BME 201 HW #10.docx - November 7th 2016 Professor Kaiming...

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November 7 th , 2016 Professor Kaiming Ye BME 201 HW #10 1. Assuming the cell is perfectly spherical, with a radius of 5mm, it has a volume of 5.23 x 10^-13 L. The cell has a molecule concentration of 10 pM, which is equivalent to 1x10^-11 M. (1x10^-11 moles) x (5.23 x 10^-13 L) = 5.23 x 10^-24 (6.02 x 10^23 molecules / 1 mole) x (5.23 x 10^-24 m^3 / 1 cell) = 3.15 molecules. 2. Scenario 1: If the replication rate, μ and differentiation rate, δ are equal, then μ = δ. If these are expressed as the value q, the expression formed is: μ = δ = q Using this expression, Xout becomes: Xout = Xin * е^ (q/q) = Xin * е¹ Scenario 1: Xout = е * Xin Based on this, when μ = δ, Xout is the value of Xin times the constant e. Scenario 2: If δ is ten times slower than μ, then μ = (δ/10). Expressed in terms of q, with μ set to equal q, the expressions for μ and δ becomes: μ = q and δ = q/10
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