unit #3 schedule: previously: – sanger sequencing – central dogma overview – mutation –...
TRANSCRIPT
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
Polypeptide subunits that make upenzymes for tryptophan synthesis
(a) Tryptophan absent, repressor inactive, operon on
DNA
mRNA 5
Protein Inactiverepressor
RNApolymerase
Regulatorygene
Promoter Promoter
trp operon
Genes of operon
OperatorStop codonStart codon
mRNA
trpA
5
3
trpR trpE trpD trpC trpB
ABCDE
Fig. 18-3b-2
(b) Tryptophan present, repressor active, operon off
Tryptophan(corepressor)
No RNA made
Activerepressor
mRNA
Protein
DNA
© 2011 Pearson Education, Inc.
Fig. 18-4b
(b) Lactose present, repressor inactive, operon on
mRNA
Protein
DNA
mRNA 5
Inactiverepressor
Allolactose(inducer)
5
3
RNApolymerase
Permease Transacetylase
lac operon
-Galactosidase
lacYlacZ lacAlacI
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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 Sequence5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACCTCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGTGCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAGTCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACCGCAACCTGCACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGGTGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGGGGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGACGTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTGCCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGTGACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATCTTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCTGCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGTCCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAGAAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCCTTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCATCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTTAAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACCCCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGTGCTGCTGTGCTCCTGGTGA 3’
mc1r Gene Sequence5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACCTCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGTGCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAGTCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACTGCAACCTGCACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGGTGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGGGGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGACGTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTGCCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGTGACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATCTTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCTGCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGTCCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAGAAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCCTTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCATCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTTAAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACCCCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGTGCTGCTGTGCTCCTGGTGA 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:MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSIPDGLFLSLGLVSLVENVLVVIAITKNRNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICGSMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIXGIWVASIFFSTLFITYYNHTAVLICLVTFFLAMLALMAXLYVHMLTRAYQHAQGIAQLQKRQGSTXQGFCLKGAXTLTIILGIFFLCWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLCSW
Amino Acid Sequence Light Fur:MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSVPDGLFLSLGLVSLVENVLVVIAITKNCNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICGSMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIVGIWVASIFFSTLFITYYNHTAVLICLVTFFLAMLALMAILYVHMLTRAYQHAQGIAQLQKRQGSTRQGFCLKGAATLTIILGIFFLCWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLCSW
Changing 1 amino acid:
• Arginine:– Strongest +charge– Very hydrophilic
• Cysteine:– Not hydrophilic– Forms disulfide bonds
Beach Mice
- Missense substitution mutation of one nucleotide CT
- Changes one amino acid: Arginine Cysteine
- Changes the function of the MC1R protein
• 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.
How is eumelanin produced?
The Melanocortin-1-ReceptorEffectively
stimulated by hormone
MC1RR67
Results in lots of cAMP production
c(tyr), ptyrp1, tyrp2
<<activated>>
LOTS of EUMELANIN produced
Ineffectively stimulated by
hormone
MC1RC67
Results in little cAMP production
c(tyr) <<activated>>
EUMELANIN not produced
p, tyrp1, tyrp2 <not activated>
Fig. 18-5
CAP: catabolite activator protein
EUKARYOTIC REGULATION OF GENE EXPRESSIONActivators and Enhancers of Transcription
Campbell 8e, Fig. 18.8
Fig. 17-8A eukaryotic promoterincludes a TATA box
3
1
2
3
Promoter
TATA box Start point
Template
TemplateDNA strand
535
Transcriptionfactors
Several transcription factors mustbind to the DNA before RNApolymerase II can do so.
5533
Additional transcription factors bind tothe DNA along with RNA polymerase II,forming the transcription initiation complex.
RNA polymerase IITranscription factors
55 53
3
RNA transcript
Transcription initiation complex
Campbell 8e, Fig. 18.9
Campbell 8e, Fig. 18.10
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
Coming Up:Friday:• Tutorial (3-5pm in C-3)
Tuesday:• Interactive Review (White-
Boards + 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) 5’-AUGAUCCCGUCCCGGGCACCUUAG-3’(B) 5’-UGAGCUGCGCCCAAUGCUUGGCAA-3’(C) 5’-CGACGACCCGGUUACGAAUCUAAC-3’(D) 5’-GGCUAAGAGUCUAGUAUCUGGAAG-3’(E) 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