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(a) Main parts of human brain are: (a) The forebrain/ Comtractile cerebrum (b) The midbrain/cerebellum (c) The vacuole hindbrain/medulla oblongata (b)(i) The cerebrum is the seat of consciousness and senses. It also Paraniyium functions in the following ways: intelligence, memory, gcanuies judgement, learning, imagination and voluntary Chicroplost actions. It collects and stores information. (ii) Hypothalamus (a) connects the pituitary gland or master gland to the thallamus (b) It contains the reflex centres. (c) It regulates body temperature. (d) It regulates water balance and osmoregulation. (e) It controls sleep and wakefulness. (f) It controls blood Of drinking (h) It is the seat of aggression and anger; pressure. (g) It is the centre for appetite, feeding and It controls the pituitary gland. It controls speech. (iii)(a) Medulla oblongata controls breathing and respiration; (a) It controls heart-beat (b) It controls digestion. (c) It controls size of blood vessels. (e) It controls the secretion of saliva. (f) It controls laughing. (g) It controls involuntary actions such as sneezing, blinking, coughing, yawning etc. (c)(i) Organs of higher plants that respond to stimuli are: (a) Flowers (b) Leaf (c) Root (d) Stem (e) Tendril (ii) Tactic movement is the locomotion, and response of the entire body of an organism in response to a stimulus. It is directional; locomotion could be positive or negative. (d)(i) Phototaxis is the movement in response to light stimulus. it is the source of light by organisms. E.g. Euglena moving away from light (ii) Hydrotaxis is the response of a whole organism to the stimulus of water or humidity. E.g. Woodlouse moves towards areas of high humidity. (iii) Chemotaxis is the movement of an organism in response to a chemical stimulus; E.g. Amoeba and Paramecium moving away from chemicals.

(a) Adaptive features of climbing animals are: (a) There is presence of prehensile tails, (e.g. monkey (b) They have opposable digits e.g. monkey (c) They possess sticky adhesive discs on fingers e.g. geckos (d) They possess grasping scales e.g. snakes (e) They have grasping pads e.g. tree frongs (f) They have long, sharp claws for climbing e.g. squirrels. (b) Important uses of water to organisms in rain forest are: (a) It is also essential for photosynthesis. (b) It helps to maintain the body temperature of organisms. (c) It is essential for plant turgidity/ mechanical support. (d) It helps in metabolism of organisms. (e) It provides moisture necessary for microbial activity. (f) It provides moisture necessary for burrowing animals. (g) It is also essential for translocation and transpiration. (h) It is also essential for movement and survival of aquatic animal. (i) It is important for plant and animal growth or germination of seeds. (c) The instrument used for measurement of temperature is the thermometer. The bulb of the instrument whose temperature is to be measured is suspended in the medium for about one minute to five minutes while the level of the mercury or liquid in the stem is recorded. About three readings are recorded and the average found. (d)(i) The following scientists have contributed to the study of solutions: (a) Larmack (b) Charles Darwin (c) De'Vries (d) A.R. Wallace ( ii) Evidence in support of evolution are: (a) Evidence from embryology (b) Evidence from Geographical distribution (c) Physiological evidence (d) Evidence of comparative morphology. (e) Evidence of vestigial organs (f) Evidence from cytology or chromosome doubling (g) Evidence from fossil records (h) Evidence form comparative anatomy.

(a) A balanced diet contains all the six classes of food which are carbohydrates, protein, fat, vitamins, water and mineral salts. (I) Examples of the classes of food vitamins fruits, vegetables such as bitter leaf, egg, milk, liver, oil etc. fat Groundnut, melon, butter, margarine, palm oil, cod liver, soya bean oil. water water in plants, water in animal, water in drinks, beverages and food mineral salt vegetables, fruits such as mango, orange, apple, cashew etc. banana,bones,egg, table salt Carbohydrates Rice, bread, cassava, yam, cocoyam, potato etc (c)(i) Functions of protein in human are: (i) It is useful for growth and formation of new cell. (ii) Protein is used for repair of worn out tissues. (iii) It functions in the formation of Hormones. (iv) It yields certain amount of energy. (v) It is also useful for formation of enzymes. (ii) A child with protein deficiency will have the. following features: (i) The child will have retarded growth. (ii) Loss of weight is also common to such child (iii) He will develop swollen legs, oedema and swollen feet (iv) He develops cracked and split skin (v) The child will have distended and swollen stomach resulted to kwashiorkor. (vi) He develops pale body. (vii) There is also a change in the hair; the hair becomes reddish brown (viii) The child also develops thin and tiny legs (ix) The child will have distended and smoker stomach resulted to kwashiorkor. (d)(i) One organelle in a living cell that produces enzymes is Ribosomes. (ii) Enzymes that act on proteins are: (i) Pepsin (ii) Trypsin (iii) Rennin.

(a) Formation of zygote in a flower takes place in the following ways: The pollen grain absorbs water, sugary fluid and glucose from sticky stigma. It swells and germinates. Pollen tube emerges, carrying a tube nucleus and a generative nucleus. The generative nucleus divides into two male nuclei; it grows down the style and ovary wall. It derives energy and food from the tissues, to reach the ovule; It enters ovule through the micropyle; to enter embryo sac; Pollen tube releases male nuclei into embryo sac; as tube nucleus degenerates. Nucleus of one male gamete fuses with the egg cell in order to achieve fertilization which resulted to the formation of zygote. (b) Importance of Amniotic sac are as follows: (i) Amniotic fluid contains the embryos. (ii) This provides a constant warm temperature for the foetus. (iii) It provides a sterile environment for development. (iv) It provides hydrostatic protection against mechanical injury and shock (v) It allows movement of the foetus. Viral disease (viruses) Bacteria diseases (Bacteria) Protozoan diseases (protozoa) yellow fever Tetanus malaria poliomyelitis Typhoid Trypanosomiasis measles tuberculosis Ameoboid dysentery common cold Leprosy small pox cholera Influenza whooping cough chicken pox

(a)(i) Respiration is a metabolic process which takes place in living cells by which organic nutrients are broken down to release energy or ATP for life activities. (ii) Two types of respiration are (a) Aerobic respiration (b) Anaerobic respiration (iii) Equation for respiration in muscle during exercise is: Glucose→ Lactic Acid + Water + Energy/ Adenosine - Triphosphate (ii) The following are the diseases of the lungs: Lung cancer; tuberculosis, pneumonia, bronchitis, asthma, common cold, .whooping cough and SARS/Influenza. (c) The structures used for gaseous exchange in these organisms are: Amoeba—Cell membrane/body surface Cockroach—trachea White mangrove—pneumatophores, breathing stomata! and tenticels Hibiscus -stomata and lenticels.

\((343)^{\frac{1}{3}} \times (0.14)^{-1} \times (25)^{-\frac{1}{2}}\) = \(\sqrt[3]{343} \times (\frac{14}{100})^{-1} \times (\sqrt{25})^{-1}\) = \(7 \times \frac{100}{14} \times \frac{1}{5}\) = 10.

Hint: Represent the question in a venn diagram, such that n(B ∩ M) = x n(B) only = 125-x n(M) only = 110-x => 125-x + 110-x + x = 220 => x = 15. n(B) only = 125 - 15 = 110

52.4 - 5.7 - 3.45 - 1.75 = 52.4 - (5.7 + 3.45 + 1.75) = 52.4 - 10.90 = 41.5

\((\sqrt{0.7} + \sqrt{70})^{2}\) = \((\sqrt{0.7} + \sqrt{70})(\sqrt{0.7} + \sqrt{70})\) = \(0.7 + 2\sqrt{0.7 \times 70} + 70\) = \(0.7 + 14 + 70 \) = 84.7

\(\frac{0.21 \times 0.072 \times 0.00054}{0.006 \times 1.68 \times 0.063}\) = \(\frac{9 \times 10^{-10}}{7 \times 10^{-8}}\) = \(1.2857 \times 10^{-2}\) = \(0.012857 \approxeq 0.01286\) (to 4 s.f)

Let the number of goats be X C.P of X goats = N4000X S.P of X goats = (75/100) x 4000X = N3000X 3000X = 180,000 => X = 180000/3000 = 60. X = 60 goats

dy/dx = 2x - 3 y = ∫2x - 3 dx => y = x2 - 3x + c 3 = (0)2 - 3(0) + c => c = 3 Thus y = x2 - 3x + 3

\(y = \sin^{2} (5x)\) Let u = sin 5x \(\frac{\mathrm d u}{\mathrm d x} = 5 \cos 5x\) \(\therefore y = u^{2}\) \(\frac{\mathrm d y}{\mathrm d u} = 2u\) \(\frac{\mathrm d y}{\mathrm d x} = 2u . 5 \cos 5x\) = \(10u \cos 5x\) = \(10 \sin 5x \cos 5x\)

y = 3x\(^2\) - 2x + 5 Slope = \(\frac{\mathrm d y}{\mathrm d x} = 6x - 2\) At x = 1, Slope : 6(1) - 2 = 4.