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(a)(i) Second law of thermodynamics: This states that when there is conversion of energy from one form to another; there is always a loss/ decreases; in the amount of useful energy. (ii) Flow of energy across different levels in a food chain Transfer of energy between trophic levels Is not 100% completely efficient Successive levels have less energy And support fewer organisms Primary producers/green plants have the highest energy When herbivores/ primary consumers feed on primary producers The energy level is reduced When carnivores/secondary consumers feed on herbivores The energy level is further reduced Part of the energy is lost as heat at each trophic level Not all parts of the preceding organisms are eaten by the predator/organism at the next trophic level; Energy is lost in the process of respiration/feeding/movement/metabolic active (b)(i) Factors that can limit human populations: - Famine / scarcity of food diseases/epidemics increased mortality/death rate by wars /natural disasters emigration increase mortality - reduced natality/birth rate. (b)(ii) Ways of overcoming the factors: Famine: - Availability of food (through improved farming techniques) Mechanized farming improved storage facilities Use of fertilizers/agro chemicals Genetic engineering Improved planting materials/seeds/stems etc. Improved finance Diseases: Health care Immunization Preventive drugs Improved diet Good sanitation. Wars: Improved/enhanced communication/dialogue Mediation. Prohibition of illegal firearms Good governance Natural disasters study of the environment promote environmental friendly activities preservation of natural habitats Accurate and prompt forecast of disaster Increased Mortality/death rate: Provision of balanced/good diet Good Medicare Adequate security Peaceful resolution of conflicts/prevention of wars Effective management of disaster Emigration: - Provide adequate food/shelter Improved health/environment/sanitation Urbanization/development Reduced natality/Birth rate Apply birth control methods Legislation against abortion Education/ awareness against sexual transmitted diseases (STD's):

(a)(i) Difference between genotype and phenotype: A person's genotype is the sum total of the person's genetic constitution; while the phenotype is the observed trait/physical characteristics. (ii) Factors which determine genotype: - Genotype of parents Mutation. Factors which determine phenotype: - Genotype Environment; Diet Health. (b) Phenotypic ratio 3 black 1 white puppies (b) Calculation: The number of white puppies using the phenotypic ratio 3 black: 1 white = 24 Total No of puppies produced = 240 . No of white puppies produced = ¼ x 240 = 60 white puppies (b)(ii) Number of black puppies = ¾ x 240 = 180 black puppies OR calculation b(ii) No of black puppies ¾ x 240 = 180 b(i) No of white puppies ¼ x 240 = 60

(a) Explanation: (I) Putrefaction: A process by which agents of decay/saprophytic/putrefying bacteria/fungi; decompose/breakdown; dead organic remains/waste products of other organisms; into ammonium compounds/nitrates. (ii) Nitrogen fixation: The process by which free atmospheric nitrogen; is converted to nitrate in the soil; by thunder storm/bacteria in root noddles of legumes/soil. (iii) Nitrification: The conversion of ammonia/ammonium compounds; to nitrites; then to nitrates; by denitrifying bacteria. (iv)Denitrification: This is the conversion of nitrates; in the soil to atmospheric nitrogen; by nitrifying bacteria/ fungi. (b)Differences between Nitrogen fixation and Denitrification Nitrogen Fixation Denitrification free atmosphere nitrogen is added to the soil Nitrates in the soil is converted into atmospheric nitrogen Nitrogen fixing bacteria are involved, it involves symbolic and non symbolic organism/bacteria Denitrifying bacteria are involved; it involves purely non-symbiotic organism /bacteria. Modern soil temperature is required High temperature is required (c) How Nitrogen from Urine gets to the roots of plants: - When a person passes urine in the soil Ammonium compound/urea present in urine; is converted into nitrite by nitrifying bacteria/nitroso monas The nitrites are oxidized to nitrates by other nitrifying bacteria /nitrobacter Roots of plants absorb the nitrates.

(a)(i) Specimens E and F belong to Class insecta. (ii) Similarities: Both specimens A and B Possess 6 legs/3 pair of legs; Both specimens A and B have 3 body d thorax and abdomen/segmented body. Both specimens A and B are bilaterally symmetrical; Both specimens A and B possess mandibles/mandibulate mouth parts; Both specimens A and B have jointed appendages/ Jointed legs; Both specimens A and B have a pair of antennae. (iii) Differences: Specimen E/ Cockroach Specimen F/Soldier Termite Small head relative to the body, Large head related to the body; Wings present Wings absent; Has compound eyes/has sight; No eyes /blind/has no sight; Dark brown colour, light brown colour; Small mandibles present Large mandibles present; (b) Economic Importance: (i) Specimen E/Cockroach Destroys clothes/ books/ boxes; - contaminates food/ reduces the quality of food; - gives foul odour, serves as a vector of diseases; (ii) Specimen F/Soldier Termite: Speeds up decaying process/by breaking down of dead trees to form humus which increases soil fertility; -They destroy wood in roofs/houses /furniture; -The mud from the termitarium is used in constructing lawn tennis courts/construction of houses; -For poultry feed/source of protein; Improves soil drainage/aeration through their channels; - Destroy crops/ timber. (c)(i) Adaptation of Specimen E/Cockroach Against Predation:- Specimen E has wings to fly away from predators; -Its brown colour matches with the environment as camouflage against predators; -the flattened shape make it easy to hide in crevices to escape. Adaptation of Specimen F Against Predation: Presence of mandibles for defense/protection. -Possesses snout-like projection that produces a poisonous substance for protection. (ii) Habitat: Specimen E/Cockroach- dark boxes/cupboards, cesspits/any other correct habitat. (ii) Specimen F/Soldier Termite-Termitarium/termite mold/hill

Specimen Part modified to form specimen Mode of propagation Stored material A stem stem water/carbohydrate B shoot bulb water/sucrose C stem rhizome water/starch D root root tuber/seed water/starch/sugar (b) Structural differences between A/Cactus and C/Ginger A (Stem of Cactus) C (Rhizome of Ginger) Green in colour, Brown/grey/yellow in colour; Flattened in shape Cylindrical shape, Leaves reduced to spines; Leaves reduced to scales Absence of nodes/internodes, Presence of nodes/internodes; Adventitious roots absent Adventitious roots present; Terminal buds absent. Terminal buds present. (C) Test for food substance(s) in Specimen D/Carrot Test (correctly recorded) Observation Inference Portion of D and two drops of iodine solution; Blue-black colour formed/obtained Presence of starch Portion of D plus 1ml of water add a few drops of sodium hydroxide/ NaOH solution and 1% Copper Sulphate solution; No colour change Blue colour retained Protein absent Crush and smear a portion of D on a filter paper and leave for five(5) mins Paper stained orange Fat absent

∠RTS = 90o (∠ in semi circle) ∠UTS = 25o (∠ in alternate segment) X + 90 + 25 = 180(sum of ∠s on a str line) X + 115 = 180 X = 180 - 115 X = 65o

TUVW is a cyclic quad 3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary) 3χ + 108 = 180 3χ = 180 - 108 3χ = 72 χ = 72/3 χ = 24o

cot\(\theta\) = \(\frac{1}{\cos \theta}\) = \(\frac{8}{15}\)(given) tan\(\theta\) = \(\frac{15}{8}\) By Pythagoras' theorem, x\(^2\) = 15\(^2\) + 8\(^2\) x\(^2\) = 225 + 64 = 289 x = \(\sqrt{289}\) = 17 Hence sin\(\theta\) = \(\frac{15}{x}\) = \(\frac{15}{17}\)

\(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}\) = \(\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\) = \(\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\) = \(\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\) =\(\frac{2}{3}\times \frac{1}{2}\) = \(\frac{1}{3}\)

\(\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}}\) = \((\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}})(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}})\) = \(\frac{2\sqrt{15} + 6 + 5 + \sqrt{15}}{5 - 3}\) = \(\frac{3\sqrt{15} + 11}{2}\)

0.21 x 0.34 = 0.0714 = 7.14 x 10-2

Let C n G = X n C only = 40 - x n G only = 30 - x 40 - X + 30 - X + X = 50 70 - X = 50 X = 70 - 50 X = 20

\(\frac{x^{3} + 3x^{2} - 10x}{2x^{2} - 8} = \frac{x(x^{2} + 3x - 10)}{2(x^{2} - 4)}\) = \(\frac{x(x^{2} - 2x + 5x - 10)}{2(x - 2)(x + 2)}\) = \(\frac{x(x - 2)(x + 5)}{2(x - 2)(x + 2)}\) = \(\frac{x(x + 5)}{2(x + 2)}\)

y ∝ \sqrtx y = K\sqrtx K = y/\sqrtx = 3/\sqrt16 = 3/4 y = 3/4\sqrtx = 3/4\sqrt64 when x = 64 = 3/4 x 8/1 = 6

x * y = x + y\(^2\) 2 * 3 = 2 + 3\(^2\) = 2 + 9 = 11 (2 * 3) * 5 = 11 + 5\(^2\) = 11 + 25 = 36