Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 23, Problem 23.2CC
The Ca2+ concentration in unstimulated cells is about 10−7 M. Why would Ca2+ be a better second messenger than sodium, given that the Na+ concentration in most cells is closer to 10−3 M?
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Students have asked these similar questions
The Graph below shows the binding curves of two proteins (A and B) for the same ligand (L). Use this Graph and determine the
dissociation constant, K, for both proteins. Which protein (A or B) has a greater affinity for ligand L? Which of the two proteins
would be more easily inhibited by an antagonist?
1.0
Y 0.5
2
A
4
6
B
8
[L] (μM)
10
12
14
16
A ligand binds two different receptors with a Kd value of 10−7 M for receptor 1 and a Kd value of 10−9 M for receptor 2. For which receptor does the ligand show the greater affinity? Calculate the fraction of receptors that have a bound ligand ([RL]/RT) in the case of receptor 1 and receptor 2 if the con- centration of free ligand is 10−8 M.
In the Nernst equation [V = 62 log10 (Co/ Ci)], the term Co represents:
the intracellular concentration of potassium
the intracellular concentration of chloride
the membrane potential (in millivolts)
the extracellular concentration of sodium
the extracellular concentration of potassium
Chapter 23 Solutions
Becker's World of the Cell (9th Edition)
Ch. 23 - One effect of the hormone insulin is to cause...Ch. 23 - You are working on a new species of sea urchin...Ch. 23 - The Ca2+ concentration in unstimulated cells is...Ch. 23 - Prob. 23.3CCCh. 23 - Prob. 23.4CCCh. 23 - Prob. 23.5CCCh. 23 - Chemical Signals and Second Messengers. Fill in...Ch. 23 - QUANTITATIVE Pure agony. Agonists are drugs that...Ch. 23 - Heterotrimeric and Monomeric G Proteins. G...Ch. 23 - Calcium Chelators and Ionophores. In addition to...
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- From the Hill Plot below, the KD of the first binding event for the receptor-ligand system under study is: 10 8 6 4 2 0 -2 4 6 8 10 (0=1) 601 log -4- -6 -6 -4 -2 0 2arrow_forwardFor a typical vertebrate cell with a membrane potential of −0.070 V (inside negative), what is the free-energy change for transporting 1 mol of Na+ from the cell into the blood at 37 °C? Assume the concentration of Na+ insidethe cell is 12 mM and in blood plasma it is 145 mM.arrow_forwardThe drug Buckeyium binds to the NMDA receptor at the orthosteric binding site (were glutamate would normally bind). Assume that glycine is already bound, and that magnesium is not blocking the NMDAR ion channel. If the channel does not open as a result of Buckeyium binding than Buckeyium would be considered aarrow_forward
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