2024 Jupeb biology

2024 Jupeb biology

BIO 001 – GENERAL BIOLOGY

A researcher conducted a field research to determine the distribution and abundance of some insects. Three students collected animals (A, B and C) for the same day across 17 areas between 8:00am and 1pm. The table below shows the results obtained.

| S/N | Animals | Site A | Site B | Site C |

|-----|---------|--------|--------|--------|

| 1 | Housefly | 16 | — | 12 |

| 2 | Grassfly | — | 21 | — |

| 3 | Fire Ant | 5 | 8 | 7 |

| 4 | Soldier Ant | 6 | — | 9 |

| 5 | Locusts | 15 | 10 | — |

| 6 | Peeping | — | 8 | — |

| 7 | Cockroach | 4 | 6 | — |

| 8 | Mosquito | — | — | — |

| 9 | Dragonfly | 8 | — | 4 |

| 10 | Bee | 29 | 13 | — |

(a) Determine the mean of the insects from each site. (3 Marks)

(b) Draw a bar chart showing the abundance of insects. (4 Marks)

© Draw a pie chart showing the percentage composition of insects in the entire zone. (20 Marks)

(d) What is likely to be the implication of insect 1 on site A? (1 Mark)

Question 7:

(9) State the position of the nitrogenous base of a nucleic acid.

(10) List TWO functions of lipids in a cell.

7. (a) Outline the following under Bioenergetics:

  1. Photosynthesis

  2. Respiration

  3. Fermentation

  4. Assimilation

  5. Crossing-over

(b) State the THREE cell theories. (3 Marks)

© State any THREE formed function statements of the following biological terminologies:

  1. Starch

  2. Mucus

  3. Nucleic acid

BIO 002 – MICROBIOLOGY

(a) Differentiate between haploid and diploid organisms.

(b) State TWO examples of diseases caused by germs.

© What is the primitive form of bacterial respiration that distances from… (partially obscured)

BIO 002 continued (Image 2):

(iii) Briefly explain the following:

  1. Facultative anaerobes

  2. Attenuation

  3. Nucleoids

  4. Lysoenzymes

  5. Plasmids (5 Marks)

4. (a) In an attempt to understand the difference between Gram positive and Gram negative bacteria, a student comes up with the table below. Fill in the missing options:

| Characteristics | Gram Positive | Gram Negative |

|----------------|---------------|---------------|

| Thickness of cell wall | | |

| Affinity of amino acids | | |

| Outer membrane | | |

| Teichoic Acid | | |

(4 Marks)

(b) Name FOUR enzymes that facilitate DNA replication in bacteria. (2 Marks)

© State the conditions for the use of the following methods for the control of microbial activities:

  1. Steam under pressure

  2. Boiling

  3. Pasteurisation

  4. Dry heat (4 Marks)

BIO 003 – BOTANY

5. (a) Draw and label the plant cell.

(b)

  • (i) List FOUR modifications in stems of plant.

  • (ii) List FOUR modifications in the roots of plant.

©

  • State TWO functions of Auxin.

  • State TWO functions of Cytokinin.

6. (a) State FOUR general characteristics of Spermatophyta.

(b) List THREE divisions of the spermatophytes and give one example each.

© Explain briefly the mechanism of transport of materials through the phloem.

(d) Highlight FOUR ethical implications of genetically modified foods.

BIO 004 – INTRODUCTORY ZOOLOGY

7. (a) Define Gametogenesis.

  • (i) Name the TWO different types of gametogenesis in organisms.

  • (ii) Describe the process of gametogenesis in male mammals.

(b) Explain the processes that occur in a nephron.

8. (a)

  • (i) Explain briefly saltatory conduction in axon.

  • (ii) Give TWO differences between a resting potential and depolarization.

(b) State TWO ways by which white blood cells carry out their activity in the body of humans.

©

  • (i) Draw a well-labelled diagram of spermatozoa.

  • (ii) State the functions of any FOUR of the labelled parts.

BIOLOGY EXAMINATION — COMPLETE SOLUTIONS

BIO 001 – GENERAL BIOLOGY

Table Data (reconstructed for calculations):

| S/N | Animals | Site A | Site B | Site C |

|-----|---------|--------|--------|--------|

| 1 | Housefly | 16 | 0 | 12 |

| 2 | Grassfly | 0 | 21 | 0 |

| 3 | Fire Ant | 5 | 8 | 7 |

| 4 | Soldier Ant | 6 | 0 | 9 |

| 5 | Locusts | 15 | 10 | 0 |

| 6 | Peeping | 0 | 8 | 0 |

| 7 | Cockroach | 4 | 6 | 0 |

| 8 | Mosquito | 0 | 0 | 0 |

| 9 | Dragonfly | 8 | 0 | 4 |

| 10 | Bee | 29 | 13 | 0 |

| Total | | 83 | 66 | 32 |

(a) Mean of insects from each site

Formula: Mean = Total ÷ Number of insects (10)

Site A:

Mean = (16+0+5+6+15+0+4+0+8+29) ÷ 10

= 83 ÷ 10

= 8.3

Site B:

Mean = (0+21+8+0+10+8+6+0+0+13) ÷ 10

= 66 ÷ 10

= 6.6

Site C:

Mean = (12+0+7+9+0+0+0+0+4+0) ÷ 10

= 32 ÷ 10

= 3.2

(b) Bar Chart — Abundance of Insects

(Draw this in your answer booklet as described below)

  • X-axis: Names of insects (Housefly, Grassfly, Fire Ant, Soldier Ant, Locusts, Peeping, Cockroach, Mosquito, Dragonfly, Bee)

  • Y-axis: Number of insects (scale 0–30)

  • Draw three bars side by side for each insect representing Site A, Site B, and Site C respectively

  • Use different shading/colours for each site

  • Include a key/legend

Total values per insect (all sites combined):

| Insect | Total |

|--------|-------|

| Housefly | 28 |

| Grassfly | 21 |

| Fire Ant | 20 |

| Soldier Ant | 15 |

| Locusts | 25 |

| Peeping | 8 |

| Cockroach | 10 |

| Mosquito | 0 |

| Dragonfly | 12 |

| Bee | 42 |

| Grand Total | 181 |

© Pie Chart — Percentage Composition

Formula: Percentage = (Individual total ÷ Grand total) × 100

Angle = (Individual total ÷ Grand total) × 360°

| Insect | Total | Percentage | Angle |

|--------|-------|------------|-------|

| Housefly | 28 | 15.47% | 55.7° |

| Grassfly | 21 | 11.60% | 41.8° |

| Fire Ant | 20 | 11.05% | 39.8° |

| Soldier Ant | 15 | 8.29% | 29.8° |

| Locusts | 25 | 13.81% | 49.7° |

| Peeping | 8 | 4.42% | 15.9° |

| Cockroach | 10 | 5.52% | 19.9° |

| Mosquito | 0 | 0.00% | 0° |

| Dragonfly | 12 | 6.63% | 23.9° |

| Bee | 42 | 23.20% | 83.5° |

| Total | 181 | 100% | 360° |

(Draw a circle, divide it using the angles above, label each segment with the insect name and percentage, and include a key)

(d) Implication of Insect 1 (Housefly) on Site A

The high population of houseflies (16) on Site A indicates poor environmental sanitation in that area. Houseflies are vectors of diseases such as typhoid fever, cholera, dysentery, and salmonellosis. Their abundance suggests the presence of decaying organic matter, exposed refuse, or poor waste management, which could pose serious public health risks to inhabitants of Site A.

Question 7 (General Biology)

(9) Position of the nitrogenous base in a nucleic acid:

The nitrogenous bases are located on the inside of the DNA double helix, projecting inward from the sugar-phosphate backbone. They are attached to the 1’ carbon of the deoxyribose (in DNA) or ribose (in RNA) sugar. In the double helix, bases of opposite strands face each other and are held together by hydrogen bonds (Adenine pairs with Thymine; Guanine pairs with Cytosine).

(10) TWO functions of lipids in a cell:

  1. Energy storage — Lipids serve as a long-term energy reserve; they yield more energy per gram than carbohydrates when oxidised.

  2. Cell membrane formation — Phospholipids form the bilayer structure of the cell membrane, regulating what enters and exits the cell.

(7a) Outline the following under Bioenergetics:

i. Photosynthesis:

Photosynthesis is the process by which green plants use sunlight, water, and carbon dioxide to manufacture glucose and release oxygen.

Equation: 6CO₂ + 6H₂O + light energy → C₆H₁₂O₆ + 6O₂

It occurs in two stages: the light-dependent reaction (in the thylakoid membrane) and the light-independent reaction/Calvin cycle (in the stroma).

ii. Respiration:

Respiration is the process by which living organisms break down glucose to release energy in the form of ATP.

  • Aerobic: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 38 ATP

  • Anaerobic: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ + 2 ATP (in yeast)

It occurs in three stages: Glycolysis (cytoplasm), Krebs cycle (mitochondrial matrix), and Electron Transport Chain (inner mitochondrial membrane).

iii. Fermentation:

Fermentation is an anaerobic process in which microorganisms (yeast, bacteria) partially break down glucose without oxygen to produce energy. Products include ethanol and CO₂ (alcoholic fermentation) or lactic acid (lactic acid fermentation). It yields only 2 ATP per glucose molecule.

iv. Assimilation:

Assimilation is the process by which absorbed digested food molecules are used by cells to build new protoplasm, repair tissues, produce energy, or synthesise complex organic compounds such as proteins, glycogen, and fats. It is the final stage of nutrition.

v. Crossing-over:

Crossing-over is the exchange of genetic material between homologous chromosomes during prophase I of meiosis. Non-sister chromatids of homologous chromosomes break at corresponding points (chiasmata) and rejoin, resulting in recombination of genetic material. This increases genetic variation in offspring.

(7b) THREE Cell Theories:

  1. All living organisms are composed of one or more cells.

  2. The cell is the fundamental structural and functional unit of all living organisms.

  3. All cells arise from pre-existing cells (Omnis cellula e cellula — Virchow, 1855).

(7c) THREE Function Statements of the following:

i. Starch:

  1. Starch serves as the primary storage carbohydrate in plants, found in seeds, tubers, and roots.

  2. It is broken down into glucose during respiration to provide energy for metabolic activities.

  3. Starch is insoluble in water, making it suitable for storage without affecting osmotic pressure.

ii. Mucus:

  1. Mucus lubricates and protects the lining of the respiratory, digestive, and reproductive tracts.

  2. It traps dust, bacteria, and foreign particles, preventing them from entering sensitive tissues.

  3. Mucus moistens surfaces and facilitates the movement of food along the alimentary canal.

iii. Nucleic Acid:

  1. DNA (deoxyribonucleic acid) carries hereditary information and controls all cellular activities.

  2. RNA (ribonucleic acid) plays a central role in protein synthesis — mRNA carries the genetic code, tRNA brings amino acids, and rRNA forms ribosomes.

  3. Nucleic acids store and transmit genetic information from one generation to the next.

BIO 002 – MICROBIOLOGY

(a) Difference between Haploid and Diploid organisms:

| Feature | Haploid | Diploid |

|---------|---------|---------|

| Chromosome number | Single set (n) | Double set (2n) |

| Example | Gametes (sperm, egg), fungi | Most body cells of animals and plants |

| Origin | Produced by meiosis | Produced by mitosis or fertilisation |

| Genetic variation | Less variation | More variation due to paired alleles |

| Occurrence | Gametophyte stage in plants | Sporophyte stage; all somatic cells |

(b) TWO examples of diseases caused by microorganisms:

  1. Tuberculosis (TB) — caused by the bacterium Mycobacterium tuberculosis

  2. Malaria — caused by the protozoan Plasmodium species (transmitted by Anopheles mosquito)

© Primitive form of bacterial respiration:

The most primitive form of bacterial respiration is anaerobic respiration (fermentation). This is considered primitive because it does not require oxygen and was the dominant form of energy production in early life on Earth before the atmosphere contained oxygen. Bacteria use electron acceptors other than oxygen (such as sulphate, nitrate, or carbon dioxide) to oxidise organic compounds and generate ATP. It yields less energy compared to aerobic respiration.

(iii) Brief explanation of the following:

1. Facultative Anaerobes:

Facultative anaerobes are microorganisms that can survive and grow in both the presence and absence of oxygen. They preferentially use oxygen for aerobic respiration when available (producing more ATP), but can switch to anaerobic respiration or fermentation when oxygen is absent. Example: Escherichia coli, Staphylococcus aureus.

2. Attenuation:

Attenuation in microbiology refers to the reduction or weakening of the virulence (disease-causing ability) of a pathogen without destroying its antigenicity. Attenuated microorganisms are still alive but cannot cause disease; they are used in live attenuated vaccines (e.g., BCG vaccine for TB, oral polio vaccine). Attenuation also refers to a regulatory mechanism in bacteria that controls gene expression based on the availability of amino acids.

3. Nucleoids:

The nucleoid is the region in a prokaryotic cell (bacteria) where the genetic material (a single circular DNA molecule) is located. Unlike eukaryotic cells, it is not enclosed by a nuclear membrane. It also contains RNA and proteins associated with DNA. It appears as an irregular, dense region within the cytoplasm.

4. Lysoenzymes (Lysozymes):

Lysozymes are antimicrobial enzymes found in secretions such as tears, saliva, mucus, and breast milk. They destroy bacteria by catalysing the hydrolysis of β-1,4-glycosidic bonds in peptidoglycan — the major component of bacterial cell walls — causing the bacteria to lyse (burst) due to osmotic pressure.

5. Plasmids:

Plasmids are small, circular, double-stranded DNA molecules found in bacteria that exist and replicate independently of the chromosomal DNA. They are not essential for normal bacterial growth but often carry genes that confer advantages such as antibiotic resistance, toxin production, and metabolic capabilities. Plasmids are widely used as vectors in genetic engineering and recombinant DNA technology.

4(a) Gram Positive vs Gram Negative Bacteria:

| Characteristics | Gram Positive | Gram Negative |

|----------------|---------------|---------------|

| Thickness of cell wall | Thick (20–80 nm) peptidoglycan layer | Thin (2–7 nm) peptidoglycan layer |

| Affinity of amino acids | High content of teichoic acids and peptidoglycan | Lower peptidoglycan content |

| Outer membrane | Absent | Present (lipopolysaccharide layer) |

| Teichoic Acid | Present | Absent |

4(b) FOUR enzymes that facilitate DNA replication in bacteria:

  1. DNA Helicase — unwinds and separates the double helix by breaking hydrogen bonds between base pairs

  2. DNA Primase — synthesises short RNA primers to provide a starting point for DNA synthesis

  3. DNA Polymerase III — adds new nucleotides to the growing DNA strand in the 5’→3’ direction

  4. DNA Ligase — joins Okazaki fragments on the lagging strand by forming phosphodiester bonds

4© Conditions for control of microbial activities:

i. Steam under pressure (Autoclaving):

  • Temperature of 121°C at 15 psi pressure for 15–20 minutes

  • Used for sterilising culture media, surgical instruments, and laboratory equipment

  • Most effective method — kills all microorganisms including endospores

ii. Boiling:

  • Temperature of 100°C for at least 10–30 minutes

  • Kills most vegetative bacteria, viruses, and fungi but NOT bacterial endospores

  • Used for disinfecting water and heat-stable materials

iii. Pasteurisation:

  • HTST (High Temperature Short Time): 72°C for 15 seconds

  • LTLT (Low Temperature Long Time): 63°C for 30 minutes

  • Used for milk, fruit juices, and beverages

  • Reduces pathogenic organisms but does not sterilise completely

iv. Dry heat:

  • Temperature of 160–180°C for 1–2 hours (hot air oven)

  • Used for glassware, metal instruments, and powders that cannot be autoclaved

  • Works by oxidation and protein denaturation

  • Less effective than moist heat; requires higher temperatures

BIO 003 – BOTANY

5(a) Plant Cell — Draw and Label

(Draw in your answer booklet. Include the following labelled parts:)

  • Cell wall

  • Cell membrane (plasma membrane)

  • Nucleus (with nucleolus and nuclear membrane)

  • Chloroplast (with grana and stroma)

  • Mitochondria

  • Vacuole (large central vacuole)

  • Endoplasmic reticulum (rough and smooth)

  • Golgi apparatus

  • Ribosomes

  • Cytoplasm

  • Plasmodesmata

5(b)(i) FOUR modifications of stems in plants:

  1. Rhizome — a horizontal underground stem that stores food and produces new shoots e.g. ginger, fern

  2. Tuber — a swollen underground stem that stores starch e.g. Irish potato

  3. Tendril — a slender coiling stem modification for climbing e.g. passion fruit

  4. Corm — a short, swollen vertical underground stem e.g. cocoyam, crocus

5(b)(ii) FOUR modifications of roots in plants:

  1. Tap root — stores food reserves e.g. carrot, radish

  2. Prop roots (aerial roots) — grow from the stem downward for support e.g. maize, mangrove

  3. Pneumatophores — grow upward from waterlogged soil for gaseous exchange e.g. mangrove

  4. Parasitic roots (haustoria) — penetrate the host plant to absorb nutrients e.g. mistletoe

5© TWO functions of Auxin:

  1. Auxin (IAA) promotes cell elongation in shoots by loosening cell walls, causing bending toward light (phototropism) and downward root growth (gravitropism).

  2. Auxin promotes apical dominance — suppressing the growth of lateral buds so the main shoot grows taller.

TWO functions of Cytokinin:

  1. Cytokinins promote cell division (cytokinesis) in plant meristems, stimulating growth and development.

  2. Cytokinins delay senescence (ageing) in leaves by preventing the breakdown of chlorophyll and proteins.

6(a) FOUR general characteristics of Spermatophyta:

  1. They produce seeds as a means of reproduction — the seed contains an embryo and stored food.

  2. They are vascular plants — they possess xylem and phloem for transport of water and nutrients.

  3. They reproduce by means of flowers (in angiosperms) or cones (in gymnosperms).

  4. They undergo double fertilisation (in angiosperms) — one sperm fertilises the egg to form a zygote; another fuses with polar nuclei to form the endosperm.

6(b) THREE divisions of Spermatophytes with one example each:

  1. Cycadophyta — Example: Cycas revoluta (Sago palm)

  2. Coniferophyta (Conifers) — Example: Pinus species (Pine tree)

  3. Angiospermae — Example: Mangifera indica (Mango tree)

6© Mechanism of transport through the phloem:

Transport in the phloem occurs by the Mass Flow Hypothesis (Pressure Flow Theory) proposed by Ernst Munch (1930).

  • Loading: Sugar (sucrose) produced in photosynthetic leaves (source) is actively loaded into sieve tube elements of the phloem, lowering the water potential.

  • Water entry: Water enters the phloem by osmosis from the xylem, creating high turgor pressure at the source end.

  • Flow: This high pressure drives the flow of phloem sap (sucrose + water) along the sieve tubes toward areas of low pressure — the sink (roots, fruits, growing regions).

  • Unloading: At the sink, sucrose is unloaded (actively or passively), water potential rises, and water moves back to the xylem. This maintains the pressure gradient driving continuous flow.

6(d) FOUR ethical implications of genetically modified (GM) foods:

  1. Health concerns — GM foods may introduce allergens or toxins into the food supply, raising questions about long-term safety for human consumption.

  2. Biodiversity loss — Widespread cultivation of GM crops may lead to the displacement of indigenous plant varieties and reduce genetic diversity in ecosystems.

  3. Unfair corporate control — Large biotechnology companies hold patents on GM seeds, potentially exploiting small-scale farmers who cannot afford licence fees, raising issues of equity and food sovereignty.

  4. Environmental contamination — Cross-pollination between GM crops and wild relatives may create herbicide-resistant “superweeds” and disrupt natural ecosystems, raising concerns about ecological balance.

BIO 004 – INTRODUCTORY ZOOLOGY

7(a) Definition of Gametogenesis:

Gametogenesis is the biological process by which diploid or haploid precursor cells undergo cell division and differentiation to form mature haploid gametes (sex cells — sperm and egg). It involves meiosis and maturation processes that occur in the gonads.

7(a)(i) TWO types of Gametogenesis:

  1. Spermatogenesis — the formation of male gametes (spermatozoa) in the testes

  2. Oogenesis — the formation of female gametes (ova/eggs) in the ovaries

7(a)(ii) Process of Gametogenesis in male mammals (Spermatogenesis):

  1. Multiplication phase: Primordial germ cells in the seminiferous tubules of the testes divide mitotically to produce many spermatogonia (diploid, 2n).

  2. Growth phase: Spermatogonia grow and differentiate into primary spermatocytes (diploid, 2n).

  3. Meiosis I: Each primary spermatocyte undergoes the first meiotic division to produce two secondary spermatocytes (haploid, n).

  4. Meiosis II: Each secondary spermatocyte undergoes the second meiotic division to produce two spermatids — a total of four spermatids per primary spermatocyte.

  5. Spermiogenesis: Spermatids undergo structural maturation (loss of cytoplasm, development of flagellum, acrosome formation) to become mature spermatozoa (sperm cells).

  6. The mature sperm are stored in the epididymis until ejaculation.

7(b) Processes that occur in a nephron:

The nephron is the functional unit of the kidney. The following processes occur:

i. Ultrafiltration (Glomerular filtration):

Blood enters the glomerulus under high pressure (due to the wider afferent arteriole vs narrow efferent arteriole). Small molecules — water, glucose, urea, salts, amino acids — are forced through the glomerular capillary walls into the Bowman’s capsule, forming the glomerular filtrate. Large molecules (proteins, blood cells) remain in the blood.

ii. Selective Reabsorption:

As filtrate passes through the proximal convoluted tubule (PCT), loop of Henle, and distal convoluted tubule (DCT), useful substances are reabsorbed back into the blood:

  • All glucose and amino acids are reabsorbed in the PCT by active transport

  • Most water is reabsorbed by osmosis

  • Salts are reabsorbed by active transport

iii. Tubular Secretion:

Additional waste products (e.g., hydrogen ions, potassium ions, drugs, creatinine) are actively secreted from the blood into the tubular filtrate, helping regulate blood pH and composition.

iv. Concentration of urine:

The loop of Henle creates an osmotic gradient in the medulla. The collecting duct, under the influence of ADH (antidiuretic hormone), allows further water reabsorption, concentrating the urine.

v. Excretion:

The concentrated filtrate (urine — containing water, urea, salts, creatinine) passes into the renal pelvis, flows down the ureter, and is stored in the urinary bladder before excretion through the urethra.

8(a)(i) Saltatory conduction in axon:

Saltatory conduction is the mode of impulse transmission in myelinated nerve fibres. The myelin sheath (produced by Schwann cells) acts as an insulator and does not allow ion exchange. Therefore, the action potential “jumps” from one Node of Ranvier (gaps in the myelin sheath) to the next, rather than travelling continuously along the entire axon membrane. This jumping movement is called saltatory conduction (from Latin saltare — to jump/leap).

Advantages:

  • Much faster impulse transmission than in unmyelinated fibres

  • Conserves energy — depolarisation occurs only at nodes, reducing ATP consumption by Na⁺/K⁺ pumps

8(a)(ii) TWO differences between resting potential and depolarisation:

| Feature | Resting Potential | Depolarisation |

|---------|-------------------|----------------|

| Charge | Inside of membrane is negative (−70mV) | Inside becomes positive (+30mV) |

| Ion movement | Na⁺ outside, K⁺ inside (maintained by Na⁺/K⁺ pump) | Na⁺ channels open; Na⁺ rushes into the cell |

| State | No nerve impulse being transmitted | Nerve impulse is being transmitted |

| Membrane permeability | Low permeability to Na⁺ | High permeability to Na⁺ |

8(b) TWO ways white blood cells (leukocytes) carry out their activity:

  1. Phagocytosis — Neutrophils and monocytes engulf and digest foreign particles, bacteria, and dead cells by surrounding and internalising them into phagosomes, where they are destroyed by lysosomal enzymes.

  2. Antibody production — B-lymphocytes (B-cells) produce specific antibodies in response to antigens. These antibodies bind to pathogens, neutralising them or marking them for destruction by other immune cells (opsonisation).

8©(i) Well-labelled diagram of spermatozoa:

(Draw in your answer booklet. Label the following parts:)

  • Head — contains the nucleus (with condensed DNA/genetic material)

  • Acrosome — cap-like structure at the tip of the head; contains hydrolytic enzymes for penetrating the egg

  • Neck — short region connecting head to middle piece; contains centriole

  • Middle piece — contains numerous mitochondria arranged in a spiral around the axoneme for energy (ATP) production

  • Tail (flagellum) — long, whip-like structure for motility; consists of axoneme (9+2 microtubule arrangement)

  • End piece — terminal thin segment of the tail

8©(ii) Functions of any FOUR labelled parts:

  1. Nucleus (Head): Contains the haploid (n) genetic material (23 chromosomes in humans) that will combine with the egg’s nucleus during fertilisation to restore the diploid number.

  2. Acrosome: Contains hydrolytic enzymes (acrosin, hyaluronidase) that are released during the acrosome reaction, digesting the zona pellucida and outer membrane of the egg to allow sperm penetration.

  3. Mitochondria (Middle piece): Carry out aerobic respiration to generate ATP, which provides the energy needed to power the movement of the flagellum for sperm motility.

  4. Tail/Flagellum: Provides the propulsive force for sperm movement by its whip-like beating action, enabling the sperm to swim through the female reproductive tract toward the egg.

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