Discussion of all aspects of biological molecules, biochemical processes and laboratory procedures in the field.
HELP! with ethidium bromide experiment
HI! I just got this assignment and have pondered over it for about 3 hours and can't get a single thing. I've looked up on journals, other chat sites and wiki and nothing helps.
Ethidium bromide is a compound that binds to DNA and makes positive (+) supercoils in circular DNA molecules. (is similar to how histones help coil DNA)
I'm thinking in the 2nd one that the space between the 2 circles may represent the perinuclear space surrounded by the outer and inner membranes of the nucleus.
Any other ideas for the other 2 questions?
Please and thanks!!!
3. Researchers grew mammalian tissue culture cells on three coverslips. Each of the coverslips was treated, under identical conditions, with detergent to remove all membranes and with high salt buffers to remove all soluble proteins including histones and non-histone proteins from chromatin. Next, one coverslip was left untreated, the second coverslip was treated with a low concentration of ethidium bromide (EtBr)* and the third coverslip was treated with a high concentration of EtBr. The coverslips were then stained with a fluorescent stain that stains the DNA blue. The results of the experiment are shown below.
No ethidium bromide = dark blue circle (nucleus?)
Low ethidium bromide = medium blue circle with lighter blue circle surrounding it (nucleus with perinuclear space?)
High ethidium bromide = dark blue circle - same as with no ethidium bromide
A. What do the results of this experiment tell the researchers about the organization of DNA in the nucleus of mammalian cells? Explain your answer. (3 points)
B. What information might be obtained if researchers measured the distance between the inner and outer circles in the low ethidium bromide sample? Explain your answer. (1 point)
C. What result would you expect if the coverslips were treated with a restriction enzyme that cut genomic DNA on average once every 50,000 to 100,000 base pairs? (1 point)
hm, that seems interestingly. Well, I would guess some euchromatin/heterochromatin (which is on the sides of nucleus, hence two circles in second experiment) detection, but they wrote they have removed all the proteins...
How does the blue dye colors the DNA? Does it compete with EtBr?
Cis or trans? That's what matters.
After searching in google a possible explanation would be that EtBR in low amounts condenses the heterochomatin (light blue) and the medium blue would be euchromatin since its slightly packed which in low amounts of EtBR most of the euchromatin is uncondensed.
No EtBR both Euchromatin and Heterochomatin are in a dissolved form ( not packed) due to the removal of histones and non histone protiens.
High amounts of EtBR causes all chomatin (euchromatin and heterochomatin) to be tighlty packed thus showing no difference in color.(hypercondensation)
so to answer the questions:
1.A. What do the results of this experiment tell the researchers about the organization of DNA in the nucleus of mammalian cells? Explain your answer.
Results indicate that DNA is organized in the form of chromatin, condensed and tightly packed by histone and non histone profiens.
Explanation: After the removal of histones and non histone protiens by the action of salt in the first culture and no EtBR shows a dark blue color showing that DNA is dispersed within the nucleus in an uncondensed form.
After the addition of a low amount of EtBR we observe an inner medium color and an outer light blue color. Etbr being similar to the role of histones condenses a part of the DNA which is heterochromatin as in normal conditions heterochromatin is tightly packed. Thus the light blue color is DNA and EtBR comparing to the medium blue in the inner part being unpacked pure DNA (no histones).
Finally adding to the third culture a high amount of EtBR condenses all DNA content showing no difference in color DNA is in a hypercondensed form.
Histone depleted DNA form loops as we know.
So in the first culture with no EtBR there isn't any supercoiled DNA since cells were treated with salt removing histone and non histone protiens so DNA is mixed throughout the nucleus.
low amounts of Etbr in the second culture causes DNA supercoils and producing loops, the light blue color are the DNA that are looped low in DNA content which is the DNA depleted histones (not packed, low amount of DNA content compared to highly dense in the inner layer).
The high amount of EtBR caused all DNA to be packed tightly thus having the same amount of DNA content throughout the nucleus, all DNA is packed tightly being very dense.
1. Results indicate that DNA is tightly packed in a condense form and Having loops which are depleted of histone and non histone protiens being loosely packed.
Explanation: (explaining the 3 cultures results) above.
2.measuring the length between the inner and outer layers would help in calculating the number of loops and the size of DNA loops (base pairs)
3. Treated with Restriction enzymes of the coverslips would result in difference of the amount and size of DNA fragments.
After reading more the second explanation seems more reasonable.
Hope this helps, please comment I also have to complete this assignment
Thank you so much! This is fantastic. It makes way more sense described like that. I definitely agree with the second explanation.
If you're in this class too, in the first part of the assignment, how do you determine the percentage of cells in G0?
Thanks so much!
Happy to help out.
Yes I am also in your class
I spent a while searching these questions are pretty hard.
43 out of 164 cells were immediately fixed so 43 cells have high DNA content (synthesis and replication of DNA) (S-Phase).
So 164-43= 121 cells that were not fixed I deduced that these cells were in G0
So % of cells in G0= 121*100/164= 73.7%
Your right. so 43 cells were immediately fixed.
0 of 12 mitotic cells were stained but then there was 16 mitotic cells?? a little confused about that.
So we know G2 starts when the first mitotic cells were stained I'm guessing if 43 cells were fixed and only 16 mitotic cells were stained so would the cells in G0 would possibly be 43-16=27
I doubt that is right, 43 cells fixed were in S-phase so cells aren't in G0.
the 121 cells left as you said could be in G1, G2 and M phase and G0.
Are we required to calculate the % or just explain how to calculate it?
Okay. That was going to be our second guess. Thanks.
Is it one of the ones we've learned then so far? Cause we were thinking of saying something like:
If you know how many cells are initially in the cell cycle, all the cells become florescent after they pass through S phase. So couldn't you just assume that it would double after 23 hours, but if there is less than double the amount of original cells after 23 hours, then you know how many cells went into G0 and therefore the percentage?
Or is it an actual experiment that we learned in tutorial?
That sounds right.
I just checked the Tutorial for week of March 7th (I'm in thursday's sections) and went through the tutorial and found that the percentage of G0 cells could be calculated using FLUORESCENCE ACTIVATED CELL SORTER (FACS)
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