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),
Each will appear in half of the double crossovers

So, expected percentage of ABC gametes = 1.2 / 2 = 0.6%

Final answers:

Gene order: A–B–C
Expected percentage of ABC gametes: 0.6%





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