In Drosophila, a cross was made between females expressing the three X-linked recessive traits: scute (sc) bristles, sable (s) body and vermillion (v) eyes, and wild-type (+) males. In the F1, all females were wild-type, while all males expressed all three recessive traits. The cross was carried to the F2 generation and 1000 offspring were counted with the results shown below.
sc s v 314
+ + + 280
+ s v 150
sc + + 156
sc + v 46
+ s + 30
sc s + 10
+ + v 14
Determine the linear sequence of these three genes.
What is the map distance between these genes?
Is there evidence of interference? If so, calculate it.
Because these traits are X-linked the genotype will be affected by the gender of the fly. For the three traits given (scute (sc) bristles, sable (s) body and vermillion (v) eyes) females can be homozygous normal (+/+ +/+ +/+), wild-type heterozygotes (sc/+ s/+ v/+), or homogygous mutant (sc/sc s/s v/v). Males on the other hand only have one X chromosome and will either be heterozygous normal (+/o +/o +/o) or heterozygous mutant (sc/o s/o v/v). In cases of X-linked genes it is handy to keep track of the males and females with the X and Y chromosomes. In the case of flies there is either two X chromosomes (female) or one X chromosome (male).
In the F1 cross the female is homozygous mutant X(sc s v) X(sc s v) while the male is heterozygous normal X(+ + +). The Punnett square would look like this:
X(sc s v) X(sc s v)
X(+ + +) X(sc s v) X(+ + +) X(sc s v) X(+ + +)
o X(sc s v) o X(sc s v) o
Half of the offspring would be heterozygous females X(sc s v) X(+ + +) and ...
Calculation of the physical order of three genes as well as distance between them and interference. Web references provided.