Genetic defects in glycolytic enzymes can have serious consequences for humans. For example, defects in the gene for pyruvate kinase can result in a condition known as hemolytic anemia. Consult a reference to learn about hemolytic anemia, and discuss why such genetic defects lead to this condition.

Short Answer

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Genetic defects in the pyruvate kinase gene impair the final stage of glycolysis, leading to insufficient ATP production. This, in turn, compromises the structural integrity of red blood cells, causing their premature destruction (hemolysis) and resulting in hemolytic anemia.

Step by step solution

01

Understanding Pyruvate Kinase

Pyruvate kinase is an enzyme, which is pivotal in the glycolysis pathway, a metabolic pathway that is crucial for cellular respiration in all living cells. It catalyzes the final step of glycolysis, converting phosphoenolpyruvate (PEP) to pyruvate, and liberating one molecule of ATP (adenosine triphosphate).
02

Role in Red Blood Cells

Red blood cells (RBCs) are largely dependent on glycolysis for their energy production, since they lack mitochondria and thus cannot carry out oxidative phosphorylation. Insufficient ATP production may result in the inability of the RBC membrane to maintain its structure, leading to hemolysis.
03

Connection to Hemolytic Anemia

In the case of defects in the gene for pyruvate kinase, ATP production is impaired. As result, RBCs destabilize and are destroyed prematurely in the spleen or liver, leading to hemolytic anemia. Hemolytic anemia is a condition characterized by the excessive destruction and inadequate production of red blood cells. Patients may present symptoms of anemia including fatigue, pallor, and shortness of breath.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pyruvate Kinase Deficiency
Pyruvate kinase deficiency is a genetic disorder characterized by a mutation in the gene responsible for producing the enzyme pyruvate kinase, which plays a fundamental role in red blood cell metabolism. This enzyme is crucial because it catalyzes the final stage of glycolysis, helping to transform phosphoenolpyruvate (PEP) into pyruvate and simultaneously produce adenosine triphosphate (ATP), the cell's principal source of energy.

When pyruvate kinase is deficient, red blood cells cannot generate adequate amounts of ATP. Since ATP is essential for maintaining the integrity of the cell membrane, a deficiency leads to the weakening and eventual breakage of red blood cells in a process known as hemolysis. This early destruction of red blood cells is a systemic issue that manifests as hemolytic anemia, characterized by symptoms such as tiredness, pale skin coloration, and shortness of breath due to the lower oxygen-carrying capacity of the blood.
Glycolytic Enzyme Abnormalities
Glycolysis is a sequence of reactions that breaks down glucose to extract energy for cellular function. Glycolytic enzyme abnormalities occur when there is a genetic defect in any of the enzymes involved in this pathway. Pyruvate kinase is one such enzyme, but others include hexokinase, phosphofructokinase, and aldolase. Any abnormality in these enzymes can hinder the process of glycolysis, affecting the production of pyruvate and ATP.

For red blood cells, which rely heavily on glycolysis due to their lack of mitochondria, these abnormalities can compromise cell survival. The premature breakdown of red blood cells leads to a shortage of healthy cells available for oxygen transport—a critical situation that impacts every organ's function. The clinical presentation of patients with glycolytic enzyme abnormalities can vary, depending on which enzyme is affected, and might include jaundice, an enlarged spleen, and chronic fatigue due to anemia.
Red Blood Cell Metabolism
Understanding red blood cell metabolism is essential to appreciate why defects in enzymes like pyruvate kinase cause hemolytic anemia. Red blood cells are unique as they carry oxygen to tissues and organs using a pigment called hemoglobin, and they do this while lacking several organelles, including mitochondria. Consequently, red blood cells depend solely on glycolysis for their energy needs.

During glycolysis, glucose is broken down into pyruvate, which yields ATP. The ATP produced is vital for maintaining the biconcave shape of the cell, ensuring flexibility to travel through narrow capillaries, and for powering the sodium-potassium pumps that regulate ion balance across the cell membrane. Any disruption in this metabolic pathway can affect the red blood cell's ability to function properly and lead to its premature breakdown, summing up to the clinical picture of hemolytic anemia. Highlighting the significance of glycolysis in red blood cells underscores why any abnormality in this process can have severe consequences for oxygen delivery throughout the body.

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Most popular questions from this chapter

(Integrates with Chapter 3 .) Enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate \(+\mathrm{H}_{2} \mathrm{O}\). The standard free energy change, \(\Delta G^{\circ},\) for this reaction is \(+1.8 \mathrm{kJ} / \mathrm{mol}\). If the concentration of 2 -phosphoglycerate is \(0.045 \mathrm{m} M\) and the concentration of phosphoenolpyruvate is \(0.034 \mathrm{m} M\), what is \(\Delta G\), the free energy change for the enolase reaction, under these conditions?

(Integrates with Chapter \(3 .)\) The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for hydrolysis of fructose- 1,6 -bisphosphate (FBP) to fructose6-phosphate (F-6-P) and \(\mathrm{P}_{\mathrm{i}}\) is \(-16.7 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{FBP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { fructose- } 6-\mathrm{P}+\mathrm{P}_{\mathrm{i}} \\] The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for ATP hydrolysis is \(-30.5 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \\] a. What is the standard free energy change for the phosphofructokinase reaction: \\[ \text { ATP + fructose- } 6 \text { -P } \longrightarrow \mathrm{ADP}+\mathrm{FBP} \\] b. What is the equilibrium constant for this reaction? c. Assuming the intracellular concentrations of [ATP] and [ADP] are maintained constant at \(4 \mathrm{m}\) Mand \(1.6 \mathrm{m} M\), respectively, in a rat liver cell, what will be the ratio of [FBP]/[fructose-6-P] when the phosphofructokinase reaction reaches equilibrium?

Determine the anticipated location in pyruvate of labeled carbons if glucose molecules labeled (in separate experiments) with \(^{14} \mathrm{C}\) at each position of the carbon skeleton proceed through the glycolytic pathway

Fructose bisphosphate aldolase in animal muscle is a class I aldolase, which forms a Schiff base intermediate between substrate (for example, fructose- 1,6 -bisphosphate or dihydroxyacetone phosphate and a lysine at the active site (see Figure 18.12 ). The chemical evidence for this intermediate comes from studies with aldolase and the reducing agent sodium borohydride, \(\mathrm{NaBH}_{4}\). Incubation of the enzyme with dihydroxyacetone phosphate and \(\mathrm{NaBH}_{4}\) inactivates the enzyme. Interestingly, no inactivation is observed if \(\mathrm{NaBH}_{4}\) is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.

Regarding phosphofructokinase, which of the following statements is true: a. Low ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. b. High ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. c. High ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. d. The enzyme is more active at low ATP than at high, and fructose- 2,6 -bisphosphate activates the enzyme. e. ATP and fructose- 2,6 -bisphosphate both inhibit the enzyme.

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