An E. coli cell of volume $10^{-12}$ cm³ contains 60 molecules of lac-repressor. The repressor has a binding affinity ($K_a$) of $10^{-8}$ M and $10^{-9}$ M with and without lactose respectively, in the medium.
To find the molar concentration of the lac-repressor protein within the E. coli cell, we need to determine the number of moles of repressor and divide it by the cell's volume in liters.
First, convert the cell volume from cubic centimeters (cm³) to liters (L), knowing that 1 cm³ = 1 mL and 1 mL = $10^{-3}$ L.
Volume in Liters = $10^{-12} \text{ cm}^3 \times \frac{10^{-3} \text{ L}}{1 \text{ cm}^3} = 10^{-15} \text{ L}$
Next, convert the number of repressor molecules to moles using Avogadro's number ($N_A \approx 6.022 \times 10^{23}$ molecules/mol).
Moles of Repressor = $\frac{60 \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}}$
Moles of Repressor $\approx 9.96 \times 10^{-23}$ mol
Calculate the molar concentration (Molarity, M) using the formula: Molarity = Moles / Volume (L).
Molar Concentration = $\frac{9.96 \times 10^{-23} \text{ mol}}{10^{-15} \text{ L}}$
Molar Concentration $\approx 9.96 \times 10^{-8}$ M
Convert the concentration from Molar (M) to Nanomolar (nM). Note that 1 M = $10^9$ nM.
Molar Concentration in nM = $(9.96 \times 10^{-8} \text{ M}) \times (10^9 \text{ nM/M})$
Molar Concentration in nM $\approx 9.96 \times 10^1$ nM
Molar Concentration in nM $\approx 99.6$ nM
This value is approximately 100 nM.
The binding affinity values ($K_a$) provided are not required for this calculation.
Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].
Assertion [a]: In multicellular organisms, cells of different lineages have different gene expression profiles.
Reason [r]: Alternative splicing is the only mechanism to generate protein diversity.
Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]
Assertion: Ab initio gene finding algorithms that predict protein coding genes in eukaryotic genomes are not completely accurate.
Reason: Eukaryotic splice sites are difficult to predict.