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: What CD spectroscopy shows.
Circular dichroism reads how a chiral protein bends left and right circularly polarised light, which reveals its secondary structure.

Step 2: Recall the alpha helix signature.
A helix gives a strong positive band near $190$ to $195$ nm and two negative dips near $208$ nm and $222$ nm.

Step 3: Test the negative bands.
At $210$ nm we are close to the $208$ nm dip, so the signal is negative, making A true. At $220$ nm we are close to the $222$ nm dip, so C is true.

Step 4: Test the positive band.
Near $195$ nm the value is clearly positive, so B is true. The claim of a negative value at $195$ nm in D is therefore wrong.

Step 5: Answer.
\[ \boxed{A,\ B,\ C} \]