Sizing up the amount of carbon in an E. coli. Make an estimate of the number of

Question

Sizing up the amount of carbon in an E. coli. Make an estimate of the number of carbon atoms needed to make an E. coli. Assume that 50% of the dry mass of E. coli is made up of carbon atoms. The dry mass of E. coli is 0.3 pg. E. coli are typically grown on minimal media which contains a starting concentration of glucose being 0.2 g/ 100 ml. Roughly how many E. coli cells can be grown on 5 ml of minimal media before the glucose runs out? The chemical formula for glucose is C6H1;06. It has a molar mass of 180 g/mol. Given that a typical bacteria (e.g. E. coli) is cylindrically shaped with a length ~mu m and diameter ~ 1 mu m. make an estimate of it's surface area and volume. What fraction (i.e. %) of the volume of an E. coli cell is occupied by the DNA of its genome? E. coli has a genome that is 4.6 million base pairs (Mbp) long. (c) Roughly 2-3 kg of bacteria are harboured in your large intestine. Make an estimate for the total number of bacteria inhabiting your intestine.

Explanation / Answer

1.
mass of E.coli = 0.3 pg = 0.3*10^-12 g
mas of carbon = 0.3/2 = 0.15

mass of 1 carbon atom = MW of carbon / Avagadro no.

mass of 1 carbon atom= (12/(6.022*10^23) )

no. of carbon atoms = mass ofcarbon of E.coli /(mnass of 1 carbon atom)

no. = (0.15*10^-12) /(12/(6.022*10^23) )
no. = 7.527*10^9

b)
mass of 5 ml solution = 0.2/100* 5 = 0.01 g

so ,
moles of Glucose = 0.01/180 = 5.556*10^-5 moles.

no . of carbon atoms = 5.556*10^-5 *(12/180)*6.022*10^23 = 2.23*10^18 atoms of carbon

as,
1 cell of carbon contains , 7.527*10^9 carbon atoms ,
so ,
no. of E .coli cells = 2.23*10^18 atoms of carbon/7.527*10^9 carbon atoms
no .of E .coli cells = 2.96*10^8 no. of cells

2. surface area of cylinder = 2*pi* r( r+h)
h = 2 micro m = 2*10^-6 m
r = 0.5 micro m =0.5*10^-6 m

so putting values ,
Surface are a= 7.85*10^-12 m^2

Volume = pi* r^2 h

putting values again ,
V = 1.57*10^-18 m^3

b)length of 4.6 milions Mbp= 4.6*10^12* 3.4*10^-10 = 1564 m

c)mass of E.coli = 0.3 pg
so ,
no of bacteria =( 2*10^15/0.3) to (3*10^15 /0.3)

no . ofbacteria = 6.667*10^15 to 10^ 16

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