NS FL #6 C/P Q48

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sdaswani
Posts: 37
Joined: Sat Jan 26, 2019 5:32 pm

NS FL #6 C/P Q48

Post by sdaswani » Tue May 14, 2019 6:42 am

Hi! I'm a little confused as to how we know delta g has decreased in this question? I understand that the mutations make it more likely to unfold, and hence a greater -ve delta g value. but isn't this considered an increase in delta g as opposed to decrease as in the correct answer choice?

Thanks for your help!
-Saakshi
NS_Tutor_Will
Posts: 766
Joined: Fri May 25, 2018 9:15 am

Re: NS FL #6 C/P Q48

Post by NS_Tutor_Will » Tue May 14, 2019 8:36 am

Thanks for the question!

So we wouldn't necessarily know if ∆∆G is referring to folding or unfolding and we'd need to go to the passage to figure that part of it out. The passage states that the "corresponding change in free energy of unfolding (∆∆G) with respect to wild-type protein" was measured. This means that ∆∆G is referring to unfolding, so a lower ∆∆G indicates more likely to unfold (because unfolding is more spontaneous and releases more energy).

Thus, when ∆∆G is negative, that means the protein is more likely to unfold (lower ∆G) and is less stable in its folded state.

I hope this helps!
sdaswani
Posts: 37
Joined: Sat Jan 26, 2019 5:32 pm

Re: NS FL #6 C/P Q48

Post by sdaswani » Tue May 14, 2019 11:43 am

Got it! so a "lower" delta G value, will be more negative, correct?

I think I took the "increase in delta G" in the answer choice (a) to mean a "larger, more negative delta g." But, I guess it's the lower the delta g (delta g decreases), the more negative the value, and hence, more spontaneous the reaction (unfolding of protein, in this case).

Does that make sense?
NS_Tutor_Will
Posts: 766
Joined: Fri May 25, 2018 9:15 am

Re: NS FL #6 C/P Q48

Post by NS_Tutor_Will » Tue May 14, 2019 2:39 pm

Yep, that sounds exactly right! :D
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