Transcript Week 5 Pre-Lecture Slidesx

Monday January 30th, 2017
Class 16 Learning Goals
Prokaryotic Gene Regulation
• After this class, you should be able to:
• Predict the strength of a promoter based on comparisons with the
consensus promoter sequence of the cell.
• Assess the likely transcription level of the lac operon based on:
• Molecular interactions within the cell
• The logical response of the cell to environmental conditions
• Classify a new regulatory mechanism as positive or negative
regulation of a particular molecule or response
• Propose a molecular interaction designed to provide a particular
expression response
Peer Instruction
Translation
rate
Protein activation
or inhibition
Onset of
transcription
RNA polymerase
Protein
Ribosome
mRNA
RNA polymerase
DNA
Transcriptional control
Translational control
Post-translational control
1) What is an evolutionary advantage of regulation?
2) Describe the three kinds of regulation of gene expression.
E. Coli Consensus -35 sequence
ATGCTATTA
(~740 transcripts per hour)
1st change:
-10% expression
ATGATATTA
Why are rates for the 1st and
2nd sequences different?
(~665 transcripts per hour)
2nd change:
___% expression
ATGATACTA
Peer Instruction
Estimate the data for the 3rd.
(~___ transcripts per hour)
Which of these sequences is part of the strongest promoter?
The lac operon has a relatively weak promoter.
Why is this a good thing for the cell?
Peer Instruction
LacI Promoter lacl
lacZ
lacY
E. coli
chromosome
Lac Operon Promoter
lacl product
-Galactosidase
Galactoside
permease
lacY product
lacZ product
1) How many promoters control the lac operon?
2) For each box, fill out Yes/No by what is most logical:
Only glucose Only lactose Glucose and lactose
is present
is present
both present?
Should the
lac operon
be on?
Peer Instruction
LacI Promoter lacl
lacZ
E. coli
chromosome
Lac Operon Promoter
lacl product
lacY
-Galactosidase
Galactoside
permease
lacY product
lacZ product
E. coli
What do the protein
products of lacZ and lacY do?
Galactoside
permease
-Galactosidase
Glucose
Galactose
Lactose
Plasma membrane
Why do the large majority of biology courses use the
Lac Operon as a model system?
Peer Instruction
Describe the negative regulation of lac operon expression.
Repressor binds to DNA.
No transcription occurs.
lacl
RNA polymerase
bound to promoter
(blue DNA)
Repressor
synthesized
lacZ
lacY
DNA
‘Polycistronic’ mRNA
Repressor
synthesized
lacl+
RNA polymerase
bound to promoter
(blue DNA)
-Galactosidase Permease
lacZ
lacY
Lactose-repressor
complex
Peer Instruction
ATP
Adenylyl
cyclase
cAMP
Two
phosphate
groups
Glucose allosterically
inhibits this enzyme
How does glucose influence levels of cAMP?
So, high [glucose] = _____ cAMP = ______ binding of RNA pol
at the lac operon
And low [glucose] = _____ cAMP = ______ binding of RNA pol
at the lac operon
Peer Instruction
Describe the positive regulation of lac operon expression.
When glucose is high,
cAMP is absent:
CAP
lacZ
CAP site
lacY
RNA polymerase bound
loosely to promoter (blue DNA)
CAP
cAMP
When glucose is low,
cAMP is present:
lacZ
CAP site
RNA polymerase bound
tightly to promoter (blue DNA)
lacY
Tuesday January 31st,
2017
Class 17 Learning Goals
Understanding the Cell Cycle
• What are the phases of the cell cycle, and how do cells decide when
to proceed through each phase?
• Why are transitions between phases controlled by checkpoints?
• What is the molecular basis of checkpoint control?
• How does this relate to prokaryotic gene regulation?
Peer Instruction
1) Label G1, G2, M, S, and
Interphase on this cell cycle.
2) When is DNA replication?
3) When do mitotic spindles need
to be correct and ready?
G1
S PHASE AND G2
MITOSIS
Chromosomes are shown partially
condensed to make them visible
Parent cell:
4 unreplicated
chromosomes
Sister
chromatids
Parent cell:
4 replicated
chromosomes
Replicated chromosomes
condense at the start of
mitosis.
1) Explain the reasons for these three checkpoints.
Peer
Instruction
Metaphase Checkpoint
checkpoint
M-phase
checkpoint
G2 GCheckpoint
2
Pass this checkpoint if:
• all chromosomes are
attached to spindle
Pass this checkpoint if:
• chromosome replication
is successfully completed
• no DNA damage
• activated enzymes present
M
G2
G1
2) What is G0?
Pass this checkpoint if:
• nutrients are sufficient
• growth factors are present
• cell size is adequate
• chromosome is undamaged
G0
S
G1 checkpoint
G1 Checkpoint
Inactive Cdk
Cyclin
Peer Instruction
Activated Cdk
(Binding)
Growth
factors
Expression of
the E2F gene
Tight
Binding
Retinoblastoma Protein
(constitutively produced)
The players involved are:
• growth factors sent from nearby cells
• CDK (an enzyme that is always present)
• Cyclin (an enzyme that is sometimes present)
• Rb (a protein with two binding sites)
• E2F (a protein with one binding site)
Decipher the molecular control of the G1 checkpoint.
Inactive Cdk
Cyclin
Activated Cdk
(Binding)
Growth
factors
Expression of
the E2F gene
Tight
Binding
Retinoblastoma Protein
(constitutively produced)
How do you think E2F initiates S-phase?
Peer Instruction
Inactive Cdk
Cyclin
Peer Instruction
Activated Cdk
(Binding)
Growth
factors
Expression of
the E2F gene
Tight
Binding
Retinoblastoma Protein
(constitutively produced)
Draw E2F, and include binding sites for other molecules.
-Where are they in relation to each other?
Does E2F binds to Rb or promoters more tightly?
Why is E2F called a ‘transcription factor’?
Wednesday February 1st,
2017
Class 18-19 Learning Goals
Eukaryotic Gene Regulation (the next two class sessions)
• After these classes, you should be able to:
• Explain the basic mechanism, including common enzyme types
and direct effects on gene expression, of:
• Enhancers and silencers
• RNA splicing
• Histone packing
• Ubiquitination
• Post-translational modification
• microRNA-based degradation
• For each of these six methods of eukaryotic gene regulation,
predict the effect on gene expression of a mutation in any
component.
Eukaryotic Gene Regulation: Data Set 1
1. Bacterial chromosome
= Location of a mutation that
lowers gene expression
= Protein-coding region
2. Eukaryotic chromosome
Enhancers (and silencers)
Peer Instruction
Co-activators
Intron
Promoter
Exon
In this example, a protein is increasing the likelihood of
the RNA polymerase binding at this promoter.
How is expression increased by the protein?
What is needed in the DNA?
Enhancers (and silencers)
Peer Instruction
Co-activators
Intron
Promoter
Exon
Here, two different proteins are used to increase expression.
How does the new protein work?
What is different about the proteins?
Could a similar system be used to decrease transcription?
Eukaryotic Gene Regulation: Data Set 2
Fig. 3
Bacterial DNA (electron micrograph)
Fig. 1
Basal
Transcription
Levels
Eukaryotic DNA (electron micrograph):
All promoters analyzed have the
same DNA sequence
Fig. 2
Basal
Transcription
Levels
Fig. 4
Homework: Gene Regulation Chart
Type of regulation
Histone acetylation
Enhancer site
Silencer-binding protein
Splicing
Ubiquitin addition to the
product
5’ Cap and 3’ Poly-A tail
Post-translational
modification
microRNA degradation
Other?
Found in
prokaryotes or
eukaryotes?
Mechanism?
Level
(dna, txn, tsl,
post-tsl?)
Increase or
decrease
expression?
Thursday February 2nd, 2017
Class 18-19 Learning Goals
Eukaryotic Gene Regulation (today and yesterday)
• After these classes, you should be able to:
• Explain the basic mechanism, including common enzyme types and
direct effects on gene expression, of:
• Enhancers and silencers
• RNA splicing
• Histone packing
• Ubiquitination
• Post-translational modification
• microRNA-based degradation
• For each of these six methods of eukaryotic gene regulation, predict the
effect on gene expression of a mutation in any component.
• Do molecular interactions decide genetic characteristics?
• Can we predict heritability from molecular actions?
Eukaryotic Gene Regulation:
Other post-translational modifications
Ubiquitination by UbiE3s
Phosphates
Addition: kinase
Removal: phosphorylase
Acetylation
Cleavage by
proteases
Addition of
carbohydrates
Data from several eukaryotic mRNAs
Gene
Molecular
Weight
Fer6
218.5+/-5.0
57 minutes
+/- 4 min
nucleus
Fer6
312.9+/-5.0
88 minutes
+/- 4 min
nucleus
Fer6
390.0+/-5.0
93 minutes
+/- 4 min
nucleus
Fer6
484.4+/-5.0
147 minutes
+/- 4 min
cytoplasm
33 minutes
+/- 4 min
cytoplasm
Tran9.1 485.1+/-5.0
Electron Micrograph Cytoplasmic
Half-life
Location found in
the cell
6. [4 points] What conclusion(s) about eukaryotic mRNA
regulation can we draw from this table?
Peer Instruction
What is this enzyme doing?
Why does this process affect
only specific mRNA molecules?
Why would a cell gain an
advantage from this?
•
•
Peer Instruction
Creating a microRNA
Does this microRNA
come from a gene?
Would the information for
this microRNA be findable
in the genome by looking
for start and stop codons?
Dominant or Recessive?
• Look to the heterozygote.
– If BC has the same phenotype as BB
• B is dominant
– If BC has the same phenotype as CC
• B is recessive
– If BC has a different phenotype than BB or CC
• Something else is going on (possible codominance)
• This is predictable using molecules
– Whereas, before, you were fed this information…
Dominant or recessive?
There are two alleles of an gene that encodes an
enzyme that determines eyelash length.
The G allele: Working eyelash enzyme makes 110% of
needed product for long eyelashes
The H allele: Non-functional eyelash enzyme makes
0% of needed product for long eyelashes
Homozygote GG Heterozygote
% of gene ‘dose’
Phenotype?
Het phenotype is
like?
Homozygote HH
Peer Instruction
This diagram ranges from the scale of a pea pod all the way
down to the scale of molecules. Label everything that you
recognize, and well discuss the process.
rr individuals:
No starch-creating
enzyme
RR individuals:
Good starchcreating enzyme
Normal peas
maintain shape
Swollen peas
eventually wrinkle
If we carefully titrate the enzyme in the peas, we see:
0%
25%
50%
75%
100%
Peer Instruction
A mutant version of the lac repressor that always
binds to DNA at the operator is a dominant mutation
compared to wild-type.
Why is this dominant?