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Hydrogen bonds form between the nitrogenous bases in one strand and the nitrogenous bases in the other strand of DNA;

This is the law of use and disuse propounded by Jean-Baptiste Lamarck.

(a) Structure of the cell relating to its function (i) Sperm cell Has a long/whip-like tail/flagellum, with which it swims to reach the egg/ovum of the female organism. Has a head with acrosome; for digesting/penetrating egg membrane Has a nucleus in its head; which fuses with the nucleus of the ovum/egg cell during fertilisation to form zygote Its middle piece contains mitochondria; for supply of energy/ATP. (ii) Palisade cell Has thin wall; for easy penetration of light Has numerous chlorosplasts in peripheral cytoplasm for maximum absorption of sunlight Has large vacuole containing cell sap; for temporary stor-age of products of photosynthesis. (b) Diagram of the front view of the female reproductive system Differences between reproduction in Mammals and Amphibians Mammals Amphibians fertilisation is internal fertilisation is external Development of offspring is internal / viparous born young ones alive development of offspring is external / oviparous/ lay eggs placenta is present placenta is absent shows parental care does not show parental care produce fewer eggs / offspring Produce numerous eggs/offspring (d) Methods of birth control in humans Abstinence Use of contraceptive pills/injection of hormonal drugs Coitus interruptus/withdrawal method Safe period/rhythm method Sterilization of men/vasectomy Sterilisation of women/tubal ligation Use of condoms/female condom/male condom Use of spermicide/foam Use of implant/Norplant Use of IUD/intra-uterine devices Use of diaphragm.

(a) Gases involved in the photosynthesis of a plant Carbon dioxide/Carbon (IV) oxide Oxygen (b) Role of gases involved in photosynthesis (i) Carbon dioxide/Carbon (IV) oxide: raw material for photosynthesis/to produce glucose/starch (ii) Oxygen: by-product/given off during photosynthesis. (c) Variegated leaf (i) A leaf with different colour parts/patches/green and non-green parts. (ii) Part of variegated leaf that would be positive when treated with iodine solution The green part/patch (iii) Reasons for the answer in 2(c) (ii) Presence of chlorophyll/chloroplast; for trapping solar energy; to manufacture starch; which turns blue-black with iodine solution (iv) Mineral elements used to manufacture chlorophyll Magnesium/iron/nitrogen. (d) Holozoic parasitic symbiotic saprophytic Autotrophic Human Tapeworm Lichen mushroom water leaf plant Housefly Rhizopus Elephant grass Venus flytrap Spyrogyra Venus flytrap

(a) (i) Renewable natural resources Materials that can be reused/replaced/ regenerated/recycled; after they have been used (ii) Non-renewable natural resources Materials that are difficult/impossible to regenerate/ recycled; after being exhausted/depleted. (b) (i) Renewable natural resources Air; soil; crops; fisheries; water; vegetation/forest; game/wildlife. (ii) Non-Renewable natural resources Rocks; coal; minerals/correctly named example; crude oil/petroleum. (c) Ways of conserving forest Prevent forest/bush fires Practice selective cutting of trees/avoid indiscriminate felling of trees Regulations/laws to control excessive clearing of forest Employ forest guards to prevent/arrest intruders Use pesticides to kill pests/control plant diseases Embark on public education on dangers of deforestation. (d) Effects of adding animal manure to garden soil Animal manure provides nutrients/nitrogenous materials /improves soil fertility Animal manure are acted upon by nitrifying bacteria To form nitrates for plants Manure improves the water-holding properties of the soil by binding the soil particles together/ improves soil structure Prevents soil from being blown away by wind/erosion (e) Possible genotypes in humans AA, AS, SS

(a) Explanation of terms (i) Gene; Is a unit of heredity/genetic material; composed of DNA; located on the chromosome. (ii) Hybrid Is an offspring; formed from a cross between parents that are genetically different but of same species. (iii) Trait Is a feature/character; that is possessed by an organism; and can be transferred from parents to offspring. (b) (c) Transmittable characters in plants Type of seed Size of fruit Colour of flower Colour of leaf Height of plant Texture of seed coat Size of plant Leaf shape Any correctly named transmittable character

(a) (i) Chemosynthesis; The process by which some organisms manufacture their food By converting carbon molecules into organic substances Using the oxidation of inorganic substances/ hydrogen sulphide/ammonia/methane As a source of energy without the use of sun light/solar energy. (ii) Examples of organisms that carry out chemosynthesis; Sulphur bacteria Nitrosomonas Nitrobacter Blue-green algae Nitrococcus (b) (i) Percentage composition of gases in the atmosphere S/N Gases in the atmosphere % composition 1 Oxygen 205 - 21.5 2 Nitrogen 78- 79 3 Noble gases / Argon 0.9 - 1.0 4 Carbon dioxide .03 - 0.05 (c) (i) Characteristics of salt marsh habitat Presence of low oxygen level; High salinity Soft muddy environment Presence of changing water levels Flat lowland Swampy/water-logged Low light'penetration (i) Modification of plants to salt marsh habitat Plants grow long/have numerous roots; to hold to the substratum; possess stilt roots for anchorage (d) (i) Bacterial diseases associated with poor food hygiene (Staphylococcal) Food poisoning Dysentery; Cholera Hepatitis Typhoid Gastritis Botulism Tuberculosis Any correctly named disease. (ii) Effects of bad food hygiene May lead to intestinal ulcer May lead to serious food poisoning Sickness/discomfort Frequent stooling/running stomach/diarrhoea Loss in economic value/loss in food value Death. (e) Restoring blood loss of an accident victim after a car accident The victim must first be rushed to the hospital Where the blood sample will be taken For agglutination test To know the blood group Then the compatible blood is transferred to the victim. (f) Differences between tillage and bush burning. Tillage Bush burning Loosens the soil compacts the soil Aerates the soil leads to poor aeration of the soil Mixes organic matter with soil/improves soil nutrients Depletes / destroys soil nutrients / leaching Enhances the activities of micro organism Micro fauna / soil organisms are killed Does not lead to soil / air pollution leads to air / soil pollution

The Chief Examiner reported that in part (a), some Candidates' carried out the requirement of BODMAS correctly tile few couldn't. However, in part (b), most Candidates' could not sketch the diagram correctly and hence failed to earn the marks allocated. In part (a), expressing as improper fraction we have (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\) Next, using BODMAS to get (\(\frac{16}{7} + \frac{2}{5} \times \frac{17}{12}) \div \frac{17}{12}\). Simplifying and inverting yields \(\frac{1198}{420} \times \frac{20}{3} = 19\frac{1}{63}\) In part (b), they were expected to sketch Then, using trigonometric ratio to have sin 48\(^o\) = \(\frac{x}{250}\) or cos 42\(^o\) = \(\frac{x}{250}\) and simplifying the vertical distance moved to be 186 m.

(a) If Musa's age = x and Manya's age = y, then x = y + 3.... (1) and (x — 7) = 2(y— 7) .......(2) Solving both equations will yield Musa's age = 13 years and Manya's age = 10 years. (b) Let n be the year when the sum of their ages will be 45, so that (10 + n) + (13 + n) = 45 and solving will give n = 11 years.

The Chief Examiner reported that quite a handful of the Candidates' got part (a) correctly but couldn't establish the relationship between km/h and m/s. (a) The distance covered = (2 x 60) + (2 x \(\frac{22}{7}\)) = 120 + 377.142 = 497.1429m so that if the athlete runs two times will covered = 2 x 497.1429 m = 994.2857m (at least one decimal place). (b) The speed m/s = \(\frac{994,2857}{200}\) and converting to km/h we will have \(\frac{994,2857}{1000} \times \frac{200}{3600}\) = 17.8973 km/h

(a) First joining XZ and WY, we have < XWY = < XZY = \(\frac{1}{2}\)(180^\circ — 80^\circ) = 50^\circ. This implies that < ZXY = 50^\circ(base angles of an isosceles triangle). So that < XWZ = 180^\circ - 80^\circ = 100^\circ. (b) The rate in every N1.00 = \(\frac{2}{100}\) so that the rate of interest charged per annum will be \(\frac{2 \times 12}{100}\) x 100 which gives 24%.

(a) The bar chart for the distribution is as shown (b) They were able to construct the frequency distribution table as Score (x) 1 2 3 4 5 6 Total Frequency (f) 2 5 13 11 9 10 50 fx 2 10 39 44 45 60 200 Mean = \(\frac{200}{50}\) = 4

(a) By the use of Pythagoras theorem, the hypotenuse side is 13 and also sin x = \(\frac{5}{13}\) and cos x = \(\frac{12}{13}\) Substituting and simplifying yields \(\frac{65}{161}\) (b) we have (11 + 2x)(14 + 2x) - (11 x 14) =186. Expanding and simplifying to get the quadratic equation 2x\(^2\) + 25x - 93 = 0. Solving the quadratic equation results in x = 3 m as the width of the path.

(a) let the original price of the shirt be x, then we have two percentages reduction leaving 60% and 70%. This implies that \(\frac{60}{100} \times \frac{70}{100} \times x\) = 35. Solving for x yields x = GH₵ 83.33. (b) (i), the first reduction = (\(\frac{40}{100}\)) x 180 = 72.00 and the second reduction = (\(\frac{30}{100}\) ) x 108 = 32.40. Then total reduction = (72.00 + 32. 40) = GH₵ 104.40. (b) (ii), the percentage reduction in the price of the article = (\(\frac{104.40}{180} \times 100\)) = 58%

(a) The trapezium PQRS is as shown (b) |QR| = (3.9 \pm 0.1) cm, < PSR = (60 \pm 1)\(^o\) and |QY| = (5.2 \pm 0.1) cm.