Gene Transcription Objectives
1. List the steps in prokaryotic vs. eukaryotic mRNA synthesis.
a. Prokaryotic mRNA synthesis:
i. Initiation (due to environmental stimuli)
ii. Termination
iii. Regulation
b. Eukaryotic mRNA synthesis:
i. Initiation in response to hormonal signals (chromatin effects, cis elements,
transcription factors)
ii. RNA processing
iii. Regulation
2. Distinguish between ‘sense’ and ‘antisense’ strands of DNA and identify the template DNA
strand in RNA synthesis.
a. The sense strand is 5’ – 3’
b. The antisense strand is 3’ – 5’ and is the template strand of DNA
c. mRNA is made complementary to the antisense aka template strand
d. The mRNA is a copy of the 5’ – 3’ DNA strand except the T’s have been replaced with
“U”
3. List the requirements for RNA polymerase activity in terms of substrates utilized, template, and
metal ions.
a. RNA polymerase catalyzes the initiation and elongation of RNA chains in the 5’ – 3’
direction
b. In prokaryotes, all three types of RNA are made by the same RNA polymerase while
eukaryotes use different RNA polymerases for each type of RNA
c. Requirements for RNA polymerase activity include: NTPs, Mg++, template DNA strand
(3’ to 5’) aka antisense, and RNA polymerase
d. ATP, GTP, CTP, and UTP have precursors: AMP, GMP, CMP, UMP polymers + PPi
i. Synthesis is driven by the hydrolysis of PPi (Pyrophosphate)
e. 5’ phosphate of the incoming NTP gets attached to the 3’OH of the previous base and
the DNA duplex is fully conserved
4. Identify the direction in which RNA is synthesized.
a. RNA is synthesized in the 5’ – 3’ direction
b. The template is read in the 3’ – 5’ direction
i. The DNA is totally conserved wand the transcription bubble moves down the
gene as the RNA is made
5. Compare the error rates of DNA and RNA synthesis.
a. RNA polymerases have fewer error correction mechanisms and so are more error prone
b. RNA polymerase makes a mistake every 10^4 – 10^5 nucleotides
c. E. Coli DNA polymerase III makes a mistake every 10^9 nucleotides
6. Define the following and identify their roles in RNA synthesis: promoter, terminator, consensus
sequence, TATA box, Pribnow box, transcription factor, sigma factor, rho factor.
a. Promoter: DNA sequences that are recognized as initiation sites for RNA synthesis
b. Terminator: RNA sequences that signal the RNA polymerase to stop synthesizing RNA
c. Consensus sequence: A given recognition sequence (DNA or RNA) or one like it that is
found in association with all of the genes influenced by a particular protein
i. These sequences aren’t totally conserved but are a consensus from which many
similar sequences deviate only marginally
d. TATA box (EUKARYOTES): Involved in determination of transcription (RNA synthesis
from DNA template)
i. Called the PRIBNOW BOX in PROKARYOTES
e. Transcription factor: A protein that binds to transcription factor binding sites on the
DNA template that promote transcription by RNA polymerase. These are generally
found upstream (early or to the left of the TATA box location of the RNA pol binding) on
the genome such that RNA pol binds afterwards and can start transcribing downstream
(+ direction)
f. Sigma factor: A part of the RNA polymerase (Holoenzyme = core subunits + sigma factor)
that allows specific promoters to be recognized and gives specificity to the core enzyme.
After RNA synthesis begins, the sigma factor falls off but the core enzyme continues to
transcribe
g.
Rho factor: A small hexamer protein that binds to and slides down RNA in an
ATP-dependent fashion to aid in the displacement of the new strand from the
template. It functions with hairpin structures
7. Identify the roles played by the following elements in prokaryotic gene regulation: operator,
promoter, inducer, repressor proteins, polycistronic mRNA.
a. Operator: This is where repressor proteins bind to block RNA polymerase
b. Promoter: Where the RNA polymerase binds to initiate mRNA synthesis
c. Inducer: Binds to repressor proteins and prevents repression and allows RNA
polymerase to make mRNA
d. Repressor proteins: Form a complex binding to the operator that prevents RNA
polymerase from transcription
e. Polycistronic RNA: A mRNA that encodes more than one protein in a sequence. It is
more common in prokaryotes than eukaryotes and eukaryotes require greater control
of gene expression.
8. Identify the mechanisms by which rifampicin, alpha-amanitin and actinimycin D inhibit gene
transcription.
a. Rifampicin: Binds to the beta subunit of prokaryotic DNA polymerase; It is not toxic to
humans because human polymerases are different than prokaryotic ones. Rifampicin is
used to treat tuberculosis
b. Alpha-Amanitin: It is synthesized by the poisonous mushroom, Amanita Phalloides and it
inhibits both RNA polymerase II and III
c. Actinomycin D: It inhibits prokaryotic and eukaryotic transcription by intercalating in
DNA and by blocking RNA polymerase progression
9. List the types of eukaryotic RNA polymerases and the type of RNA that each synthesize.
a. RNA polymerase I: 45S rRNA precursor that is located in the nucleolus
b. RNA polymerase II: hnRNA (heterogenous nuclear), mRNA synthesis; located in
nucleoplasm
c. RNA polymerase III: tRNA and some snRNA synthesis; located in nucleoplasm
10.
11.
12.
13.
14.
d. mtRNA polymerase: all types of RNA in the mitochondria
Describe the manner in which DNA is compacted in the chromatin structure and the role of
histones.
a. DNA is wrapped around histones (2x H2A, H2B, H3, and H4) that form nucleosomes
which can compact DNA by up to 7x the amount. Linker histones (H1) link the
nucleosomes together and scaffolding proteins bind to nucleosomes and form a twisted
helical structure that further compacts the DNA
Identify the characteristics of actively transcribed chromatin.
a. Chromatin structure surrounding the actively transcribed genes allows greater
accessibility to regulatory proteins.
b. The chromatin structure is less condensed and the H1 histone is frequently missing or
depleted (it falls off when phosphorylation).
c. Topoisomerases are found in association with active genes to relieve torsional stress of
transcription.
d. Core histones are generally more extensively modified, frequently by acetlyations which
add a negative charge to the histone, making them repulsive to the negatively charged
DNA. Actively transcribed genes tend to be hypomethylated.
Define the roles of acetylation and methylation of histones during gene regulation.
a. C5 of cytosine can be methylated by specific methylases and methylations occur
primarily at GC dinucleotides in eukaryotes
b. Methylation is responsible for GENOMIC IMPRINTING and actively transcribed
mammalian genes contain fewer methylated CG’s than inactive genes (so
hypomethlyated)
c. Acetylation induces a conformational change in the core histones
i. Example: Lysine rich tails bind tightly to DNA and repress transcription by
blocking access of factors to the DNA template
1. Acetylation of the lysine residues alters chromatin structure and allows
the binding of transcription factors
Identify the location on the base and the base that is primarily involved in DNA methylation in
eukaryotes.
a. The carbon-5 of cytosine can be methylated by specific methylases. Methylation occurs
mainly in CG dinucleotides as mentioned above and in eukaryotes, they are called the
CPG island hotspots for methylation. Methylation blocks transcription factors and
renders the DNA transcriptionally inactive.
Identify the roles of the TATA box, promoter cis elements, trans-acting factors, coregulators,
enhancers and silencers, general transcription factors and the preinitiation complex in
eukaryotic gene expression.
a. The TATA box (in eukaryotes) is a consensus sequence that is specifically recognized by a
binding protein
b. The TATA box is located -25 downstream in eukaryotes and the Pribnow box is located 10 downstream in prokaryotes
c. Transcription factors are protein factors that activate or repress transcription (called
activators or repressors)
i. The factor that binds to DNA is considered to be the TRANS-ACTING factor
d. These transcription factors may bind to specific DNA sequences
i. The specific DNA sequence is considered to be CIS element
e. These transcription factors also may act as adaptor molecules to mediate
communications between other factors and RNA polymerase (called co-activators or corepressors)
f. General transcription factors:
i. TFII (transcription factors that guide RNA polymerase II)
1. TFIIA, TFIIB, TFIID, TFIIE, TFIIF, TFIIH
2. Each is a complex of proteins
ii. The TATA binding protein (TBP) is a 30 kD component of the larger TFIID
complex
iii. The TBP binds to the tata box (-25 in euks) and the DNA is unwound and bent
g. Enhancers and silencers are DNA sequences (cis elements) that are not part of the basal
promoter
i. They can function close to a promoter, or at a distance from the promoter, in
either orientation, or even within an intron
ii. They bind transcription factors (activators or repressors) that are frequently
cell-type specific
1. Activators bind to enhancers and repressors bind to silencers
15. Describe the regulation of steroid receptors and the mechanisms of action of estrogen and
tamoxifen.
a. Steroid hormone receptors are transcription factors activated by binding of steroid
hormones
i. The estrogen receptor (ER) exists in an inactive form in the cytoplasm
ii. After binding estrogen, the ER enters the nucleus and binds to specific DNA
regions to activate transcription
iii. Most types of breast cancer rely upon estrogen receptors and estrogen for
growth
iv. TAMOXIFEN inhibits the effects of estrogen and the estrogen receptor and is a
primary treatment for breast cancer
1. The tamoxifen/receptor complex recruits corepressors
v. Enhancers mediate the action of steroid hormones. The hormone/receptor
complex binds to the DNA allowing for transcription to begin and the binding of
the complex allows other transcription factors to bind (HRE = hormone response
element)
16. Define homeotic genes and homeobox proteins and identify their roles in development.
a. Each segment of an insects body will develop into a specific body part
b. HOMEOTIC genes control the functional architecture plan of the embryos
i. The sequence of homeo box genes in both flies and mammals corresponds to
their order of action along the anterior-posterior axis of the embryo
c. The products of homeotic genes are called homeobox proteins
i. These proteins contain a conserved 61 amino acid sequence that is very basic
and can bind to DNA
ii. The primary function of these proteins is to regulate transcription
iii. The individual homeotic genes are conserved from flies to humans
iv. The arrangements of homeotic genes on individual chromosomes are also
conserved
v. Examples include the Drosophila bithorax mutation from a T3 Haltere into a T2
d. Waardenburg Syndrome
i. Pigmentation disturbances (Frontal white patch of the hair)
ii. Cochlear deafness and heterochromia iridis
iii. Wide bridge of the nose due to lateral displacement of the inner canthus of
each eye
iv. Mutations in the human homeobox gene PAX3 (formerly HuP2) produce the
phenotypes of this syndrome
v. Similar mutations in the mouse homolog, PAX3 can produce similar effects in
mice
17. Identify the main features of eukaryotic mRNA, including the 5’ and 3’ ends.
a. Eukaryotic mRNA has a 5’ 7-methylguanylate cap that is attached by a triphosphate
linkage. Prokaryotic RNA’s are NOT CAPPED.
i. The 5’ cap helps to mediate efficient translation and a cap binding protein is
part of the translation initiation complex. It helps to stabilize the mRNA and may
be involved in subsequent mRNA splicing or export of mRNA to the cytoplasm
ii. Methyl groups can be found at the 2’ positions of riboses in both the first and
second nucleotides
b. The 3’ end of the mRNA is cleaved by a specific endonuclease and then a polyA
polymerase adds a Poly A tail through the use of ATP hydrolysis (ATP -> PPi)
c. Exons are the regions of a gene that are represented by the mRNA and may contain
coding and noncoding RNA sequences
d. Introns are the regions of a gene that are missing from the mRNA and almost all human
genes are split (meaning they have introns) and introns may be much larger than exons
e. Not all mRNA has a polyA tail
i. mRNA for histones don’t have a polyA tail because the histone proteins are
relatively stable and do not require continuous synthesis from mRNA
f. hnRNA has introns which are non-protein coding and must be spliced out from the
protein-coding exon regions to form the mature mRNA that is ready for translation
g. Not all genes have introns although many euk. genes do
18. List the steps involved in the maturation of an initial pre-RNA into an mRNA.
a. The initial transcript of a gene produced in the nucleus is called the primary transcript or
pre-mRNA that is processed to yield mRNA
b. The 5’ end is capped and this enhances the binding of the mRNA to ribosomes
c. The 3’ end is usually modified by the addition of A-residues to form the “poly-A” tail
d. Portions of the hnRNA are removed in a complex process called splicing
e. The intervening sequences that are removed are called introns and the portions of the
sequence that are retained in the mature mRNA are called exons
19. Describe how defective mRNA splicing results in thalassemia.
a.
Humans have a lot more genes than flies or yeast. More than ½ of human genes
produce alternatively spliced mRNA’s and thus have a greatly increased the number of
potential proteins encoded by the genome
b. Thalassemia syndromes:
i. A group of hereditary anemias characterized by absence or near absence of
alpha or beta chains
ii. Alpha and beta-thalassemias involve errors in the synthesis of the alpha and
beta chains of hemoglobin
iii. Mutations that result in thalassemia include:
1. Nonsense mutations due to single nucleotide mutations or frameshifts
(insertions or deletions)
2. mRNA processing defects due to intron mutations when the mRNA is
unspliced or improperly spliced and is degraded in the nucleus that
eventually leads to a truncated protein being made
20. Describe how differential splicing can result in different mRNAs from the same original
transcript.
a. Some genes may overlap on the mRNA transcript and alternate splicing can result in
different mature mRNA’s that can code for different protein products. Thus, a single
mRNA transcript can produce several different variant protein products
RNA Transcription 1 & 2, Protein Translation
Post Translational Modification Dr. Robert Trumbly
RNA Translation Objectives
1. Identify the reactions catalyzed by amino-acyl-tRNA synthetases.
a. Amino acids are the substrates for protein synthesis
b. The starting point in protein synthesis involves an activation of amino acid followed by
attachment of the amino acid to a tRNA
i. Aminoacyl tRNA synthetases catalyze these processes
ii. There is at least one aminoacyl tRNA synthetase for each amino acid and each
one recognizes its own amino acid and appropriate tRNAs
c. Amino acid + ATP aminoacyl tRNA synthetase aminoacyl adenylate + PPi
d. Aminoacyl Adenylate + tRNA aminoacyl tRNA synthetase aminoacyl tRNA (charged
tRNA)
2. Identify the roles played by the following factors in protein synthesis: initiation factors, the Cap
binding complex, elongation factors and release factors.
a. Initiation factors recruit the cap binding protein complex that then binds to the 5’ head
of the mRNA. Additional initiation factors with helicase activity then bind and denature
mRNA secondary structures to iron out the mRNA for translation.
b. After translation is initiated, elongation factors recruit specific amino acid-tRNA’s to
bind in the A site of the ribosomes. Elongation factors then help mediate the
translocation of the amino acid on the P-site amino acid-tRNA to the A-site with the
hydrolysis of GTP, moving forward along the mRNA.
c. When the chain reaches a stop codon, release factors come and peptidyl transferase on
the ribosome accepts H2O instead of another amino acid-tRNA. This results in the
hydrolysis of the tRNA-peptide bond, releaseing the tRNA, peptide chain, and the
release factor. The ribosome then dissociates from the mRNA.
3. From a diagram of tRNA, identify the positions where the amino acid is linked to it, where the
anticodon is located, and what three nucleotide residues are at the 3’-end of each tRNA.
a. The amino acid is linked to the 3’terminus at the CCA sequence. Each tRNA recognizes a
specific tRNA and the anticodon is located at the loop at the bottom of the “T”
4. Describe the architecture of ribosomes and list the steps required for the initiation, elongation
and termination of protein biosynthesis.
a. Ribosomes have two subunits: A 40S and 60S in eukaryotes (30S and 50S in
prokaryotes). The two subunits are linked together such that they form interconnected
tunnels for the mRNA and translated protein. The 40S subunits contain the tRNA binding
sites. The 60S subunit contain the peptidyl transferase that catalyzes the peptitde bond
formation as well as a GTPase that powers the movement of mRNA during translation.
b. Initiation
i. Aminoacyl-tRNA transferase attaches the amino acids to tRNA using a two-step,
ATP dependent mechanism
ii. In prokaryotes, the mRNA positions itself on the 30S subunit and the sequence
in the 5’ leader base pairs with the 3’ end of the 16S rRNA on the ribosome
iii. In prokaryotes, the sequence on the mRNA is 7-10 nucleotides upstream of the
AUG or GUG start codon and is called the Shine Dalgarno Sequence
iv. In eukaryotes, initiation factors recruit the cap binding protein to attach to the
5’ end of the RNA
v. This allows binding of the other initiation factors with helicase activity to
denature secondary structures in mRNA
vi. After this is complete, the initiation of tRNA binds to the 40S subunit to form the
40S preinitiation complex. The 40S preinitiation complex slides down the RNA
until the first initiation AUG codon is detected
vii. If the AUG codon is flanked by an optimal arrangement of nucleotides (called a
KOZAK consensus), translation will be initiated; otherwise, the pre-initiation
complex continues to move to the next AUG
viii. If the correct Kozak sequence is found, the 60S subunit comes and binds to the
small subunit and mRNA, with the initiation tRNA in the P-site
c. Elongation
i. Elongation begins with elongation factors that recruit the next appropriate tRNA
with an anticodon that matches the next codon
ii. The next tRNA binds in the A-site and a new peptide bond is formed via peptidyl
transferase with the ester bond of the first amino acid-tRNA breaking to attach
to the amino group of the second amino acid
iii. Elongation factors mediate the translocation of the A-site tRNA to the P-site,
kicking out the first tRNA and moving forward on the mRNA
iv. This process continues until a stop codon is reached
d. Termination
5.
6.
7.
8.
9.
10.
i. When a stop codon is recognized in the A-site, termination factors force
peptidyl transferase to accept an H2O rather than another tRNA
ii. This hydrolyzes the peptide chain from the tRNA; both of which are released
iii. The ribosomes dissociate from the mRNA
Match a list of inhibitors of protein biosynthesis with their mechanism of action.
a. Streptomycin – Prokaryote – Formation of initiation complex
b. Tretracyclin – Prokaryote – Aminoacyl tRNA binding at A-site
c. Erythromycin – Prokaryote – Binds 50S, inhibits translocation
d. Fusidic Acid – Prok/Euk – Inhibits elongation, binds eEF2/GDP
e. Diphtheria Toxin – Euk – Inactivates eEF2 by polyriboslyation; NOT an antibiotic
f. Puromycin – Prok/Euk – Aminoacyl-tRNA analog that acts as a peptidyl acceptor
Identify what “wobble” means with respect to codon: anticodon recognition.
a. The wobble refers to the third codon base on the mRNA or first anticodon base on the
tRNA being promiscuous about its binding and allowing different pairings to be
accepted. This contributes to the degeneracy of the genetic code as a separate tRNA is
not required for every codon due to the wobble.
b. First anticodon: C  Third Base codon: G
c. First anticodon: A  Third Base codon: U
d. First anticodon: U  Third Base codon: A or G
e. First anticodon: G  Third Base codon: U or C
f. First anticodon: I  Third Base codon: U, C, or A
Identify where Shine-Dalgarno and Kozak sequences are found and what function they perform.
a. Shine-dalgarno sequences in prokaryotes and Kozak sequences in eukaryotes are
sequences on mRNA that function in signal where the ribosome should bind to start
translation. Prokaryotes use either AUG or GUG as start codons while eukaryotes use
AUG flanked by a kozak consensus sequence
Identify the mechanism of action of diphtheria toxin.
a. Diphtheria toxin binds to cell membranes and is cleaved by a protease to form an active
enzyme that is transported to the cytoplasm. This toxin catalyzes the ADP-riboslyation
of elongation factor 2 and effectively inactivates the elongation factor. A single molecule
of diphtheria toxin is sufficient to kill a cell because it is an enzyme.
Describe the general characteristics of the genetic code, including the number of nucleotides in
a codon, the degeneracy of the code, and the nature of initiation and termination codons.
a. The genetic code translates 3 nucleotide codons to the 20 amino acids. Because there
are more 3 nucleotide codes than amino acids, some amino acids can be represented by
multiple codes. This is the reason for the degeneracy of the code.
b. Initiation codons in prokaryotes are either AUG or GUG and code for Met or Val.
c. Initiation codons in eukaryotes are AUG and code for Met
d. Three codons code for termination of translation: TAA, TAG, and TGA
Interpret the effects of mutations in coding regions on translation of the resulting mRNA.
a. A nucleotide substitution leads to an amino acid substitution in the hemoglobin beta
chain
b. Sickle cell HbA has Glu in 6th position
c. Sickle cell HbS has Val replace Glu in 6th position and leads to a beta substitution in the
hemoglobin beta chain
d. Null mutation: completely eliminates the function of the gene
e. Nonsense mutation: any change in DNA that results in a termination codon replacing an
amino acid codon
f. Point mutation: change involving a single base pair
g. Frameshift mutation: deletion or insertion of bases that alters the reading frame
11. Match the following terms with their definitions:
Wobble: Wobble refers to the third base in the codon and the first base in the anticodon being more
flexible in base pairing
Degeneracy: this refers to amino acids being able to be coded by multiple codons
initiation codon: AUG or GUG for Met or Val in prokaryotes. AUG for Met in eukaryotes.
triplet code: All amino acids are coded as a 3 nucleotide sequence
anticodon: Three nucleotide sequence on the tRNA that binds complementary to codons on mRNA
codon: Three nucleotide sequence on the mRNA that binds complementary to anticodons on tRNA
open reading frame: Reading of every 3 non-overlapping nucleotides as coding for proteins starting from
the start codon to the stop codon
peptide site: P-site of the ribosome where the initiation tRNA binds and where tRNA holding the peptide
chain moves to after each peptide bond formation step
initiator tRNA: the first tRNA holding the methionine that binds to the 40S subunit to form the 40S preinitiation complex
aminoacyl site: A-site of the ribosome where subsequent tRNA’s bind to and to which the growing
peptide bond attaches to the next amino acid
deletion mutation: mutation resulting in the loss of nucleotides
insertion mutation: mutation resulting in the addition of nucleotides
nonsense mutation: any change in the DNA that results in a termination codon replacing an amino acid
codon
null mutation: a mutation that completely eliminates the function of a gene
point mutation: a mutation involving a change in a single base pair
peptidyl transferase: enzymatic activity found on the 60S subunit of a ribosome that catalyzes the
formation of peptide bonds
polysomes: mRNA + a lot of attached ribosomes
release factors: factors that force peptidyl transferase to accept an H2O when it hits a stop codon,
ending translation.
stop codon: sequence on the mRNA that signals the end of translation: TAA, TAG, TGA
Shine-Dalgarno: AUG or GUG sequence on prokaryotic mRNA that tells the ribosome where to bind
frameshift mutation: Mutation that alters the reading frame of mRNA.
cap binding protein: recruited by initiation factors to bind to the 5’ end of the mRNA. This protein allows
the subsequent binding of other initiation factors that denature secondary structures on the mRNA in
preparation for translation
termination codon: AGA, and AGG are termination codons in eukaryotes and UGA is Trp instead of
termination and AUA Is Met instead of Ile; UGA (Trp) for Prokaryotes.
aminoacyl-tRNA: enzyme that activates amino acids with ATP and then attaches them to the appropriate
tRNA
synthetase: catalyst that attaches an amino acid to a tRNA
Kozak sequence: AUG flanked by specific sequences that signals where to bind; Eurkaryotes.
signal peptide: Directs synthesis of the secreted proteins to the ER membrane. The pre-sequence is
usually the signal peptide leader sequence (insulin and transforming growth factor beta – TGF-beta)
Post Translational Modification Objectives
1. Identify the covalent modifications that allow for the attachment of proteins on the inner and
outer surfaces of the cell membrane
a. Covalent attachment of fatty acids such as palmitoyl CoA or myristyl CoA to proteins
adds a hydrophobic component that allows proteins to be associated with cell
membranes.
b. The GPI or glycosylphosphatidyl inositol anchors can also be attached to proteins. Unlike
just hydrophobic domains on the protein, GPI anchors recognize specific targets in
membrane domains (lipid rafts) that are important for cellular signaling
2. Identify the sites of linkage of N-linked and O-linked oligosaccharides to proteins.
a. N-linked oligosaccharides: single species of oligosaccharide is transferred to an amino
group of an ASPARAGINE amino acid in the growing peptide chain
b. O-linked oligosaccharides: oligosaccharides are added to the –OH of serine or threonine
residues
3. List the potential pairs of amino acids that are the targets of cleavage by proteolytic processing
enzymes.
a. Proteolytic processing enzymes typically cleave next to pairs of basic residues such as
Arg-Arg or Lys-Arg
4. Describe the manner in which insulin is generated by proteolytic processing.
a. Insulin is synthesized as preproinsulin. Proteolytic enzymes first cleave off the NH3+ end
and form proinsulin.
b. The proinsulin is then cleaved at 2 sites, forming insulin. The insulin has 2 protein
components that are linked by two disulfide bridges
5. Describe the role of ubiquitin in protein degradation.
a. Protein degradation by ubiquitin is an ATP-dependent mechanism.
b. The ubiquinating enzyme complex binds ubiquitin to the amino group on a LYSINE of the
targeted cytosolic protein.
c. After the first ubiquitin binds, subsequent ubiquitins bind to the ubiquitin and form an
ubiquitin-protein complex.
d. Each ubiquitin addition uses ATP.
e. The ubiquinated protein is then recognized by ubiquitin-dependent proteases that cut
up the protein to constituent amino acids and then recycle the previously attached
ubiquitin molecules.