AP Biology: Cellular Respiration and Respirometry Practice Questions
Interpret original seed-respiration experiments using stated assumptions and simple data. Distinguish gas measurements from ATP production, choose useful controls, and recognize limits on causal conclusions. This is independent study practice, not an official AP resource.
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1. A sealed seed respirometer contains an effective carbon dioxide absorbent. Temperature is stable, leaks are absent, and changes in other gases are negligible. What does a decrease in gas volume primarily indicate?
Explanation. With released carbon dioxide removed and the other stated conditions controlled, loss of gas mainly tracks oxygen uptake. The apparatus does not count ATP molecules; converting oxygen uptake to ATP production would require additional assumptions and measurements.
2. A seed chamber and a matched chamber containing only inert glass beads share a water bath. Both have carbon dioxide absorbent and the same initial gas volume. Why record the bead chamber alongside the seed chamber?
Explanation. The beads do not respire. A matched chamber can reveal background changes associated with the apparatus and physical conditions. Its reading can support a correction when the drift is shared; it does not measure ATP or replace biological replication.
3. An instrument reports gas-volume decrease as a positive value. Over 8 minutes, the seed chamber shows 0.46 mL and its matched bead control shows 0.06 mL. Assume additive shared drift. What is the corrected oxygen uptake rate?
Explanation. Subtract the control in the same sign convention: 0.46 - 0.06 = 0.40 mL over the interval. Dividing by 8 minutes gives 0.050 mL per minute. Adding the control or using the uncorrected reading would retain rather than remove the assumed drift.
4. A 2.5 g seed sample has a corrected oxygen uptake of 0.75 mL over 6 minutes. What is its average mass-specific oxygen uptake rate?
Explanation. Normalize by both elapsed time and sample mass: 0.75 / (6 x 2.5) = 0.050 mL per gram per minute. Dividing only by time gives the whole sample's rate, while dividing only by mass gives uptake per gram over the entire six-minute interval.
5. In a model respirometer without a carbon dioxide absorbent, a sample consumes 0.30 mmol of oxygen and releases 0.30 mmol of carbon dioxide into the gas space. Temperature and pressure are constant and other gas changes are negligible. What net gas-volume change is predicted?
Explanation. The gas loses 0.30 mmol of oxygen and gains 0.30 mmol of carbon dioxide, so the net change in gas amount is zero. At fixed temperature and pressure the volume therefore stays approximately constant. No net volume change does not imply no respiration.
6. A class compares oxygen uptake at two temperatures using independently prepared seed chambers. Which design best isolates temperature as the explanatory variable?
Explanation. Comparable samples, random assignment, and controls at each temperature reduce alternative explanations for a rate difference. Species, mass, leaks, and unequal unnormalized durations could otherwise confound the comparison. Independent replicate chambers also help estimate variability.
7. Corrected cumulative oxygen uptake is 0.00, 0.10, 0.24, and 0.42 mL at 0, 2, 4, and 6 minutes, respectively. In which measured interval is the average uptake rate greatest?
Explanation. The interval rates are 0.10/2 = 0.050, 0.14/2 = 0.070, and 0.18/2 = 0.090 mL per minute. The largest is from 4 to 6 minutes. Rates come from changes in cumulative uptake over time, not from dividing every reading by the same endpoint time.
8. After correction with a sealed bead control, a seed respirometer shows almost no volume change. Inspection finds that its stopper is loose and outside air can enter. Which conclusion is justified?
Explanation. A leak allows gas exchange that is not caused by the sample's respiration and can mask a volume decrease. A sealed control does not correct a leak unique to the sample chamber. The near-zero reading therefore cannot establish the seeds' respiration rate.
9. A sealed respirometer has corrected oxygen uptake of 0.080 mL per minute. Which additional claim cannot be determined from this measurement alone?
Explanation. Oxygen uptake alone does not specify an exact ATP production rate. Coupling efficiency and the contributions of different metabolic processes would also matter. The other statements concern units, matched comparisons, or the arithmetic of an explicitly constant rate.
10. Each temperature treatment uses four independently prepared seed chambers. Each chamber is read five times during one trial. How many independent biological replicate chambers are there per treatment?
Explanation. The independent experimental units are the four separately prepared chambers per treatment. Five observations from one chamber describe its trajectory but remain repeated measurements of that unit. Counting all twenty readings as independent chambers would overstate replication.
11. Two groups of germinating seeds are tested with matched controls. Group A consumes 0.60 mL of oxygen in 10 minutes and has a mass of 3.0 g. Group B consumes 0.48 mL in 10 minutes and has a mass of 1.5 g. Both uptake values are corrected. Which comparison is correct?
Explanation. Total rates are 0.060 mL per minute for A and 0.048 for B. Dividing by mass gives 0.020 mL per gram per minute for A and 0.032 for B. Thus A has higher whole-sample uptake while B has higher mass-specific uptake; the comparison depends on the stated measure.
12. In a well-controlled seed experiment, mean mass-specific oxygen uptake is higher at 24 degrees C than at 14 degrees C. Only these two temperatures and this seed species were tested. Which conclusion stays within the evidence?
Explanation. The results support the observed comparison in the tested system. Two temperatures cannot locate a full optimum or establish a trend at untested temperatures, and one species cannot establish a universal response. Additional conditions and species require their own measurements.
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FAQ
Why correct a seed respirometer reading with a matched control?
A nonliving control helps estimate shared physical drift. Apply the correction using the instrument's sign convention and matched conditions; it cannot repair a leak unique to one chamber.
How do I calculate a mass-specific oxygen uptake rate?
Divide corrected oxygen uptake by elapsed time and sample mass. For example, 0.36 mL over 6 minutes from a 2 g sample is 0.030 mL per gram per minute.
Does oxygen uptake reveal an exact ATP production rate?
No. Oxygen uptake measures one aspect of respiration. An exact ATP estimate also requires information about coupling and other metabolic contributions.