If mitochondria exist at random within a cell, isn't there a possibility that cell division will result in a daughter cell with no mitochondria? If not, what is the process for guaranteeing at least one is present in each daughter cell? If so, what happens to that cell?
$\begingroup$
$\endgroup$
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
Isn't there a possibility that cell division will result in a daughter cell with no mitochondria?
Yes, there is always the possibility. However, there must be a strong negative selection pressure against eukaryotic life that cannot achieve the proper partitioning of mitochondria, so you can imagine that there are mechanisms in place to prevent this case.
Mitochondria are both passively and actively partitioned to daughter cells. This is understood to occur through the cytoskeleton and with the control of mitochondrial fusion and fission at key stages of the cell cycle, prior to mitosis and cytokinesis!
Here is a great review from several years ago that addresses your question well.
-
$\begingroup$ The review is just 5 years old. If it's "several years old" then what would you say about the waiting time to be eligible for an academic job? 😁 $\endgroup$– WYSIWYGCommented May 20, 2019 at 21:33
-
1$\begingroup$ I’m not sure there needs to be any strong selective pressure for this whatsoever. Stochastically, it’s quite likely (actually, virtually assured) that both daughter cells end up with at least a few mitochondria and if not, well, then one of the daughter cells dies and the other one simply divides again. Little harm done, except in gametes. And gametes mis-segregate all the time. One more error in a centillion isn’t going to make a difference. $\endgroup$ Commented May 22, 2019 at 10:11
-
1$\begingroup$ I thought about this too. But mitochondria are no ideal gas and do not follow such rules. Also, not all mitochondria-hosting life is multi-cellular and tolerant to such errors. I would agree with your randomness point, given the following suppositions: 1. mitochondria are randomly distributed in their host cells (untrue); 2. the number of mitochondria in the daughter cell is always unimportant (untrue); 3. cytokinesis is symmetrical (very often untrue); 4. cells without mitochondria are not selected against (which cannot be universally true in all ecological scenarios). $\endgroup$– S PrCommented May 22, 2019 at 11:43
$\begingroup$
$\endgroup$
4
In addition to S Pr's excellent example, I wanted to point out that some very recent research describes some special behavior in oocyte development specifically related to mitochondria selection.
Here's a easy-to-read version: https://www.sciencedaily.com/releases/2019/05/190515131741.htm
Here's the original version in Nature: https://www.nature.com/articles/s41586-019-1213-4
Specifically, during meiosis, the oocyte specifically "puts the mitochondria to the test" by separating all of them (fragmentation) and having each of them operate independently. (Typically mitochondria act in concert, each one potentially making up for deficiencies in their peers). Any that do not "make the cut" are eliminated, and the result is an egg cell that has the best mitochondria to pass along to the next generation.
-
$\begingroup$ This is really interesting! I'd like to understand more about what "mitochondria act in concert" means. Perhaps one is better at one function while its "peer" is better at a different function? Or do you just mean that stronger ones make up for weaker ones that aren't "pulling their own weight"? $\endgroup$– uhohCommented May 21, 2019 at 1:08
-
4$\begingroup$ @uhoh Probably both. Mitochondria have many different functions, and they're very susceptible to genetic damage (since they store their genome in an oxygen-rich environment). Over the lifetime of an organism, many mitochondria fail to perform all or some of their functions, and one typical thing that happens with cancerous cells is that mitochondria replicate far more than usual. In the end, these rarely cause significant issues (beyond the effects of aging etc.) - with one major exception, which is the production of ova. There a single broken mitochondrion can "doom" the offspring. $\endgroup$– LuaanCommented May 21, 2019 at 7:35
-
$\begingroup$ Can you also cite the original research? $\endgroup$– WYSIWYGCommented May 23, 2019 at 12:31
-
$\begingroup$ @WYSIWYG I added a link to the Nature article... Is that acceptable? (I'm not sure how to find the link to the original research.) $\endgroup$ Commented May 23, 2019 at 12:43
$\begingroup$
$\endgroup$
3
A typical animal cell has 1000-2000 mitochondria. From a statistical point of view, assuming a random distribution of the mitochondria and that the cell splits in half, the probability of having 0 mitochondria is (1/2)^1000 or 9e-302. This makes it an impossibility for all practical purposes.
With enough mitochondria, a process to ensure the cell splits roughly in half and a somewhat random distribution of mitochondria would be sufficient to get at least one mitochondria in each daughter cell.
To address the assumptions:
- Random distribution of mitochondria - assumed in the question
- Cells split roughly in half - source on dividing assymetries "In somatic divisions, however, cell size asymmetry is mild and, only rarely, one daughter cell is more than double the size of the other."
-
$\begingroup$ I think both of your assumptions are highly suspect, unless you can cite research to back them up. $\endgroup$ Commented May 21, 2019 at 8:06
-
1$\begingroup$ @Mike While I agree that some citation would increase the quality of this answer, I really like the perspective of it. Other answers regard the high evolutionary pressure on cells without mitochondria and refer to mechanisms to actively prevent that case by distributing mitochondria to the daughter cells. This one brings perspective to the importance of these mechanisms: It certainly is beneficial to actively control the mitochondria distribution, but is this a necessity or just a "nice to have" feature? The actual numbers certainly are not very representative, but I like the idea of it. $\endgroup$ Commented May 21, 2019 at 8:47
-
1$\begingroup$ @NiklasMertsch Exactly. I'm simply trying to demonstrate that in a certain class of cells, there are enough mitochondria that a random process will sufficiently guarantee distribution to the daughter cells; no active process necessary. $\endgroup$ Commented May 21, 2019 at 16:14