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In the book Lehninger Principles of Biochemistry, there is a metal stick analogy to explain why an enzyme being compliment to its substrate would actually make a terrible enzyme.

The metal anaology

If the enzyme is complimentary to the substrate, it will form many weak interactions, hence the enzyme-substrate (ES) complex is very stable, meaning it will require more energy to reach the transition state, so very few products are made

Note: The green drawing represents the enzyme

If the enzyme isn't entirely compliment to its substrate, it will form a few weak interactions (that allows it to bind to its substrate). This isn't its lowest energy state, as there are still more weak interactions that can occur. The weak interactions will mould the substrate into its transition state. This would mean that the transition state is the most stable state in this environment (I would assume). The thing that I am confused about is that if the transition state would be the most stable state, then why would it go on to form a product? Once the weak interactions mould the substrate, then energy would be released from the substrate, forming a transition state, therefore making the transition state have lower energy than the substrate, not higher, so the reaction profile in the final diagram doesn't really make sense to me. Is there anything that I am missing or not understanding?

enter image description here

I would really appreciate someone clearing up any confusion about this. Thanks.

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  • $\begingroup$ I'd recommend starting from the diagrams here, since it seems like you understand them. Compare the 2nd and 3rd plots, and use them to explain to yourself why an enzyme that works like the 3rd plot is drawn is better than the 2nd in both the ES and transition states. Then reread the text and think about it as a comparison between those two diagrams rather than the imaginary other case you're thinking of. $\endgroup$
    – Bryan Krause
    Aug 7 at 22:54
  • $\begingroup$ I think I've got it now. The energy required to reach the transition state is lower because of 2 things: -The ES complex is less stable in diagram 3 then diagram 2 (higher energy) -Once the ES complex has formed, there are still more weak interactions that can occur which provide some energetic driving force towards the transition state, so require less energy to reach there. Is this right? $\endgroup$
    – Pencilcase
    Aug 8 at 9:01
  • $\begingroup$ That sounds accurate to me! I think it's worth keeping in mind that the "transition state" is energetically not very favorable: it's at the top of all of these charts. So, when describing "weak interactions (that) mould the substrate into its transition state", you're making the transition state more favorable than it would be without those interactions, not necessarily more favorable than a different, favorable state. If you have to jump over a rope, and someone else puts their hand on it to make it lower to the ground, there's still a rope to get over. $\endgroup$
    – Bryan Krause
    Aug 9 at 15:56

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