Apply differential elongation data to shoot tropisms
Calculate growth increments and ratios before predicting curve direction.
Extended only (Supplement); Biology 0610, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked
Before you begin
- Identify the root and shoot of a seedling; recall growth, cell structure and the variables in a controlled investigation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which calculation gives the shaded-to-lit growth ratio?
What you will learn
- Use increments rather than final lengths for growth ratios
- Apply the same geometry to light and gravity models
Apply differential elongation data to shoot tropisms
Invented cell-length models: lit/upper increment 10 µm, shaded/lower increment 30 µm, ratio 3. Balanced increments 20 µm each. No curvature angle is supplied. Original teaching model.
| Model and side | Initial / µm | Final / µm |
|---|---|---|
| Light · Lit | 20 µm | 30 µm |
| Light · Shaded | 20 µm | 50 µm |
| Gravity · Upper | 20 µm | 30 µm |
| Gravity · Lower | 20 µm | 50 µm |
| Balanced · Left | 20 µm | 40 µm |
| Balanced · Right | 20 µm | 40 µm |
| Feature | Meaning |
|---|---|
| Light model | Lit 20→30; shaded 20→50 µm |
| Gravity model | Upper 20→30; lower 20→50 µm |
| Balanced model | Both sides 20→40 µm |
| Interpretation | Compare increments; shoot scope; no invented angle or universal dose rule |
Use increments rather than final lengths for growth ratios
The invented elongation table supplies matching cell lengths, not a directly measured whole-shoot angle. In the light model, both sides begin at 20 µm; lit-side cells finish at 30 µm and shaded-side cells at 50 µm. The elongation increments are 10 and 30 µm. Shaded-side elongation is therefore three times the lit-side elongation in this particular model.
In the horizontal shoot-gravity model, upper-side cells grow from 20 to 30 µm and lower-side cells from 20 to 50 µm. The upper increment is 10 µm and lower increment is 30 µm, again a ratio of three. The lower side becomes longer, so the shoot bends towards the upper side and grows away from gravity. The ratio is supplied-model arithmetic, not a universal auxin dose rule.
Apply the same geometry to light and gravity models
A balanced comparison has cells on each side growing from 20 to 40 µm. Both increments are 20 µm, giving a ratio of one and straight elongation in the simplified model. Comparing final lengths instead would answer a different question: 50/30 is about 1.67, not the ratio of the 30 and 10 µm growth increments.
Interpret the data through the full chemical account: auxin from the shoot tip becomes unequal under directional light or gravity and stimulates unequal shoot cell elongation. The data illustrate growth consequences but do not directly measure auxin concentration. They also do not provide sufficient whole-region dimensions to calculate a curvature angle or establish an unchanged root response. Retain these limits when explaining an unfamiliar record.
Pause and explain
What does the balanced 20→40 µm comparison predict?
Worked example
Calculate shaded-to-lit elongation ratio for 20→50 and 20→30 µm.
Show the worked solution
- Subtract initial lengths: 30 and 10 µm.
- Divide increments: 30/10 = 3.
Answer Shaded-to-lit elongation ratio is 3; the shoot bends towards the less-elongating lit side.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the lit and shaded increments and their shaded-to-lit ratio.
Give me a hint
Subtract 20 µm from both final lengths.
Compare my reasoning
- Lit: 30−20 = 10 µm; shaded: 50−20 = 30 µm.
- Ratio: 30/10 = 3.
Lit increment 10 µm; shaded increment 30 µm; shaded-to-lit elongation ratio 3.
Look for these in your work
- Use increments with µm
- Calculate the ratio
Calculate the upper and lower increments in the horizontal shoot model and predict its response.
Give me a hint
The lower side is the longer outside of the curve.
Compare my reasoning
- Upper adds 10 µm; lower adds 30 µm.
- Greater lower elongation bends the shoot up away from gravity.
Upper increment 10 µm; lower increment 30 µm. The shoot bends up away from gravity, showing negative gravitropism.
Look for these in your work
- Calculate both increments
- Apply the correct growth geometry
A report uses 50/30 to claim a growth ratio, an exact curvature angle and an identical root response. Evaluate all three claims.
Give me a hint
Separate final lengths, geometry and plant-part scope.
Compare my reasoning
- Growth ratio uses increments, so it is 30/10 = 3.
- No angle geometry is supplied; the stated auxin rule concerns shoots.
50/30 compares final lengths, not growth increments; the elongation ratio is 3. No exact angle follows from the supplied lengths alone, and the shoot auxin rule cannot be transferred unchanged to roots.
Look for these in your work
- Correct the growth calculation
- Reject unsupported angle and root claims
Common mistakes
- Using 50/30 as the elongation ratio.
- Calculating an angle from cell lengths without geometry.
What can you explain now?
What does the balanced 20→40 µm comparison predict?
Compare with the explanation
Equal increments and straight growth in the model
Both sides elongate by 20 µm.
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Make it stick
Calculate differential elongation ratios and connect them to light and gravity responses with the model limits retained.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.