This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.
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Here is the solution to question 6. QUESTION 6 a) State the Hardy-Weinberg equation and define each term. (04 marks) The Hardy-Weinberg equations are: p + q = 1 p^2 + 2pq + q^2 = 1 Where: p: Represents the frequency of the dominant allele (e.g., A). q: Represents the frequency of the recessive allele (e.g., a). p^2: Represents the frequency of the homozygous dominant genotype (e.g., AA). q^2: Represents the frequency of the homozygous recessive genotype (e.g., aa). 2pq: Represents the frequency of the heterozygous genotype (e.g., Aa). b) In a population of 1000 individuals, 360 are homozygous dominant (AA), 480 are heterozygous (Aa), and 160 are homozygous recessive (aa). Calculate the: Total individuals = 1000 Number of AA = 360 Number of Aa = 480 Number of aa = 160 i. Allele frequencies of A and a. (04 marks) Step 1: Calculate the total number of alleles in the population. Total alleles = 2 × Total individuals = 2 × 1000 = 2000 Step 2: Calculate the number of A alleles. Number of A alleles = (2 × Number of AA) + (1 × Number of Aa) Number of A alleles = (2 × 360) + (1 × 480) = 720 + 480 = 1200 Step 3: Calculate the frequency of allele A (p). p = (1200)/(2000) = 0.6 Step 4: Calculate the frequency of allele a (q). q = 1 - p = 1 - 0.6 = 0.4 The allele frequencies are: Frequency of A (p): 0.6 Frequency of a (q): 0.4 ii. Genotype frequencies of AA, Aa, and aa. (04 marks) Step 1: Calculate the frequency for each genotype. Frequency of AA = (360)/(1000) = 0.36 Frequency of Aa = (480)/(1000) = 0.48 Frequency of aa = (160)/(1000) = 0.16 The genotypic frequencies are: Frequency of AA: 0.36 Frequency of Aa: 0.48 Frequency of aa: 0.16 iii. Phenotype frequencies of dominant and recessive traits. (04 marks) Step 1: Calculate the frequency of the dominant phenotype (AA + Aa). Frequency of dominant phenotype = Frequency of AA + Frequency of Aa = 0.36 + 0.48 = 0.84 Step 2: Calculate the frequency of the recessive phenotype (aa). Frequency of recessive phenotype = Frequency of aa = 0.16 The phenotype frequencies are: Frequency of dominant trait: 0.84 Frequency of recessive trait: 0.16 iv. Number of homozygous dominant, heterozygous, and homozygous recessive individuals in the next generation, assuming the population is in Hardy-Weinberg equilibrium. (04 marks) Using the allele frequencies p = 0.6 and q = 0.4 from part (i) and the total population of 1000 individuals: Step 1: Calculate the expected frequency of homozygous dominant (AA) individuals (p^2). p^2 = (0.6)^2 = 0.36 Number of AA individuals = 0.36 × 1000 = 360 Step 2: Calculate the expected frequency of heterozygous (Aa) individuals (2pq). 2pq = 2 × 0.6 × 0.4 = 0.48 Number of Aa individuals = 0.48 × 1000 = 480 Step 3: Calculate the expected frequency of homozygous recessive (aa) individuals (q^2). q^2 = (0.4)^2 = 0.16 Number of aa individuals = 0.16 × 1000 = 160 The number of individuals in the next generation, assuming Hardy-Weinberg equilibrium, are: Homozygous dominant (AA): 360 individuals Heterozygous (Aa): 480 individuals Homozygous recessive (aa): 160 individuals Last free one today — make it count tomorrow, or type /upgrade for unlimited.