Question

part i – enzyme kinetics calculations

1. if vmax = 120 μmol/min and velocity = 60 μmol/min, determine substrate concentration relative to km.
2. explain effect of doubling enzyme concentration on km and vmax.

part ii – lineweaver-burk graph interpretation

1. identify inhibition type if lines intersect at y-axis.
2. identify inhibition type if lines are parallel.

part iii – clinical scenario

1. patient experiences flushing after alcohol consumption. explain biochemical cause.

part iv – hemoglobin

1. explain molecular basis of cooperativity.
2. explain bohr effect using proton concentration.

part v – carbohydrates

1. identify anomeric carbon in glucose.
2. explain difference between alpha and beta glucose.

Explanation:

Step1: Use Michaelis-Menten equation

The Michaelis-Menten equation is $v = \frac{V_{max}[S]}{K_m + [S]}$, where $v$ is reaction velocity, $V_{max}$ is maximum velocity, $[S]$ is substrate concentration, and $K_m$ is Michaelis constant.

Step2: Substitute given values

Substitute $v=60\ \mu\text{mol/min}$, $V_{max}=120\ \mu\text{mol/min}$:
$60 = \frac{120[S]}{K_m + [S]}$

Step3: Solve for $[S]$ relative to $K_m$

Multiply both sides by $K_m + [S]$:
$60(K_m + [S]) = 120[S]$
$60K_m + 60[S] = 120[S]$
$60K_m = 60[S]$
$[S] = K_m$

1. $K_m$ is an enzyme-substrate affinity constant, independent of enzyme concentration. $V_{max}$ is directly proportional to enzyme concentration, so doubling enzyme concentration doubles $V_{max}$ while $K_m$ stays unchanged.
2. In Lineweaver-Burk plots, lines intersecting at the Y-axis indicate competitive inhibition; this inhibition competes with substrate for the active site, leaving $V_{max}$ unchanged while increasing $K_m$.
3. Parallel lines in Lineweaver-Burk plots indicate uncompetitive inhibition; this inhibition binds only to the enzyme-substrate complex, decreasing both $V_{max}$ and $K_m$ proportionally, resulting in parallel lines.
4. This is due to aldehyde dehydrogenase (ALDH) deficiency. Alcohol is first converted to acetaldehyde by alcohol dehydrogenase; without functional ALDH, acetaldehyde accumulates, triggering vasodilation (flushing) and other symptoms.
5. Hemoglobin has 4 subunits. Binding of one oxygen molecule to a subunit induces a conformational change (from T-state to R-state) that increases the affinity of the remaining subunits for oxygen, creating cooperative binding.
6. The Bohr effect describes that high proton concentration (low pH, e.g., in tissues) stabilizes hemoglobin's low-affinity T-state. Protons bind to hemoglobin residues, reducing oxygen affinity, so hemoglobin releases more oxygen to oxygen-depleted tissues.
7. In glucose (an aldohexose), the anomeric carbon is carbon 1 (C1), which is the carbonyl carbon in the open-chain form that becomes chiral when the molecule cyclizes.
8. Alpha and beta glucose differ in the orientation of the hydroxyl group (-OH) on the anomeric carbon (C1). In alpha glucose, the -OH on C1 is oriented below the plane of the ring; in beta glucose, the -OH on C1 is oriented above the plane of the ring.

Answer:

1. $[S] = K_m$
2. $K_m$ remains unchanged; $V_{max}$ doubles.
3. Competitive inhibition
4. Uncompetitive inhibition
5. Aldehyde dehydrogenase (ALDH) deficiency leads to acetaldehyde accumulation, causing vasodilation (flushing).
6. Subunit conformational change (T→R state) after O₂ binding increases remaining subunits' O₂ affinity.
7. High proton concentration stabilizes hemoglobin's low-affinity T-state, reducing O₂ affinity and promoting O₂ release.
8. Carbon 1 (C1) of glucose
9. Alpha glucose has C1 -OH below the ring plane; beta glucose has C1 -OH above the ring plane.