The distance between the genes A and B is 15 map unit, B and C 8 map unit and A and C 23 map unit. In an individual of genotype AbC/aBc, what will be the order of gene? What will be the expected percentage of gametes with the genotype ABC?
Question: The distance between the genes A and B is 15 map unit, B and C 8 map unit and A and C 23 map unit. In an individual of genotype AbC/aBc, what will be the order of gene? What will be the expected percentage of gametes with the genotype ABC? Given: A–B = 15 map units B–C = 8 map units A–C = 23 map units Check if A–B–C fits: A–B + B–C = 15 + 8 = 23 So, gene order is: A–B–C Genotype of individual: AbC / aBc This is a double heterozygote and the arrangement of alleles shows coupling and repulsion between different loci. Parental chromosomes: AbC (from one parent) aBc (from another parent) Now we determine the expected frequency of ABC type gamete. To get ABC, recombination must occur in both segments: Between A and B Between B and C So, this is a double crossover product. Double crossover frequency = (distance A–B) × (distance B–C) = (15/100) × (8/100) = 0.15 × 0.08 = 0.012 = 1.2% But since there are two possible double crossover gametes (ABC and abc)...