Question

You plan to expose your chlamydomonas culture to 100 nM colchicine to inhibit new protein synthesis. You need to treat 1 x 108cells. The density of your culture is 5 x 107 cells/ml. The molecular weight of colchicine is 399.44. Explain what you need to do. Show your calculations.

Answer 1

Answer:

mass(g) of colchicine = 3.99 × 10⁻⁵g

Explanation:

Given that;

the number of moles colchicine =  100 nM = 100 × 10⁻⁹M

Density of chlamydomonas culture = 5 × 10⁷ cells/mL

∴ In 1mL of chlamydomonas culture there are 5 × 10⁷ cells present.

To decide the number of one cell, we need to use $\frac{1}{5*10^7}$ mL of the cellulose

However, 1 × 10⁸ cells will be present in $1*10^8*\frac{1}{5*10^7}$ mL , which in turn give us;

$\frac{10^8}{5*10^7}$mL

Afterwards, to get the required 1 × 10⁸ cells, 2mL(molarity, since molarity=  no of moles/litre) has to be taken

If colchicine has to be treated, we need to determine the mass of colchicine that is required in the process as well;

since,  the number of moles colchicine =  100 nM = 100 × 10⁻⁹M

And, the given molecular weight = 399.44;

we can determine the mass of colchicine as;

∴ $numberofmoles = \frac{mass(g)}{molar mass(g)}$

substituting the parameters given, we have:

$100*10^{-9}= \frac{mass(g)}{399.4}$

mass(g) = 399.4 × 100 × 10⁻⁹

            = 3.99 × 10⁻⁵g

Hence, the mass of colchicine that is required in the process to make 100 nM dissolve in the in 2mL of the culture in one Litre of water  is 3.99 × 10⁻⁵g.