Sunday, November 18, 2012

How to convert a white mutant (white-eyed) Gal4 driver line into a white-plus (red-eyed) background

With Drosophila melanogaster becoming an increasingly popular model organism for neuroscience, many of us are finding it necessary to become more familiar with basic genetic techniques. The popular Gal4/UAS system is often used to express a particular useful gene (sometimes called an effector) in a specific set of cells (provided by a driver line). The problem is that sometimes the driver lines or effector lines have unwanted other mutant genes that were used in creating the line to begin with. For instance, I am currently doing experiments on a driver line (which expresses Gal4 in an interesting set of neurons) that has mutant eye color. That is, not only does it have a transgene (Gal4) inserted into the third chromosome, but it is missing the gene, white, which is ordinarily on the first chromosome. The problem is that white does not only encode eye color, but it is important for development of the eye in general and has other functions in the nervous system. Anne Sustar explains: "The white mutant is a problem for behavioral assays because in addition to being important for production of pigment in the eye, it is an amine transporter that is also likely important for the production of neuromodulators in the brain. The mini-white that's on all transgenes uses an eye-specific promoter.  Thus it rescues eye color, but does not rescue the potential neural and behavioral defects.  More on this story here: http://www.sdbonline.org/fly/genebrief/white.htm." Kain and colleagues recently published a paper that confirms that white-eyed flies can phototax, but have other behavioral abnormalities: Phototactic personality in fruit flies and its suppression by serotonin and white. Kain JS, Stokes C, de Bivort BL. Proc Natl Acad Sci U S A. 2012 Nov 27;109(48):19834-9. doi: 10.1073/pnas.1211988109. Epub 2012 Nov 13.

Now this isn't necessarily a problem if the effector line I cross these flies to has a functional version of the white gene. But if the effector line is also mutant for white, my eyes will have abnormal vision, totally independent of any effects of the effector gene in my neurons of interest. In order to do experiments that involve vision (which is important for a large proportion of behaviors) I need to get my driver line to have a normal version of the white gene.

So, we have a Gal4 driver line with Gal4 on the third chromosome and white eyes. The males have a genotype something like this:

1. w-/Y;+/+;G/G

Where G is the Gal4 gene and promoter.

We want to cross this to a virgin female with wild-type, (i.e. white-plus, or red-colored) eyes (that is, a line with the wild-type allele for the gene white on the first chromosome. On the third chromosome we will need balancer chromosomes. This could be something like this:

2. w+/w+;+/+;Sb/Tb

Here Sb stands for stubble, a gene with a dominant phenotype of short bristles. Tb stands for tubby, a dominant phenotype for a short "tubby" pupal case and larval body shape. This line is stable because each of these genes are homozygous lethal: a fly homozygous for either one does not grow to adulthood. Hence, all the flies in this bottle have the above genotype.

When we cross these two flies, we can get two possible male genotypes in the offspring:

3. w+/Y;+/+;Sb/G
4. w+/Y;+/+;Tb/G

Both have red eyes. 3 are stubble (have stubble? are stubbly?). 4 are tubby. For our purpose we could use either 3 or 4. Let's say we pick 3--we select males that have stubble. We now cross these with a virgin female of genotype 2 (above).

The female offspring could be any of four possible genotypes:

5. w+/w+;+/+;Sb/Sb
6. w+/w+;+/+;Sb/G
7. w+/w+;+/+;Tb/G
8. w+/w+;+/+;Sb/Tb

Now, remember stubble is homozygous lethal, so there will actually be zero offspring with genotype 5. For our purposes, either 6 or 7 would do. Let's say we pick 6. We have to select flies with stubble but that are not tubby. Now we can either just leave those in a bottle together, and when we choose experimental flies down the line make sure to pick flies that do not have stubble. It is also possible that stubble will be eliminated on its own, through a slight fitness cost. In order to make a true homozygous stock, however, it would probably be best to cross some of genotype 6 together, then select flies that do not have stubble. These we can be sure have the genotype

9. w+/w+;+/+;G/G

which is homozygous for both white and our Gal4 gene and promoter.

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