Question

How would experimental results of the rest cross establish linkage between two autosomal fruit fly genes?

Answer 1

Test crosses are used to test an individual's genotype by crossing it with an individual of a known genotype.A test cross involves mating an unknown genotypic individual with a known homozygous recessive.

The test cross for detection of linkage is somewhat different from a regular test cross. In the former case, the genotypes of both of the parents are known. The parent with the dominant phenotype is known to be a heterozygote (for example,  AaBb) and the tester parent is known to be completely homozygous recessive (aabb). The heterozygous parent with the dominant phenotype, if the genes showed independent assortment, should make four gametes AB, Ab,  aB and ab in equal frequencies. Linkage will be detected if there are more testcross progeny who got the parental gametes from the dominant individual that you would expect by chance alone.

SUPPOSE CONSIDERED A CASE OF DROSOPHILA WITH HAVING BLACK COLOUR GENE = A DOMINATED OVER YELLOW COLOUR = a and C FOR CURLY WINGS AND c for plain wings

Consider the lowly dust rhino again. In the dust rhino one horn (A) is dominant to two horns (b) and  Wrinkled Knees (C) are  dominant to smooth knees (c). We generate a known heterozygote from an F1 cross.

P1	BLACK COLOUR AND CURLY WING (AACC)	X	WHITE COLOUR PLAIN WINGS (aacc)
All F1 cross progeny are AaCc

Since the gametes that make up the double heterozygote are AC and ac, these are the parental gametes. The AaCc individual has the potential to make Ac and aC gametes. These would be the recombinant gametes.

The double heterozygote is crossed to the completely recessive individual with genotype aacc

Test Cross:	(AaCc)BLACK COLOUR AND CURLY WING	X	WHITE COLOUR PLAIN WINGS (aacc)
GAMETE	AC,Ac.aC.ac		ac

GAMETE	ac	observed frequency	expected frequency
AC	AaCc	40	25
Ac	Aacc	10	25
aC	aaCc	10	25
ac	aacc	40	25

The distance or map units is given by the frequency of the recombinant gametes denoted in re in table

10 + .10 = .20   In this case the genes for black colour and curly wings is are 20 map units apart.