The peptide below is placed into a buffer with a pH = 1.54 -NH-CH-C HO- H;N-CH- CH2 -NH -CH-C -NH-CH-C -NH-CH-C ČH2 CH2 ČH3 CH2 C: :O CH2 H2G NH3 NH2 OH What direction will the peptide move? If it will move towards the positive electrode give your answer as 1 If it will move towards the negative electrode give your answer as -: If it will not move give your answer as 0
The peptide below is placed into a buffer with a pH = 1.54 -NH-CH-C HO- H;N-CH- CH2 -NH -CH-C -NH-CH-C -NH-CH-C ČH2 CH2 ČH3 CH2 C: :O CH2 H2G NH3 NH2 OH What direction will the peptide move? If it will move towards the positive electrode give your answer as 1 If it will move towards the negative electrode give your answer as -: If it will not move give your answer as 0
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The peptide below is placed into a buffer with a pH = 1.54
OH
-NH-CH-C
-NH—CH— с—ОН
-NH-CH-C
H;N-CH-
CH2
-NH -CH-C
ČH2
CH2
ČH3
CH2
C:
:O
NH2
CH2
OH
H2G
NH3
What direction will the peptide move?
If it will move towards the positive electrode give your answer as 1
If it will move towards the negative electrode give your answer as -:
If it will not move give your answer as 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F488cca3e-5c0d-407a-bbe3-202f45494ccf%2Fff1f9291-c254-4833-827f-6c01ed8996fc%2Fe6ujy8_processed.png&w=3840&q=75)
Transcribed Image Text:The peptide below is placed into a buffer with a pH = 1.54
OH
-NH-CH-C
-NH—CH— с—ОН
-NH-CH-C
H;N-CH-
CH2
-NH -CH-C
ČH2
CH2
ČH3
CH2
C:
:O
NH2
CH2
OH
H2G
NH3
What direction will the peptide move?
If it will move towards the positive electrode give your answer as 1
If it will move towards the negative electrode give your answer as -:
If it will not move give your answer as 0
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