Question

There are two similar epitopes of a bacterial toxin. They both have turns with leucine extending...

There are two similar epitopes of a bacterial toxin. They both have turns with leucine extending from the protein surface. The turns are identical, except for the substitution of an aspartic acid (Turn B) with a glutamic acid (Turn C). They are recognized by the same antibody. The parent antibody recognizes the acidic residues of Turns B&C with a lysine. In both of the new antibodies, a nearby glycine is mutated to a serine to potentially pick-up another hydrogen bond with the acidic residue of the target Turn. Given that the average strength of an H-bond is –20 kJ/mol, is the increase in affinity for the Turn C antigen observed with anti-Turn C compared to the parent antibody (Ka = 1.0•106 M-1 ) consistent with the formation of an additional hydrogen bond?

Homework Answers

Answer #1

Answer:

   _    ΔG / R*T

= - (- 20/8.314472*310)

= 0.0776.

ΔG is the free energy decides affinity towards additional hydrogen bond.

Specificity of antibody binding probably depends on the complementarity of surfaces for hydrogen bonding and polar bonding as well as van-derWaals contacts, and hydrophobic interactions and the exclusion of water from the interacting surfaces of proteins may contribute a large but nonspecific component to the energy of binding.

With small change in free energy shows differences in affinity. ΔG is larger higher is the affinity.

Note: The provided answer is as per my knowledge may be or may not be 100% correct, but definitely relevant to your question thank you so much.

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