29. A protein polypeptide chain exists in α-helical conformation in a solvent and it has a value of −30,000 deg cm² dmol⁻¹ for the mean residue ellipticity at 222 nm ([Θ]₂₂₂) in the temperature range 20–50°C. On raising the temperature above 50°C, [Θ]₂₂₂ increases and reaches a value of −2,000 deg cm² dmol⁻¹ at 70°C and the value of [Θ]₂₂₂ remains unchanged above 70°C. The observed value of [Θ]₂₂₂ is −14,000 deg cm² dmol⁻¹ at 60°C. If one assumes that the heat-induced denaturation is a two-state process, the fraction of α-helix at 60°C is (A) 0.40 (B) 0.43 (C) 0.50 (D) 0.57

29. A protein polypeptide chain exists in α-helical conformation in a solvent and it has a value of −30,000 deg cm² dmol⁻¹ for the mean residue ellipticity at 222 nm ([Θ]₂₂₂) in the temperature range 20–50°C. On raising the temperature above 50°C, [Θ]₂₂₂ increases and reaches a value of −2,000 deg cm² dmol⁻¹ at 70°C and the value of [Θ]₂₂₂ remains unchanged above 70°C. The observed value of [Θ]₂₂₂ is −14,000 deg cm² dmol⁻¹ at 60°C. If one assumes that the heat-induced denaturation is a two-state process, the fraction of α-helix at 60°C is

(A) 0.40

(B) 0.43

(C) 0.50

(D) 0.57

Fraction of Alpha-Helix from Circular Dichroism During Two-State Protein Denaturation

Correct Answer

Option (2): 0.43

Explanation

Circular Dichroism (CD) spectroscopy is extensively used to study the secondary structure of proteins because different structural elements produce characteristic CD signals. The mean residue ellipticity at 222 nm is particularly sensitive to the presence of α-helices. A highly negative value of [θ]222 indicates a greater proportion of α-helical structure, whereas values that become less negative indicate progressive unfolding of the protein.

In this problem, the protein exhibits a mean residue ellipticity of -30,000 deg cm2 dmol-1 between 20°C and 50°C, indicating that the protein is completely folded in its α-helical conformation within this temperature range. As the temperature increases beyond 50°C, thermal energy begins to disrupt the hydrogen bonds stabilizing the α-helix. Consequently, the protein gradually unfolds, causing the ellipticity at 222 nm to become less negative. At 70°C, the ellipticity reaches -2,000 deg cm2 dmol-1, which represents the completely unfolded state, and no further structural changes occur beyond this temperature.

The question states that protein unfolding follows a two-state transition. Under this assumption, only two conformations exist: the completely folded α-helical state and the completely unfolded state. Any observed ellipticity between these two extremes represents a mixture of folded and unfolded molecules. Therefore, the fraction of α-helix can be calculated using a simple linear relationship between the observed ellipticity and the ellipticities of the folded and unfolded states.

Calculation

The fraction of α-helix is calculated using the expression

Fraction of α-helix = (θobs − θunfolded) / (θfolded − θunfolded)

Substituting the given values,

θfolded = -30,000

θunfolded = -2,000

θobserved = -14,000

Fraction of α-helix = [(-14,000) − (-2,000)] / [(-30,000) − (-2,000)]

= (-12,000) / (-28,000)

= 0.4286

0.43

Thus, approximately 43% of the protein molecules remain in the α-helical conformation at 60°C, while the remaining 57% have undergone thermal unfolding.

Why Option (1) is Incorrect

Option (1) suggests that the fraction of α-helix is 0.40. This value does not satisfy the linear relationship expected for a two-state transition using the given ellipticity values. Substituting the experimental data into the appropriate equation gives a value closer to 0.43 rather than 0.40.

Why Option (2) is Correct

Option (2) is correct because the observed ellipticity at 60°C lies between the ellipticity of the fully folded and fully unfolded protein. Applying the two-state denaturation equation yields a fraction of 0.4286, which rounds to 0.43. This indicates that nearly half of the original α-helical structure has been retained at this temperature.

Why Option (3) is Incorrect

Option (3) assumes that exactly half of the protein remains folded. However, the observed ellipticity of -14,000 deg cm2 dmol-1 is not midway between -30,000 and -2,000. The calculated fraction is approximately 0.43, making 0.50 inconsistent with the experimental measurements.

Why Option (4) is Incorrect

Option (4) overestimates the amount of α-helical structure present at 60°C. A fraction of 0.57 would correspond to a more negative observed ellipticity than the value provided in the question. Therefore, this option is not supported by the CD data.

Understanding the Two-State Denaturation Model

The two-state model assumes that a protein exists only in two distinct conformations during thermal unfolding: a completely folded native state and a completely unfolded denatured state. Intermediate partially folded conformations are considered negligible. Because the experimentally observed CD signal is the weighted average of these two populations, the fraction of folded protein can be determined directly from the measured ellipticity. This model is widely used for proteins that unfold cooperatively over a relatively narrow temperature range.

Importance of Mean Residue Ellipticity at 222 nm

The wavelength of 222 nm is particularly useful for monitoring α-helical content because peptide bonds within an α-helix generate a strong negative CD signal at this wavelength. As thermal denaturation progresses, the ordered hydrogen-bonded helical structure is disrupted, causing the negative ellipticity to decrease in magnitude. Monitoring changes at 222 nm therefore provides a direct and sensitive method for following protein unfolding and estimating the fraction of α-helical structure present under different conditions.

Conclusion

The protein undergoes a cooperative two-state thermal denaturation in which the ellipticity at 222 nm changes linearly between the folded and unfolded states. Using the observed ellipticity at 60°C, the calculated fraction of α-helical structure is 0.4286, which is approximately 0.43. Therefore, Option (2) is the correct answer.

Leave a Reply

Your email address will not be published. Required fields are marked *

Latest Courses