The Recombination of Linked Genes
Linked genes do not assort independently since they are found on the same chromosome. If a gray bodied normal winged fruit fly was mated with a black bodied vestigial winged fruit fly, one would expect an equal number of 4 different types of offspring. ( wild type, black-vestigial, gray-vestigial, and black normal). These results would be consistent with the law of independent assortment proposed by Mendel. In actuality the results did not conform to the 1:1:1:1 ratio. Instead the wild type and the black vestigial winged flies accounted for well over 80 per cent of the offspring and the gray-vestigial and black-normal accounted for the remaining 17 percent. The ratios within each group worked out to be 1:1. Independent assortment did not explain these results. These genes are not on different chromosomes as once thought. It appears that the gene for wing type and body color are on the same chromosome. The only explanation that could explain what had happened is a phenomena called crossing over.
Crossing over is a random process that occurs during prophase I of meiosis. It occurs in the tetrads when pieces of two homologous chromosomes switch places with each other. The new chromosomal set up is different from the original homologs found in the cell before prophase I began. Geneticists can use recombination data to map a chromosomes genetic loci. A genetic map can be constructed from the crossing over data. This map is called a linkage map. Let us look at an example of how this process works. A test cross must be conducted in order to demonstrate the process of crossing over. A test cross is the mating of a heterozygous fly for both characteristics with a homozygous recessive fly for both characteristics.
Test cross ( heterozygous X homozygous recessive )
Wild Type ( b+ vg+) Female
Gray -Normal Wings ( b vg)
Black- Vestigial wing ( b vg) Male
( b vg)
b+ vg+ b -vg b+ vg b -vg+ b- vg b- vg b- vg b- vg 961Wild type 949 Black-vestigial 199 Gray-vestigial 192 Black- Normal Parental Types Recombinants
In the above chart a heterozygous wild type heterozygous normal winged female is mated with a black -vestigial winged male. The genotypes of the 4 main types of offspring are given as a combination of the gametes b+vg+ and b vg for the wild type, b vg and b vg for the black vestigial winged ( parental types or the original fly types before recombination)., and b+ vg and b vg gray- vestigial, b vg+ , b vg for the black-normal flies ( recombinants). Looking at the ratios of the parental flies we see it is 1:1. The ratio for the recombinants is also 1:1. But the ratio of the parental types and the recombinants is not 1:1. This indicates that crossing over has occurred.
A linkage map can be constructed based on recombination frequencies. The map is not a to scale map of the chromosome but a sequence of genes along the chromosome. The units do not represent an absolute size and it does not give a precise location of the genes.
Determining recombination frequencies:
1. List the 2 types of offspring ( parental types and recombinants).
2. Place the number of each type of offspring produced under each of the offspring's labels.
3. Total up the offspring produced by that cross.
4. Divide the total into the number of recombinants and move the decimal 2 places to the right to form a percent.
Wild type _______Black- vestigial ______Gray-vestigial _______ Black-normal
---961 _____________949_______________ 199 __________ -------192
391 recombinants X 100 = 17 % 2300 total offspring
The eye color gene is 17 map units away from the wing type gene.