I'm having trouble with this biochemistry problem. Could someone help?
- hcbiochemLv 73 weeks ago
i. In case A, the deprotonated form of histidine will be preferred because of the uncharged, non-polar environment. In Case B, the protonated form will be preferred because the positively charged, protonated form of histidine will be stabilized by the negative charge or the aspartate.
ii. In Case A, because the unprotonated state is stabilized, the result would be a decrease in the pKa of the histidine side chain. In Case B, because the protonated state is stabilized, this would increase the value of the histidine pKa.
iii. Basically did this within the two other points.
- Anonymous3 weeks ago
Sorry, my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category, Most 'Anonymous' posts are trolls, maybe yours is not, but the high likelihood is there. If you post as yourself, I would be glad to help you with any problem you have.
I spend a lot of time on each problem, I just want to make sure it's for a worthwhile cause.