Follow a molecular switch after its input ends

1. Frame the biological question

A G-protein-coupled receptor links extracellular recognition to an intracellular nucleotide switch. An activated receptor promotes exchange of GDP for GTP on a G-protein subunit. The GTP-bound state can regulate an effector; GTP hydrolysis returns it toward an inactive GDP-bound state. Turning the switch off does not require destroying every protein molecule.

2. Explain mechanisms and assumptions

An effector may produce a second messenger such as cyclic AMP, which can activate downstream proteins. Messenger production, degradation and target regulation introduce additional time scales. The present model stops at the active G-protein fraction; it deliberately does not equate that fraction with a measured cAMP concentration or transcriptional response.

Our fixed pool begins inactive. During a two-minute input, activation converts inactive material at rate kact(1−f), while hydrolysis removes active material at rate khyd f. The sustained-input limit is kact/(kact+khyd). After the input ends, activation is zero and the remaining active fraction declines exponentially with hydrolysis rate.

A prolonged signal after ligand removal could reflect slow switch termination, persistent ligand, receptor retention or a downstream feedback loop. The modeled contrast isolates only one explanation. To discriminate mechanisms, retain upstream input and downstream time courses rather than interpreting a single late endpoint as proof that receptors remain occupied.

3. Work a quantitative example

For kact=khyd=1 min⁻¹, the sustained-input limit is 0.5. At the two-minute pulse end, f=0.5(1−exp(−4))≈0.49084. Two minutes later it is 0.49084exp(−2)≈0.06643. The post-pulse half-life is ln2≈0.693 min.

4. Predict and investigate

Predict first: If hydrolysis becomes slower, will the signal necessarily disappear immediately when receptor activation ends?

Compare hydrolysis rates 0.5 and 2 at one activation rate. Separate pulse-end fraction from the post-pulse decay. Use the sustained-input dashed curve only as a modeled counterfactual, not as a performed control experiment.

Retain the initial prediction, settings, raw quantities and interpretation in the lesson notebook. These are supplied or calculated teaching records, not observations from a performed biological experiment.

5. Apply the reasoning to a changed case

Apply the mechanism to a changed case

A switch has active fraction 0.8 when activation stops and hydrolysis rate 0.5 min⁻¹. Predict its active fraction two minutes later and its half-life.

Retain your calculation, units and assumptions in the lesson notebook before opening this public practice response.

Compare a worked response

The remaining fraction is 0.8exp(−1)≈0.2943. Half-life is ln2/0.5≈1.386 min. This predicts the specified switch, not every downstream messenger.

6. Reflect and retain evidence

Which observation would distinguish the proposed mechanism from a measurement artifact or an omitted process? Define the biological unit, controls, counting boundary and uncertainty. A correct calculation or submitted plan does not certify supervised practical skill or independently demonstrated mastery.

Sources and course continuation

Scientific supporting reference, checked 30 October 2026. Original prose, numerical cases and diagrams; no source exercise or figure reproduced. Continue through General Biology I for mixed written evidence and the retained multistage project.