In: Biology
In Drosophila, the recessive dp allele of the dumpy gene produces short, curved wings, while the recessive allele bw of the brown gene causes brown eyes. In a testcross using females heterozygous for both of these genes, the following results were obtained:
wild-type wings, wild-type eyes 178
wild-type wings, brown eyes 185
dumpy wings, wild-type eyes 172
dumpy wings, brown eyes 181
In a testcross using males heterozygous for both of these genes, a different set of results was obtained:
wild-type wings, wild-type eyes 247
dumpy wings, brown eyes 242
a. What can you conclude from the first testcross?
b. What can you conclude from the second testcross?
c. How can you reconcile the data shown in parts a and b ? Can you exploit the difference between these two sets of data to devise a general test for synteny in Drosophila?
d. The genetic distance between dumpy and brown is 91.5 m.u. How could this value be measured?
a) and b) looking at the number of progeny with different phenotypic characteristics of testcross, first test cross suggest there is a recombination phenomenon took place in female Drosophila heterozygous for both the trait. whereas there is no recombination in male Drosophila.
c) From the given data, one can predict that both the genes are in case of female Drosophila may be present on the same chromosome but they may (distantly) or may not be linked (Ratio is pretty much near to 1:1:1:1). whereas in male Drosophila both the genes are very closely linked (very close linkage result in less chance of recombination). Therefore, general test in which a panel of cell lines, each containing just a few Drosophila chromosomes, can use for synteny testing in which the presence or absence of a specific gene product is correlated with the presence or absence of each chromosome. for synteny in Drosophila.
d) genetic distance between two genes can be calculated by = (no of recombinant progeny)/total number of progeny
However in case of linkage number of recombinant progeny is always less than or equal to 50% of total population. If the number of recombinants is more than 50% than it is considered as they are independently assorted.
even in the example of the first testcross the genetic distance between dumpy and brown is (185+172)/(178+185+172+181) = 0.498 i.e is 49.8mu
so if we consider that dumpy and brown is 91.5m.u. then we have to find the genetic distance between of those gene who lies between these two genes, then by adding those distance, we can calculate the distance between such distantly apart gene if present in the same chromosome.