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The matching of Charles Darwin → Preservation of some food with heat, Gregor Mendel → Natural selection and Louis Pasteur → Independent assortment of genes, were wrong but the only correct matching was Jean Baptiste Lamarck → Inheritance of acquired characteristics

Scientist Famous work I. Charles Darwin → Natural selection III. Jean Baptiste Lamarck → Inheritance of acquired characteristics

Functions of the parts of the nephron (a)(i)Glomerulus carries out ultrafiltration or it filters the blood passing through it. (ii)Bowman's capsule receives blood plasma from the glomerulus and filters it to initiate urine formation. (iii)Proximal convoluted tubule is responsible for reabsorbing most of the glucose, amino acids, and small peptides that enter the ultrafiltrate back into the bloodstream. (iv)Distal convoluted tubule regulates electrolytes and water balance, and helps maintain pH. (b) Organism Structure for excretion Form in which waste is excreted by organism Amoeba/Paramecium contractile vacuole Gaseous/Liquid Flatworm Flame cell Liquid Earthworm Nephridium Liquid Cockroach Malpighian tubule Solid Human Skin/kidney/lung Liquid Plants(Green) Stomata pore Gaseous (c) Excretory products of plants include the following: Carbon dioxide, water vapor, oxygen, resins, gum, tannins, alkaloids, oils, and nitrogenous compounds.

(a)Roles played by the organs involved: (i) Mouth: contains saliva that moistens or softens the food. (ii)Skin excretes excess salt and eliminates excretory waste from the body. (iii)Tongue tastes the food and rolls the food into bolus. (iv)Kidney excretes excess water and salt or carries out osmoregulation. (v)Stomach temporarily stores the food/churning of the ingested food. (vi)Small intestine absorbs and digests the food as the enzyme maltase converts maltose into glucose while lipase converts fats and oil to fatty acids and glycerol. (vii)Large intestine moves waste to the rectum as faeces to be egested out of the body through the anus. (viii)Liver converts excess glucose to glycogen for storage/ deaminates excess amino acids. (b) Three Organs that would contain enzymes needed to break down food include: - Mouth - Stomach - Small intestine (c)Three digestive enzymes that would be involved in the digestion of the meal. - salivary amylase/ptyalin - pancreatic amylase - Lipase (d)Two end products of the meal after digestion. - amino acids - glucose - fatty acids and glycerol (e)(i)one effect that the excess salt in the meal might have on the individual. - headache - stroke - dehydration - heart failure - high blood pressure - enlarges heart muscles - kidney disease (ii)Three advantages that the individual would derive from the meal. - The immune system will be strengthened. - Digestion will be improved. - Energy rate will be increased. - Immune system wil be strengthened. - Cardiovascular disease would be prevented.

a(i)four likely changes that would be observed in the meat. - bad or offensive odour - presence of maggot - microbial growth - colour change - change in texture (ii)The biological process that led to the changes that are observed in the meat is decay or putrefaction. (iii) Four methods of preserving the meat One principle involved in the preservation method Salting Salt inhibits microbial growth or kills microorganisms by increasing the osmotic pressure in the meat Frying Involves the immersion of the meat in the boiling oil. It dehydrates the food and reduces microbial growth Freezing Reduces microbial growth by storing food at very low temperature. Drying Removal of moisture content in order to inhibit microbial growth. Canning Involves heating to destroy microorganisms (iv)Environmental factors that may be responsible for decay - Moisture content - Temperature - Humidity - Oxygen (b) Method of preserving (i)Dried maize grain: Store in jute bag; sun drying; storage chemical treatment; pack in silos barn (ii)Fresh tomatoes: Canning; drying; freezing (iii)Vegetables: refrigeration; drying; curing (iv)Beans: Low humid container; chemical treatment; sun drying (v)Milk: Pasteurizatio;canning

a(i) Cell Type of cell division set of chromosomes Onion cell mitosis diploid Sperm cell meiosis haploid Pollen grain meiosis haploid Ovum meiosis haploid Guard cell mitosis diploid Cheek cell mitosis diploid (ii)Mitosis is a process in which a cell divides into two identical daughter cells. It's a type of cell division that occurs in eukaryotic cells. Mitosis is responsible for growth and replacing worn-out cells. A cell replicates its chromosome and segregates them. (b) Parents with blood groups A or B can have an offspring with blood group O if both parents are "heterozygous" for their blood type, meaning they carry both the dominant A or B allele and the recessive O allele, and each parent passes on the O allele to the child, resulting in a type O blood group for the offspring; this is possible because the O allele is recessive to both A and B alleles. (c) Genes are the fundamental unit of inheritance, and their variations, primarily caused by mutations, are the raw material for evolution; meaning that changes in the frequencies of different gene variants within a population over time, driven by natural selection, lead to the observable changes we see as evolution. Genetic variations that alter gene activity or protein function can introduce different traits in an organism. If a trait is advantageous and helps the individual survive and reproduce, the genetic variation is more likely to be passed to the next generation (a process known as natural selection).

a(i) Competition is an interaction between organisms or species in which both require one or more resources that are in limited supply (such as food, water , space or territory). (ii)two types of competition: - intra-specific competition - inter-specific competition (iii) factors that organisms compete for in a habitat. - Space - food - sunlight - nutrient - water - mate - inorganic substances (iv)Relationship between Competition and Succession Competition is a key driving force behind ecological succession, as the struggle for limited resources between different species leads to the replacement of less competitive species with more competitive ones, causing the community composition to change over time and progress through different stages of succession until a climax community is reached. For changes to occur in a community, there must be competition between the old population and the new one. b(i)Structural adaptation Structural adaptation refers to a physical characteristic of an organism that has evolved to help it survive in its environment, meaning it's a change in the organism's outward appearance or body structure that provides an advantage for its survival; examples include a giraffe's long neck for reaching high leaves, a duck's webbed feet for swimming, or a polar bear's thick fur for insulation in cold climates. (ii) Types of structural adaptations in animals - adaptation for offence and defence - adaptation for securing mate/partner - adaptation for feeding - adaptation for water conservation - adaptation for temperature control (c) Features of variation in humans that are used in crime detection. - DNA, - Fingerprints, - Blood groups - Eye colour - Facial expression - Skin colour - Peculiar body features and - skeletal traits are used in crime detection. (d) Three types of adaptive colouration in animals One example of animal that show the type of colouration Counter-shading/disruptive colouration Fishes in water Camouflage Chameleon Alluring Colouration Carnivorous spider Protective/cryptic colouration Lizard, lady bird, green snakes in green grasses. Mimicry Grasshopper, Moth (e) Examples of plants found in a tropical Rainforest. - Rubber tree - Cacao tree - Heliconia - Passion flower - Amazon water lily - Orchids - Mahogany - Obeche - Oil Palm tree

Specimen B - whole liver from a domestic fowl Specimen D - gizzard Specimen E - Intestine (a) Phylum: Chordata / Vertebrata Class: Aves (b) Observable features of specimen B - reddish/dark red colour - has a greenish sac attached to it (presence of gall bladder) - cone shaped - soft and spongy to touch c(i) Functions of specimen B to the organism from which it was obtained - regulation of blood sugar level - deamination of proteins/excretion of nitrogenous waste - formation of fibrinogen - produce heat for maintaining constant body temperature - stores Vitamin A, B, D and E. (ii) one function each of specimens D and E to the organism from which they were obtained. Specimen D - help domestic fowl break down food into smaller, more digestible pieces. Specimen E - digest food and absorb nutrients from it. (iii) one adaptation each of specimens D and E to the function stated in 1c(ii) D: a thick muscular wall, a hardened cuticle, and sometimes small stones called gastroliths. E: have a short, efficient digestive system that's adapted for fast digestion. (d)(i) Test Observation Inferences Red litmus paper + fluid fluid turns litmus paper blue Basic or alkaline Blue litmus paper + fluid Blue litmus paper remains unchanged Not acidic but basic or alkaline (ii) Functions of Bile - breaks down fat globules into smaller pieces so they can be digested - helps the body absorb fat, cholesterol, and fat-soluble vitamins - neutralizes the acidic chyme from the stomach - carries bilirubin, a waste product from red blood cells, out of the body in faeces - carries excess cholesterol out of the body - has antibacterial properties that help fight microorganisms in food. e(i) The biological system in the body of vertebrates that specimens D and E belongs to is the Digestive system. (ii) Specimen D (gizzard) is absent in human f(i) The mode of feeding of the organism that possesses specimen E(intestine) is heterotrophic/ holozoic mode of nutrition. (ii) The type of food that the organism that possesses specimen D consumes. - grains - seeds - crops - insects

Specimen F - Ginger rhizome Specimen H - Leaf of Boplytum Specimen K - Cassava tuber (a) Functions of specimen H to the plant from which it was obtained. - site for photosynthesis - for exchange of gases - for transpiration - asexual reproduction - water storage b(i) one function each of specimen F and K to the plants from which they were obtained. F: stores food, helps the plant reproduce and survive unfavorable seasons. K: storage of food (ii) one reason each for the answers in 2b(i) F: presence of buds / rhizome located underground K: presence of swollen root (iii) Part of the plant modified for the function stated in 2b(i)? F: stem K: root c(i)Four observable differences between specimens F and K. F(Ginger rhizome) K(Cassava) presence of buds absence of buds presence of nodes and internodes Absence of nodes and internodes presence of adventitious roots presence of lateral roots smaller in size Bigger in size (ii) Similarity between specimens F and K. - both are modified underground stems. - both storage organs. - both swollen - both are carbohydrates (d) Labelled diagram of Specimen H (see diagram above)

∠SOR = 2 \times 28^\circ = 56^\circ (angle at the centre is twice the angle at the circumference) From ∆SOR |OS| = |OR| (radii) So, ∆SOR is isosceles. ∠ORS = \(\frac{180^0 - 56^0}{2} = \frac{124^0}{2}\) ( base angles of isosceles triangle are equal) ∴ ∠ORS = 62^\circ

Reflex ∠MOR = 2 \times 124^\circ = 248^\circ (angle at the centre is twice the angle at the circumference) ∠MOR = 360^\circ - 248^\circ = 112^\circ (sum of angle at a point is 360^\circ) ∠OMN = 360^\circ - (124^\circ+ 48^\circ + 112^\circ) (sum of angles in a quadrilateral is 360^\circ) = 360^\circ - 284^\circ ∴ ∠OMN = 76^\circ

∠RMN = 90^\circ (angles in a semicircle is a right angle) ∠SRN = ∠RMN + ∠MNR (ext. angle of a ∆ is equal to the sum of two opp. int. angles) ⇒ 5x + 20 = 90 + x ⇒ 5x - x = 90 - 20 ⇒ 4x = 70 x = \(\frac{70}{4}\) = 17.4^\circ therefore, 2x = 2\(\times17.5\) = 35^\circ

t = ∠NTS (the angle between a tangent and a chord is equal to the angle in the alternate segment). Therefore, t = 36^\circ

a. 2m = 2 \times n (the angle at the centre is twice the angle at the circumference) ∴ m = n ∠SOR = ∠QRO = 54^\circ (alternate angles are equal) From ∆ROQ Since |OQ| = |OR| (radii), therefore, ∆ROQ is an isosceles triangle 2m = 180^\circ - 2(54^\circ) = 180^\circ - 108^\circ (sum of angles in a ∆ is 180^\circ) 2m = 72^\circ m = \(\frac{72}{2}\) = 36^\circ Recall: m = n ∴ m = n = 36^\circ b. Let the length of the rectangle be L and the width be W L = W + 4 (given) Perimeter = 2(L + W) 40 = 2(L + W) 40 = 2((W + 4) + W) (Recall: L = W + 4) 40 = 2(2W + 4) 40 = 4W + 8 40 - 8 = 4W 32 = 4W W = \(\frac{32}{4}\) W = 8cm L = W + 4 = 8 + 4 L = 12cm Area of a rectangle = L x W = 12 \times 8 = \(96 cm^2\)

\(413_7\) to base 5 convert first to base 10 \(417_7 = 4 \times 7^2 + 1 \times 7^1 + 3 \times 7^0\) = 4 \times 49 + 1 \times 7 + 3 \times 1 = 196 + 7 + 3 = \(206_{10}\) convert this result to base 5 5 206 5 41R1 5 8R1 5 1R3 0R1 \(∴ 413_7 = 1311_5\)