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Glycolysis Quiz

12 questions biology Grades 9-12

The question sheet

Reveal any answer as you study
  1. In which part of the cell does glycolysis occur?

    • Nucleus
    • Cytoplasm
    • Mitochondria
    • Ribosome
    Reveal answer

    Answer: Cytoplasm

    Source evidence

    PDF page 296: You have read that nearly all of the energy used by living cells comes to them in the bonds of the sugar, glucose. Glycolysis is the first step in the breakdown of glucose to extract energy for cellular metabolism. Nearly all living organisms carry out glycolysis as part of their metabolism. The process does not use oxygen and is therefore anaerobic. Glycolysis takes place in the cytoplasm of both prokaryotic and eukaryotic cells. Glucose enters heterotrophic cells in two ways. One method is through secondary active transport in which the transport takes place against the glucose concentration gradient. The other mechanism uses a group of integral proteins called GLUT proteins, also known as glucose transporter proteins. These transporters assist in the facilitated diffusion of glucose. Glycolysis begins with the six carbon ring-shaped structure of a single glucose molecule and ends with two molecules of a three-carbon sugar called pyruvate. Glycolysis consists of two distinct phases. The first part of the glycolysis pathway traps the glucose molecule in the cell and uses energy to modify it so that the six-carbon sugar molecule can be split evenly into the two three-carbon molecules. The second part of glycolysis extracts energy from the molecules and stores it in the

  2. What is the net gain of ATP molecules from glycolysis of one glucose?

    • Four
    • One
    • Three
    • Two
    Reveal answer

    Answer: Two

    Source evidence

    PDF page 300: Glycolysis starts with glucose and ends with two pyruvate molecules, a total of four ATP molecules and two molecules of NADH. Two ATP molecules were used in the first half of the pathway to prepare the six-carbon ring for cleavage, so the cell has a net gain of two ATP molecules and 2 NADH molecules for its use. If the cell cannot catabolize the pyruvate molecules further, it will harvest only two ATP molecules from one molecule of glucose. Mature mammalian red blood cells are not capable of aerobic respiration—the process in which organisms convert energy in the presence of oxygen—and glycolysis is their sole source of ATP. If glycolysis is interrupted, these cells lose their ability to maintain their sodium-potassium pumps, and eventually, they die. The last step in glycolysis will not occur if pyruvate kinase, the enzyme that catalyzes the formation of pyruvate, is not available in sufficient quantities. In this situation, the entire glycolysis pathway will proceed, but only two ATP molecules will be made in the second half. Thus, pyruvate kinase is a rate-limiting enzyme for glycolysis.

    PDF page 295: hydrogen atoms pass to NAD , reducing it to NADH. Although two molecules of ATP were invested to destabilize glucose at the beginning of the process, four molecules of ATP are formed by substrate-level phosphorylation, resulting in a net gain of two ATP and two NADH molecules for the cell. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in

  3. Glycolysis begins with glucose and ends with two molecules of what?

    • Ethanol
    • Lactate
    • Pyruvate
    • Acetyl-CoA
    Reveal answer

    Answer: Pyruvate

    Source evidence

    PDF page 296: You have read that nearly all of the energy used by living cells comes to them in the bonds of the sugar, glucose. Glycolysis is the first step in the breakdown of glucose to extract energy for cellular metabolism. Nearly all living organisms carry out glycolysis as part of their metabolism. The process does not use oxygen and is therefore anaerobic. Glycolysis takes place in the cytoplasm of both prokaryotic and eukaryotic cells. Glucose enters heterotrophic cells in two ways. One method is through secondary active transport in which the transport takes place against the glucose concentration gradient. The other mechanism uses a group of integral proteins called GLUT proteins, also known as glucose transporter proteins. These transporters assist in the facilitated diffusion of glucose. Glycolysis begins with the six carbon ring-shaped structure of a single glucose molecule and ends with two molecules of a three-carbon sugar called pyruvate. Glycolysis consists of two distinct phases. The first part of the glycolysis pathway traps the glucose molecule in the cell and uses energy to modify it so that the six-carbon sugar molecule can be split evenly into the two three-carbon molecules. The second part of glycolysis extracts energy from the molecules and stores it in the

  4. How many ATP molecules are invested in the first half of glycolysis?

    • Zero
    • Four
    • One
    • Two
    Reveal answer

    Answer: Two

    Source evidence

    PDF page 297: Figure 7.6 The first half of glycolysis uses two ATP molecules in the phosphorylation of glucose, which is then split

    PDF page 295: All organisms, from simple bacteria and yeast to complex plants and animals, carry out some form of cellular respiration to capture and supply free energy for cellular processes. Although cellular respiration and photosynthesis evolved as independent processes, today they are interdependent. The products of photosynthesis, carbohydrates and oxygen gas, are used during cellular respiration. Likewise, the byproduct of cellular respiration, CO2 gas, is used during photosynthesis. Glycolysis is the first pathway used in the breakdown of glucose to extract free energy. Used by nearly all organisms on earth today, glycolysis likely evolved as one of the first metabolic pathways. It is important to note that glycolysis occurs in the cytoplasm of both prokaryotic and eukaryotic cells. (Remember that only eukaryotic cells have mitochondria.) Like all metabolic pathways, glycolysis occurs in steps or stages. In the first stage, the six-carbon ring of glucose is prepared for cleavage (“splitting”) into two three-carbon molecules by investing two molecules of ATP to energize the separation. (Don’t worry; the cell will get the investment of ATP back. It’s like the stock market: You have to invest money to, hopefully, make money!) As glucose is metabolized further, bonds are rearranged through a series of enzyme-catalyzed steps, and free energy is released to form ATP from ADP and free phosphate molecules. The availability of enzymes can affect the rate of glucose metabolism. Two molecules of pyruvate are ultimately produced. High-energy electrons and

  5. Glycolysis does not use oxygen and is therefore described as:

    • Oxidative
    • Anaerobic
    • Aerobic
    • Phototrophic
    Reveal answer

    Answer: Anaerobic

    Source evidence

    PDF page 296: You have read that nearly all of the energy used by living cells comes to them in the bonds of the sugar, glucose. Glycolysis is the first step in the breakdown of glucose to extract energy for cellular metabolism. Nearly all living organisms carry out glycolysis as part of their metabolism. The process does not use oxygen and is therefore anaerobic. Glycolysis takes place in the cytoplasm of both prokaryotic and eukaryotic cells. Glucose enters heterotrophic cells in two ways. One method is through secondary active transport in which the transport takes place against the glucose concentration gradient. The other mechanism uses a group of integral proteins called GLUT proteins, also known as glucose transporter proteins. These transporters assist in the facilitated diffusion of glucose. Glycolysis begins with the six carbon ring-shaped structure of a single glucose molecule and ends with two molecules of a three-carbon sugar called pyruvate. Glycolysis consists of two distinct phases. The first part of the glycolysis pathway traps the glucose molecule in the cell and uses energy to modify it so that the six-carbon sugar molecule can be split evenly into the two three-carbon molecules. The second part of glycolysis extracts energy from the molecules and stores it in the

  6. How many NADH molecules are the net gain from glycolysis of one glucose?

    • One
    • Four
    • Two
    • Three
    Reveal answer

    Answer: Two

    Source evidence

    PDF page 300: Glycolysis starts with glucose and ends with two pyruvate molecules, a total of four ATP molecules and two molecules of NADH. Two ATP molecules were used in the first half of the pathway to prepare the six-carbon ring for cleavage, so the cell has a net gain of two ATP molecules and 2 NADH molecules for its use. If the cell cannot catabolize the pyruvate molecules further, it will harvest only two ATP molecules from one molecule of glucose. Mature mammalian red blood cells are not capable of aerobic respiration—the process in which organisms convert energy in the presence of oxygen—and glycolysis is their sole source of ATP. If glycolysis is interrupted, these cells lose their ability to maintain their sodium-potassium pumps, and eventually, they die. The last step in glycolysis will not occur if pyruvate kinase, the enzyme that catalyzes the formation of pyruvate, is not available in sufficient quantities. In this situation, the entire glycolysis pathway will proceed, but only two ATP molecules will be made in the second half. Thus, pyruvate kinase is a rate-limiting enzyme for glycolysis.

    PDF page 295: hydrogen atoms pass to NAD , reducing it to NADH. Although two molecules of ATP were invested to destabilize glucose at the beginning of the process, four molecules of ATP are formed by substrate-level phosphorylation, resulting in a net gain of two ATP and two NADH molecules for the cell. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in

  7. Which enzyme catalyzes the first step of glycolysis?

    • Aldolase
    • Enolase
    • Pyruvate kinase
    • Hexokinase
    Reveal answer

    Answer: Hexokinase

    Source evidence

    PDF page 297: Step 1. The first step in glycolysis (Figure 7.6) is catalyzed by hexokinase, an enzyme with broad specificity that catalyzes the phosphorylation of six-carbon sugars. Hexokinase phosphorylates glucose using ATP as the source of the phosphate, producing glucose-6-phosphate, a more reactive form of glucose. This reaction prevents the phosphorylated glucose molecule from continuing to interact with the GLUT proteins, and it can no longer leave the cell because the negatively charged phosphate will not allow it to cross the hydrophobic interior of the plasma membrane. Step 2. In the second step of glycolysis, an isomerase converts glucose-6-phosphate into one of its isomers, fructose-6-phosphate. An isomerase is an enzyme that catalyzes the conversion of a molecule into one of its isomers. (This change from phosphoglucose to phosphofructose allows the eventual split of the sugar into two three-carbon molecules.). Step 3. The third step is the phosphorylation of fructose-6-phosphate, catalyzed by the enzyme phosphofructokinase. A second ATP molecule donates a high-energy phosphate to fructose-6-phosphate, producing fructose-1,6-bisphosphate. In this pathway, phosphofructokinase is a rate-limiting enzyme. It is active when the concentration of ADP is high; it is less active when ADP levels are low and the concentration of ATP is high. Thus, if there is “sufficient” ATP in the system, the pathway slows down. This is a type of end product inhibition, since ATP is the end product of glucose catabolism. Step 4. The newly added high-energy phosphates further destabilize fructose-1,6-bisphosphate. The fourth step in glycolysis employs an enzyme, aldolase, to cleave fructose-1,6-bisphosphate into two three-carbon isomers: dihydroxyacetonephosphate and glyceraldehyde-3-phosphate. Step 5. In the fifth step, an isomerase transforms the dihydroxyacetone-phosphate into its isomer, glyceraldehyde-3-phosphate. Thus, the pathway will continue with two molecules of glyceraldehyde-3-phosphate. At this point in the pathway, there is a net investment of energy from two ATP molecules in the breakdown of one glucose molecule.

  8. What product does hexokinase form from glucose in step 1?

    • Glyceraldehyde-3-phosphate
    • Pyruvate
    • Glucose-6-phosphate
    • Fructose-6-phosphate
    Reveal answer

    Answer: Glucose-6-phosphate

    Source evidence

    PDF page 297: Step 1. The first step in glycolysis (Figure 7.6) is catalyzed by hexokinase, an enzyme with broad specificity that catalyzes the phosphorylation of six-carbon sugars. Hexokinase phosphorylates glucose using ATP as the source of the phosphate, producing glucose-6-phosphate, a more reactive form of glucose. This reaction prevents the phosphorylated glucose molecule from continuing to interact with the GLUT proteins, and it can no longer leave the cell because the negatively charged phosphate will not allow it to cross the hydrophobic interior of the plasma membrane. Step 2. In the second step of glycolysis, an isomerase converts glucose-6-phosphate into one of its isomers, fructose-6-phosphate. An isomerase is an enzyme that catalyzes the conversion of a molecule into one of its isomers. (This change from phosphoglucose to phosphofructose allows the eventual split of the sugar into two three-carbon molecules.). Step 3. The third step is the phosphorylation of fructose-6-phosphate, catalyzed by the enzyme phosphofructokinase. A second ATP molecule donates a high-energy phosphate to fructose-6-phosphate, producing fructose-1,6-bisphosphate. In this pathway, phosphofructokinase is a rate-limiting enzyme. It is active when the concentration of ADP is high; it is less active when ADP levels are low and the concentration of ATP is high. Thus, if there is “sufficient” ATP in the system, the pathway slows down. This is a type of end product inhibition, since ATP is the end product of glucose catabolism. Step 4. The newly added high-energy phosphates further destabilize fructose-1,6-bisphosphate. The fourth step in glycolysis employs an enzyme, aldolase, to cleave fructose-1,6-bisphosphate into two three-carbon isomers: dihydroxyacetonephosphate and glyceraldehyde-3-phosphate. Step 5. In the fifth step, an isomerase transforms the dihydroxyacetone-phosphate into its isomer, glyceraldehyde-3-phosphate. Thus, the pathway will continue with two molecules of glyceraldehyde-3-phosphate. At this point in the pathway, there is a net investment of energy from two ATP molecules in the breakdown of one glucose molecule.

  9. Which enzyme is the rate-limiting enzyme active when ADP is high?

    • Isomerase
    • Hexokinase
    • Phosphofructokinase
    • Aldolase
    Reveal answer

    Answer: Phosphofructokinase

    Source evidence

    PDF page 297: Step 1. The first step in glycolysis (Figure 7.6) is catalyzed by hexokinase, an enzyme with broad specificity that catalyzes the phosphorylation of six-carbon sugars. Hexokinase phosphorylates glucose using ATP as the source of the phosphate, producing glucose-6-phosphate, a more reactive form of glucose. This reaction prevents the phosphorylated glucose molecule from continuing to interact with the GLUT proteins, and it can no longer leave the cell because the negatively charged phosphate will not allow it to cross the hydrophobic interior of the plasma membrane. Step 2. In the second step of glycolysis, an isomerase converts glucose-6-phosphate into one of its isomers, fructose-6-phosphate. An isomerase is an enzyme that catalyzes the conversion of a molecule into one of its isomers. (This change from phosphoglucose to phosphofructose allows the eventual split of the sugar into two three-carbon molecules.). Step 3. The third step is the phosphorylation of fructose-6-phosphate, catalyzed by the enzyme phosphofructokinase. A second ATP molecule donates a high-energy phosphate to fructose-6-phosphate, producing fructose-1,6-bisphosphate. In this pathway, phosphofructokinase is a rate-limiting enzyme. It is active when the concentration of ADP is high; it is less active when ADP levels are low and the concentration of ATP is high. Thus, if there is “sufficient” ATP in the system, the pathway slows down. This is a type of end product inhibition, since ATP is the end product of glucose catabolism. Step 4. The newly added high-energy phosphates further destabilize fructose-1,6-bisphosphate. The fourth step in glycolysis employs an enzyme, aldolase, to cleave fructose-1,6-bisphosphate into two three-carbon isomers: dihydroxyacetonephosphate and glyceraldehyde-3-phosphate. Step 5. In the fifth step, an isomerase transforms the dihydroxyacetone-phosphate into its isomer, glyceraldehyde-3-phosphate. Thus, the pathway will continue with two molecules of glyceraldehyde-3-phosphate. At this point in the pathway, there is a net investment of energy from two ATP molecules in the breakdown of one glucose molecule.

  10. Which enzyme cleaves fructose-1,6-bisphosphate into two three-carbon isomers?

    • Mutase
    • Aldolase
    • Hexokinase
    • Enolase
    Reveal answer

    Answer: Aldolase

    Source evidence

    PDF page 297: Step 1. The first step in glycolysis (Figure 7.6) is catalyzed by hexokinase, an enzyme with broad specificity that catalyzes the phosphorylation of six-carbon sugars. Hexokinase phosphorylates glucose using ATP as the source of the phosphate, producing glucose-6-phosphate, a more reactive form of glucose. This reaction prevents the phosphorylated glucose molecule from continuing to interact with the GLUT proteins, and it can no longer leave the cell because the negatively charged phosphate will not allow it to cross the hydrophobic interior of the plasma membrane. Step 2. In the second step of glycolysis, an isomerase converts glucose-6-phosphate into one of its isomers, fructose-6-phosphate. An isomerase is an enzyme that catalyzes the conversion of a molecule into one of its isomers. (This change from phosphoglucose to phosphofructose allows the eventual split of the sugar into two three-carbon molecules.). Step 3. The third step is the phosphorylation of fructose-6-phosphate, catalyzed by the enzyme phosphofructokinase. A second ATP molecule donates a high-energy phosphate to fructose-6-phosphate, producing fructose-1,6-bisphosphate. In this pathway, phosphofructokinase is a rate-limiting enzyme. It is active when the concentration of ADP is high; it is less active when ADP levels are low and the concentration of ATP is high. Thus, if there is “sufficient” ATP in the system, the pathway slows down. This is a type of end product inhibition, since ATP is the end product of glucose catabolism. Step 4. The newly added high-energy phosphates further destabilize fructose-1,6-bisphosphate. The fourth step in glycolysis employs an enzyme, aldolase, to cleave fructose-1,6-bisphosphate into two three-carbon isomers: dihydroxyacetonephosphate and glyceraldehyde-3-phosphate. Step 5. In the fifth step, an isomerase transforms the dihydroxyacetone-phosphate into its isomer, glyceraldehyde-3-phosphate. Thus, the pathway will continue with two molecules of glyceraldehyde-3-phosphate. At this point in the pathway, there is a net investment of energy from two ATP molecules in the breakdown of one glucose molecule.

  11. How many ATP molecules are formed by substrate-level phosphorylation in glycolysis?

    • Two
    • Six
    • Four
    • Three
    Reveal answer

    Answer: Four

    Source evidence

    PDF page 295: hydrogen atoms pass to NAD , reducing it to NADH. Although two molecules of ATP were invested to destabilize glucose at the beginning of the process, four molecules of ATP are formed by substrate-level phosphorylation, resulting in a net gain of two ATP and two NADH molecules for the cell. Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in

  12. During step 6, high-energy electrons are picked up by which carrier?

    • NAD
    • ATP
    • ADP
    • FAD
    Reveal answer

    Answer: NAD

    Source evidence

    PDF page 297: So far, glycolysis has cost the cell two ATP molecules and produced two small, three-carbon sugar molecules. Both of these molecules will proceed through the second half of the pathway, and sufficient energy will be extracted to pay back the two ATP molecules used as an initial investment and produce a profit for the cell of two additional ATP molecules and two even higher-energy NADH molecules. Step 6. The sixth step in glycolysis (Figure 7.7) oxidizes the sugar (glyceraldehyde-3-phosphate), extracting high-energy

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