Two strains of Bacterium sweetans, A and B, use sucrose (table sugar) as a sole

ID: 72783 • Letter: T

Question

Two strains of Bacterium sweetans, A and B, use sucrose (table sugar) as a sole carbon source. The first step in the process of sucrose utilization is the passage of sucrose through a sucrose transporter protein in the membrane. The characteristics of the two transport proteins are as follows (assuming [E]tot is the same in both):

Strain A B km 1000mM 10 mM

Vmax 1000 mM/min 100mM /min

Assuming that the rate of sucrose uptake is the rate limiting step in growth, choose the correct statement if the concentration of sucrose is 10 mM?

Options(

Strain A will grow faster with a v0 of 99m mol/min

B will grow faster with a v0 of 99 m mol/min

B will grow faster with a v0 of 50 mmol/min

A will grow faster with a v0 of 50 mmol/min

A will grow faster with a v0 of 9.9 m mol/min

Explanation / Answer

By using the Michaelis-Menten equation, the initial velocity of sucrose uptake for each strain under the given conditions can be calculated. The given data can be considered at the initial rate of sucrose because if sucrose is depleted in the medium, the rate would decrease. Therefore, it is assumed that the transport protein is always running at Vo.

[sucrose] mM

Vo, enzyme of A:

Km=1000mM

Vmax=1000mmol/min

Vo, enzyme of B:

Km= 10mM

Vmax= 100mmol/min

Fast growing strain

9.9mmol/min

50mmol/min

100

91mmol/min

Neither A or B strain grow faster

1000

500mmol/min

99mmol/min

At 10mM concentration of sucrose, for strain B V0 = ½ Vmax , therefore the strain B grows faster.

At 100mM sucrose concentration, the V0 = Vmax, therefore neither strain A nor strain B grows faster.

At 1000mM of sucrose concentration, the V0 = ½ Vmax for strain A hence, the strain A grows faster.

Therefore, the option 3 is correct.