Use the concept of specific activity in calculations. A researcher wishes to estimate the rate of uptake of the sugar galactose by carrot cells in a suspension culture. She prepares \(250 \mathrm{~mL}\) of the cell culture medium containing \(10^{7}\) cells per \(\mathrm{mL}\) and unlabelled galactose at a concentration of 5 mmol \(L^{-1}\). She then 'spikes' this with \(5 \mu \mathrm{L}\) (regard this as an insignificant volume) of radioactive standard containing \(55 \mathrm{MBq}\) of \({ }^{14} \mathrm{C}\)-labelled galactose (regard as an insignificant concentration). Answer to two significant figures. (a) Calculate the specific activity of the galactose in the culture solution in \(\mathrm{Bq} \mathrm{mol}^{-1}\). (b) If the total cell sample takes up \(79.2 \times 10^{5} \mathrm{~Bq}\) in a 2-h period, calculate the galactose uptake rate in \(\mathrm{mol} \mathrm{s}^{-1} \mathrm{cell}^{-1}\).

Step by step solution

01

Calculate the total volume of the galactose in the culture

We know that the total volume of the cell culture medium is \(250 \mathrm{~mL}\). Given that there are \(10^{7}\) cells per \(\mathrm{mL}\), the total number of cells in the cell culture will be \(250 \mathrm{~mL} \times 10^{7} \mathrm{cells/mL} = 2.5 \times 10^{9} \mathrm{cells}\).

02

Calculate the concentration of galactose

The galactose concentration is given to be 5 mmol \(L^{-1}\). We have to convert this to mol \(L^{-1}\), hence, the concentration of galactose will be \(5 \times 10^{-3} \mathrm{mol~L^{-1}}\). Since we have 250 mL or \(0.25 \mathrm{L}\) of the medium, the total amount of galactose in the medium will be \(5 \times 10^{-3} \mathrm{mol~L^{-1}} \times 0.25 \mathrm{L} = 1.25 \times 10^{-3} \) mol.

03

Calculating the Specific Activity

The radioactive standard contains \(55 \mathrm{MBq}\) of \({ }^{14} \mathrm{C}\)-labelled galactose, or \(55 \times 10^{6} \mathrm{Bq}\). The specific activity is defined as the activity per unit quantity of the substance. Hence, the specific activity of the galactose in the culture will be \( \frac{55 \times 10^{6} \mathrm{Bq}}{1.25 \times 10^{-3} \mathrm{mol}} = 4.4 \times 10^{10} \mathrm{Bq~mol^{-1}} \).

04

Calculate the Uptake Rate

We are given that the total cell sample takes up \(79.2 \times 10^{5} \mathrm{Bq}\) in a 2-hour period which is equivalent to \(7200 \mathrm{s}\). Since this is the amount of \(^{14}\)C-galactose taken up, we can use the specific activity to find the amount of galactose this corresponds to. This will be \( \frac{79.2 \times 10^{5} \mathrm{Bq}}{4.4 \times 10^{10} \mathrm{Bq~mol^{-1}}} = 1.8 \times 10^{-6} \mathrm{mol} \). Finally, the rate of uptake per cell per second will be \( \frac{1.8 \times 10^{-6} \mathrm{mol}}{2.5 \times 10^{9} \mathrm{cells} \times 7200 \mathrm{s}} = 1.0 \times 10^{-16} \mathrm{mol~s^{-1}~cell^{-1}}\).

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