Imagine that in red blood cells (RBCS) thefollowing concentrations were noted for the"adenylate pool": [ATP] = 2000 uM; [ADP] = 270uM; and [AMP] = 30 uM. ASsume that the pH is7 and remains constant.a) calculate the Energy change (EC) (to fourdecimal places for RBCS given the above values.Show workb) If the [ATP] decreases by 5%, calculate thenew [ATP], [ADP], and [AMP] given that theconcentrations of all three species are"connected" by the adenylate kinase reaction,which is shown below. You will need to use thequadratic equation for this one. show your work.%3|%3DADP + ADPATP + AMP Keq = 0.44

Answered: Imagine that in red blood cells (RBCS)…

Human Physiology: From Cells to Systems (MindTap Course List)

9th Edition

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Author:Lauralee Sherwood

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Chapter3: The Plasma Membrane And Membrane Potential

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Problem 1SQE: Using the Nernst equation, calculate the equilibrium potential for Ca2 and for C1 from the following...

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Imagine that in red blood cells (RBCS) the
following concentrations were noted for the
"adenylate pool": [ATP] = 2000 uM; [ADP] = 270
uM; and [AMP] = 30 uM. ASsume that the pH is
7 and remains constant.
a) calculate the Energy change (EC) (to four
decimal places for RBCS given the above values.
Show work
b) If the [ATP] decreases by 5%, calculate the
new [ATP], [ADP], and [AMP] given that the
concentrations of all three species are
"connected" by the adenylate kinase reaction,
which is shown below. You will need to use the
quadratic equation for this one. show your work.
%3|
%3D
ADP + ADP
ATP + AMP Keq = 0.44

Expert Solution

Step 1

The energy change or Gibb's energy change is difference in energy of products and reactants ( $∆ G$ ) in a reaction.

Step 2

(a) In R.B.C, concentration of adenylate pool is given as [ATP]= 2000 $μ M$ , [ADP]=270 $μ M$ , and [AMP]= 30 $μ M$ .

Formula for free energy change is given as:

$∆ G = ∆ G ° + R T \ln Q w h e r e; ∆ G = e n e r g y c h a n g e f o r r e a c t i o n a t g i v e n c o n c e n t r a t i o n s o f r e a c \tan t s a n d p r o d u c t s ∆ G ° = s \tan d a r d e n e r g y c h a n g e f o r r e a c t i o n a t e q u i l i b r i u m c o n c e n t r a t i o n Q = R e a c t i o n q u o t e i n t = \frac{[P r o d u c t s}}{[R e a c \tan t s]} R = g a s c o n s \tan t = 8.314 J / K . m o l e T = 37 ° C = 310 K$

Now at equilibrium $∆ G$ =0, so:

$∆ G ° = - R T \ln K_{e q} w h e r e K_{e q} = e q u i l i b r i u m c o n s \tan t = 0.44 (g i v e n) ∆ G ° = - 8.314 \times 310 \times \ln (0.44) ∆ G ° = - 2115.9460 J / m o l$

So $∆ G$ $= - 2115.9460 + 8.314 \times 310 \times \ln \frac{2000 \times 30}{270 \times 270} = - 2115.9460 - 502.0634 = - 2618.0094 J / m o l$

or $∆ G$ =-2.6180 KJ/mol

Hence, energy change for R.B.C for given adenylate pool is -2.6180 KJ/mol or for converting from ADP to ATP and AMP, there is release of 2.6180 KJ/mol of energy and reaction is spontaneous.

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