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29 June, 21:38

An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterized by ultracentrifugation. In the absence of any ligand, the s20, w = 11.7S, density of solvent is 1g/cm3and viscosity 1.005 cP; the specific volume of the protein is 0.732 cm3/g. a) Find the radius of the protein, assuming it is sphericalb) Upon binding of the substrate the sedimentation coefficient increases by 3.5%; what is the radius of the bound enzyme?

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  1. 29 June, 21:51
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    (a) r = 6.26 * 10⁻⁷cm

    (b) r₂ = 6.05 * 10⁻⁷cm

    Explanation:

    Using the sedimentation coefficient formula;

    s = M (1-Vρ) / Nf; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

    s = M (1 - Vρ) / N*6πnr

    making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs

    Note: S = 10⁻¹³ sec, 1 KDalton = 1 * 10³ g/mol, I cP = 0.01 g/cm/s

    r = { (3.1 * 10⁵ g/mol) (1 - (0.732 cm³/g) (1 g/cm³) } / { (6.02 * 10²³) (6π) (0.01 g/cm/s) (11.7 * 10⁻¹³ sec)

    r = 6.26 * 10⁻⁷cm

    b. Using the formula r₂/r₁ = s₁/s₂

    s₂ = 0.035 + 1s₁ = 1.035s₁

    making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

    r₂ = 6.3 * 10⁻⁷cm / 1.035

    r₂ = 6.05 * 10⁻⁷cm
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