Question

The biological roles of complex organic molecules are determined by their shape -- the way atoms and electrons create charge distributions on the molecule's surface. As a result of those charge distributions, the molecules can interact in specific ways, attracting and sticking to each other or to other organic molecules. For some proteins, their shape can be changed by the addition of a small amount of energy. Since different shapes can result in different biological roles, the same molecule can play different biological roles if the energy needed to change their shape is not too large.
Since the probability of a molecule having a higher energy than its ground state in a thermal bath is proportional to the Boltzmann factor, e=ΔEkBT, we will use this to generate estimates of the probability that a protein has a changed shape.

Let's call the normal (ground state) shape of a molecule α (alpha). Suppose that adding an energy ΔE to a molecule will change it to a new shape, β (beta). For the purposes of this problem we'll assume that this is a two-state system. The molecule has to be in either state α or the state β. If the molecule is immersed in a thermal bath of temperature T, find an expression for the probability, Pα that the molecule will be in the state α and the probability Pβ that the molecule will be in the state β. Since there are only two states, your two probabilities should add to 1 (100%). Express each probability as a function of the symbol z = e-ΔEkBT.

a. First, find Pα.
b. If the thermal bath (cellular fluid) is at 300 K, and the excitation energy ΔE needed to change the molecule's shape is 50 meV (milli-electron volts), find the probability that the molecule will be in shape β. Give your answer as a decimal, not a percent. (Hint: At 300 K, the value of kBT is about 1/40 eV.)
c. If the temperature were small (close to 0 K), what would the probability be that the molecule is in its ground state?
d. What value of ΔE would a molecule in the thermal bath from part (c) need to have to be in shape β one quarter of the time? Give your answer in meV.

Answer 1

Answer:

a) P_α =  exp (-ΔE / kT),  b)   P_β = 0.145 , d)  ΔE = 309.7 meV

Explanation:

The expression for the number of molecules or particles in a given state in Boltzmann's expression

            n = n₀ exp (-ΔE / kT)

Where k is the Bolztmann constant and T the absolute temperature

The probability is defined as the number of molecules in a given state over the total number of particles

          P = n / n₀ = exp (- ΔE / kT)

Let's apply this expression to our case

a) P_α = n_α / n₀ = exp (-ΔE / kT)

b) the Boltzmann constant

       k = 1,381 10⁻²³ J / K (1 eV / 1.6 10⁻¹⁹ J) = 8.63 10⁻⁵ eV / K

       kT = 8.63 105 300 = 2,589 10⁻² eV

       P_β = exp (- 50 10⁻³ /2.589 10⁻² = exp (-1.931)

       P_β = 0.145

c) If the temperature approaches absolute zero, the so-called is very high, so there is no energy to reach the excited state, therefore or all the molecules go to the alpha state

d) For molecules to spend ¼ of the time in this beta there must be ¼ of molecules in this state since the decay is constant.

        P_β = ¼ = 0.25

     

       P_β = exp (- ΔE / kT)

       ΔE = -kT ln P_β

       ΔE = - 2,589 10⁻² ln 0.25

       ΔE = 0.3097 eV

       ΔE = 309.7 meV