1. Synthesis of Gastric Juice: EnergeticsBiological Membranes : Biological Membranes : Problems and solutions Problems and solutions

y gGastric juice (pH 1.5) is produced bypumping HCl from blood plasma (pH 7.4)into the stomach Calculate the amount ofinto the stomach. Calculate the amount offree energy required to concentrate the H+

in 1 L of gastric juice at 37 C. Underll l diti h l fcellular conditions, how many moles of

ATP must be hydrolyzed to provide thisamount of free energy? The free-energygy gychange for ATP hydrolysis under cellularconditions is about - 58 kJ/mol. Ignore theeffects of the transmembrane electricaleffects of the transmembrane electricalpotential.(Problem # 7, page # 415-416, Michael M.C d D id L N l P i i l fCox and David L. Nelson: Principles ofBiochemistry, Fifth Edition)

SolutionpH = log [H] [H] = 10-pH.pH = log [H], [H] = 10 p

At pH 1.5, [H] = 10 - 1.5 = 3.2 x 10 - 2 M.At pH 7.4, [H] = 10-7.4 = 4.0 x 10-8 M.Gt = RT ln (C2/C1)at 37 C, RT = 2.58 kJ/mol,

Gt = (2.58 kJ/mol) ln (3.2 x 10 - 2 / 4.0 x 10-8 ) = 35 kJ/mol.

The amount of ATP required to provide 35 kJ is 0 60 mol (35 kJ/58kJ)is 0.60 mol (35 kJ/58kJ)

2 Energetics of the Na+K+ ATPase2. Energetics of the Na K ATPaseFor a typical vertebrate cell with amembrane potential of 0.070 V (inside

ti ) h t i th f hnegative), what is the free-energy changefor transporting 1 mol of Na+ from the cellinto the blood at 37oC? Assume theconcentration of Na+ inside the cell is 12mM, and that in blood plasma is 145 mM.(Problem # 8 page # 416 Michael M Cox (Problem # 8, page # 416, Michael M. Cox and David L. Nelson: Principles of Biochemistry, Fifth Edition)

S l ti Solution:

Gt = RT ln (C2/C1) + Z Э ∆ψt

Z = charge on the ion Э = Faraday constant (96,480 J/v.mol) Э Faraday constant (96,480 J/v.mol) ∆ψ=Transmembrane electrical potential (in volts)

Gt = (2.58 kJ/mol) ln (145/12) + (1)(96,480 J/V mol)(0.070 V)= 6.4 kJ/mol + 6.8 kJ/mol = 13 kJ/mol

6.8 kJ/mol is the membrane potential portion

3. Water Flow through an Aquaporing q pA human erythrocyte has about 2 x 105,AQP-1 monomers. If water molecules flowthrough the plasma membrane at a rate ofthrough the plasma membrane at a rate of5x108 per AQP-1 tetramer per second, andthe volume of an erythrocyte is 5x10-11 mL,h idl ld th t h l ithow rapidly could an erythrocyte halve itsvolume as it encountered the highosmolarity (1 M) in the interstitial fluid of they ( )renal medulla? Assume that the erythrocyteconsists entirely of water.(Problem # 16 page # 416 Michael M Cox(Problem # 16, page # 416, Michael M. Coxand David L. Nelson: Principles ofBiochemistry, Fifth Edition)

Solution 1. Calculate the number of water molecules

that must leave the erythrocyte tohalve its volume. The volume of the cell is 5 x 10-11 ml. The volume of the cell is 5 x 10 ml. For [H2O] = 55 M

The number of water molecules in the cell =The number of water molecules in the cell =(5 x 10-11 ml/cell )(6.02 x 1020

molecules/mmol) (55 mmol H2O/ml) molecules/mmol) (55 mmol H2O/ml) =1.7 x 1012

Half of these molecules (8.5 x 1011) must l t h l th ll lleave to halve the cell volume.

2. Calculate how fast the cell can loset l lwater molecules.

The cell has 2 x 105 aquaporinmonomers, or 5 x 104 tetramers.,Each tetramer allows passage of 5 x 108

H2O molecules per second,The flux of water molecules through theThe flux of water molecules through theplasma membrane =(5 x 108 H2O molecules/s/aquaporin t t ) (5 104 i tetramer) (5 x 104 aquaporin tetramers/cell)= 2.5 x 1013 H2O molecules/s2Time required to remove half the volumeof water =(8.5 x 1011 H2O molecules)/(2.5 x 1013(8.5 x 10 H2O molecules)/(2.5 x 10H2O molecules/s) = 3 x 10-2 s

4. Surface Density of a Membrane Protein4. Surface Density of a Membrane ProteinE. coli can be induced to make about 10,000copies of the lactose transporter (Mr31 000) per cell Assume that E coli is a31,000) per cell. Assume that E. coli is acylinder 1 µm in diameter and 2 µm long.What fraction of the plasma membranesurface is occupied by the lactosetransporter molecules? Explain how youarrived at this conclusion.a ed at t s co c us o(Problem # 21, page # 417, Michael M. Coxand David L. Nelson: Principles ofBiochemistry Fifth Edition)Biochemistry, Fifth Edition)

Solution Calculate E. coli surface area.Calculate E. coli surface area.The surface area of a cylinder =where , r= radius, d = diameter, andh h i ht h = height. For a cylinder 2 µm high and 1 µm in diameter, the surface area =2 π (0.5 µm)2 + π (1 µm) (2 µm) = 8 µm2.

To estimate the cross-sectional area of a To estimate the cross sectional area of a globular protein of Mr 31,000, Use the dimensions for hemoglobin (Mr 64,500; diameter 5 5 nm) th s a protein ofdiameter 5.5 nm), thus a protein ofMr 31,000 has a diameter of about 3 nm, assuming the proteins have the same density.

The crosssectional area of a sphere of pdiameter 3 nm (0.003 µm)—or of a single transporter molecule = π r2 = 3 14(1 5 x 10-3 µm)2 = 7x10-6 µm2π r = 3.14(1.5 x 10 µm) = 7x10 µm

The total cross-sectional area of 10,000 t t l l 7 10 2 2 transporter molecules = 7x10-2 µm2. The fraction of an E. coli cell surface covered by transporter molecules=y p(7x10-2 µm2)/(8 µm2) = 0.009, or about 1%.

5. Molecular Species in the E. coli MembraneThe plasma membrane of E. coli is about75% protein and 25% phospholipid byweight. How many molecules of membraneg ylipid are present for each molecule ofmembrane protein? Assume an averageprotein Mr of 50 000 and an averageprotein Mr of 50,000 and an averagephospholipid Mr of 750. What more wouldyou need to know to estimate the fractionof the membrane surface that is covered bylipids?(Problem # 23, page # 417, Michael M. Cox( , p g ,and David L. Nelson: Principles ofBiochemistry, Fifth Edition)

SolutionConsider a sample that contains 1 g ofConsider a sample that contains 1 g ofmembrane, of which 0.75 g is protein (Mr50,000) and 0.25 g is phospholipid (Mr 750).

4 5(0.75 g protein)(1 mol/5 x104 g) = 1.5 x10-5 mol protein in 1 g of membrane(0.25 g phospholipid)(1 mol/750 g) = 3.3x 10-4( g p p p )( g)mol phospholipid in 1 g of membrane3.3x 10-4 mol phospholipid/ 1.5 x10-5 molprotein = 22 mol phospholipid/mol proteinTo estimate the percentage of the surfaceco ered b phospholipid o o ld need tocovered by phospholipid, you would need toknow (or estimate) the average cross-sectional area of a phospholipid in a bilayerand the average crosssectional area of a 50kDa protein