Fig. 12.3b Dogma Central de la Biología Molecular en células

11/30/2010
Capítulo 11: Estructura de ácidos
nucleicos, replicación del ADN y
estructura de los cromosomas
• DNA from the beginning http://www.dnaftb.org/
– Classical genetics (temas 1-14)
Criterios que debe cumplir el
material genético:
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Dogma
Central de
la Biología
Molecular
en células
eucariotas
Fig. 12.3b
Cytosol
Nucleus
DNA
Transcription
RNA processing
Polypeptide
Pre-mRNA
mRNA
Entry into cytosol
1.
2.
3.
4.
Información
Replicación
Transmisión
Variación
Translation
Ribosome
1
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DNA from the beginning http://www.dnaftb.org/
• Molecules of genetics (temas 15-19) – dar
énfasis a las contribuciones de los siguientes
científicos:
–
–
–
–
–
–
–
–
Friedrich Miescher
Archibald Garrod
George Beadle y Edward Tatum
James Watson
Francis Crick
Erwin Chagaff
Rosalind Franklin
Maurice Wilkins
• 1869- F. Miescher aisla nucleina a
partir de núcleos de células blancas.
• 1900- Nucleina es una molécula larga
compuesta de: azúcar de 5 carbonos,
fosfato y 5 tipos de bases
nitrogenadas (A, T, C, G y U)
• 1920- Se describen 2 tipos de ácidos
nucléicos: RNA y DNA.
¿Proteínas o Ácidos Nucléicos?
• A finales del 1800, científicos
postularon una base bioquímica.
• Investigadores estaban convencidos de
que los cromosomas contenían la
información genética.
• Entre 1920 y 1940 esperaban confirmar
que la porción proteica de los
cromosomas era el material genético.
1920s, Frederick Griffith
Transformación de bacterias
2
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Strains of Streptococcus
penumoniae
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Type S cells
are virulent.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Type R cells
are benign.
Heat-killed
type S cells
are benign.
3
11/30/2010
Pregunta 1
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• ¿Por qué se aislaron bacterias de la sepa
S de la sangre del ratón si en la mezcla
inyectada solo había baterias de la sepa R
vivas y bacterias de la sepa S muertas?
• RESUSITARON ?????
Virulent type S
strain in dead
mouse’s blood
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Treatment
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Virulent type S
strain in dead
mouse’s blood
Living type
R cells have
been
transformed
into virulent
type S cells
by a
substance
from the
heat-killed
type S cells.
Result
Conclusion
1 Control:
Injected living
type S bacteria
into mouse.
Type S cells
are virulent.
2 Control:
Injected living
type R bacteria
into mouse.
Type R cells
are benign.
3 Control:
Injected heatkilled type S
bacteria into
mouse.
Heat-killed
type S cells
are benign.
4
Injected living
type R and
heat-killed
type S bacteria
into mouse.
Virulent type S
strain in dead
mouse’s blood
Living type
R cells have
been
transformed
into virulent
type S cells
by a
substance
from the
heat-killed
type S cells.
4
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• 1944 –Oswald Avery. Colin M. MacLeod
y Maclyn McCarthy
ØGriffith
Descubre el proceso de transformación
bacteriana.
Define un factor transformador ®
ümolécula que continen la información
genética
ües hereditaria
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
± DNase
± RNase
± Protease
± DNase
+ Type R cells
± RNase
± Protease
A
B
C
D
+ DNA
+ DNase
+ DNA
+ RNase
E
+ Type R cells
A
B
C
D
E
A
B
C
D
E
Add
antibody
Control
A
A
B
C
D
E
Control
+ DNA
+ DNA
+ DNase
+ DNA
+ RNase
+ DNA
B
C
D
+ DNA
+ Protease
E
+ DNA
+ Protease
A
B
C
D
E
5
11/30/2010
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Type S cells
in supernatant
Type R cells
in pellet
THE DATA
B
D
A
C
DNA extract
Control
E
DNA extract + RNase
DNA extract + DNase
DNA extract + protease
Centrifuge
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
HYPOTHESIS A purified macromolecule from type S bacteria, which functions as the genetic material, will be able to convert type
R bacteria into type S.
KEY MATERIALS Type R and type S strains of Streptococcus pneumoniae.
Purify DNA from a type S strain.
This involves breaking open cells
and separating the DNA away from
other components by
centrifugation.
± DNase
± RNase
± Protease
+ Type R cells
2 Mix the DNA extract with type R
bacteria. Allow time for the DNA
to be taken up by the type R cells,
converting a few of them to type S.
Also, carry out the same steps but
add the enzymes DNase, RNase, or
protease to the DNA extract, which
digest DNA, RNA, and proteins,
respectively. As a control, don’t add
any DNA extract to some type R cells.
3
Add an antibody, a protein made by
the immune system of mammals,
that specifically recognizes type R
cells that haven’t been
transformed. The binding of the
antibody causes the type R cells to
aggregate.
4
Remove type R cells by
centrifugation. Plate the remaining
bacteria (if any) that are in the
supernatant onto petri plates.
Incubate overnight.
• 1944 –Oswald Avery. Colin M. MacLeod
y Maclyn McCarthy
Conceptual level
Experimental level
1
A
A
B
B
C
C
D
D
E
Add
antibody
A
B
C
D
Control
+ DNA
+ DNA
+ DNase
+ DNA
+ RNase
A
B
C
E
– Purifican el factor transformador.
– Lo identifican como DNA.
+ DNA
+ Protease
D
E
E
Type S cells
in supernatant
Type R cells
in pellet
Centrifuge
5
THE DATA
B
D
A
C
DNA extract
Control
E
DNA extract + RNase
DNA extract + DNase
DNA extract + protease
6
CONCLUSION DNA is responsible for transforming type R cells into type S cells.
7
SOURCE Avery, O.T., MacLeod, C.M., and McCarty, M. 1944. Studies on the Chemical Nature of the Substance Inducing
Transformation of Pneumococcal Types. Journal of Experimental Medicine 79:137–156.
6
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1952, Alfred Hershey y Martha
Chase
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DNA
• Avery, MacLeod and
McCarty/Hershey and
Chase: DNA Wins! • http://www.5min.com/Video/A
very-MacLeod-andMcCartyHershey-and-ChaseDNA-Wins-150948625
• Hershey and Chase
experiment http://highered.mcgrawhill.com/olc/dl/120076/bio21.s
wf
üBacteriófago = virus que infecta bacterias.
üComposición del bacteriófago T2:
Proteína y DNA.
Phage head
(capsid)
Sheath
Tail fiber
Base plate
E. coli cell
T2 genetic
material being
injected into
E. coli
(a) Schematic drawing of
T2 bacteriophage
50 nm
(b) An electron micrograph of T2 bacteriophage
infecting E. coli
b: © Eye of Science/Photo Researchers
Hershey y Chase
Base para desarrollar estrategia
experimental
• Las proteínas presentes en la cubierta
contienen azufre y no contienen fósforo.
• El DNA contiene fósforo y no contiene
azufre.
7
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•Marcar proteínas del fago
•Marcar DNA del fago
35S
+
35S
35S
32P
35S
35S
+
32P
32P
32P
32P
32P
35S
Ø Hershey & Chase
•Proteínas
marcadas con
Ø Hershey & Chase
•DNA marcado con
32P
35S
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Experiment 1
E. coli cells were
infected with
35S-labeled phage
and subjected to
blender treatment.
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Experiment 2
E. coli cells were
infected with
32P-labeled phage
and subjected to
blender treatment.
Bacterial cell
Bacterial cell
Phage DNA
32P-labeled
35S-labeled
sheared
empty phage
phage DNA
Sheared empty phage
Transfer to tube
and centrifuge.
Transfer to tube
and centrifuge.
Supernatant
has 35S-labeled
empty phage.
Supernatant has
unlabeled empty
phage.
Pellet has
E. coli cells
infected with
unlabeled
phage DNA.
Pellet has
E. coli cells
infected with
32P-labeled
phage DNA.
8
11/30/2010
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Extracellular 35S
Total isotope in supernatant (%)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Extracellular 32P
100
80
Using a Geiger counter,
determine the amount of
radioactivity in the supernatant.
Blending removes 80%
of 35S from cells.
60
40
Geiger (radioisotope)
counter
Most of the 32P (65%)
remains with intact cells.
20
0
0
1
2
3
4
5
6
7
8
Agitation time in blender (min)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1 E. coli cells were
infected with
35S-labeled phage
and subjected to
blender treatment.
Bacterial cell
Phage DNA
35S-labeled
sheared
empty phage
2 Transfer to tube
Experiment 2
E. coli cells were
infected with
32P-labeled phage
and subjected to
blender treatment.
Using a Geiger counter,
determine the amount of
radioactivity in the supernatant.
Conclusiones:
Geiger (radioisotope)
counter
Bacterial cell
1. Las proteinas virales se quedan fuera de la
célula huesped.
32P-labeled
phage DNA
Sheared empty phage
and centrifuge.
Transfer to tube
and centrifuge.
Supernatant
has 35S-labeled
empty phage.
Supernatant has
unlabeled empty
phage.
Pellet has
E. coli cells
infected with
unlabeled
phage DNA.
3
Pellet has
E. coli cells
infected with
32P-labeled
phage DNA.
4 THE DATA
Total isotope in supernatant (%)
Experiment 1
100
Extracellular
35S
Extracellular
32P
2. El DNA viral es inyectado dentro de la célula
huesped.
80
Blending removes 80%
of 35S from cells.
60
40
Most of the 32P (65%)
remains with intact cells.
20
0
0
1
2
3
4
5
Agitation time in blender (min)
6
7
8
3. El DNA inyectado dirige la producción de
nuevas partículas de virus.
4. Material hereditario = DNA.
9
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Pregunta 2
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o Para demostrar que el DNA (y no las proteínas)
era el material genético, bacteriófagos T4 fueron
marcados con 32P o 35S y antes de infectar
bacterias E. coli.
o ¿Qué marca el 32 P?
o ¿Qué marca el 35S?
o ¿Cómo este experimento demostró que el
DNA era el material genético?
Evidencia circunstancial
1947, Erwin Chargaff
10
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Pregunta 3
üAnálisis de la composición de DNA de distintos
organismos.
üLa composición del DNA varía de especie en especie.
o ¿Qué revelan estos datos?
üDiversidad molecular = diversidad de especies.
üLas bases nitrogenadas están presentes en una razón
característica.
üRegla de Chargaff
A=T
G=C
A+G = C+T
purinas = pirimidinas
A+T ≠ C+G
Niveles de
estructura
How Genetics Got a Chemical
Education
ERWIN CHARGAFF
Annals of the New York Academy of
Sciences (1979) 325, 345-360.
http://post.queensu.ca/~forsdyke/bioinfo1.
htm
1.Monómeros - nucleótidos
2.Cadena sencilla
3.Cadena doble (doble
hélice)
4.Arreglo de DNA +
proteínas en células
compone los
cromosomas.
– Cromatina
5.Genoma – todo el
material genético de un
organismo
11
11/30/2010
RESOLVIENDO
LA ESTRUCTURA
DEL DNA
45
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Pyrimidines
(single ring)
Purines
(double ring)
Base
O–
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O
P
NH2
O
N
O
O
Phosphate
CH2
H H
OH
CH6
N
H
N
H
O
N
H H
H
N
H
H
Adenine (A)
H
Thymine (T)
NH2
O
Deoxyribose
O
N
N
N
H
H
NH2
H
N
H
N
N
H
Guanine (G)
N
O
H
Cytosine (C)
Nucleótidos en DNA
12
11/30/2010
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Base
O–
O
P
CH3 5
NH2
N
O
O
Phosphate
CH2
H H
OH
N
H H
H
N
H
6
H
Uracil (U)
NH2
O
N
N
H
N
O–
H
Adenine (A)
OH
Ribose
N
H
H
NH2
H
H
Guanine (G)
N3
H
H
N
H
O
O
4
P
O
O–
N
N
O
O
H
Cytosine (C)
Nucleótidos en RNA
Phosphate
CH
O
5´
4´
H
H
3´
OH
1N
H
Thymine
2 O
1´
H
2´
H
H
Deoxyribose
Pregunta 4
o Describe los componentes de un nucleótido.
o ¿Cuántos nucleótidos pueden encontrarse
en el DNA, cuántos en el RNA?
Menciónalos.
o ¿En qué se diferencian las bases
nitrogenadas?
Monómero
Polímero
Nucleótidos
Acido nucléico
ØAcido nucléico = Cadenas lineales de
nucleótidos unidos por enlaces fosfodiestéricos .
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Formación del enlace fosfodiestérico
Ø1950
üDNA es aceptado como el material
genético.
üSe conoce la composición de los
ácidos nucléicos.
üNo se conoce la estructura
tridimensional.
ØLinus Pauling (California)
ØMaurice Wilkins y Rosalind Franklin
(Londres)
~1951- Linus Pauling – physical
chemist
~1951- Wilkins
and Franklin
• nature of the chemical bond
– resonance theory proposed
that some molecules
"resonate" between different
structures
• use of X-rays to examine the
molecular structure of
crystals
• X-ray diffraction photographs of DNA
– Description of a-helical
polypeptide structure.
http://www.achievement.org/aut
odoc/photocredit/achievers/pa
u0-029
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Descubrimientos de Franklin
Rosalind Franklin (1920 - 1958)
• Esqueleto de azúcar-fosfato hacia
afuera de la molécula
• Dos configuraciones distintas (A y B)
• Doble hélice
• Datos cuantitativos de la forma y
tamaño de la doble hélice (3.4 nm
cada vuelta 2 nm de ancho).
• http://www.accessexcellence.org/RC/AB/
BC/Rosalind_Franklin.html
Secret of photo 51
• http://www.pbs.org/wgbh/nova/photo51/
• ¿Cómo parean las bases?
2-nm
15
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Evidencia circunstancial
1947, Erwin Chargaff
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3´ end
5 ´ end
HO
H
HH
Adenine
N
H
O
O
O P O CH2
–
5´ phosphate
O
H
N
O
H H H
H
–
N
N
H H
O
O
CH2 O P O
O
H
–
H
HH
O
N
N
H
Guanine H H
CH3
H
N
H
Thymine
H
O
O
N
H
O P O CH2 O
N
N
H
O
HH H H
HH
N
HH
O
N
O
–
CH2 O P O
H
N
O
N
O
H
–
Cytosine
H
O
O P O CH2
O
–
O
HH
3´ hydroxyl
OH
N
N
O
N
Guanine
H
H
N
H
N
N
O
CH2 O P O
O
O
H
N
H
H
HH
H
–
–
N
O
H
HH
H
H
Cytosine
H
Hydrogen
bond
3´ end
5 ´ end
(b) Base pairing
Key Features
• Two strands of DNA form a double helix.
• The bases in opposite strands hydrogenbond according to the AT/GC rule.
• The 2 strands are antiparallel.
• There are ~10 nucleotides in each
strand per complete turn of the helix.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
5´ end
4
Bases
3´ end
fosfato
2
O
pentosa
5
1
5 ´ end
3´ end
O
O P O CH2
2
O
H
N
N
N
O
H
H
H
O P O CH2
–
3´ hydroxyl
H
N
O
–
CH2 O P O
O
O
H
Guanine
H
H
N
N
N
H
H
N
N
O
HH
O
N
H
O
HH
N
H
H
H
O
O
H
H
N
Cytosine
H
N
H
O
N
One nucleotide
0.34 nm
5´ end
1
3´ end
2 nm
(a) Double helix
O
–
CH2 O P O
H
O
–
N
O
H
H
HH
H
OH
H
Cytosine
H
HH
H
Hydrogen
bond
3´ end
Base Nitrogenada
3
OH
1
4
O
pentosa
5
5
H
N
H
Thymine
O P O CH2
Complete turn
of the helix
3.4 nm
H
Guanine
CH3
O
fosfato
fosfato
2
Base Nitrogenada
O
pentosa
2
H
–
O
N
H H
O
H H H
O
CH2 O P O
N
–
3
3
1
N
O
H
O
H
–
4
4
O
pentosa
O
HH
H
N
N
H
O
O
Base Nitrogenada
5
1
fosfato
5
Base Nitrogenada
Sugar-phosphate
backbone
fosfato
3
2
2
O
pentosa
1
5´ phosphate
3
4
O
pentosa
4
5
N
H
Hydrogen bond
–
fosfato
HO
H
H
Adenine
Base Nitrogenada
Base Nitrogenada
3
OH
Cadenas
antiparalelas
5 ´ end
(b) Base pairing
Key Features
• Two strands of DNA form a double helix.
• The bases in opposite strands hydrogenbond according to the AT/GC rule.
• The 2 strands are antiparallel.
• There are ~10 nucleotides in each
strand per complete turn of the helix.
16
11/30/2010
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Apareamiento Watson y Crick
Apareamiento por complementaridad
Major groove
Minor groove
• Explica la regla de Chargaff, A=T y C=G.
• No hay restricción en cuanto la secuencia de
bases
• Estabilidad entre las cadenas de DNA =
– Puentes de hidrógeno
Major groove
• Sugiere mecanismo de replicación.
– Cadenas son complementarias.
– Mecanismo de templado.
Minor groove
Replicación del DNA
Watson y Crick proponen:
• Puentes de H se rompen
de manera secuencial.
• Separación de las hebras.
• Cada cadena puede servir
como templado.
• Genes son copiados en
base a un pareo
específico entre las bases
complementarias.
• Modelo semiconservativo.
1953- Watson y Crick
• Watsongenética
• Crick- físico,
experto en
cristalografía
de rayos X
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5. Resume en una oración la
contribución de los siguientes
científicos
– Friedrich Miescher
– Erwin Chagaff
– Alfred Hershey y Martha Chase
– Frederick Griffith
– Oswald Avery. Colin M. MacLeod y
Maclyn McCarthy
18