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MBB 222 > Midterm 3 > Flashcards

Midterm 3 Flashcards

1
Q

.

Five functional classes of proteins

A

1)Metabolic Enzymes
2)Structural proteins
3)Transport proteins
4)Cell signalling
5)Genomics caretakers

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2
Q

Heme

A

Fe 2+ poryphyrin complex in myoglobin and hemoglobin located inside the molecule

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3
Q

Hemoglobin

A

A protein that transports o2 from the lungs to the tissues composed of four polypeptide units (a1,a2,b1,b2) with one heme per subunit (Tetramer)

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4
Q

Myoglobin

A

A protein that is concentrated in tissue cells to store O2 comprised of a single peptide chain and heme group (Monomer)

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5
Q

Globin fold

A

A special type of 8 alpha helix fold found in myoglobin and hemoglobin that prevents the oxidization of Fe2+

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6
Q

Draw the shape of a Heme

A
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7
Q

Proximal Histidine

A

A Histidine residue below the Fe2+ plane that links the Heme group to the main protien via a covalent bond with the iron Also Called F8

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8
Q

Distal Histidine

A

A histidine above the Fe2+ that when the myoglobin or hemoglobin are oxygenated helps to stablazie the Fe2+ bond via a hydrogen bond with the O2 Called E7

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9
Q

Naming convention of the Globin fold

A

Helices are labelled A-H with numbers assigned afterwards that denotate the position of the bound molecule

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10
Q

Puckered position

A

Fe2+ is below the plane of nitrogen’s because it’s fat ass cannot keep itself straight and puckers down towards the proximal hisitidine

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11
Q

Planar position

A

Fe2+ is in the plane of nitrogen when an oxygen has become attatched to it the electrons shrink just enough, pulling out out of its depressed puckered state

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12
Q

Dissociation Constant (Kd)

A

Inverse of the association constant where higher Kd indicates more dissociated species or less affinity between species

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13
Q

Fractional saturation

A

Number of occupied binding sites/ number binding sites or [L]/[L]+Kd

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14
Q

Which line represents a higher Kd and how do you know?

A

Line B has the higher Kd, with higher Kd fractional saturation decreases depressing the curve

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15
Q

What is Kd equal to when Theta is 0.5?

A

Kd will equal the number of Ligands present, you can use this to find Kd of any graph by determining what concentraion of ligands gives you a theta of 0.5

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16
Q

Shape of Myoglobin binding curve vs Hemoglobin binding curve

A

Myoglobin is parabolic, Hemoglobin is Sigmoidal

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17
Q

Cooperative binding

A

The binding of one molecule to a macromolecule lowers the energy of subsequent molecules to bind to the ame marcomolecule

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18
Q

Two main ligands of Hemoglobin

A

O2 and 2-3 Biphosphoglycerate

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19
Q

Effector protiens

A

Control the equilibrium of the ligands for a given molecule

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20
Q

T-state

A

The deoxygenated stated of hemoglobin (deoxyhemoglobin), triggered by Low Ph and Low O2 concentrations

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21
Q

R-State

A

The oxygenated state of hemoglobin (Oxyhemoglobin) when it is bound to 4 oxygen molecules

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22
Q

Allosteric effector

A

Molecules that inhibit or activate a conformation

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23
Q

Positive effector

A

Promote the binding of a molecule

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24
Q

Negative Effector

A

Reduce the ability of a molecule to bind

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25
Q

Homotropic effector

A

An effector that binds to the same spot as the molecule of intrest ex.O2 acting as an effector for more O2 to bind

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26
Q

Heterotropic effector

A

An effector that doesn’t bind to the site of interest but to a secondary site which induces a shift in the conformation of the first site

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27
Q

The four effectors in Hemoglobin

A

H+(Neg,Het),Co2(Neg, Het),O2(Pos,Homo), and 2-3 BPG(Neg, Het)

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28
Q

What is the name of this strucutre and its function in myoglobin?

A

protoporphyrin IX, Holds the Fe2+ of Heme down

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29
Q

Difference from the R and T state of hemoglobin

A

A1B1 are rotatated 15 degrees of axis with A2B2 leading to a bring the sub units together

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30
Q

Describe how H+ acts as an effector for Hemoglobin

A

It causes the blood to become more acidic which decreases the affinity for O2 on hemoglobin because several residues on Hemoglobin are able to be protonated and deprotonanted

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31
Q

Describe how 2-3 BPG acts as an effector for hemoglobin

A

Binds to the center of the RBC causing conformational changes in the subunits that block O2 from binding to it. Can only occur when the RBC is in the T state as the R states central cavity is too small

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32
Q

Describe how Co2 acts as an effector for hemoglobin

A

It binds to Hemoglobin at the N-terminus stabilizing the T state and also drops off protons into the blood decreasing ph while doing so

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33
Q

Difference between fetal and adult hemoglobin

A

Fetal Hemoglobin has y subunits instead of b subunits that substitute Hist 143 with Serine which lowers the affinity for 2-3 BPG for Fetal hemoglobin. It essentially can steal dropped off oxygen from the mothers cells!

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34
Q

Importance of B1 His 146

A

Is one of the salt bridges that stabalize the T or R state depending on it’s protonation state, when protonated Favours T, when deprotonated favours R

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35
Q

Bohr effect

A

The decrease in O2 affinity of hemoglobin in the tissues due to high concentrations of H+
and CO2, and the increase in O2 affinity in the the lungs due to low concentrations of H+
and CO2,

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36
Q

What causes sickle cell anemia

A

Glu6 (negative) is swapped for valine(neutral) so the different sub units stick together forming sickle shapes, normally since they are hydrophilic they will not bind together but when becoming hydrophobic aggregate to get away from water!

37
Q

Job of Alkoxide in proteins

A

Perform Neuclophillic attacks

38
Q

Job of Chymotrypsin

A

Cleaves the peptide bond between COO- and NH2 via the addition of protons to both molecules

39
Q

How does Chymotrypsin bind to a substrate?

A

By having a hydrophobic tip it can insert itself into hydrophobic areas of the molecule its attacking

40
Q

Catalytic Trio

A

Histidine 157, Serine 195, Asparagine 102

41
Q

Summary of the Catalytic trios interaction

A

1)Histidine takes proton from Serine
2)Histidine is then stablaized by Asparagine
3)Serine is now a O- capable of covalent bonds and hydrolyzing Peptide bonds

42
Q

Peptide bond breaking reaction

A

E+S->E-P2 + P1 -> E +P2

43
Q

Anion Hole

A

Special pocket created by Glycine and Serine that helps to stabilize any negative charges

44
Q

Steps To get from E + S -> E-P2 +P1

A

1)Polypetide substrate binds to enzymes active site by slotting itself into the enzyme pocket
2)Histidine57 steals the proton from Serine 195 causing it to covalently bond to the carbonyl group, this will cause the double bond oxygen to convert into a single bond which is stabalized by the anion hole
3)His57 then hands its stolen proton to the Amino end of the petide bond, causing the bond between it and the carboxyl group to break allowing it to leave the enxyme

45
Q

Which amino acids will be targeted by Chrymotrypsin?

A

Tryptophan,Phenolaline, Tyrosine, Methionine, any other amino acids that are aromatic of have large bulky side chains

46
Q

Steps to get from EP2->E + P2

A

1)Water floods the enzyme, and His57 steals another proton from it, this OH- now attacks the carbonyl group of the intermediate
2) Now with a OH tacked onto the carbonyl group, His57 gives the stolen water proton to serine, causing it to break it’s bond with the carbonyl group
3)Now the N terminus gets kicked out of the enzyme and the catalytic triad has reset to its original state

47
Q

scissile bond

A

Bond between Carboxyl and amino group that is cleaved

48
Q

Draw the shape of the Peptide bond cleavage reaction

A
49
Q

Different types of Substrate Specific pockets

A

Chrymotrypsin-Large binding pocket good for aromatics,Trypsin-Negative binding pocket good for positive charges,Elastase-Small binding pocket only small side chains can enter

50
Q

Which carbon do the following attach: Base, Phosphate, Nucleotide

A

1’=Base
3’=Next nucleotide
5’=Phosphate group

51
Q

Name this structure

A

Deoxyribose

52
Q

Name this structure

A

Ribose

53
Q

Name this structure and point out which number is the Pentose binding site

A

Pyrimidine, 1

54
Q

Name this structure and point out which number is the Pentose binding site

A

Purine, 9

55
Q

B-Glycoside bond

A

a covalent bond that is formed above the plane of the molecule between the sugar and base in DNA

56
Q

Oligonucleotide

A

short, single- or double-stranded DNA or RNA molecules

57
Q

Phosphodiester linkage

A

ester bonds that form between sugar and phosphate to form the backbone of nucleic acids

58
Q

How are DNA strands read?

A

5’->3’

59
Q

Which base pairs are these?

A

Guanine and Cytosine

60
Q

Which Base pairs are these?

A

Adenine and Thymine

61
Q

2 mechanisms that form the DNA double helix

A

Base stacking: Bases align flat atop one another to minimize hydrophobic areas interacting with water and using van der waals induced dipoles to do so
H bonding: H bonding is done between the nitrogen and oxygen in the bases to stabilize the structure

62
Q

Major groove

A

More exposed side of the DNA double helix, when reading the left strand its oxygen points up when reading the right strand its oxygen point down

63
Q

Minor Groove

A

Les exposed side of the DNA double helix, when its on the left strand its oxygen points down and on the right strand its oxygen points up

64
Q

Right handed vs left handed helix

A

Pretend the back bone is the direction of the screw turns, based on which way you turn the screw Counterclockwise(left) or Clockwise (right) you will know the direction of your helix

65
Q

Which peak single and double strand DNA

A

Blue=Single strand, more absorbtion
Pink=Double helix

66
Q

Hyper chromatic shift

A

An increase in the absorbtion of ultraviolet light by a solution of DNA as these molecules are subjected to heat, alkaline conditions, etc. The shift is caused by the disruption of the hydrogen bonds of each DNA duplex to yield single-stranded structures.

67
Q

Tm

A

Temp at which half the DNA is double stranded and half the DNA is single stranded

68
Q

Salt’s effect on Tm

A

Salt ions shield the
negatively-charged phosphates on the DNA backbones, which can repel each other
when unshielded, thereby stabilizing the ds structure increasing Tm

69
Q

Sequence Length on Tm

A

Greater number of H bonds which need to be broken increasing Tm for DNA

70
Q

Importance of the Major and Minor groove of DNA

A

allow proteins to bind to and recognize DNA sequences from the outside of the helix

71
Q

Tomoisomers

A

DNA molecules with the same base pairings but are arranged in different coil configurations

72
Q

Supercoiling

A

When DNA overlaps with its own helices

73
Q

Linking Number

A

Number of times one strand needs to be passed through the other for the two strands to seperate

74
Q

Lk=

A

Lk=Tw+Wr where Tw is Twist and Wr is writhe

75
Q

Twist

A

The number of helical turns of one strand of DNA about the other, LK=Tw when flat. Twist is Positive when right handed but negative when left handed

76
Q

Writhe

A

The stress of a DNA molecule in which the DNA double helix will cross itself in 3D, Right handed is negative, Left handed is poisitive

77
Q

Lk0

A

Lk0=Number of base pairs/10.5

78
Q

ΔLk

A

ΔLk=Lk-Lk0, where a positive ΔLk is positively supercoiled and a negative ΔLk is a negatively supercoiled

79
Q

Over winding

A

Left handed, positive ΔLk and difficult to seperate

80
Q

Underwinding

A

Right handed, negative ΔLk and easy to seperate

81
Q

Topoisomerase 1

A

Enzyme which cuts the DNA at one point and unwinds the coil, endothermically favourable and changes linking number by 1

82
Q

Topoisomerase 2

A

Enzyme which will cut both strands and pass them through one another then reseal the strands, this can be used to either relax the DNA or for the cost of ATP coil the DNA

83
Q

How does DNA polymerase ensure correct base pairs are connected together?

A

For base pairs to join together in the active site they must be the correct paring, incorrect pairings will have steric class against the polymerase active site meaning that no bond can form.

84
Q

How does DNA polymerase know where to begin the origin of replication?

A

In Prokaryotes AT rich regions signify where the ori will begin

85
Q

dNMP

A

deoxyribonucleoside monophosphate, formed after dNTP is broken from its two phosphates

86
Q

Topoisomerase role in DNA replication

A

preventing DNA from getting tangled, as well as relieving pressure in supercoiled DNA during DNA replication.

87
Q

BRLO

A

B-form, right-hand helix, left-hand writhe coiling, overwinding

88
Q

BRRU

A

B-form, right-hand helix, right-hand writhe coiling, underwinding

89
Q
A

MBB 222 (2 decks)

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