Three point test cross

**Three point test cross in Drosophila:**

Wild-type Male Drosophila was crossed with female Drosophila homozygous for three recessive X-linked mutations—scute (sc) bristles, echinus (ec) eyes, and crossveinless (cv) wings to obtain F1 progeny.
Wild Male Drosophila= (sc+, ec+, cv+)
Mutated female Drosophila = (sc, ec, cv)
Then F1 progeny were intercrossed to produce F2 flies, which are then classified and counted.
The F1 males carried the three recessive mutations on their single X chromosome. Thus, this intercross was equivalent to a testcross with all three genes in the F1 females present in the homozygous form.
The F2 progeny flies from the intercross comprised eight phenotypically distinct classes, two of them are parental and six recombinant.

Class	Phenotypes	Characters	Genotypes	Counts
1.	Scute, echinus, crossveinless	Parental	Sc, ec, cv	1158
2.	Wildtype (non scute, non echinus, crossvein)	Parental	Sc+, ec+, cv+	1455
3.	scute	Recombinant	Sc, ec+, cv+	163
4.	Echinus, crossveinless	Recombinant	Sc+, ec, cv+	130
5.	Scute, echinus	Recombinant	Sc, ec, cv+	192
6.	Crossveinless	Recombinant	Sc+, ec+, cv	148
7.	Scute, crossveinless	Recombinant	Sc, ec+, cv	1
8.	echinus	Recombinant	Sc+, ec, cv+	1
Total				3248

The parental classes were by far the most numerous (1158+1455=2613). The less numerous recombinant classes each represented a different kind of crossover chromosome.
To figure out which crossovers were involved in producing each type of recombinant, we must first determine how the genes are ordered on the chromosome.
There are three possible gene orders :

Four of the recombinant must have come from a single crossover in one of the two regions of the genes. The other two recombinant must have come from double crossing over—one exchange in each of the two regions. Because a double crossover switches the gene in the middle with respect to the genetic markers on either side of it, it is used for determining the gene order.
Again, intuitively, double crossover occur much less frequently than a single crossover. Therefore, among the six recombinant classes, the two rare ones must represent the double crossover chromosomes
From the given example, the double crossover must have occurred in class 7 (sc ec+cv) and class 8 (sc+ec cv+), each containing a single recombinant F2 progeny.
Comparing these rare recombinant to parental class 1 (sc ec cv) and class 2 (sc+ec+Cv+), the echinus allele has been switched with respect to scute and crossveinless.
Consequently, the echinus gene must be located between the other two.
Therefore the correct gene order is sc–ec–cv.

It is the distance between each pair of gene and it is obtained by estimating the average number of crossovers.
Total map distance between these three genes is map distance between sc and ec plus map distance between ec and

We can obtain the length of the region between sc and ec by identifying the recombinant classes that involved a crossover between these genes.
There are four such classes: class 3 (sc ec+cv+), class 4 (sc+ec cv), class 7 (sc ec+cv), and class 8 (sc+ec cv+).
Classes 3 and 4 involved a single crossover between sc and ec, and classes 7 and 8 involved two crossovers, one between sc and ec and the other between ec and
We can therefore use the frequencies of these four classes to estimate the average number of crossovers between sc and ec:
Average crossover between sc and ec =(163+130+1+1) /3248

=0.091 Morgan

=9.1 centiMorgan or Map unit

Thus, in every 100 chromosomes coming from meiosis in the F1 females, 9.1 had a crossover between sc and ec.
The distance between these genes is therefore 9.1 map units.

In a similar way, we can obtain the distance between ec and cv.
Four recombinant classes involved a crossover in this region: class 5 (sc ec cv+), class 6 (sc+ec+cv), class 7 and class 8.
The double recombinants are also included here because one of their two crossovers was between ec and cv.
The average cross between ec and cv =(192+148+1+1)/3248

=0.105 morgan

= 10.5 centiMorgans or map unit

Total map distance:

Directly calculating the average number of crossovers between these genes:
Recombination frequency (RF)= Non–crossover + Single crossover + Double crossover

= (0)*(1158+1455)/3248 + 1 (163+130+192+148)/3248 + 2 (1+1)/3248

= 0 + 0.195 + 0.0006

= 0.196 Morgan

= 19.6 CentiMorgan

Inference is the phenomenon of inhibition of crossover of by another crossover nearby.
For example, the crossover frequency between sc and ec in region I was (163 +130 +1+1)/3248 =0.091, and crossover frequency between ec and cv in region II was (192+148 +1 +1)/3248 =0.105.
If we assume both crossover are independence of each other, the expected frequency of double crossovers in the interval between sc and cv would be 0.091 *0.105 = 0.0095.
But actual observed frequency of double crossover is (1+1)/3248 = 0006
Double crossovers between sc and cv were much less frequent than expected.
The result suggest one crossover inhibited the occurrence of another nearby, a phenomenon called interference
The extent of the interference is measured by the coefficient of coincidence (C).
Coefficient of coincidence is the ratio of observed frequency to double cross to expected frequency to double cross.
C= (observed frequency of double crossovers)/(expected frequency of double crossovers)

=0.0006/0.0095

C=0.063

Level of inference = 1-C
=1-0.063
=0.937
Because in this example the coefficient of coincidence is close to zero, its lowest possible value, interference was very strong (I is close to 1).
cases:
if a coefficient of coincidence equal to 1; no interference between crossover at all which means the crossovers occurred independently of each other.
If a coefficient of coincidence is equal to 0; very strong inference between crossover therefore double cross do not occur.
** map distance less than 20cM has very strong inference. Thus, double crossovers seldom occur in short chromosomal regions.
The strength of interference is therefore a function of map distance