New & Noteworthy

Strung Out Chromosomes and Coyotes

January 29, 2018


As Wile E. Coyote knows, when setting a bear trap, you need to make sure the tension is just right. As you can see in this video, if there isn’t enough, then the Road Runner gets away:

And of course, Wile E. ends up getting caught in the trap!

Like our super-genius coyote friend, Saccharomyces cerevisiae cells need to make sure there is the right amount of tension sometimes, too. One of those times, for example, is tension on their chromosomes at the metaphase stage of cell division.

If the chromosomes aren’t attached to the spindle in just the right way, the cell will not let mitosis move from metaphase to anaphase. The cell won’t put down the birdseed until it “knows” the trap is set properly.

While Wile E. Coyote probes the tension by eyeballing it, yeast does it in a much more thorough way. Two key pieces of the sensing machinery in yeast are the tension sensing motif (TSM) from histone H3, 42KPGT, and the Shugoshin 1 protein, Sgo1p.

The current model is that the TSM of histone H3 acts as an anchoring point or a landing pad for Sgo1p, the protein responsible for sensing that there is the right amount of tension between chromosomes and the spindle/kinetochore. When the TSM is mutated with, for example, a G44S allele, the idea is that Sgo1p can no longer bind causing chromosomes to sometimes segregate incorrectly.

Previous work had shown that mutating Gcn5p can partially rescue the effects of the G44S allele. In a new study in GENETICS, Buehl and coworkers provide evidence that the reason it can has to do with the tail of histone H3. Basically, this tail can act as a secondary anchor if the TSM, the first anchor point, is mutated. But the tail can only work if it is not acetylated by Gcn5p.

CoyoteRoadrunner

Bad things can happen when plans go astray for Wile E. Coyote and for chromosome segregation in yeast. (Wikimedia)

It is as if Wile E. Coyote’s bear trap is broken so he has to fashion a new trigger to get it to snap. When the TSM of H3 is gone, yeast can fashion a new sensor by eliminating the acetylation on the tail of the H3 histone.

Buehl and coworkers focused on the tail because it is a known target of Gcn5p acetylation. In particular, there are four lysine residues in the tail—K9, K14, K18, and K23—that are acetylated by Gcn5p to varying degrees. The most acetylated is K14 followed by K23 and K18. K9 is hardly acetylated at all.

When they deleted the histone tail, they found that the effects of the TSM mutation, G44S, were more severe. But when they mutated the four lysines all at once, the effects of G44S weren’t as bad. This is consistent with the unacetylated tail serving as a landing pad for Sgo1p in the absence of the TSM. No tail, no binding, leading to chromosome mis-segregation.

As might be expected if this was linked to Gcn5p, each individual lysine mutant had effects that corresponded to their level of acetylation by Gcn5p. So K14A did the best job of partially rescuing the G44S allele, K18A and K23A did an okay job and K9A had very little effect.

Buehl and coworkers provide three additional sets of experiments to show that G44S causes chromosome mis-segregation and that K14A can partially rescue it. There is only time to discuss one of these, but you can read the paper for the other two.

This strategy uses the idea that diploids cannot mate because of a repression function on each copy of chromosome III (chrIII). If a diploid loses one of its chrIII’s, then it will be able to mate.

As expected, the authors found that both wild type and the K14A mutant had very few diploids that could mate. But it was a different story for the G44S mutant—it had many more diploids able to mate, presumably due to the loss of one chrIII. The double mutant K14A/G44S had more diploids mate than wild type but less that G44S on its own.

This is what we would expect if K14A can only affect tension sensing in a TSM mutant. It would do little on its own but could partially rescue the TSM mutant G44S.

ROADrunner

If Wile E. Coyote had a backup plan like yeast does for chromosome segregation, he might get to dine on the Road Runner. (Rosenfeld Media)

The authors next used chromatin immunoprecipitation (ChIP) assays to directly show that the K14A mutation in histone H3 allowed Sgo1p to bind when the TSM was mutated. There was one hitch though—Sgo1p bound most everywhere and not just at the pericentromeres where it usually does.

In a final set of experiments, the authors used GST-pulldowns to show a direct interaction between the H3 tail and Sgo1p. To their surprise, they found that acetylating the tail did not affect Sgo1p binding. They suggest that perhaps the acetylated tail interacts with something that blocks Sgo1p from binding.

Yeast has a backup just in case its main way of sensing tension between the spindle/kinetochore and chromosome, the TSM of histone H3, breaks down. In that case, it can use the tail of histone H3, assuming it is not acetylated.

If only Wile E. Coyote included backups like this in his crazy plans! Then he could finally dine on that tasty Road Runner he is always chasing.

by Barry Starr, Ph.D., Director of Outreach Activities, Stanford Genetics

Categories: Research Spotlight

Tags: acetylation , Histone H3 , chromosome segregation , metaphase

Sweet or Salty? It’s Hard to Tell Just By Looking

March 05, 2015


Just as you need to be careful when adding any white granulated substance to your cereal, you should also be careful assuming that orthologs from related species do the exact same thing. Image via Wikimedia Commons

If you have ever accidentally added salt to your coffee, you know that sugar and salt are very different things even though they look pretty much the same. Turns out that genes can sometimes be this way too. They can look similar at the DNA level but have very different functions.

A great example of this can be found in a new study in GENETICS by Varshney and coworkers. They found that a protein kinase in Candida albicans, Sch9, is important for ensuring that chromosomes end up in the right place when this yeast reproduces by budding.

Turns out that the same is not true for the Sch9 ortholog in our favorite yeast Saccharomyces cerevisiae. There is no evidence that Sch9 has anything to do with chromosome segregation there, even though the Sch9 sequences in these two yeasts look very similar.

C. albicans Sch9 is very important for keeping filamentous growth at bay under certain conditions (hypoxia and high levels of carbon dioxide). To understand better how Sch9 does this, Varshney and coworkers used chromatin immunoprecipitation (ChIP) to figure out where the protein binds in the genome. They were surprised when they found that it bound mostly to centromeres.

Despite this binding, the authors saw no evidence that Sch9 was involved in stabilizing the kinetochore, the protein structure that forms at the spindle of sister chromatids. When a kinetochore is destabilized, a cell’s nuclear morphology changes, its centromeres decluster during the cell cycle, and the centromeric histone Cse4 delocalizes away from its centromeres. The authors saw none of these things in a C. albicans strain in which the SCH9 gene was deleted.

They did, however, find that C. albicans cells lacking Sch9 had anywhere from a 150 to a 750-fold increase in chromosome loss. They found this by using a strain of C. albicans that had an arginine marker on one copy of its chromosome 7 and a histidine marker on the other, and looking for how often cells lost one of the two markers. From this the authors concluded that like many other kinetochore associated proteins, Sch9 is involved in chromosome segregation.

As a final experiment, Varshney and coworkers used ChIP to see if the Sch9 protein bound to centromeres in S. cerevisiae. It did not. While the authors did not directly test whether Sch9 had any effect on chromosome segregation in S. cerevisiae, the presumption is that it didn’t, as it doesn’t appear to interact with centromeres and no such effect has been seen previously.

But Sch9 isn’t completely different in the two yeasts. A close look at the ChIP data showed that Sch9 bound the rDNA locus in both C. albicans and S. cerevisiae.

How did orthologous proteins in two budding yeasts end up with such different functions? One idea is that the ancestral gene to Sch9 was important for rDNA regulation and that it later gained a function in chromosome segregation in C. albicans. Another possibility is that the ancestral gene had both functions and that centromere binding was lost in S. cerevisiae. More work will need to be done to tell the difference.

Whichever explanation is correct, this study reminds us that, just like sugar and salt, even if two genes look similar they may have quite different functions. Assuming that similar appearance means identical function may lead to an experimental result that is just as unpleasant as salty coffee!

by D. Barry Starr, Ph.D., Director of Outreach Activities, Stanford Genetics

Categories: Research Spotlight

Tags: ortholog , Saccharomyces cerevisiae , chromosome segregation , Candida albicans

As Good as the Original

December 07, 2012

Unlike the rest of us, cells are happy with either a copy or the original.

If you got to choose between an original and a copy of that original, you’d undoubtedly choose the original. Because of mistakes during the copying process, the original is bound to be superior.

Cairns was the first scientist to try to apply the same logic to cells like mother cells of the budding yeast S. cerevisiae, or stem cells. The idea is that since these cells make lots of copies of themselves, they might have some mechanism for keeping the original DNA and sending the copies to the new cells. This would protect the original DNA from building up mutations.

While this seems reasonable, the many studies done to date have failed to provide any compelling evidence to support the idea. When scientists look at a cell’s DNA in bulk, they see it randomly dividing between mother and daughter. But this is not the end of the story.

Another idea people have had is that one of the strands of the double helix is kept in its original form in the mother while the other is free to be passed on to the daughter. There is mounting evidence in yeast and mammalian cells that there is some strand-specific selection going on. But in a new study out in GENETICS, Keyes and coworkers show that this selection is not dependent on a strand being an original as opposed to a copy.

If mother cells preferentially keep one of the strands, then the red ones should only end up in the MD and DM cell types in this experiment. The real results were that all four cell populations had the red DNA. (Image from Keyes et al. used with permission from GENETICS)

They use a couple of cool tricks to be able to follow specific strands of DNA in specific cell types in these experiments. First, they engineered a strain that would incorporate BrdU (bromodeoxyuridine) into newly synthesized DNA, so that they could distinguish between the original and copied strands. Second, they used a technique to separate mother cells from daughter cells that involves arresting the mother cells with alpha factor, labeling them with biotin, and then allowing them to divide: the mothers can be pulled away from the daughters by their biotin labels.

As you can see from the image on the right, Keyes and coworkers let a population of biotin-labeled mother cells divide once in the presence of BrdU. Next they separated mothers (M) from daughters (D) and biotinylated the daughters.

Next they let the two populations divide separately one more time in unlabeled thymidine (TdR) instead of BrdU, and separated mothers from daughters again. As shown in the figure, this allowed them to isolate four different cell populations:

1) MM: the original mother cells
2) MD: daughters of the original mothers
3) DM: mother cells derived from the first daughter
4) DD: daughters of the first daughter

The image shows the prediction if the Watson strand (W) is preferentially kept by the mother. As you follow along, remember that the red strand represents the labeled DNA.

If mother cells inherit the Watson strand specifically (W) and the daughter cell always gets the Crick strand (C), then we would predict that only MD and DM cells should have BrdU DNA. The other two cell types should only have unlabeled DNA.

Let’s follow the mother side to see why. The mother cell would inherit the unlabeled Watson strand and a labeled Crick strand because she would keep the original from the first division. The labeled and so copied Crick strand would then be passed preferentially to the daughter.

The same sort of logic applies to the daughter cell. The daughter inherited the labeled Crick strand. Since this daughter now becomes the mother in the next division, the labeled strand now becomes the Watson strand. She would keep this labeled strand and pass the unlabeled Crick strand to her daughter.

These are not the results they got. Instead, all the cell types had pretty close to the same amounts of BrdU. The moms did not preferentially hang on to either the original Watson or Crick strand.

They then followed this up with a separate, similar experiment that looked at each individual chromosome on a microarray. The results were that there was no bias for either strand for any of the chromosomes.

It seems that mom doesn’t hang onto the original DNA even at a single chromosome. In the cases of strand-specific selection that are now being studied, there are most likely other ways to pick a strand that have nothing to do with whether it is an original or a copy. Another great idea foiled by data…

by D. Barry Starr, Ph.D., Director of Outreach Activities, Stanford Genetics

Categories: Research Spotlight

Tags: Saccharomyces cerevisiae , chromosome segregation

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