2D Hydrophobic-polar protein folding model

Biology Level 2

The above image contains two proteins and their corresponding random conformations based on a hydrophobic-polar protein folding model. The total energy E 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 , E ( H / H ) = 1 , E ( H / P ) = − 2 , E ( P / P ) = − 1 , 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 H and P ) . 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 E_1 and E 2 E_2 for both of the above protein conformations? ( ( Hint: ∣ E 1 + E 2 ∣ = 28. ) |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?

The former is more stable than the latter The latter is more stable than the former Both are equally stable

Zeeshan Ali by Zeeshan Ali , 25, Pakistan

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1 solution

Zeeshan Ali
May 18, 2018

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

The former protein conformation contains; four E ( H − P ) E(H-P) , three E ( H / H ) E(H/H) , one E ( H / P ) E(H/P) , and four E ( P / P ) E(P/P) interactions. That is, E 1 = 4 × E ( H − P ) + 3 × E ( H / H ) + 1 × E ( H / H ) + 4 × E ( P / P ) = 4 × ( − 3 ) + 3 × ( + 1 ) + 1 × ( − 2 ) + 4 × ( − 1 ) = − 15 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 ) E(H-P) , four E ( H / H ) E(H/H) , one E ( H / P ) E(H/P) , and three E ( P / P ) E(P/P) interactions. That is, E 2 = 4 × E ( H − P ) + 4 × E ( H / H ) + 1 × E ( H / H ) + 3 × E ( P / P ) = 4 × ( − 3 ) + 4 × ( + 1 ) + 1 × ( − 2 ) + 3 × ( − 1 ) = − 13 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 |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 E_1<E_2 or − 15 < − 13 -15<-13 therefore the former is more stable than the latter \text{the former is more stable than the latter} .

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