The above image contains two proteins and their corresponding random conformations based on a hydrophobic-polar protein folding model. The total energy $E$ of the proteins can be estimated by summing up the energy values for each type of interactions among the residues using the following rules: $E(H-P)=-3,\quad E(H/H)=1,\quad E(H/P)=-2,\quad E(P/P)=-1,$ where the dashes $(-)$ represent the indirect linear (horizontal or vertical) interactions and the slashes $(/)$ represent the (left or right) diagonal interactions among the residues $($ i.e. $H$ and $P).$ For clarification, note that the directed interactions between two residues that are directly connected (as shown by arrows) do not play a part in the estimation of the total energy. Also, note that the interaction between a unique pair of residues can only be considered once. The two lines in the image are hints, not connections.

Can you calculate the total energy values
$E_1$
and
$E_2$
for both of the above protein conformations?
$($
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Hint:
**
$|E_1+E_2|=28.)$

The lower the total energy value of a protein conformation, the greater the stability.

What do you conclude about the stability of the above protein conformations?

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Given that $E(H-P)=-3$ , $E(H/H)=+1$ , $E(H/P)=-2$ , and $E(P/P)=-1$ .

The former protein conformation contains; four $E(H-P)$ , three $E(H/H)$ , one $E(H/P)$ , and four $E(P/P)$ interactions. That is, $E_1=4 \times E(H-P)+3 \times E(H/H)+1 \times E(H/H)+4 \times E(P/P)=4\times{(-3)}+3\times{(+1)}+1\times{(-2)}+4\times{(-1)}=-15$ .

The latter one contains; four $E(H-P)$ , four $E(H/H)$ , one $E(H/P)$ , and three $E(P/P)$ interactions. That is, $E_2=4 \times E(H-P)+4 \times E(H/H)+1 \times E(H/H)+3 \times E(P/P)=4\times{(-3)}+4\times{(+1)}+1\times{(-2)}+3\times{(-1)}=-13$ .

Did my hint help? Since, $|E_1+E_2|=|(-15)+(-13)|=|-28|=28$

The lower the total energy value of a protein conformation, the greater the stability. Since $E_1<E_2$ or $-15<-13$ therefore $\text{the former is more stable than the latter}$ .