Life Sciences Grade 11 September 2025 – KZN

KWAZULU-NATAL PROVINCE
EDUCATION
REPUBLIC OF SOUTH AFRICA
NATIONAL SENIOR CERTIFICATE

GRADE 11

LIFE SCIENCES

SEPTEMBER 2025 — QUESTION PAPER AND MARKING GUIDELINES

MARKS: 50TIME: 1 hour

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 answer and write only the letter (A–D) next to the question number in the ANSWER BOOK.

1.1.1 Total number of organisms of a particular species that can be supported by resources in the environment:

A. Ecological niche
B. Census
C. Carrying capacity
D. Population density
Show answer
Answer: C — Carrying capacity.

Carrying capacity is the largest population of a species that an environment can support over time with the resources available, such as food, water, space and shelter.

1.1.2 The part of the brain that stimulates heart muscles when a person is exposed to an environment with a high level of carbon dioxide is the …

A. cerebellum
B. medulla oblongata
C. hypothalamus
D. cerebrum
Show answer
Answer: B — Medulla oblongata.

The medulla oblongata helps control involuntary activities, including breathing and heart rate. It responds to changes such as increased carbon dioxide levels in the blood.

1.1.3 The table below shows rate and depth of breathing in a healthy individual during a three-minute strenuous exercise.

Table 1: Rate and depth of breathing during exercise
Time of exercise (minutes)Rate of breathing (breaths per minute)Depth of breathing (litres per minute)
11020
21530
32040

Which ONE statement best describes the relationship between time of exercise, rate and depth of breathing?

A. Time of exercise increases as rate and depth of breathing decrease.
B. Depth and rate of breathing decrease as time of exercise increases.
C. Depth and rate of breathing increase as time of exercise increases.
D. Time of exercise increases as rate and depth of breathing increase.
Show answer
Answer: C.

As exercise continues from minute 1 to minute 3, the breathing rate rises from 10 to 20 breaths per minute and the depth of breathing rises from 20 to 40 litres per minute, as shown in the table. During strenuous exercise, the body needs more oxygen and must remove more carbon dioxide, so both the rate and depth of breathing increase.

(3 × 2) (6)

1.2 Give the correct biological term for each description. Write only the term next to the question number in the ANSWER BOOK.

1.2.1 Specialised cells in the inner wall of the Bowman's capsule. (1)

Show answer
Answer: Podocytes.

Podocytes are specialised cells associated with the inner layer of Bowman’s capsule. They help form the filtration barrier that allows the kidney to filter blood.

1.2.2 The process of eliminating metabolic waste product. (1)

Show answer
Answer: Excretion.

Excretion is the removal of metabolic waste products made by the body’s cells, such as urea and excess salts.

(2 × 1) (2)

1.3 Indicate whether each description in COLUMN I applies to A ONLY, B ONLY, BOTH A AND B or NONE of the items in COLUMN II. Write the appropriate response next to each question number.

Table 2: Column I and Column II
Question / Column IColumn II
1.3.1 Blood vessel that carries deoxygenated blood from the heart to the lungs
Show answer
Answer: A only.

The pulmonary artery carries deoxygenated blood from the heart to the lungs. The pulmonary vein carries oxygenated blood back to the heart, so only A is correct.

A: Pulmonary artery
B: Pulmonary vein
1.3.2 Sodium level increases in the blood
Show answer
Answer: B only.

When sodium levels in the blood increase, less aldosterone is secreted. Aldosterone normally promotes sodium reabsorption by the kidneys, so reducing its secretion helps limit further sodium retention.

A: More aldosterone secreted
B: Less aldosterone secreted
1.3.3 The air passages
Show answer
Answer: Both A and B.

The nostrils and bronchi are both parts of the respiratory air passages through which air travels.

A: Nostrils
B: Bronchus

(3 × 2) (6)

1.4 The diagram below shows a part of the human urinary system.

Human urinary system diagram with labels A, B, C and D
Figure 1: Part of the human urinary system

1.4.1 Give the LETTER and NAME of the part that:

(a) Carries blood with nitrogenous waste to the kidney for filtration. (2)

Show answer
Answer: B — Renal artery.

The renal artery carries blood containing nitrogenous waste to the kidney, where the blood is filtered.

(b) Carries urine from the kidney to the bladder. (2)

Show answer
Answer: D — Ureter.

The ureter is the tube that transports urine from the kidney to the bladder.

(c) Carries deoxygenated and filtered blood to the heart. (2)

Show answer
Answer: C — Renal vein.

The renal vein carries filtered blood away from the kidney towards the heart.

TOTAL SECTION A: 20

SECTION B

QUESTION 2

2.1 The diagram below shows the human lungs.

Human lungs diagram with labels A to E
Figure 2: Human lungs

2.1.1 Identify part:

(a) B (1)

Show answer
Answer: Bronchus (plural: bronchi).

A bronchus is an air passage that carries air from the trachea into a lung.

(b) E (1)

Show answer
Answer: Alveolus.

An alveolus is a tiny air sac where oxygen and carbon dioxide are exchanged between the air and the blood.

2.1.2 Describe the role of part C and D during exhalation. (4)

Show answer

During exhalation, the diaphragm relaxes and returns to a dome-shaped position. The external intercostal muscles also relax, causing the rib cage to move downwards and inwards. These movements reduce the volume of the thoracic cavity, increasing pressure inside the lungs and pushing air out.

  • The diaphragm relaxes and becomes arched / dome-shaped.
  • The external intercostal muscles relax.
  • The rib cage moves downwards and inwards.
  • The volume of the thoracic cavity decreases, pushing air out of the lungs.

2.1.3 Bronchoalveolar carcinoma is a rare form of lung cancer that originates and damages the cells lining part E. Explain how this disease may result in death if it is not detected earlier. (3)

Show answer

Bronchoalveolar carcinoma damages cells lining the gas-exchange region of the lungs. This can reduce the surface area available for gas exchange or obstruct air passages. As a result, less oxygen enters the blood and reaches tissues, including the heart muscle. Carbon dioxide may also build up because it cannot be exhaled effectively, potentially contributing to heart failure.

  • The surface area for gaseous exchange is reduced / air passages may be blocked by tumours.
  • Less oxygen enters the blood or reaches body tissues and heart muscles.
  • Carbon dioxide may be trapped and not exhaled, potentially resulting in heart failure.

2.2 The bar graph below shows a comparison between youth unemployment in different countries in 2024 and 2025.

Bar graph comparing youth unemployment in 2024 and 2025
Figure 3: Comparison of youth unemployment in different countries in 2024 and 2025

2.2.1 Which country above is classified as a developing country? (1)

Show answer
Answer: South Africa.

South Africa is the developing country identified in the graph.

2.2.2 Explain ONE reason for your answer to QUESTION 2.2.1. (2)

Show answer
Answer:

South Africa has the highest youth unemployment rate shown in the graph. A high rate of youth unemployment is one indicator of the country’s economic challenges.

2.2.3 Calculate the percentage increase in youth unemployment in France between 2024 and 2025. Show ALL working. (3)

Show answer
Answer: 33.33%.

2024 value = 15; 2025 value = 20.

Increase = 20 − 15 = 5.

Percentage increase = (increase ÷ original value) × 100 = (5 ÷ 15) × 100 = 33.33%.

The 2024 value is the starting value, so divide the increase by 15 and multiply by 100.

[15]

QUESTION 3

3.1 Read the extract below.

THE ECOLOGICAL ROLES OF LIONS

Lions are the apex predator of the African savanna since they hunt in packs. They play the critical role of managing large herbivore populations such as buffaloes, impalas, zebras, and even elephants and giraffes. In a balanced ecosystem, the number of herbivores needs to be kept in balance. If there are too many herbivores, the ecosystem will inevitably degrade.

3.1.1 Name the type of relationship between lions and the herbivores. (1)

Show answer
Answer: Predator–prey relationship.

Lions are predators because they hunt and kill herbivores for food; the herbivores are the prey.

3.1.2 Explain why:

(a) lions are successful predators in the savanna. (2)

Show answer
Answer:

Lions hunt in packs. Working together allows them to surround or pursue prey and increases their chances of a successful hunt.

(b) the number of herbivores needs to be kept in balance in the ecosystem. (2)

Show answer
Answer:

Herbivore numbers must remain balanced to prevent overgrazing and land degradation. If too much vegetation is eaten, the ecosystem may support fewer organisms and its carrying capacity can decrease.

3.1.3 Name TWO symbiotic interactions that benefit one species. (2)

Show answer
Answers: Commensalism and parasitism.

In commensalism, one species benefits while the other is not significantly affected. In parasitism, the parasite benefits while the host is harmed.

3.2 Grade 11 learners conducted an investigation to determine the effect of drinking water on urine production.

A healthy athlete was requested not to drink water or eat food for 24 hours before the investigation. The investigation was conducted over a period of five days.

The following procedure was followed:

  • Day 1 and 2: the athlete was given 500 mL of water to drink each day.
  • Day 3: the athlete was given 660 mL of water to drink.
  • Day 4: the athlete was given 870 mL of water to drink.
  • Day 5: the athlete was given 990 mL of water to drink.

On each day of the investigation, the amount of urine produced by the athlete was measured and recorded over a five-hour period after drinking water.

Results were recorded in the table below:

Table 3: Water consumed and urine produced during the investigation
DayAmount of water consumed (mL)Amount of urine produced (mL/day)
1500410
2500410
3660604
4870823
5990902

3.2.1 State ONE:

(a) way in which the reliability of the investigation was ensured. (1)

Show answer
Answer:

The investigation was conducted over five days. Repeating observations across several days helps show whether the pattern is consistent.

(b) reason why the participant was asked not to consume water or food for 24 hours before the investigation. (1)

Show answer
Answer:

The participant was asked not to eat or drink for 24 hours to reduce the amount of water in the blood before the investigation and help improve the validity of the results.

(c) planning step the learners considered for this investigation. (1)

Show answer
Any ONE answer:
  • Ask the participant’s permission.
  • Decide which instrument will be used to collect urine.
  • Decide how the results will be recorded.
  • Decide the time or venue of the investigation.

3.2.2 Name the hormone responsible for water balance in the body. (1)

Show answer
Answer: ADH (antidiuretic hormone).

ADH helps regulate water balance by controlling how much water the kidneys reabsorb.

3.2.3 Explain the role of the hormone mentioned in QUESTION 3.2.2 in the homeostatic response after day 3. (4)

Show answer
Answer:

After the athlete drinks more water, less ADH is secreted into the blood. The walls of the distal convoluted tubule and collecting tubule become less permeable, so less water leaves the tubules by osmosis and returns to the blood. More water therefore remains in the tubules and is excreted as a larger volume of more dilute urine.

  • Less ADH is secreted into the blood.
  • The distal convoluted tubule and collecting tubule become less permeable.
  • Less water is reabsorbed into the blood by osmosis.
  • More water remains in the tubules and is released as a larger volume of dilute urine.

TOTAL SECTION B: 30

GRAND TOTAL: 50

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