WEST AFRICAN EXAMINATIONS COUNCIL (WAEC) GCE STYLE

Paper 3: Alternative to Practical Work

Questions and Answers with Marking Guide

Independently Produced Practice Paper for Revision Purposes

This is a 100% original mock examination prepared strictly with reference to the general structure and scope of the WAEC Biology Alternative to Practical paper, for practice purposes only. It is not a WAEC past question paper, it does not reproduce any official WAEC document, and it does not claim to predict or represent the actual WAEC examination in any way.

General Instructions

1. This is an Alternative to Practical Work paper, designed for candidates who are unable to sit for the laboratory- and field-based Biology practical examination. Since real specimens and apparatus cannot be issued in this format, detailed written descriptions and specimen results have been provided in place of actual specimens, diagrams and experimental set-ups. Study each question carefully before answering.

2. Answer ALL five questions.

3. Time allowed: 2 hours.

4. Total marks obtainable: 100 marks, as shown against each question.

5. Write clearly, number your answers correctly, and show all necessary working where calculations are required.

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QUESTIONS (100 MARKS)

Answer ALL questions.

Question 1: Specimen Identification and Classification (20 marks)

Study the descriptions of Specimens A to E below, then answer the questions that follow.

Specimen A: A green, broad, flat plant part with a network of veins radiating from a central midrib, attached to a stem by a short stalk.

Specimen B: A small, dry, hard structure with a tough outer coat. When the coat is removed, a plumule, a radicle, and two large, fleshy cotyledons are visible inside.

Specimen C: A segmented invertebrate with three pairs of jointed legs, one pair of antennae, one or two pairs of wings, and a body clearly divided into a head, a thorax and an abdomen.

Specimen D: A soft-bodied invertebrate with a coiled, hard, calcareous shell, a broad muscular foot on which it glides, and a pair of retractable tentacles bearing eyes.

Specimen E: A swollen underground stem bearing several small ‘eyes’ (buds) on its surface, covered by a thin outer skin, and serving as a food storage organ for the plant.

(a) Identify Specimens A to E. (5 marks)

(b) State the group (for example, plant organ, phylum or class) to which each specimen belongs. (5 marks)

(d) Name the type of underground food storage structure represented by Specimen E, and state one of its functions to the plant. (3 marks)

(e) State the class (Monocotyledonae or Dicotyledonae) to which the plant that produced Specimen B belongs, giving one reason for your answer. (3 marks)

Question 2: Food Tests (Biochemistry) (20 marks)

A candidate carried out food tests on four food samples, labelled P, Q, R and S. The results obtained are shown in the table below.

Food Sample	Test with Iodine Solution	Test with Biuret Reagent	Test with Benedict’s Solution (heated)
P	Blue-black colouration	No colour change (remains blue)	No colour change (remains blue)
Q	No colour change (remains brown/yellow)	Purple/violet colouration	No colour change (remains blue)
R	No colour change	No colour change	Brick-red precipitate formed
S	Blue-black colouration	Purple/violet colouration	Brick-red precipitate formed

(a) State the food nutrient tested for by: (i) iodine solution; (ii) Biuret reagent; (iii) Benedict’s solution. (3 marks)

(b) Identify the food nutrient(s) present in each of the food samples P, Q, R and S, based on the results given. (8 marks)

(d) State the colour change that indicates a positive test for a reducing sugar in the Benedict’s test. (2 marks)

(e) Explain why food sample S gave a positive result in three different tests. (3 marks)

(f) State one precaution that should be observed when carrying out the Benedict’s test. (2 marks)

Question 3: Osmosis Experiment (20 marks)

Six equal-sized cylinders of fresh potato tuber, each weighing 5.0 g at the start, were placed separately into six different concentrations of sucrose solution for one hour. Each cylinder was then removed, blotted dry and reweighed. The results are shown in the table below.

Sucrose Concentration (mol/dm3)	Initial Mass (g)	Final Mass (g)
0.0 (distilled water)	5.0	5.8
0.2	5.0	5.4
0.4	5.0	5.0
0.6	5.0	4.6
0.8	5.0	4.2
1.0	5.0	3.8

(a) Copy and complete the table by calculating the change in mass and the percentage change in mass for each concentration. (6 marks)

(b) Define osmosis. (2 marks)

(c) State, with a reason, the concentration of sucrose solution that is isotonic with (has no net osmotic effect on) the potato tissue. (3 marks)

(d) Explain why the potato cylinder gained mass in distilled water. (3 marks)

(e) Explain why the potato cylinder lost mass in the 1.0 mol/dm3 sucrose solution. (3 marks)

(f) State one practical application of osmosis in everyday life. (3 marks)

Question 4: Ecological Study (20 marks)

A student used a 1 m by 1 m quadrat to sample plant species in two different habitats: Habitat I, an open grassland, and Habitat II, a shaded area near a stream. The quadrat was thrown 10 times in each habitat, and the number of times each plant species was present within the quadrat was recorded, as shown in the table below.

Species	Number of Quadrats Present in Habitat I (out of 10)	Number of Quadrats Present in Habitat II (out of 10)
Grass	9	3
Broad-leaved weed	4	8
Fern	0	7
Moss	1	6

(a) Calculate the percentage frequency of each species in Habitat I and in Habitat II. (8 marks)

(b) Identify the dominant species in Habitat I and the dominant species in Habitat II. (2 marks)

(d) Suggest two possible reasons for the difference in the distribution of the fern between the two habitats. (4 marks)

Question 5: Structure of a Mammalian Tooth (20 marks)

The description below represents a longitudinal section through a mammalian incisor tooth.

Structure W is a bone-like tissue covering the root of the tooth, anchoring it to the jawbone by means of fibres.

Structure X is a hard, white substance forming the outermost covering over the crown of the tooth.

Structure Y lies beneath Structure X and forms the bulk of the body of the tooth.

Structure Z is a soft tissue occupying the central cavity of the tooth, containing blood vessels and nerve endings.

(a) Identify Structures W, X, Y and Z. (4 marks)

(b) State one function each of Structures W, X, Y and Z. (8 marks)

(d) State the dental formula of an adult human, showing the teeth present on one side of the upper and lower jaw. (4 marks)

DETAILED MARKING GUIDE

Award marks as guided below, giving credit for any other correct, relevant and biologically sound identification, calculation, or explanation not listed.

Question 1

(a) Identification (1 mark each): A, a leaf; B, a (dicotyledonous) seed, such as a bean seed; C, an insect, such as a grasshopper or locust; D, a land snail; E, a potato tuber.

(b) Grouping (1 mark each): A is a vegetative/photosynthetic organ of a plant (a leaf); B is the reproductive structure (seed) of a flowering plant; C belongs to Phylum Arthropoda, Class Insecta; D belongs to Phylum Mollusca, Class Gastropoda; E is a modified underground stem (a plant storage organ).

(c) Adaptive features of an insect for terrestrial life (any two at 2 marks each): A hard exoskeleton (cuticle) that reduces water loss from the body; jointed legs that allow efficient movement on land; wings that allow flight for escape from predators and dispersal; spiracles and tracheae that allow gaseous exchange in air; a waterproof/waxy covering on the cuticle that reduces desiccation.

(d) Specimen E is a tuber (a swollen underground stem used for food storage) (1.5 marks). Function: it stores food (mainly starch) for the plant, which can be used for growth during the next growing season, and the buds (eyes) on it can also give rise to new plants (vegetative propagation) (1.5 marks).

(e) The seed (Specimen B) belongs to the class Dicotyledonae (1.5 marks). Reason: it possesses two cotyledons, which is the defining feature of dicotyledonous seeds, as opposed to monocotyledonous seeds which have only one cotyledon (1.5 marks).

Question 2

(a) Nutrients tested for (1 mark each): (i) Iodine solution tests for starch. (ii) Biuret reagent tests for protein. (iii) Benedict’s solution tests for reducing sugar.

(b) Nutrients present (2 marks each): Sample P contains starch only (positive iodine test only). Sample Q contains protein only (positive Biuret test only). Sample R contains reducing sugar only (positive Benedict’s test only). Sample S contains starch, protein and reducing sugar (positive result in all three tests).

(c) A positive test for starch is indicated by the iodine solution changing from its original brown/yellow colour to a blue-black colouration. (2 marks)

(d) A positive test for a reducing sugar in the Benedict’s test is indicated by the blue colour of the solution changing to a green, yellow, or brick-red precipitate on heating, with brick-red typically indicating a high concentration of reducing sugar. (2 marks)

(e) Food sample S gave a positive result in all three tests because it contains a mixture of nutrients, that is, starch (detected by the iodine test), protein (detected by the Biuret test), and reducing sugar (detected by the Benedict’s test), all present together in the same food sample. (3 marks)

(f) Any one precaution at 2 marks: Heat the mixture of food sample and Benedict’s solution gently, preferably in a water bath, rather than directly over a naked flame, to avoid uneven heating or a bump/splash hazard; avoid overheating, which could cause the test tube to crack or its contents to spill; use clean, dry test tubes and equipment to avoid contaminating the results.

Question 3

(a) Completed table (6 marks, allocated across the six rows): Concentration 0.0, change in mass = +0.8 g, percentage change = +16%. Concentration 0.2, change in mass = +0.4 g, percentage change = +8%. Concentration 0.4, change in mass = 0.0 g, percentage change = 0%. Concentration 0.6, change in mass = -0.4 g, percentage change = -8%. Concentration 0.8, change in mass = -0.8 g, percentage change = -16%. Concentration 1.0, change in mass = -1.2 g, percentage change = -24%. (Percentage change in mass = (final mass – initial mass) / initial mass x 100.)

(b) Osmosis is the movement of water (solvent) molecules from a region of higher water concentration (lower solute concentration) to a region of lower water concentration (higher solute concentration), through a semi-permeable/partially permeable membrane, until equilibrium is reached. (2 marks)

(c) The 0.4 mol/dm3 sucrose solution is isotonic with the potato tissue (1.5 marks), because the potato cylinder placed in it showed no change in mass (0% change), indicating that there was no net movement of water into or out of the potato cells at this concentration (1.5 marks).

(d) In distilled water, the water potential (concentration of water) outside the potato cells was higher than the water potential inside the cells (which contain dissolved solutes). Water therefore moved into the potato cells by osmosis, causing the cells to become turgid and the potato cylinder to gain mass. (3 marks)

(e) In the 1.0 mol/dm3 sucrose solution, the water potential outside the potato cells was lower than the water potential inside the cells. Water therefore moved out of the potato cells by osmosis into the surrounding solution, causing the cells to lose turgidity (become flaccid/plasmolysed) and the potato cylinder to lose mass. (3 marks)

(f) Any one valid application at 3 marks: Osmosis explains the uptake of water by plant roots from the soil; it is applied in the preservation of foods by salting or sugaring, which draws water out of microorganisms by osmosis and prevents their growth; it explains water absorption and reabsorption processes in animal kidneys/intestines; it is used in the medical setting to prepare isotonic solutions/intravenous drips that do not damage blood cells.

Question 4

(a) Percentage frequency (1 mark for each of the 8 values, using percentage frequency = (number of quadrats present / total number of quadrats thrown) x 100): Habitat I, Grass = 90%, Broad-leaved weed = 40%, Fern = 0%, Moss = 10%. Habitat II, Grass = 30%, Broad-leaved weed = 80%, Fern = 70%, Moss = 60%.

(b) The dominant species in Habitat I is grass, with the highest percentage frequency of 90% (1 mark). The dominant species in Habitat II is the broad-leaved weed, with the highest percentage frequency of 80% (1 mark).

(c) Definitions (2 marks each): (i) A habitat is the natural home or environment in which an organism lives, and which provides the conditions and resources needed for its survival. (ii) A quadrat is a sampling device, usually a square frame of a known fixed area, used to estimate the abundance and distribution of organisms, particularly plants, within a habitat. (iii) A population is the total number of individuals of the same species living together in a particular habitat at a given time.

(d) Any two reasons at 2 marks each: Ferns generally require a moist, shaded environment to thrive, which is more available in Habitat II (near the stream, under shade) than in the more open, possibly drier Habitat I; ferns reproduce by spores which require moisture for successful germination and growth, favouring the damper Habitat II; the grassland habitat (Habitat I) may experience more direct sunlight and lower humidity, conditions less favourable to ferns; competition from the dominant grass in Habitat I may also limit the establishment of ferns there.

Question 5

(a) Identification (1 mark each): W is the cement (cementum); X is the enamel; Y is the dentine; Z is the pulp (pulp cavity).

(b) Functions (2 marks each): Cement (W) anchors the root of the tooth firmly to the jawbone by means of periodontal fibres. Enamel (X) forms a hard, protective covering over the crown, protecting the tooth against wear and decay, as it is the hardest substance produced by the body. Dentine (Y) forms the bulk of the tooth, supporting the enamel and transmitting sensations (such as temperature and pressure) to the nerves within the pulp. Pulp (Z) contains blood vessels and nerve endings, which supply nutrients and oxygen to the tooth and provide sensation.

(c) Any two differences at 2 marks each: A carnivore has well-developed, sharp and pointed canine teeth for seizing, killing and tearing flesh, while a herbivore has small or absent canines; a carnivore has sharp, cutting/shearing (carnassial-type) molars and premolars adapted for cutting meat, while a herbivore has broad, flat, ridged molars adapted for grinding plant material; a herbivore often has a gap (diastema) between the incisors and the cheek teeth to aid manipulation of plant food and chewing (cud chewing in ruminants), which is typically absent or less developed in carnivores.

(d) Dental formula of an adult human, for one side of the upper and lower jaw: Incisors 2/2, Canines 1/1, Premolars 2/2, Molars 3/3 (4 marks, 1 mark for each correctly stated tooth type and number). This gives a total of 32 teeth in the full adult human dentition.

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