Download DNA: The Molecule of Inheritance

DNA: The Molecule of Inheritance
 Mendel’s experiments proved that molecules from
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the parents were transferred to offspring
These molecules could store genetic info, be
replicated, and undergo mutations
But what type of molecule? Could it be a lipid,
protein, nucleic acid, carbohydrate?
Scientists narrowed it down to 2 possibilities:
nucleic acids or proteins
What did they look like? What was their
structure?
Early DNA Experiments: Griffith
 Inject mice with live R bacteriamice live, no live R cells in blood
 Inject mice with live S bacteriamice die, live S cells in blood
 Inject mice with dead S bacteriamice live, no live S cells in blood
 Inject mice with live R bacteria + dead S bacteriamice die, live S cells in
blood
 What happened in the last experiment?
Early Experiments:
Griffith
 Interpretation of results
 Heat killed S cells, but not hereditary material
 This material was transferred to R cells transforming them
into S cells  Transformation
 Permanent change, after 100s of generations transformed R
cells still contained the instructions on causing infection
Early Experiments: Avery (1944)
Early Experiments:
Avery
 Interpretation of results
 If add protein digesting enzymes, protein destroyed, but
DNA intactR cells transformed into S cells, mice die
 If add DNA digesting enzymes, DNA destroyed, protein
intactR cells did not transform, mice live
Early Experiments:
Hershey and Chase (1950s)
• Used T4 bacteriophage-
bacterial virus
• Made of a outer protein
coat which protects its
inner genetic material
• In order to infect its
hosts its genetic material
must be introduced into
the host first
• Was it the protein coat
or the inner genetic
material that caused
infection?
They knew that 1) T2 phages were 50:50
Protein:DNA, 2) Viral reproduction occurs inside
bacterial cell
Early Experiments:
Hershey and Chase (1950s)
 Radiolabel protein coat
with 35S, after T4
infection of bacteria
found label outside
 Radiolabel inner
material with 32P, after
T4 infection of
bacteria found label
inside
Early Experiments:
Hershey and Chase
 Interpretation of results
 Infection of bacteria cells is caused by the transfer of
it genetic material to the inside
 Whichever molecule was found on inside must be
molecule of inheritance
 Proved beyond any doubt that DNA was the molecule
of inheritance
 But what did it look like?
Discovery of the
Structure of DNA
 DNA consisted of only 4 nucleotides: Adenine, Thymine,
Cytosine, Guanine deoxyribose sugar, phosphate group,
base
 T, C  pyrimidines (one ring)
 A, G  purines ( 2 rings)
Discovery of the
Structure of DNA
 Chargaff’s Laws
(1949): A=T, G=C,
DNA was species
specific
 Structure solved; it
was a helix
Discovery of the Structure of DNA: X-Ray
Diffraction (Wilkins, Franklin)
 X-ray diffraction of DNA fibers gave a regular repeating
pattern of atoms in the DNA
Discovery of the Structure of DNA: Watson and
Crick Experiments (1953)
• Using all previous
data from other
scientists and
experiments,
Watson and Crick
published a 1 page
paper on solving the
structure of DNA
• Watson, Crick,
Wilkins received
Nobel Prize in 1962
Structure of DNA
 2 nm in diameter
 Distance between each base pair= 0.34nm
 Double stranded helix, “twisted ladder”
 A base pairs with T, C base pairs with G, 2 H bonds for
A-T, 3 H bonds for C-G
 One strand runs 5’3’ (right side up), the other strand
runs 3’5’ (upside down)
DNA Replication
• Duplication of DNA. When
does it occur?
• Semiconservative-part of
the original DNA molecule
is conserved during
replication.
• 1 DNA molecule2 DNA
molecules each consisting
of a new and old strand
• New strand is made using
old strand as template and
base pair rules
DNA Replication
 What would be the complementary strand of the
following DNA sequence?
 ACGTATACGTGC
 The following piece of DNA is to be replicated. Give
the correct daughter strand.
 TTACCGGTTC
DNA Replication and Complementary
Strands
 ACGTATACGTGC original
 TGCATATGCACG complementary
 TTACCGGTTC original
 AATGGCCAAG replicated
DNA Replication: Enzymes Involved
• Topoisomerase and helicase-unwind and uncoil
DNA (break H bonds)
• DNA polymerase adds the correct nucleotide in the
5’-3’ direction only
• DNA pol moves in the forward direction on one
strand, moves in the reverse direction on the other
strand, however, always moving in the 5’-3’
direction
DNA Replication
• Replication is
continuous on
leading strand;
discontinuous
on the lagging
strand
• DNA pol only
adds 5’3’
• Lagging strand
is composed of
Okazaki
fragments that
must be linked
together by
ligase
Prokaryotic vs Eukaryotic Replication
• Bacteria have a circular
chromosome  replication
occurs in 2 directions at the
origin of replication
• Replication is fast  minutes
• Eukaryotes have long, linear
chromosomes  replication
begins at many locations 
replication bubbles,
replication forks
• Replication is slower  hours
DNA Repair
• What happens if the incorrect base is added and a
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mismatch base pair occurs? Ex. A-G or C-T
DNA Pol has a proofreading function
It will cut out the incorrect base and put back the
correct base
Ligase comes in and repairs the “cut” in the DNA
Results in a very low error rate  1 out of 100 million
base pairs
From DNA to Proteins
 DNA is the genetic instructions for life,
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but DNA itself does not do work to
sustain life.
Which type of molecule is responsible
for all of a cell’s processes?
Flow of genetic information
instructions translator
worker
DNA
RNA
Proteins
transcription translation
nucleus
cytoplasm
Where does transcription take place in a
prokaryotic cell?
Comparison of DNA to RNA
RNA
 Single stranded
 Ribose sugar
 Bases A, Uracil, C, G
 A-U, C-G base pairs
 3 types/functions
 Unstable, easily degrades
DNA
 Double stranded
 Deoxyribose sugar
 Bases A, T, C, G
 A-T, C-G base pairs
 1 type/function
 Very stable over time
Types of RNAs
 Messenger RNA (mRNA)-carries the protein building
instructions:
 1 gene=1 mRNA=1 protein
 Ribosomal RNA (rRNA)-this RNA along with other
proteins make up ribosomes (site of protein synthesis)
 Transfer RNA (tRNA)-brings correct amino acid to
the ribosome; binds to mRNA sequence
The Genetic Code
 DNARNAAmino Acids (proteins)
 How many different nucleotides are there?
 How many different amino acids are there?
 How would the nucleotides specify each of the
amino acids?
The Genetic Code Solved!
 There are only 4 nucleotides to specify 20 different amino
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acids
If use 1 nucleotide per amino acid, only can specify 4
Using 2 nucleotides, only can specify 16
If use 3 nucleotides per amino acid, can specify 64 amino
acids
3 nucleotides = 1 codon, 1 codon per amino acid
The Genetic
Code
 The code is universal
 The code is degenerative
 most amino acids have
more than 1 codon. Why?
 Each codon has only 1
meaning, CCA is different
than CUA, and can
specify a different amino
acid
 Has 1 start signal and 3
stop signals
Transcription: DNARNA
• Initiation of Transcription
• RNA pol binds to specific DNA
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sequence called a promoter
which is usually at the
beginning of a gene
DNA unwinds using enzymes
RNA pol adds nucleotides in
5’3’ direction: A-U, C-G
RNA pol falls off at end of
gene, releasing mRNA
transcript
Many RNA pols work
simultaneously to produce
mRNAs
RNA Transcript Processing and
Modification
• Newly synthesized
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transcripts are not ready to
make a protein
3 Modifications are made:
Introns are cut out by
“Splicing”
Genes  exons and introns
Exons  coding sequences
that make a protein
Introns  noncoding (junk)
sequences
RNA Transcript Processing and Modification
5’ cap is added
 Provides an docking area for the ribosome to bind
during translation
 Prevents degradation
3’ Poly A (adenine) Tail is added
 100-200 As added to prevent degradation by enzymes
Similarities Between Replication and
Transcription
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Occur in the nucleus
Use DNA as template to build new strands
Add nucleotides in the 5’3’ direction
Add nucleotides according to base pair rules: A-T,U or
C-G
 Both use helicase and topoisomerase to unwind DNA
Differences Between Replication and
Transcription
Replication
 DNADNA copy
 Copy all of the DNA
 Copy is double stranded
 Use DNA pol
 Use A-T, C-G
Transcription
 DNARNA copy
 Copy only part of DNA
 Copy is single stranded
 Use RNA pol
 Use A-U, C-G
Transcription: DNARNA
The following sequence is to be transcribed:
AATCGGTCGATGG
What is the sequence of the transcript?
AATCGGTCGATGG DNA
UUAGCCAGCUACC RNA
Translation: RNAProteins
 Occurs in the cytoplasm,
inside of ribosomes
(attached to ER)
 Requires a mRNA with a
start/stop codon
 Requires a tRNA
 Requires amino acids
Translation: RNAProteins
Role of tRNA
• tRNA-2 Binding sites
• mRNA binding site
called the anticodon
• UAA CGC AAC mRNA
• AUU GCG UUG tRNA
• amino acid binding site,
to bring the correct
amino acid to growing
polypeptide chain
Translation:
Role of Ribosomes
• Site of protein synthesis
• Composed of large and small
subunits
• During translation:
• Small subunit binds mRNA,
then combines with large
subunit to form intact
ribosome (initiation)
• When small and large
subunits combine, form E
site (exit), P site (peptide),
and an A site (amino acid)
Translation
 Initiation: initiator tRNA
+ small ribosome unit,
mRNA + large ribosome
unit
 Elongation: initiator
tRNA (start codon)  P
site
 Where does the 2nd tRNA
add?
Translation: RNAProteins
 2nd tRNA  A site
 peptide bond  1st +
2nd aa
 Initiator tRNA released
 E site
 2nd tRNA  P site
 3rd tRNA  A site
Translation: Elongation
 Peptide bond  2nd + 3rd
aa
 This is repeated many
times until a stop codon is
reached
 Proteins can have as little
as 30 aa or up to 1000 aa
 12-17 aa are added every
second!!!!
Translation
• Termination: stop codon
is reached, mRNA released,
and ribosome subunits
separate, polypeptide chain
is released
• What happens to the newly
synthesized protein?
• Golgi for processing and
shipping by exocytosis
• Used in the cell it was made
Reading the Genetic Code
• There are 64 codons (61 specify
amino acids, 3 are stops)
• Some amino acids have more
than one codon, ex. arginine,
leucine
• For codon GAC, first find the
first base on code, second, and
finally third; where all lines
intersect that is the amino acid
specified. GAC = aspartate
Genetic Code Problem
What would be the amino acid sequence specified by the
following DNA sequence?
TAC GCT ATA CCC ATT
How many amino acids would be made?
Genetic Code Problem
TAC GCT ATA CCC ATT DNA
AUG CGA UAU GGG UAA RNA
start-arginine-tyrosine-glycine-stop amino acid
(methionine)
Structure of Eukaryotic
Chromosomes
 Chromosomes
(made of DNA) have
proteins that help
tightly package DNA
in the nucleus 
Histones
 They arrange the
DNA around “beads”
called
nucleosomes
 heterochromatin