The missing graph is in the attachment.
Answer: (a) [S] = 0.0016M
(b) Vmax = 3V; Vmax = [tex]\frac{3V}{2}[/tex]; Vmax = [tex]\frac{11V}{10}[/tex]
(c) Enzyme A: black graph; Enzyme B = red graph
Explanation: Enzyme is a protein-based molecule that speed up the rate of a reaction. Enzyme Kinetics studies the reaction rates of it.
The relationship between substrate and rate of reaction is determined by the Michaelis-Menten Equation:
[tex]V=\frac{V_{max}[S]}{K_{M}+[S]}[/tex]
in which:
V is initial velocity of reaction
Vmax is maximum rate of reaction when enzyme's active sites are saturated;
[S] is substrate concentration;
Km is measure of affinity between enzyme and its substrate;
(a) To determine concentration:
[tex]0.25V_{max}=\frac{V_{max}[S]}{0.005+[S]}[/tex]
[tex]0.25V_{max}(0.005+[S])=V_{max}[S][/tex]
[tex]0.00125+0.25[S]=[S][/tex]
0.75[S] = 0.00125
[S] = 0.0016M
For a Km of 0.005M, substrate's concentration is 0.0016M.
(b) Still using Michaelis-Menten:
[tex]V=\frac{V_{max}[S]}{K_{M}+[S]}[/tex]
Rearraging for Vmax:
[tex]V_{max}=\frac{V(K_{M}+[S])}{[S]}[/tex]
(b-I) for [S] = 1/2Km
[tex]V_{max}=\frac{V(K_{M}+0.5K_{M})}{0.5K_{M}}[/tex]
[tex]V_{max}=\frac{V(1.5K_{M})}{0.5K_{M}}[/tex]
[tex]V_{max}=[/tex] 3V
(b-II) for [S] = 2Km
[tex]V_{max}=\frac{V(K_{M}+2K_{M})}{2K_{M}}[/tex]
[tex]V_{max}=\frac{V(3K_M)}{2K_M}[/tex]
[tex]V_{max}=\frac{3V}{2}[/tex]
(b-III) for [S] = 10Km
[tex]V_{max}=\frac{V(K_{M}+10K_M)}{10K_M}[/tex]
[tex]V_{max}=\frac{V(11K_{M})}{10K_{M}}[/tex]
[tex]V_{max}=\frac{11V}{10}[/tex]
(c) Being the affinity between enzyme and substrate, the lower Km is the less substrate is needed to reach half of maximum velocity.
Km of enzyme A is 2μM and of enzyme B is 0.5μM.
Enzyme B has lower Km than enzyme A, which means the first will need a lower concnetration of substrate to reach half of Vmax.
Analyzing each plot, notice that the red-coloured graph reaches half at a lower concentration, therefore, red-coloured plot is for enzyme B, while black-coloured plot is for enzyme A
