3.1 Enzymes

1. AIt speeds up the reaction by joining with the substrate to become the product.
   A student who thinks enzymes are used up in reactions picks this. The enzyme binds the substrate only temporarily; the products are released and the enzyme is left unchanged.
2. BIt speeds up the reaction and is left unchanged after the reaction is over. Correct
   An enzyme is a catalyst: it increases the rate of the reaction by lowering its activation energy, and it is not used up. After the products are released, the same enzyme molecule can catalyze the reaction again.
3. CIt makes possible a reaction that could not take place at all without it.
   A student who thinks an enzyme makes a reaction possible picks this. An enzyme speeds up a reaction that can also occur without it, usually much more slowly.
4. DIt speeds up the reaction by increasing the amount of energy that the reaction releases.
   A student who thinks an enzyme changes the energy a reaction releases picks this. An enzyme lowers the activation energy but leaves the energy of the reactants and products, and so the energy released, unchanged.

CED 3.1.A.1 · Read this in Fix

1. AMolecule W
   A student who thinks binding depends only on shape picks the wedge with positive charges. It fits the active site's shape, but its positively charged groups would sit against the active site's positively charged groups and be repelled.
2. BMolecule X
   A student who thinks a substrate has the same shape and charges as its active site picks the notched block with positive charges. A substrate's shape is the reverse of the active site's, and its charged groups are opposite to those of the active site, not the same.
3. CMolecule Y
   A student who thinks opposite charges alone make a molecule bind picks the square block with negative charges. Its flat top does not fit into the V-shaped active site, so its charged groups cannot come close to those of the active site.
4. DMolecule Z Correct
   This molecule's wedge shape is the reverse of the V-shaped active site, so it fits into it, and its two negatively charged groups line up with the two positively charged groups of the active site and are attracted to them. Both its shape and its charges are compatible with the active site.

CED 3.1.A.2 · Read this in Fix

1. AGraph 1
   A student who thinks an enzyme changes how much energy a reaction releases picks the graph in which the dashed line ends at a lower level. An enzyme leaves the energy of the products unchanged; only the peak is lowered.
2. BGraph 2
   A student who thinks an enzyme supplies energy to the reactants picks the graph in which the dashed line starts at a higher level. An enzyme does not add energy to the reactants; it lowers the peak that they must reach.
3. CGraph 3 Correct
   An enzyme lowers the activation energy, so the peak of the dashed line is lower. It does not change the energy of the reactants or of the products, so both lines start and end at the same levels, and the reaction releases the same amount of energy with or without the enzyme.
4. DGraph 4
   A student who thinks an enzyme removes the activation energy altogether picks the graph in which the dashed line has no peak. An enzyme lowers the activation energy, but some is still needed, so a lower peak remains.

CED 3.1.A.1 · Read this in Fix

1. A33% Correct
   Activation energy is the rise from the reactants to the peak: 120 − 30 = 90 kJ/mol without the enzyme and 90 − 30 = 60 kJ/mol with it. The decrease is 30 kJ/mol, and 30/90 × 100 = 33%.
2. B25%
   A student who reads each activation energy as the value at the top of the peak picks this: (120 − 90)/120 × 100 = 25%. Activation energy must be measured from the reactants' level of 30 kJ/mol.
3. C50%
   A student who divides the change by the new value picks this: 30/60 × 100 = 50%. A percent change is calculated from the original value, 90 kJ/mol.
4. D30%
   A student who reports the change of 30 kJ/mol as a percentage picks this. The change must be divided by the original activation energy: 30/90 × 100 = 33%.

Working Activation energy is measured from the energy of the reactants (30 kJ/mol) up to the peak. Without the enzyme: 120 − 30 = 90 kJ/mol. With the enzyme: 90 − 30 = 60 kJ/mol. Percent decrease = (90 − 60)/90 × 100 = 33%. Distractors: reading each activation energy as the peak's value from zero gives (120 − 90)/120 × 100 = 25%; dividing the change by the new value gives (90 − 60)/60 × 100 = 50%; reporting the change of 30 kJ/mol as a percentage gives 30%.

CED 3.1.A.1 · Read this in Fix

1. AThe enzyme converted all of the S within minutes but very little of the T. Correct
   With the enzyme, all 100 μmol of S had been converted by about 6 minutes, but only about 3 μmol of product had formed from T after 10 minutes. The enzyme acts on its substrate, S, but not effectively on T, even though T is similar in size and shape.
2. BWith the enzyme, S and T were converted at about the same rapid rate.
   A student who thinks an enzyme acts on any molecule similar to its substrate picks this. The graph shows that S was converted rapidly but T hardly at all, so the enzyme is specific to S.
3. CWithout the enzyme, none of S was converted during the 10 minutes.
   A student who thinks a reaction cannot occur without its enzyme picks this. Without the enzyme, product from S rose slowly to about 10 μmol by 10 minutes: the reaction occurred, only far more slowly.
4. DWith the enzyme, product formation from S stopped as the enzyme ran out.
   A student who thinks enzymes are used up picks this. Product formation leveled off at 100 μmol, the amount of S at the start: it stopped because all of the substrate had been converted, not because the enzyme ran out.

CED 3.1.A.2 · Read this in Fix

1. ASucrase's active site binds the sugars the body needs, and the body does not need lactose.
   A student who thinks enzymes act on what the body needs picks this. Whether sucrase's active site binds a sugar depends on whether the sugar fits it, not on the body's needs.
2. BLactose's shape and pattern of partial charges do not fit sucrase's active site. Correct
   An enzyme acts only on molecules that can bind in its active site. Lactose differs from sucrose in shape and in the arrangement of its polar groups, which carry partial charges, so it cannot form an enzyme–substrate complex with sucrase.
3. CSucrase's active sites are used up hydrolyzing sucrose, so none are left for lactose.
   A student who thinks enzymes are used up in the reactions they catalyze picks this. Sucrase's active sites are not used up; the enzyme is released unchanged after each reaction. It does not act on lactose because lactose does not bind in its active site.
4. DSucrase supplies the activation energy for splitting sucrose but not for splitting lactose.
   A student who thinks an enzyme speeds up a reaction by supplying energy picks this. Sucrase does not supply activation energy to its substrate; it lowers the activation energy, and lactose is not hydrolyzed because it does not bind in sucrase's active site.

CED 3.1.A.2 · Read this in Fix

1. AB no longer forms at all, as A cannot change into B without enzyme E.
   A student who thinks a reaction cannot happen without its enzyme picks this. The uncatalyzed reaction still occurs, only far more slowly.
2. BB forms at the normal rate, as the bacterium makes a new enzyme because it needs B.
   A student who thinks organisms make new enzymes because they need a product picks this. A mutant bacterium does not produce a replacement enzyme because B is needed, and no other enzyme acts on A.
3. CB forms at the normal rate, as ATP supplies the energy that enzyme E used to supply.
   A student who thinks enzymes supply energy to reactions picks this. Enzyme E did not supply energy; it lowered the activation energy, and nothing in the cell takes over that role.
4. DB still forms, but far more slowly, as the reaction is no longer catalyzed. Correct
   An enzyme speeds up a reaction by lowering its activation energy; it does not make the reaction possible. Without enzyme E, A is still converted to B, but at the much slower uncatalyzed rate, so far less B is made in a given time.

CED 3.1.A.1 · Read this in Fix

1. AIt should stay colorless, or else the investigation has failed and must be repeated.
   A student who thinks a control must show no change picks this. P may break down slowly on its own, so tube 2 may turn yellow eventually; what matters is how its time compares with tube 1's.
2. BIt measures the independent variable, the yellow color, in a tube with no extract.
   A student who swaps the independent and dependent variables picks this. The color change is measured, so it is the dependent variable; the presence of extract is the independent variable.
3. CIt shows how fast P breaks down with no extract, for comparison with tube 1. Correct
   Tube 2 differs from tube 1 only in having no extract, so it shows the rate at which P breaks down on its own. If tube 1 turns yellow much sooner than tube 2, the faster breakdown can be attributed to the extract.
4. DIt keeps the two tubes at 25 °C, which makes temperature a controlled variable.
   A student who confuses a control tube with a controlled variable picks this. Temperature is held constant by keeping both tubes at 25 °C; tube 2 is there to show the result without the extract.

CED 3.1.A.1 · Read this in Fix

1. ARapid conversion needs the molecule's charges to be compatible with the active site. Correct
   The original enzyme, with a positive group in its active site, converted X (negative group) rapidly and Y (positive group) hardly at all; the mutant, with a negative group, did the reverse. X and Y have the same shape, so the difference comes from whether the charges are compatible.
2. BBinding depends only on shape, so the charge change had no effect on either molecule.
   A student who thinks binding depends only on shape picks this. X and Y have the same shape, yet each enzyme converted one far faster than the other, and the mutation reversed which one.
3. CBinding is strongest when the substrate and active site carry the same charge.
   A student who thinks a substrate's charges match those of the active site picks this. The positively charged original active site converted the negatively charged X, and the negatively charged mutant active site converted the positively charged Y: opposite charges, not like charges, went together.
4. DAny molecule with a charge opposite to the active site's binds, whatever its shape.
   A student who thinks opposite charges alone make a molecule bind picks this. X and Y have the same shape, so these data cannot show that shape does not matter; a molecule must both fit the active site and carry compatible charges.

CED 3.1.A.2 · Read this in Fix

1. AIt supplies the activation energy to H₂O₂ molecules, so more of them can react.
   A student who thinks an enzyme supplies energy to its substrate picks this. Catalase does not supply energy to H₂O₂; it lowers the activation energy that the molecules must reach.
2. BIt increases the energy each H₂O₂ molecule releases, so the reaction runs faster.
   A student who thinks an enzyme changes the energy a reaction releases picks this. Catalase does not change the energy of the reactants or products, so each H₂O₂ molecule releases the same energy with or without it.
3. CIt makes the H₂O₂ molecules move faster, so they collide and react more often.
   A student who thinks an enzyme works like heating picks this. The temperature, and so the speed of the molecules, is the same in both cases; catalase lowers the activation energy.
4. DIt lowers the activation energy, so more H₂O₂ molecules have enough energy to react. Correct
   Catalase provides an active site in which H₂O₂ reacts with a lower activation energy. At the same temperature, a larger share of the H₂O₂ molecules can then react, so O₂ is produced faster.

CED 3.1.A.1 · Read this in Fix