LECTURE 1 Introduction to DNA and RNA (Chapter 09) 1

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

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4 H 1 He

Fusion fueled by GRAVITY

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It’s why the sun shines!

• The source of all life’s “problems” is ENTROPY• The “solutions” are stored as genes in DNA

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DnA iS DiGiTal Information

A,C,G,T

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• DNA is a highly ORDERED molecule that…• Fights entropy to perpetuate itself• By…Building its own survival machines

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• And…passing them on• Geneticists view this as…

• a “blind” undirected process

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Life = Problems

“I had trouble in getting to Solla Sollew where they never have troubles at least very few…” –Dr. Suess

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There is no happy retirement from life. Forget it!

I hope we can get little

Davey off meth…

I hate my thighs…

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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”

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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

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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

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• 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

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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

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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

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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

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Afterseveraldays

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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

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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”

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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!

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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

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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

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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.

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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

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• 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

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34

Infects E. coli cells by binding to the outside and injecting their DNA

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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

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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

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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

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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

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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

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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

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4′ 1′

2′3′

5′

4′ 1′

3′ 2′

N

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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

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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

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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

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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

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H

N

NN

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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

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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

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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

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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

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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

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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

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Chargaff pioneered many of the biochemical techniques for the isolation, purification and measurement of nucleic acids from living cells

Erwin Chargaff (1950)

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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

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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

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Figure 9.13

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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

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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

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Sugar

–O

O O

OP

Base

Sugar

Base

Sugar

O

OOP

Base

Sugar

BaseO–

Mg2+

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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

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(b) Original model of the DNA double helix

(a) Watson and Crick

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© 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

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Figure 9.16

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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

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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

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(b) Space-filling model of DNA(a) Ball-and-stick model of DNA

Minorgroove

Majorgroove

Minorgroove

Majorgroove

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© 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

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E.g. Restriction enzymes like Eco RI

Palindrome

Primitive immune system

4 + 2 Rule

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“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

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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

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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

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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

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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