Question #a4d61

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Answer 1 · 2015-01-08T14:46:57

Here's my (very rough) estimate of what that number could be.

The human gastrointestinal tract (GI Tract) has recently been discovered to cover less than previously believed. At some point, the surface area of the GI Tract was estimated at ${250-300 m}^{2}$ $"250-300 m"^2$ ; however, recent studies have brought forth a more conservative number, ${30-40 m}^{2}$ $"30-40 m"^2$ .

![http://www.sciencedaily.com/releases/2014/04/140423111505.htm]( useruploads.socratic.org )

So, let's assume we have an average GI tract size of ${35.0 m}^{2}$ $"35.0 m"^2$ . Since we're talking about the digestive tract, we'll use Escherichia coli, or E coli.

![http://www.livescience.com/3092http://-human-gut-loaded-bacteria-thought.html](https://useruploads.socratic.org/uyKSXBnAT7mFcg9jehDY_070926_e_coli_02.jpg)

The average E coli bacterium has a lenght of about $2.0$ $"2.0"$ $μ m$ $mum$ and a diameter of about $0.25-1$ $"0.25-1"$ $μ m$ $mum$ . I'll treat each cell as an ellipse in order to get an estimate of its surface area. The estimated area of an average E coli bacterium will be

$area 1 bacterium = π \cdot 2.0$ $"area 1 bacterium" = pi * 2.0$ $μ m \cdot 0.50$ $mum * 0.50$ $μ m = 3.14 μ m^{2}$ $mum = 3.14mum^2$

fF you set each of these ellipses side by side, you could get a rough number of

$35.0$ $35.0$ $m^{2} \cdot \frac{1 bacterium}{3.14 \cdot 10^{- 12} m^{2}} = 1.12 \cdot 10^{13}$ $"m"^2 * ("1 bacterium")/(3.14 * 10^(-12) "m"^2) = 1.12 * 10^13$ $bacteria$ $"bacteria"$

The number of bacteria that would cross would be

$1.12 \cdot 10^{13}$ $1.12*10^13$ $bacteria \cdot \frac{0.001}{100} = 1.12 \cdot 10^{8}$ $"bacteria" * (0.001)/100 = 1.12 * 10^8$ $bacteria$ $"bacteria"$

I believe the actual number is bigger, not that this one is particularly small. By comparison, the number of stars in our galaxy alone is estimated at 400 billion, $4 \cdot 10^{11}$ $4*10^11$ .

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