HC12 protein synthesis, structure and purification

Question 1

coding strand of DNA

Answer 1

the strand whose base sequence corresponds to the base sequence of RNA transcript produced (expect that T is replaced by U).

Thus: RNA pol II binds the non-coding strand and transcribed it to synthesise mRNA.

Question 2

synthesis of RNA and DNA direction + why?

Answer 2

5’-3’. This is needed for the proofreading function of the polymerases and the energy needs of synthesis.

Question 3

translation peptide bond + energy?

Answer 3

formed between NH2 terminus of amino acid bound to tRNA and the COO terminus of nascent chain.

Energy for the formation of peptide bond is provided by the cleavage of the ester bond between tRNA + COO terminus of aa (aminoacyl-tRNA)

Question 4

inter and intra-molecular interactions in proteins determine…..

Answer 4

protein folding.

Question 5

secondary modifications determine……

Answer 5

if interactions between proteins are possible, so they determine the activity of the protein

Question 6

adding of urea in lab?

Answer 6

protein folding can be reversed. urea denatures proteins. this is reversible.

Question 7

disulfide bridges: oxidative/reductive environment

Answer 7

oxidative environment: electrons (and protons) will be removed from the -SH group of cysteine –> disulfide bridges will form (S-S).

Reductive environment: disulfide bridges will be reduced.

OILRIG: Oxidation is losing electrons, reduction is gaining electrons.

Question 8

incorrectly folded proteins

Answer 8

Hsp60 can assist.
misfolded protein is captured by hydrophobic interactions with exposed opening of Hsp60 –> initial binding helps to unfold misfolded protein –> ATP binds to GroEs cap –> release of protein in enclosed space of Hsp60 –> bidning of other ATPs –> cap lets go –> correctly folded protein is released again

Question 9

levels of ubiquitination, resulting in?

Answer 9

poly- :proteasomal degradation/ DNA repair
mono- :histone regulation
multi- :endocytosis

Question 10

degradation signal

Answer 10

recognised by F-box proteins –> protein is targeted to E2 and E3 –> place an ubiquitin molecule on protein –> E1 activates these ubiquitin molecules (costs ATP)

Question 11

how can antibiotics inhibit protein synthesis?

Answer 11

by binding to ribosomes (EPA sites).

note: ribosomes of eukaryotic mitochondria can be insensitive to inhibitiors (just like bacteria)

Question 12

When do reactions run spontaneously?

Answer 12

when gibson free energy (delta G) < 0. then the reaction is exergonic/exothermic and energy is released

Question 13

delta G >0

Answer 13

reaction is endergonic/endothermic, and energy is required for the reaction to occur in that reaction.

Question 14

nature of reactants: formula

Answer 14

delta G 0’ reaction = delta G 0’product - delta G 0’ substrate

Question 15

delta G 0’

Answer 15

free energy change of a reaction under standard conditions:
25oC
initial concentration of all reactions is 1.0M

Question 16

formula concentration of reactants

Answer 16

delta G reaction = delta G 0’reaction + RT ln ( [product]/[substrate])

Question 17

Catalytic triad:

Answer 17

aspartic acid (D):
Histidine (H)
Serine (S)

Aspartic acid side chain –> induces histidine to remove proton of a serine –> this activates serine (-) –> enables it to form a covalent bond with enzyme substrate –> hydrogen bond rearrangements.

Question 18

reacrtion between enzyme and a substrate is described by two parameters

Answer 18

1. binding/interaction between substrate and enzyme
2. conversion of subtrate to the product = the catalysis.

Question 19

Km + formula for km

Answer 19

substrate concentration at which the reaction rate is half the maximum rate (1/2 Vmax)

Km= (k2 + k-1)/k1

k1: E + S –> ES
k-1: ES –> E + S
k2: ES –> E + P = Kcat.

Question 20

Km is low:

Answer 20

at a low substrate concentration, the reaction is already at half maximum –> enzyme has a high affinity for the substrate.

Question 21

kcat

Answer 21

= k2. measures how efficiently the enzyme can convert the subtrate into the product.

Question 22

rate of reaction according to michaelis menten

Answer 22

v = vmax x ([S]/ (Km + [S]))

Question 23

competitive inhibitors + new michaelis menten equation

Answer 23

Km will increase, without changing the Vmax.

v = vmax x ([S]/ ([S] + Km x (1 + [I]/Ki))

Question 24

protein purification: methods

Answer 24

• ion exchange chromatography: proteins purfication based on charge
• gel-filtration chromatography: based on size, the smallest are the slowest.
• affinity chromatography: based on affinity. so beads covered in antibodies for instance.

Question 25

typical prufication scheme of protein

Answer 25

charge and size separation ion-exchange –> increase [salt ions] –> salt ions compete with proteins –> proteins will be eluted out of column.
proteins with low net charge eluted first.

Activity essay: pool fractions of proteins together –> apply gel filtration column –> pool fractions together aggain –> affinity chromatography –> confirm purified protein on SDS-PAGE

Question 26

SDS-PAGE

Answer 26

SDS: all proteins become negatively charged
Beta-mercaptoethanol: breaks down disulfide bridges –> denaturation of proteins

separation based only on size!

Question 27

non-denaturing gel electrophoresis

Answer 27

separates proteins based on charge. pH gradient through the gel, proteins will stop when pH =pKa (the isoelectric point of protein, they have no net charge at this pH)

Question 28

Mass spectrometry

Answer 28

purify protein of interest from gel –> incubate with trypsin –> trypsin cleaves C-terminally from lysine and arginine (+ charged) –> peptides are relased –> masses are measured by mass spectrometry –> peptide patterns.

Double MS: peptides are further fragmented –> aa sequence can be determined.