In a linear pathway \(A \to B \to C \to D\), each arrow is an enzyme-catalyzed step. At steady state, the concentration of each intermediate stays constant because the rate at which it is formed equals the rate at which it is consumed. When the enzyme that converts \(B\) to \(C\) is missing, \(B\) can still be formed from \(A\) but can no longer be consumed. Its formation now exceeds its consumption, so \([B]\) rises. Because \(C\) is no longer being produced, the reactions that consume \(C\) to make \(D\) draw down the existing pool of \(C\), so \([C]\) falls. \(D\), the final product, becomes depleted because its supply chain is cut. The pattern is directional: everything upstream of the block accumulates, everything downstream is depleted. The magnitude of accumulation depends on how fast the upstream enzyme supplies \(B\) and how quickly \(B\) can be removed by any side reaction; if no side reaction exists, \([B]\) rises until the upstream enzyme itself is limited by substrate or product inhibition.
Learn MotionWhen a Key Metabolic Enzyme Is Missing: Pathways, Regulation, and Physiological Consequences
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Watching Accumulation and Depletion at a Blocked Step
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Start with all four enzyme activities at their normal levels and watch the concentrations settle into steady state — each intermediate stays flat because it is made as fast as it is used. Now lower the activity of the enzyme that converts B to C. B is still being produced from A, but it is no longer being consumed, so its concentration climbs. C, meanwhile, is no longer being made, and the step that uses C keeps running, so C drains away. D follows C downward. Notice the direction of the effect: everything upstream of the block rises, everything downstream falls. Try raising the activity of the upstream enzyme — B accumulates faster. Then switch on the side reaction that removes B, and the rise levels off, because B now has an escape route.
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