Question

(e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 µM, how many glucose molecules are present per hexokinase molecule?

          (e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 µM, how many glucose molecules are present per hexokinase molecule?
        
(e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 µM, how many glucose molecules are present per hexokinase molecule?

Added by Alexander T.

Close

Biology for AP Courses
Biology for AP Courses
Julianne Zedalis, John Eggebrecht
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
(e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20mu M, how many glucose molecules are present per hexokinase molecule? (e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukaryotic cell is 20 uM,how many glucose molecules are present per hexokinase molecule?
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn Kathleen Carty
Ivan Kochetkov verified

Adi S and 82 other subject Biology educators are ready to help you.

Ask a new question

*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
in-glycolysis-the-km-of-hexokinase-is-004-mm-under-physiological-conditions-the-glucose-concentration-at-the-cellular-level-is-between-4-mm-to-15-mm-based-on-this-information-which-of-the-fo-71722

In glycolysis, the KM of hexokinase is 0.04 mM. Under physiological conditions, the glucose concentration at the cellular level is between 4 mM to 15 mM. Based on this information, which of the following is true of hexokinase under physiological conditions? (MULTIPLE ANSWER) a. Hexokinase is saturated with its substrate b. The enzyme works at its Vmax c. Such a low KM value ensures that the glycolysis process will continue even when the glucose level drops to its lowest level of 4 mM. d. The enzyme has low affinity for its substrate e. Glucose phosphorylation occurs slowly

Adi S.

hexokinase-catalyzes-the-phosphorylation-of-glucose-using-atp-to-form-a-molecule-of-glucose-6-phosphate-this-is-the-first-step-of-glycolysis-in-the-cytoplasm-of-animal-cells-in-this-reaction-46744

Hexokinase catalyzes the phosphorylation of glucose using ATP to form a molecule of glucose-6-phosphate. This is the first step of glycolysis in the cytoplasm of animal cells. In this reaction, hexokinase [Choose ] ATP [Choose ] glucose [Choose ] glucose-6-phosphate [Choose ] is a product is the enzyme is a substrate is an inhibitor

Adi S.

hexokinase-is-an-enzyme-that-binds-specifically-to-glucose-and-converts-it-into-glucose-6-phosphate-the-activity-of-hexokinase-is-suppressed-by-glucose-6-phosphate-which-binds-to-hexokinase-19706

Hexokinase is an enzyme that binds specifically to glucose and converts it into glucose 6-phosphate. the activity of hexokinase is suppressed by glucose 6-phosphate, which binds to hexokinase at a location that is distinct from the active site. this is an example of what?

Sri K.


*

Recommended Textbooks

-
Biology for AP Courses

Biology for AP Courses

Julianne Zedalis, John Eggebrecht
achievement 1,642 solutions
Objective Biology for NEET

Objective Biology for NEET

Rajiv Vijay 1st Edition
achievement 1,426 solutions
Introduction to General, Organic and Biochemistry

Introduction to General, Organic and Biochemistry

Frederick A. Bettelheim, William H. Brown, Mary K. Campbell 12th Edition
achievement 1,053 solutions

*

Transcript

-
00:01 So what is given in this question is we are given some information regarding glycolysis process.
00:06 Glycolysis.
00:07 So what is given in this question is that the km value of enzyme hexokinness which is used in glycolysis process.
00:15 Okay.
00:16 Km value for enzyme hexokinines is given to us and this value is 0 .04 millimolar.
00:23 This is the value of km which is given.
00:25 And another information which is mentioned here already that you can all say it.
00:30 Set as by default that it is all around physiological condition that means ph will be around 7 ph will be around 7 the temperature will be around 37 degrees celsius so that enzyme can work at their best okay this is the value which is given to us another point which is given to us about this is that glucose concentration at cellular level so concentration or glucose is given to us at cellular level and this concentration of glucose is between 4 millimolar to 15 millimolar.
01:05 This is a value which is given to us about glucose concentration at cellular level.
01:11 So what we have to understand what we have to find is which of the following statement is true for hexokinous enzyme.
01:17 We have to identify the true statement.
01:20 Okay.
01:21 So according to the michael's method equation, what is it say that v0, that is velocity is equals to v -machyman.
01:27 V max plus v max v max into substrate right into substrate concentration upon km value plus s.
01:38 This is the michaelis menter equation which is given to us.
01:40 Now look at the values of these which is given to us because if values are given to us that there might be some role.
01:47 Otherwise there is no need for the extra information that they will give it just to confuse us.
01:52 There is a information which is given to us indirectly.
01:55 And what is this equation information which is given to us.
01:57 That s, the s, that is the substrates, is very, very, very less than km, right? this is the point we have to understand from here is that the substrate, the glucose at which reaction is being carried out, has a concentration much, much, much larger than the km of the hexokinous because we can clearly see that it is around 50 millimolar, but it is only around 0 .04 millimolar.
02:21 So we have to interpretate this information by observing the question, right? so it is given to us...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever