Transcript Chapter 14 2015 - Franklin College

What is a gene?
• A sequence of DNA nucleotides that
specifies the primary structure of a
polypeptide chain (tells the cell how to
make it)
• Genes-made of nucleotides
• Proteins-made of amino acids
• How does a nucleotide code (in the
nucleus) specify an amino acid sequence
(in the cytoplasm)?
The Central Dogma
• DNA is transcribed into RNAcharacteristics of RNA
• RNA is translated into protein
• Advantages
• Exceptions
LE 17-4
Gene 2
DNA
molecule
Gene 1
Gene 3
DNA strand
(template)
5
3
TRANSCRIPTION
mRNA
5
3
Codon
TRANSLATION
Protein
Amino acid
The Genetic code-characteristics
• Triplet (3 nucleotides=codon=info for a
specific amino acid);64 different codons (3
are stop codons)
• Universal
• Redundant (61 codons-20 amino acids)variability in third nucleotide of codon.
Advantages of a redundant code?
• Non-overlapping
• Exceptions (ciliates; mito/chloroplasts)
Third mRNA base (3 end)
LE 17-5
Second mRNA base
Figure 17-06
Gene Expression
• If a gene is transcribed and the m-rna is
translated (the gene is expressed); a protein is
made. This often changes the phenotype of the
cell that produces the protein.
• Differential gene expression is involved in
embryonic development and cell specialization.
• Totipotency-each cell has the genetic
information for an entire organism.
• Differential gene expression results in cell
specialization (differentiation)
• Hormones often play a role in gene expression
Transcription
•
•
•
•
•
•
•
The first step in gene expression
Takes place in the nucleus
Requirements
A. RNA nucleotides
B. DNA template (gene)
C. Enzymes (RNA polymerase)
Only one of the two DNA strands is copied
(template strand)
LE 17-7a-2
Promoter
Transcription unit
5
3
Start point
RNA polymerase
DNA
3
5
Initiation
5
3
RNA Template strand
Unwound tran- of DNA
DNA
script
3
5
LE 17-7a-3
Promoter
Transcription unit
5
3
Start point
RNA polymerase
3
5
DNA
Initiation
5
3
3
5
RNA Template strand
Unwound tran- of DNA
DNA
script
Elongation
Rewound
DNA
5
3
3
5
RNA
transcript
3
5
LE 17-7a-4
Promoter
Transcription unit
5
3
Start point
RNA polymerase
3
5
DNA
Initiation
5
3
3
5
RNA Template strand
Unwound tran- of DNA
DNA
script
Elongation
Rewound
DNA
5
3
3
5
3
5
RNA
transcript
Termination
5
3
3
5
5
Completed RNA transcript
3
LE 17-7b
Elongation
Non-template
strand of DNA
RNA nucleotides
RNA
polymerase
3
3 end
5
Direction of transcription
(“downstream”)
5
Newly made
RNA
Template
strand of DNA
LE 17-8
Promoter
Eukaryotic promoters
5
3
3
5
TATA box
Start point
Template
DNA strand
Several transcription
factors
Transcription
factors
5
3
3
5
Additional transcription
factors
RNA polymerase II
5
3
Transcription factors
3
5
5
RNA transcript
Transcription initiation complex
Transcription-some important
details
• Rate-30-60 nucleotides/second
• RNA polymerase (Many forms in eucaryotes, 3
basic types in bacteria: type I transcribes r-rna,
type II-mrna, types III-trna)
• Promotors-(approximately 100 nucleotides)strong and weak promotors
• Eukaryotes-transcription factors needed to help
RNA polymerase to bind to TATA box (region of
promotor 25 nucleotides upstream from initiation
site).
RNA products of transcription
•
•
•
•
•
•
m-rna
t-rna
r-rna
sn-RNA (small nuclear)
mi-Rna (micro)
Si-rna (small interfering)
Not all genes code for proteins (m-rna)-Rrna and
t-rna are obvious examples
• Actually, recent discoveries indicate that a large
part of the eukaryotic genome is non-coding
RNA-Introns
• Small rna (micro rna and small interfering rna)play a crucial role in the regulation of gene
expression involving both transcription and
translation. Rna interference (Rnai)
• We’ll talk about regulation of gene expression in
Chapter 15.
Ribosomal RNA and ribosomes
• R-rna; one of two important components of
ribosomes (other is protein-some of the proteins
are enzymes). 60% r-rna; 40% protein.
• Ribosomes consist of 2 subunits
• Ribosomes needed to translate proteins
• “workbench of protein synthesis”
• Position t-rna (which is attached to a specific
amino acid) on the codon of a m-rna
• Result is the synthesis of a protein (whose
amino acid sequence is specified by the m-rna;
which is transcribed from a gene)
LE 17-16b
P site (Peptidyl-tRNA
binding site)
A site (AminoacyltRNA binding site)
E site
(Exit site)
E
P
A
mRNA
binding site
Schematic model showing binding sites
Large
subunit
Small
subunit
LE 17-16a
tRNA
molecules
Growing
polypeptide
Exit tunnel
Large
subunit
E P
A
Small
subunit
5
3
mRNA
Computer model of functioning ribosome
Ribosomal –rna processing
T-rna
• Single polynucleotide chain folded into a
complex 3-D shape (inter-chain H
bonding). 75-80 nucleotides in length
• Binds a specific amino acid (involvement
of amino-acyl-trna-synthetase
• Attaches to a specific m-rna codon via its
anticodon
• How many different t-rna’s are there? 61?
Actually only 45 (wobble)
LE 17-14a
3
Amino acid
attachment site
5
Hydrogen
bonds
Anticodon
Two-dimensional structure
Amino acid
attachment site
5
3
Hydrogen
bonds
3
Anticodon
Three-dimensional structure
5
Anticodon
Symbol used in this book
“Charging” t-rna with its specific
amino acid
• “charging” enzyme-amino acyl t-rna
synthetase (20 different enzymes)
• Requires ATP
LE 17-15
Amino acid
Aminoacyl-tRNA
synthetase (enzyme)
Pyrophosphate
Phosphates
tRNA
AMP
Aminoacyl tRNA
(an “activated
amino acid”)
Messenger Rna (m-rna)
• Contains the information for the primary
sequence of a polypeptide chain
• Consists of codons
• Binds to ribosomes
• T-rna binds to m-rna (codon/anticodon)
LE 17-13
Amino
acids
Polypeptide
tRNA with
amino acid
attached
Ribosome
tRNA
Anticodon
Codons
5
mRNA
3
Translation
•
•
•
•
•
•
•
Codons (m-rna) read by ribosomes/t-rna
Polypeptide chain produced
3 steps in translationA. initiation
B. elongation
C. termination
Translation is a process that consumes a
tremendous amount of energy (ATP and GTP)
LE 17-16c
Amino end
Growing polypeptide
Next amino acid
to be added to
polypeptide chain
E
tRNA
mRNA
5
3
Codons
Schematic model with mRNA and tRNA
Translation-Initiation
• Initiation codon is AUG
• T-rna that bonds to AUG has an anticodon
UAC-this carries the amino acid
methionine
• Requires a GTP molecule
• Requires proteins called initiation factors.
LE 17-17
Large
ribosomal
subunit
P site
Initiator tRNA
GTP
GDP
E
A
mRNA
5
3
5
3
Start codon
mRNA binding site
Small
ribosomal
subunit
Translation initiation complex
Translation-Elongation
• The elongation cycle takes about 60
milliseconds
• During elongation, one m-rna codon is
read and then the ribosomes moves down
the message to the next codon.
• Binding of incoming t-rna to the A site of
the ribosome requires a GTP
• Translocation-requires a GTP
LE 17-18
Amino end
of polypeptide
E
3
mRNA
Ribosome ready for
next aminoacyl tRNA
P A
site site
5
2
GTP
2 GDP
E
E
P
A
P
GDP
GTP
E
P
A
A
Translation-Termination
• When the ribosome reaches a termination
codon, it causes the m-rna/ribosome
complex to separate
• No t-rna binds to the termination codon.
• Release factors
• Newly made polypeptide chain is released
(folds into its characteristic 3-D shape)
LE 17-19
Release
factor
Free
polypeptide
5
3
3
3
5
5
Stop codon
(UAG, UAA, or UGA)
When a ribosome reaches a stop
codon on mRNA, the A site of the
ribosome accepts a protein called
a release factor instead of tRNA.
The release factor hydrolyzes the
bond between the tRNA in the
P site and the last amino acid of the
polypeptide chain. The polypeptide
is thus freed from the ribosome.
The two ribosomal subunits
and the other components
of the assembly dissociate.
Summary of energy demands for
protein synthesis
•
A.
B.
C.
D.
A rough estimate is that for every amino acid
incorporated into a polypeptide chain, 3
ATP/GTP are consumed
Charging the amino acid (1 ATP)
Binding of incoming t-rna into the A site (1
GTP)
Translocation (1 GTP)
So a small protein (120 amino acids in length)
would cost the cell 360 ATP/GTP to make (the
equivalent of 12 glucose molecules going
through aerobic cell respiration)
Polyribosomes
• A single ribosome can translate an
average-sized polypeptide in about 1
minute
• Several ribosomes can translate the same
message one after the other.
• Increases the efficiency of protein
production
LE 17-20a
Growing
polypeptides
Completed
polypeptide
Incoming
ribosomal
subunits
Start of
mRNA
(5 end)
End of
mRNA
(3 end)
An mRNA molecule is generally translated simultaneously
by several ribosomes in clusters called polyribosomes.
LE 17-20b
Ribosomes
mRNA
0.1 m m
This micrograph shows a large polyribosome in a prokaryotic cell (TEM).
M-rna modifications
• Eukaryotic M-rna is modified extensively
after transcription (while its still in the
nucleus)
• These modifications include
A.Polyadenylation-added to 3’ end of m-rna
B. 5’ cap
C. Intron removal
M-RNA modifications
•
•
•
•
Poly A tail
A. added to the 3’ end of the m-rna
B.30-200 Adenine nucleotides
C. roles-regulation of transport of m-rna
out of the nucleus; regulation of
degradation of m-rna in the cytoplasm;
helps m-rna attach to small ribosomal
subunit
M-RNA modifications (continued)
• 5’ cap
• A. Modified guanine nucleotide stuck onto
5’ end of m-rna
• B. Roles- positioning of m-rna on small
ribosomal subunit in initiation; protects
m-rna from degradation
LE 17-9
Protein-coding segment
Polyadenylation signal
5
5 Cap
5 UTR
Start codon
Stop codon
3 UTR
Poly-A tail
Introns
• Many eukaryotic genes have nucleotide
sequences that don’t code for amino acids
(Introns)
• Introns separate coding sequences (exons).
Split genes
• Introns must be removed from the m-rna before
it is translated (introns have nucleotide
sequences that indicate splicing sites)
• Splicesomes are molecular machines that
remove introns from m-rna
LE 17-11-1
RNA transcript (pre-mRNA)
5
Exon 1
Intron
Exon 2
Protein
Other proteins
snRNA
snRNPs
Spliceosome
LE 17-11-2
Spliceosome
5
Spliceosome
components
Cut-out
intron
mRNA
5
Exon 1
Exon 2
Significance of introns
•
Why would chromosomes carry around
extra DNA that isn’t used in the final mrna?
A. Expensive to maintain (energy).
B. Splicing out introns is a risky business
(what if it’s done incorrectly)
C. With these disadvantages, there must be
an advantage or natural selection would
not favor this arrangement
Benefits of Introns
• Evolution of protein diversity
• One gene can be alternatively spliced in a
number of different ways to form several
different types of m-rna (alternative
splicing)
• Human antibody genes-about 500 genes
can code for billions of different antibody
molecules because of alternative splicing.
Figure 15.12
Exons
DNA
2
1
3
5
4
Troponin T gene
Primary
RNA
transcript
2
1
3
4
5
RNA splicing
mRNA
1
2
3
5
or
1
2
4
5
Summary of Transcription and Translation
Mutation
• An alteration in the nucleotide sequence of
a DNA molecule (chromosome)
• Chromosomal mutations (duplications;
deletions; inversions)
• Point mutations-alterations of one or a few
nucleotides in a gene
Point mutations
•
•
•
•
•
•
Spontaneous mutations
Induced mutations
Consequences of mutationsA. no effect-”silent mutations”
B. harmful mutations-(may be lethal)
C. beneficial mutations (rare)
Spontaneous mutations
• Base pairing errors; why aren’t they
corrected by DNA repair enzymes?
• Effects:
• A. no effect-silent mutation (redundancy of
genetic code; alteration of a non-critical
amino acid)
• B. Positive effect-rare
• C. negative effect-missense mutations;
nonsense mutations
LE 17-24a
Wild-type
mRNA
5
Protein
Amino end
3
Stop
Carboxyl end
LE 17-24b
Base-pair substitution
No effect on amino acid sequence
U instead of C
Stop
Missense
A instead of G
Stop
Nonsense
U instead of A
Stop
Sickle cell anemia
• Results of a spontaneous missense
mutation
• Result-altered hemoglobin molecule
• Effect-Depends on the environmental
conditions and number of copies of the
defective gene you inherited.
LE 17-23
Wild-type hemoglobin DNA
3
Mutant hemoglobin DNA
5
3
mRNA
5
mRNA
3
Normal hemoglobin
5
5
3
Sickle-cell hemoglobin
Induced mutations
• Caused by environmental damage
• Radiation (UV)- T-T dimers; excision
repair enzymes; xerdoerma pigmentosa
• Chemicals-Common result-base pair
addition or deletion
• Result of addition or deletion (frame shift
mutation)-missense or nonsense
• Worst scenario-addition/deletion of 1 or 2
nucleotides at the beginning of a gene
LE 17-25
Wild type
mRNA 5
Protein
3
Stop
Carboxyl end
Amino end
Base-pair insertion or deletion
Frameshift causing immediate nonsense
Extra U
Stop
Frameshift causing
extensive missense
Missing
Insertion or deletion of 3 nucleotides:
no frameshift but extra or missing amino acid
Missing
Stop
Mutations and Cancer
• Many mutations make cells cancerous
• 90% of known carcinogens are mutagens
• Ames test-screens potential chemicals for
being carcinogens by seeing if they are
mutagens
• Bacteria are the test subjects in the Ames
test.
Employee Resource Manual
• Plasmids
• Restriction Endonucleases
• Agarose Gel Electrophoresis
Plasmids
• Small extrachromosomal pieces of DNA
found in some bacterial species
• May carry additional genes (such as
antibiotic resistance)
• Can be genetically modified and used as
vectors for genetic engineering
PUC 18-Plasmid
Restriction Endonucleases
• Produced by some bacteria as a defense
against virus infection
• Cleave DNA at specific bases sequences
(different recognition site for each different
enzyme)
• Can be used to join DNA from 2 different
sources (plasmid DNA and genomic DNA)
ECOR1
Agarose Gel Electrophoresis
• Separates DNA based upon size
differences
• DNA is pulled through a gel by an electric
current
• (-) charged DNA is pulled to the positive
pole of the apparatus.
• Smaller pieces of DNA migrate through
the gel faster than larger pieces of DNA
Agarose Gel Electrophoresis
Procell
Procell in Action
What is your first job
assignment?
• Clone the H gene (use a bacteria to make
copies of the gene for us)
What kind of bacteria do we use to
clone the H gene?
• E.coli (lacZ(-), amp sensitve)
Where is the H gene located?
• Lambda virus
How do you get the cloned gene into the
bacteria so the bacteria can copy it?
• Transform E.coli (lacZ(-), amp sensitve)
with PUC 18-lambda plasmid (heat shock
and osmotic shock)
Lambda virus genes have
been inserted into the plasmids here
How do you get lambda genes into
PUC 18 plasmid?
• Incubate PUC-18 and lambda with EcoRI,
ligate products
How many different plasmids do you get when you
mix PUC 18 and lambda, both of which have been
ECOR1 and then ligated?
7
Would all 7 of the plasmids be
recombinant (have lambda DNA)?
No!
How do you tell if bacteria have been
transformed successfully with PUC-18
plasmid?
• They will grow on amp agar.
How can you distinguish whether plasmids that
transformed bacteria were recombinant (lambda
and PUC-18) or nonrecombinant (pUC-18 only)?
Plate the transformed cells on Xgal-amp
agar
E.Coli transformed
With nonrecombinant
Plasmids (PUC-18)
E.Coli transformed
With recombinant
Plasmids (PUC-18/lambda)