Answer:
50 micro mol/min
Explanation:
So, we have the following parameters which are going to help us in solving this particular Question or problem;
=> For the wild type enzyme, the value for the Maximum velocity = 100 micromol/min.
=> For the mutant type enzyme, the value for the Maximum velocity = 1 micromol/min.
So, we can determine or calculate the value for the initial velocity for each of the enzyme type by using the formula below;
Initial velocity = (substrate concentration × maximum velocity) / substrate concentration + Km.
Therefore, for the wild type enzyme; the Initial velocity = (substrate concentration × maximum velocity) / substrate concentration + Km.
Initial velocity =( 10mM × 100micromol/min) ÷ ( 10mM + 10 mM ) = 50 micro mol/min.
Initial velocity for the wild type enzyme = 50 micro mol/min.
C. Costal surface.
Please please pleaseeee tell me if this is wrong.
If the alcohol concentration in PEA agar was increased, it would inhibit both Gram - and Gram + bacteria by breaking down membrane absorbency barrier, tolerating influx of substances that are generally blocked and leakage of cellular potassium. High alcohol concentrations would lead to disruption or halting of DNA synthesis in both Gram + and Gram - bacteria
A, the others are unrealistic speeds.
Answer:
ATP is produced most efficiently by structure C.(mitochondria)
