Download Chapter 16 The Molecular Basis of Inheritance

The Molecular Basis of
Inheritance
Chapter 16
A. P. Biology
Mr. Knowles
Liberty Senior High
• Concept 16.1: DNA is the genetic material
• Early in the 20th century
– The identification of the molecules of inheritance
loomed as a major challenge to biologists
Hammerling’s Acetabularia
Exp.
• Danish biologist, Joachim Hammerling
in the 1930’s.
• Used a unicellular green algae,
Acetabularia.
• Proved that the hereditary material is in
the nucleus.
DNA is a “Transforming
Principle”
• Frederick Griffith, 1928, showed
that dead bacteria could be
transformed into living cells.
• Used 2 strains of Pneumococcus
bacteria, one pathogenic and the
other nonpathogenic.
Fig. 16.2 - Transformation
Identified the “Transforming
Agent”
• Oswald Avery- in 1944, he
separated and purified the different
organic compounds of the bacteria
and identified the DNA as
responsible for the transformation
effect.
Fig. 16.3
Support for Avery
• Alfred Hershey and Martha Chasein 1952, used bacteriophage (T2) as a
model.
• They radiolabeled the protein coat
with 35 S and the DNA with 32P in
order to “see” which of the molecules
actually entered the cell and produced
more phage.
A Bacteriophage
Fig. 16.4: Hershey-Chase Exp.
Other Contributors
• Friedrich Miescher- 1869,
extracted “nuclein”- nucleic acid
from human cells and fish sperm.
• Erwin Chargaff- A = T and the G =
C, called Chargaff’s Rule; equal
proportion of purines and
pyrimidines; amt. varied from
species to species.
The Chemistry of DNA
• A nucleotide = 1. PO4,
2. a five carbon sugar,
3. a nitrogencontaining base.
• Phosphodiester Bonds- 2 P-O-C
bonds link nucleotides.
What is a Nucleotide?
• Subunits of DNA/RNA are
Nucleotides = nitrogenous base +
deoxy- or ribose sugar (5 carbons)
+ PO4
• Purines: Adenine and Guanine
• Pyrimidines: Cytosine, Thymine,
Uracil (in RNA)
CUT = Py
AG = Pur.
Fig. 16.8
Monosaccharides of Nucleic
Acids
Adenosine Monophosphate
• Base = adenine
• In DNA, sugar = deoxyribose
(In RNA, sugar = ribose)
• A phosphate group, PO4
• The Nucleotide = AMP
Adenosine Monophosphate
Adenosine Triphosphate (ATP)
Fig. 16.5
• Maurice Wilkins and Rosalind Franklin
– Were using a technique called X-ray crystallography
to study molecular structure
• Rosalind Franklin
– Produced a picture of the DNA molecule using this
technique
Figure 16.6 a, b
(a) Rosalind Franklin
(b) Franklin’s X-ray diffraction
Photograph of DNA
• Franklin had concluded that DNA
– Was composed of two antiparallel sugar-phosphate
backbones, with the nitrogenous bases paired in the
molecule’s interior
• The nitrogenous bases
– Are paired in specific combinations: adenine with
thymine, and cytosine with guanine
DNA is Antiparallel
5 end
O
OH
Hydrogen bond
P
–O
3 end
OH
O
O
A
T
O
O
O
CH2
P
–O
O
H2C
O
–O
P
O
O
G
O
C
O
O
CH2
P
O
O–
O
P
H2C
O
O
C
O
G
O
O
O
CH2
P
–O
O–
O
O
O–
O
P
H2C
O
O
A
O
T
O
CH2
OH
3 end
O
O–
P
O
Figure 16.7b
(b) Partial chemical structure
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
O
5 end
Three Dimensional Structure of
DNA
• Rosalind Franklin- X-ray crystallography
of DNA- showed that DNA was in a helix
with PO4 and sugars to the outside.
• James Watson and Francis Crick- took
Franklin’s data- in April 23, 1953, and
deduced the structure of DNA. Won the
Nobel Prize along with Maurice Wilkins.
Fig. 16.6
CUT = Py
AG = Pur.
Fig. 16.8
• Watson and Crick reasoned that there must be
additional specificity of pairing
– Dictated by the structure of the bases
• Each base pair forms a different number of
hydrogen bonds
– Adenine and thymine form two bonds, cytosine
and guanine form three bonds
Watson and Crick
Fig. 16.7
Characteristics of DNA
• All chains of DNA and RNA
have a 5’ PO4 end and a 3’ OH
end.
• Base sequences are written in a
5’ to 3’ direction.
• Ex. 5’ pGpTpCpCpApT-OH 3’
Characteristics of DNA
• Base pairs stabilize the
molecule by forming H-bonds.
• Antiparallel Strands5’----------------3’
3’----------------5’
• Strands are complementary.
• In DNA replication
– The parent molecule unwinds, and two new daughter
strands are built based on base-pairing rules
T
A
T
A
T
A
C
G
C
G
C
T
A
T
A
T
A
A
T
A
T
A
T
G
C
G
C
G
C
G
A
T
A
T
A
T
C
G
C
G
C
G
T
A
T
A
T
A
T
A
T
A
T
C
G
C
G
C
A
G
(a) The parent molecule has two
complementary strands of DNA.
Each base is paired by hydrogen
bonding with its specific partner,
A with T and G with C.
Figure 16.9 a–d
(b) The first step in replication is
separation of the two DNA
strands.
(c) Each parental strand now
serves as a template that
determines the order of
nucleotides along a new,
complementary strand.
(d) The nucleotides are connected
to form the sugar-phosphate
backbones of the new strands.
Each “daughter” DNA
molecule consists of one parental
strand and one new strand.
• DNA replication is semiconservative
– Each of the two new daughter molecules will have
one old strand, derived from the parent molecule,
and one newly made strand
Parent cell
(a)
(b)
(c)
Figure 16.10 a–c
Conservative
model. The two
parental strands
reassociate
after acting as
templates for
new strands,
thus restoring
the parental
double helix.
Semiconservative
model. The two
strands of the
parental molecule
separate,
and each functions
as a template
for synthesis of
a new, complementary strand.
Dispersive
model. Each
strand of both
daughter molecules contains
a mixture of
old and newly
synthesized
DNA.
First
replication
Second
replication
• Experiments performed by Meselson and Stahl
– Supported the semiconservative model of DNA
replication
EXPERIMENT Matthew Meselson and Franklin Stahl cultured E. coli bacteria for several generations
on a medium containing nucleotide precursors labeled with a heavy isotope of nitrogen, 15N. The bacteria
incorporated the heavy nitrogen into their DNA. The scientists then transferred the bacteria to a medium with
only 14N, the lighter, more common isotope of nitrogen. Any new DNA that the bacteria synthesized would be
lighter than the parental DNA made in the 15N medium. Meselson and Stahl could distinguish DNA of different
densities by centrifuging DNA extracted from the bacteria.
1
Bacteria
cultured in
medium
containing
15N
2
Bacteria
transferred to
medium
containing
14N
RESULTS
3
DNA sample
centrifuged
after 20 min
(after first
replication)
4
DNA sample
centrifuged
after 40 min
(after second
replication)
Less
dense
More
dense
The bands in these two centrifuge tubes represent the results of centrifuging two DNA samples from the flask
Figure 16.11 in step 2, one sample taken after 20 minutes and one after 40 minutes.
CONCLUSION
Meselson and Stahl concluded that DNA replication follows the semiconservative
model by comparing their result to the results predicted by each of the three models in Figure 16.10.
The first replication in the 14N medium produced a band of hybrid (15N–14N) DNA. This result eliminated
the conservative model. A second replication produced both light and hybrid DNA, a result that eliminated
the dispersive model and supported the semiconservative model.
First replication
Conservative
model
Semiconservative
model
Dispersive
model
Second replication
Fig. 16.11- Meselson and Stahl
DNA Replication Experiment
Getting Started: Origins of
Replication
• The replication of a DNA molecule
– Begins at special sites called origins of replication
(special sequence, AT rich), where the two strands
are separated.
Bidirectional Replication in Bacteria- One
Origin
Eukaryotic DNA Replication
• Replicates the DNA on a chromosome
in a discrete section- Replication Unit.
(about 100 kbp long).
• Prokaryotic Replication: 500
nucleotides/ second.
• Eukaryotic Replication: 50
nucleotides/second.
Problem?
• If eukaryotic replication is 100 N/
sec. and there are 3.0 X 108 N/
chromosome, how long would it take
to replicate one human genome?
• Ans: 3 X 106 sec. = 34.7 days! How
long does it actually take to go
through S phase?
• S phase = 8 hours
• How?
Multiple Origins of
Replication
• Each replication unit on a
chromosome has its own origin of
replication.
• Multiple units can be replicating at
any given time.
• Each chromosome has numerous
replication forks.
• A eukaryotic chromosome
– May have hundreds or even thousands of replication
origins
Origin of replication
Parental (template) strand
0.25 µm
Daughter (new) strand
1
Replication begins at specific sites
where the two parental strands
separate and form replication
bubbles.
Bubble
Replication fork
2 The bubbles expand laterally, as
DNA replication proceeds in both
directions.
3
Eventually, the replication
bubbles fuse, and synthesis of
the daughter strands is
complete.
Two daughter DNA molecules
(a) In eukaryotes, DNA replication begins at many sites along the giant
DNA molecule of each chromosome.
Figure 16.12 a, b
(b) In this micrograph, three replication
bubbles are visible along the DNA of
a cultured Chinese hamster cell (TEM).
A eukaryotic chromosomes - have hundreds or even thousands of
replication origins.
Fig. 16.12
• Elongation of new DNA at a replication fork
– Is catalyzed by enzymes called DNA polymerases,
which add nucleotides to the 3 end of a growing
strand
Fig. 16.13
The Problem with DNA
Replication
• DNA Polymerase can only build in
the 5’ to 3’ direction.
• Since the parent strands are
antiparallel, the new strands are
synthesized in opposite directions.
• DNA polymerases add nucleotides
– Only to the free 3end of a growing strand.
• Along one template strand of DNA, the leading
strand:
– DNA polymerase III can synthesize a
complementary strand continuously, moving toward
the replication fork
• To elongate the other new strand of DNA, the
lagging strand:
– DNA polymerase III must work in the direction
away from the replication fork
• The lagging strand
– Is synthesized as a series of segments called
Okazaki fragments, which are then joined together
by DNA ligase
Fig. 16.14
DNA Replication
• Leading Stand- elongates
toward the fork, 5’ to 3’
• Lagging Strand- elongates
way from fork; synthesized
discontinuously in short
segments-Okazaki Fragments.
Priming DNA Synthesis
• DNA polymerases cannot initiate the synthesis
of a polynucleotide
– They can only add nucleotides to the 3 end
• The initial nucleotide strand
– Is an RNA or DNA primer made by – Primase.
• Only one primer is needed for synthesis of the
leading strand
– But for synthesis of the lagging strand, each
Okazaki fragment must be primed separately.
1
Primase joins RNA nucleotides
into a primer.
3
5
5
3
Template
strand
RNA primer
3
5
3
DNA pol III adds DNA nucleotides to the
primer, forming an Okazaki fragment.
2
5
3
1
After reaching the next
RNA primer (not shown),
DNA pol III falls off.
Okazaki
fragment
3
3
5
1
5
4
After the second fragment is
primed. DNA pol III adds DNA
nucleotides until it reaches the
first primer and falls off. 5
3
5
3
2
5
1
DNA pol 1 replaces the
RNA with DNA, adding to
the 3 end of fragment 2.
5
3
6
5
1
DNA ligase forms a bond
between the newest DNA
and the adjacent DNA of
fragment 1.
5
3
Figure 16.15
3
2
7
The lagging strand
in this region is now
complete.
3
2
1
Overall direction of replication
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
5
Single-stranded Binding (SSB)
Proteins
Helicase
Fig. 16.16
• A summary of DNA replication
Overall direction of replication
1 Helicase unwinds the
parental double helix.
2 Molecules of singlestrand binding protein
stabilize the unwound
template strands.
3 The leading strand is
synthesized continuously in the
5 3 direction by DNA pol III.
DNA pol III
Lagging
Leading
strand Origin of replication strand
Lagging
strand
OVERVIEW
Leading
strand
Leading
strand
5
3
Parental DNA
4 Primase begins synthesis
of RNA primer for fifth
Okazaki fragment.
5 DNA pol III is completing synthesis of
the fourth fragment, when it reaches the
RNA primer on the third fragment, it will
dissociate, move to the replication fork,
and add DNA nucleotides to the 3 end
of the fifth fragment primer.
Replication fork
Primase
DNA pol III
Primer
4
DNA ligase
DNA pol I
Lagging
strand
3
2
6 DNA pol I removes the primer from the 5 end
of the second fragment, replacing it with DNA
nucleotides that it adds one by one to the 3 end
of the third fragment. The replacement of the
last RNA nucleotide with DNA leaves the sugarphosphate backbone with a free 3 end.
Figure 16.16
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
1
3
5
7 DNA ligase bonds
the 3 end of the
second fragment to
the 5 end of the first
fragment.
Replication Fork
Some Animations of DNA
Replication
•http://www.wiley.com/college/pratt/0471393878/student/animatio
ns/dna_replication/index.html - tutorial and self quiz
•http://www.wiley.com/legacy/college/boyer/0470003790/animatio
ns/replication/replication.htm - tutorial and self quiz
•http://www.youtube.com/watch?v=5VefaI0LrgE&feature=related
DNA Replication
Several Enzymes (+12) involved:
Helicase + ATP - unwinds the DNA and stabilizes it.
Single-stranded binding Proteins-stabilize the DNA and
prevent base pairing.
Primase -makes an RNA primer.
DNA Polymerase III- builds new complementrary strands
in a 5’--->3’ direction.
DNA Polymerase I – removes the primer from the 5’ end
of Okazaki fragment and replaces it with DNA.
DNA Ligase – bonds the 3’ end of the second Okazaki
fragment with the 5’ end of the first, and so on.
Proofreading and Repairing
DNA
• DNA polymerases proofread newly made DNA
– While the initial pairing errors are high as nucleotides
come in.
– Proofreading replaces any incorrect nucleotides; error
rate of about 1 in 10 billion nucleotides.
• In mismatch repair of DNA
– Repair enzymes correct errors in base pairing
– Defective repair enzymes – xeroderma pigmentosum
Nucleotide Excision Repair
Fig. 16.17
Replicating the Ends of DNA Molecules
• The ends of eukaryotic chromosomal DNA
– Get shorter with each round of replication
5
End of parental
DNA strands
Leading strand
Lagging strand
3
Last fragment
Previous fragment
RNA primer
Lagging strand
5
3
Primer removed but
cannot be replaced
with DNA because
no 3 end available
for DNA polymerase
Removal of primers and
replacement with DNA
where a 3 end is available
5
3
Second round
of replication
5
New leading strand 3
New lagging strand 5
3
Further rounds
of replication
Figure 16.18
Shorter and shorter
daughter molecules
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Fig. 16.18
• Eukaryotic chromosomal DNA molecules
– Have at their ends nucleotide sequences that repeat
(100 – 1,000), called telomeres, that postpone the
erosion of genes near the ends of DNA molecules
Figure 16.19
1 µm
Show me the Replication!
DNA Packaging
• Eukaryotic DNA is packaged into
nucleosomes- 146 bp of DNA
wrapped around histones.
• Condenses the DNA into a
nucleus.
• Protects the DNA from exposure.
How many genes are on a
chromosome?
• Human Chromosome #22composed of 33.4 megabases
(Mb), identified 545 genes, 298 of
these were unknown.
• 100-200 more genes may be
identified!
• The rest is “junk” DNA.
Chromosomes have...
• Humans have about 3 X 109 bp of
DNA.
• Conservative estimate is 30,000
genes.
• Therefore, the average gene is
100,000 bp long.
One Gene-One Enzyme
Hypothesis
• 1941- Beadle and Tatum- created mutants
in the fungus Neurospora using X-rays.
• Exposed wild-type spores to the mutagen.
• Change growth from a complete to minimal
media.
• Mutants no longer able to synthesize
essential organics (e.g. amino acids)
One Gene-One Enzyme
• Beadle and Tatum- found several mutants.
• Found a different site for each enzyme.
• Each mutant had a different mutation at a
different site (enzyme) on the chromosome.
• Each mutant had a defect in a different
enzyme.
• Concluded: one gene encodes one enzyme.
One Gene-One Enzyme
Hypothesis
• Genetic traits are expressed as a result of
the activities of enzymes.
• Many enzymes contain multiple subunits.
• Each subunit encoded by a differernt
gene.
• Now, referred to as one gene-one
polypeptide.
How does DNA Encode
Proteins?
QuickTime™ and a
Video decompressor
are needed to see this picture.
QuickTime™ and a
Animation decompressor
are needed to see this picture.
QuickTime™ and a
Animation decompressor
are needed to see this picture.
Some Differences
between Prokaryotic and
Eukaryotic Protein
Synthesis
Prokaryotes
Eukaryotes
• Have introns• Lack introns- no
mRNA is spliced.
mRNA
•
mRNA
completely
processing.
formed before
• Begin translation
translation.
before
• mRNA has a 5’
transcription is
methyl G cap
finished.
added
Prokaryotes
Eukaryotes
• mRNA begins at • mRNA has a poly
an AUG start
A tail added at the
codon, no cap.
3’.
• Multiple genes on • A single gene on
one mRNA.
one mRNA.
• 70S ribosome
•
80S
ribosome
used.
used.
Other Differences in Gene Organization
• Tandem Clusters- several hundred
genes organized together; Ex. rRNA
genes.
• Multigene Families- genes very
different from each other, but encode
similar proteins; Ex. globin genes.
• Pseudogenes- silent copies of genes
inactivated by mutations.
Transposons
Acute Promyelocytic
Leukemia- Several
Translocation Events; one is
between #7 and #15.
DNA Replication
Prokaryotes
Eukaryotes
• single, circular
• multiple, linear
chromosome.
chromosomes.
• one single origin • several origins on
on the
each chromosome.
chromosome.
• rate of about 50• rate of over 1,000 100 nt/second
nt/second.