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 ellipiticity 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 cm 2 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 cm 2dmol-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
To calculate the fraction of α-helix at 60°C, we can use the following formula:
Fraction of α-helix = ([θ]obs - [θ]random) / ([θ]helix - [θ]random)
Given values:
[θ]obs = -14,000 deg cm2 dmol-1 at 60°C
[θ]random = 0 deg cm2 dmol-1 (assuming completely random coil conformation)
[θ]helix = -30,000 deg cm2 dmol-1 (at 20-50°C)
Plug in the values:
Fraction of α-helix = (-14,000 - 0) / (-30,000 - 0) = 14,000 / 30,000 ≈ 0.467
Therefore, the fraction of α-helix at 60°C is approximately 0.467 or 46.7%.
Thanks.
To determine the fraction of α-helix at 60°C, we can use the information provided regarding the change in mean residue ellipticity ([θ]222) with temperature.
First, let's define the two states involved in the heat-induced denaturation process:
State 1: Native (α-helical conformation)
State 2: Denatured (non-α-helical conformation)
Given that the heat-induced denaturation is a two-state process, we can assume that the transition from State 1 to State 2 occurs at a specific temperature, which we'll refer to as Tm.
Now, let's analyze the provided information step by step:
1. At 20-50°C:
- [θ]222 is -30,000 deg cm2 dmol-1.
- This represents the mean residue ellipticity for the α-helical conformation (State 1).
2. At 50°C:
- [θ]222 is not specified, but we know that the value changes as we raise the temperature above 50°C.
- This indicates that the transition from State 1 to State 2 occurs at a temperature higher than 50°C.
3. At 70°C:
- [θ]222 is -2,000 deg cm2 dmol-1.
- At this temperature, the denaturation process is complete, and the protein is fully in the denatured state (State 2).
4. At 60°C:
- [θ]222 is -14,000 deg cm2 dmol-1.
- We need to determine the fraction of α-helix at this temperature.
Now, let's calculate the fraction of α-helix at 60°C using the following equation:
Fraction of α-helix = ([θ]222 - [θ]222,d) / ([θ]222,n - [θ]222,d)
Where:
- [θ]222 is the observed mean residue ellipticity at 60°C (-14,000 deg cm2 dmol-1).
- [θ]222,d is the mean residue ellipticity of the denatured state at 70°C (-2,000 deg cm2 dmol-1).
- [θ]222,n is the mean residue ellipticity of the native state at 20-50°C (-30,000 deg cm2 dmol-1).
Plugging in the known values:
Fraction of α-helix = (-14,000 - (-2,000)) / (-30,000 - (-2,000))
Fraction of α-helix = -12,000 / -28,000
Fraction of α-helix = 0.43 (rounded to two decimal places)
Therefore, the fraction of α-helix at 60°C is approximately 0.43 or 43%.