Question

In circular dichroism (CD) spectroscopy, the difference in molar extinction coefficients (\(\Delta \varepsilon\)) is plotted as a function of wavelength \(\lambda\) (in nm). In a CD spectrum of an alpha helical protein, \(\Delta \varepsilon\) will have a

• A. negative value at \(210\) nm
• B. positive value at \(195\) nm
• C. negative value at \(220\) nm
• D. negative value at \(195\) nm

Hint:

An \(\alpha\)-helix in CD spectroscopy typically shows one positive peak near \(190\text{--}195\) nm and two negative peaks near \(208\) nm and \(222\) nm.

Answer 1

The Correct Option is A, B, C

Step 1: Understand circular dichroism (CD) spectroscopy.
Circular dichroism spectroscopy measures the differential absorption of left and right circularly polarized light by chiral molecules such as proteins. It is widely used to study protein secondary structure.

Step 2: Recall the characteristic CD spectrum of an \(\alpha\)-helix.
An alpha helical protein shows characteristic CD signals:
\[ \text{Positive band near }190\text{--}195\ \text{nm} \] and two negative bands near
\[ 208\ \text{nm} \quad \text{and} \quad 222\ \text{nm}. \]

Step 3: Analyze option (A).
Since alpha helices show a negative CD band near \(208\) nm, the value at \(210\) nm is also negative.
Hence, option (A) is correct.

Step 4: Analyze option (B).
Alpha helices exhibit a strong positive band around \(190\text{--}195\) nm.
Therefore, \(\Delta \varepsilon\) is positive at \(195\) nm.
Hence, option (B) is correct.

Step 5: Analyze option (C).
Another characteristic negative band for alpha helices occurs near \(222\) nm. Thus, the value at \(220\) nm is also negative.
Hence, option (C) is correct.

Step 6: Analyze option (D).
At \(195\) nm, the CD signal is positive rather than negative. Therefore, option (D) is incorrect.

Step 7: Final conclusion.
Thus, the correct statements are
\[ \boxed{(A),\ (B)\ \text{and}\ (C)} \]