Energy, Stability and the Rosetta Score

How does a computer measure a protein's stability?

Lecture 2 — From Physical Forces to a Single Number

"If you unfold a protein — stretch it out into a straight line — does that require energy, or does energy get released? Why?"

Why does stability matter for drug design?

Stable protein ✓

  • Maintains its 3D shape under physiological conditions
  • Binding site geometry is consistent and predictable
  • Drug binds reliably every time
  • Good drug target

Unstable protein ✗

  • Fluctuates between multiple shapes
  • Binding site geometry changes unpredictably
  • Drug may bind sometimes but not others
  • Unreliable drug target
Connect to Project 1: Before designing a drug to fit a binding site, we need to know that binding site holds its shape. Measuring stability is step zero of drug design.

Four forces hold every protein together

1. Van der Waals interactions

Weak attractive forces between any two atoms 3.5–5.0 Å apart. Like cling film — individually almost nothing, but thousands add up to something substantial.

2. Hydrogen bonds

Attraction between a hydrogen on N or O, and another N or O nearby (1.8–3.5 Å). Like a firm handshake — directional, specific, and relatively strong.

3. Electrostatic interactions

Attraction between oppositely charged amino acids (e.g. Arginine + and Aspartate −). Like opposite poles of a magnet — act over a longer range.

4. The hydrophobic effect

Non-polar amino acids cluster together to avoid water. Like oil in water — they group together not because they attract each other, but because water prefers to exclude them.

Proteins fold because the folded shape is more stable

  • Lower energy = more stable. The folded protein sits in an energy valley — it needs energy input to unfold.
  • The depth of the valley = stability. A very stable protein has a deep valley. A marginally stable one has a shallow valley and unfolds more easily.
  • Mutations can change the valley depth. In Project 1 we ask: does this mutation make the valley deeper or shallower?

Scientists measure stability by free energy ΔG

ΔG (delta G) tells you whether a process is favourable or not.

ΔG = ΔH − TΔS
ΔG Meaning Example
Negative
ΔG < 0
Favourable — happens spontaneously Protein folding, drug binding to its target
Positive
ΔG > 0
Unfavourable — requires energy input Protein unfolding, pulling a drug out of its binding site
Important: Rosetta does not calculate true thermodynamic free energy — it uses an approximation. The scores correlate well with real stability measurements but are not exact free energies. This is why we use REU (Rosetta Energy Units) rather than kJ/mol.

Rosetta's scoring function

"Rosetta adds up all non-covalent interactions across every atom in the protein and produces a single number representing overall stability."
Component What it measures
fa_atrAttractive van der Waals forces between atom pairs
fa_repRepulsive forces when atoms are too close (clashes)
hbond_scHydrogen bonds between side chains
fa_elecElectrostatic interactions between charged residues
fa_solSolvation — how much each residue interacts with water

You do not need to memorise these terms — just the concept that the score is a sum of many physical contributions.

REU — the unit in every project

Lower REU = more stable protein  
  More negative = better
Protein size Typical Rosetta score
Small (~100 residues, e.g. Lysozyme)−1,000 to −2,000 REU
Medium (~300 residues)−3,000 to −6,000 REU
Large (~500+ residues)−8,000 REU and below
Connect to Project 1: When you run Cell 3, you will see a number like −1,742.38 REU. That single number is the sum of every van der Waals contact, every hydrogen bond, every electrostatic interaction in Lysozyme.

Score per residue:
making fair comparisons

The problem: a larger protein always has a more negative raw score — simply because it has more atoms.

Structure Raw score Residues Score per residue
Lysozyme (1LYZ) −1,800 REU 129 −13.95
Lysozyme variant (2LZT) −2,400 REU 200 −12.00
2LZT looks better by raw score — but per residue it is actually weaker. Normalisation tells the true story.

Connect to Project 2: You will download five Lysozyme structures and rank them by score per residue. This normalisation step is what makes the comparison valid.

Every measurement needs a reference point

A baseline is the original, unmodified measurement that all results are compared against — the same idea as a control group in an experiment.

In Project 1 — before mutating any amino acid:

original_score = scorefxn(pose)  ← the baseline

difference = mutant_score − original_score

Positive difference → mutation made the protein less stable (destabilising)

Negative difference → mutation made the protein more stable (stabilising)
Analogy: Measuring your running speed after changing shoes only means something if you know your original time. The original time is your baseline. The difference tells you whether the new shoes helped or hurt.

Project 1 in one picture

Load Lysozyme (1LYZ)
Score with get_fa_scorefxn()
Original score (baseline)
Clone protein → Mutate one residue to Alanine
Score the mutant → Calculate difference
Positive → Destabilising
 | 
Negative → Stabilising
Repeat for 7 positions → Plot bar chart

Summary & Preparation

  • Proteins are stable because the folded state has lower free energy than the unfolded state.
  • Four non-covalent forces hold proteins together: van der Waals, hydrogen bonds, electrostatics, and the hydrophobic effect.
  • ΔG < 0 means a process is favourable — both protein folding and drug binding have negative ΔG.
  • The Rosetta scoring function sums all non-covalent interactions across every atom to produce a single stability score in REU.
  • Lower REU = more stable — more negative is always better in Rosetta.
  • Dividing by residue count gives a normalised score that allows fair comparisons between proteins of different sizes.
Preparation for Lecture 3:
"Ibuprofen is a non-polar molecule with a large hydrophobic region. The COX-2 binding site is also largely hydrophobic. Why does this chemical match matter? What would happen if you tried to fit a highly charged, polar drug into a hydrophobic pocket?"