Design evidence that separates enzyme explanations

One observation can fit several settings

At S = 2 and E = 1 the model rate is 5 µmol/min. At S = 1 and E = 1.5, capacity is 15 and the saturation fraction is 1/3, again giving 5 µmol/min. One rate therefore does not uniquely identify both substrate and enzyme amounts.

A tempting comparison, S = 10 and E = 0.5, instead gives 5 × 10/12, about 4.167 µmol/min. Similar-looking settings are not necessarily equivalent. Calculate the prediction and compare the difference before claiming a match.

Design a substrate series

Choose several substrate concentrations while holding enzyme amount, reaction volume, pH and temperature comparable. Define the short time interval used to estimate initial rate. Include a zero-enzyme preparation to detect a signal that occurs without the enzyme, and a zero-substrate preparation to investigate background. Plan independent preparations and retain all observations.

A control can reveal an alternative explanation; it does not guarantee all alternatives are removed. For example, a measurement signal might be nonlinear with product concentration. Calibration of the measuring instrument is a separate task.

In the browser, repeated clicks at the same settings yield identical results. They are not independent experimental replicates and cannot estimate measurement variation. Distinguish the deterministic prediction from the future measurements your procedure would collect.

Plan a model challenge

Reserve a concentration or condition before fitting a model. Predict its rate and explain what discrepancy would matter relative to measurement uncertainty. A mismatch may indicate poor measurements, incorrect concentrations or an inadequate kinetic model. Do not automatically blame the specimen.

Write a procedure another learner could follow, a table layout with units, and an alternative explanation. Specify which real-world work would require an educator-approved protocol. This lesson does not ask you to handle chemicals or biological samples; your evidence can combine the controlled browser comparison, the recorded-data investigation below and your written plan.

Compare a model with recorded enzyme rates

The Puromycin dataset distributed with R contains 23 observations attributed to Treloar (1974), reproduced through Bates and Watts (1988). Substrate concentration is in ppm; rate is derived from radioactive-product counts, in counts/min/min. The documentation describes one treated and one untreated experiment. It does not establish that repeated rows are independent treatment replicates. Read the R dataset description.

These are recorded observations, not regenerated simulator outputs. Do not convert their rate units to µmol/min without an appropriate calibration. Our earlier simulator’s numerical parameters and units do not transfer to this experiment. This activity uses an existing data record; it does not involve handling puromycin or radioactive material.

Predict before calculating: if substrate concentration doubles from 0.11 to 0.22 ppm, will the mean treated rate double? Record your prediction in the lesson notebook. Then calculate each mean using both observations, compare the ratio, and retain the individual measurements when explaining the result.

Inspect all observations

The source row is a record identifier, not an asserted specimen identifier or measurement time. The untreated series has only one observation at 1.1 ppm; do not invent a partner or treat absent replication as zero variation. Download the complete CSV to make a scatterplot. Use shared axes from 0 to 1.2 ppm and 0 to 240 counts/min/min, and distinct marker shapes for the two series. Plot every row rather than connecting observations as a time sequence.

Check the doubling prediction after calculating

The treated means are (123 + 139)/2 = 131 at 0.11 ppm and (159 + 152)/2 = 155.5 at 0.22 ppm. The rate ratio is about 1.187, an increase of about 18.7%, while concentration increased by 100%. These observations challenge proportionality over this interval. They are compatible with a flattening response, but do not by themselves prove a unique molecular mechanism.

Challenge two explicit predictions

Compare two proposed descriptions of the treated series. They are teaching candidates, not fitted estimates or accepted biological constants. Let S mean substrate concentration in ppm and v mean predicted rate in counts/min/min.

• Proportional candidate: v = 1,200S. The slope has rate units per ppm.
• Saturation candidate: v = 220S/(0.06 + S). The capacity 220 has rate units; 0.06 has concentration units.

At S = 0.06, the candidates predict 72 and 110 respectively. The observed values are 97 and 107. Define a residual as observed minus predicted: the saturation residuals are −13 and −3, whereas the proportional residuals are 25 and 35. A negative residual means the prediction was too high. Keep both residuals rather than replacing the observations with their mean.

Your task: calculate both candidates at 0.22 and 1.1 ppm. Which has smaller absolute residuals at those concentrations? Then identify a concentration where even the better candidate shows disagreement. Explain why a better match among these two choices does not establish that its assumptions are true.

Compare predictions and residuals

At 0.22 ppm, proportional prediction is 264 and saturation prediction is about 172.86. Saturation residuals are about −13.86 and −20.86; proportional residuals are −105 and −112. At 1.1 ppm, predictions are 1,320 and about 208.62. Saturation residuals are about −1.62 and −8.62 for observations 207 and 200. At 0.02 ppm, saturation predicts 55, between observations 47 and 76. The remaining discrepancies could involve measurement variability, concentration uncertainty or model inadequacy; this table alone cannot separate them.

What would strengthen the conclusion?

At 0.56 ppm, calculate the mean of each series and their difference. Describe that difference before making a causal claim. The record does not supply preparation identities, random assignment, run order or calibration uncertainty. Repeated values at one concentration do not supply independent replication of the treatment experiment.

Check the comparison and its limit

The means are 196 for treated and 151 for untreated, a difference of 45 counts/min/min. That is a descriptive comparison of these records. Independent preparations, allocation information and measurement controls would help distinguish a reproducible treatment effect from other differences between the experimental runs.

In your saved reflection, include the dataset identity r-puromycin-frozen-v1, source row numbers, your original prediction, one calculation, one residual comparison and a limitation. Propose a new independently collected concentration series to test the candidate. Because all these observations and feedback are visible, reusing this table is practice, not a previously unseen validation study. A reviewer should assess the argument and proposed design, not just whether text was entered.

The local bundle preserves the original R data source, provenance and checksums, copyright notices and GPL version 2 license. CSV conversion preserves source order and all values, with explicit column names and an added row identity. The dataset bundle is distributed under GPL version 2 without warranty.