| Literature DB >> 34100646 |
Sachel M Villafañe1, Vicky Minderhout2, Bruce J Heyen3, Jennifer E Lewis4, Andrew Manley2, Tracey A Murray5, Heather Tienson-Tseng6, Jennifer Loertscher2.
Abstract
Metabolic systems form the very foundation of life and as such are broadly taught in the molecular life sciences. Here, we describe the biochemistry educator community-based development and use of an assessment instrument designed to evaluate students' ideas about metabolic pathway dynamics and regulation in undergraduate biochemistry courses. Analysis of student responses showed that most students were able to interpret visual representations in an unfamiliar metabolic pathway and that many could make basic predictions about how the system would be expected to respond to changes. However, fewer students generated nuanced responses that accounted for both microscopic changes at the protein level and macroscopic changes in pathway product outputs. These findings identify some of the challenges of meaningfully assessing students' understanding of metabolic pathways and could inform how instructors think about teaching and assessing metabolism in undergraduate biochemistry and beyond. The results also suggest future avenues for biochemistry education research.Entities:
Mesh:
Year: 2021 PMID: 34100646 PMCID: PMC8715806 DOI: 10.1187/cbe.20-04-0078
Source DB: PubMed Journal: CBE Life Sci Educ ISSN: 1931-7913 Impact factor: 3.325
Metabolic pathway dynamics and regulation threshold concept (adapted from Loertscher )
|
|
Mapping targeted learning objectives to instrument items and rubric
| Learning objectives | Items | Rubric statements |
|---|---|---|
| Interpret visual representations related to regulation and dynamics in pathways. | Q1 Q2 | Multiple choice—no rubric |
| Predict the effect of an allosteric modifier on the rate of product formation. | Q4 | R4A: Increase rate of delta production.R4B: Foxtrot is an inhibitor of X2 activity.R4C: Loss of allosteric binding leads to increased rate of delta production. |
| Predict the effect of including a downstream product in the growth medium. | Q5 | R5A: Foxtrot production initially increases due to inhibition of Y.R5B: Over time there will be little change in foxtrot production. |
| Create a strategy to maximize production of an intermediate through enzyme mutation. | Q7Q8/9 | R8/9A: Regulatory site of Y is mutated.R8/9B: Mutation will relieve the inhibitory effect of echo on Y.R8/9C: The mutation will prevent echo from binding Y.R8/9D: The mutation creates unregulated (high) production of echo. |
| Explain the role of isoenzymes in metabolic pathways. | Q11 | R11A/B: Isoenzymes allow for fine-tuning of metabolites under fluctuating conditions (A); isoenzymes are regulated differently under different conditions (B).R11C: X1 and X2 have different regulatory properties. |
FIGURE 1.Cover page of the assessment instrument showing a bacterial amino acid biosynthetic pathway and explanation of visual conventions. The figure shown here has been modified from the version received by students to preserve the security of the instrument. Pathway intermediates (shown in boxes) have been assigned names from the International Civil Aviation Organization phonetic alphabet. Enzymes (shown in circles) were assigned letter names in the original version of the instrument and have been assigned different letter names in this modified version.
FIGURE 2.Timeline summarizing the iterative instrument development process.
Instrument structure and item types
| Section 1 | |
|---|---|
| Q1 and Q2 | Multiple-choice questions probing understanding of visual representations used |
| Q3 | Multiple-choice confidence question |
Sample student responses to Q4 (enzyme X2 was mutated to eliminate the effects of foxtrot as an allosteric modifier)
| Category | Rubric statement(s) | Sample student responses |
|
|---|---|---|---|
| Fully correct prediction and justificationMost students omitted a reference to binding | Delta production will increase because foxtrot will be unable to inhibit X2 activity because it cannot bind to its regulatory site.It will increase the rate of delta production because the mutation allows X2 to catalyze the creation of delta unhindered by any allosteric inhibition from foxtrot. | 168 | |
| Correct prediction, but limited justification | The rate of delta will increase because X2 is no longer being inhibited by product so it won’t know when to stop and keep catalyzing the reaction. | 33 | |
| Incorrect prediction, but recognition of foxtrot as an inhibitor | The mutation will allow delta production to continue even with high concentrations of foxtrot. This occurs because foxtrot is no longer able to allosterically inhibit X2. | 16 | |
| Incorrect ideas | None of the possible statements | X2 was mutated at an allosteric site not the active site of X2. This allosteric site mutation will probably have no effect on the rate of delta production because the active site is still intact and able to catalyze the reaction. | 31 |
aN = total number of responses fitting a given pattern.
Sample student responses to Q11 (explain why organisms have multiple isoenzymes in pathways and use X1 and X2 to explain the advantage of having two isoenzymes in E. coli)
| Category | Rubric statement(s) | Sample student responses |
|
|---|---|---|---|
| Fully correct response | Isoenzymes allow for more precise, complex regulation. This is especially important for branch point enzymes, because a single metabolite can form multiple products, allowing for partial inhibition. X1 and X2 allows this step to be only partially inhibited by foxtrot so that it can continue to produce other products such as foxtrot. | 116 | |
| General statement about role of isoenzymes without specific consideration of X1 and X2 | Having multiple isoenzymes enables the same reaction to be allosterically regulated by different products without affecting the rates of other pathways that use the same intermediates. | 37 | |
| Considered X1 and X2 specifically and did not make a generalization about the role of isoenzymes | Multiple isoenzymes allows for different inhibition of the use of metabolites as a result of different things. X1 allows for delta to be made still even if there is excess foxtrot. X2 allows production of delta because it does not need to make as much to produce foxtrot if there are high levels of foxtrot. | 53 | |
| Incorrect ideas | None of the possible statements | Isoenzymes can catalyze the same reaction using different enzymes, if one enzyme is mutated, the cell would be able to continue the pathway if another enzyme that catalyzes the same reaction is available. | 42 |
aN = total number of responses fitting a given pattern.
Sample student responses to Q5 (predict what would happen to the overall production of foxtrot in the presence of high concentrations of echo and offer an explanation)
| Category | Rubric statement(s) | Sample student responses |
|
|---|---|---|---|
| Fully correct prediction and justification | In high concentrations of echo, enzyme Y would be inhibited, therefore all the delta substrate will be used to produce foxtrot, increasing the overall production of foxtrot until accumulation and then foxtrot will inhibit the enzymes in the foxtrot pathway. | 41 | |
| Only considered initial effect | I would expect foxtrot production to increase because more delta is available to enzyme Z since enzyme Y has been allosterically inhibited. | 129 | |
| Did not mention initial effect, but made correct prediction about foxtrot production | It will remain constant because excess foxtrot will inhibit enzyme Z preventing the accumulation of excess foxtrot. | 6 | |
| Incorrect ideas | None of the possible statements | Foxtrot production would not be affected since echo only inhibits its own production (only inhibits enzyme Y and not enzyme catalyzing foxtrot). | 72 |
aN = total number of responses fitting a given pattern.
Sample student responses to Q8/9 (explain how a proposed mutation would affect the enzyme and why the mutation would be effective for overproducing echo)
| Category | Rubric statement(s) | Sample student responses |
|
|---|---|---|---|
| Fully correct prediction and justificationShows evidence of molecular and systems level thinking. | Mutate the regulatory site. This wouldn’t allow echo to bind to inhibit the enzyme. By mutating the regulatory site, you prevent echo from binding, thus preventing inhibition of echo production. This would cause a theoretically unlimited production of echo.I would mutate the regulatory site on enzyme Y so that echo would no longer be able to bind in the correct orientation to Y and allosterically inhibit it. Thus, Y would not be able to be affected by the presence of echo. | 88 | |
| Correct predictionJustification shows evidence of systems thinking, but lacks evidence of molecular level thinking | Mutate the allosteric inhibition site so echo cannot allosterically inhibit the enzyme. It would be effective because now even if there is a high concentration of echo the enzyme won’t be inhibited and will still make echo. | 63 | |
| Correct prediction, but limited justification | Allosteric regulation site; not active site; still want enzyme Y to catalyze reaction of delta to echo. | 16 | |
| Incorrect proposal |
| Echo is a competitive inhibitor of enzyme Y (product inhibition). Mutate the active site so it has lesser affinity for echo. Echo would be less likely to bind enzymes so its inhibition will be less. | 27 |
aN = total number of responses fitting a given pattern.