ASper ion released =RIn [ Cb/Ce Cb=ions on DNA(2M)-[ions] in solution R=0.00199 kcal/K mol Kdiss=10(AG/2.303RT ASionrelease =0.00653 kcal/Kmol 4Gionrelease =-0.00653 kcal/KmoIx 298K= 1.95 kcallmol x 4 ions =-7 8 kcal/mol Kds=1042.303Rn=1xl06 b) By how much does the dissociation constant change as you raise the salt from 75 mM to 400 0.0032 kcal/Kmol AGion release =0.0032 kcal/Kmo x 298K=0.95 kcalmol x 4 ions =3. 8 kcal/mol Kdiss=10-/4G/2.303RT/=1.6x 10-3 The dissociation constant changes by 1600 fold. 8. You measure the Kdiss of YFDB on specific and non-specific DNA using footprinting and you get I x1o- and lx l0-o, respecitively. Being a thorough iochemist, you repeat the experiment specific Kdiss is 5 x 10 a) Explain the essence of each method. See lecture notes b) Why do you think the methods are more different for the not non-specific Kdiss than for the specIfIc Filter binding is a non-equilibrium method. As soon as you bind the protein DNA complex to the filter, it will begin to dissociate, and it will continue to dissociate during the washes. The magnitude of this effect will be particularly pronounced for the weak binding to non-specific DNA where your off-rate is high. The interaction with specific DNA is much tighter. Therefore, dissociation is not such a problem and the nvo techniques give essentially the same result c)Which result do you think is more reliable? footprinting since it's an equilibrium method. 9. Estimate the molecular weight in Daltons, of the nucleosome core particle This question requires you to know the composition of the particle and the approximate MWs of the proteins(partial credit for composition alone 6x14,000D田H2A.H2B,H3) 2x10.000D(H4 1x146x660=96000DNA) Total= 200 kDSper ion released = R ln [Cb/Cf] Cb = [ions] on DNA (~2 M) Cf = [ions] in solution R= 0.00199 kcal / K mol Kdiss = 10[G /2.303 RT] Sion release = 0.00653 kcal / Kmol Gion release = -0.00653 kcal / Kmol x 298 K = 1.95 kcal/mol x 4 ions =-7.8 kcal/mol Kdiss = 10[G /2.303 RT] = 1 x 10-6 b) By how much does the dissociation constant change as you raise the salt from 75 mM to 400 mM? Sion release(400mM) = 0.0032 kcal / Kmol Gion release = 0.0032 kcal / Kmol x 298 K = 0.95 kcal/mol x 4 ions =3.8 kcal/mol Kdiss = 10-[G /2.303 RT] = 1.6 x 10-3 The dissociation constant changes by 1600 fold. 8. You measure the Kdiss of YFDB on specific and non-specific DNA using footprinting and you get 1 x 10-10 and 1x 10-6 , respecitively. Being a thorough biochemist, you repeat the experiment using using filter binding. Using this approach, your specific Kdiss is 2 x 10-10, and your non￾specific Kdiss is 5 x 10-4 . a) Explain the essence of each method. See lecture notes b) Why do you think the methods are more different for the not non-specific Kdiss than for the specific Kdiss? Filter binding is a non-equilibrium method. As soon as you bind the protein DNA complex to the filter, it will begin to dissociate, and it will continue to dissociate during the washes. The magnitude of this effect will be particularly pronounced for the weak binding to non-specific DNA where your off-rate is high. The interaction with specific DNA is much tighter. Therefore, dissociation is not such a problem and the two techniques give essentially the same result. c) Which result do you think is more reliable? footprinting, since it’s an equilibrium method. 9. Estimate the molecular weight in Daltons, of the nucleosome core particle. This question requires you to know the composition of the particle and the approximate MWs of the proteins (partial credit for composition alone). 6 x 14, 000 D (H2A. H2B, H3) 2 x 10,000 D (H4) 1 x 146 x 660 = 96,000 (DNA) Total = 200 kD