Solution Found!
Answer: The monoanion of adenosine monophosphate (AMP) is
Chapter , Problem 102IE(choose chapter or problem)
The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism:
where A = adenosine. If the \(\mathrm p K_{a}\) for this anion is 7.21, what is the ratio of \(\mathrm {{\left[A M P-O H^{-}\right]}}\) to \(\mathrm {{\left[A M P-O^{2-}\right]}}\) in blood at pH 7.4?
Equation Transcription:
Text Transcription:
O-
A-O-P-OH=AMP-OH^-
pK_a
[AMP-OH^{-}]
[AMP-O^{2-}]
Questions & Answers
QUESTION:
The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism:
where A = adenosine. If the \(\mathrm p K_{a}\) for this anion is 7.21, what is the ratio of \(\mathrm {{\left[A M P-O H^{-}\right]}}\) to \(\mathrm {{\left[A M P-O^{2-}\right]}}\) in blood at pH 7.4?
Equation Transcription:
Text Transcription:
O-
A-O-P-OH=AMP-OH^-
pK_a
[AMP-OH^{-}]
[AMP-O^{2-}]
ANSWER:
Step 1 of 4
Formation of 
