Now fill in the two by eight Punnett square containing the male and female gametes from the F1 parents with the expected genotypes of the F2 progeny resulting.

At this stage it will be clear that the three genes are linked to each other on the X chromosome. The male gametes will bear either an X chromosome or the Y chromosome. On the other hand, the female gametes will bear X chromosomes carrying one of eight combinations of alleles of the three genes. Two of the eight will be parental (P) combinations, and six recombinant (R) combinations. Label the genotypes that correspond to recombinant combinations on your Punnett square. The latter have arisen from meioses in which cross-overs have occurred between the loci of the three genes in the female F1 flies.  (Remember we are considering the products of meiosis from the triply heterozygous F1 female, and "parental" relates to P1 and P2 parents).

IMPORTANT: When completing this table, make sure that you follow these instructions:

1. List the alleles in the order of yellow (y) forked (f) cross veinless (cv) as in the example "yfcv".  
2. Indicate the wild type allele by using a (+), such as y+f+cv would indicate wild type yellow and forked, mutant cross-veinless. 
3. When completing the progeny genotype, always list the female X genotype first.
4. Don't use any spaces or other symbols in your answers