Are cells guaranteed to get at least one mitochondrion when they divide?Does every mitochondrion in a cell contain the same DNA?How are different types of cells created from zygote?Mitochondria - are they really separate organisms that once merged into eukaryotic cells?Are mitochondria transferable between cell types, individuals and species?In cell division, are daughter cells identical?Can mitochondria be regenerated by the cell?Why isn't meiosis II called mitosis (as the chromosome number doesn't half)?Why do cells divide (or copy themselves)?Cell organels during cell reproductionHow does the zygote divide much faster than other cells?
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Are cells guaranteed to get at least one mitochondrion when they divide?
Does every mitochondrion in a cell contain the same DNA?How are different types of cells created from zygote?Mitochondria - are they really separate organisms that once merged into eukaryotic cells?Are mitochondria transferable between cell types, individuals and species?In cell division, are daughter cells identical?Can mitochondria be regenerated by the cell?Why isn't meiosis II called mitosis (as the chromosome number doesn't half)?Why do cells divide (or copy themselves)?Cell organels during cell reproductionHow does the zygote divide much faster than other cells?
$begingroup$
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?
cell-biology mitochondria mitosis
asked May 20 at 12:39
sourcebugsourcebug
14115
$endgroup$
add a comment |
$begingroup$
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?
cell-biology mitochondria mitosis
asked May 20 at 12:39
sourcebugsourcebug
14115
$endgroup$
add a comment |
$begingroup$
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?
cell-biology mitochondria mitosis
asked May 20 at 12:39
sourcebugsourcebug
14115
$endgroup$
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?
cell-biology mitochondria mitosis
cell-biology mitochondria mitosis
asked May 20 at 12:39
sourcebugsourcebug
14115
asked May 20 at 12:39
sourcebugsourcebug
14115
asked May 20 at 12:39
sourcebugsourcebug
14115
asked May 20 at 12:39
sourcebugsourcebug
14115
asked May 20 at 12:39
sourcebugsourcebug
14115
14115
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered May 20 at 13:29
S PrS Pr
2,130213
$endgroup$
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 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.
$endgroup$
– Konrad Rudolph
May 22 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 Pr
May 22 at 11:43
add a comment |
$begingroup$
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.
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
$endgroup$
$begingroup$
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"?
$endgroup$
– uhoh
May 21 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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
add a comment |
$begingroup$
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."
answered May 20 at 20:08
UnderminerUnderminer
20117
$endgroup$
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 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.
$endgroup$
– Underminer
May 21 at 16:14
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered May 20 at 13:29
S PrS Pr
2,130213
$endgroup$
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 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.
$endgroup$
– Konrad Rudolph
May 22 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 Pr
May 22 at 11:43
add a comment |
$begingroup$
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.
answered May 20 at 13:29
S PrS Pr
2,130213
$endgroup$
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 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.
$endgroup$
– Konrad Rudolph
May 22 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 Pr
May 22 at 11:43
add a comment |
$begingroup$
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.
answered May 20 at 13:29
S PrS Pr
2,130213
$endgroup$
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.
answered May 20 at 13:29
S PrS Pr
2,130213
edited May 20 at 14:07
answered May 20 at 13:29
S PrS Pr
2,130213
answered May 20 at 13:29
S PrS Pr
2,130213
answered May 20 at 13:29
S PrS Pr
2,130213
2,130213
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 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.
$endgroup$
– Konrad Rudolph
May 22 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 Pr
May 22 at 11:43
add a comment |
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 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.
$endgroup$
– Konrad Rudolph
May 22 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 Pr
May 22 at 11:43
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 at 21:33
$begingroup$
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? 😁
$endgroup$
– WYSIWYG
May 20 at 21:33
1
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.
$endgroup$
– Konrad Rudolph
May 22 at 10:11
$begingroup$
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.
$endgroup$
– Konrad Rudolph
May 22 at 10:11
1
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 Pr
May 22 at 11:43
$begingroup$
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 Pr
May 22 at 11:43
add a comment |
$begingroup$
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.
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
$endgroup$
$begingroup$
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"?
$endgroup$
– uhoh
May 21 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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
add a comment |
$begingroup$
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.
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
$endgroup$
$begingroup$
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"?
$endgroup$
– uhoh
May 21 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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
add a comment |
$begingroup$
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.
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
$endgroup$
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.
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
edited 2 days ago
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
answered May 20 at 17:50
Reginald BlueReginald Blue
39129
39129
$begingroup$
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"?
$endgroup$
– uhoh
May 21 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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
add a comment |
$begingroup$
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"?
$endgroup$
– uhoh
May 21 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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
$begingroup$
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"?
$endgroup$
– uhoh
May 21 at 1:08
$begingroup$
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"?
$endgroup$
– uhoh
May 21 at 1:08
4
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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
@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.
$endgroup$
– Luaan
May 21 at 7:35
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
Can you also cite the original research?
$endgroup$
– WYSIWYG
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
$begingroup$
@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.)
$endgroup$
– Reginald Blue
2 days ago
add a comment |
$begingroup$
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."
answered May 20 at 20:08
UnderminerUnderminer
20117
$endgroup$
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 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.
$endgroup$
– Underminer
May 21 at 16:14
add a comment |
$begingroup$
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."
answered May 20 at 20:08
UnderminerUnderminer
20117
$endgroup$
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 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.
$endgroup$
– Underminer
May 21 at 16:14
add a comment |
$begingroup$
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."
answered May 20 at 20:08
UnderminerUnderminer
20117
$endgroup$
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."
answered May 20 at 20:08
UnderminerUnderminer
20117
edited May 21 at 15:51
answered May 20 at 20:08
UnderminerUnderminer
20117
answered May 20 at 20:08
UnderminerUnderminer
20117
answered May 20 at 20:08
UnderminerUnderminer
20117
20117
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 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.
$endgroup$
– Underminer
May 21 at 16:14
add a comment |
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 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.
$endgroup$
– Underminer
May 21 at 16:14
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
I think both of your assumptions are highly suspect, unless you can cite research to back them up.
$endgroup$
– Mike Scott
May 21 at 8:06
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 at 8:47
$begingroup$
@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.
$endgroup$
– Niklas Mertsch
May 21 at 8:47
1
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.
$endgroup$
– Underminer
May 21 at 16:14
$begingroup$
@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.
$endgroup$
– Underminer
May 21 at 16:14
add a comment |
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