1, blank.bmp
"An enzyme catalyses a reaction by decreasing the activation energy." 
"T"
"Enzymes increase the rate at which a reaction comes to equilibrium by decreasing the activation energy of the reaction."
"An enzyme catalyses a reaction by increasing the activation energy." 
"F"
"Enzymes increase the rate at which a reaction comes to equilibrium by decreasing the activation energy of the reaction."
2, blank.bmp
"An enzyme does not affect the position of the equilibrium of the reaction catalysed."
"T"
"An enzyme increases the speed at which the reaction comes to equilibrium, but cannot alter the position of the equilibrium."
"An enzyme changes the position of the equilibrium of the reaction catalysed."
"F"
"An enzyme increases the speed at which the reaction comes to equilibrium, but cannot alter the position of the equilibrium."
3, blank.bmp
"The active site of an enzyme consists of both a substrate binding site and a catalytic site." 
"T"
"Together the substrate binding site and the catalytic site make up the active site of the enzyme."
"The active site of an enzyme consists of both a substrate binding site and a catalytic site." 
"T"
"Together the substrate binding site and the catalytic site make up the active site of the enzyme."
4, blank.bmp
"The active site of an enzyme is made up from the side-chains of amino acids that may be some distance apart in the primary sequence." 
"T"
"The folding of the protein chain brings amino acids that may be some distance apart in the primary sequence into close proximity; the side-chains of these amino acids then form the active site of the enzyme."
"Only amino acids that are adjacent to each other in the primary sequence of the protein can form the active site of an enzyme."
"F"
"The folding of the protein chain brings amino acids that may be some distance apart in the primary sequence into close proximity; the side-chains of these amino acids then form the active site of the enzyme."
5, blank.bmp
"The enzyme is always unchanged at the end of the reaction." 
"T"
"However, the enzyme may undergo chemical modification, with the formation of a covalent intermediate, during the reaction sequence."
"The enzyme itself may undergo a chemical change during the reaction."
"T"
"However, at the end of the reaction the enzyme is unchanged, even if there is a covalently-bound intermediate step in the reaction."
6, blank.bmp
"If the enzyme is saturated with substrate, then the rate of reaction is proportional to the amount of enzyme present." 
"T"
"At high concentrations of substrate, when the enzyme is more or less saturated, then the limiting factors in the rate of formation of product are the amount of enzyme protein present and the innate activity of the enzyme."
"If the enzyme is saturated with substrate, then the rate of reaction is proportional to the amount of substrate present." 
"F"
"At high concentrations of substrate, when the enzyme is more or less saturated, then the limiting factors in the rate of formation of product are the amount of enzyme protein present and the innate activity of the enzyme."
7, blank.bmp
"If an enzyme requires a coenzyme or prosthetic group, then the enzyme protein alone is known as the holo-enzyme." 
"F"
"The holo-enzyme is the active enzyme, consisting of both the protein and the prosthetic group. The protein alone, without its prosthetic group, is the apo-enzyme, which is catalytically inactive."
"If an enzyme requires a coenzyme or prosthetic group, then the enzyme protein alone is known as the apo-enzyme." 
"T"
"The holo-enzyme is the active enzyme, consisting of both the protein and the prosthetic group. The protein alone, without its prosthetic group, is the apo-enzyme, which is catalytically inactive."
8, blank.bmp
"The apo-enzyme is catalytically inactive." 
"T"
"For an enzyme that requires a prosthetic group, the enzyme protein without the prosthetic group is known as the apo-enzyme, and is catalytically inactive."
"The apo-enzyme is catalytically active." 
"F"
"For an enzyme that requires a prosthetic group, the enzyme protein without the prosthetic group is known as the apo-enzyme, and is catalytically inactive."
9, blank.bmp
"The holo-enzyme is catalytically inactive." 
"F"
"The holo-enzyme is the enzyme protein plus its prosthetic group, and is catalytically active."
"The holo-enzyme is catalytically active." 
"T"
"The holo-enzyme is the enzyme protein plus its prosthetic group, and is catalytically active."
10, blank.bmp
"Hydrolases catalyse the cleavage of carbon-carbon bonds." 
"F"
"Hydrolases catalyse the cleavage of ester and amide bonds by hydrolysis." 
"Hydrolases catalyse the cleavage of ester and amide bonds."
"T"
"Hydrolases catalyse the cleavage of ester and amide bonds by hydrolysis." 
11, blank.bmp
"Isomerases, unlike most other enzymes, have only a single substrate."
"T"
"Isomerases catalyse the interconversion of isomers, by transferring reactive groups from one position to another within the same substrate molecule." 
"Isomerases catalyse the cleavage of ester and amide bonds."
"F"
"Isomerases catalyse the interconversion of isomers, by transferring reactive groups from one position to another within the same substrate molecule. While they will catalyse cleavage of carbon-carbon bonds, they do not catalyse hydrolysis."
12, blank.bmp
"Kinases catalyse the transfer of phosphate from ATP onto a substrate." 
"T"
"Kinases are a sub-class of transferases, and catalyse the transfer of phosphate between ATP (or sometimes GTP) and a substrate."
"Kinases catalyse the transfer of phosphate from a substrate onto ADP, forming ATP."
"T"
"Kinases are a sub-class of transferases, and catalyse the transfer of phosphate between ATP (or sometimes GTP) and a substrate."
13, blank.bmp
"Ligases catalyse the cleavage of carbon-carbon double bonds." 
"F"
"Ligases catalyse the formation of bonds between two substrates, frequently coupled to the utilization of ATP, and involving the formation of a phosphorylated intermediate."
"Ligases catalyse the transfer of a reactive group from one position to another within a substrate molecule." 
"F"
"Ligases catalyse the formation of bonds between two substrates, frequently coupled to the utilization of ATP, and involving the formation of a phosphorylated intermediate."
14, blank.bmp
"Ligases may catalyse the formation of ester or amide bonds." 
"T"
"Ligases catalyse the formation of bonds between two substrates, frequently coupled to the utilization of ATP, and involving the formation of a phosphorylated intermediate."
"Ligases catalyse dehydrogenation to form carbon-carbon double bonds." 
"F"
"Ligases catalyse the formation of bonds between two substrates, frequently coupled to the utilization of ATP, and involving the formation of a phosphorylated intermediate."
15, blank.bmp
"Oxidoreductases always involve reaction with molecular oxygen." 
"F"
"Not all oxidoreductases react with molecular oxygen - oxygenases are a sub-class of oxidoreductases which do react with molecular oxygen."
"Oxidoreductases never involve reaction with molecular oxygen." 
"F"
"Oxygenases are a sub-class of oxidoreductases which do react with molecular oxygen."
16, blank.bmp
"Transferases catalyse the transfer of a reactive group from one position to another within a substrate molecule." 
"F"
"Transferases catalyse the transfer of a reactive group from one substrate to another." 
"Hydrolases catalyse the transfer of phosphate from ATP onto a substrate." 
"F"
"Hydrolases catalyse the cleavage of ester and amide bonds by hydrolysis. They may act to remove a phosphate from a substrate, releasing inorganic phosphate."
17, blank.bmp 
"Oxidoreductases catalyse oxidation reactions." 
"T"
"They catalyse the oxidation of one substrate, at the expense of the other, which is reduced."
"Oxidoreductases catalyse reduction reactions." 
"T"
"They catalyse the oxidation of one substrate, at the expense of the other, which is reduced."
18, blank.bmp 
"Transferases catalyse the transfer of a reactive group from one position to another within a substrate molecule." 
"F"
"Transferases catalyse transfer of a reactive group from one substrate to another. It is isomerases that catalyse transfer of reactive groups within the same molecule."
"Transferases always catalyse the transfer of phosphate from ATP (or GTP) onto a substrate." 
"F"
"Transferases catalyse transfer of a reactive group from one substrate to another. Kinases are a sub-class of transferases which catalyse the transfer of phosphate between ATP (or GTP) and a substrate."
19, blank.bmp 
"Oxidoreductases always involve transfer of hydrogen onto NAD+." 
"F"
"One class of oxidoreductase uses NAD+ as the hydrogen acceptor, but others use other coenzymes, and some use oxygen as the substrate to be reduced."
"Oxidoreductases always involve transfer of hydrogen onto a flavin." 
"F"
"One class of oxidoreductase uses flavins as the hydrogen acceptor, but others use other coenzymes, and some use oxygen as the substrate to be reduced."
20, blank.bmp 
Km is half Vmax." 
"F"
"Km is concentration of substrate at which the enzyme achieves half Vmax." 
"Km is half the substrate concentration required to saturate the enzyme."
"F"
"Km is the concentration of substrate required to achieve half Vmax. Although the enzyme is saturated with substrate at Vmax, this requires a very much higher concentration of substrate than 2 x Km."
21, blank.bmp 
"Km is the substrate concentration at which the enzyme achieves half Vmax." 
"T"
"This is a useful working definition of Km. A concentration of substrate several-fold higher than Km is required to achieve saturation of the enzyme." 
"Vmax is the rate of reaction when the substrate concentration = 2 x Km." 
"F"
"Although Km is the concentration of required  to achieve half Vmax, a substrate concentration twice Km does not saturate the enzyme, so does not permit it to achieve Vmax."
22, blank.bmp 
"Vmax is the rate of reaction when the substrate concentration = half Km." 
"F"
"Km is concentration of substrate at which the enzyme achieves half Vmax." 
"Vmax is the rate of reaction when the substrate concentration = half Km." 
"F"
"Km is concentration of substrate at which the enzyme achieves half Vmax." 
23, blank.bmp 
"The effect of any reversible inhibitor can be overcome by increasing the concentration of substrate."
"F"
"Only competitive inhibitors are overcome by increasing the concentration of substrate. Non-competitive inhibitors are also reversible, but their effects are not overcome by increasing the concentration of substrate."
"The effect of an irreversible inhibitor can be overcome by increasing the concentration of substrate."
"F"
"By definition, an irreversible inhibitor permanently inhibits the enzyme, forming a covalent bond; its effects cannot be reversed."
24,  blank.bmp 
"A competitive inhibitor increases Vmax but does not affect Km." 
"F"
"A competitive inhibitor increases Km without affecting Vmax."
"An non-competitive inhibitor increases Km and Vmax." 
"F"
"A non-competitive inhibitor decreases Vmax, and has no effect on Km."
25, blank.bmp
"Km is the substrate concentration at which the enzyme achieves Vmax." 
"F"
"Km is the substrate concentration at which the enzyme achieves half Vmax." 
"Vmax is the rate when the enzyme is half saturated with substrate." 
"F"
"Vmax is the rate when the enzyme is more or less completely saturated with substrate."
26, blank.bmp 
"A non-competitive inhibitor lowers Km but does not affect Vmax." 
"F"
"A non-competitive inhibitor decreases Vmax, and has no effect on Km."
"A competitive inhibitor increases the activation energy of the reaction." 
"F"
"A competitive inhibitor increases Km without affecting Vmax."
27, blank.bmp 
"A competitive inhibitor lowers Km but does not affect Vmax." 
"F"
"A competitive inhibitor increases Km but does not affect Vmax."
"A non-competitive inhibitor increases Km but does not affect Vmax." 
"F"
"A non-competitive inhibitor decreases Vmax, and has no effect on Km."
28, blank.bmp 
"The effect of a competitive inhibitor can be overcome by increasing the concentration of substrate." 
"T"
"Because a competitive inhibitor increases apparent Km, addition of more substrate will permit the enzyme to achieve the same Vmax as when no inhibitor is added."
"The effect of a non-competitive inhibitor can be overcome by increasing the concentration of substrate." 
"F"
"Because a non-competitive inhibitor decreases Vmax, addition of more substrate will not affect the activity of the enzyme on the presence of inhibitor."
29, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region A the enzyme is acting at or near Vmax." 
"F"
"Vmax is the maximum rate of enzyme reaction - through region A there is an increase in the rate of reaction, so the enzyme cannot be acting at Vmax."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the enzyme is acting at or near Vmax."
"T"
"As the concentration of substrate rises high enough, so the enzyme becomes saturated with substrate and acts at or near its maximum rate - this is Vmax."
30, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). There is a linear increase in the formation of product as the concentration of substrate increases." 
"F"
"Although at low concentrations of substrate the increase in rate is almost linear with increasing [substrate], this is a hyperbolic curve, and no part is linear."
"The point marked C is the Vmax of the enzyme." 
"T"
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). This is the (extrapolated) maximum rate at which the enzyme can act when it is saturated with substrate."
31, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The point marked D is the Vmax of the enzyme." 
"F"
"The (extrapolated) maximum rate at which the enzyme can act when it is saturated with substrate (Vmax) is point C. Point D is half Vmax."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). At low concentrations of substrate the enzyme acts at a constant rate." 
"F"
"At low concentrations of substrate, the rate of activity increases with increasing substrate. It is only at high concentrations of substrate, when the enzyme is more or less saturated, that it acts at a constant rate."
32, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). At high concentrations of substrate the enzyme acts at a constant rate." 
"T"
"At low concentrations of substrate, the rate of activity increases with increasing substrate, however, at high concentrations of substrate, when the enzyme is more or less saturated, it acts at a constant rate."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the rate of formation of product is approximately proportional to the concentration of substrate." 
"F"
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the enzyme is more or less saturated, and changing the concentration of substrate has little or no effect on the rate."
33, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region A the rate of formation of product is approximately proportional to the concentration of substrate." 
"T"
"In region A the enzyme is far below saturation, and increasing the concentration of substrate permits a considerable increase in the rate of reaction."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the rate of formation of product is approximately proportional to the concentration of substrate." 
"F"
"In region B the enzyme is more or less saturated, and increasing the concentration of substrate has little effect. The rate of formation of product is proportional to the amount of enzyme present in this region"
34, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The intercept on the x axis is the Km of the enzyme." 
"F"
"The intercept of this curve on the x axis is zero activity in the presence of zero substrate."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The intercept on the x axis is the Vmax of the enzyme." 
"F"
"The intercept of this curve on the x axis is zero activity in the presence of zero substrate."
35, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The point marked D is the Km of the enzyme." 
"F"
"Point D is on the y axis (rate of reaction). It is half Vmax. Km is the concentration of substrate required to achieve half Vmax."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The point marked D is the Vmax of the enzyme." 
"F"
"Point D is on the y axis (rate of reaction), and is half the Vmax of the enzyme. Vmax is shown by point C"
36, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The point marked E is half the Vmax  of the enzyme." 
"F"
"Point E is on the x axis (concentration of substrate). This is the concentration of substrate required to achieve half Vmax - this is the Km."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). The point marked E is the Km of the enzyme." 
"T"
"Point E is on the x axis (concentration of substrate). This is the concentration of substrate required to achieve half Vmax - this is the Km."
37, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region A the concentration of substrate is above the Km of the enzyme." 
"F"
"In region A the concentration of substrate is below the Km of the enzyme."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region A the concentration of substrate is below the Km of the enzyme." 
"T"
"In region A the concentration of substrate is below the Km of the enzyme."
38, svplot1.bmp
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the concentration of substrate is above the Km of the enzyme." 
"T"
"In region B the concentration of substrate is considerably above the Km of the enzyme."
"The graph shows the rate of reaction for an enzyme-catalysed reaction with increasing concentration of substrate (the s/v curve). In region B the concentration of substrate is below the Km of the enzyme." 
"F"
"In region B the concentration of substrate is considerably above the Km of the enzyme."
39, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. At low values of 1 / [S] the enzyme acts at or near its maximum rate."
"T"
"A low value of 1 / [S] means a high value of [S]. When the substrate concentration is high enough, the enzyme will act at its maximum rate (Vmax)."
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. At high values of 1 / [S] the enzyme acts at or near its maximum rate."
"F"
"A high value of 1 / [S] means a low value of [S]. When the substrate concentration is low, the enzyme will act very much below its maximum rate (Vmax)."
40, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The intercept on the x axis can be used to calculate the Vmax of the enzyme." 
"F"
"The intercept on the x axis can be used to calculate the Km of the enzyme, not Vmax."
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The intercept on the y axis can be used to calculate the Vmax of the enzyme." 
"T"
"The intercept on the y axis can indeed be used to calculate the Vmax of the enzyme."
41, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The intercept on the x axis can be used to calculate the Km of the enzyme."
"T"
"The intercept on the x axis can indeed be used to calculate the Km of the enzyme."
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The intercept on the y axis can be used to calculate the Km of the enzyme."
"F"
"The intercept on the y axis can be used to calculate the Vmax of the enzyme, not Km."
42, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The value of 1 / [S] at point B = -1 / Km." 
"T"
"This is the intercept on the 1 / concentration axis; since Km is a concentration, this intercept must be related to Km." 
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The value of 1 / v at point B =1 / Km." 
"F"
"At point B the value of 1 / v = 0."
43, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. There is a linear relationship between the rate of reaction and the concentration of substrate, 
"F"
"This is a double reciprocal plot. There is a hyperbolic relationship between the rate of reaction and the concentration of substrate."
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The value of 1 / [S] at point A = Km." 
"F"
"This is the intercept on the 1 / (rate of reaction) axis; Km is a concentration, so cannot be determined from this axis."
44, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The value of 1 / v at point A =1 / Vmax." 
"T"
"This is the intercept on the 1 / (rate of reaction) axis, and the lowest value of 1 / rate of reaction must be the highest value of V = Vmax." 
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The concentration of substrate at point A is infinite."
"T"
"At point A the value of 1 / [substrate] is zero, so the value of [substrate] must be infinite."
45, lbplot1.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The enzyme has maximum activity at an infinite concentration of substrate."
"T"
"At point A, which represents 1 / Vmax, the value of 1 / [substrate] is zero, so the value of [substrate] must be infinite."
"The graph shows the Lineweaver-Burk double reciprocal plot (1/ rate of reaction : 1/[substrate]) for an enzyme incubated with different concentrations of substrate. The enzyme has maximum activity at an infinite concentration of substrate."
"T"
"At point A, which represents 1 / Vmax, the value of 1 / [substrate] is zero, so the value of [substrate] must be infinite."
46, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line Q represents incubation in the presence of the highest concentration of inhibitor." 
"F"
"Line Q shows the lowest values for 1 / rate of reaction, and therefore the highest values for rate of reaction. This must be the line in the presence of no inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line P represents incubation in the presence of the highest concentration of inhibitor." 
"T"
"Line P shows the highest values for 1 / rate of reaction, and therefore the lowest values for rate of reaction. This must be the line in the presence of the highest concentration of inhibitor."
47, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line Q represents incubation in the absence of inhibitor."
"T"
"Line Q shows the lowest values for 1 / rate of reaction, and therefore the highest values for rate of reaction. This must be the line in the presence of no inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line P represents incubation in the absence of inhibitor."
"F"
"Line P shows the highest values for 1 / rate of reaction, and therefore the lowest values for rate of reaction. This must be the line in the presence of the highest concentration of inhibitor."
48, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Km of the enzyme is increased by the inhibitor."
"T"
"In the presence of maximum inhibitor the value of -1 / Km (the x intercept) is lowest - therefore the value of Km must be higher in the presence of the inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Km of the enzyme is decreased by the inhibitor."
"F"
"In the presence of maximum inhibitor the value of -1 / Km (the x intercept) is lowest - therefore the value of Km must be higher in the presence of the inhibitor."
49, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Vmax of the enzyme is increased by the inhibitor."
"F"
"An inhibitor reduces the rate of reaction. Therefore it cannot possibly increase Vmax."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Vmax of the enzyme is decreased by the inhibitor."
"F"
"Vmax is the intercept on the 1 / v axis. This is unchanged by the inhibitor."
50, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. This is competitive inhibition." 
"T"
"The Km of the enzyme is increased by the inhibitor, but Vmax is unchanged."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. This is non-competitive inhibition
"F"
"The Km of the enzyme is increased by the inhibitor, but Vmax is unchanged. This is competitive inhibition."
51, lbplot2.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The enzyme has a lower affinity for the substrate in the presence of the inhibitor." 
"T"
"Because the enzyme has a higher Km, this means that its affinity for the substrate is lower in the presence  of the inhibitor. Km is inversely proportional to affinity of the enzyme for its substrate."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The enzyme has a higher affinity for the substrate in the presence of the inhibitor." 
"F"
"Because the enzyme has a higher Km, this means that its affinity for the substrate is lower in the presence of the inhibitor, not higher. Km is inversely proportional to affinity of the enzyme for its substrate."
52, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line Q represents incubation in the presence of the highest concentration of inhibitor." 
"F"
"Line Q shows the lowest values for 1 / rate of reaction, and therefore the highest values for rate of reaction. This must be the line in the presence of no inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line P represents incubation in the presence of the highest concentration of inhibitor." 
"T"
"Line P shows the highest values for 1 / rate of reaction, and therefore the lowest values for rate of reaction. This must be the line in the presence of the highest concentration of inhibitor."
53, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line Q represents incubation in the absence of inhibitor."
"T"
"Line Q shows the lowest values for 1 / rate of reaction, and therefore the highest values for rate of reaction. This must be the line in the presence of no inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. Line P represents incubation in the absence of inhibitor."
"F"
"Line P shows the highest values for 1 / rate of reaction, and therefore the lowest values for rate of reaction. This must be the line in the presence of the highest concentration of inhibitor."
54, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Km of the enzyme is increased by the inhibitor."
"F"
"The intercept on the x axis, which is used to determine Km, is unchanged in the presence of the inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Km of the enzyme is decreased by the inhibitor."
"F"
"The intercept on the x axis, which is used to determine Km, is unchanged in the presence of the inhibitor."
55, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Vmax of the enzyme is increased by the inhibitor."
"F"
"An inhibitor reduces the rate of reaction. Therefore it cannot possibly increase Vmax."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The Vmax of the enzyme is decreased by the inhibitor."
"T"
"Vmax is the intercept on the 1 / v axis. This is at a higher value of 1 / v in the presence of the inhibitor, and therefore at a lower value of v (the rate of reaction) by the inhibitor."
56, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. This is competitive inhibition." 
"F"
"The Km of the enzyme is unchanged by the inhibitor, and Vmax is decreased. This is non-competitive inhibition."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. This is non-competitive inhibition
"T"
"The Km of the enzyme is unchanged by the inhibitor, and Vmax is decreased."
57, lbplot3.bmp
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The enzyme has a lower affinity for the substrate in the presence of the inhibitor." 
"F"
"The affinity of the enzyme for its substrate is inversely proportional to Km, which is unchanged in the presence of the inhibitor."
"The graph shows the Lineweaver-Burk double reciprocal plot (1 / v against 1 / [S]) for an enzyme catalysed reaction in the presence of varying concentrations of an inhibitor. The enzyme has a higher affinity for the substrate in the presence of the inhibitor." 
"F"
"The affinity of the enzyme for its substrate is inversely proportional to Km, which is unchanged in the presence of the inhibitor."
58, blank.bmp
"In a ping-pong bisubstrate reaction the enzyme binds both substrates at the same time, and in a specific order." 
"F"
"In a ping-pong bisubstrate reaction the enzyme reacts with one substrate, undergoes modification, then reacts with the second substrate." 
"In a ping-pong bisubstrate reaction the enzyme reacts with one substrate, undergoes modification, then reacts with the second substrate." 
"T"
"In a ping-pong bisubstrate reaction the enzyme reacts with one substrate, undergoes modification, then reacts with the second substrate." 
59, blank.bmp
"In an ordered bisubstrate reaction the enzyme binds both substrates at the same time, and in a specific order." 
"T"
"This is the definition of an ordered bisubstrate reaction."
"In an ordered bisubstrate reaction the enzyme reacts with one substrate, undergoes modification, then reacts with the second substrate." 
"F"
"In an ordered bisubstrate reaction the enzyme binds both substrates at the same time, and in a specific order." 
60, blank.bmp
"In a ping-pong bisubstrate reaction the enzyme (or coenzyme) undergoes a chemical modification in the first half of the reaction."
"T"
"In a ping-pong bisubstrate reaction the enzyme reacts with one substrate, undergoes modification (either to itself or to the coenzyme), then reacts with the second substrate. In this second half of the reaction the enzyme (or coenzyme) undergoes chemical modification back to its original form." 
"In an ordered bisubstrate reaction the enzyme (or coenzyme) undergoes a chemical modification in the first half of the reaction."
"F"
"In an ordered bisubstrate reaction the enzyme binds both substrates at the same time, and in a specific order. There is no modified form of the enzyme or coenzyme that does not have substrate bound to it." 
61,linear.bmp
"The diagram shows a step in a reaction sequence that is inhibited by an enzyme. If the aim of the drug is to increase the concentration of intermediate B in the cell then a competitive inhibitor will be appropriate."
"T"
"If the enzyme catalysing the reaction from B to C is inhibited then there will be an increase in the concentration of B in the cell regardless of the type of inhibitor used."
"The diagram shows a step in a reaction sequence that is inhibited by an enzyme. If the aim of the drug is to increase the concentration of intermediate B in the cell then a non-competitive inhibitor will be appropriate."
"T"
"If the enzyme catalysing the reaction from B to C is inhibited then there will be an increase in the concentration of B in the cell regardless of the type of inhibitor used."
62, linear.bmp
"The diagram shows a step in a reaction sequence that is inhibited by an enzyme. If the aim of the drug is to decrease the concentration of product D in the cell then a competitive inhibitor will be appropriate."
"F"
"If a competitive inhibitor is used, then as the concentration of intermediate B increases, so it will compete with the inhibitor, so that there will be the same rate of formation of product D in the presence or absence of the inhibitor."
"The diagram shows a step in a reaction sequence that is inhibited by an enzyme. If the aim of the drug is to decrease the concentration of product D in the cell then a non-competitive inhibitor will be appropriate."
"T"
"A non-competitive inhibitor reduces the rate at which the reaction proceeds, regardless of the concentration of substrate. Therefore, even though the concentration of intermediate B will increase, this will not increase the rate of formation of C and D. A non-competitive inhibitor both increases the concentration of substrate of the inhibited enzyme and decreases the rate of formation of product."
63, branched.bmp
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then when there is a relatively low supply of substrate A, most of the intermediate D will be converted to product R."
"T"
"When there is a low supply of substrate A there will be a low concentration of intermediate D. Therefore the enzyme with the lower Km (enzyme A), which has a higher affinity for intermediate D, will be able to bid its substrate more readily, and most of the intermediate D will be converted to product R rather than product Z."
"The diagram shows a branched metabolic pathway. If enzyme A has a high Km, and enzyme B a low Km, compared with the usual intracellular concentration of intermediate D, then when there is a relatively low supply of substrate A, most of the intermediate D will be converted to product R."
"F"
"When there is a low supply of substrate A there will be a low concentration of intermediate D. Therefore the enzyme with the lower Km (enzyme B), which has a higher affinity for intermediate D, will be able to bid its substrate more readily, and most of the intermediate D will be converted to product Z rather than product R."
64, branched.bmp
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then when there is a relatively low supply of substrate A, most of the intermediate D will be converted to product Z."
"F"
"When there is a low supply of substrate A there will be a low concentration of intermediate D. Therefore the enzyme with the lower Km (enzyme A), which has a higher affinity for intermediate D, will be able to bid its substrate more readily, and most of the intermediate D will be converted to product R rather than product Z."
"The diagram shows a branched metabolic pathway. If enzyme A has a high Km, and enzyme B a low Km, compared with the usual intracellular concentration of intermediate D, then when there is a relatively low supply of substrate A, most of the intermediate D will be converted to product Z."
"T"
"When there is a low supply of substrate A there will be a low concentration of intermediate D. Therefore the enzyme with the lower Km (enzyme B), which has a higher affinity for intermediate D, will be able to bid its substrate more readily, and most of the intermediate D will be converted to product Z rather than product R."
65, branched.bmp
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then as there is an increase in the supply of substrate A, more of the intermediate D will be converted to product Z."
"T"
"When there is a low availability of substrate A,  enzyme A, with the lower Km, and hence higher affinity, will bind most of intermediate D, leading to formation of product R. As the amount of A increases (and hence the amount of D increases), so enzyme A will become saturated and there will be no further increase in the amount of product R formed. As the concentration of intermediate D increases, so enzyme B becomes more active, and an increasing amount of intermediate D will be converted to product Z."
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then as there is an increase in the supply of substrate A, more of the intermediate D will be converted to product R."
"F"
"When there is a low availability of substrate A,  enzyme A, with the lower Km, and hence higher affinity, will bind most of intermediate D, leading to formation of product R. As the amount of A increases (and hence the amount of D increases), so enzyme A will become saturated and there will be no further increase in the amount of product R formed. As the concentration of intermediate D increases, so enzyme B becomes more active, and an increasing amount of intermediate D will be converted to product Z."
66, branched.bmp
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then as there is an increase in the supply of substrate A, less of the intermediate D will be converted to product Z."
"F"
"When there is a low availability of substrate A,  enzyme A, with the lower Km, and hence higher affinity, will bind most of intermediate D, leading to formation of product R. As the amount of A increases (and hence the amount of D increases), so enzyme A will become saturated and there will be no further increase in the amount of product R formed. As the concentration of intermediate D increases, so enzyme B becomes more active, and an increasing amount of intermediate D will be converted to product Z."
"The diagram shows a branched metabolic pathway. If enzyme A has a low Km, and enzyme B a high Km, compared with the usual intracellular concentration of intermediate D, then as there is an increase in the supply of substrate A, less of the intermediate D will be converted to product R."
"F"
"When there is a low availability of substrate A,  enzyme A, with the lower Km, and hence higher affinity, will bind most of intermediate D, leading to formation of product R. As the amount of A increases (and hence the amount of D increases), so enzyme A will become saturated and there will be no further increase in the amount of product R formed. As the concentration of intermediate D increases, so enzyme B becomes more active, and an increasing amount of intermediate D will be converted to product Z."
-999
Copyright David A Bender 2002
