Which of these statements about enzymes is not true?
- if enough substrate is added, the normal vmax of a reaction can be attained even in the presence of a competitive inhibitor.
- when [s] < < km, the reaction is second order and v0 depends on [s] and [et].
- their kcat is a second order rate constant.
- the lower their km, the better they recognize their substrate, but the lower their reaction rate.
- when [s] < < km, v0 depends on [s] and [et].

1. True. 2. True. 3. Not true. 4. True. 5. True

Explanation:

1. Yes, because if the amount of substrate i much greater than of competitive inhibitor then the probability of substrate to bind to ferment is much higher than of inhibitor (if we have noncompetitive inhibitor it damages the structure of active site and the substrate concentration does not have a role in reaction rate).

2. Yeah, because then the michaelis-menten equation will transform into [tex} V0=(kcat*[E]*[S])/Km [/tex] and it is a second order equation.

3. No, because it is measured in sec-1 and that means it is 1 rate constant.

4. True, if the lower Km the better is binding and due to that rate is slower because it's harder for substrate to unbind.

5. The same as question two.

The correct option is;

The sum of angles A and B are supplementary to angle C

Step-by-step explanation:

The statements are analysed as follows

1. Angle A is congruent to itself reflective property

Which shows that ΔABC and ΔADE have a common and equal angle

2. Segment ED and CB are parallel

From the transversal line passing EB and CB which shows that the angles ∠ADE and ∠ABC are equal and also ∠AED and ∠ACB are equal

The statement is used to prove similarity between the ΔABC and ΔADE

3. The sum of angles A and B are supplementary to angle C

The above statements relates to only ΔABC and i does not show similarity between ΔABC and ΔADE.