transcription

Report
Unit #3 Schedule:
• Previously:
– Sanger Sequencing
– Central Dogma Overview
– Mutation
– Transcription, RNA Processing, Translation
• Last Class:
– Central Dogma Sculpting
• Today:
– Regulation of Gene Expression + Trivia
– StudyNotes 9 Due
• Tutorial (Apr 5)
• Review (Apr 9)
• EXAM 3 (Apr 11)
• Homework 6 Due
Regulation of Gene Expression
There are at least
300 different kinds
of cells in the
human body.
Most of them have
identical DNA.
Regulation of Gene Expression
We examine this at a very general level.
Prokaryotes vs. Eukaryotes
Definition: Operon
• A region of DNA that codes for a series of
functionally related genes and is transcribed
from a single promoter into mRNA.
Negative Control and Positive Control
• Transcription can be regulated via negative
control or positive control.
• Negative control occurs when a regulatory
protein binds to DNA and shuts down
transcription.
• Positive control occurs when a regulatory protein
binds to DNA and triggers transcription.
© 2011 Pearson Education, Inc.
© 2011 Pearson Education, Inc.
Negative Control
 Negative control occurs when a regulatory
protein binds to DNA and shuts down
transcription.
Fig. 18-2
PROKARYOTIC REGULATION OF GENE EXPRESSION:
Campbell Fig. 18-3a
trp operon
Promoter
Promoter
Genes of operon
DNA
trpR
Regulatory
gene
mRNA
Protein
5
trpE
3
Operator
Start codon
mRNA 5
RNA
polymerase
Inactive
repressor
(a) Tryptophan absent, repressor inactive, operon on
E
trpD
trpC
trpB
trpA
B
A
Stop codon
D
C
Polypeptide subunits that make up
enzymes for tryptophan synthesis
Fig. 18-3b-2
DNA
No RNA made
mRNA
Protein
Active
repressor
Tryptophan
(corepressor)
(b) Tryptophan present, repressor active, operon off
© 2011 Pearson Education, Inc.
Fig. 18-4b
lac operon
DNA
lacZ
lacI
3
mRNA
5
lacA
RNA
polymerase
mRNA 5
-Galactosidase
Protein
Allolactose
(inducer)
lacY
Inactive
repressor
(b) Lactose present, repressor inactive, operon on
Permease
Transacetylase
© 2011 Pearson Education, Inc.
Negative Control
Transcription of Trp Operon in the absence of
Tryptophan.
Tryptophan activates the repressor.
Transcription of the Lac Operon in the
presence of Lactose.
Lactose deactivates the repressor.
Positive Control
 Positive control occurs when a regulatory
protein binds to DNA and triggers transcription.
Fig. 18-5
CAP: catabolite
activator protein
What is cAMP?
© 2011 Pearson Education, Inc.
mc1r Gene Sequence
5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACC
TCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGT
GCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAG
TCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACCGCAACCTG
CACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGG
TGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGG
GGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGAC
GTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTG
CCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGT
GACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATC
TTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCT
GCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGT
CCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAG
AAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCC
TTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCA
TCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTT
AAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACC
CCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGT
GCTGCTGTGCTCCTGGTGA 3’
mc1r Gene Sequence
5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACC
TCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGT
GCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAG
TCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACTGCAACCTG
CACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGG
TGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGG
GGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGAC
GTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTG
CCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGT
GACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATC
TTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCT
GCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGT
CCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAG
AAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCC
TTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCA
TCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTT
AAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACC
CCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGT
GCTGCTGTGCTCCTGGTGA 3’
Consequence of Mutation
• A single nucleotide mutation from a Cytosine to a
Thymine leads to…
• An amino acid change from an Arginine to a Cysteine
Amino Acid Sequence Dark Fur:
MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSIPDGLFLSLGLVSLVENVLVVIAI
TKNRNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICG
SMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIXGIWVASIFFSTLFITYYNHTAVL
ICLVTFFLAMLALMAXLYVHMLTRAYQHAQGIAQLQKRQGSTXQGFCLKGAXTLTIILGIFFL
CWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLC
SW
Amino Acid Sequence Light Fur:
MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSVPDGLFLSLGLVSLVENVLVVIAI
TKNCNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICG
SMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIVGIWVASIFFSTLFITYYNHTAVL
ICLVTFFLAMLALMAILYVHMLTRAYQHAQGIAQLQKRQGSTRQGFCLKGAATLTIILGIFFL
CWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLC
SW
Changing 1 amino acid:
• Arginine:
– Strongest +charge
– Very hydrophilic
• Cysteine:
– Not hydrophilic
– Forms disulfide bonds
Beach Mice
-
-
Missense substitution mutation of
one nucleotide CT
Changes one amino acid:
Arginine  Cysteine
Changes the function of the MC1R
protein
How is
eumelanin
produced?
• When the MC1R protein
is stimulated, cAMP is
produced
• Lots of cAMP within a
melanocyte cell will
facilitate the expression
of at least four genes:
c(tyr), Tyrp1, Tyrp2, p
How is
eumelanin
produced?
• When cAMP is plentiful,
c(tyr), Tyrp1, Tyrp2 and
p are all expressed and
their enzymes facilitate
the biosynthetic
pathway that leads to
eumelanin production.
How is
eumelanin
produced?
• When cAMP is scarce,
c(tyr), Tyrp1, Tyrp2 and p
are not as readily
expressed.
• If only small amounts of
cAMP are present, c(tyr)
may still be expressed
and its enzyme may
facilitate the
biosynthetic pathway
leading pheomelanin
production.
How is
eumelanin
produced?
• If c(tyr) is not
adequately expressed it
is possible that neither
biosynthetic pigment
production pathway may
occur. This would result
in no pigment
production.
The Melanocortin-1-Receptor
Effectively
stimulated by
hormone
Ineffectively
stimulated by
hormone
MC1RR67
MC1RC67
Results in lots of
cAMP production
Results in little
cAMP production
c(tyr), p
tyrp1, tyrp2
<<activated>>
c(tyr) <<activated>>
p, tyrp1, tyrp2
<not activated>
LOTS of EUMELANIN
produced
EUMELANIN not
produced
Fig. 18-5
CAP: catabolite
activator protein
EUKARYOTIC REGULATION OF GENE EXPRESSION
Activators and Enhancers of Transcription
Campbell 8e, Fig. 18.8
Fig. 17-8
1
Promoter
A eukaryotic promoter
includes a TATA box
Template
5
3
TATA box
Start point
2
Transcription
factors
3
5
Template
DNA strand
Several transcription factors must
bind to the DNA before RNA
polymerase II can do so.
5
3
3
5
3
Additional transcription factors bind to
the DNA along with RNA polymerase II,
forming the transcription initiation complex.
RNA polymerase II
Transcription factors
5
3
3
5
5
RNA transcript
Transcription initiation complex
Campbell 8e, Fig. 18.9
Controlling Gene Expression:
Enhancers and Activators

Provide a way to turn on specific
genes in specific cells

Different genes have different
enhancers

Different cells have different
activators
Campbell 8e, Fig. 18.10
Controlling Gene Expression:
Enhancers and Activators

Tissue- and cell-type specific gene
expression

Liver cells make albumin, but
not crystallin

Lens cells make crystallin, but
not albumin
Campbell 8e, Fig. 18.10
Coming Up:
Friday:
• Tutorial (3-5pm in C-3)
Tuesday:
• Interactive Review (WhiteBoards + Clickers)
Thursday:
• Midterm Exam 3
Review Part 1
Clicker Review
With respect to nucleotide bonds:
(A)A-T is stronger than C-G
(B) C-G is stronger than A-T
(C) A-T and C-G have approximately equal
strength
Which mode of information
transfer usually does not occur?
(A)DNA to DNA
(B) DNA to RNA
(C) DNA to protein
(D)RNA to protein
(E) All occur in a working cell
In replication of DNA, the helix is
opened and untwisted by
(A)DNA polymerase
(B) ligase
(C) helicase
(D) telomerase
(E) topoisomerase
_______________ joins DNA
fragments to the lagging strand.
(A) Telomere
(B) DNA Polymerase I
(C) Helicase
(D) DNA Polymerase III
(E) Ligase
In a nucleic acid, the phosphate group,
nitrogenous base and free hydroxyl group are
attached to the _______________ carbons of
ribose (respectively).
(A) 1', 3', 5'
(B) 5', 3', 1'
(C) 3', 5', 1'
(D) 5', 1', 3'
(E) 3', 1', 5'
DNA polymerase III is thought to add
nucleotides
(A) to the 5' end of the RNA primer
(B) to the 3' end of the RNA primer
(C) in the place of the primer RNA after it is
removed
(D) on single stranded templates without need
for an RNA primer
(E) in the 3' to 5' direction
Considering the structure of double
stranded DNA, what kinds of bonds hold
one complementary strand to the other?
(A)peptide
(B) covalent
(C) hydrogen
(D) phosphodiester
(E) ionic
The nitrogenous base adenine is found in
all members of which group?
(A)proteins, ATP, and DNA
(B) proteins, carbohydrates and ATP
(C) glucose, ATP and DNA
(D) ATP, RNA and DNA
(E) proteins, glycerol and hormones
Where and how are Okazaki fragments
synthesized?
(A)on the leading strand, in a 5’  3’ direction
(B) on the leading strand, in a 3’  5’ direction
(C) on the lagging strand, in a 5’  3’ direction
(D)on the lagging strand, in a 3’  5’ direction
Which of the following types of mutation,
resulting in an error in the mRNA just after the
AUG start of translation, is likely to have the
most serious effect on the polypeptide product?
(A) insertion of a codon
(B) deletion of two codons
(C) substitution of the third nucleotide in an ACC codon
(D) deletion of a nucleotide
(E) insertion of 9 nucleotides
Regarding beach mice:
The C  T substitution at position 199 of
the mc1r gene:
A. arose by a mutation in the beach mouse populations in
response to a need for protection from predation.
B. leads to the failure of melanocytes to make an MC1R
protein.
C. arose by a mutation then increased in frequency because
it was selectively advantageous in the beach mouse
populations.
D. had no effect on the beach mouse populations.
E. produced an alternate allele that was detrimental to mice
on the white sand beaches
Regarding beach mice:
What was the reason for the lighter coat colors of
the mice on the white sand beaches?
(A) Owls and other carnivores prey on beach mice that do
not carry the mutant allele.
(B) A substitution of cysteine for arginine at position 67
of the MC1R protein.
(C) A substitution of thymine for cystosine at position 199
of the mc1r gene nucleotide sequence.
(D) The failure of melanocytes to lay down melanin
pigment in the cortex of hairs of the lighter colored
beach mice.
(E) The poorer binding affinity for α-MSH and the lower
amount of cAMP produced by individuals with the
mutated MC1R protein.
The template strand of DNA at the beginning of a protein-coding
region has the sequence:
5'–TACTGGGATAGCC*TACAT–3'
The “*” indicates the position of a point mutation: a T originally
present at this location has been deleted.
This deletion will most likely result in _____.
(A) mRNA codons preceding the mutation being misread
(B) mRNA codons following the mutation being misread
(C) no change in the polypeptide coded by this gene
(D)the AUC triplet functioning as a chain terminator
Biologists use the terms transcription and
translation to describe the two steps in genetic
information flow from DNA to protein. Which of
the following is correct?
(A) Transcription is the synthesis of protein from mRNA by
ribosomes; translation is the synthesis of mRNA from DNA by
RNA polymerase.
(B) Transcription is the synthesis of mRNA from DNA by
ribosomes; translation is the synthesis of protein from mRNA
by RNA polymerase.
(C) Transcription is the synthesis of protein from mRNA by RNA
polymerase; translation is the synthesis of mRNA from DNA by
ribosomes.
(D) Transcription is the synthesis of mRNA from DNA by RNA
polymerase; translation is the synthesis of protein from mRNA
by ribosomes.
Transcription in eukaryotes requires which
of the following in addition to RNA
polymerase
(A) several transcription factors
(B) the protein product of the promoter
(C) start and stop codons
(D) ribosomes and tRNA
(E) a signal recognition particle
Transcription occurs along a ____ template
forming an mRNA in the ____ direction.
(A) 5' to 3'; 5' to 3'
(B) 5' to 3'; 3' to 5'
(C) 3' to 5'; 5' to 3'
(D) 3' to 5'; 3' to 5’
(E) All of the above could be correct depending
on the orientation.
Assume that RNA polymerase transcribes a gene
containing the section of DNA shown below:
5'–GATGCGAATCGT–3'
3'–CTACGCTTAGCA–5'
If the top strand were the template strand, the RNA
corresponding to this section would be _____.
(A) 5'–GATGCGAATCGT–3’
(B) 5'–GAUGCGAAUCGU–3’
(C) 5'–ACGATTCGCATC–3’
(D) 5'–ACGAUUCGCAUC–3'
What happens when RNA
polymerase reads a stop codon?
(A)The RNA transcript is cleaved off.
(B) RNA polymerase detaches from the DNA.
(C) Both of the above.
(D)Neither of the above.
How is transcription terminated in
eukaryotes?
(A)A stop codon is read.
(B) A hairpin loop forms.
(C) A polyadenylation sequence is read.
(D)A termination sequence is read.
(E) None of the above.
The chicken ovalbumin gene is 7,700 base pairs in
length, yet the mature messenger RNA is only 1872
nucleotides.
Which of the following correctly describes the
composition of the mature messenger RNA:
(A) It has, in part, the complementary sequence to the poly-A
tail of the gene.
(B) It lacks the complementary sequence to the exons of the
gene.
(C) It has, in part, the complementary sequence to the TATA
box of the promoter.
(D) It has, in part, the complementary sequence to the 5' cap
of the gene.
(E) It lacks the complementary sequence to the introns of the
gene.
A spliceosome:
(A)Splices out introns from mRNA
(B) Splices out exons from pre-mRNA
(C) Splices out exons from mRNA
(D)Splices out introns from pre-MRNA
(E) None of the above
Why is a 5’ poly-A tail added to an
mRNA transcript in eukaryotes?
(A)To help the mRNA bind to the ribosome.
(B) To prevent the mRNA from degrading on its
journey from the nucleus to the cytosol.
(C) To promote the binding of a signal
recognition particle.
(D)Both (A) and (B)
(E) A 5’ poly-A tail is not added to an mRNA
transcript in eukaryotes.
What modifications are made to an mRNA
transcript in prokaryotes prior to
translation?
(A)A 5’ modified guanine cap is added.
(B) A poly-A tail is added.
(C) Introns are spliced out.
(D)All of the above.
(E) None of the above.
Which site in a ribosome accepts a
charged tRNA?
(A)The aminoacyl tRNA binding site
(B) The peptidyl tRNA binding site
(C) The exit site
(D)The initiation site
(E) The start codon
A particular triplet of bases in the template strand of
DNA is 5'–TGA–3'. Which of the following is the
anticodon component of the tRNA that binds the mRNA
codon transcribed from this DNA?
(Note: By convention, the 3' end of each anticodon is written on
the left, and the 5' on the right.)
(A) ACU
(B) AGU
(C) AGT
(D) UGA
What type of bond forms between
amino acids in a growing polypeptide
chain:
(A)peptide
(B) covalent
(C) hydrogen
(D) phosphodiester
(E) ionic
How is translation terminated?
(A)A stop codon is read by the ribosome.
(B) A hairpin loop forms in the polypeptide
chain.
(C) A signal recognition particle cleaves the chain
off.
(D)A release factor adds water to the growing
polypeptide chain.
(E) The small and large subunit of the ribosome
break apart.
Which of the following pieces of mRNA
would be successfully translated into a
polypeptide chain in a prokaryote?
(mRNA shown in its entirety)
(A)
(B)
(C)
(D)
(E)
5’-AUGAUCCCGUCCCGGGCACCUUAG-3’
5’-UGAGCUGCGCCCAAUGCUUGGCAA-3’
5’-CGACGACCCGGUUACGAAUCUAAC-3’
5’-GGCUAAGAGUCUAGUAUCUGGAAG-3’
Any/all of the above
Which component is not directly
involved in translation?
(A)mRNA
(B) tRNA
(C) ribosomes
(D)GTP
(E) All of the above are directly involved.
What is the function of a signal
recognition particle?
(A)To help mRNA bind to the ribosome.
(B) To help mRNA find a ribosome.
(C) To help a ribosome bind to the RER.
(D)To block transcription inhibitors.
(E) None of the above.
Muscle cells differ from nerve cells
mainly because they
(A)express different genes
(B) contain different genes
(C) use different genetic codes
(D)have unique ribosomes
(E) have different chromosomes

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