Question

Question #a1b58

Answer 1

Since the yeast divides every two hours, its population (and hence yeast mass) doubles every two hours. Because the doubling time is a constant it is exponential growth.

The mass of yeast grows exponentially with time as :
`m(t)=m_o\exp(\lambdat); \qquad => \lambda = \ln(2)/\tau`,
where `\lambda` is the growth factor and `\tau` is the doubling time.
We are given the doubling time (`\tau`) and the initial mass `m_0` and are asked to estimate `t` for `m(t)` to reach the value of the mass of a typical human. Let us assume `75` kg as the mass of a typical human, so `m(t)=75` kg.

`\tau=2 hrs \quad => \quad \lambda = 0.3465 hr^{-1}; \qquad t=?`
`m_0=60 pg = 6.0\times10^{-14} kg, \qquad m(t) = 75 kg`

# m(t) = m_o\exp(\lambda t); \quad => \quad t=\ln((m(t))/m_o)/\lambda = (\ln((m(t))/m_o))/(\ln(2))\tau#

If your calculator struggles to evaluate the logarithm of such a large number you may have to assist it. Remember the following results,

`\ln(ab)=\ln(a)+\ln(b)` and `\ln(x^y)=y\ln(x)`.

Use these to write `\ln(12.5\times10^{14})` as `\ln(12.5)+14\ln(10)`
`\ln((m(t))/m_o)=34.7619; \qquad t=(34.7619)/(\ln(2)) \times 2 hrs=100.3` hrs.

So it takes `100.3` hrs, which is approximately, `4` days and `4` hours for the yeast to grow to the mass of a typical human.