Life Sciences Grade 10 June 2025 – KZN

Grade 10 Life Sciences — June 2025

ACE Learner Q&A Teaching Guide • Answers and teaching explanations are hidden until opened.

NATIONAL SENIOR CERTIFICATE

GRADE 10

LIFE SCIENCES

JUNE COMMON TEST — 2025

MARKS: 150
TIME: 2½ Hours

The Question Paper content is reproduced in the learner guide, with the teaching answer layer added beneath each question.

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

  1. Answer ALL the questions.
  2. Write ALL the answers in the ANSWER BOOK.
  3. Start the answers to each question at the top of a NEW page.
  4. Number the answers correctly according to the numbering system used in this question paper.
  5. Present your answers according to the instructions of each question.
  6. Do ALL drawings in pencil and label them in blue or black ink.
  7. Draw diagrams, tables or flow charts only when asked to do so.
  8. The diagrams in this question paper are NOT necessarily drawn to scale.
  9. Do NOT use graph paper.
  10. You may use a non-programmable calculator, protractor and a compass.
  11. Write neatly and legibly.

SECTION A

QUESTION 1

1.1

Various options are provided as possible answers to the following questions. Choose the correct answer and write only the letter (A – D) next to the question number (1.1.1 to 1.1.10) in your ANSWER BOOK, for example 1.1.1 D.

1.1.1 (2)

The blood vessel that carries blood to the kidneys is the …

  1. renal vein.
  2. hepatic vein.
  3. hepatic artery.
  4. renal artery.
SHOW ANSWER

Answer

D — renal artery.

The renal artery carries blood to the kidneys.

The renal artery carries blood from the heart towards the kidneys for filtration and other kidney functions.

1.1.2 (2)

The first scientist to see living cells using microscope is …

  1. Robert Hooke.
  2. Gregor Mendel.
  3. Antonie Van Leeuwenhoek.
  4. Charles Darwin.
SHOW ANSWER

Answer

C — Antonie Van Leeuwenhoek.

Antonie Van Leeuwenhoek is associated with the early observation of living microscopic cells using microscopes.

1.1.3 (2)

The chemical used to test for glucose is …

  1. ether.
  2. Benedict's solution.
  3. iodine solution.
  4. Millon's reagent.
SHOW ANSWER

Answer

B — Benedict's solution.

Benedict's solution is used to test for glucose and other reducing sugars.

1.1.4 (2)

The following elements are found in the organic compounds:

  1. Hydrogen
  2. Nitrogen
  3. Oxygen
  4. Carbon

Which ONE of the following is the CORRECT combination of elements found in carbohydrates?

  1. (i) only
  2. (ii) and (iii) only
  3. (i), (ii), (iii) only
  4. (i), (iii), (iv) only
SHOW ANSWER

Answer

D — (i), (iii), (iv) only.

Carbohydrates contain carbon, hydrogen and oxygen, represented by statements (iv), (i) and (iii).

1.1.5 (2)

Which ONE of the following parts form the axial skeleton?

  1. Skull, pectoral girdle, sternum, and lower limbs
  2. Sternum, arms, ribs and legs
  3. Ribs, upper limbs, lower limbs and sternum
  4. Skull, vertebral column, ribs and sternum
SHOW ANSWER

Answer

D — Skull, vertebral column, ribs and sternum.

The axial skeleton forms the central framework of the body and includes the skull, vertebral column, ribs and sternum.

1.1.6 (2)

The type/s of skeleton/s found in snail is/ are …

  1. endoskeleton.
  2. exoskeleton.
  3. hydrostatic skeleton.
  4. exoskeleton and endoskeleton.
SHOW ANSWER

Answer

B/C — exoskeleton and endoskeleton.

The answer shown preserves the source's stated alternatives for the type or types of skeleton in a snail.

1.1.7 (2)

The diagram below shows a phase in mitosis.

Phase in mitosis diagram
Original question visual for 1.1.7.
1234
ANuclear membraneSpindle fibresChromosomeCentrosome
BCell membraneChromosomeCentrosomeSpindle fibres
CNuclear membraneCentrosomeChromosomeSpindle fibres
DNuclear membraneSpindle fibreCentrosomeChromosome
SHOW ANSWER

Answer

No answer is provided for this item in the supplied source. No answer.

No answer is provided for this item in the supplied source.

1.1.8 (2)

The instrument used to measure the rate of transpiration is called …

  1. thermometer.
  2. potometer.
  3. photometer.
  4. hectometre.
SHOW ANSWER

Answer

B — potometer.

A potometer is used to estimate the rate of water uptake associated with transpiration.

1.1.9 (2)

Which of the following consist of sclerenchyma cap, phloem, cambium, and xylem?

  1. Pith
  2. Epidermis
  3. Vascular bundle
  4. Cortex
SHOW ANSWER

Answer

C — Vascular bundle.

A vascular bundle contains conducting tissues including xylem and phloem, together with associated tissues.

1.1.10 (2)

Which of the following parts form part of phloem tissue?

  1. Tracheids and sieve tubes.
  2. Xylem vessels and tracheids.
  3. Sieve tubes and companion cells.
  4. Companion cells and tracheids.
SHOW ANSWER

Answer

C — Sieve tubes and companion cells.

Sieve tubes and companion cells are the characteristic conducting components identified as part of phloem tissue.

(10 × 2) (20)

1.2

Give the correct biological term for each of the following descriptions. Write only the term next to the question number (1.2.1 to 1.2.9) in your ANSWER BOOK.

1.2.1(1)
A large molecule formed when more than two monosaccharides are joined together.
SHOW ANSWER

Answer

Polysaccharides

Polysaccharides are large carbohydrate molecules formed when many monosaccharide units are joined together.

1.2.2(1)
Specialised cell in ciliated columnar epithelial tissue which produce mucus.
SHOW ANSWER

Answer

Goblet cells

Goblet cells are specialised cells that produce mucus, helping to protect and lubricate the epithelial surface.

1.2.3(1)
Monomers of protein.
SHOW ANSWER

Answer

Amino acids

Proteins are built from smaller units called amino acids, which are joined together to form protein molecules.

1.2.4(1)
The plastid that gives red, yellow and orange pigment to plants.
SHOW ANSWER

Answer

Chromoplast

Chromoplasts contain pigments that give plant parts red, yellow and orange colours.

1.2.5(1)
Part of human skull that encloses and protects the brain.
SHOW ANSWER

Answer

Cranium

The cranium forms the part of the skull that surrounds and protects the brain.

1.2.6(1)
The upper chambers of the heart.
SHOW ANSWER

Answer

Atrium / atria

The atria are the upper chambers of the heart and receive blood returning to the heart.

1.2.7(1)
Connective tissue containing a dense network of collagen fibres.
SHOW ANSWER

Answer

Fibrous / connective tissue

The dense collagen fibres give fibrous connective tissue strength and support.

1.2.8(1)
Conducting tissue that transport manufactured food from the leaves to all parts of the plant.
SHOW ANSWER

Answer

Phloem

Phloem transports manufactured food from the leaves to other parts of the plant.

1.2.9(1)
Plant organ that has vascular bundles arranged in a ring.
SHOW ANSWER

Answer

Stem

In the stem, the vascular bundles are arranged in a ring, which is a characteristic used to identify this plant organ.

(9 × 1) (9)

1.3 (4 × 2) (8)

Indicate whether each of the statements in COLUMN I applies to A ONLY, B ONLY, BOTH A AND B or NONE of the items in COLUMN II. Write A only, B only, both A and B, or none, next to the question number (1.3.1 to 1.3.4) in the ANSWER BOOK.

Column IColumn II
1.3.1 Tissue that joins bone to boneA: Ligaments
B: Tendons
1.3.2 Provides mechanical support to the plantA: Sclerenchyma
B: Collenchyma
1.3.3 Contains digestive enzymesA: Lysosomes
B: Golgi body
1.3.4 Contains bone cells in hard matrixA: Cartilage
B: Ligaments
1.3.1
Tissue that joins bone to bone — A: Ligaments; B: Tendons.
SHOW ANSWER

Answer

A only

Ligaments join bone to bone, whereas tendons connect muscle to bone.

1.3.2
Provides mechanical support to the plant — A: Sclerenchyma; B: Collenchyma.
SHOW ANSWER

Answer

Both A and B

Both sclerenchyma and collenchyma contribute mechanical support to plant tissues.

1.3.3
Contains digestive enzymes — A: Lysosomes; B: Golgi body.
SHOW ANSWER

Answer

A only

Lysosomes contain digestive enzymes that break down materials inside cells.

1.3.4
Contains bone cells in hard matrix — A: Cartilage; B: Ligaments.
SHOW ANSWER

Answer

None

Neither cartilage nor ligaments contains bone cells embedded in a hard bone matrix.

(4 × 2) (8)

1.4 (7)

The diagram below shows part of a human skeleton.

Part of human skeleton
1.4.1(a)(1)
Identify part B.
SHOW ANSWER

Answer

B — Scapula

The scapula is the shoulder blade and forms part of the pectoral region of the appendicular skeleton.

1.4.1(b)(1)
Identify part C.
SHOW ANSWER

Answer

C — Humerus

The humerus is the long bone of the upper arm and connects the shoulder region to the elbow.

1.4.2(a)(2)
Identify the LETTER and the NAME of the part that provide base for finger movement and flexibility.
SHOW ANSWER

Answer

A — metacarpals

The metacarpals form the framework of the hand between the wrist bones and the finger bones.

1.4.2(b)(1)
Identify the part of the skeleton which the above diagram represents.
SHOW ANSWER

Answer

Appendicular skeleton

The appendicular skeleton includes the limb bones and the girdles that attach the limbs to the axial skeleton.

1.4.2(c)(2)
Identify any TWO joints found in the above diagram.
SHOW ANSWER

Answer

No answer is provided for this item.

No answer is provided for this item in the supplied source.

1.5 (6)

The diagrams below show cross sections through three types of blood vessels.

Three types of blood vessels
1.5.1(a)(1)
Identify blood vessel W.
SHOW ANSWER

Answer

W — Vein

A vein carries blood towards the heart and has structural features suited to return blood at relatively low pressure.

1.5.1(b)(1)
Identify blood vessel X.
SHOW ANSWER

Answer

X — Capillary

Capillaries are tiny vessels with very thin walls that allow exchange between blood and surrounding tissues.

1.5.2(a)(2)
Give the LETTER and the NAME of the blood vessel that has thick muscles and elastic.
SHOW ANSWER

Answer

V — Artery

An artery has thick muscular and elastic walls because it carries blood away from the heart under relatively high pressure.

1.5.2(b)(2)
Give the LETTER and the NAME of the blood vessel that carries blood to the heart.
SHOW ANSWER

Answer

W — Vein

Veins carry blood towards the heart.

SECTION B

QUESTION 2

2.1 (11)

The diagrams below show animal and plant tissues.

Animal and plant tissues
2.1.1(a)(1)
Identify tissue Y.
SHOW ANSWER

Answer

Y — parenchyma

Parenchyma is a plant tissue whose relatively unspecialised cells can perform storage and other basic tissue functions.

2.1.1(b)(1)
Identify tissue Z.
SHOW ANSWER

Answer

Z — columnar epithelium

Columnar epithelium consists of elongated cells arranged in a layer and is found in locations such as the stomach and intestines.

2.1.2(a)(2)
Describe the structure and location of tissue W.
SHOW ANSWER

Answer

Thin / flat / single layer of cells lining blood vessels / lungs.

A thin single layer gives a short diffusion distance, which suits tissue lining areas where substances need to move across the surface.

2.1.2(b)(2)
Describe the structure and location of tissue Z.
SHOW ANSWER

Answer

Long / elongated / single layer of cells lining the stomach / intestines / parts of respiratory tracts.

The elongated cells form a single epithelial layer in locations where this tissue carries out its specialised lining function.

2.1.3(1)
State ONE function of tissue Y.
SHOW ANSWER

Answer

  • Tissue Y acts as a packing tissue.
  • It stores food, water and other substances.
  • It allows movement of water and gases through the intercellular spaces.

Parenchyma can fill spaces within plant organs, store substances and allow movement through the spaces between its cells.

2.1.4(4)
Explain TWO ways in which tissue X is structurally suited to perform its function.
SHOW ANSWER

Answer

  • The cells are elongated and the cross walls disappear to form continuous tubes for the transport of water.
  • Xylem tissue does not contain cytoplasm / contains no living tissue so that the water is not blocked by the cytoplasm.
  • The cells are strengthened with lignin, protecting them from collapsing under strong suction pressure.

These structural features allow xylem to form strong, continuous pathways for water movement while reducing obstruction and preventing the vessels from collapsing.

2.2 (12)

Grade 10 learners conducted an investigation to illustrate the movement of water across a visking tubing. (Visking tubing allows water to move in and out like the cell membrane of the cell).

The learners:

  • Filled the visking tubing with strong sucrose solution.
  • Placed the filled visking tubing in a beaker with tap water.
  • Recorded the level of water in the beaker at the beginning of investigation.
  • The set up was left in the laboratory for 2 hours.
  • After 2 hours there was a change in the level of water in the beaker and the solution in the visking tubing.
Strong sucrose solution before and after

(a) ONE reason for recording the water levels in the beaker at the beginning of the investigation. (2)

(b) TWO changes that are shown in the level of the strong sucrose solution after 2 hours. (4)

(c) Why the sugar molecules did not move from the visking tubing to the beaker. (2)

2.2.1(1)
Identify the process illustrated in the above investigation.
SHOW ANSWER

Answer

Osmosis.

Osmosis is the movement of water across a selectively permeable membrane, as demonstrated by the change in the visking tubing setup.

2.2.2(3)
The process shown above is a passive transport. Tabulate ONE difference between passive transport and active transport.
SHOW ANSWER

Answer

Passive transportActive transport
Does not require energy / carrier moleculeRequires energy / carrier molecule
Movement along concentration gradientMovement against concentration gradient

Passive transport moves substances without the stated energy requirement, while active transport requires energy and can move substances against the concentration gradient.

2.2.3(a)(2)
Explain ONE reason for recording the water levels in the beaker at the beginning of the investigation.
SHOW ANSWER

Answer

To serve as a baseline / reference point where the results will be compared after two hours, so that the change in the water level can be determined.

A starting measurement provides a reference against which the later water level can be compared to determine the change.

2.2.3(b)(4)
Explain TWO changes that are shown in the level of the strong sucrose solution after 2 hours.
SHOW ANSWER

Answer

There is an increase in the amount of solution in the visking tubing. Since the water has moved from the beaker into the visking tubing through the selectively permeable membrane, the level of solution in the visking tubing increases.

Water moves into the visking tubing through the selectively permeable membrane, increasing the amount and level of solution inside it.

2.2.3(c)(2)
Explain why the sugar molecules did not move from the visking tubing to the beaker.
SHOW ANSWER

Answer

The sugar molecules are too large to pass through the selectively permeable membrane of the visking tubing.

The selectively permeable membrane allows water to pass but prevents the larger sugar molecules from passing through.

2.3 (13)

The diagram below shows a type of a cell.

Type of a cell
2.3.1(a)(1)
Identify the type of cell in the diagram above.
SHOW ANSWER

Answer

Plant cell

The diagram represents a plant cell, which has characteristic structures such as a cell wall, chloroplasts and a large vacuole.

2.3.1(b)(1)
Identify part B.
SHOW ANSWER

Answer

B — Vacuole

The vacuole is a large fluid-filled compartment in a typical plant cell.

2.3.1(c)(1)
Identify part C.
SHOW ANSWER

Answer

C — Cell wall

The cell wall forms a rigid outer boundary that supports and protects the plant cell.

2.3.2(2)
Give TWO visible reasons for your answer in 2.3.1(a).
SHOW ANSWER

Answer

  • Cell wall
  • Chloroplasts
  • Regular shape
  • Large vacuole

These visible features distinguish a typical plant cell from an animal cell. The complete set of valid points is shown so the learner can study the available evidence.

2.3.3(2)
State TWO functions for part B.
SHOW ANSWER

Answer

  • Provides support to the plant cell.
  • Serves as a storage organ / stores water / salts.

The vacuole stores substances and helps maintain the internal pressure that supports the plant cell.

2.3.4(a)(2)
Explain how part D ensures the cell functions properly.
SHOW ANSWER

Answer

During cellular respiration part D / mitochondrion produces / stores energy, which will be used by the cell for its functions / metabolic processes.

The mitochondrion is involved in cellular respiration, providing energy for the cell's functions and metabolic processes.

2.3.4(b)(4)
Explain what will happen to the cell if part A becomes impermeable.
SHOW ANSWER

Answer

There will be no movement of substances in and out of the cell; useful substances will become a shortage inside the cell; waste material will also accumulate inside the cell, resulting in the death of the cell / cell becoming dysfunctional.

If the membrane becomes impermeable, substances cannot be exchanged normally. Useful substances become limited and wastes accumulate, so normal cell function is eventually disrupted.

2.4 (14)

Grade 10 learners conducted an investigation to determine the effect of temperature on the action of pepsin (protein-digestive enzyme).

The procedure was as follow:

  • 20 ml of stomach fluid and 10 grams of protein were placed in each of five test tubes.
  • The tubes were then kept at different temperatures.
  • After 24 hours, the contents of each tube were tested to determine the amount of protein that had been digested.
Temperature (°C)Amount of protein (g)
50.5
101.0
204.0
379.5
850.0
2.4.1(1)
Identify the dependent variable.
SHOW ANSWER

Answer

Amount of protein digested.

The dependent variable is the result being measured in response to the temperature conditions.

2.4.2(1)
Give the optimum temperature of the pepsin enzyme.
SHOW ANSWER

Answer

37°C.

The greatest amount of protein was digested at 37°C, so this temperature is the optimum shown by the investigation.

2.4.3(2)
State TWO ways which can be done to increase the reliability of this investigation.
SHOW ANSWER

Answer

  • Increase the sample size / amount of stomach fluid and proteins.
  • Repeat investigation / several times.

Increasing the amount of material and repeating the investigation provide more observations and help make the result more reliable.

2.4.4(3)
Calculate the percentage increase between 20°C and 37°C.
SHOW ANSWER

Answer

Percentage increase = ((9.5 − 4.0) ÷ 4.0) × 100 = 137.5%.

The percentage increase compares the increase from the 20°C value with the original 20°C value, then expresses that change as a percentage.

2.4.5(a)(2)
State the outcome of the investigation with starch instead of protein.
SHOW ANSWER

Answer

The starch would not be digested.

The stated investigation concerns pepsin, a protein-digestive enzyme, so the outcome given for starch is that it would not be digested.

2.4.5(b)(2)
Give ONE reason for your answer in QUESTION 2.4.6.
SHOW ANSWER

Answer

No answer is provided for this item.

No answer is provided for this item in the supplied source.

2.4.6(3)
Explain the reason for the absence of protein at 85°C.
SHOW ANSWER

Answer

No answer is provided for this item.

No answer is provided for this item in the supplied source.

QUESTION 3

3.1 (14)

The diagram below shows a root hair.

Root hair diagram
3.1.1(1)
Identify part A.
SHOW ANSWER

Answer

A — Vacuole.

The vacuole is the labelled internal space in the root hair cell and contains cell sap.

3.1.2(1)
Give ONE function of part B.
SHOW ANSWER

Answer

B — Controls all the activities of the cell / contains DNA which controls hereditary characteristics.

The nucleus contains genetic material and controls the activities of the cell.

3.1.3(1)
State the region of the root where the root hairs is found.
SHOW ANSWER

Answer

Root hair region.

Root hairs occur in the specialised root hair region, where they provide the surface used for absorption.

3.1.4(3)
Name THREE forces that are responsible for the upward movement of water from the roots to the leaves.
SHOW ANSWER

Answer

  • Root pressure
  • Transpiration pull
  • Capillarity

These three forces contribute to the upward movement of water through the plant from the roots towards the leaves.

3.1.5(a)(4)
Explain ONE difference related to permeability of part C and D.
SHOW ANSWER

Answer

Part C / cell wall of a root hair is fully permeable and allows both salt and water to pass through, whereas part D / cell membrane of a root hair is semi-permeable and does not allow molecules of salt to pass through.

The cell wall allows substances to pass through freely, while the selectively permeable cell membrane controls which substances can cross.

3.1.5(b)(4)
Explain TWO ways in which the root hair cell is structurally suitable to perform its function.
SHOW ANSWER

Answer

  • It has finger-like projections, which increase surface area for absorption of water.
  • It has a thin cell wall, allowing it to absorb water from the soil.
  • It has low water potential in the cell sap, increasing the water absorption potential.

The long projection increases surface area, the thin wall shortens the pathway for water, and the low water potential supports water absorption from the soil.

3.2 (11)

The diagram below shows the human heart.

Human heart diagram
3.2.1(1)
Give only the LETTER of the part that transport deoxygenated blood from the lower parts of the body.
SHOW ANSWER

Answer

D.

Part D is the vessel carrying deoxygenated blood from the lower parts of the body back towards the heart.

3.2.2(5)
Describe the path of blood flow from the lungs until it reaches part A.
SHOW ANSWER

Answer

Oxygenated blood returns from the lungs → enters the heart through the pulmonary veins → passes into the left atrium → passes the bicuspid/mitral valve → enters the left ventricle → which pumps the blood to part A / aorta.

The blood returns oxygenated from the lungs, enters the left side of the heart, passes through the mitral valve into the left ventricle, and is then pumped into the aorta.

3.2.3(2)
Describe ONE structural difference between part C and E.
SHOW ANSWER

Answer

Part C / left ventricle is thicker and more muscular / larger, while part E / right ventricle is thinner and less muscular.

The left ventricle has a thicker muscular wall because it must pump blood out to the body, whereas the right ventricle pumps blood to the lungs.

3.2.4(3)
Explain the consequences to the heart if part B becomes blocked.
SHOW ANSWER

Answer

The heart is starved of oxygen; since the blood with oxygen cannot enter, this leads to serious heart damage such as heart attack / death of heart muscle.

A blockage reduces the oxygen supply to heart muscle. Prolonged oxygen shortage can damage the muscle and may result in a heart attack or death of heart muscle.

3.3 (17)

Grade 10 learners were given the set up to use as shown below.

Potometer experimental setup

Learners had to:

  • Give the leafy shoot some time to adapt to a new environmental condition before taking readings.
  • Introduce an air bubble into the capillary tubing and time how long it takes to move from point A to B.
  • To repeat the procedure several times.
  • Use the reservoir with the tap to move the air bubble back to point A before it was timed again.
  • The table below shows the average results learners obtained after repeating the procedure three times.
Environmental conditionsAverage time taken for air bubble to move 5 cm (seconds)
Laboratory155
Bright light125
Wind / fan135
Plastic bag160
3.3.1(2)
State the aim of the above experiment.
SHOW ANSWER

Answer

To measure the rate of transpiration under different environmental conditions.

The investigation compares transpiration under different environmental conditions by measuring how quickly the air bubble moves.

3.3.2(1)
Under what condition was the rate of transpiration the lowest.
SHOW ANSWER

Answer

Plastic bag.

The plastic-bag condition had the longest average time for the air bubble to move 5 cm, corresponding to the lowest transpiration rate in the investigation.

3.3.3(a)(2)
Give ONE reason for your answer in QUESTION 3.3.3.
SHOW ANSWER

Answer

No answer is provided for this item.

No answer is provided for this item in the supplied source.

3.3.3(b)(1)
Give ONE factor that was kept constant during the investigation.
SHOW ANSWER

Answer

Same leafy twig was used.

Keeping the same leafy twig constant helps ensure that the environmental condition is the factor being compared.

3.3.4(a)(2)
Explain ONE precautionary measure when cutting the leaf under water.
SHOW ANSWER

Answer

Leafy twig shoot should be cut at oblique angle to increase surface area for water absorption.

Cutting the shoot obliquely increases the cut surface area available for water absorption.

3.3.4(b)(3)
Explain what would happen when the underside of each leaf was covered with Vaseline.
SHOW ANSWER

Answer

The stomata would be clogged / blocked, causing the leaf to retain water and stopping transpiration.

Blocking the stomata reduces the route through which water vapour leaves the leaf, so transpiration is reduced or stopped.

3.3.5(6)
Draw a bar graph to represent the information on the table.
SHOW ANSWER

Answer

Model answer graph:

Model bar graph for environmental conditions and transpiration

Graph requirements: use a bar graph; include a caption containing both variables; label the X-axis and Y-axis correctly with units; use equal-width bars with equal spaces; and plot all four values correctly.

The model graph shows the four environmental conditions and their average times, with the axes, units and bars presented so the data can be compared.

3.4 (8)

Read the extract below.

Leukemia and Bone Marrow Transplant

Leukemia is type of cancer which affects blood cells. This disease (as well as many others) is caused by abnormal stem cells in the bone marrow producing defective or immature blood. These abnormal stem cells are usually treated by chemotherapy.

Unfortunately, the large doses of chemotherapy required also destroy healthy bone marrow. A bone marrow transplant is necessary to replace the diseased and damaged bone marrow.

3.4.1(2)
Define the term cancer.
SHOW ANSWER

Answer

Cancer is a disease caused by abnormal and uncontrolled cell division in the body.

Cancer involves abnormal, uncontrolled cell division, which can disrupt normal body tissues and functions.

3.4.2(a)(1)
From the extract, state ONE effect of large doses of chemotherapy.
SHOW ANSWER

Answer

Destroys healthy bone marrow.

Large doses of chemotherapy can destroy healthy bone marrow as well as the abnormal cells being treated.

3.4.2(b)(1)
From the extract, state why bone marrow is necessary.
SHOW ANSWER

Answer

Bone marrow replaces the diseased and damaged bone marrow; it produces the blood cells and platelets; it contains stem cells; and it stores fat.

Bone marrow is important because it can replace diseased or damaged marrow and is associated with blood-cell and platelet production, stem cells and fat storage.

3.4.2(c)(2)
From the extract, state how leukemia is caused.
SHOW ANSWER

Answer

It is caused by abnormal stem cells in the bone marrow producing defective / immature blood cells.

The extract links leukemia to abnormal stem cells in bone marrow that produce defective or immature blood cells.

3.4.3(2)
Give TWO other ways of treating cancer other than the one mentioned in the extract.
SHOW ANSWER

Answer

  • Surgery
  • Radiotherapy

Surgery can remove affected tissue, while radiotherapy uses radiation as another treatment approach.

Learner guide complete. Open each answer individually or use SHOW ALL ANSWERS.

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