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.

Gene order:

• 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 :

1. sc—ec—cv
2. ec—sc—cv
3. ec—cv—sc

• 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.

Map distance:

• 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

i. Map distance between sc and ec:

• 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.

ii. Map distance between ec and cv:

• 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:

• Combining the data for the two regions, the map is sc—9.1— ec—10.5— cv
• Thus map distances between sc and cv= 9.1 cM +10.5 cM =19.6 cM

Alternative way of calculating 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 and coefficient of coincidence:

• 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):

• 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

Three point test cross: gene order, map distance, inference and coefficient of coincidence and level of inference

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• Gene mapping is the process of determining the genes and their location along the length of chromosome.
• T. D Morgan pave the foundation of gene map by identifying gene for white eye Drosophila on X-chromosome of mutant. Later his students able to locate other X-linked gene on X-chromosome.
• The procedure of gene mapping was developed by Alfred H Sturtevent. His procedure is based on the principle of linkage. The gene located on same chromosome inherits together known as linked gene. However, some gene on same chromosome could separate during meiosis and new combination of genes are formed. The phenomenon of recombination is due to crossover and chaismata formation during meiosis.
• Gene map is by counting the number of crossovers that occur during meiosis. However, because the actual crossover events cannot be seen, they cannot count them directly. So, recombination frequency is calculated to estimate the crossover.
• Chiasmata are counted through cytological analysis, whereas recombinant chromosomes are counted through genetic analysis.

Gene map distance:

• Gene map distance is the distance between points on a chromosome which can be estimated by counting the number of crossovers between them. Therefore, the distance between two points on the genetic map of a chromosome is the average number of crossovers between them. Genetic map distances are, in fact, based on such averages.
• Points that are far apart should have more crossovers between them than points that are close together. However, the number of crossovers must be understood in a statistical sense.
• In any particular cell, the chance that a crossover will occur between two points may be low, but in a large population of cells, this crossover will probably occur several times simply because there are so many independent opportunities for it. Thus, the quantity that we really need to measure is the average number of crossovers in a particular chromosome region.
• let us consider 100 oogonia undergoes gametogenesis (meiosis).
• In some cells, no crossovers will occur between sites A and B; in others, one, two, or more crossovers will occur between these loci.
• At the end of meiosis, there will be 100 gametes, each containing a chromosome with either zero, one, two, or more crossovers between A and B.
• We estimate the genetic map distance between these loci by calculating the average number of crossovers in this sample of chromosomes.
• The result from the data is 0.42.

Two point test cross in Drosophila:

• When wild-type Drosophila Females were mated to Males homozygous for two autosomal mutations—vestigial (vg) short wings, and black (b) body coloration. Ie. Female (vg+vg+ , b+b+) and male (vg vg, b b)
• All the F1 flies had long wings and gray bodies; thus, the wild-type alleles (vg+ and b+) are dominant.
• The F1 female progeny were then testcrossed to vestigial winged black body males (vg b), and the F2 progeny were obtained and there are classified on the basis of phenotypic characters and counted.
• There were four phenotypic classes, two abundant and two rare. The abundant classes had the same phenotypes as the original parents (vestigial wing black body and long wing and grey body), and the rare classes had recombinant phenotypes (ie. vestigial wings with grey body and long wings with black body).

• Number of F2 Progeny with vestigial wing and black body (vg vg, b b)= 405
• Number of F2 progeny with long wing and grey body (vg+ vg+, b+ b+)= 415
• Number of f2 progeny with vestigial wing and grey body (vg vg, b+ b)= 92
• Number of F2 progeny with long wing and black body (vg+ vg, b b)=88
• The genes for vestigial wings and black body are linked because the number of recombinants are much fewer than 50 percent of the total progeny counted in F2 generation. Therefore, these genes must be on the same chromosome.

Map distance:

• Map distance is the distance between genes.
• To determine the distance between the genes for vestigial wing and black body, we must estimate the average number of crossovers in the gametes of the doubly heterozygous F1 females (vg+vg, b+b)
• Average crossover is estimated by calculating the frequency of recombinant F2 progeny
• The average number of crossovers in the whole sample of progeny is therefore,

Frequency of recombination (RF) = (0) (415+405)/1000 + 1 (92+88)/1000

=0.82+0.18

=0.18 Morgan

=18 centimorgan or map unit

• This simple analysis indicates that, on average, 18 out of 100 chromosomes recovered from meiosis had a crossover between vg and b.
• Thus, vg and b are separated by 18 units
• 100 centiMorgans equal one Morgan (M). Therefore vg and b are 18 cM (or 0.18 M) apart.
• The map distance is equal to the frequency of recombination.

Gene mapping: two point test cross, map distance and frequency of recombination

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