Passage
A newborn girl is diagnosed with a rare genetic disorder that causes premature lysis of red blood cells. Through genome sequencing, physicians identified a specific mutation (R510Q) in both copies of her pyruvate kinase gene, which encodes the enzyme that catalyzes the final step of glycolysis. Pyruvate kinase produces pyruvate and ATP from phosphoenolpyruvate (PEP) and ADP.PEP + ADP ↔ Pyruvate + ATPReaction 1Because red blood cells lack mitochondria, pyruvate in erythrocytes does not enter the citric acid cycle and the electron transport chain. Therefore, red blood cells rely almost exclusively on glycolysis to produce sufficient ATP for cellular functions. In patients with pyruvate kinase deficiency, it is thought that a lack of ATP compromises the integrity of the red blood cell membrane, leading to premature lysis in the spleen. Although these observations are well established, the exact mechanism by which low ATP alters the cell membrane is unclear.Scientists interested in analyzing this particular patient's gene variant expressed and purified recombinant R510Q pyruvate kinase. They performed several biochemical tests to characterize the R510Q mutant with respect to wild-type (WT) pyruvate kinase. In the first experiment, WT and R510Q pyruvate kinases were each incubated with a constant amount of ADP and varying amounts of PEP. The activity of the enzyme was measured at each concentration of PEP and plotted. They then repeated the experiment in the presence of ATP and found that ATP is a noncompetitive inhibitor of the WT enzyme, but a mixed inhibitor of the R510Q enzyme. The results are shown in Figure 1. All experiments were done at the same enzyme concentration.
Figure 1 Pyruvate kinase activity with and without ATP as a function of [PEP]. Open markers are shown at ½V_max for each plot.
Adapted from Wang C, Chiarelli LR, Bianchi P, et al. Human erythrocyte pyruvate kinase: characterization of the recombinant enzyme and a mutant form (R510Q) causing nonspherocytic hemolytic anemia. Blood. 2001;98(10) :3113-20.
-Scientists predicted that gene copy number is proportional to protein expression for pyruvate kinase. If this hypothesis is correct, which kinetic parameter would be expected to double when the gene copy number doubles?

Question

Passage
A newborn girl is diagnosed with a rare genetic disorder that causes premature lysis of red blood cells. Through genome sequencing, physicians identified a specific mutation (R510Q) in both copies of her pyruvate kinase gene, which encodes the enzyme that catalyzes the final step of glycolysis. Pyruvate kinase produces pyruvate and ATP from phosphoenolpyruvate (PEP) and ADP.PEP + ADP ↔ Pyruvate + ATPReaction 1Because red blood cells lack mitochondria, pyruvate in erythrocytes does not enter the citric acid cycle and the electron transport chain. Therefore, red blood cells rely almost exclusively on glycolysis to produce sufficient ATP for cellular functions. In patients with pyruvate kinase deficiency, it is thought that a lack of ATP compromises the integrity of the red blood cell membrane, leading to premature lysis in the spleen. Although these observations are well established, the exact mechanism by which low ATP alters the cell membrane is unclear.Scientists interested in analyzing this particular patient's gene variant expressed and purified recombinant R510Q pyruvate kinase. They performed several biochemical tests to characterize the R510Q mutant with respect to wild-type (WT) pyruvate kinase. In the first experiment, WT and R510Q pyruvate kinases were each incubated with a constant amount of ADP and varying amounts of PEP. The activity of the enzyme was measured at each concentration of PEP and plotted. They then repeated the experiment in the presence of ATP and found that ATP is a noncompetitive inhibitor of the WT enzyme, but a mixed inhibitor of the R510Q enzyme. The results are shown in Figure 1. All experiments were done at the same enzyme concentration.

Figure 1 Pyruvate kinase activity with and without ATP as a function of [PEP]. Open markers are shown at ½V_max for each plot.
Adapted from Wang C, Chiarelli LR, Bianchi P, et al. Human erythrocyte pyruvate kinase: characterization of the recombinant enzyme and a mutant form (R510Q) causing nonspherocytic hemolytic anemia. Blood. 2001;98(10) :3113-20.
-Scientists predicted that gene copy number is proportional to protein expression for pyruvate kinase. If this hypothesis is correct, which kinetic parameter would be expected to double when the gene copy number doubles?

Quizplus · Accepted Answer

11ecba2d_0e52_450f_a3bf_4334690d9db3_MD0008_00 The rate of an enzyme-catalyzed reaction plateaus at high substrate concentrations ([S] >> K_m) because the active sites of enzymes are completely saturated with substrate; adding more substrate fails to increase the rate of catalysis. Adding more enzyme, on the other hand, provides additional active sites, allowing the reaction to proceed more quickly as more substrate can be bound. The maximum velocity V_max of the reaction is the product of the number of active sites (total enzyme concentration [E_tot]) and the rate at which each site converts substrate to product (turnover number k_cat). Mathematically, this is described as V_max = k_cat × [E_tot]. Therefore, an increase in the [E_tot] will increase V_max. The question states that scientists predicted that the gene copy number is proportional to protein expression. If they are correct, doubling the gene copy number should result in double the total enzyme purified from a given number of cells. As a result, doubling pyruvate kinase's gene copy number should also double the V_max of the pyruvate kinase reaction. (Choice A) Catalytic efficiency is the ratio of k_cat to the Michaelis constant K_m. Both are intrinsic properties of the enzyme and will not be altered by a change in enzyme concentration. (Choice C) Catalytic turnover k_cat is an intrinsic property of the enzyme and will not be altered by a change in enzyme concentration. (Choice D) [E_tot] (and therefore gene copy number) has no effect on the equilibrium constant K_eq of a reaction. Enzymes increase the rate at which equilibrium is achieved but not the equilibrium itself. Educational objective: At high substrate concentrations ([S] >> K_m), the active sites of enzymes are completely saturated with substrate, and adding more substrate does not increase the reaction rate. Adding more enzyme increases the number of active sites, and therefore the reaction velocity. The maximum velocity at any enzyme concentration is calculated as V_max = k_cat× [E_tot].

Passage A Newborn Girl Is Diagnosed with a Rare Genetic Disorder