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Biology 8 Online
OpenStudy (abb0t):

Estimate number of bacteria

OpenStudy (abb0t):

|dw:1426743331228:dw| i think

OpenStudy (kainui):

How many cells are there if 85% is not cells in the whole volume and a single cell is 1fL? \[\Large Volume = 1 cm^2 * 50 \mu m = 5*10^9\mu m^3\] since \[\Large 10^2 cm = 1 m = 10^6 \mu m \\ \Large (10^2 cm)^2= (10^6 \mu m )^2 \\ \Large 10^4 cm^2 = 10^{12} \mu m^2 \\ \Large 1 = 10^8 \frac{\mu m^2}{cm^2}\] now we can do something similar with fL to um^3 but I prefer Wolfram Alpha so yeah. \[\Large 1 fL = 1 \mu m^3\] We want total number of cells so we have number of cells per volume is \[\Large \frac{volume}{cell} = 1 \frac{ fL}{cell}\] that's the same thing if we flip it upside down as \[\Large 1 \frac{cell}{\mu m^3}\] So we know how many cells per volume of cells, how many cells do we have, well since 85% is not cells 100%-85%=15% is cells so the total volume we got was \[\Large (5*10^9 \mu m^3)*(.15) = total \ volume*\frac{volume \ of \ cells}{total \ volume}\]so you multiply that by the final thing, which ends up just being 1. I guess we can memorize this obscure conversion to save us time in the future while we're here that 1fL=1um^3 lol. So that's how I got \[\Large 7.5*10^8 cells\] sorry I'm kinda tired and rambling I didn't have enough time to make it shorter and I just sorta hit a random bunch of spots at this problem since I don't really know where the problem was but it as hopefuly somewhere that I shot with this shotgun approach lol

OpenStudy (abb0t):

tmv virus is a singly stranded RNA virus that infects plants, particularly tobacco. Find the weight of this virus.

OpenStudy (abb0t):

in fg

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