1. Unit #3 Schedule: Previously: Today: Tutorial (Apr 5) Review (Apr 9)
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

2. Regulation of Gene Expression
There are at least 300 different kinds of cells in the human body. Most of them have identical DNA.

3. Regulation of Gene Expression
We examine this at a very general level. Prokaryotes vs. Eukaryotes

4. Definition: Operon
A region of DNA that codes for a series of functionally related genes and is transcribed from a single promoter into mRNA.

5. 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.

8. Negative Control
 Negative control occurs when a regulatory protein binds to DNA and shuts down transcription.

9. PROKARYOTIC REGULATION OF GENE EXPRESSION:
Fig. 18-2
PROKARYOTIC REGULATION OF GENE EXPRESSION:
Figure 18.2 Regulation of a metabolic pathway

10. Polypeptide subunits that make up enzymes for tryptophan synthesis
Campbell Fig. 18-3a
trp operon
Promoter
Promoter
Genes of operon
DNA
trpR
trpE
trpD
trpC
trpB
trpA
Regulatory
gene
Operator
Start codon
Stop codon 3
mRNA 5
mRNA
RNA
polymerase 5 E D C B A
Protein
Inactive
repressor
Polypeptide subunits that make up
enzymes for tryptophan synthesis
Figure 18.3 The trp operon in E. coli: regulated synthesis of repressible enzymes
(a) Tryptophan absent, repressor inactive, operon on

11. (b) Tryptophan present, repressor active, operon off
Fig. 18-3b-2
DNA
No RNA made
mRNA
Protein
Active
repressor
Tryptophan
(corepressor)
(b) Tryptophan present, repressor active, operon off
Figure 18.3 The trp operon in E. coli: regulated synthesis of repressible enzymes

14. (b) Lactose present, repressor inactive, operon on
Fig. 18-4b
lac operon
DNA
lacI
lacZ
lacY
lacA
RNA
polymerase 3
mRNA
mRNA 5 5
-Galactosidase
Permease
Transacetylase
Protein
Figure 18.4 The lac operon in E. coli: regulated synthesis of inducible enzymes
Allolactose
(inducer)
Inactive
repressor
(b) Lactose present, repressor inactive, operon on

16. 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.

17. Positive Control
 Positive control occurs when a regulatory protein binds to DNA and triggers transcription.

18. CAP: catabolite activator protein
Fig. 18-5
Figure 18.5 Positive control of the lac operon by catabolite activator protein (CAP)
CAP: catabolite activator protein

19. What is cAMP?

21. mc1r Gene Sequence
5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACCTCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGTGCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAGTCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACCGCAACCTGCACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGGTGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGGGGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGACGTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTGCCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGTGACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATCTTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCTGCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGTCCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAGAAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCCTTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCATCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTTAAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACCCCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGTGCTGCTGTGCTCCTGGTGA 3’

22. mc1r Gene Sequence
5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACCTCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGTGCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAGTCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACTGCAACCTGCACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGGTGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGGGGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGACGTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTGCCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGTGACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATCTTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCTGCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGTCCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAGAAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCCTTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCATCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTTAAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACCCCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGTGCTGCTGTGCTCCTGGTGA 3’

23. 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

24. Changing 1 amino acid: Arginine: Cysteine: Strongest +charge
Very hydrophilic
Cysteine:
Not hydrophilic
Forms disulfide bonds

25. Beach Mice Missense substitution mutation of one nucleotide CT
Changes one amino acid: Arginine  Cysteine
Changes the function of the MC1R protein

26. 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
The following sequence of slides details how eumelanin is produced.
[Note: in the above schematic, the gene c(tyr) produces the enzyme Tyrosinase that is shown as the catalyst to facilitate the conversion of Tyrosine to Dopaquinone.]
Eumelanin production occurs as follows:
Activation of the MC1R protein occurs when it successfully binds to a molecule called an α melanocyte stimulating hormone (α-MSH). If instead the MC1R protein binds to a different molecule, an agouti stimulating peptide, a biosynthetic pathway begins that produces pheomelanin rather then eumelanin. However, assuming that α-MSH binding occurs, cAMP synthesis begins and in turn stimulates the expression of four genes: c(Tyr), Tryp1, Tyrp2 and p. The c(Tyr) gene codes for an enzyme called tyrosinase that catalyzes a reaction in a biosynthetic pathway that converts the amino acid tyrosine into a molecule called dopaquinone. The Tyrp1 and Tryp2 genes code for tyrosine-related enzymes that catalyze reactions in a subsequent pathway that converts dopaquinone into eumelanin. The p gene codes for a protein that also aids in this process. Eumelanosomes are then made in the golgi apparatus of the melanocyte that are then transported out of the melanocyte to the hair cortex.
If there is not enough cAMP, dopaquinone binds to cysteine and the product, cysteinyldopa, continues on a biosynthetic pathway that results in pheomelanin.
If there is lots of cAMP, the four genes [c(tyrp), Tyrp1, Tyrp2 and p] are all expressed and dopaquinone is converted into leucodopachrome. This results in a biosynthetic pathway that results in melanin.
Without the successful binding of an α-MSH molecule by the MC1R protein the multistep biosynthetic pathway that results in eumelanin synthesis is not initiated. Therefore, structural changes to the MC1R protein as a result of a genetic mutation could make it non-functional and result in a different coat color due to a change in pigment production. This is often the case in coastal populations of beach mice where coat colors are lighter than their inland neighbors.
Mechanism taken from: Barsh, G.S. (1996) The genetics of pigmentation: from fancy genes to complex traits, Trends in Genetics 12 (8),
“Agouti and α-MSH are extracellular proteins that regulate MClR signaling at the cell membrane to depress and elevate, respectively, intracellular levels of cyclic AMP (cAMP). Tyrosine is oxidized by the enzyme tyrosinase (Tyr) encoded by the c locus, to produce dopaquinone, a common precursor of pheomelanin and eumelanin. Intermediate enzymatic steps from dopaquinone to pheomelanin have not been identified, but the tyrosinase-related enzymes encoded by the Tyrp2 and TyrpI genes catalyze oxidative steps in the pathway from dopaqninone to eumelanin. All three enzymes and their products are localized within melanosomes because the oxidative intermediates of melanin synthesis are cytotoxic. The product of the pink-eyed dilution gene (p) is an integral membrane protein specific to eumelanosomes and appears to help stabilize tyrosinase and the two tyrosinase-related enzymes as part of a multiprotein complex. Increased levels of intracellular CAMP stimulate expression of the c, Tyrp1, Tyrp2 and p genes (indicated by the dashed arrows), which provides a mechanism whereby MClR signaling, regulated positively or negatively at the extracellular level by α-MSH or Agouti protein, respectively, determines whether dopaquinone is diverted into the eumelanin or the pheomelanin pathway.”

27. 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.

28. 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.

29. 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.

30. 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

31. CAP: catabolite activator protein
Fig. 18-5
CAP: catabolite activator protein
Figure 18.5 Positive control of the lac operon by catabolite activator protein (CAP)

32. EUKARYOTIC REGULATION OF GENE EXPRESSION
Activators and Enhancers of Transcription
Campbell 8e, Fig. 18.8

34. Several transcription factors must bind to the DNA before RNA
Fig. 17-8 1
A eukaryotic promoter
includes a TATA box
Promoter
Template 5 3 3 5
TATA box
Start point
Template
DNA strand 2
Several transcription factors must
bind to the DNA before RNA
polymerase II can do so.
Transcription
factors 5 3 3 5 3
Additional transcription factors bind to
the DNA along with RNA polymerase II,
forming the transcription initiation complex.
Figure 17.8 The initiation of transcription at a eukaryotic promoter
RNA polymerase II
Transcription factors 5 3 3 5 5
RNA transcript
Transcription initiation complex

35. Campbell 8e, Fig. 18.9

37. 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

38. 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

39. Coming Up: Friday: Tuesday: Thursday: Tutorial (3-5pm in C-3)
Interactive Review (White-Boards + Clickers)
Thursday:
Midterm Exam 3

40. Review Part 1
Clicker Review

41. With respect to nucleotide bonds:
A-T is stronger than C-G
C-G is stronger than A-T
A-T and C-G have approximately equal strength

42. Which mode of information transfer usually does not occur?
DNA to DNA
DNA to RNA
DNA to protein
RNA to protein
All occur in a working cell

43. In replication of DNA, the helix is opened and untwisted by
DNA polymerase
ligase
helicase
telomerase
topoisomerase

44. _______________ joins DNA fragments to the lagging strand.
(A) Telomere
(B) DNA Polymerase I
(C) Helicase
(D) DNA Polymerase III
(E) Ligase

45. 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'

46. 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

47. Considering the structure of double stranded DNA, what kinds of bonds hold one complementary strand to the other?
peptide
covalent
(C) hydrogen
(D) phosphodiester
(E) ionic

48. The nitrogenous base adenine is found in all members of which group?
proteins, ATP, and DNA
(B) proteins, carbohydrates and ATP
(C) glucose, ATP and DNA
(D) ATP, RNA and DNA
(E) proteins, glycerol and hormones

49. Where and how are Okazaki fragments synthesized?
on the leading strand, in a 5’  3’ direction
on the leading strand, in a 3’  5’ direction
on the lagging strand, in a 5’  3’ direction
on the lagging strand, in a 3’  5’ direction

50. 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

51. Regarding beach mice: The C  T substitution at position 199 of the mc1r gene:
arose by a mutation in the beach mouse populations in response to a need for protection from predation.
leads to the failure of melanocytes to make an MC1R protein.
arose by a mutation then increased in frequency because it was selectively advantageous in the beach mouse populations.
had no effect on the beach mouse populations.
produced an alternate allele that was detrimental to mice on the white sand beaches

52. Regarding beach mice: What was the reason for the lighter coat colors of the mice on the white sand beaches?
Owls and other carnivores prey on beach mice that do not carry the mutant allele.
A substitution of cysteine for arginine at position 67 of the MC1R protein.
A substitution of thymine for cystosine at position 199 of the mc1r gene nucleotide sequence.
The failure of melanocytes to lay down melanin pigment in the cortex of hairs of the lighter colored beach mice.
The poorer binding affinity for α-MSH and the lower amount of cAMP produced by individuals with the mutated MC1R protein.

53. mRNA codons preceding the mutation being misread
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 _____.
mRNA codons preceding the mutation being misread
mRNA codons following the mutation being misread
no change in the polypeptide coded by this gene
the AUC triplet functioning as a chain terminator

54. 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?
Transcription is the synthesis of protein from mRNA by ribosomes; translation is the synthesis of mRNA from DNA by RNA polymerase.
Transcription is the synthesis of mRNA from DNA by ribosomes; translation is the synthesis of protein from mRNA by RNA polymerase.
Transcription is the synthesis of protein from mRNA by RNA polymerase; translation is the synthesis of mRNA from DNA by ribosomes.
Transcription is the synthesis of mRNA from DNA by RNA polymerase; translation is the synthesis of protein from mRNA by ribosomes.

55. 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

56. 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.

57. 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 _____.
5'–GATGCGAATCGT–3’
5'–GAUGCGAAUCGU–3’
5'–ACGATTCGCATC–3’
5'–ACGAUUCGCAUC–3'

58. What happens when RNA polymerase reads a stop codon?
The RNA transcript is cleaved off.
RNA polymerase detaches from the DNA.
Both of the above.
Neither of the above.

59. How is transcription terminated in eukaryotes?
A stop codon is read.
A hairpin loop forms.
A polyadenylation sequence is read.
A termination sequence is read.
None of the above.

60. 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.

61. A spliceosome: Splices out introns from mRNA
Splices out exons from pre-mRNA
Splices out exons from mRNA
Splices out introns from pre-MRNA
None of the above

62. Why is a 5’ poly-A tail added to an mRNA transcript in eukaryotes?
To help the mRNA bind to the ribosome.
To prevent the mRNA from degrading on its journey from the nucleus to the cytosol.
To promote the binding of a signal recognition particle.
Both (A) and (B)
A 5’ poly-A tail is not added to an mRNA transcript in eukaryotes.

63. What modifications are made to an mRNA transcript in prokaryotes prior to translation?
A 5’ modified guanine cap is added.
A poly-A tail is added.
Introns are spliced out.
All of the above.
None of the above.

64. Which site in a ribosome accepts a charged tRNA?
The aminoacyl tRNA binding site
The peptidyl tRNA binding site
The exit site
The initiation site
The start codon

65. 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.)
ACU
AGU
AGT
UGA

66. What type of bond forms between amino acids in a growing polypeptide chain:
covalent
(C) hydrogen
(D) phosphodiester
(E) ionic

67. How is translation terminated?
A stop codon is read by the ribosome.
A hairpin loop forms in the polypeptide chain.
A signal recognition particle cleaves the chain off.
A release factor adds water to the growing polypeptide chain.
The small and large subunit of the ribosome break apart.

68. 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

69. Which component is not directly involved in translation?
mRNA
tRNA
ribosomes
GTP
All of the above are directly involved.

70. What is the function of a signal recognition particle?
To help mRNA bind to the ribosome.
To help mRNA find a ribosome.
To help a ribosome bind to the RER.
To block transcription inhibitors.
None of the above.

71. Muscle cells differ from nerve cells mainly because they
express different genes
contain different genes
use different genetic codes
have unique ribosomes
have different chromosomes