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(a) Test cross; This is the crossing/mating of an organism of unknown genotype with a homozygous recessive/double recessive organism;to determine the genotype of an organism. (ii) monohybrid cross; Is a genetic cross between parents that differ in the alleles they posses; for one particular gene

(a) Kidney diseases Diseases One effect One remedy kidney stones Abdominal pains; decreased frequency of urination, pains during urination; presence of blood in the urine; increased blood pressure. Drink plenty of water kidney transplant; avoid foods rich in calcium/red meat/meat; consult a physician. Nephritis Oedema/swelling of feet and ankle; high blood pressure; blood in urine/dizziness/back pain/fatigue/general weakness; inflammation of the kidney tubules consult a physician; kidney transplant; dialysis/all food items must be washed/cooked properly; drinking water boiled/filtered/sterilized Diuresis Thirst/dehydration occurs; high blood pressure may occur; loss of appetite;weakness; fatigue; nausea; loss of serum electrolyte. Consult a physician; avoid fans/air conditions in cold weather, avoid excessive drinking of alcohol (c) (i) Estuarine habitat; is the place point where a river enters to ocean/sea, into which the tides flow; fresh water mix with salt water; to form brackish water. (ii) food chain typical of an estuarine; phytoplanktons - branacles => fish => brid OR Detritus => worm => molluse => bird OR Detritus => shrimp => fish => bird (d)(i) Pests that can attack plant crops : Stem borer army worms weevils black tea-thrips aphids root mealybugs variegated grasshopper beetle; rodents squirrel locust birds mites cotton stainer any correctly named examples. NOTE : Spelling must be correct to score. (ii) Ways of controlling plant diseases Chemical control/ use of fungicides/ nematicides/ bactericides/ pesticides Biological control/ use of predators and parasites of insects/ pests to keep them in check Breeding resistant varieties of crop Planting genetically modified crops that can resist viruses/ bacteria transmitted by pests Destruction of infested crops by burying/ deep burial/pruning of affected parts Destruction of alternate host Adopt good management/ crop rotation. (e). Functions of a public health authority ensures cleanliness of public places/markets/schools/playgrounds concerned with the proper disposal of refuse provides maternity clinics/health centres/nursing homes/family planning provides child welfare services provides medical inspection of schools registers birth/death prevents controls infectious diseases/ administers vaccination/ immunization/ inoculation informs World Health Organization of the outbreak of infectious diseases ensures adequate provision of clean water to the public provides ambulance services provides quarantine services for plants/ animals/ humans gives health certificate to travelers provides public health education/awareness of diseases inspects and certifies health of animals and cleanliness of slaughter house/ abattoir. (f). Ways of maintaining food hygiene : - proper harvesting/ storage methods must be ensured to prevent damage to food items food items must be properly washed/ cleaned to prevent contamination food items must be properly cooked at appropriate temperatures kitchen/cooking areas must be kept clean always/free from germs perishable food items must be properly preserved/ refrigerated proper personal hygiene must be observed by those handling food cooking utensils must be properly washed/ cleaned before and after use cooked/prepared food must be covered at all time proper disposal of unused/leftover food.

(a) Division:Angio-spermatophyta / Angiospermophyta /magniliophyta (ii) Class: Dicotyledoneae / magniliopsida (b)(i) Types of leaf arrangements: Specimen A - Alternate/Spiral, Specimen B - Decussate/opposite (ii) Leaf Shape: Specimen A- Lanceolate/ Sword-Iike, Specimen B - Oval/Ovate. (c)(i) Type of leaf margin of specimen A: Entire/Smooth/Wavy. (ii) Types of venation found in Specimens A and B :Specimen A - Net/reticulate, Specimen B - Net/reticulate. (d)(i)Observable similarities between Specimens A and B: - Leaves have leaf stalk/petiole/petiolate; - Leaf stalk/ (d) petiole swollen at the base; - Leaves contain chlorophyll /are green in color; - Leaves have entire/smooth margins; - Leaves have net/reticulate venation; -Leaves have pointed apex; - Leaves have prominent midrib; - Both leaves are simple. (ii) Observable differences between Specimens A and B: Specimen A (mango) Specimen B (Ixora) - Petiole/leaf stalk is long Petiole/leaf stalk is short; - Leaves have elongated apexlaccuminate/acuminate Leaves have short apex/acute; -Leaves have a narrow lamina/leaf blade Leaves have a broad lamina/leaf blade; - Spiral/alternate leaf arrangement Decussate/ opposite leaf arrangement - Leaves are bigger. Leaves are smaller. (e) Diagram of Specimen A (leaf)

(a)(i) Phylum of specimens G and H -Arthoropoda (ii) Reasons: -presence of jointed legs/appendages, - presence of chitinous exoskeleton; - Metamerically seamented: - Bilateral symmetry; - Have three body layers triploblastic. (b) uructures of specimens G and H used for defence: G - Sting/Ovipositor; H - (Strong and powerful/biting) mandible/jaw. (C)(i) Habitats of specimens G and H G- (Bee) Hive H-Termitarium/anthill/termite mound/nest/soil/old logs of trees/timber of wooden buildings. (ii) Economic importance of Specimens G and H: G-It pollinates flowers - it produces honey, - produces wax; - sting causes pain/death: - bee keeping is a source of employment/ income, H-it destroys wood/furniture/books; - it destroys crops/plants; The termitarium is used as tennis lawb court - source of food/ protein. - aids soil aeration through tunnelling; - enhances penetration of water into the Soil/increases porosity; - introduces vegetable/plant matter into the soil/ increases humus contents, -their mound/ termitarium forms impediments to land use. (d)(i)Class of organisms to which specimens M, N and P belong: M-Aves, N - Osteichthyes, P - Mammalia. Common features of specimens N, N and P: protection of the organism (against mechanical injury/dessication) (ii) Observable differences between specimens M and P Specimen M/Quill feather Specimen P/ Mammalian skin with hair Presence of inferior umbilicus absence of inferior umbilicus, Presence of superior umbilicus absence of superior umbilicus - Presence of after shaft absence of after shaft; Presence of shaft/rachis absence of shaft/rachis; - Presence of Quil/calamus absence of Quill/calamus; - Presence of barbs absence of barbs; Absence of hair/fur presence of hair/fur Texture is hard when touched texture is soft when touched; - Barbs are restricted to certain parts of the Quill Hairs/ fur cover the whole surface of the specimen. (e) Structures of biological importance in Specimen M: - Presence of Quill feather, serves as water proof/protection keeps the bird warm/insulation; - vane is flat/broad; to provide surface area for flapping during flight/against desiccation; - rachis/shaft which is centrally placed; to provide point of attachment to barbs/strengthen the feather - barbs with hooks; interlock to form the vane; - the overlapping arrangements of Quill feathers over the body of birds; allows smooth movement/li resistance flight; - colouration of the Quill feather; enhances courtship display/camouflage.

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< URT = < TRQ (angle alternate a tangent and a chord equal to angle in the alternate segment) < RUT = xo In \(\bigtriangleup\) URT < RUT + < RUT + < UTR = 180o (sum of int. < s of \(\bigtriangleup\)) < URT + x + x = 180o < URT = 180o - 2x

Apparent cubic expansivity = \(\frac{\text{mass of hot liquid expelled}}{\text{mass of hot liquid left x temp. rise}}\) = \(\frac{2}{8 \times 40} = 0.00625\)

The superscript is the atomic mass = 35 (The mass number also called atomic mass number or nucleon number, is the total number of protons and neutrons) The subscript is the atomic or proton number = 17 The number of electrons in a neutral atom is equal to the number of protons.

The colour of a neutral litmus solution is purple. Litmus solution is made up of purple dye which is extracted from Lichens. It is a natural Indicator and it is used to indicate the presence of an acid or a base. A litmus paper will turn red in the presence of an acid and it turns blue in the presence of a base.

At the beginning of periods 2 and 3 of the Periodic Table, there are several cases where an element at the top of one group has some similarities with an element in the next group. For example, boron is a non-metal with some properties rather like silicon. Unlike the rest of Group 2, beryllium has some properties resembling aluminium. And lithium has some properties which differ from the other elements in Group 1, and in some ways resembles magnesium. There is said to be a diagonal relationship between these elements. There are several reasons for this, but each depends on the way atomic properties like electronegativity vary around the Periodic Table.

The period 4 transition metals are scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), and zinc (Zn). Diamagnetism results from the lack of unpaired electrons. Diamagnetic substances are substances that are repelled by magnetic fields. The examples of diamagnetic materials are copper, gold, antimony, silver, lead and hydrogen.

The neutral Iron atom has the electric configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d6.to write the electronic config. of Fe3+ we have to subtract 3 electrons from the outermost shell which is 4s and 3d orbitals. Thus we obtain the electronic configuration for Fe3+ as: 1s2 2s2 2p6 3s2 3p6 4s0 3d5 OR [AR] 4s0 3d5.

The gas used in gas bottles is LPG – liquefied petroleum gas – which are either propane or butane, flammable hydrocarbons used as fuel for home heating, cooking, hot water and LPG cars and vehicles.

To find \( 110111_2 + 10100_2 \): Convert to decimal: \(110111_2 = 55_{10}, \quad 10100_2 = 20_{10}\) Add: \(55 + 20 = 75\) Convert back to binary: \(75_{10} = 1001011_2\) Final result: \(110111_2 + 10100_2 = 1001011_2\) OR \(\begin{array}{r} 110111_2 \\ + 010100_2 \\ \hline 1001011_2 \end{array}\) \(\therefore \; 110111_2 + 10100_2 = 1001011_2\)

SP = Nx CP = N60,000 Percentage loss = 15% Using percentage loss... \(\frac{CP - SP}{CP} = \text{100%}\) \(\text{15%} = \frac{60000 - x}{60000} \times \text{100%}\) \(\text{15%} \times 60000 = (60000 - x)\text{100%}\) \(60000 - x = \frac{\text{15%} \times 60000}{\text{100%}}\) 60000 - x = 3 x 3000 60000 -x = 9000 x = 60000 - 9000 x = N51,000