Phases You Can't See with Spindle Inhibition

  • Thread starter Thread starter nobahar
  • Start date Start date
  • Tags Tags
    Formation
AI Thread Summary
In a discussion about spindle inhibition and observable phases of cell division, participants debated which phases would be visible under a microscope. It was concluded that prometaphase, metaphase, and anaphase would not be observable due to the lack of spindle formation. Some argued that early prometaphase could still be seen, but the consensus leaned towards the idea that significant events in prometaphase could not complete without spindles. The importance of spindle fibers in aligning chromosomes at the metaphase plate was emphasized, clarifying that without spindles, proper alignment cannot occur. Overall, the conversation highlighted the critical role of spindle formation in cell division phases.
nobahar
Messages
482
Reaction score
2
Hello!

Homework Statement


If a chemical is used to inhibit spindle formation, what phases would you not see [under a microscope]?
2. Possible answers
Prophase
Prometaphase
Metaphase
Anaphase
Telophase

The Attempt at a Solution


I figured prometaphase, metaphase and anaphase.
I know some people don't use prometaphase, by this phase I don't mean prophase. Instead I mean the stage in which the connection of the micotubules with the kinetochores begins to occur, and there is the 'back and forth' movement to attempt to align the chromosomes along the metaphase plate (when this occurs it is obviously metaphase). However, apparently I'm wrong and prometaphase isn't one. There's supposed to be three answers. I guess the other is going to be telophase, but I can't see how prometaphase can be ruled out...
Thanks in advance.
 
Physics news on Phys.org
You would probably observe the first parts of prometaphase (e.g. disappearance of the nuclear envelope) but because spindle formation could not occur, you could not progress to the end of prometaphase and into metaphase.
 
Thanks for the reply,
I think that what you have said supports my thinking, since important stages of prometaphase cannot be complete, then I would say that it is perfectly reasonable to suggest that this particular stage would not be observable. Very peculiar, but, I suppose that if early prometaphase can be observed then strictly this does indeed count. A little annoying.
Thanks.
 
i may be wrong but i think that you would see up to metaphase. because, if i remember my biology well, the chromatids will line up at the equator of the cell. It is only in anaphase that the spindle come into action to pull apart the sister chromatids to opposite poles. the absence of spindles will not prevent the chromatids to line up during metaphase!
 
No. The spindle is required to align the chromosomes up at the metaphase plate.
 
I was thinking of it from the point of view that the microtubules as they 'reach' out, and the asters, are what constitute the spindle structure. Perhaps it only counts as a spindle structure once the chromosomes are aligned along the metaphase plate. I'll find some sources and check back, if anyone else is interested... unless someone else already knows.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

Similar threads

Back
Top