Diatoms genome sequencing You have an average number of diatoms/field of view at

Question

Diatoms genome sequencing You have an average number of diatoms/field of view at 400x We used 20 mu l of H_2O that contained an unknown number of diatoms The total area at 400x magnification was pi* 25^2 = 1963.495 mu m^2 The average mass of a diatom is 0.00043 g. Of the total mass. 0.06% is nuclear DNA.0.008% is chloroplast DNA and 0.01 % is mitochondrial DNA As a helpful tip...1963.495 mu m^2 = 0.9817 mu l How many diatoms are present in 1 of H_2O? How much H_2O do you need to acquire 1.8 mu g chloroplast DNA? How much H_2O do you need to acquire 20 mu g of nuclear DNA. ?

Explanation / Answer

Answer:

Since the average number of diatoms/field view at 400X is not mentioned, the calculations have been done considering the average number/field view = 4.

For 400X view, area = 1963.495 micro m2 = 0.9817 microL

So, number of views covered by 20 microL of water = 20/0.9817 = 20.37 views

Now, 1 view contains 4 diatoms (assumed number of diatoms/field view)

Therefore, for 20.37 views (20 microL) = 4 * 20.37 diatoms = 81.48 diatoms

1 L (1000000 microL) contains (81.48 *106) / 20 = 4 * 106 diatoms

So, number of diatoms per liter = 4 * 106

Chloroplast DNA:

Mass of 1 diatom = 0.00043 gm = 430 microgram

So, amount of chloroplast DNA from 1 diatom = (0.008/100) * 430 microgram = 0.0344 microgram

For acquiring 1.8 microgram of chloroplast DNA, we require (1.8/0.0344) = 52.32 diatoms

4 diatoms = 0.9817 microL

Therefore, 52.32 diaoms = 0.9817/4 * 52.32 = 12.84 microL of water

Nuclear DNA:

Mass of 1 diatom = 0.00043 gm = 430 microgram

So, amount of nuclear DNA from 1 diatom = (0.06/100) * 430 microgram = 0.258 microgram

For acquiring 20 microgram of chloroplast DNA, we require (20/0.258) = 77.5 diatoms

4 diatoms = 0.9817 microL

Therefore, 77.5 diaoms = 0.9817/4 * 77.5 = 19.02 microL of water

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