Upload
wesley-richard
View
226
Download
0
Tags:
Embed Size (px)
Citation preview
LECTURE 1
Introduction to DNA and RNA
(Chapter 09)
1
Colored background: Review on your own
A BRIEF INTRODUCTION TO LIFE
• Genetics: Study of the structure, function, and transmission of genes
• Only living organisms have genes• To understand genetics, we will start with the
question: “What is Life?”– Characteristics shared by all living forms but not
by non-living forms
2
Group Exercise
What is the role of genes in life?
• DNA carries information to create order• Life “got going” as a negative entropy machine• Why? Because it did! • It’s a marvel!
3
HIGH ENTROPY (DISORDER)
Arrow of time
LOW ENTROPY(ORDER)
4
Time
22 deg C(surrounding
air)
100º C 22º C
22.5 deg C(surrounding
air)
ENERGY
5
HIGH ENTROPY (DISORDER)
Arrow of time
LOW ENTROPY(ORDER)
Can’t turn back
time so use
ENERGY instead!
Where does the energy come from?
6
High potential energy; Low entropy
Short mean λ
Longer mean λ
Life!
• Fusion reactions in sun’s core– Hydrogen to helium with release of photons
– Mass converted to energy (E = mc2)– 600 million tons hydrogen fused to 596 million tons helium
per second– 4 million tons hydrogen converted to 3.9 x 1026 Joules
7
4 H 1 He
Fusion fueled by GRAVITY
8
It’s why the sun shines!
• The source of all life’s “problems” is ENTROPY• The “solutions” are stored as genes in DNA
9
DnA iS DiGiTal Information
A,C,G,T
10
11
• DNA is a highly ORDERED molecule that…• Fights entropy to perpetuate itself• By…Building its own survival machines
12
• And…passing them on• Geneticists view this as…
• a “blind” undirected process
13
Life = Problems
“I had trouble in getting to Solla Sollew where they never have troubles at least very few…” –Dr. Suess
14
There is no happy retirement from life. Forget it!
I hope we can get little
Davey off meth…
I hate my thighs…
15
Practice Problem:(Identify all correct answers)
From a geneticist’s point-of-view, which of the following statements is true?a. Life is trying to evolveb. A bee is a survival machine for its genesc. DNA is a digital information moleculed. Life requires an energy source to fight entropye. When you graduate from college, all your problems
will disappear
MOLECULAR GENETICS
• The study of the structure and function of DNA at the molecular level– Highly ordered digital information– Survives to perpetuate itself
• “Selfish gene”
16
To do this DNA must…
• 1. Carry the information necessary to encode its survival machine– Reverse entropy– Perpetuate itself
• 2. Be able to transmit this information to daughter cells– Reproduction
• 3. Be capable of change– Subject to mutation
17
18
Could life exist if DNA did not have these three properties?
What would happen to this primitive cell if its DNA did not encode its structures/functions, it could not reproduce, and/or it’s DNA was not capable of change?
IDENTIFICATION OF DNA AS THE GENETIC MATERIAL
• Early humans noticed that traits were inherited between parents and offspring
19
• The substance that held this “heritable information” was mysterious
• Mendel’s work in mid-1800s revealed patterns (“laws”) that govern its inheritance
• In 1901, Hugo de Vries and Carlos Correns linked these laws to chromosome behavior during meiosis
20
We’ll study Mendel’s Laws and their connection to meiosis later on this semester
• Early biochemists studied the chemical components in the nucleus to try to identify the heritable information molecule
• Nucleus contains proteins and nucleic acids– Proteins are more chemically diverse than nucleic
acids– Known to do almost everything in the cell– Most biologists hypothesized that the heritable
material was a protein
• Starting in 1928, three seminal experiments shattered this hypothesis– Revealed DNA as the heritable molecule
21
1. Frederick Griffith Experiments with Streptococcus pneumonia (1928)
• Griffith studied a bacterium (pneumococci) now known as Streptococcus pneumoniae
• S. pneumoniae comes in two strains– S Smooth
• Secrete a polysaccharide capsule– Protects bacterium from the immune system
of animals
• Produce smooth colonies on solid media– R Rough
• Unable to secrete a capsule• Produce colonies with a rough
appearance22
• Experimental tools:– R cells– S cells– Mice– Agar plating
23
Figure 9.1
Living type S bacteria wereinjected into a mouse.
Mouse died
Deadtype S
Livetype R
Mouse died Mouse survived Mouse survived
Living type R bacteria wereinjected into a mouse.
Heat-killed type S bacteriawere injected into a mouse.
Living type R and heat-killedtype S bacteria were injectedinto a mouse.
Type S bacteria were isolatedfrom the dead mouse.
No living bacteria were isolatedfrom the mouse.
No living bacteria were isolatedfrom the mouse.
Type S bacteria were isolatedfrom the dead mouse.
(a) Live type S (b) Live type R (c) Dead type S (d) Live type R + dead type S
Afterseveraldays
Afterseveraldays
Afterseveraldays
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Afterseveraldays
24
Griffith concluded that something from the dead type S bacteria was transforming type R bacteria into type S Change was stable and heritable He called this process transformation
Provided an experimental model for studying the molecule that carries heritable information Griffith did not know what it was His experiment suggested that it was resistant to heat
First experimental indication that the heritable molecule was not a protein
Most proteins are sensitive to heat
25
We will be transforming bacteria in lab
As an aside… We now know that the formation of the
capsule is guided by the bacteria’s DNA Transformed R bacteria acquired a gene from
the dead lysed S bacteria that directs the cell how to make a polysachharide capsule
Variation exists in ability to make capsule (R cells carry a non-functional allele for the gene)
The gene required to create a capsule is replicated and transmitted from mother to daughter cells
Once R cells acquired it they passed the trait to their “offspring”
26
27
28
Practice Problem(Only one correct answer):
If Griffith had mixed heat-killed R cells with living S cells and injected the mixture into his mice, what would have happened?a. Mice alive; only smooth colonies on plateb. Mice dead; only smooth colonies on platec. Mice alive; only rough colonies on plated. Mice dead; only rough colonies on platee. Mice dead; Both smooth and rough colonies on plate
2. The Experiments of Avery, MacLeod and McCarty (1944)
• Avery, MacLeod and McCarty realized that Griffith’s observations could be used to identify the genetic material
• They carried out their experiments in the 1940s– At that time, it was known that DNA, RNA, proteins and
carbohydrates are the major constituents of living cells
• They prepared cell extracts from type S cells and purified each type of macromolecule
• Then repeated Griffith’s experiment without the mice!
29
X
• Replaced the mice with an antibody that binds to R cells
• Served the same purpose as the mouse’s immune system– Rather than killing the cells it allowed them to pull them out of
the cell mixtures prior to plating
• Experimental tools:– R cells– S cells– Antibody to R cells– DNase– Rnase– Proteinase– Agar plating
30
Figure 9.2 Avery et al. conducted the following experiments
To further verify that DNA, and not a contaminant (RNA or protein), is the genetic material
Mix Mix Mix Mix
Type Rcells
Type Rcells
Type SDNA
extract
Type Rcells
Type SDNA
extract+
DNase
Type Rcells
Type SDNA
extract+
RNase
Type Rcells
Type SDNA
extract+
protease
Transformed Transformed Transformed
1 2 3 4 5
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Allow sufficient time for the DNA to be taken up by the type R bacteria. Only a small percentage of the type R bacteria will be transformed to type S.
Add an antibody that aggregates type R bacteria (that have not been transformed). The aggregated bacteria are removed by gentle centrifugation.
Plate the remaining bacteria on petri plates. Incubate overnight.
31
32
Practice Problem(Circle all correct answers):
Why did Avery, McLeod, and McCarty use an antibody to R cells in their experiment?a. To protect their mice from dyingb. So they could detect smooth colonies if they were presentc. So they could repeat the essential features of Griffith’s
experiment without having to sacrifice animalsd. Because only R cells have a polysaccharide capsulee. Because only S cells have a polysaccharide capsule
3. Hershey and Chase Experiment with Bacteriophage T2 (1952)
• Alfred Hershey and Marsha Chase provided further evidence that DNA is the genetic material
33
• Experimental tools:
– Bacteriophage T2– S35 (labels proteins)– P32 (labels DNA)– E. coli– A Waring brand kitchen blender
Head
Tail fiber
Base plate
Sheath
DNA(inside thecapsid head)
Made up of protein
Inside the
capsid
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
34
Infects E. coli cells by binding to the outside and injecting their DNA
35
Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.
36
37
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
80%Blending removes 80%of 35S from E. coli cells.
35%Most of the 32P (65%)remains with intactE. coli cells.
Tota
l iso
top
e in
su
per
nat
ant
(%)
Agitation time in blender (min)
100
80
60
40
20
00 1 2 3 4 5 6 7 8
Data from A. D. Hershey and Martha Chase (1952) Independent Functions of Viral Protein and Nucleic Acid in Growth ofBacteriophage. Journal of General Physiology 36, 39–56.
Extracellular 35S
Extracellular 32P
The Data Most of the 35S was found in the supernatant
But only a small percentage of 32P
These results suggest that DNA is injected into the bacterial cytoplasm during infection
Data is not conclusive since less than 100% of the DNA or protein ended up in the cell or supernatant
Data is consistent with the hypothesis that DNA is the genetic material38
39
Practice Problem(Circle all correct answers):
Which of the following were used in the Hershey-Chase experiment?a. E. colib. Bacteriophage T2c. S cellsd. A kitchen blendere. 35S
In 1956, A. Gierer and G. Schramm isolated RNA from the tobacco mosaic virus (TMV), a plant virus Purified RNA caused the same lesions as intact TMV
viruses Therefore, the viral genome is composed of RNA
Since that time, many RNA viruses have been found Refer to Table 9.1
RNA Functions as the Genetic Material in Some Viruses
40
41
9.2 NUCLEIC ACID STRUCTURE – On Your Own
• DNA and RNA are large macromolecules with several levels of complexity
– 1. Nucleotides form the repeating unit of nucleic acids– 2. Nucleotides are linked to form a linear strand of RNA
or DNA– 3. Two strands can interact to form a double helix– 4. The 3-D structure of DNA results from folding and
bending of the double helix. Interaction of DNA with proteins produces chromosomes within living cells
– Refer to fig. 9.6
42
Figure 9.6
CC
G
AA
GA
TG
GT
T TA
A
GC
AT
AT
T TA
GC
Nucleotides
Single strand
Double helix
Three-dimensional structure
A T CG
CA A
T CA TG
CA
AT C
GC
A
G CA T
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
43
The nucleotide is the repeating structural unit of DNA and RNA
It has three components A phosphate group A pentose sugar A nitrogenous base
Refer to Figure 9.7
Nucleotides
44
9-24Figure 9.7
Phosphate group Sugars
D-Deoxyribose (in DNA)
Purines(double ring)
Pyrimidines(single ring)
Bases
OO
O–
O–
P
H
H
H
HO
OHOHOCH2
HH
D-Ribose (in RNA)
H
OH
H
HO
OHOHOCH2
HH
Uracil (U) (in RNA)Thymine (T) (in DNA)
Cytosine (C)
Adenine (A)
Guanine (G)
NH2
N
HH
H
H
H
O
N
43
2
156
7
89
43
21
5
6
O
CH3 H
43
2
156
7
8
9
5′
O
NH2
H
HN
N
N
N
NH2
N
N
H
N
N
N
H
H O
N43
21
5
6
O
H
H O
43
21
5
6
N
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
4′ 1′
2′3′
5′
4′ 1′
3′ 2′
N
45
HH
H
OCH2
Base
(a) Repeating unit of deoxyribonucleic acid (DNA)
(b) Repeating unit of ribonucleic acid (RNA)
Phosphate
Deoxyribose
HH
OH
OCH2
Base
Phosphate
Ribose
5′
OH OH
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
4′ 1′
3′ 2′
5′
4′ 1′
3′ 2′
OO
O
P
O–
O
O
P
O–
O
HH HH
Figure 9.8 The structure of nucleotides found in (a) DNA and (b) RNA
A, G, C or T
These atoms are found within individual nucleotides However, they are removed when nucleotides join together to make
strands of DNA or RNA
A, G, C or U
46
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
Base + sugar nucleoside Example
Adenine + ribose = Adenosine Adenine + deoxyribose = Deoxyadenosine
Base + sugar + phosphate(s) nucleotide Example
Adenosine monophosphate (AMP) Adenosine diphosphate (ADP) Adenosine triphosphate (ATP)
Refer to Figure 9.9
47
Figure 9.9
Base always attached here
Phosphates are attached here
Adenosine
Adenosine monophosphate
Adenosine diphosphate
Adenine
Phosphate groups
Phosphoester bond
Ribose
H
OP CH2
O–
OO P
O–
O O O
–O P
O–
H
OHHO
O
H
2′3
1′4′5′
Adenosine triphosphate
NH2
N
H
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
H
N
NN
48
Nucleotides are covalently linked together by phosphodiester bonds A phosphate connects the 5’ carbon of one nucleotide to
the 3’ carbon of another Therefore the strand has directionality
5’ to 3’ In a strand, all sugar molecules are oriented in the same
direction
The phosphates and sugar molecules form the backbone of the nucleic acid strand The bases project from the backbone
49
Figure 9.10
NH2
N O
N
O
N O
N
Adenine (A)
Guanine (G)
Thymine (T)
BasesBackbone
Cytosine (C)
O
HH
H
H
HH
OOO
O–
P CH2
O–
HH
H
H
H
HH
OOO
O
P CH2
O–
NH2
N
N
H
N
N
HH
H
HH
OOO
O
P CH2
O–
NH2
HN
N
N
H
N
HH
HOH
HH
OOO
O
P CH2
O–Singlenucleotide
Phosphodiesterlinkage
Sugar (deoxyribose)
Phosphate
3′
5′
5′4′ 1′
2′3′
5′4′ 1′
2′3′
5′4′ 1′
2′3′
5′4′ 1′
2′3′
CH3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
50
In 1953, James Watson and Francis Crick elucidated the double helical structure of DNA
The scientific framework for their breakthrough was provided by other scientists including Linus Pauling Rosalind Franklin and Maurice Wilkins Erwin Chargaff
Key Events Leading to the Discovery of the Double Helix
51
In the early 1950s, he proposed that regions of protein can fold into a secondary structure a-helix
Linus Pauling
To elucidate this structure, he built ball-and-stick models
52
53
Figure 9.11
CC
C
CCC
O
C
O
OC
OH
H
H
H
N
CN
NN
CC
CC O
O
HHNN
C
C
C
CCC
OC
O
C
O
OC
OH
HH
N
NN
H
N
H
N
Carbonyl oxygenAmide hydrogen
Hydrogen bond
(a) An a helix in a protein
Wet DNA fibers
X-ray beam
The pattern represents theatomic array in wet fibers.
X rays diffractedby DNA
(b) X-ray diffraction of wet DNA fibers
She worked in the same laboratory as Maurice Wilkins
She used X-ray diffraction to study wet fibers of DNA
Rosalind Franklin (early 1950s)
The diffraction pattern is interpreted (using mathematical theory)
This can ultimately provide information concerning the
structure of the molecule
Figure 9.12b
54
She made marked advances in X-ray diffraction techniques with DNA
The diffraction pattern she obtained suggested several structural features of DNA Helical More than one strand 10 base pairs per complete turn
55
56
Chargaff pioneered many of the biochemical techniques for the isolation, purification and measurement of nucleic acids from living cells
Erwin Chargaff (1950)
57
It was known that DNA contained the four bases: A, G, C and T
• Experimental tools:– E. coli and many other organisms– DNA extraction protocol– Protease– Perchloric acid– Paper chromatography– Spectroscopy
58
The Hypothesis
An analysis of the base composition of DNA in different species may reveal important features about the structure of DNA– Chargaff analyzed the base composition of
DNA isolated from many different species
Testing the Hypothesis
Refer to Figure 9.13
59
Figure 9.13
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Experimental level Conceptual level
For each type of cell, extract the chromosomal material. This can bedone in a variety of ways, including theuse of high salt, detergent, or mild alkalitreatment. Note: The chromosomescontain both DNA and protein.
1.
Remove the protein. This can be done in several ways, including treament withprotease.
2.
Hydrolyze the DNA to release the bases from the DNA strands. A common wayto do this is by strong acid treatment.
3.
Separate the bases by chromatography. Paper chromatography provides an easyway to separate the four types of bases.(The technique of chromatography isdescribed in the Appendix.)
4.
Extract bands from paper into solutions and determine the amounts of each baseby spectroscopy. Each base will absorblight at a particular wavelength. Byexamining the absorption profile of asample of base, it is then possible tocalculate the amount of the base.(Spectroscopy is described in theAppendix.)
5.
Compare the base content in the DNA from different organisms.
6.
Solution of chromosomalextract
DNA
DNA + proteins
Individual bases
Origin
Protease
AcidA
A
AAA A
AA
CC
C CC CC
GG GG G GG
TT TT T
T
T
T
G
G
G
C
C
C
y
60
The Data
61
Interpreting the Data
The data shown in Figure 9.13 are only a small sampling of Chargaff’s results
The compelling observation was that Percent of adenine = percent of thymine Percent of cytosine = percent of guanine
This observation became known as Chargaff’s rule It was a crucial piece of evidence that Watson and Crick
used to elucidate the structure of DNA
62
Familiar with all of these key observations, Watson and Crick set out to solve the structure of DNA They tried to build ball-and-stick models that
incorporated all known experimental observations
A critical question was how the two (or more strands) would interact An early hypothesis proposed that DNA strands interact
through phosphate-Mg++ cross-links Refer to Figure 9.14
Watson and Crick (1953)
63
Figure 9.14
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Sugar
–O
O O
OP
Base
Sugar
Base
Sugar
O
OOP
Base
Sugar
BaseO–
Mg2+
64
WRONG! (But why is it appealing?)
65
Watson and Crick weren’t the only ones making incorrect guesses for the structure of DNA!
Triple helix
Watson and Crick went back to the ball-and-stick units
They then built models with the Sugar-phosphate backbone on the outside Bases projecting toward each other
They first considered a structure in which bases form H bonds with identical bases in the opposite strand ie., A to A, T to T, C to C, and G to G
Model building revealed that this also was incorrect
66
They then realized that the hydrogen bonding of A to T was structurally similar to that of C to G So they built ball-and-stick models with AT and CG
interactions between the two DNA strands These were consistent with all known data about DNA
structure Refer to Figure 9.15
Published model in April 1953 (Nature) Watson, Crick and Maurice Wilkins were
awarded the Nobel Prize in 1962 Rosalind Franklin died in 1958, and Nobel prizes
are not awarded posthumously
67
(b) Original model of the DNA double helix
(a) Watson and Crick
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
© Barrington Brown/Photo Researchers
© Hulton|Archive by Getty Images
Figure 9.15 68
69
Practice Problem(One correct answer):
A species of frog has a genome that contains 12% G. What percent T does it contain?a. 12%b. 38%c. 44%d. 76%e. 88%
General structural features (Figures 9.16 & 9.17)
The DNA Double Helix
Two strands are twisted together around a common axis
There are 10 bases and 3.4 nm per complete turn of the helix
The two strands are antiparallel One runs in the 5’ to 3’ direction and the other 3’ to 5’
The helix is right-handed As it spirals away from you, the helix turns in a
clockwise direction70
General structural features (Figures 9.16 & 9.17)
The double-bonded structure is stabilized by 1. Hydrogen bonding between complementary
bases A bonded to T by two hydrogen bonds C bonded to G by three hydrogen bonds
2. Base stacking Within the DNA, the bases are oriented so that the
flattened regions are facing each other
71
Figure 9.16
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
H
N HN
N
N
G
NH2
H
P
S
P
SP
5¢end
3¢end
H NH2
N
N
HH
H
HH
OOO
O
P CH2
O
HH
H
OH
HH
OOO
O
P CH2
O
HH
H
HH
OOO
O
P CH2
O
CH2
HH
H
HH
OOO
O
P
O
T
Key Features
• Two strands of DNA form a right-handed double helix.
• The bases in opposite strands hydrogen bond according to theAT/GC rule.
• The 2 strands are antiparallel with regard to their 5′ to 3′ directionality.
• There are ~10.0 nucleotides in each strand per complete 360° turn ofthe helix.
2 nm
One nucleotide0.34 nm
One complete turn 3.4 nm
TA
G C
T AP
P
P
P
P
S
S
S
S
S
S
S
S
S
A
C
G
C G
C G
G C
G C
GC
G C
C
P
PS
P
P
P
P
P
S
S
S
S
S
PP
P
P
P
P
P
P
P
P
S
S S
S
S
S
S
S
S
P
S
S
P
P
P
S
S
S
S
P
3¢
5¢
G
S3¢5¢
S
A
P
P
C
T A
O
N
N
N
NA
H
H NH2
N
O
H
N
CH3
H
T
H
HH2N
N
N C
O
3¢end
5¢end
H
H
HH
OOO
O
PCH2
O
H
H
H
H
HH
OOO
O
PCH2
O
H
H
HH
OOO
O
PCH2
O
H
H
HH
OOO
O
PCH2
O
HO
N
O
H
N
CH3
H
T
O
NH N
N
N
G
H2N
H
H
H
N
N
N
N A
H
H2N H
-
-
-
-
-
-
-
-
-
-
72
73
Please note that due to differing operating systems, some animations will not appear until the presentation is viewed in Presentation Mode (Slide Show view). You may see blank slides in the “Normal” or “Slide Sorter” views. All animations will appear after viewing in Presentation Mode and playing each animation. Most animations will require the latest version of the Flash Player, which is available at http://get.adobe.com/flashplayer.
General structural features (Figures 9.16 & 9.17)
There are two asymmetrical grooves on the outside of the helix
1. Major groove 2. Minor groove
74
(b) Space-filling model of DNA(a) Ball-and-stick model of DNA
Minorgroove
Majorgroove
Minorgroove
Majorgroove
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
© Laguna Design/Photo Researchers
Figure 9.17 75
Proteins can bind within these grooves They can thus interact with a particular sequence of bases to
“read” the DNA sequence
76
E.g. Restriction enzymes like Eco RI
Palindrome
Primitive immune system
4 + 2 Rule
77
“Homodimerizes” then binds in successive major grooves to cut each DNA strand
78
To calculate average distance between cut sites:
4(# bp in recognition site)
For Eco RI = 46 = 4,096 bp/cut
To calculate number of cuts/genome:
(bp in genome)/4(# bp in recognition site)
Eco RI cuts the human genome:
(3.2 x 109 bp)/4096 bp/cut = 780,000 cuts
To fit within a living cell, the DNA double helix must be extensively compacted into a 3-D conformation This is aided by DNA-binding proteins
Refer to 9.20
The Three-Dimensional Structure of DNA
79
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 9.20
DNA wound around histone
proteins
Radial loops(300 nm in diameter)
Metaphasechromosome
DNA(2 nm in diameter)
Nucleosomes(11 nm in diameter)
Histoneprotein
Each chromatid(700 nm in diameter)
30 nm fiber
80
81
1 meter DNA/cell1013 cells/human
1013 meters of DNA/human3.8 x 108 meters from Earth to Moon
If uncoiled, DNA from a single human would stretch from earth to moon and back ~13,000 times!!
The primary structure of an RNA strand is much like that of a DNA strand Refer to Figure 9.21 vs. 9.10
RNA strands are typically several hundred to several thousand nucleotides in length
In RNA synthesis, only one of the two strands of DNA is used as a template
RNA Structure
82
Figure 9.21
Adenine (A)
Guanine(G)
Uracil (U)
BasesBackbone
Cytosine (C)
O
HHHH
OOO
O–
P CH2
O–
HHHH
OOO
O
P CH2
O–
NH2
H
N
HHHH
OOO
O
P CH2
O–
H
H
HH
OH
HH
OOO
O
P CH2
O–
Sugar (ribose)
Phosphate
5′
4′ 1′
2′3′
5′
4′ 1′
2′3′
5′
4′ 1′
2′3′
5′
4′ 1′
2′3′OH
OH
OH
OH
RNAnucleotide
Phosphodiesterlinkage
3′
5′
NH2
OH
H
H
O
NH O
N
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
NN
N
N
N
N
N N
N
NH2
H
H
83
Although usually single-stranded, RNA molecules can form short double-stranded regions This secondary structure is due to complementary base-
pairing A to U and C to G
This allows short regions to form a double helix
RNA double helices typically Are right-handed Have the A form with 11 to 12 base pairs per turn
Different types of RNA secondary structures are possible Refer to Figure 9.22
84
A U
A U
U A
G C
C G
C G
A U
U A
U A
C G
C G
C G
C G
C G
A
A
U
U
G
G
C
C
C
(a) Bulge loop (b) Internal loop (c) Multibranched junction (d) Stem-loop
G
C
C
G
U
AA
U
G
C
G
C
C
GA
UA U
A U
G C
A U
U A
G C
C G
C G
G
C
G
C
C
GA
U
A
U
G
C
C
G
C
GA
U
A
U
A
U
U
G
G
C
C
CA U
A U
G C
C G
A
AAU
U
U
U
G
G
C
C
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Figure 9.22
Also called hair-pin
Complementary regions
Non-complementary regions
Held together by hydrogen bonds
Have bases projecting away from double stranded regions
85
Many factors contribute to the tertiary structure of RNA For example
Base-pairing and base stacking within the RNA itself
Interactions with ions, small molecules and large proteins
Figure 9.23 depicts the tertiary structure of tRNAphe
The transfer RNA that carries phenylalanine
Molecule contains single- and double-stranded regions
These spontaneously fold and interact to
produce this 3-D structure
Figure 9.23(a) Ribbon model
3′ end(acceptorsite)5′ end
Double helix
Double helix
Anticodon
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
86