VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:Part 3. Ribonuclease Mechanism e (6pts) Draw the curved arrows for each step of the hydrolysis of the RNA in the scheme below. e (2 pts) Draw the missing intermediate in this mechanism. 7 : ] o=P-00 VY 7 o o0=p-00 UY 0. SNP . INTUNH 07 _o. DA — H-N7"NH Oo OH =, B — he A is 12 5 o \={ 0=pP-00 NS A His 12 HO—P- © a SyH O70 od )=/® >, HNTSN ° 19H; 7 )=/ 0=P-00 119His’ © OH S 0=p-00 A ¢ R RB | i Q o=p-0© VY o ® |S oJ H-N7 NH - oP His 12 0" 0® Ape SN HNN on 119His” A HO _o O OH 0=p-00 Q R R ji 9 o=p-00 UY o_o or NZ NH 0 on \={ =P His 12 or 0G nd HSH H I 119His’
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
Part 3. Ribonuclease Mechanism
e (6pts) Draw the curved arrows for each step of the hydrolysis of the RNA in the
scheme below.
e (2 pts) Draw the missing intermediate in this mechanism.
7
: ]
o=P-00 VY 7
o o0=p-00 UY
0. SNP .
INTUNH 07 _o. DA
— H-N7"NH
Oo OH =, B —
he A is 12 5 o \={
0=pP-00 NS A His 12
HO—P- ©
a SyH O70 od
)=/® >, HNTSN °
19H; 7 )=/
0=P-00 119His’ © OH
S 0=p-00
A ¢
R
RB |
i
Q
o=p-0© VY
o ®
|S oJ H-N7 NH
- oP His 12
0" 0®
Ape
SN
HNN on
119His” A
HO _o
O OH
0=p-00
Q
R
R
ji
9
o=p-00 UY
o_o
or NZ NH
0 on \={
=P His 12
or 0G nd
HSH H I
119His’
This is AI generated Response.
Asked by: Firdos
Created at: 2025-12-07 15:32:33
Asked by: Firdos
Created at: 2025-12-07 15:32:33
Final Answer:
Rendered
Raw Text
Did you like this answer?
## Step 1: First Nucleophilic Attack and Intermediate Formation
**Explanation:**
- His12 acts as a general, abstracting the '-hydroxyl proton from theose, making the 2'-oxygen a nucleophile.
- The 2'-oxygen attacks the adjacent phosphorus atom, forming a pentavalent transition state and breaking the bond between phosphorus and the 5'-oxygen of the adjacent nucleotide.
- His119 acts as a general acid, donating a proton to the 5'-oxygen leaving group, resulting in cleavage of the phosphodiester bond and formation of a **2',3'-cyclic phosphate intermediate**.
**Intermediate Structure (to be drawn in the blank box):**
```
O U
/ \ |
R-O-P O---(ribose ring)
|| | |
O O |
/ \ |
O O
|
(cyclic phosphate linkage between 2' and 3')
```
- The 2',3'-cyclic phosphate remains attached to the ribose ring.
- Histidine 12 is now protonated and Histidine 119 is deprotonated.
---
## Step 2: Hydrolysis of the Cyclic Intermediate
**Explanation:**
- Water, activated by His119 (now a base), attacks the phosphorus atom of the cyclic phosphate.
- The ring opens, leading to the formation of a 3'-phosphate and a free 2'-hydroxyl group.
- His12 returns to its base form, and His119 is reprotonated.
**Final Products:**
- RNA fragment with a 3'-phosphate end.
- RNA fragment with a 5'-hydroxyl end.
- Both histidines return to their original states.
---
## Final Answer
**Step 1:** His12 and His119 facilitate nucleophilic attack by the 2'-OH, forming a 2',3'-cyclic phosphate intermediate.
**Step 2:** Water, activated by His119, attacks the cyclic intermediate to yield the final hydrolyzed products.
**The missing intermediate is the 2',3'-cyclic phosphate (with histidine residue charges appropriately shown).**
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :