A wild-type fruit fly that is heterozygous for both the black body ( b ) and vestigial wings ( vg )

Let R represents the vg gene.

The first species has the genotype: BbRr

The second species is bbrr

Black body has the genotype: bb
Vestigial wings has the genetype: rr


Of the 500...

199 / 500 are: BB/Bb and RR/Rr
42 / 500 are: BB/Bb and rr
48 / 500 are: bb and RR/Rr
211 / 500 are: bb and rr.

Let's look at the punnet squares:

[attached]

Notice, 25% are BbRr -- Wild-type body, wild-type eyes, but in the statistics it's 39.8% (199 / 500). If they were found on separate chromosomes, the numbers would be more lose to theoretical ones.

For example, if the genes Aa and Bb occur on the same chromosome then a double heterozygote AaBb formed from Ab and aB gametes would yield a significantly larger number of Ab and aB gametes than AB or ab gametes. Similarly, if the parental gametes were AB and ab, the heterozygote would generate AB and ab gametes frequently, and Ab and aB gametes rarely. This is in complete contrast to independent assortment where the genes are on different chromosomes and therefore produce all types of gamete equally frequently. When genes are linked, the two less frequent gametes are produced as a result of crossing over. As crossing over is relatively rare the genes on one chromosome tend to be inherited together.

Hope this helps!