LECTURE 1 Introduction to DNA and RNA (Chapter 09)

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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

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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! LOW ENTROPY (ORDER)

HIGH ENTROPY (DISORDER)

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22 deg C (surrounding air)

22.5 deg C (surrounding air)

Time

100º C

22º C

4

LOW ENTROPY (ORDER)

HIGH ENTROPY (DISORDER)

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Where does the energy come from?

Short mean λ Longer mean λ High potential energy; Low entropy

Life! 6

• 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

Fusion fueled by GRAVITY

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

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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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I hate my thighs…

I hope we can get little Davey off meth…

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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. b. c. d. e.

Life is trying to evolve A bee is a survival machine for its genes DNA is a digital information molecule Life requires an energy source to fight entropy When you graduate from college, all your problems will disappear

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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 17

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?

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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 We’ll study Mendel’s Laws and their connection to meiosis later on this semester

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• 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 appearance 22

• Experimental tools: – – – –

R cells S cells Mice Agar plating

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Living type S bacteria were injected into a mouse.

Living type R bacteria were injected into a mouse.

Heat-killed type S bacteria were injected into a mouse.

Living type R and heat-killed type S bacteria were injected into a mouse. Live type R

Dead type S

After several days

Mouse died

After several days

Mouse survived

After several days

Mouse survived

After several days

Mouse died

Type S bacteria were isolated from the dead mouse.

No living bacteria were isolated from the mouse.

No living bacteria were isolated from the mouse.

Type S bacteria were isolated from the dead mouse.

(a) Live type S

(b) Live type R

(c) Dead type S

(d) Live type R + dead type S

Figure 9.1

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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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

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

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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. b. c. d. e.

Mice alive; only smooth colonies on plate Mice dead; only smooth colonies on plate Mice alive; only rough colonies on plate Mice dead; only rough colonies on plate Mice dead; Both smooth and rough colonies on plate 28

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!

X 29

• 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



Avery et al. conducted the following experiments

Figure 9.2



To further verify that DNA, and not a contaminant (RNA or protein), is the genetic material

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1

Type R cells

2

3

Type S DNA extract

Type R cells

Mix

4

Type S DNA extract + DNase

Type R cells

Mix

5

Type S DNA extract + RNase

Type R cells

Mix

Type R cells

Type S DNA extract + protease Mix

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.

Transformed

Transformed

Transformed

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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 dying b. So they could detect smooth colonies if they were present c. So they could repeat the essential features of Griffith’s experiment without having to sacrifice animals d. Because only R cells have a polysaccharide capsule e. Because only S cells have a polysaccharide capsule

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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

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DNA (inside the capsid head) Head Sheath

Inside the capsid

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

Made up of protein

Tail fiber

Base plate 34

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The Data

Most of the 35S was found in the supernatant

But only a small percentage of 32P

Total isotope in supernatant (%)

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100 80

Extracellular 35S

80% Blending removes 80% of 35S from E. coli cells.

Extracellular 32P

35% Most of the 32P (65%) remains with intact E. coli cells.

60 40 20 0 0

1

2

3

4

5

6

7

8

Agitation time in blender (min) Data from A. D. Hershey and Martha Chase (1952) Independent Functions of Viral Protein and Nucleic Acid in Growth of Bacteriophage. Journal of General Physiology 36, 39–56.



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 material 38

Practice Problem (Circle all correct answers): Which of the following were used in the HersheyChase experiment? a. b. c. d. e.

E. coli Bacteriophage T2 S cells A kitchen blender 35S

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RNA Functions as the Genetic Material in Some Viruses 

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 40

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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 42

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C Nucleotides A

TC

G

C

AA

TC

G

C

AA

TC

G

C

A A T

Single strand

C G

C A T GT T A A

G T A C GA T

C A T GT T A A

G

Double helix

Figure 9.6

Three-dimensional structure

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Nucleotides 

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 44

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Bases Phosphate group

Sugars

Purines (double ring) NH2

5′

HOCH2 H

O

H

H

P

H

5

7

H

N

9

4

O CH3

1N

5

2

6

3

H

N

D-Deoxyribose (in DNA)

1′

4′

H

H 3′

HO

N

OH

O H

H

2′

OH

D-Ribose (in RNA)

6 5

7

H

8

N

9

5

2

6

O

4

H

4

1

N

3N

2

O

H

Thymine (T) (in DNA)

O HOCH2

1

H

3N

H

Adenine (A)

5′

4

O

N

H

O–

Figure 9.7

6

8

H

HO O

N

2′

3′

O

OH 1′

4′

O–

Pyrimidines (single ring)

Uracil (U) (in RNA)

NH 2 H 1N

H 5

2 3

N

H

6

NH2

H

4

1

3N 2

N

O

H Guanine (G)

Cytosine (C)

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9-24



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 T

A, G, C or U

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O

O P

O

Base O

O–

Phosphate

CH2

5′ 4′

H

O

O

H 3′

1′

H

H

2′

OH H Deoxyribose (a) Repeating unit of deoxyribonucleic acid (DNA) Figure 9.8

P

Base O

O–

Phosphate

CH2

5′

4′

H

H 3′

O H

1′

H

2′

OH OH Ribose (b) Repeating unit of ribonucleic acid (RNA)

The structure of nucleotides found in (a) DNA and (b) RNA 46



Base + sugar  nucleoside 



Base + sugar + phosphate(s)  nucleotide 



Example  Adenine + ribose = Adenosine  Adenine + deoxyribose = Deoxyadenosine

Example  Adenosine monophosphate (AMP)  Adenosine diphosphate (ADP)  Adenosine triphosphate (ATP)

Refer to Figure 9.9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

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Adenosine triphosphate Adenosine diphosphate

Adenosine monophosphate Adenosine Adenine

Phosphoester bond

NH2

N

N

H O –O

P O–

O O

P

O O

O–

P

N O

O–

CH2

5′ 4′

Phosphate groups Phosphates are attached here

Figure 9.9

O 1′

H

H

N

H

H

Base always attached here

2′

3

HO

OH Ribose 48



Nucleotides are covalently linked together by phosphodiester bonds 



Therefore the strand has directionality  



A phosphate connects the 5’ carbon of one nucleotide to the 3’ carbon of another 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

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Backbone

Bases O

5′

CH3

N

Thymine (T) H

–

O O

P

O –

O

CH2 5′ 4′ H H

O

N

O 1′ H 2′ H H

3′

NH2 N

Phosphodiester linkage

N

Adenine (A)

H N

O O

P O–

O

CH2 5′ 4′ H H

O 1′ H 2′ H H

3′

P O–

NH2 H

N

H

O O

N

O

CH2 5′ 4′ H H

Cytosine (C) O

N

O 1′ H 2′ H H

3′

Guanine (G) O H

N

N

H N

O

Single nucleotide

O

P

O

CH2 5′ O– 4′ H Phosphate H 3′ OH

Figure 9.10

3′

N

NH2

O 1′ H 2′ H H

Sugar (deoxyribose)

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Key Events Leading to the Discovery of the Double Helix 



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

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Linus Pauling 

In the early 1950s, he proposed that regions of protein can fold into a secondary structure 



a-helix

To elucidate this structure, he built ball-andstick models

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C

H

C N O C H H C N C C N O Carbonyl C oxygen O H C H Amide C N O C N hydrogen O C H C N C C NO C Hydrogen O H C H bond N C O C N O C H H C N C C NO H

O

Figure 9.11

(a) An a helix in a protein

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Rosalind Franklin (early 1950s) 

She worked in the same laboratory as Maurice Wilkins

X rays diffracted by DNA Wet DNA fibers X-ray beam

The pattern represents the atomic array in wet fibers.



She used X-ray diffraction to study wet fibers of DNA

(b) X-ray diffraction of wet DNA fibers

Figure 9.12b

The diffraction pattern is interpreted (using mathematical theory) This can ultimately provide information concerning the structure of the molecule 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

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Erwin Chargaff (1950) 

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



It was known that DNA contained the four bases: A, G, C and T

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• 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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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. y Conceptual level

Experimental level

1. For each type of cell, extract the chromosomal material. This can be done in a variety of ways, including the use of high salt, detergent, or mild alkali treatment. Note: The chromosomes contain both DNA and protein.

Solution of chromosomal extract

DNA + proteins

Protease

DNA

2. Remove the protein. This can be done in several ways, including treament with protease.

Acid

Individual bases

T

A

3. Hydrolyze the DNA to release the bases from the DNA strands. A common way to do this is by strong acid treatment.

G

A C

C

G

T

C G

4. Separate the bases by chromatography. Paper chromatography provides an easy way to separate the four types of bases. (The technique of chromatography is described in the Appendix.)

5. Extract bands from paper into solutions and determine the amounts of each base by spectroscopy. Each base will absorb light at a particular wavelength. By examining the absorption profile of a sample of base, it is then possible to calculate the amount of the base. (Spectroscopy is described in the Appendix.)

A A A A C

C

G

G G

C

G

A

A

C C C C G

G

G

Origin T T

T T

T

T

6. Compare the base content in the DNA from different organisms.

Figure 9.13

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The Data

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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

Watson and Crick (1953) 

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 63

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Base

Base Sugar

Sugar O

Base

O P

O Sugar

–O

Mg 2+

O–

O

Base

P

O

O Sugar

Figure 9.14

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

Watson and Crick weren’t the only ones making incorrect guesses for the structure of DNA!

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



Watson and Crick went back to the ball-and-stick units They then built models with the  



They first considered a structure in which bases form H bonds with identical bases in the opposite strand 



Sugar-phosphate backbone on the outside Bases projecting toward each other

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

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© Barrington Brown/Photo Researchers

(a) Watson and Crick

© Hulton|Archive by Getty Images

Figure 9.15

(b) Original model of the DNA double helix

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Practice Problem (One correct answer): A species of frog has a genome that contains 12% G. What percent T does it contain? a. b. c. d. e.

12% 38% 44% 76% 88%

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The DNA Double Helix 

General structural features (Figures 9.16 & 9.17) 





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 direction 70



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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2 nm

Key Features

5 P 3 S P P A S

• Two strands of DNA form a right-handed double helix. • The bases in opposite strands hydrogen bond according to the AT/GC rule. • The 2 strands are antiparallel with regard to their 5′ to 3′ directionality.

P

S

G G

S

S

P

C

C

O P O

C P

O P

S

P

P S

G

C P

P S

S A

T P

S

G C

S

P

-

CH2

NH2

-

CH2 O P

C

O

H

H N

N

O

O H

-

O

N

G

H2N

H

P S

O

O

O

H

T

H

N

H2N N

H

H H N

A N

-

O

O

H

H

H

H H

H

CH2 O P

N

H

S

O P

-

O

H

G

O

H

H

N H

H

H2N

N

CH2

O

O

O

H

One nucleotide 0.34 nm

S

-

CH2 O P

CH3

O

C

O

O

O

P

P

H H

H H

H

CH2 H

H H

N

N O P

N

O

O

P

A

N H

N

NH2

O

C

H

H N

OH

H

H H

H H

O

-

O CH2 O P O

3 end

G

H

N

O

G

C

S

O

O

H

H

S

P S PS P S C

H

G S

S

T G

O

H

H H

H

O

H O

H

A

P

S

N

N

HO

N

H

O

P

P

H

A

O

P

T

P

N

O

H

S

One complete turn 3.4 nm

-

CH2

T

NH2

H

H

G

S

O

N

H

H

P S P S P S A T S S

3 end H

CH3 O

S

P

5 end

P

C

P S S

• There are ~10.0 nucleotides in each strand per complete 360° turn of the helix.

S

P

-

O

5 end

P

S

5

3

Figure 9.16

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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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Minor groove

Minor groove

Major groove

Major groove © Laguna Design/Photo Researchers

(a) Ball-and-stick model of DNA

Figure 9.17

(b) Space-filling model of DNA

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

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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“Homodimerizes” then binds in successive major grooves to cut each DNA strand

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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 78

The Three-Dimensional Structure of DNA 

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

79

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Radial loops (300 nm in diameter)

Metaphase chromosome 30 nm fiber

Nucleosomes (11 nm in diameter) DNA (2 nm in diameter)

Histone protein

DNA wound around histone proteins

Each chromatid (700 nm in diameter)

Figure 9.20

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81

RNA Structure 

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 82

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Backbone

Bases

5′

O H

H

N

Uracil (U)

O– O

P O–

O

5′ 4′

CH2

H

H

O

N

O 1′ H 2′ OH

H

H

3′

NH2 N

Phosphodiester linkage

N

Adenine (A)

H N

O O

P O–

O

5′ 4′

CH2

H

N

O 1′ H 2′ OH

H

H

3′

NH2

H

N

H

O

O

H

P

O –

O

5′ 4′

CH2

O

N

O

H

H

Cytosine (C)

H

1′ H

Guanine (G)

2′ OH

3′

O

H

N

N

H N

O

RNA nucleotide

O

P

O –

O

Phosphate

5′ 4′

CH2

H

N

NH2

O

H

3′ OH

H

1′ H

2′ OH

Sugar (ribose)

Figure 9.21

3′

83



Although usually single-stranded, RNA molecules can form short double-stranded regions 

This secondary structure is due to complementary basepairing 





This allows short regions to form a double helix

RNA double helices typically  



A to U and C to G

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

Complementary regions Held together by hydrogen bonds

Figure 9.22

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A U

U A

A U

A U U A

A

U

C

G

C

C U

C

C G A

A U (a) Bulge loop

C G

U

C

G

U A

C G A A G G C U U C

C C G A A G G C U U

A C

G C

C G

A U

C G

A U (c) Multibranched junction

Non-complementary regions

Have bases projecting away from double stranded regions

C G

U

G C A U A U

C G

G

C G

(b) Internal loop

C

G

C G

G

C

U

G C C G

C G

U

A U

U A A

G

U

G C

A

A C G U U G C A

G G C A C C G U

(d) Stem-loop

Also called hair-pin

85

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Double helix



Many factors contribute to the tertiary structure of RNA 

5′ end

Molecule contains single- and doublestranded regions

For example 



3′ end (acceptor site)

Base-pairing and base stacking within the RNA itself

Double helix

Interactions with ions, small molecules and large proteins

These spontaneously fold and interact to produce this 3-D structure

Anticodon

(a) Ribbon model



Figure 9.23

Figure 9.23 depicts the tertiary structure of tRNAphe 

The transfer RNA that carries phenylalanine 86