Relative fraction of population of different enzyme conformations, knowing the rate constants
If we use the equilibrium constants:

$$\ce{A &lt;=&gt;[$K_1$] B &lt;=&gt;[$K_2$] C &lt;=&gt;[$K_3$] D}$$

And we call $[\text{E}]$ the total enzyme concentration, we have:

$$ [\text{A}] + [\text{B}] + [\text{C}] + [\text{D}] = [\text{E}]\label{a}\tag{1}$$

It's not hard to see that $\displaystyle K_1 K_2 K_3 = \frac{[\text{D}]}{[\text{A}]}$. From here, we have that $\displaystyle [\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3}$. We also get:

$$[\text{B}] = K_1[\text{A}] = \frac{[\text{D}]}{K_2 K_3}$$
$$[\text{C}] = K_2[\text{B}] = \frac{[\text{D}]}{K_3}$$

Hence, substituting these last three equalities in $\ref{a}$, we get:

$$ \frac{[\text{D}]}{K_1 K_2 K_3} + \frac{[\text{D}]}{K_2 K_3} + \frac{[\text{D}]}{K_3} + [\text{D}] = [\text{E}]$$

So:

$$[\text{D}] = \frac{K_1 K_2 K_3 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

From here we can also express the concentrations of the other species in terms of $[\text{E}]$:

$$[\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3} = \frac{[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{B}] = K_1 [\text{A}] = \frac{K_1 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{C}] = K_2 [\text{B}] = \frac{K_1 K_2[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$

With these equations we may compare the concentrations of each species at equilibrium, thus we are able to determine which one is the highest based on the values for the equilibrium constants; and we can also obtain their mole fractions (divide each by $[\text{E}]$, which will eliminate this term from all the numerators).