Question 2 of 6: Protein Structure, Function and Enzyme Kinetics

Question 2: Protein Structure, Function and Enzyme Kinetics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Tertiary structure is the overall three-dimensional fold of a single polypeptide chain, produced when the secondary-structure elements (α-helices, β-sheets, loops) pack together and are stabilized by hydrophobic burial of nonpolar side chains, hydrogen bonds, ionic (salt-bridge) interactions and, in some proteins, disulfide bonds. This folding is what creates a catalytic domain: an enzyme's primary sequence typically places the amino-acid side chains that will ultimately form the active site far apart in the linear chain, and it is only the specific three-dimensional fold that brings those residues together in space to form a contiguous binding pocket with the correct geometry, electrostatics and chemical functionality (acid/base groups, nucleophiles, metal-binding residues) to bind the substrate and stabilize the reaction's transition state. Without the correct tertiary fold — for example, after thermal or chemical denaturation, which disrupts the tertiary (and secondary) structure while leaving the primary sequence intact — those same catalytic residues are scattered and mobile in an unfolded chain and can no longer form a functional active site, so catalytic activity is lost even though every amino acid is still present. Tertiary structure therefore converts a one-dimensional sequence of building blocks into a precisely shaped, chemically functional catalytic machine, and it is this shape-function relationship (not the sequence per se) that determines substrate specificity and catalytic efficiency.

(b) Parts (i)–(iii) apply the Michaelis–Menten equation directly to compute and then graph the enzyme's rate-vs-substrate-concentration behaviour.

Given.

Find. The reaction rate $v$ at $[S]=0.8$ mM and at $[S]=1.2$ mM, and a labelled graph of $v$ vs $[S]$ over a range that shows the saturating (hyperbolic) shape of the curve together with $V_{max}$ and $K_M$.

Approach. Substitute each given $[S]$ into the Michaelis–Menten equation to get $v$ directly; then evaluate the same equation over a range of $[S]$ to plot the full saturation curve and mark the asymptote at $V_{max}$ and the half-saturation point that defines $K_M$.

1. Michaelis–Menten rate law. For a simple one-substrate enzyme reaction, the initial rate is $$ v = \frac{V_{max}[S]}{K_M + [S]} $$ with $V_{max}=100\ \mu\text{mol/s}$ and $K_M = 1\ \text{mM}$.
2. (i) Rate at $[S]=0.8$ mM. $$ v(0.8) = \frac{(100)(0.8)}{1 + 0.8} = \frac{80}{1.8} = \boxed{44.4\ \mu\text{mol/s}} $$ Because $[S] < K_M$, the enzyme is well below half-saturation and the rate is correspondingly below $V_{max}/2$.
3. (ii) Rate at $[S]=1.2$ mM. $$ v(1.2) = \frac{(100)(1.2)}{1 + 1.2} = \frac{120}{2.2} = \boxed{54.5\ \mu\text{mol/s}} $$ Here $[S] > K_M$, so the enzyme is past half-saturation and the rate is correspondingly above $V_{max}/2 = 50\ \mu\text{mol/s}$, but still well short of $V_{max}$ because saturation is approached only asymptotically.
4. (iii) Full curve and key features. Evaluating $v([S])$ over $[S]=0$ to $6$ mM traces the characteristic rectangular-hyperbola shape of Michaelis–Menten kinetics (figure below). Two features are labelled: the horizontal asymptote at $v=V_{max}=100\ \mu\text{mol/s}$, which the curve approaches but never reaches because full saturation of every enzyme active site would require infinite substrate; and $K_M$, defined algebraically as the substrate concentration at which $v=V_{max}/2$ — substituting $[S]=K_M$ into the rate law gives $v=V_{max}K_M/(K_M+K_M)=V_{max}/2=50\ \mu\text{mol/s}$, so $K_M$ is read off the graph as the $x$-value directly below the point where the curve crosses the $v=50$ line. The two computed points from (i) and (ii) are marked on the curve and fall exactly where their $[S]$ and $v$ values place them relative to that half-saturation point, which is itself a useful check that the numbers in (i)/(ii) are internally consistent with the shape of the curve.