9A plant mesophyll cell with a solute potential (Ψs) of -0.80 MPa is placed in a beaker of pure distilled water (Ψ = 0 MPa). At dynamic equilibrium, what will be the cell's pressure potential (Ψp) and physiological state?
A.Ψp = -0.80 MPa; the cell becomes plasmolyzed
B.Ψp = +0.80 MPa; the cell becomes fully turgid
C.Ψp = 0 MPa; the cell lyses due to lack of a rigid cell wall
D.Ψp = +1.60 MPa; water continues to enter indefinitely
Explanation: Water moves down a water potential gradient from higher (0 MPa) to lower (-0.80 MPa) potential into the plant cell. As water enters, the protoplast pushes against the rigid cellulose cell wall, generating an outward hydrostatic pressure potential (Ψp). At equilibrium, the cell's total water potential (Ψ = Ψs + Ψp) matches the external solution (0 MPa), meaning -0.80 MPa + Ψp = 0, so Ψp = +0.80 MPa, and the cell is fully turgid.
10What internal microenvironment enables lysosomal acid hydrolases to efficiently degrade worn-out cellular organelles during autophagy without damaging the surrounding cytoplasm?
A.An acidic lumen (pH ~4.5–5.0) maintained by active proton (H+) pumping across the lysosomal membrane
B.An alkaline matrix (pH ~8.5) buffered by high bicarbonate ion concentrations
C.A completely anhydrous interior created by hydrophobic lipid monolayers
D.High concentrations of molecular oxygen that catalyze non-enzymatic spontaneous combustion
Explanation: Lysosomes maintain an internal acidic pH of approximately 4.5 to 5.0 through membrane-bound V-type H+ ATPase pumps that actively pump protons from the cytosol into the lumen. Lysosomal enzymes (acid hydrolases) have optimal catalytic activity at this acidic pH. If a lysosome ruptures, the neutral pH of the cytosol (~7.2) protects the cell because the enzymes are largely inactive at that pH.

About the FY13CE Biology Exam

The Fiji Year 13 Certificate Examination (FY13CE) in Biology, the successor to the Fiji Seventh Form Examination, tests advanced biological principles in a single 3-hour paper worth 100 marks. This practice bank covers all three official strands: Structure and Life Processes, Living Together, and Biodiversity, Change and Sustainability.

Exam sponsor: Examinations and Assessment Unit, Ministry of Education, Fiji. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Single three-hour written paper with 10 minutes reading time, marked out of 100: Strand 1 Structure and Life Processes (28 marks), Strand 2 Living Together (16), Strand 3 Biodiversity, Change and Sustainability (36), and two compulsory essays (20).

Time Limit

3 hours plus 10 minutes reading time

Passing Score

Not published as a fixed percentage; subject marks are reported out of 100 and are scaled by the Ministry of Education from the 2026 academic year.

Exam / Certification Fees

No candidate entry fee is published for school candidates; a recount costs $5 FJD per subject and a remark $25 FJD per subject during the 30-day provisional results period.

Exam sponsor website

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.

50 questions

Structure and Life Processes

Cell biology, enzymes, cellular respiration, photosynthesis, protein synthesis, and physiology.

25 questions

Living Together

Ecosystem energetics, trophic structures, nutrient cycles, and community interactions.

25 questions

Biodiversity, Change and Sustainability

Inheritance, molecular genetics, evolution, and Pacific marine and terrestrial conservation.

Preparing for the FY13CE Biology Exam

What You Need to Know

  • Passing score: Not published as a fixed percentage; subject marks are reported out of 100 and are scaled by the Ministry of Education from the 2026 academic year.
  • Assessment: Single three-hour written paper with 10 minutes reading time, marked out of 100: Strand 1 Structure and Life Processes (28 marks), Strand 2 Living Together (16), Strand 3 Biodiversity, Change and Sustainability (36), and two compulsory essays (20).
  • Time limit: 3 hours plus 10 minutes reading time
  • Exam / certification fees: No candidate entry fee is published for school candidates; a recount costs $5 FJD per subject and a remark $25 FJD per subject during the 30-day provisional results period. 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

FY13CE Biology: Suggested Study Strategy

1Master the stages and biochemical equations of aerobic respiration and photosynthesis.
2Understand DNA replication enzymes including helicase, DNA polymerase, and ligase.
3Practice monohybrid, dihybrid, and sex-linked genetic crosses and pedigree analysis.
4Review Pacific ecological case studies involving coral reefs, mangrove ecosystems, and endemic biodiversity.

Frequently Asked Questions

What topics are examined in FY13CE Biology?

The examination is divided into three strands: Structure and Life Processes (cellular and physiological biology), Living Together (ecology and ecosystems), and Biodiversity, Change and Sustainability (genetics, evolution, and conservation).

Is there a practical or coursework component?

No separate practical or internal assessment mark is published for FY13CE Biology. The subject result comes from the single 3-hour written paper marked out of 100, which includes multiple-choice, short-answer and two compulsory essay questions.

Is this practice bank the same format as the real paper?

No. This bank is an English-language multiple-choice study adaptation built from the official strand weightings. It cannot replace practice at the short-answer and extended essay writing the real paper requires.

What is the passing score for FY13CE Biology?

The Ministry of Education does not publish a fixed percentage pass mark. Marks are reported out of 100 and, from the 2026 academic year, Cabinet has approved the reintroduction of scaling across national examinations.

What language is the exam administered in?

English. All FY13CE science papers are set and answered in English; separate language subjects exist for Hindi, Urdu, Vosa Vakaviti, Rotuman and French.