/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 In the first step of glycolysis,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the first step of glycolysis, what is glucose transformed into? a. glucose----phosphate b. fructose- \(1,6\) -bisphosphate c. dihydroxyacetone phosphate d. phosphoenolpyruvate

Short Answer

Expert verified
a. glucose----phosphate

Step by step solution

01

Understand Glycolysis

Glycolysis is the process of breaking down glucose into pyruvate to release energy. It consists of many steps, each transforming the molecule further.
02

Identify the First Step of Glycolysis

The first step of glycolysis involves the phosphorylation of glucose. Here, a phosphate group is added to glucose to form glucose-6-phosphate.
03

Analyze the Answer Choices

Look at the provided options: a. glucose----phosphate b. fructose-1,6-bisphosphate c. dihydroxyacetone phosphate d. phosphoenolpyruvate. Out of these, the correct product formed in the first step is glucose-6-phosphate.
04

Select the Correct Answer

Based on the analysis, the closest answer provided in the list is a. glucose----phosphate. This refers to glucose-6-phosphate.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Glucose Phosphorylation
In the first step of glycolysis, glucose undergoes phosphorylation. This is a crucial initial step in the metabolic process that breaks down glucose to release energy. During glucose phosphorylation, a phosphate group from ATP is transferred to the glucose molecule. This process is catalyzed by the enzyme hexokinase. Hexokinase facilitates the transfer by binding both ATP and glucose and helping the phosphate group make the jump. As a result of this transfer, ATP (adenosine triphosphate) is converted into ADP (adenosine diphosphate). The phosphorylated glucose molecule is now known as glucose-6-phosphate.
This modification is significant because it helps trap the glucose molecule inside the cell, as glucose-6-phosphate cannot easily pass through the cell membrane. It is also key because it makes the glucose molecule more reactive, facilitating its further breakdown in the subsequent steps of glycolysis.
Glucose-6-Phosphate
After glucose is phosphorylated, it is transformed into a molecule known as glucose-6-phosphate. This compound plays a pivotal role not only in glycolysis but also in other metabolic pathways like the pentose phosphate pathway and glycogenesis. Glucose-6-phosphate is a six-carbon sugar with a phosphate group attached to the sixth carbon atom. This phosphorylation is strategically essential:
  • It prevents the molecule from leaving the cell.
  • It marks the molecule for further processing in the glycolysis sequence.
  • It increases the energy potential of the molecule, making it more amenable to subsequent enzymatic actions.
The enzyme that catalyzes this reaction, hexokinase, has a high affinity for glucose, ensuring the efficient processing of even small amounts of glucose that enter the cell. Additionally, this step uses one molecule of ATP, showcasing the role of ATP as an energy-investing molecule early in glycolysis.
Energy Release in Glycolysis
Energy release is the primary purpose of glycolysis, which includes multiple enzyme-catalyzed steps to systematically break down glucose. Though the pathway starts with an energy investment phase, where ATP is consumed, the later stages lead to a net gain of energy. This breakdown process results in the production of two molecules of pyruvate, along with a net gain of two ATP molecules and two NADH molecules.
The energy release occurs in the latter stages through substrate-level phosphorylation, where a phosphate group is directly transferred to ADP, forming ATP. This occurs twice in one cycle of glycolysis, doubling the ATP yield from the initial investment. Also, NAD+ is reduced to NADH, capturing high-energy electrons, which can produce more ATP in the electron transport chain, though this occurs outside glycolysis. Therefore, glycolysis serves as a fundamental mechanism for energy extraction from glucose, providing quick ATP while setting the stage for more efficient energy production in cellular respiration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

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.

Glucose catabolism pathways are sequential and lead to the production of ATP. What is the correct order of the pathways for the breakdown of a molecule of glucose as shown in the formula? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}+\) energy \(\begin{aligned} \text { a. } & \text { oxidative phosphorylation } \rightarrow \text { citric acid cycle } \\ & \rightarrow \text { oxidation of pyruvate } \rightarrow \text { glycolysis } \end{aligned}\) \(\begin{aligned} \text { b. the oxidation of pyruvate } & \rightarrow \text { citric acid cycle } \\ & \rightarrow \text { glycolysis } \rightarrow \text { oxidative phosphorylation } \end{aligned}\) c. glycolysis \(\rightarrow\) oxidation of pyruvate \(\rightarrow\) citric acid cycle \(\rightarrow\) oxidative phosphorylation d. citric acid cycle \(\rightarrow\) glycolysis \(\rightarrow\) oxidative phosphorylation \(\rightarrow\) oxidation of pyruvate

What evidence provides the strongest support that glycolysis is an older and more conserved pathway than the citric acid cycle? a. Glycolysis is the primitive pathway as it is found in all three domains. It also occurs in anaerobic conditions and in the cytosol. b. This pathway occurs in the cytosol, is found in all animals and plants, and does not require oxygen. c. Glycolysis takes place in anaerobic conditions, can metabolize cholesterol and fatty acids, and occurs even in methanogens. d. This pathway only occurs in the mitochondria. It is highly flexible because it is found in almost all organisms.

Which of the following statements most directly supports the claim that different species of organisms use different metabolic strategies to meet their energy requirements for growth, reproduction, and homeostasis? a. During cold periods, pond-dwelling animals can increase the number of unsaturated fatty acids in their cell membranes while some plants make antifreeze proteins to prevent ice crystal formation in their tissues. b. Bacteria lack introns while many eukaryotic genes contain many of these intervening sequences. c. Carnivores have more teeth that are specialized for ripping food while herbivores have more teeth specialized for grinding food. d. Plants generally use starch molecules for storage while animals use glycogen and fats for storage.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.