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4.2 The number of phage on Earth Please cite your sources and show your work for

ID: 193119 • Letter: 4

Question

4.2 The number of phage on Earth

Please cite your sources and show your work for the volume of water on Earth.

(a) An estimate for the number of phages on Earth can be obtained using data from Bergh et al. (1989). By taking water samples from lakes and oceans and counting the various phages, one arrives at counts of the order of 10^6-10^7 phages/mL. Using reasonable assumptions about the amount of water on Earth, make an estimate of the number of phages.
(b) Work out the mass of all of the phage particles on the Earth using your result from (a).

Explanation / Answer

Answer:

a) From Bergh et al.(1989), it was estimated that the amount of virus particles in natural waters is 2.5X10^8 virus particle/mL.

Now, from this estimate :

Total volume of water on earth = 1.386 billion Km3 which is equal to 1.368 X 109 Km3 .

Now, 1 Km3 = 1000 m

1 Km3 = 1000 X 1000 X 1000 m3 = 109 m.

1m3 = 100 cm which is equal to 100 X 100 X 100 cm3 = 106 cm3 .

And morever, 1cm3 = 1 mL.

Therefore,

1 Km3 = 109X 106 cm3

= 1015 cm3 = 1015 mL

Now, total volume of water on earth is 1.386 X 109 km3 = 1.368 X 109 X 1015 mL = 1.386 X 1024 mL.

So, total no. of Phages on earth is equal to

= no. of phages X mL of water on earth

= 2.5 X 108 X 1.386 X 1024

= 3.465 X 1032 no. of phage particles on earth's natural waters.

b) According to an article : Peterson et al,. journal of structural biology, 2001, the mass of an bacteriophage Psi 29 , which is a Phage of Bacillus subtilis, is 33.6 MDa (Mega Dalton). Therefore the mass of other phages must be near this value.

So, an approximate mass of all the phage particles on earth from the result of (a) will be :

= 3.405 X 1032 X 33.6 MDa

= 114.408 X 1032 MDa.

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