2. (18 pts) Answer the following questions based on your understanding of the se
ID: 274930 • Letter: 2
Question
2. (18 pts) Answer the following questions based on your understanding of the secondary structure of a 5HT3 receptor protein described in "Cis-trans isomerization at a proline opens the pore of a neurotransmitter-gated ion channel", Nature 438, 2005, p248-252. The paper can be found in the Exam folder. (a) What structural motif discussed in class is exhibited by the ? helical domain? (b) Many of these helices appear to be antiparallel. How might the helix dipole play a role in assembling this molecule? (c) The ? helical domain is described as being transmembrane. Calculate the thickness of a membrane spanned by this structural unit given your knowledge of the regular dimensions of an a helix. (d) Draw the structures of both cis and tran tripeptide LPA with the Dmp form of proline in the middle. Show the steric configuration of each amino acid residue. e)Explain why the cis conformer of tripeptide LPA with Dmp in the middle is favored over the trans conformer based on the structures in (d).Explanation / Answer
Answering any 3 of the 3 questions:
a.The transmembrane helices (M2 and M3) of HT-3 receptor is being discussed in this paper. Over here, importance of proline in the loop between M2 and M3 is shown to connect the binding to overall conformational change of the ion channels (act as a gate).
b. The cumulative dipole effect from C to N terminus increases in case of helix but tends to be lesser in transmembrane helices in vacuum. However, Helix dipole depends upon the length of helix and the electrostatic interactionwith the solvent. In case of transmembrane helices dipole effect is less with longer helices( increased number of residues). Thus tranmembrane helices stabilises the net dipole effect with the solvent or the environment in which it is present.
c. transmembrane domain has approximately 4-10 residues and distance between each residue is 1.5 Ang. Thus thickness of the membrane considering minimum distancee 4*1.5=6Ang to 15Ang
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