Question

In a bacterium, a mutation resulted in an increase of \( K_s \) (substrate-specific constant) for ammonium from 50 \( \mu M \) to 5000 \( \mu M \) without affecting \( \mu_{max} \). The specific growth rate (\( \mu \)) of the mutant growing on 0.5 mM ammonium in the medium decreases by a factor of ..........

Hint:

The specific growth rate is inversely proportional to \( K_s \), and the Monod equation can help in calculating growth rate changes due to substrate concentration variations.

Answer 1

Correct Answer: 10

Step 1: Understanding the relationship between \( K_s \) and \( \mu \).
The specific growth rate \( \mu \) for a bacterium is given by the Monod equation: \[ \mu = \mu_{max} \frac{[S]}{K_s + [S]} \] where \( [S] \) is the concentration of the limiting substrate and \( K_s \) is the substrate-specific constant.
Step 2: Calculating the change in \( \mu \).
For the initial condition, \( K_s = 50 \, \mu M \) and \( [S] = 0.5 \, mM = 500 \, \mu M \), the ratio \( \frac{[S]}{K_s + [S]} \) is: \[ \frac{500}{50 + 500} = \frac{500}{550} = 0.909 \] For the final condition, \( K_s = 5000 \, \mu M \) and \( [S] = 500 \, \mu M \), the ratio is: \[ \frac{500}{5000 + 500} = \frac{500}{5500} = 0.091 \] The specific growth rate \( \mu \) decreases by a factor of: \[ \frac{0.909}{0.091} = 10 \] Step 3: Conclusion.
The specific growth rate decreases by a factor of 100.