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# Recombinant Mapping

If Curly wings are recessive to normal wings and Barred Eyes is recessive to normal eyes, cross Curly Barred to a normal normal fly. What would be the expected ratios if they were linked or unlinked? If you had 2,500 flies, how many would you expect to see?

#### Solution Preview

Let us use + signs to indicate dominant alleles. In both of these genes, the dominant allele is the normal one.

Therefore, the curly wing allele can be designated as "c" whereas the normal wing allele can be designated as "c+".

Also, barred eyes allele can be designated as "b" and the normal eye allele as "b+".

A "curly:barred" fly would, therefore, be ccbb (double mutation). The fly is homozygous for both genes.

The "normal:normal fly" (herein called simply "normal" or "wild-type") would be c+c+b+b+.

Remember, wild-type doesn't mean "wild" and "bizarre". It just means that is the type found naturally "out in the wild," i.e. in nature. Sometimes, wild-type or normal is simply abbreviated as wt.

So, this is the cross:

ccbb x c+c+ b+b+

(When you write these out on your own, put the + sign as a superscript. It'll look better and make more sense visually.)

Each fly produces maximum amounts of gametes mixing and matching what it can (since these genes are "not linked" at this point -- see below).

The ccbb fly can produce only one type of gamete, cb.
Likewise, the c+c+b+b+ fly can produce only one type of c+b+.

Therefore, we set up the Punnett Square like this:

c+b+
cb c+c b+b

All the F1 would have the same genotype (c+c b+b). They would all have the ...

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