When vesicles of the cardiac sarcoplasmic reticulum (CSR) are incubated with ATP

Question

When vesicles of the cardiac sarcoplasmic reticulum (CSR) are incubated with ATP, Mg2+ and Ca2+, they take up Ca2+ and reach a pseudo steady-state. This is a steady state that changes, but slowly. The uptake of Ca2+ is mediated by the SERCA2a CaATPase. The uptake reaction can be quenched by adding EGTA to the external solution, which binds the Ca2+ outside of the vesicles, or by adding glucose plus hexokinase, that converts the ATP to ADP and glucose-6 phosphate. When the uptake reaction is stopped, Ca2+ that was already taken up by the vesicles leaks out passively.

A. The amount of Ca2+ taken up by the vesicles is generally normalized to the amount of CSR protein in mg rather than being expressed as a concentration. A typical steadystate Ca2+ uptake is 40 nmol mg-1. In separate experiments, the enclosed volume of the CSR vesicles was determined to be 5 µL mg-1. What is the approximate concentration of Ca2+ inside the vesicles at steady-state? The total Ca2+ concentration is its amount divided by the volume in which it is distributed. In this case the amount is 40 nmol mg CSR protein-1 and the volume is 5 µL mg CSR protein-1. The concentration is estimated as

B. The average vesicle size determined by electron microscopy is about 150 nm. What is the volume and surface area of a vesicle this size, assuming it is a sphere?

C. Given that the enclosed volume of the aggregate vesicles is 5 ìL mg-1, how many vesicles are there per mg of CSR protein? How much surface area is there per mg of CSR protein?

D. The initial passive efflux at a load of 40 nmol mg-1 when the pump is stopped is 16 nmol min-1mg-1. Convert this to a flux in units of nmol cm-2 s-1 by dividing by the surface area per mg of CSR protein and converting min to s.

E. What is the passive permeability to Ca2+ in cm s-1?

Explanation / Answer

A. Per mg enclosed volume of vesicle is 5 uL; we need to calculate molar concentariton of vesicle by using the following formula

C = (m÷V) × (1÷Mw)

Here, C is the concentration, m =mass of the molecule, V = volume, Mw is the molecular weight.

We have been provided m =1 mg, V =1uL, Mw = 38 (Ca2+)

Convert all units into nano-units; m =1 x10-6 ng, V = 10-3 nL

Therefore, concentration C is-

C = 1( x10-6 ÷10-3) × (1÷ 38)

C = 0.026 × 10-3

C = 2.6 x 10-5 nmole

B. For 150nm vesicle size i.e. the diameter, the volume and surface area are following-

Diameter D =150nm

Radius = D/2 = 150/2 =75nm

Volume V = 4/3 r3

V = 4/3 × 3.14 × (75)3

V = 1.77 x 106 nm3

Surface area S = 4 r2

S = 4 × 3.14 × (75)2

S = 7.065 × 104 nm2

C. if enclosed volume of aggregate vesicles = 5 ìL mg-1 = 50 uL mg1

Numbers of vesicles N = Enclosed volume of aggregate vesicles ÷ Per vesicle volume

N = 50 uL mg1 ÷5 uL mg1

N = 10

Surface area St =7.065 × 104 x 10 nm2

St = 7.065 × 105 nm2

D. Flux F= Initial passive efflux÷ (Surface Area)

F = 16 nmol min-1mg-1 ÷ 7.065 × 105 nm2

F = 16 / 60 ÷ 7.065 × 105 (10-7)2                     [ 1 minute =60seconds and 1 nm = 10-7 cm ]

F = 0.27 ÷ 7 × 105 (10-14)

F = 0.0385 × 1011

F = 385 × 108 nmol cm-2 s-1

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