10-7. A classic paper studied the behavior of lipids in the two monolayers of a membrane by labeling individual molecules with nitroxide groups, which are stable free radicals (Figure Q10-2). These spin-labeled lipids can be detected by electron spin-resonance (ESR) spectroscopy, technique that does not harm living cells. Spin-labeled lipids are introduced into small lipid vesicles which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane. The two spin- labeled phospholipids shown in Figure Q10-2 were incorporated into intact human red cell membranes in this way. To determine whether they were introduced equally into the two monolayers of the bilayer, ascorbic acid (vitamin C), which is a water-soluble reducing agent that does not cross membranes, was added to the medium to destroy any nitroxide radicals exposed on the outside of the cell. The ESR signal was followed as a function of time in the presence and absence of ascorbic acid as indicated in Figure Q10-3A and B. (A) PHOSPHOLIPID 1- RED CELLS (B) PHOSPHOLIPID 2-RED CELLS -ascorbate 100 100 nitroxide radical 75 75 -ascorbate + ascorbate 50 25 + ascorbate 10 20 30 (C) PHOSPHOLIPID 1-GHOSTS (D) PHOSPHOLIPID 2-GHOSTS phospholipid 1 -ascorbate 100 75 + ascorbate 50 phospholipid 2 10 30 20 time (minutes) time (hours) Figure Q10-2. Structure of two nitroxide- labeled lipids (Problem 10-7) Then nitroxide radical is shown at the top, and its position of attachment to the phospholipids is shown below. Figure Q10-3. Decrease in ESR signal intensity as a function of time in intact red cells and red cell ghosts in the presence and absence of ascorbate (Problem 10- 7). (A and B) Phospholipid 1 and phospholipid 2 in intact red cells (C and D) Phospholipid 1 and phospholipid 2 in red cell ghosts. A. Ignoring for the moment the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1 (Figure Q10-3A) reacts faster with ascorbate than does phospholipid 2 (Figure Q10-3B). Note that phospholipid 1 reaches a plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour. signal intensity (%) signal intensity (%) 25 -ascorbate ascorbate 8 100 75 50 25

Biology 2e
2nd Edition
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark
Chapter4: Cell Structure
Section: Chapter Questions
Problem 17RQ: Which of the following sequences correctly lists in order the steps involved in the incorporation of...
icon
Related questions
Question

Help!

10-7. A classic paper studied the behavior of lipids in the two monolayers of a membrane by labeling individual molecules
with nitroxide groups, which are stable free radicals (Figure Q10-2). These spin-labeled lipids can be detected by electron
spin-resonance (ESR) spectroscopy, technique that does not harm living cells. Spin-labeled lipids are introduced into small
lipid vesicles which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane. The two spin-
labeled phospholipids shown in Figure Q10-2 were incorporated into intact human red cell membranes in this way. To
determine whether they were introduced equally into the two monolayers of the bilayer, ascorbic acid (vitamin C), which is
a water-soluble reducing agent that does not cross membranes, was added to the medium to destroy any nitroxide radicals
exposed on the outside of the cell. The ESR signal was followed as a function of time in the presence and absence of ascorbic
acid as indicated in Figure Q10-3A and B.
(A) PHOSPHOLIPID 1- RED CELLS
(B) PHOSPHOLIPID 2- RED CELLS
- ascorbate
* 100.
100
nitroxide
radical
75
75
- ascorbate
+ ascorbate
50
50
25
25
+ ascorbate
10
20
30
(C) PHOSPHOLIPID 1- GHOSTS
(D) PHOSPHOLIPID 2 - GHOSTS
phospholipid 1
- ascorbate
- ascorbate
E 100
100
75
75
+ ascorbate
+ ascorbate
50
50
phospholipid 2
25
25
10
20
30
1
time (minutes)
time (hours)
Figure Q10-2. Structure of two nitroxide-
labeled lipids (Problem 10-7) Then
nitroxide radical is shown at the top, and
its position of attachment to
phospholipids is shown below.
Figure Q10-3. Decrease in ESR signal intensity as a
function of time in intact red cells and red cell ghosts
in the presence and absence of ascorbate (Problem 10-
7). (A and B) Phospholipid 1 and phospholipid 2 in
intact red cells (C and D) Phospholipid 1 and
phospholipid 2 in red cell ghosts.
the
A. Ignoring for the moment the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1
(Figure Q10-3A) reacts faster with ascorbate than does phospholipid 2 (Figure Q10-3B). Note that phospholipid 1
reaches a plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour.
signal intensity (%)
signal intensity (%)
Transcribed Image Text:10-7. A classic paper studied the behavior of lipids in the two monolayers of a membrane by labeling individual molecules with nitroxide groups, which are stable free radicals (Figure Q10-2). These spin-labeled lipids can be detected by electron spin-resonance (ESR) spectroscopy, technique that does not harm living cells. Spin-labeled lipids are introduced into small lipid vesicles which are then fused with cells, thereby transferring the labeled lipids into the plasma membrane. The two spin- labeled phospholipids shown in Figure Q10-2 were incorporated into intact human red cell membranes in this way. To determine whether they were introduced equally into the two monolayers of the bilayer, ascorbic acid (vitamin C), which is a water-soluble reducing agent that does not cross membranes, was added to the medium to destroy any nitroxide radicals exposed on the outside of the cell. The ESR signal was followed as a function of time in the presence and absence of ascorbic acid as indicated in Figure Q10-3A and B. (A) PHOSPHOLIPID 1- RED CELLS (B) PHOSPHOLIPID 2- RED CELLS - ascorbate * 100. 100 nitroxide radical 75 75 - ascorbate + ascorbate 50 50 25 25 + ascorbate 10 20 30 (C) PHOSPHOLIPID 1- GHOSTS (D) PHOSPHOLIPID 2 - GHOSTS phospholipid 1 - ascorbate - ascorbate E 100 100 75 75 + ascorbate + ascorbate 50 50 phospholipid 2 25 25 10 20 30 1 time (minutes) time (hours) Figure Q10-2. Structure of two nitroxide- labeled lipids (Problem 10-7) Then nitroxide radical is shown at the top, and its position of attachment to phospholipids is shown below. Figure Q10-3. Decrease in ESR signal intensity as a function of time in intact red cells and red cell ghosts in the presence and absence of ascorbate (Problem 10- 7). (A and B) Phospholipid 1 and phospholipid 2 in intact red cells (C and D) Phospholipid 1 and phospholipid 2 in red cell ghosts. the A. Ignoring for the moment the difference in extent of loss of ESR signal, offer an explanation for why phospholipid 1 (Figure Q10-3A) reacts faster with ascorbate than does phospholipid 2 (Figure Q10-3B). Note that phospholipid 1 reaches a plateau in about 15 minutes, whereas phospholipid 2 takes almost an hour. signal intensity (%) signal intensity (%)
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Biology 2e
Biology 2e
Biology
ISBN:
9781947172517
Author:
Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:
OpenStax
Anatomy & Physiology
Anatomy & Physiology
Biology
ISBN:
9781938168130
Author:
Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:
OpenStax College