Enzymes, metabolism and the central dogma

What the exam asks

Expect a graph or a table and a mechanism described in the stem. The commonest items are: identify an inhibitor's type from what happened to Vmax and Km, predict the effect of adding substrate, work out which comparison isolates the variable, and trace a described mutation to its protein. The trap that catches most candidates is reading a lower rate as a lower ceiling: a lower curve at one substrate concentration is consistent with every kind of inhibition, so it discriminates nothing. Two readings decide these questions, never one, and the second reading is always at the far right of the graph.

This topic covers what enzymes do to reaction rates, how their behaviour is described and altered, and how information moves from DNA to protein. It is the largest single block of Section III biology, because almost any biological scenario can be posed as an enzyme working faster or slower than it should.

The exam never asks you to recall a pathway. It gives you one, in the stem, and asks what happens when a step is blocked, an inhibitor is added, or a product accumulates. What you carry in is the reasoning: rate is set by the slowest step, an enzyme lowers activation energy without touching the position of equilibrium, and a system with feedback resists being pushed.

The central dogma half is examined the same way. You are given a mutation and asked what it does downstream, or given a protein and asked which step must have failed. The skill is tracing a consequence through a chain, not naming the chain.

What to hold

• An enzyme speeds a reaction by lowering activation energy. It does not change the equilibrium position, the free energy change, or which direction is favourable.
• Vmax is the ceiling the rate approaches at saturating substrate; Km is the substrate concentration giving half of Vmax, and a smaller Km means the enzyme gets there on less substrate.
• A competitive inhibitor binds the active site, so substrate can outcompete it: apparent Km rises, Vmax is unchanged.
• An inhibitor binding elsewhere cannot be outcompeted by substrate, so the ceiling itself falls.
• The rate of a pathway is set by its slowest step, so speeding any other step changes nothing.
• Feedback inhibition means the end product shuts off an earlier step, which is why a blocked pathway often makes an upstream intermediate accumulate.
• Transcription copies DNA to mRNA in the nucleus; translation reads mRNA to protein at the ribosome. Information runs that way and not back.
• A point mutation may be silent, may swap one amino acid, or may stop translation early. Which one depends on the codon, and the stem will give you the code.
• An insertion or deletion that is not a multiple of three shifts the reading frame, and everything downstream of it is wrong.
• Initial rates are measured before substrate is appreciably consumed, which is precisely why any argument resting on substrate running out is answering a later question.