46 average diatoms per field of view Diatoms genome sequenci

4.6 average diatoms per field of view

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 L 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. ?

Solution

4.6 average diatoms per field of view (1963.495 micro m square or pi*25 square or 0.9817 micro litre and not in 20 micro litre)

or 4.6*0.00043g in 0.9817 micro litre

or in 0.9817 micro litre average 4.6 diatoms

or in 1 micro litre 4.6/0.9817 diatoms

or in 1000,000 micro litre 4.6/0.9817*1000,000 diatoms

or in 1 litre 4685749.21055 diatoms

to acquire 1.8 micro gram chloroplast DNA which is only 0.008%

1/0.008*1.8=125*1.8=225 micro gram diatoms

in 0.9817 micro litre average 4.6 diatoms or 4.6*0.00043g or 0.001978g

1 micro litre 0.001978/0.9817g=0.00201487216g or 2.01487216 micro gram in 1 micro lit

1 micro gram in 1/2.01487216 micro lit

225 micro gram diatoms in 225/2.01487216 micro lit

Ans. or 111.669615 micro lit water needed to acquire 1.8 micro gram chloroplast DNA

Similarly to acquire 20 micro gram nuclear DNA which is 0.06% of diatom mass it is required 100/0.06 times more diatom mass or 2000/0.06=33333.33 micro gram

1 micro gram in 1/2.01487216 micro lit

33333.33 micro gram in 33333.33/2.01487216 micro lit

Ans. or 16543.6467 micro lit water you need to acquire 20 micro gram nuclear DNA