Protein Design Partition Tournament

Started by bkoep

actiasluna Lv 1

I'm eager to see how this works… I like the idea and enjoyed the first round, was interesting to work with the patterns that never scored well in monomers and play around with "off-book" ideas.

Susume Lv 1

Looking at the two energy landscape graphs in Blog Part 2, it looks like in the left hand one the 2nd best solution (orange dot) was about 25 energy points (250 foldit points) away from the best (blue dot). If our decoy score is 250 foldit points off of the original design, we can expect so tiny a slice of the partition function that it won't matter. Am I reading that right?

In the right hand graph, the 2nd best dot (orange) is less than one energy point (10 foldit points) away from the best (blue), so if we can get within 10 foldit points we can expect to capture a big slice (in this example, 32%) of the partition function. The next best dot (green) is right about one energy point (10 foldit points) from the best (blue), for a still respectable 12% of the partition function. The 4th best dot (red) is about 3 energy points (30 foldit points) from the best (blue) and gives only a 1.3% slice of the partition function.

I realize every energy landscape is different, but does it make sense as a ballpark figure to say that we can only capture a decent slice of the partition function if we get within 30 or so foldit points of the original score?

jeff101 Lv 1 

https://fold.it/portal/node/2005638 discussed two different proteins:

One protein had a blue structure making up ~100% of the population and an 
orange structure making up < 10^(-10) % of the population. If one defines 
y as the difference in Foldit points between 2 structures, the ratio of 
their populations (probability factor) is p, which equals 10^(y/14). 
Thus, one has p=10^(y/14), log(p)=log(10^(y/14))=(y/14)log(10)=y/14, and 
y=14log(p). If one sets the orange structure's Foldit score and Rosetta 
Energy to 0 and finds the ratio of populations p using the orange 
structure's population in the denominator, one gets the following chart:

                                   ratio of
                                   populations              change in
                       fraction    or                       Rosetta
           percent of  of          probability  change in   Energy in
structure  population  population  factor p     Foldit pts  kcal/mol
----------------------------------------------------------------------
blue       ~100%       ~1 or 10^0  >10^12       >168        <-16.8
orange     <10^(-10)%  <10^(-12)    1              0           0

The other protein had a blue state with 54% of the population, an orange 
one with 32% of the population, a green one with 12% of the population, 
and a red one with 1.3% of the population. These %'s total to 99.3% of 
the population. If one sets the red state's Foldit score and Rosetta 
Energy to 0 and uses the red state's population as the denominator in 
all population ratio calculations, one gets the following chart:

                                   ratio of
                                   populations              change in
                       fraction    or                       Rosetta
           percent of  of          probability  change in   Energy in
structure  population  population  factor p     Foldit pts  kcal/mol
----------------------------------------------------------------------
blue       54%         0.54        41.538462    22.658306   -2.2658306
orange     32%         0.32        24.615385    19.476893   -1.9476893
green      12%         0.12         9.230769    13.513331   -1.3513331
red         1.3%       0.013        1            0           0

Bruno Kestemont Lv 1

on the blog, I would interpret that a visible (small) partition slice would appear if we get less than 300 Foldit points from the original pose.

From -300 to 0 and more, we get a higher slice (more and more challenging to the pose).

jeff101 Lv 1 

Using the probability factors p and Rosetta Energy changes U (in kcal/mol) 
from the above calculations with the formula p=exp(-U/KT), 
one can get KT in kcal/mol and the temperature T as below:

First, p=exp(-U/KT) gives ln(p) = -U/KT, KT = -U/ln(p), and T = -U/(K ln(p)).

With the above formulas, 
p=10^12 for U = -16.8 kcal/mol gives KT = 0.608012 kcal/mol.

Furthermore, the pairs 
p=41.538462 for U = -2.2658306 kcal/mol, 
p=24.615385 for U = -1.9476893 kcal/mol, and
p= 9.230769 for U = -1.3513331 kcal/mol 
all give KT = 0.608012 kcal/mol as well.

Next, using K = 1.9872041 x 10^(-3) kcal/(mol K) from 
https://en.wikipedia.org/wiki/Boltzmann_constant#Value_in_different_units
gives T = 305.96 K, which using the formulas at
https://en.wikipedia.org/wiki/Conversion_of_units_of_temperature
converts to 32.81 deg C and 91.06 deg F, somewhere 
between room temperature and human body temperature.

Finally, since -16.8 kcal/mol of Rosetta Energy gives 168 Foldit points, 
one could also write KT as 6.08012 Foldit points.