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How does citrate from the citric acid cycle affect glycolysis? a. Citrate and ATP are negative regulators of phosphofructokinase-1. b. Citrate and ATP are negative regulators of hexokinase. c. Citrate and ATP are positive regulators of phosphofructokinase-1. d. Citrate and ATP are positive regulators of hexokinase.

Short Answer

Expert verified
a. Citrate and ATP are negative regulators of phosphofructokinase-1.

Step by step solution

01

Identify the Enzyme involved in Glycolysis

Glycolysis involves different enzymes, but key regulators include hexokinase and phosphofructokinase-1 (PFK-1).
02

Understand the Role of Citrate in Metabolism

Citrate is an intermediate of the citric acid cycle that can inhibit glycolysis by affecting certain enzymes.
03

Determine the Regulatory Effect of Citrate

Citrate typically acts as a negative regulator, meaning it inhibits the activity of enzymes rather than promoting it.
04

Analyze the Choices

Based on steps 2 and 3, identify that citrate and ATP inhibit an enzyme involved in glycolysis. Both are negative regulators of phosphofructokinase-1 (option a).

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Key Concepts

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

phosphofructokinase-1 regulation
Phosphofructokinase-1 (PFK-1) is one of the key regulatory enzymes in glycolysis. This enzyme is responsible for converting fructose-6-phosphate to fructose-1,6-bisphosphate. Because it catalyzes a crucial step, its regulation ensures the efficiency and control of the whole glycolytic pathway.

PFK-1 is heavily regulated by various molecules, both inhibitors and activators. For example, ATP and citrate are negative regulators, meaning they inhibit PFK-1's activity. When the cell has plenty of ATP, it signals that there is enough energy, so glycolysis is slowed down. Similarly, citrate, which is an intermediate in the citric acid cycle, signals that the cell's energy needs are met and further glycolysis is not required.

On the other hand, AMP acts as a positive regulator for PFK-1. When cellular energy levels are low, AMP concentration increases, promoting PFK-1 activity and thus accelerating glycolysis to produce more ATP.
  • Negative Regulators: ATP, citrate
  • Positive Regulators: AMP
By balancing these regulatory signals, PFK-1 plays a vital part in maintaining cellular energy homeostasis.
citric acid cycle
The citric acid cycle, also known as the Krebs cycle, is a series of chemical reactions that take place in the mitochondria. It plays a fundamental role in cellular respiration, where cells extract energy from nutrients. This cycle fully oxidizes acetyl-CoA to CO鈧 and reduces NAD鈦 to NADH and FAD to FADH鈧, which are used in the electron transport chain to produce ATP.

One of the notable intermediates of the citric acid cycle is citrate. Citrate's primary role is in the biosynthesis of fatty acids and can be transported out of the mitochondria to the cytosol when required.

Besides its role as an intermediate, citrate also provides critical regulatory feedback. High levels of citrate signal that the cell has sufficient metabolic intermediates, thereby inhibiting glycolysis through the regulation of PFK-1. This type of feedback mechanism ensures that the cell doesn't unnecessarily produce more energy intermediates than needed.
The overall objectives of the citric acid cycle are:
  • Convert acetyl-CoA into CO鈧
  • Generate NADH and FADH鈧 to fuel ATP synthesis
  • Provide intermediates for other biosynthetic pathways
negative regulation in metabolism
Negative regulation in metabolism involves inhibiting enzymes or pathways to prevent the overaccumulation of end products or intermediates. This is crucial for maintaining cellular balance and efficiency.

For instance, in glycolysis, PFK-1 is negatively regulated by ATP and citrate. When ATP levels are high, it signals the cell to reduce further ATP production by inhibiting PFK-1. Similarly, citrate, as an intermediate from the citric acid cycle, signals ample energy availability and inhibits PFK-1 to slow down glycolysis.

Negative regulation ensures that pathways do not operate unnecessarily or wastefully. It also allows the cell to switch between different metabolic pathways according to energy demands. This is important for dynamic cellular conditions, like changing energy requirements or nutrient availability.
Some key points on negative regulation:
  • Promotes metabolic efficiency
  • Prevents excess product formation
  • Responds to cellular signals to adjust pathway activity
Negative regulatory mechanisms like these underpin the cell's ability to maintain homeostasis and respond adaptably to various metabolic contexts.

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Most popular questions from this chapter

What role does \(\mathrm{NAD}^{+}\) play in redox reactions? a. \(\mathrm{NAD}^{+}\) , an oxidizing agent, can accept electrons and protons from organic molecules and get reduced to NADH. b. \(\mathrm{NAD}^{+}\) , a reducing agent, can donate its electrons and protons to organic molecules. c. \(\mathrm{NAD}^{+}\) , an oxidizing agent, can accept electrons from organic molecules and get reduced to NADH. d. \(\mathrm{NAD}^{+}\) , a reducing agent, can donate its electrons and protons to inorganic molecules.

E. coli are enteric (gut-dwelling) facultative anaerobic bacteria. (Facultative anaerobes can grow either with or without free oxygen. Obligatory anaerobes grow only in the absence of free oxygen.) Researchers planned to grow cultures of \(E .\) coli under a range of conditions to model the transition from strictly anaerobic to aerobic respiration. The oxygen content of atmospheres at constant total pressure will be controlled by volumes of nitrogen and oxygen gases. Ratios of volume, \(r=\mathrm{V}_{\mathrm{O}_{2}} / \mathrm{V}_{\mathrm{N}_{2}}\) between 0 and 0.25 of shaken growth flasks can be measured in terms of optical density, which is the percent of transmission of light through a sample of the growing \(E\) . coli culture. A rule of thumb is that the range of strict anaerobes is when r \(<0.01,\) and the boundary for aerobic respiration is when \(\mathrm{r}\) \(=0.05 .\) A large number of flasks that can be constantly shaken at fixed temperature, and from which samples can be taken without atmospheric contamination, are available for this study. These results of the experiment will be used to infer growth rates of \(E\) . coli along the entire 7.5 \(\mathrm{m}\) length of the average human intestine (small intestine and large intestine), where the oxygen content varies from atmospheric to anaerobic conditions. The retention time of food in the small intestine, whose average length is \(2.5 \mathrm{m},\) is approximately four hours. The retention time of food over the entire length of the intestine is between 24 and 72 hours. A. Describe and apply a mathematical model that can be used to represent the variation of oxygen environments of a bacterium that is being transported with the food along the length of the intestine. B. Design the experimental sampling times in terms of growth intervals of interest in this study: i) the time when the bacteria is passing the small-large intestine boundary; ii) the time when the bacteria reaches the end of the large intestine; and iii) the time when the bacterium reaches facultative anaerobic conditions, r \(<0.05 .\)C. C. Sketch a graph that predicts the distribution of aerobic, facultative anaerobic and obligatory anaerobic bacteria along the length of the entire intestine based on these parameters. Keep in mind that anaerobes have a lower respiration rate.

A. [Extension] Living systems require free energy to carry out cellular functions, and employ various strategies to capture, use, and store free energy. Explain the advantage that the higher energy efficiency per kg of the Krebs cycle provides to you compared to a metabolism based on glycolysis alone. Your explanation should make use of all the following facts: \(\bullet\)\triangle \mathrm{G}\( for glycolysis is \)-135 \mathrm{kJ}\( per mole of glucose \)\bullet\( \triangle G\) for aerobic respiration is - 2880 \(\mathrm{kJ}\) per mole glucose \(\bullet\) the basal metabolic rate of mammals is often represented as \(-300 \mathrm{kJ} / \mathrm{day} \cdot \mathrm{m}^{0.75}\) \(\bullet\) the molar mass of glucose is 180 \(\mathrm{g} / \mathrm{mole}\) B. Explain the bioenergetic difference between aerobic and anaerobic respiration in terms of the difference between free-energy production and power. Your explanation should make use of all the following facts: \(\cdot\) power is the rate of free-energy production \(\cdot\) cancer cells derive most of their free energy from glycolysis \(\cdot\) enzymes of the citric acid (Kreb's) cycle form coordinate complexes on the cytoskeleton within the mitochondria C. The life cycle of the human parasite Trypanosoma brucei is divided between the body of the tsetse fly and the human blood stream. The parasite causes 鈥渟leeping sickness鈥 in Sub-Saharan Africa. Within the human bloodstream, the parasite depends on glycolysis, with enzymes compartmentalized in a membrane-bound organelle called the glycosome. In the insect host, the parasite utilizes glycolysis as well as substrate-level and oxidative phosphorylation. Explain the advantage of a life cycle in the human host that employs anaerobic respiration with a rate of free-energy production that is enhanced by compartmentalization in the glycosome and a life cycle in the insect host that is aerobic. D. Predict the advantages of a biological system that uses both glycolysis and oxidative phosphorylation. Your prediction should make use of all the following facts: \(\cdot\) signaling can be used to detect low-oxygen environments and to regulate response \(\cdot\) some cells, such as muscle and blood cells, must function in both low- and high-oxygen environments \(\cdot\) glycolysis is reversible \(\cdot\) the citric acid cycle is not reversible \(\cdot\) thermoregulation is needed for homeostasis

Where in a cell does glycolysis take place in both prokaryotes and eukaryotes? a. the cytosol b. the mitochondria c. the plasma membrane d. the nucleus

What three steps are included in the breakdown of pyruvate? a. Pyruvate dehydrogenase removes a carboxyl group from pyruvate producing carbon dioxide. Dihydrolipoyl transacetylase oxidizes a hydroxyethyl group to an acetyl group, producing NADH. Lastly, an enzyme-bound acetyl group is transferred to CoA, producing a molecule of acetyl-CoA. b. Pyruvate dehydrogenase oxidizes hydroxyethyl group to an acetyl group, producing NADH. It further removes a carboxyl group from pyruvate producing carbon dioxide. Lastly, dihydrolipoyl transacetylase transfers enzyme-bound acetyl group to CoA forming an acetyl-CoA molecule. c. Pyruvate dehydrogenase transfers enzymebound acetyl group to CoA forming an acetyl CoA molecule. It then oxidizes a hydroxyethyl group to an acetyl group, producing NADH. Dihydrolipoyl transacetylase removes a carboxyl group from pyruvate producing carbon dioxide. d. Pyruvate dehydrogenase removes carboxyl group from pyruvate producing carbon dioxide. Dihydrolipoyl dehydrogenase transfers enzymebound acetyl groups to CoA forming an acetylCoA molecule. Lastly, a hydroxyethyl group is oxidized to an acetyl group, producing NADH.

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