Hog's Folly Revealed
A short Walkthrough of Hog's Folly Simplified
And Tips about Multiloops
This Doesn’t Work
This Does Work
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Lots of slides here
Lots of slides here
Lots of slides here
Lots of slides here
Lots of slides here
Lots of slides here
Lots of slides here
Your perseverance is rewarded with the original slides
Please Click the Link above to be Directed to the Puzzle
Hog's Folly Simplified Constraints
Part 1: Stabilization
As can be seen in the previous slide, the puzzle only allows one G-C pair, and 8 A-U pairs are required. No G-U pairs are allowed.
So, with all puzzles, the most important thing we have to do is make sure every part of the puzzle can work on it’s own.
Remember, when it comes to solving a puzzle, if a section of the puzzle will not work on it’s own, it probably will not work when it is a part of the puzzle as a whole.
Example, the stack on the right with the large hairpin loop. There are only 2 possible ways to stabilize that stack with only A-U pairs.
Part 1.2: Basic Stabilization of Right Stack
Option 1: No Guanine Dangling End Bonus
Option 2: Guanine Dangling End Bonus
Total Kcal = -0.21
Total Kcal = -0.01
Part 1.2 Cont'd
Both option 1 and option 2 of the previous slide are valid for basic stability.
However, since the gap between the two stacks is only one nucleotide, Option 2 will not have a low enough free energy for the penalty that occurs there.
So, why not put the G-C on the right?
Required Use of G-C pairs off Multiloops
Any stack that is 2 or fewer base pairs long, off the open loop, or a 4+ stack multi branch loop, will need at least 1 G-C pair, even for tetraloops.
Required Use of G-C pairs off Multiloops Cont'd
This rule of two does not always apply to multiloops with 3 branches though, so be careful
Someday, we'll know if this is stable or not
(my money is on red)
Part 1.3: Finish filling in the pairs
Let's go with Option 1
Part 1.4: Determining how far away from stability using free energy differences
For any puzzle, an important rule thing to remember, is the free energy of the target molecule (Total Kcal in top left corner) has to be the lowest. This means the natural mode, cannot be lower than the target, under most circumstances.
This is most easily done, by switching to natural mode, and comparing it's total free energy, to the total free energy of the target mode.
Part 1.4: Determining how far away from stability using free energy differences, Cont'd
So, here is Option 1 at it's most basic.
Target Mode Natural Mode
Total: -1.91 Kcal
Total: -2 Kcal
>
This is NOT what we want
Total: -1.91 Kcal
Total: -2 Kcal
>
How Can We Fix This?
Hog's Folly Simplified Conundrum
Based on the previous slides, we see that we only need the target fold to be 0.1 kilocalories lower in order to be stable (Or natural mode higher)
However, all the bases on the right side are locked
(As pointed out earlier, there are only 2 solutions).
As for the left side, you can flip the closing base pair of the open loop (Pair 1-8), which is the only section that will directly influence the difference in free energy. It won't get much better than that A-U pair though.
So, what is left that we can mutate to fix this problem?
Part 2: How to manipulate the single base between two stacks to solve puzzles
Part 2 Explanation
As it turns out, base number 9 in the puzzle Hog's Folly Simplified is the key to solving.
Mutation of this single base can both affect the free energy of the target fold, and the natural fold.
There are some very simple rules for mutating this base. Let's start with guanine.
Part 2.1: The use of Guanine between two stacks separated by 1 base
Wow, now how did the addition of one Guanine Stabilize the puzzle without lowering the free energy of the target fold?
Open Loop Closed,