Now my question is-
How will I get to know whether my two genes are tightly linked or not?
For example, if I tell you the genomic positions of my genes are 10.0 and -9.0 Does that mean they are not closely linked?
And from this information how can we determine how many animals we have to pick to get the double mutant phenotype?
If two genes are positioned on the genetic map at 10.0 and -9.0 on the same chromosome, then the first gene is 10 map units to the right of zero and the second is 9 map units to the left of zero. Thus, the total distance between them is 19 map units.
You might want to read some relevant parts of WormBook, especially David Fay’s chapters on “Forward genetics and genetic mapping”. From the first of his chapters,:
The genetic distance separating two genes (or any two points on a chromosome) is determined by the frequency of meiotic recombination that takes place between them. The nearer the two genes are to each other, the less likely that a recombination event will occur in that span. One (1.0) map unit (also sometimes called a centiMorgan; cM) is equal to a 1% meiotic recombination frequency. In other words, if on average 1% of all gametes (sperm or oocytes) have experienced a recombination event between two particular genes, then these genes are considered to be 1.0 map unit apart
For your example, if we call your genes gene-2 and gene-1 and you start with a trans-heterozygote (genotype gene-1(lf) gene-2(+)/gene-1(+) gene-2(lf)), and as I’ve said the markers are 19 map units (19% recombination frequency) apart, and you pick out the gene-1(lf) homozygotes from the self-progeny, meaning that you’re looking only at those progeny, which are 1/4 of the total progeny, then your progeny classes are:
(1-0.19)(1-0.19) = 65% received two parental gene-1(lf) gene-2(+) chromosomes
2(1-0.19)(0.19) = 31% received one parental (gene-1(lf) gene-2(+)) and one recombinant (gene-1(lf) gene-2(lf)) chromosome
(0.19)(0.19) = 4% received two recombinant (gene-1(lf) gene-2(lf)) chromosomes
(This may be easier to visualize using a Punnett Square).
So, you have two choices: you can pick enough gene-1(lf) homozygotes that one of them will probably be homozygous for gene-2(lf) as well, or you can pick enough so that one will be heterozygous for gene-2(lf) and look among the progeny of each animal you picked until you find one 1/4 of whose progeny are double homozygotes.
If you pick n gene-1(lf) animals, the chance that none of the n will be double homozygotes for gene-1(lf) and gene-2(lf) is (1-0.04)^n. The chance that none will carry the gene-2(lf) mutation at all is about (0.65)^n. So if you pick a dozen gene-1(lf) animals, there’s about a 60% chance none of them will be a double homozygote, but there’s a 99.5% chance one will be heterozygous or homozygous for gene-2(lf) and so will give at least 1/4 doubly homozygous progeny.