This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.
ScanSolve AI Answer
Hey bilgate, good to see you again. Here are the solutions to the questions from the exam paper: SECTION A 1. a) What are enzymes? Enzymes are biological catalysts, which are proteins that speed up the rate of biochemical reactions in living organisms without being consumed in the process. They are highly specific to the reactions they catalyze. 1. b) List 5 characteristics of enzymes. • Enzymes are biological catalysts that speed up reactions. • They are proteinaceous in nature. • They are highly specific, meaning each enzyme typically catalyzes only one or a few specific reactions. • They are sensitive to temperature, with an optimal temperature range, and can be denatured by high temperatures. • They are sensitive to pH, with an optimal pH range, and can be denatured by extreme pH values. • They are reusable and remain unchanged at the end of the reaction. • They are required in small quantities. 1. c) What is the importance of proteins to man? Proteins are vital for numerous functions in the human body: • Structural Components: They form essential components of tissues like muscles, hair, nails, skin, and connective tissues. • Enzymes: All enzymes, which catalyze nearly all biochemical reactions in the body, are proteins. • Hormones: Many hormones, such as insulin and growth hormone, are proteins that regulate various bodily processes. • Antibodies: Proteins form antibodies, which are crucial for the immune system to fight off infections. • Transport: Proteins like hemoglobin transport oxygen in the blood, while others transport nutrients across cell membranes. • Repair and Growth: They are essential for the repair of tissues and the growth of new cells. • Energy Source: While not the primary energy source, proteins can be broken down to provide energy when carbohydrates and fats are insufficient. 1. d) How would you demonstrate that soaked maize grain contain reducing sugar? To demonstrate the presence of reducing sugars in soaked maize grain, you can use the Benedict's test: Materials: Soaked maize grains, mortar and pestle, distilled water, test tubes, Benedict's solution, water bath, dropper. Procedure: 1. Crush a few soaked maize grains using a mortar and pestle. 2. Add a small amount of distilled water to the crushed maize to make an extract. 3. Filter the extract into a clean test tube to remove solid particles. 4. Add an equal volume of Benedict's solution to the maize extract in the test tube. 5. Place the test tube in a boiling water bath for 5-10 minutes. Observation: If reducing sugars are present, the Benedict's solution will change color from blue to green, yellow, orange, or brick-red precipitate, depending on the concentration of the sugar. Conclusion: A color change to green, yellow, orange, or brick-red indicates the presence of reducing sugars in the soaked maize grain. 2. a) Draw a large labeled diagram to show a section through the mammalian skin. As an AI, I cannot draw diagrams. However, a labeled diagram of mammalian skin would typically show three main layers: • Epidermis: The outermost layer, composed of stratified squamous epithelium. It includes the stratum corneum (dead, keratinized cells) and the stratum germinativum (living cells that divide). • Dermis: The layer beneath the epidermis, made of connective tissue. It contains blood vessels, nerve endings, hair follicles, sebaceous glands (producing sebum), and sweat glands (producing sweat). • Subcutaneous Layer (Hypodermis): The innermost layer, primarily composed of adipose tissue (fat) and loose connective tissue, anchoring the skin to underlying muscles. 2. b) How does the skin function in: i) Protection The skin provides protection in several ways: • It acts as a physical barrier against mechanical injury, harmful chemicals, and the entry of pathogens (bacteria, viruses). • The keratinized cells of the epidermis make it waterproof, preventing excessive water loss from the body and absorption of water from the environment. • Melanin pigment in the skin protects against harmful ultraviolet (UV) radiation from the sun. ii) Secretion The skin secretes substances through glands: • Sweat glands secrete sweat, which is primarily water, salts, and some urea, helping in excretion and cooling. • Sebaceous glands secrete sebum, an oily substance that lubricates the skin and hair, keeping them soft and pliable, and also has some antimicrobial properties. iii) Temperature regulation during over cooling When the body is over-cooling, the skin helps to conserve heat: • Vasoconstriction: Blood vessels in the dermis constrict, reducing blood flow near the skin surface. This minimizes heat loss from the blood to the environment. • Piloerection: Tiny muscles attached to hair follicles contract, causing hairs to stand upright. This traps a layer of insulating air close to the skin, reducing heat loss (though less effective in humans than in furry animals). • Shivering: If cooling persists, muscles throughout the body contract rapidly and involuntarily (shivering), generating heat through increased metabolic activity. 3. a) Distinguish between the following pairs of terms: i) Diploid and Haploid • Diploid cells contain two complete sets of chromosomes (2n), one set inherited from each parent. Most somatic (body) cells in sexually reproducing organisms are diploid. • Haploid cells contain only one complete set of chromosomes (n). Gametes (sperm and egg cells) are haploid. ii) Gamete and Zygote • A gamete is a mature haploid reproductive cell (sperm or egg) that is capable of fusing with another gamete of the opposite sex during sexual reproduction. • A zygote is a diploid cell formed by the fusion of two haploid gametes (sperm and egg) during fertilization. It is the first cell of a new organism. 3. b) A poultry farmer crossed pure breed white fowls with breed black fowls and all offspring fowls produced were blue. When two of these blue fowls were crossed; it was observed that out of the 492 F2 fowls produced, 121 were black, 125 were white and 246 were blue. i) Using appropriate genetic symbols and diagrams, explain the genetics of the crosses. Step 1: Analyze the F1 generation. When pure breed white fowls are crossed with pure breed black fowls, and all F1 offspring are blue, this indicates incomplete dominance. Neither black nor white is fully dominant; instead, they blend to produce an intermediate phenotype (blue). Step 2: Assign genetic symbols. Let C^B represent the allele for black feathers. Let C^W represent the allele for white feathers. The heterozygous genotype C^BC^W will result in blue feathers. Step 3: Parental (P) Cross. Pure breed black fowls have genotype C^BC^B. Pure breed white fowls have genotype C^WC^W. Parental Genotypes: C^BC^B (Black) × C^WC^W (White) Parental Gametes: C^B C^W Punnett Square for P cross: |c|c| & C^W \\ C^B & C^BC^W \\ F1 Genotypes: All C^BC^W F1 Phenotypes: All Blue fowls Step 4: F1 Self-cross (or cross between two F1 blue fowls). Two F1 blue fowls (C^BC^W) are crossed. F1 Genotypes: C^BC^W (Blue) × C^BC^W (Blue) F1 Gametes: C^B, C^W C^B, C^W Punnett Square for F1 self-cross: |c|c|c| & C^B & C^W \\ C^B & C^BC^B & C^BC^W \\ C^W & C^BC^W & C^WC^W \\ F2 Genotypes: 1C^BC^B : 2C^BC^W : 1C^WC^W F2 Phenotypes: 1 Black : 2 Blue : 1 White Step 5: Compare expected F2 ratio with observed F2 results. Total F2 fowls produced = 492. Observed: 121 black, 125 white, 246 blue. Expected ratio for incomplete dominance is 1:2:1. Expected number of Black = (1)/(4) × 492 = 123 Expected number of Blue = (2)/(4) × 492 = 246 Expected number of White = (1)/(4) × 492 = 123 The observed numbers (121 black, 125 white, 246 blue) are very close to the expected 1:2:1 ratio, confirming the pattern of incomplete dominance. ii) What is the expected result of a cross between a blue and black fowl? This cross is between a blue fowl (C^BC^W) and a black fowl (C^BC^B). Parental Genotypes: C^BC^W (Blue) × C^BC^B (Black) Parental Gametes: C^B, C^W C^B Punnett Square: |c|c|c| & C^B & C^W \\ C^B & C^BC^B & C^BC^W \\ Offspring Genotypes: 1C^BC^B : 1C^BC^W Expected Phenotypes: 1 Black : 1 Blue The expected result is a 1:1 ratio of black fowls to blue fowls. **4. a) What do you understand by the following