7 03 2005 LECTURE 21 EUKARYOTIC GENES AND GENOMES II In the last lecture we considered the structure of genes in eukaryotic organisms and went on to figure out a way to identify S cerevisiae genes that are transcriptionally regulated in response to a change in environment The ability to regulate gene expression in response to environmental cues is a fundamental requirement for all living cells both prokaryote and eukaryote We considered how many genes each organism has about 4 000 for E coli 6 000 for yeast and a little over 20 000 for mouse and humans But only a subset of these genes is actually expressed at any one time in any particular cell For multicellular organisms this becomes even more apparent it is obvious that skin cells must be expressing a different set of genes than liver cells although of course there must be a common set of genes that are expressed in both cell types these are often called housekeeping genes There are a number of ways that gene regulation in eukaryotes differs from gene regulation in prokaryotes Eukaryotic genes are not organized into operons Eukaryotic regulatory genes are not usually linked to the genes they regulate Some of the regulatory proteins must ultimately be compartmentalized to the nucleus even when signaling begins at the cell membrane or in the cytoplasm Eukaryotic DNA is wrapped around nucleosomes Today we will consider how one can use genetics to begin to dissect the mechanisms by which gene transcription can be regulated For this we will take the example of the yeast GAL genes in S cerevisiae GALACTOSE METABOLISM IN YEAST Reaction Enzyme Gene Galactokinase GAL1 Galactose transferase GAL7 Galactose epimerase GAL10 D galactose D galactose 1 phosphate UDP D galactose UDP D glucose D glucose 1 phosphate GAL1 encoded UDP glucose Phosphorylase Phosphoglucomutase D glucose 6 phosphate GLYCOLYSIS GAL1 GAL7 GAL10 transcription all induced in the presence of glucose How is this achieved Once a gene has been identified as being inducible under certain inducing conditions in this case in the presence of galactose we can begin to dissect the regulatory mechanism by isolating mutants i e mutants that constitutively express the GAL genes even in the absence of galactose and mutants that have lost the ability to induce the GAL genes in the presence of galactose If we were studying galactose regulation today we would probably use a lacZ reporter system as we discussed in the last lecture However when the Gal regulatory system was fist genetically dissected it was done by actually measuring the induction of Gal1 encoded galactokiase activity so this is how we will discuss the genetic dissection of the system Mutagenized GAL1 Tn7lacZ fusion strain grown on GLYCEROL Another approach is to simply measure galactokinase activity in the presence or absence of Galactose GALACTOSE X Gal GLYCEROL X Gal Constitutive Uninducible What we know is that Gal4 mutants are uninducible and that Gal80 and Gal81 mutants constitutively express the Gal1 galactokinase gene along with the other Gal genes Let s analyze each mutant in turn Gal4 mutant It was first established that like Gal1 the Gal4 mutant phenotype is recessive because heterozygous diploids generated by mating Gal4 to wild type have normal regulation It was then established that the mutation in the Gal4 strain lies in a new gene and not simply in the GAL1 galactokinase gene Gal1 mutants don t express galactokinase activity in the presence of galactose just as was seen for the Gal4 mutant That and Gal4 mutants have Gal1 mutations in different genes was shown by complementation analysis diploids from mating Mat Gal4with Mata Gal1 behave like wild type and the fact that the GAL4 and GAL1 genes are unlinked was established by tetrad analysis You should think about what the tetrads from the aforementioned diploids would look like Put together the simplest model is that Gal4 is a positive regulator of Gal1 and the other Gal genes The sign indicates that Gal4 increases Gal expression but does not indicate whether this is direct or indirect Gal80 mutant The next useful regulatory mutant isolated was Gal80 in which the Gal1 encoded galactokinase is expressed even in the absence of galactose and is not further induced in its presence Again heterozygous diploids Gal80 wt showed that Gal80 is recessive Tetrad analysis showed that Gal80 is not linked to Gal1 Gal4 or any of the Gal genes If a mutant Gal80 results in constitutive Gal1 expression the simplest model is that Gal80 negatively regulates the Gal genes Since Gal4 positively regulates and Gal80 negatively regulated Gal1 expression we have to figure out how these two gene products work together to achieve such regulation Assuming that Gal4 and Gal80 act in series there are two formal possibilities Model 1 Model 2 Model 1 is that Gal4 positively regulates Gal1 and that Gal80 negatively regulates Gal4 the presence of galactose somehow inhibits Gal80 function thus releasing Gal4 to positively activate Gal1 expression Model 2 is that Gal80 negatively regulates Gal1 and Gal4 negatively regulates Gal80 here the presence of galactose positively activates Gal4 which in turn negatively regulates Gal80 thus relieving inhibition of Gal1 expression We can distinguish between these two models by doing what s called an epistasis test to establish the epistatic relationship between Gal4 and Gal80 This involves making a double Gal4 Gal80 mutant strain The phenotype of the double mutant will indicate which of the two models is most likely to be true take a look at the two models to predict what phenotype the double mutant should have For Model 1 the double mutant would become uninducible for Model 2 the double mutant should be constitutive We could make the Gal4 Gal80 double mutant strain using molecular biological approaches but an easier way is to let yeast meiosis do the job for you If we mate the Gal4 Gal80 haploid strain with the Gal4 Gal80haploid strain we know these two genes are unlinked we should obtain double mutants among the tetratype and non parental ditype tetrads that result from this cross Parental Ditype Tetratype Non Parental Ditype Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 Gal4 Gal80 These results clearly support Model 1 i e because the double mutant is uninducible rather than constitutive Gal4 liklely behaves as a positive activator of Gal1 expression and in the absence of
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