What is the substrate concentration [S] when the Vo is 1/4 Vmax? Hint: start with Vo = (Vmax * [S]) / (Km + [S]) substitute in for the Vo and solve for [S]

Step-by-step explanation:

Given that,

Vo = ¼ Vmax

We are given the equation

Vo = (Vmax × [S]) / (Km + [S])

We want to find substrate [S]

Let substitute Vo = ¼Vmax into the given equation, so we have

Vo = (Vmax × [S]) / (Km + [S])

¼ Vmax = (Vmax × [S]) / (Km + [S])

Cross multiply

Vmax × (Km + [S]) = 4 × Vmax × [S] Divide both side by Vmax

(Km + [S]) = 4 × [S]

Km + [S] = 4•[S]

Make [S] subject of formula

4•[S] — [S] = Km

3•[S] = Km

Divide both sides by 3

[S] = Km / 3

[S] = ⅓ Km

So the substrate [S] is ⅓ Km when Vo is ¼Vmax

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3