question 1: name, pdb codes
DESCRIPTION
Question 1: Name, PDB codes. I chose*: 3onz HUMAN TETRAMERIC HEMOGLOBIN: PROXIMAL NITRITE LIGAND AT BETA Function: Oxygen transport Resolution: 2.09 Å R = 0.218 Rfree = 0.280 “ WORSE THAN AVERAGE at this resolution”. I was assigned: 1d66 DNA-binding domain of Gal4. - PowerPoint PPT PresentationTRANSCRIPT
Question 1: Name, PDB codes• Eric Martz• [email protected]• (no research lab)• Macromolecular visualization
I chose*:• 3onz• HUMAN TETRAMERIC HEMOGLOBIN: PROXIMAL NITRITE LIGAND AT BETA• Function: Oxygen transport• Resolution: 2.09 Å• R = 0.218• Rfree = 0.280 “WORSE THAN AVERAGE at this resolution”
I was assigned:• 1d66• DNA-binding domain of Gal4.• Transcriptional regulation.• Resolution: 2.70 Å• R = 0.230• Rfree not published
* If you chose a homology model, skip this and see the next slide.
Question 1 (continued): Homology Model(Do this slide only if you chose a homology model instead of a PDB code)
This is provided only as an example.I did not use this homology model in the slides below.
I chose a homology model:Sequence: MNGTEGPNFYVPFSNATGVVRSPFEYPQYYLAEPWQFSMLAAYMFLLIVLGFPINFLTLYVTVQHKKLRTPLNYILLNLAVADLFMVLGGFTSTLYTSLHGYFVFGPTGCNLEGFFATLGGEIALWSLVVLAIERYVVVCKPMSNFRFGENHAIMGVAFTWVMALACAAPPLAGWSRYIPEGLQCSCGIDYYTLKPEVNNESFVIYMFVVHFTIPMIIIFFCYGQLVFTVKEAAAQQQESATTQKAEKEVTRMVIIMVIAFLICWVPYASVAFYIFTHQGSNFGPIFMTIPAFFAKSAAIYNPVIYIMMNKQFRNCMLTTICCGKNPLGDDEASATVSKTETSQVAPA
Length of sequence: 348Source of sequence: UniProt P08100Name and function: Human rhodopsin. Detects light in the eye.Template: 1u19 (Bovine rhodopsin)Modeled residue range: 1-348Sequence identity: 93.4%
Question 1 (continued): Homology Model(Do this slide only if you chose a homology model instead of a PDB code)
Sequence Alignment from Swiss-ModelWhat is important here is the number and sizes of gaps. Because this is a long alignment, I deleted the secondary structure lines to make it more compact.
(Edit, Paste Special, Styled Text.) Continue on another slide if needed.
TARGET 1 MNGTEGPN FYVPFSNATG VVRSPFEYPQ YYLAEPWQFS MLAAYMFLLI1u19_1 1 mngtegpn fyvpfsnktg vvrspfeapq yylaepwqfs mlaaymflli TARGET 49 VLGFPINFLT LYVTVQHKKL RTPLNYILLN LAVADLFMVL GGFTSTLYTS1u19_1 49 mlgfpinflt lyvtvqhkkl rtplnyilln lavadlfmvf ggftttlyts TARGET 99 LHGYFVFGPT GCNLEGFFAT LGGEIALWSL VVLAIERYVV VCKPMSNFRF1u19_1 99 lhgyfvfgpt gcnlegffat lggeialwsl vvlaieryvv vckpmsnfrf TARGET 149 GENHAIMGVA FTWVMALACA APPLAGWSRY IPEGLQCSCG IDYYTLKPEV1u19_1 149 genhaimgva ftwvmalaca applvgwsry ipegmqcscg idyytpheet TARGET 199 NNESFVIYMF VVHFTIPMII IFFCYGQLVF TVKEAAAQQQ ESATTQKAEK1u19_1 199 nnesfviymf vvhfiipliv iffcygqlvf tvkeaaaqqq esattqkaek TARGET 249 EVTRMVIIMV IAFLICWVPY ASVAFYIFTH QGSNFGPIFM TIPAFFAKSA1u19_1 249 evtrmviimv iaflicwlpy agvafyifth qgsdfgpifm tipaffakts TARGET 299 AIYNPVIYIM MNKQFRNCML TTICCGKNPL GDDEASATVS KTETSQVAPA1u19_1 299 avynpviyim mnkqfrncmv ttlccgknpl gddeasttvs ktetsqvapa
ABOVE IS FOR THE FIRST CHAIN. BELOW IS THE SAME ALIGNMENT FOR THE SECOND CHAIN OF THE HOMODIMER.
TARGET 349 - MNGTEGP NFYVPFSNAT GVVRSPFEYP QYYLAEPWQF SMLAAYMFLL1u19_1 1 ---mngtegp nfyvpfsnkt gvvrspfeap qyylaepwqf smlaaymfll *
Question 2, 3onz: Number of chains
Protein: 2 chains (2 distinct)DNA: 0 chainsRNA: 0 chains
Chain A:HEMOGLOBIN SUBUNIT ALPHA141 residues (1 missing) of Protein.Source: Homo sapiens (Human).Other_details: Blood.
Chain B:HEMOGLOBIN SUBUNIT BETA146 residues (12 missing) of Protein.Source: Homo sapiens (Human).Other_details: Blood.
3orn:2 HEM: Protoporphyrin ix containing Fe2 NO2: Nitrite ion4 MBN: Toluene
1d66:4 CD : Cadmium ion
Question 3, 3onz: Ligands and non-standard residues
Chain A:Crystal: 146 amino acidsFull length: 147 amino acids
Chain B:Crystal: 141 amino acidsFull-length: 142 amino acids
Question 4, 3onz: Full length vs. crystallized sequence
This structure is all alpha helices (red) and “coil” (white). It has no beta strands.
80.7% alpha helices0% beta strands19.3% neither
Question 5, 3onz: Secondary structure
I found no patches of all positive or all negative charges.
For 3onz chain A:pI = 8.69.Charges at pH• 4.0: +20.7• 7.0: +4.3• 10.0: -7.0
Question 6: Charge distribution
Glu43 in chain B has a high temperature.
Question 7, 3onz: Local uncertainty
Question 8 - Hydrophobic cores
Yes. Each of the two chains in 3onz has a hydrophobic core (circled in red).
Question 9 - Water solubility
3onz appears to be soluble because there are polar residues everywhere on the surface.
3onz is hemoglobin which is known to be a soluble protein.
Question 10, 3onz - Disulfide bonds
(Methionine sulfurs are not shown in this snapshot.)
Question 11, 3onz - Missing residues
13 Missing Residues including 2-, 3+ charged amino acids!
All sidechains are complete.
Question 12, 3onz - Non-covalent interactions
• Top: hydrophobic van der Waals interaction between two carbon atoms.
• Bottom: histidine nitrogen interacting with negatively charged oxygen (carboxyl). Partial salt bridge since His will have partial positive charge at pH 7, or hydrogen bond since His N epsilon has a hydrogen to donate..
Question 13 - Biological unit
Asymmetric unit: 2 chains Biological unit: 4 chains
Question 14, 3onz - Evolutionary Conservationhttp://consurf.tau.ac.il/results/1368001022/output.php
Top: Lys 61 unexpected.
Bottom: Lys 66 expected because it forms salt bridges with the carboxyls on the heme ligand.
Question 17a - Cation-pi interaction
There are no significant cation-pi interactions in 3onz.1d66 contains 3 significant cation-pi interactions. One is Lys25 with Tyr40 in
chain B. See next slide for snapshot.Here is the report from CaPTURE:
(see previous slide for explanation)
Question 17b, 1d66 - Cation-pi interaction
Question 18 - Polyview-3D static imageThis image is for 1d66.
Deoxyguanosine 26 in chain E is red.
Question 19 - Polyview-3D animation for PowerpointThis animation is for 1d66.
Deoxyguanosine 26 in chain E is red.
Question 20a - Intrinsically disordered regions
3onz chain B is 141 amino acids. The full length sequence has 142. Below are results for the full length sequence.
FoldIndex predicts no intrinsically disordered regions, despite the presence of 3 segments of missing residues in the crystal model. Therefore I analyzed 2gry which is more interesting (next slide).
Question 20b - Intrinsically disordered regions
2GRY is an X-ray crystallographic structure with resolution 2.35 Å. The full length sequence is 679 amino acids. The crystal includes residues 126-526 (plus an N-terminal His tag). Thus, 1-125 and 527-679 (length 153) were removed before crystallization. FoldIndex predicts that the removed portions are intrinsically disordered.
Continued on next slide …
Question 20c - Intrinsically disordered regionsFoldIndex predicts that 5 segments of the crystallized sequence will be disordered.
Segments predicted to be disordered:
1. All but 7 of these 69 N terminal residues are missing in 2GRY.
2. The middle 4 of these 6 residues are missing in 2GRY.
3, 4, 5: None of these predictions overlap with the four additional segments of missing residues in 2GRY. Continued on next slide …
Question 20d - Intrinsically disordered regions
The two segments of amino acids missing in 2GRY are at the N terminus and near the N terminus. The flanking residues (marked with yellow halos) have mostly low temperatures, while other missing segments (not predicted by FoldIndex) have higher temperatures.
2GRY colored by temperature showing “empty baskets”, regions with missing residues. Yellow halos mark the alpha carbons flanking missing residues predicted to be disordered by FoldIndex.