Download Chapter 16

Chapter 16:
The Molecular Basis of Inheritance
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.1 Watson and Crick with their DNA model
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.2 Can the genetic trait of pathogenicity
be transferred between bacteria?
EXPERIMENT Bacteria of the “S” (smooth) strain of Streptococcus pneumoniae are pathogenic because they
have a capsule that protects them from an animal’s defense system. Bacteria of the “R” (rough) strain lack a capsule
and are nonpathogenic. Frederick Griffith injected mice with the two strains as shown below:
Living S
(control) cells
Living R
Heat-killed
(control) cells (control) S cells
Mixture of heat-killed S cells
and living R cells
RESULTS
Mouse dies
Mouse healthy
Mouse healthy
Mouse dies
Living S cells
are found in
blood sample.
CONCLUSION Griffith concluded that the living R bacteria had been transformed into pathogenic S bacteria by an
unknown, heritable substance from the dead S cells.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.3 Viruses infecting a bacterial cell
Phage
head
Tail
Tail fiber
Bacterial
cell
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
100 nm
DNA
Figure 16.4 Is DNA or protein the genetic material
of phage T2?
EXPERIMENT
In their famous 1952 experiment, Alfred Hershey and Martha Chase used radioactive sulfur
and phosphorus to trace the fates of the protein and DNA, respectively, of T2 phages that infected bacterial cells.
1
2 Agitated in a blender to 3
Mixed radioactively
separate phages outside
labeled phages with
the bacteria from the
bacteria. The phages
bacterial cells.
infected the bacterial cells.
Phage
Centrifuged the mixture 4 Measured the
so that bacteria formed
radioactivity in
a pellet at the bottom of
the pellet and
the test tube.
the liquid
Radioactivity
(phage protein)
in liquid
Radioactive Empty
protein
protein shell
Bacterial cell
Batch 1: Phages were
grown with radioactive
sulfur (35S), which was
incorporated into phage
protein (pink).
DNA
Phage
DNA
Centrifuge
Radioactive
DNA
Batch 2: Phages were
grown with radioactive
phosphorus (32P), which
was incorporated into
phage DNA (blue).
Pellet (bacterial
cells and contents)
Centrifuge
Pellet
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Radioactivity
(phage DNA)
in pellet
RESULTS
Phage proteins remained outside the bacterial cells during infection, while phage DNA entered the cells.
When cultured, bacterial cells with radioactive phage DNA released new phages with some radioactive phosphorus.
CONCLUSION
Hershey and Chase concluded that DNA, not protein, functions as the T2 phage’s genetic material.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.5 The structure of a DNA strand
Sugar-phosphate
backbone
5 end
5
O P O CH2
Nitrogenous
bases
CH3
O–
O–
4 H
H
3
O
H
O
1
N
N
H
H
O
Thymine (T)
H
2
H
O
O
P O
CH2
O–
H
H
N
O
H
H
H
H
Adenine (A)
H
H
O
P O
H
N
N
H
O
N
N
H
CH2
O–
H
O
H
N H
N
H
N
H
H
O
Cytosine (C)
H
O
O
5
P O CH2
H
O
1
O– 4 H
H
Phosphate H
H
2
3
H
OH
Sugar (deoxyribose)
3 end
N
O
N
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
N
N H
N H
H
Guanine (G)
DNA nucleotide
Figure 16.6 Rosalind Franklin and her X-ray
diffraction photo of DNA
(a) Rosalind Franklin
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(b) Franklin’s X-ray diffraction
Photograph of DNA
Figure 16.7 The double helix
G
5 end
C
O
A
T
–O
T
A
P
OH
O
H2C
Hydrogen bond
O
O
C
C
3.4 nm
G
–O
P
O
O
G
T
A
–O
A
T
O
CH2
G
O
P
O
O
H2C
O
O
A
T
P
O–
O
T
A
OH
3 end
G
C
0.34 nm
A
CH2
O
O–
P
O
O
O
C
O
O
O
O
O–
P
C
O
H2C
A
CH2
O
O
H2C
O
G
T
O
O
T
A
C
–O
P
A
T
1 nm
G
3 end
OH
T
(a) Key features of DNA structure
O
CH2
O
O–
P
O
O
5 end
(b) Partial chemical structure
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(c) Space-filling model
Unnumbered Figure p. 298
Purine + Purine: too wide
Pyrimidine + pyrimidine: too narrow
Purine + pyrimidine: width
Consistent with X-ray data
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.8 Base pairing in DNA
H
N
N
N
N
Sugar
O
H
H
CH3
N
N
N
O
Sugar
Thymine (T)
Adenine (A)
H
O
N
N
Sugar
N
H
H
N
N
N
N
N
H
Guanine (G)
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
H
O
Sugar
Cytosine (C)
Figure 16.9 A model for DNA replication: the basic
concept (layer 1)
A
T
C
G
T
A
A
T
G
C
(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.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.9 A model for DNA replication: the basic
concept (layer 2)
A
T
A
T
C
G
C
G
T
A
T
A
A
T
A
T
G
C
G
C
(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.
(b) The first step in replication is
separation of the two DNA
strands.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Figure 16.9 A model for DNA replication: the basic
concept (layer 3)
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
C
G
T
A
A
C
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.
(b) The first step in replication is
separation of the two DNA
strands.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
T
(c) Each parental strand now
serves as a template that
determines the order of
nucleotides along a new,
complementary strand.
Figure 16.9 A model for DNA replication: the basic
concept (layer 4)
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.
(b) The first step in replication is
separation of the two DNA
strands.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(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.
Figure 16.10 Three alternative models of DNA replication
Parent cell
(a) Conservative
model. The two
parental strands
reassociate
after acting as
templates for
new strands,
thus restoring
the parental
double helix.
(b) Semiconservative model.
The two strands
of the parental
molecule
separate,
and each functions
as a template
for synthesis of
a new, complementary strand.
(c) Dispersive
model. Each
strand of both
daughter molecules contains
a mixture of
old and newly
synthesized
DNA.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
First
replication
Second
replication
Figure 16.11 Does DNA replication follow the
conservative, semiconservative, or dispersive model?
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
2 Bacteria
cultured in
medium
containing
15N
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
in step 2, one sample taken after 20 minutes and one after 40 minutes.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
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
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
Second replication
Figure 16.12 Origins of replication in eukaryotes
Origin of replication
1 Replication begins at specific sites
where the two parental strands
separate and form replication
bubbles.
Bubble
Parental (template) strand
Daughter (new) strand
0.25 µm
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.
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
(b) In this micrograph, three replication
bubbles are visible along the DNA of
a cultured Chinese hamster cell (TEM).
Figure 16.14 Synthesis of leading and lagging
strands during DNA replication
1 DNA pol Ill elongates
DNA strands only in the
5
3 direction.
3
5
Parental DNA
2 One new strand, the leading strand,
can elongate continuously 5
3
as the replication fork progresses.
5
3
Okazaki
fragments
2
1
3
5
3 The other new strand, the
lagging strand must grow in an overall
3
5 direction by addition of short
segments, Okazaki fragments, that grow
5
3 (numbered here in the order
they were made).
DNA pol III
Template
strand
3
Leading strand
Lagging strand
2
Template
strand
1
DNA ligase
Overall direction of replication
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
4 DNA ligase joins Okazaki
fragments by forming a bond between
their free ends. This results in a
continuous strand.
Figure 16.15 Synthesis of the lagging strand
1
Primase joins RNA nucleotides
into a primer.
3
5
5
3
Template
strand
2
RNA primer
3
5
3
DNA pol III adds DNA nucleotides to the
primer, forming an Okazaki fragment.
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
3
2
DNA ligase forms a bond
between the newest DNA
and the adjacent DNA of
fragment 1.
5
3
5
1
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
Figure 16.17 Nucleotide excision repair of DNA damage
1 A thymine dimer
distorts the DNA molecule.
2 A nuclease enzyme cuts
the damaged DNA strand
at two points and the
damaged section is
removed.
Nuclease
3 Repair synthesis by
a DNA polymerase
fills in the missing
nucleotides.
DNA
polymerase
DNA
ligase
Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings
4 DNA ligase seals the
Free end of the new DNA
To the old DNA, making the
strand complete.