DNA: The Genetic Material Chapter 14 1. Learning Objectives 14.1 The Nature of the Genetic Material...

DNA: The Genetic MaterialChapter 14

1

Learning Objectives

14.1 The Nature of the Genetic Material

•Understand experiments of

– Griffith & Avery

– Avery, MacLeod, and McCarty

– Hershey and Chase

•For both experiments, know/understand major findings

2

Frederick Griffith – 1928 • Studied Streptococcus pneumoniae, a

pathogenic bacterium causing pneumonia

• 2 strains of Streptococcus

– S strain is virulent

– R strain is nonvirulent

3

http://o.quizlet.com/i/GEJK81oHlTEYutTzmSQK6Q.jpg

Griffith’s Experiment

• Griffith infected mice with these strains hoping to understand the difference between the strains

4

5

Live NonvirulentStrain of

S. pneumoniae

Mice live

b.

Live VirulentStrain of S. pneumoniae

Mice die

Polysaccharidecoat

a.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


6

Heat-killed VirulentStrain of S. pneumoniae

Mice live

c.

+

Mixture of Heat-killed Virulentand Live Nonvirulent

Strains of S. pneumoniae

Their lungs contain livepathogenic strain of

S. pneumoniae

Mice die

d.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


7

• Griffith’s results

– Live S strain cells killed the mice

– Live R strain cells did not kill the mice

– Heat-killed S strain cells did not kill the mice

– Heat-killed S strain + live R strain cells killed the mice

8

• Transformation

– Information specifying virulence passed from the dead S strain cells into the live R strain cells

• Our modern interpretation is that genetic material was actually transferred between the cells

9

Avery, MacLeod, & McCarty – 1944

• Repeated Griffith’s experiment using purified cell extracts

http://biology.kenyon.edu/courses/biol114/KH_lecture_images/How_DNA_works/FG11_02.JPG

10

Avery, MacLeod, & McCarty – 1944

• Removal of all protein from the transforming material did not destroy its ability to transform R strain cells

• DNA-digesting enzymes destroyed all transforming ability

• Supported DNA as the genetic material

11

Hershey & Chase –1952

• Investigated bacteriophages

– Viruses that infect bacteria

• Bacteriophage was composed of only DNA and protein

• Wanted to determine which of these molecules is the genetic material that is injected into the bacteria

12

• Bacteriophage DNA was labeled with radioactive phosphorus (32P)

• Bacteriophage protein was labeled with radioactive sulfur (35S)

• Radioactive molecules were tracked


13

+

+

Phage grown in radioactive 35S,which is incorporated into phage coat

Virus infectbacteria

Blender separatesphage coat from bacteria

Centrifuge formsbacterial pellet

35S in supernatant

35S-Labeled Bacteriophages

Phage grown in radioactive 32P.which is incorporated into phage DNA

Virus infectbacteria

Blender separatesphage coat from bacteria

Centrifuge formsbacterial pellet

32P in bacteria pellet

32P-Labeled Bacteriophages

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

Hershey & Chase Experiment

14

• Only the bacteriophage DNA (as indicated by the 32P) entered the bacteria and was used to produce more bacteriophage

• Conclusion: DNA is the genetic material!


Question 16Mixing a killed virulent strain of bacteria and a living strain of benign bacteria together produces virulent bacteria. What does this demonstrate?

a. Genes are inactivated when a cell dies

b. DNA can only be passed on during reproduction

c. Cells can pick up genes from the environment

d. All bacteria are virulent

e. None of the above

Learning Objectives

14.2 DNA Structure

•Understand the contributions of the following people in elucidating DNA’s structure

– Chargaff

– Wilkins & Franklin

– Watson & Crick

16

17

DNA Structure

• DNA is a nucleic acid• Composed of nucleotides

– 5-carbon sugar called deoxyribose

– Phosphate group (PO43-)

• Attached to 5′ carbon of sugar– Nitrogenous base

• Adenine, thymine, cytosine, guanine– Free hydroxyl group (—OH)

• Attached at the 3′ carbon of sugar

Nucleotide Subunits of DNA and RNA

18


Pu

rin

esP

yrim

idin

es

Adenine Guanine

NH2CC

NN

N

C

H

N

C

CH

O

H

H

OC

NC

H

N

C

NH2

H

CH O

O

C

NC

H

N

CH3C

CH

H

O

O

C

NC

H

N

CH

CH

NH2

CC

NN

N

C

H

N

C

CHH

Nitrogenous Base

4′

5′

1′

3′ 2′

2

8

7 6

39

4

5

1Phosphate group

Sugar

Nitrogenous base

CH2

N N

O

NNH2

OH in RNA

Cytosine(both DNA and RNA)

Thymine(DNA only)

Uracil(RNA only)

OH

H in DNA

O

P

O–

–O O

• Phosphodiester bond– Bond between

adjacent nucleotides– Formed between the

phosphate group of one nucleotide and the 3′ —OH of the next nucleotide

• The chain of nucleotides has a 5′-to-3′ orientation

19


BaseCH2

O

5′

3′

O

P

O

OH

CH2

–O O

C

Base

O

PO4

Phosphodiesterbond

Phosphate group

hydroxyl group

Chargaff’s Rules

• Erwin Chargaff determined that

– Amount of adenine = amount of thymine

– Amount of cytosine = amount of guanine

– Always an equal proportion of purines (A and G) and pyrimidines (C and T)

20

Representation of Chargaff’s Data Table (1952)

Organism

%A %G %C %T A/T G/C %GC %AT

φX174 24.0 23.3 21.5 31.2 0.77 1.08 44.8 55.2

Maize 26.8 22.8 23.2 27.2 0.99 0.98 46.1 54.0

Octopus 33.2 17.6 17.6 31.6 1.05 1.00 35.2 64.8

Chicken 28.0 22.0 21.6 28.4 0.99 1.02 43.7 56.4

Rat 28.6 21.4 20.5 28.4 1.01 1.00 42.9 57.0

Human 29.3 20.7 20.0 30.0 0.98 1.04 40.7 59.3

Grasshopper

29.3 20.5 20.7 29.3 1.00 0.99 41.2 58.6

Sea Urchin

32.8 17.7 17.3 32.1 1.02 1.02 35.0 64.9

Wheat 27.3 22.7 22.8 27.1 1.01 1.00 45.5 54.4

Yeast 31.3 18.7 17.1 32.9 0.95 1.09 35.8 64.4

E. coli 24.7 26.0 25.7 23.6 1.05 1.01 51.7 48.3

http://en.wikipedia.org/wiki/Chargaff%27s_rules

22

Rosalind Franklin

• Performed X-ray diffraction studies to identify the 3-D structure– Discovered that DNA is helical– Using Maurice Wilkins’ DNA

fibers, discovered that the molecule has a diameter of 2 nm and makes a complete turn of the helix every 3.4 nm

a.

b.


Courtesy of Cold Spring Harbor Laboratory Archives

James Watson and Francis Crick – 1953

• Deduced the structure of DNA using evidence from Chargaff, Franklin, and others

• Did not perform a single experiment themselves related to DNA

• Proposed a double helix structure

23

Double helix

• 2 strands are polymers of nucleotides

• Phosphodiester backbone – repeating sugar and phosphate units joined by phosphodiester bonds

• Wrap around 1 axis• Antiparallel

24

5´

3

O

O

O

O

4

5

1

3 2

4

5

1

3 2

4

5

1

3 2

4

5

1

3 2

5-carbon sugar

Nitrogenous base

Phosphodiester bond

Phosphate group

OH

P

P

P

P


Antiparallel Nature of DNA

25http://academic.brooklyn.cuny.edu/biology/bio4fv/page/molecular%20biology/dsDNA.jpg

C

C

C

G

G

G

G

G

T

T

T

T

A

A

A

2nm5′ 3′

3.4nm

0.34nm

Minorgroove

Majorgroove

5′3′


Majorgroove

Minorgroove

26

• Complementarity of bases

• A forms 2 hydrogen bonds with T

• G forms 3 hydrogen bonds with C

• Gives consistent diameter

27


A

H

Sugar

Sugar

Sugar

Sugar

T

G C

N

H

N O

H

CH3

H

HN

N N H N

N

N

H

H

H

N O H

H

H N

N H

N

N HN N

Hydrogenbond

Hydrogenbond

Question 7

Chargaff’s rule states that

a. DNA strands are in antiparallel alignment

b. G matches with C, and T matches with A

c. The DNA molecule is a double helix

d. DNA transformation occurs when an organism incorporates DNA from the environment

e. The nuclei of cells are totipotent

Question 11

If a strand of DNA had the sequence 5’- AGTCCA- 3’, which of the following would be the complementary DNA strand?

a. 3’- CATGGT- 5’

b. 3’- TCAGGT- 5’

c. 3’- TCAAAU- 5’

d. 3’- AGTCCA- 5’

e. 3’- GGTTCA- 5’

Question 12

If a DNA molecule contains 40% thymine, how much guanine will it contain?

a. 10%

b. 20%

c. 30%

d. 40%

Learning Objectives

14.3 Basic Characteristics of DNA Replication

•How did Meselson and Stahl figure out DNA replication?

•What is required for DNA replication?

31

32

DNA Replication

3 possible models1. Conservative model

2. Semiconservative model

3. Dispersive model

33


Conservative

34


Conservative Semiconservative

35


Conservative Semiconservative Dispersive

36

Meselson and Stahl – 1958

• Bacterial cells were grown in a heavy isotope of nitrogen, 15N

• All the DNA incorporated 15N

• Cells were switched to media containing lighter 14N

• DNA was extracted from the cells at various time intervals

37

Meselson and Stahl – 1958

• Bacterial cells were grown in a heavy isotope of nitrogen, 15N

• All the DNA incorporated 15N

• Cells were switched to media containing lighter 14N

• DNA was extracted from the cells at various time intervals

38

Samples are centrifuged

E. coli

0

1

2

0 rounds 1 round 2 rounds

Bottom

15N medium

14N medium

E. coli cells grownin 15N medium

Cells shifted to14N medium andallowed to grow

DNA

Samples taken atthree time pointsand suspended incesium chloridesolution

Rounds ofreplication

Top

0 min0 rounds

20 min1 round

40 min2 rounds

From M. Meselson and F.W. Stahl/PNAS 44(1958):671


Meselson and Stahl’s Results

• Conservative model = rejected– 2 densities were not observed after round 1

• Semiconservative model = supported– Consistent with all observations– 1 band after round 1– 2 bands after round 2

• Dispersive model = rejected– 1st round results consistent– 2nd round – did not observe 1 band

39

40

DNA Replication

• Requires 3 things

– Something to copy

• Parental DNA molecule

– Something to do the copying

• Enzymes

– Building blocks to make copy

• Nucleotide triphosphates

41

• DNA replication includes

– Initiation – replication begins

– Elongation – new strands of DNA are synthesized by DNA polymerase

– Termination – replication is terminated

42

P

P

P

P

P

P

P

P

P

P

P

Pyrophosphate

3′

3′

5′

5′

New StrandTemplate Strand

O

HO

OH

O

O

O

O

O

O

O

O

O

O

C

C

T

T

T

A

A

A

G

G

A

P

P

P

P

P

PP P

P P

P

P

P

P

3′

3′

5′

5′

New StrandTemplate Strand

O

HO

OH

OH

O

O

O

O

O

O

O

O

O

C

C

T

T

A

A

A

G

G

A

Sugar–phosphatebackbone

DNA polymerase III

TO

P


• DNA polymerase– Matches existing DNA bases with

complementary nucleotides and links them– All have several common features

• Add new bases to 3′ end of existing strands• Synthesize in 5′-to-3′ direction• Requires a primer of RNA

43


5

3

5

5 5 3

3

RNA polymerase makes primer DNA polymerase extends primer

Question 5

The Meselson-Stahl experiment demonstrated that DNA replication is

a. Conservative

b. Semi-conservative

c. Disruptive

d. Differentiated

Learning Objectives

14.4 Prokaryotic Replication

•How many DNA pol’s does E. coli have and what are their functions?

•What other enzymes are needed for DNA replication (in E. coli)

•What occurs at the replication fork?

•How is DNA replication semidiscontinous?

•What are the leading and lagging strands?

45

Prokaryotic Replication

• E. coli model

• Single circular molecule of DNA

• Replication begins at one origin of replication

• Proceeds in both directions around the chromosome

• Replicon – DNA controlled by an origin

46

47


Replisome

Replisome

TerminationOrigin

Termination

Origin

Origin Origin

Origin

Termination TerminationTermination

48

• E. coli has 3 DNA polymerases– DNA polymerase I (pol I)

• Acts on lagging strand to remove RNA primers and replace them with DNA

– DNA polymerase II (pol II)• Involved in DNA repair processes

– DNA polymerase III (pol III)• Main replication enzyme

– All 3 have 3′-to-5′ exonuclease activity – proofreading

– DNA pol I has 5′-to-3′ exonuclase activity

• Unwinding DNA causes torsional strain– Helicases – use energy from ATP to unwind

DNA– Single-strand-binding proteins (SSBs) coat

strands to keep them apart– Topoisomerase prevent supercoiling

• DNA gyrase is used in replication49


Supercoiling

Replisomes

No Supercoiling

Replisomes

DNA gyrase

Semidiscontinous

• DNA polymerase can synthesize only in 1 direction!!!!!

• Leading strand synthesized continuously from an initial primer

• Lagging strand synthesized discontinuously with multiple priming events

– Okazaki fragments

50

51

RNA primer

Open helixand replicate

First RNA primer

Open helix andreplicate further

Lagging strand(discontinuous)

Second RNA primer

Leading strand(continuous)

RNA primer

5′

3′

3′

5′

5′

3′

3′

5′

5′

3′

5′

3′

5′

3′


52

• Partial opening of helix forms replication fork

• DNA primase – RNA polymerase that makes RNA primer

– RNA will be removed and replaced with DNA

Leading-strand synthesis

– Single priming event

– Strand extended by DNA pol III

• Processivity – subunit forms “sliding clamp” to keep it attached

53


a-b: From Biochemistry by Stryer. © 1975, 1981, 1988, 1995 by Lupert Stryer. Used with permission of W.H. Freeman and Company

a. b.

Lagging-strand synthesis– Discontinuous synthesis

• DNA pol III– RNA primer made by primase for each Okazaki

fragment– All RNA primers removed and replaced by DNA

• DNA pol I– Backbone sealed

• DNA ligase•Termination occurs at specific site

– DNA gyrase unlinks 2 copies

54

55


5′

3′

Primase

RNA primer

Okazaki fragmentmade by DNApolymerase III

Leading strand(continuous)

DNA polymerase I

Lagging strand(discontinuous)

DNA ligase

56

Replication forkCopyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

5 3

New bases β clamp (sliding clamp) Leading strand

Single-strand bindingproteins (SSB)

DNA gyrase

ParentDNA

PrimaseHelicase

3 5

Clamp loader

Open β clamp

Lagging strandOkazaki fragment

5 3

DNA ligase

polymerase IDNA

RNA primer

New bases

polymerase IIIDNA

57

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.

Question 3

Where are the Okazaki fragments found?

a. On the lagging strand

b. On the leading strand

c. At the replication origin

d. In the cytoplasm

e. On both strands

Question 4

Name the enzyme that links Okazaki fragments

a. DNA polymerase I

b. RNA primase

c. DNA ligase

d. Helicase

e. ATP synthase

Question 6

During DNA replication, what enzyme is responsible for untwisting the DNA helix?

a. DNA polymerase I

b. RNA primase

c. DNA ligase

d. DNA polymerase III

e. Helicase

Question 9

Why does replication proceed in opposite directions on the leading and lagging strands?

a. The polymerase enzyme needs a primer

b. DNA polymerase III can only add to the 3´ end of a strand

c. The Okazaki fragments are only on the leading strands

d. The parent strands are oriented in the same direction

e. Helicase only allows for replication of one strand at a time

Question 15

What would be the immediate consequence of a non-functional primase enzyme?

a. The strands would break due to the torsional strain from rapid untwisting

b. The helix could be opened

c. The DNA polymerase III enzyme would have nothing to bind to

d. The Okazaki fragments would not be linked together

e. The single DNA strands could not be held open

Learning Objectives

14.5 Eukaryotic Replication

•What are differences between Prok and Euk replication?

•What are telomeres and how are they replicated?

14.6 DNA Repair

•What are the three forms of DNA repair?

•Why is DNA repair important for the cell?63

64

Eukaryotic Replication

• Complicated by

– Larger amount of DNA in multiple chromosomes

– Linear structure

• Basic enzymology is similar

– Requires new enzymatic activity for dealing with ends only

Telomeres • Specialized structures found on the ends

of eukaryotic chromosomes

• Protect ends of chromosomes from nucleases and maintain the integrity of linear chromosomes

• Gradual shortening of chromosomes with each round of cell division

65

http://www.scientificamerican.com/media/inline/telomeres-telomerase-and_1.jpg

66

Leading strand (no problem)Lagging strand (problem at the end)

Last primer

Replication first round

Shortened template

Origin

5´

3´

3´

5´

3´

5´

5´

3´

3´

5´5´

3´

5´

3´

5´

3´

3´

5´

3´

5´

Removed primercannot be replaced

Leadingstrand

Laggingstrand


Primer removal

Replication second round

67

• Telomeres composed of short repeated sequences of DNA

• Telomerase – enzyme makes telomere section of lagging strand using an internal RNA template (not the DNA itself)– Leading strand can be replicated to the end

• Telomerase developmentally regulated– Relationship between senescence and telomere length

• Cancer cells generally show activation of telomerase

68

G

GGGGG

T T

TTT TG

TT

GGG

GG

T

TTT

CCCCC AAAA

CCCCC AAAA

Telomere extendedby telomerase

Template RNA ispart of enzyme

Telomerase

Now readyto synthesizenext repeat

5 ́

3 ́

5 ́

3 ́

5 ́

3 ́

Synthesis by telomerase

Telomerase moves andcontinues to extend telomere


69

DNA Repair

• Errors due to replication– DNA polymerases have proofreading ability

• Mutagens – any agent that increases the number of mutations above background level– Radiation and chemicals

• Importance of DNA repair is indicated by the multiplicity of repair systems that have been discovered

70

DNA Repair

Falls into 2 general categories1. Specific repair

– Targets a single kind of lesion in DNA and repairs only that damage

2. Nonspecific– Use a single mechanism to repair multiple

kinds of lesions in DNA

71

Photorepair

• Specific repair mechanism• For one particular form of damage caused

by UV light• Thymine dimers

– Covalent link of adjacent thymine bases in DNA

• Photolyase– Absorbs light in visible range– Uses this energy to cleave thymine dimer

72

T

A

T

A

A A

A A

TA

TA

TT

T T

Thymine dimercleaved

Photolyase

Helix distorted bythymine dimer

Thymine dimer

DNA with adjacent thymines

UV light

Visible light

Photolyase bindsto damaged DNA


Excision repair

• Nonspecific repair

• Damaged region is removed and replaced by DNA synthesis

• 3 steps1. Recognition of damage

2. Removal of the damaged region

3. Resynthesis using the information on the undamaged strand as a template

73

74


Damaged or incorrect base

Uvr A,B,C complexbinds damaged DNA

DNA polymerase

Excision of damaged strand

Resynthesis by DNA polymerase

Excision repair enzymes recognize damaged DNA

Question 13

The enzyme telomerase attaches the last few bases on the lagging strand. As cells age, telomerase activity drops. What would happen to the chromosomes in the absence of telomerase activity?

a. Chromosome replication would be terminated

b. Okazaki fragments would not be linked together

c. Chromosomes would shorten during each division

d. The leading strand would become the lagging strand

e. The cells would become cancerous

Question 10

Mutations can be caused by copying mistakes, and by exposure to chemicals or electromagnetic radiation.

a. This is True

b. This is False

Date post:	02-Jan-2016
Category:	Documents
Upload:	georgia-hamilton
View:	220 times
Download:	0 times

DNA: The Genetic Material Chapter 14 1. Learning Objectives 14.1 The Nature of the Genetic Material...

Documents