Unit 3 practice test: Cellular Energetics

Isoleucine lowers the rate from 100% to 4% across the range; valine leaves it at 97-100%.

• Correct: Its rate falls steeply as isoleucine rises; the same amounts of valine barely change it.: Steep fall with isoleucine, almost none with valine.
• Its rate falls by about the same amount whether isoleucine or valine is added.: Valine leaves the rate at 97-100%, while isoleucine cuts it to 4%.
• Its rate rises as isoleucine rises and falls as valine rises.: Isoleucine lowers the rate; it does not raise it.
• Its rate falls just at 5.0 mM isoleucine, and at lower amounts isoleucine has no clear effect on it.: The rate already falls to 82% at 0.1 mM and 45% at 0.5 mM.

Changing one amino acid at the allosteric site makes enzyme 1 ignore isoleucine (93-99%), so isoleucine must act at that site. Valine, though similar, does not fit it.

• Correct: Isoleucine binds enzyme 1's allosteric site and changes its shape; the changed site does not bind it.: The changed-enzyme column ties isoleucine's effect to the allosteric site.
• Isoleucine competes with threonine for enzyme 1's active site, since both are amino acids of a similar size and shape.: The changed enzyme differs only at the allosteric site, yet isoleucine stops working on it, so the active site is not where isoleucine acts.
• Isoleucine denatures enzyme 1, since the rate drops close to zero at the highest amount added.: Denaturation would not be undone by a change at the allosteric site, and a substance that denatures proteins would affect the changed enzyme too.
• Isoleucine is broken down by enzyme 1 instead of threonine, so less threonine is converted.: The experiments had plenty of threonine, and the allosteric-site result points to regulation, not a new substrate.

The normal strain's pathway shuts down when isoleucine is plentiful. The changed enzyme 1 ignores isoleucine, so the pathway keeps running and wastes raw materials and energy.

• Correct: It keeps making isoleucine it does not need, spending threonine and ATP the normal strain saves.: Without feedback inhibition, the pathway runs whether or not the product is needed.
• It stops making isoleucine sooner, because its enzyme 1 is more sensitive to isoleucine.: The data show the changed enzyme is far less sensitive (93% at 5.0 mM against 4%).
• It makes the same amount of isoleucine, because the medium supplies the isoleucine it needs.: The normal strain switches its pathway off; the changed strain cannot, so the two differ.
• It makes no isoleucine, because a change at the allosteric site also destroys the shape of the active site.: The changed enzyme works at full rate with nothing added (100%), so its active site is intact.

If the last step were blocked, the first four would keep running and intermediates would pile up. Blocking the first step saves the starting material and the energy spent on every step.

• Correct: Stopping step 1 keeps threonine and energy from being spent on intermediates that are not needed.: Shutting the entrance saves material and energy along the whole pathway.
• The first enzyme is the one enzyme of the pathway that has an active site, so it is the one that can be inhibited.: Every enzyme in the pathway has an active site.
• The last enzyme makes isoleucine, so inhibiting it would raise the amount of isoleucine in the cell.: Inhibiting the last enzyme would lower isoleucine production, not raise it, but would still waste the earlier steps.
• Inhibiting the first enzyme lowers the ΔG of each later step, so the pathway runs faster.: Inhibition slows a pathway; it does not change ΔG or speed anything up.

Respiring peas take in O₂ and give out CO₂. If the CO₂ stayed as gas the two would nearly cancel; KOH absorbs the CO₂, so the shrinking volume measures O₂ used.

• Correct: It removes the CO₂ the peas release, so the change in gas volume shows O₂ taken up alone.: Removing CO₂ lets the volume change track O₂ uptake.
• It supplies O₂ to the peas, so that they can keep respiring for the full 20 minutes.: KOH absorbs CO₂; it does not produce O₂.
• It keeps the temperature inside each vial steady, so that the water bath does not warm the gas.: The water bath holds the temperature; KOH has a chemical job.
• It kills bacteria on the peas, so that just the peas' own respiration is measured in the vial.: KOH is there to absorb CO₂; nothing in the setup is aimed at bacteria.

Beads do not respire, so any movement in their tube comes from physical changes. Subtracting it corrects every vial at that temperature.

• Correct: They show volume changes not caused by respiration, such as shifts in temperature or pressure.: Beads-only vials are a control for physical changes in gas volume.
• They show how much O₂ the glass beads themselves use, so that the beads can be subtracted from the dry-pea vial.: Glass beads are not alive and use no O₂; any reading is a physical change.
• They give the peas in the other vials extra space, so that the peas do not run out of O₂.: Each vial is separate; beads in one vial do nothing for peas in another.
• They show the O₂ use of germinating peas at a lower rate, for comparison with the dry peas.: The beads-only vials contain no peas at all.

A germinating seed is growing: it needs ATP, so it oxidizes its stored food quickly, using O₂. A dry seed's cells are nearly inactive.

• Correct: Germinating peas break down stored starch for ATP to grow; dry peas are dormant, with little metabolism.: Active growth needs ATP, which comes from respiration using O₂.
• Dry peas carry out photosynthesis inside the vial, which releases O₂ and hides their uptake of O₂.: The vials are sealed and the peas are seeds; dry peas have no green leaves to photosynthesize, and the dry-pea vial shows a little uptake, not a release.
• Germinating peas are larger than dry peas, so more of their surface absorbs KOH and makes the drop move.: The vials hold the same total volume, and peas do not absorb KOH; the drop moves because O₂ is used.
• Dry peas lack mitochondria until they absorb water, so they are unable to carry out cellular respiration in the vial.: Dry seeds do have mitochondria; their respiration is very slow, not absent (0.06 mL in 20 minutes).

Lower temperature means slower molecules and fewer energetic collisions, so enzyme-catalyzed respiration slows (to about 39%). Below the enzymes' optimum, warming speeds it up.

• Correct: Respiration's enzymes run slower at 10 °C, so at 30 °C the peas would likely use O₂ faster than at 25 °C.: Enzyme rates rise with temperature up to their optimum, so 30 °C would likely be faster.
• Cold denatures the enzymes of respiration, so the peas at 10 °C would not recover if they were warmed again.: Cold slows enzymes but does not denature them; warming restores the rate.
• Peas at 10 °C need less ATP because they are colder, so they switch from respiration to photosynthesis.: Seeds in a dark sealed vial do not photosynthesize; they simply respire more slowly.
• The KOH absorbs CO₂ more slowly at 10 °C, so the cold vials show less gas taken up than was really used.: KOH works well at both temperatures; the beads-only correction already accounts for physical effects.