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Sinauer Associates, Inc. W. H. Freeman and Company Lecture Notebook to accompany Copyright © 2012 Sinauer Associates, Inc. Cover photograph © Fred Bavendam/Minden Pictures. This document may not be modified or distributed (either electronically or on paper) without the permission of the publisher, with the following exception: Individual users may enter their own notes into this document and may print it for their own personal use.

Lecture Notebook to accompany Principles of Life · 2013. 10. 30. · This document may not be modified or distributed (either electronically or on paper) without the permission of

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  • Sinauer Associates, Inc. W. H. Freeman and Company

    Lecture Notebook to accompany

    Copyright © 2012 Sinauer Associates, Inc. Cover photograph © Fred Bavendam/Minden Pictures.

    This document may not be modified or distributed (either electronically or on paper) without the permission of the publisher, with the following exception: Individual users may enter their own notes into this document and may print it for their own personal use.

  • © 2012 Sinauer Associates, Inc.

    DNA and Its Role in Heredity 9

    2

    To add your own notes to any page, use Adobe Reader’s Typewriter feature, accessible via the Typewriter bar at the top of the window. (Requires Adobe Reader 8 or newer. Adobe Reader can be downloaded free of charge from the Adobe website: http://get.adobe.com/reader.)

    POL HillisSinauer AssociatesMorales Studio Figure 09.01 Date 07-23-10

    Most cells are in G1 because it is the longest period of the cell cycle.

    These cells are in S, G2, and M phases and have replicated their DNA but not yet divided.

    Num

    ber

    of c

    ells

    0 32 64 96 128 160 192 224 256Amount of DNA in each cell

    400

    800

    1200

    1600

    2000

    2400

    (B)(A)

    FIGURE 9.1 DNA in the Nucleus and in the Cell Cycle (Page 166)

    Stained chromosomes 5 µmIN-TEXT ART (Page 166)

    http://get.adobe.com/readerRGreeneTypewritten TextDNA Review:Present in the cell nucleus and in condensed chromosomesDoubled during S phase of the cell cycleTwice as abundant in the diploid cells as in the haploid cells of a given organism.Showed the same patterns of transmission as the genetic information it was supposed to carryDNA first isoloted in 1868 by Swiss researcher Friedric Miescher. Genes were not yet viewed.

    RGreeneTypewritten TextStained DNA cells were passed through flow cytometer.This allowed for phases of mitosis to be viewed.These experiments also showed that after meiosis gametes had half the amount of DNA as somatic cells.

    RGreeneTypewritten Text(body)

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    Chapter 9 | DNA and Its Role in Heredity 3

    POL HillisSinauer AssociatesMorales Studio Figure 09.02 Date 07-23-10

    1 Bacteriophage T2 attaches to the surface of a bacterium and injects its DNA. Its protein coat stays outside the cell.

    2 Viral genes take over the host’s machinery, which synthesizes new viruses.

    3 The bacterium bursts, releasing about 200 viruses.

    DNA

    DNA

    Bacteriophage T2

    Proteincoat

    0.1 µm

    FIGURE 9.2 Viral DNA and Not Protein Enters Host Cells (Page 167)

    RGreeneTypewritten TextThe Transmission of DNAChromosomes in eukaryotic cells contain DNA and proteins. It was difficult for scientists to rule outthat genetic information might be carried on proteins.Bacteriophages (viruses that only infect bacteria) were used to show that genetic material is carried on the DNA and not the proteins. This experiment was observed by an electron scanning microscope.

    RGreeneHighlight

  • © 2012 Sinauer Associates, Inc.

    Chapter 9 | DNA and Its Role in Heredity 4

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.03 Date 07-22-10

    HYPOTHESIS

    CONCLUSION

    INVESTIGATION

    Go to yourBioPortal.com for original citations, discussions,and relevant links for all INVESTIGATION figures.

    The cells were transformed by DNA.

    DNA can transform eukaryotic cells.

    METHOD

    FIGURE 9.3 Transformation of Eukaryotic Cells The use of a marker gene shows that mammalian cells can be genetically transformed by DNA. Usually, the marker gene is carried on a larger DNA molecule.

    RESULTS

    ANALYZE THE DATA

    Isolate mammalian cells that lack the gene for thymidine kinase. (They cannot use thymidine in the growth medium.)

    1

    Add DNA with the marker gene for thymidine kinase.

    2a 2b Add control DNA without the gene for thymidine kinase.

    Cells with the thymidine kinase gene grow in thymidine.

    3a

    Cells without thethymidine kinasegene cannot use the thymidine inthe growthmedium and donot grow.

    3b

    Transformation was achieved by adding the DNA in a solution of calcium chloride (CaCl2) at pH 7. In other experiments, the type or amount of DNA, pH, or CaCl2 concentration was varied. The transformation efficiency was calculated as the percentage of cells that produced colonies on a medium containing thymidine. Explain these data.

    Transformation conditions Efficiency (%)

    Mammalian DNA with TK gene 10 µg 15 20 µg 55 30 µg 10 40 µg 5 20 µg, no CaCl2 10 20 µg, pH 6.5 0 20 µg, pH 7.5 0Bacterial virus DNA with TK gene 20 µg 0

    (Page 168

    RGreeneTypewritten TextExperimental evidence confirmed that DNA is the genetic material.Biologists also used model organisms such as bacteria in transformation experiments.The addition of DNA from one strain of bacterium could genetically transform another strain of bacterium.

    RGreeneTypewritten TextSome cells cannot grow in a medium that contains thymidine as the only source for dTTP. When the cells were incubated with DNA containing the gene for thymidine kinase, some cells transformed with the the TK gene and were able to grow on the thymidine-containing medium.

    RGreeneTypewritten Text---> optimal conditions for DNA growth with TK gene

    RGreeneTypewritten Text--->pH too low

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    Chapter 9 | DNA and Its Role in Heredity 5

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.04 Date 07-22-10

    These spots are caused by diffracted X rays.

    Photographic plate

    DNA sample

    Beam of X rays

    Lead screenX ray source

    (A) (B)

    FIGURE 9.4 X-Ray Crystallography Helped Reveal the Structure of DNA (Page 169)

    POL 9E Sadava Sinauer AssociatesMorales Studio POL_INTXT09.02 Date 07-22-10

    In DNA, A + G… …is always equal to T + C.

    =

    =

    T

    C

    Purines Pyrimidines =

    G

    A

    IN-TEXT ART (Page 169)

    RGreeneTypewritten TextThe Discovery of the three-dimensional structure of DNA.Rosalind Franklin used X-Ray crystallography to reveal the structure of DNA. This x-ray allows some substances, when they are isolated and purified can form crystals. Her discoveries led the way for Watson and Crick to discover the double helix of DNA.

    RGreeneTypewritten Text1920-1958

    RGreeneTypewritten Text------> Discovered by Erwin ChargaffChargaff's rule

    RGreeneTypewritten Text

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    Chapter 9 | DNA and Its Role in Heredity 6

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.05 Date 07-22-10

    The blue bands represent the two sugar–phosphate backbones, which run in opposite directions:

    5′

    3′

    3′

    5′

    (A)

    (B) Phosphorus

    Hydrogen

    Oxygen

    Carbon insugar–phosphatebackbone

    Majorgroove

    Minorgroove

    Bases

    3.4 nm

    3′

    5′

    A

    A

    AT

    T

    T

    T

    T

    A

    A

    A

    A

    G

    G

    G C

    C

    C

    C

    C

    G

    3′5′

    3′5′

    FIGURE 9.5 DNA Is a Double Helix (Page 170)

    RGreeneTypewritten TextDNA molecule must be helicalStrands run in opposite directions-antiparallel.How are nucleotides oriented?1. nucleotide bases on the two strands2. sugar to phosphate backbone on the outside3. Purines always equal pyrimidines. (Chargaff's rule)DNA structure1. Double helix2. It is right handed- Z-DNA do not have grooves and is a tighter helix.3. Antiparallel- built in a 5' to 3' direction but the two strands run in opposite directions4. Outer edges have minor and major grooves.Double helix is essential to function.1. Storage of genetic information2. Precise replication during the cell division cycle3. Susceptibility to mutations.4. Expression of coded information of phenotypes.

    RGreeneTypewritten Text-spiral shaped.

    RGreeneTypewritten Text

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    Chapter 9 | DNA and Its Role in Heredity 7

    POL HillisSinauer AssociatesMorales Studio Figure 09.06 Date 08-04-10

    H H

    H H

    H

    N

    O

    N

    N

    N

    N

    N

    N

    OH

    A T

    N H

    H H

    H

    H H

    H

    N

    N

    N

    N

    N

    N

    N

    O

    O

    G C

    H

    NH

    HH

    H

    HH

    H

    N

    N

    N

    N

    N

    N

    N

    O

    O

    GC

    H

    HH

    HH

    H

    N

    O

    N

    N

    N

    N

    N

    N

    O H

    AT

    CH3

    CH3

    The shaded atoms are available for hydrogen bonding to other molecules.

    The sugar–phosphate backbone is on the outside of the double helix.

    Major groove

    Minor groove

    Major groove

    Minor groove

    Major groove

    Minor groove

    Major groove

    Minor groove

    FIGURE 9.6 Base Pairs in DNA Can Interact with Other Molecules (Page 171)

    RGreeneTypewritten TextThis diagram shows 4 possibleconfigurations of base pairswithin the double helix.

    RGreeneTypewritten TextThis allows proteins to recognize specific DNA sequences.

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    Chapter 9 | DNA and Its Role in Heredity 8

    POL Sadava Sinauer AssociatesMorales Studio POL_INTXT09.03 Date 07-22-10

    Original DNAAfter one roundof replication

    IN-TEXT ART (Page 172)

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.07 Date 07-22-10

    Nucleotides are addedto the 3′ end.

    Bonds linking the phosphate groups arebroken, releasing energy to drive the reaction.

    C pairs with G.

    The enzyme DNA polymerase adds the next deoxyribonucleotideto the —OH group at the 3′ end of the growing strandand releases pyrophosphate.

    5′ end

    OH

    OH

    OH

    OH

    OH

    3′ end

    3′ end

    3′ end

    5′ end

    3′ end

    3′ end 5′ end

    Phosphate

    Sugar

    Base

    Pyrophosphate ion

    Growing strand

    5′ end Growing strandTemplate strand

    Phosphate ions

    5′ end

    T

    T

    A

    A

    G

    G

    GG

    C

    C

    T

    T

    A

    A

    G

    G

    C

    C

    C

    C

    CC

    DNA polymerase

    FIGURE 9.7 Each New DNA Strand Grows by the Addition of Nucleotides to Its 3′ End (Page 172)

    RGreeneTypewritten TextDNA Replicates SemiconservativelyThis means that each strand of parental DNA acts as a templatefor a new strand, which is formed by base pairing.The new resulting strands of DNA are "hybrids", one labeled and one unlabeled. Or, one old and one new.

    RGreeneTypewritten TextDNA replication involves a number of different enzymes and other proteins. It takes place in two steps.1. The DNA double helix is unwound to separate the two template strands and get them ready for base pairing.2. New nucleotides are formed with the template DNA, they are covalently linked together by phosphodiester bonds, forming a polymer whose base sequence is complementary to the bases in the template strand. The template DNA reads in the 3' to 5' direction. Growing strand 5' to 3'.3. Materials for DNA synthesis: 4 nucleotidesdeoxyadenosine triphosphate(dATP), deoxythymidine triphosphate(dTTP), deoxyguanine triphosphate(dGTP), deoxycytidine triphosphate(dCTP), (deoxyribonucleoside triphosphates dNTPs)

    RGreeneTypewritten Text

    RGreeneTypewritten Text

    RGreeneTypewritten TextDNA polymerase catalyzes the addition of nucleotides as the newDNA chain grows. All chromosomes have ori.

    RGreeneHighlight

    RGreeneTypewritten Text*during DNA synthesisthe 2 outer phosphates groups are released in an exothermic reaction*

    RGreeneTypewritten TextATP--->ADP

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    Chapter 9 | DNA and Its Role in Heredity 9

    POL HillisSinauer AssociatesMorales Studio Figure 09.08 Date 07-28-10

    1 The ori sequence binds the replication complex.

    There are multiple origins of replication.

    Replication forks move away from each other.

    2 Two replication forks grow away from one another.

    ori

    orioriori

    ori

    oriori

    Pre-replicationcomplex

    (A) Prokaryotic

    (B) Eukaryotic

    Initiation of replication

    FIGURE 9.8 The Origin of DNA Replication (Page 173)

    RGreeneHighlight

    RGreeneHighlight

    RGreeneTypewritten Text***Ori-region of replication

    RGreeneHighlight

    RGreeneHighlight

    RGreeneHighlight

    RGreeneTypewritten Text

    RGreeneTypewritten Text-protein

    RGreeneTypewritten TextBinding occurs when proteins in the complexrecognize DNA sequenceswithin the ori.

    RGreeneTypewritten TextReplication rates of E. Coli is 1000 bp (base pairs) per second.

    RGreeneTypewritten Text*

    RGreeneTypewritten Text**

    RGreeneTypewritten TextEukaryotic chromosomes are longer and up to a billion base pairs. They are linear not circular. They also have multiple origins of replication.

  • © 2012 Sinauer Associates, Inc.

    Chapter 9 | DNA and Its Role in Heredity 10

    POL HillisSinauer AssociatesMorales Studio Figure 09.09 Date 07-23-10

    1 Primase binds to the template strand and synthesizes an RNA primer.

    2 When the primer is complete, primase is released. DNA polymerase binds and synthesizes new DNA.

    New strand

    Template strand

    RNA primer

    RNA primer

    5′5′3′

    Primase

    5′ 5′3′

    5′3′

    DNA polymerase

    3′5′

    Primase

    FIGURE 9.9 DNA Forms with a Primer (Page 174)

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.10 Date 07-22-10

    (A) (B)

    “Thumb”

    “Fingers”

    RNA primer

    Template strand

    New strand

    DNA polymerase

    DNA polymerase

    5′

    3′

    3′

    5′

    DNA

    FIGURE 9.10 DNA Polymerase Binds to the Template Strand (Page 174)

    RGreeneTypewritten TextDNA Replication begins with a Primer*DNA polymerase can link an old and a new strand of nucleotides together. But, it has to have a primer to start the process.*A primer is usually a short single strand of RNA.The primer is complementary to the DNA. It issynthesized by an enzyme called primase.Once the DNA replication has taken place, the primase is released and breaks down. Only a new strand of DNA exists.DNA polymerases are large and bigger than theirsubstrates, nucleotides and the template strand.

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    Chapter 9 | DNA and Its Role in Heredity 11

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.11 Date 07-22-10

    Synthesis of the leading strand is continuous.

    The lagging strand is synthesized as Okazaki fragments.

    The replication fork grows.

    3′

    5′

    5′

    5′

    3′

    3′

    5′5′

    3′

    3′

    5′

    5′3′

    3′

    5′

    3′

    5′

    5′3′

    5′

    3′

    Okazaki fragments

    Primer

    FIGURE 9.11 The Two New Strands Form in Different Ways (Page 175)

    RGreeneTypewritten TextThe two DNA strands grow differentlyat the fork.

    RGreeneTypewritten TextThe leading strand grows forward in a 5' to 3' direction.

    RGreeneTypewritten TextThe lagging strand grows away

    RGreeneTypewritten Textgrows in shorter, backward stretches, with gaps.

    RGreeneTypewritten TextOkazaki fragments are 100 to 200 nucleotides long. DNA polymerasesynthesizes the fragments.A different DNA polymerase removesthe old primer and replaces it with DNA.DNA ligase links the the fragments together,making the fragment whole.A DNA polymerase can add thousands ofof nucleotides before it detaches from DNA.It is processive!

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    Chapter 9 | DNA and Its Role in Heredity 12

    POL HillisSinauer AssociatesMorales Studio Figure 09.12 Date 07-23-10

    1 Primase forms an RNA primer.

    3 A different DNA polymerase hydrolyzes the primer and replaces it with DNA.

    2 DNA polymerase adds nucleotides to the new Okazaki fragment only at the 3′ end, continuing until it encounters the primer on the previous Okazaki fragment.

    4 DNA ligase then catalyzes theformation of the phosphodiesterlinkage that finally joins the twoOkazaki fragments.

    DNA polymerase

    DNA polymerase

    Lagging strand RNA primer RNA primer

    Lagging strand template

    3′

    3′5′

    5′

    Okazaki fragment

    Gap

    3′

    3′

    3′

    5′

    5′5′

    Primase

    3′

    3′5′

    5′

    DNA ligase(open)

    DNA ligase(closed)

    3′

    3′5′

    5′

    FIGURE 9.12 The Lagging Strand Story (Page 176)

    RGreeneTypewritten TextDNA ligase catalyzes the phosophidester linkage tojoin the fragments together and make a whole strand.

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    Chapter 9 | DNA and Its Role in Heredity 13

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.13 Date 09-13-10

    Removal of the RNA primer leads to the shortening of the chromosome after each round of replication. Chromosome shortening eventually leads to cell death.

    An RNA sequence in telomerase acts as a template for DNA. This enzyme adds the telomeric sequence to the 3′ end of the chromosome.

    The original length of the chromosomal DNA has been restored. Note the gap where the primer for DNA replication has been removed.

    3′

    3′5′

    5′

    Parent DNA

    Newstrands

    (A)

    3′

    3′

    Telomere

    Telomerase

    3′

    3′

    RNA template(C)

    (B)

    Gap

    Telomeres

    FIGURE 9.13 Telomeres and Telomerase (Page 177)

    RGreeneTypewritten TextNew chromosomes can have a bit of a single strandof DNA at each end.A mechanism in the cell can cut off a region along witha double strand of DNA. This can cause a shortenedchromosome or death.Telomeres are repetitive sequences that are at these ends of chromosomes. In humans and vertebrates thesesequences are TTAGGG . In humans it is repeated about2500 times. These telomeres protect the chromosomes.

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    Chapter 9 | DNA and Its Role in Heredity 14

    POL HillisSinauer AssociatesMorales Studio Figure 09.14 Date 07-23-10

    4 In the last step, DNA ligase repairs the remaining nick.

    321

    During DNA replication, an incorrect base may be added to the growing chain.

    321 The DNA polymerase immediately excises the incorrect base…

    …and replaces it with the correct base before proceeding with replication.

    During DNA replication, a base was mispaired and missed in proofreading.

    The mismatch repair proteins excise the mismatched base and some adjacent bases.

    DNA polymerase I adds the correct bases.

    (A) DNA proofreading

    (B) Mismatch repair

    DNA polymerase

    DNA polymerase

    FIGURE 9.14 DNA Repair Mechanisms (Page 178)

    RGreeneTypewritten TextErrors in DNA replication can be repaired.

    RGreeneTypewritten Text-occurs after DNA polymerase inserts a nucleotide

    RGreeneTypewritten Text- backup to proofreading.

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    Chapter 9 | DNA and Its Role in Heredity 15

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.15 Date 07-22-10

    RESEARCH TOOLSFIGURE 9.15 The Polymerase Chain Reaction The steps in this cyclic process are repeated many times to produce millions

    of identical copies of a DNA fragment. This makes enough DNA for chemical analysis and genetic manipulations.

    A DNA molecule with a target sequence to be copied is heatedto 90°C to denature it.

    1 When the mixture cools, artificially synthesized primers bond to the single- stranded DNA.

    2 dNTPs and heat-resistant DNA polymerase synthesize two new strands of DNA.

    3 The process is repeated, doubling the amount of DNA.

    4 By repeating the process, many copies of the original DNA can be produced in a short time.

    5

    Primer New DNA

    3′ 3′ 5′5′

    Target sequence

    New DNA

    (Page 179)

    RGreeneTypewritten TextDNA replication can be used to amplify DNA in a lab.

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    Chapter 9 | DNA and Its Role in Heredity 16

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.16 Date 07-22-10

    DNA

    Normal allele: Codes for a functional protein

    mRNA

    Protein

    Mutation

    Silent mutation: Codes for a functional protein

    Loss-of-function mutation: Codes for a nonfunctional protein

    Gain-of-function mutation: Codes for a protein with new function

    FIGURE 9.16 Mutation and Phenotype (Page 180)

    IN-TEXT ART (Page 180)

    RGreeneTypewritten TextMutations are Heritable Changes in DNAMutations are changes in the nucleotide sequence of DNA that are passed on fromone cell, or organism, to another.Some errors escape the proofreading ofDNA polymerase and are passed on to daughter cells.2 types of mutations:Somatic mutations- occurs in body cells. Example: a mutation could occur in a single skin cell that can cause a patch of skincells that have the same mutation.Germline mutations-occurs in cells that produce gametes.

    RGreeneTypewritten TextMutations may or may not have phenotypic effects.Mutations can have an effect on protein-coding genes and their functions. * Silent mutations-Do not affect gene function. They also do not have an effect on protein coding.* Loss of function mutations-can result in the loss of expression of a gene or the production of a nonfunctional protein or RNA. Loss of function mutations almost always show recessive inheritance in a diploid organism. Example: Mendel's wrinkled seeds. Also known as a leaky mutation. * Gain of function mutations usually shows a dominant inheritance. This type of mutation is also common in cancer. Normally a receptor for a growth factor normally requires binding to activate the cell cycle. Some cancer are caused by mutations in genes that code the receptors to stay on. This causes an overgrowth of the cancer cells.*Conditional mutations- cause their phenotypes only under restrictive conditions. Manyconditional mutants are temperature-sensitive in that certain enzymes that cause a geneto be expressed may become denatured if temps rise. Example would be in some cats,the enzyme tyrosinase protein becomes denatured(>35* C) and the fur of the cat is pale. However,if their extremities are cooler, the fur on feet, tails and ears can be darker.

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    Chapter 9 | DNA and Its Role in Heredity 17

    POL Sadava Sinauer AssociatesMorales Studio POL_INTXT09.05 Date 07-22-10

    Sickled cell

    Normal cell

    2.5 µm

    IN-TEXT ART (Page 181)

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.17 Date 07-22-10

    Duplication and deletion result when homologouschromosomes break at different points…

    …and swap segments.

    Inversion results when a broken segment is inserted in reverse order.

    Deletion is the loss of a chromosome segment.

    Reciprocal translocation results when nonhomologous chromosomes exchange segments.

    (lost)

    (A)

    (B)

    (C)

    (D)

    D

    G

    A B C D E F G A B E F G

    C

    A B C D E F G A B E F G

    A B C D E F G A B C D E FC D

    A B C D E F G A B E D C F G

    A B C D E F G A B L M N O

    H I J K L M ON H I C D E FJ K G

    FIGURE 9.17 Chromosomal Mutations (Page 181)

    RGreeneTypewritten TextA point mutation results in the gain, loss, or substitution of a single nucleotide. If a point mutationoccurs within a single gene, it results in a new allele of that gene. It may or may not result in a new phenotype.An example of a point mutation with a significant effecton phenotype is the one that causes sickle cell disease.The sickle allele differs from the normal allele by one base pair, resulting in a polypeptide that differs by oneamino acid from the normal protein. Individuals have tobe homozygous recessive to have the blood disorder.A p53 protein normally functions to inhibit cell cycle but a point mutation can cause the protein to promote the cellcycle and prevent cell death. This mutation causes a gainof oncogenic function.

    RGreeneTypewritten TextChromosomal mutations can cause the most dramaticchanges. Whole chromosomes break and rejoin, causing huge disruptions in gene sequences.4 types:*deletions ( Cri du chat syndrome- deletion on # 5 chromosome)*duplication- can be produced at the same time as a deletion.(dup15 chromosome) * inversion-the breaking and rejoining of chromosomes. A section of DNA may be removed but reinserted and flipped.* translocation-results when segments of chromosomes break off and become joined to different chromosomes. (Translocation Downs Syndrome) different from Trisomy 21

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    Chapter 9 | DNA and Its Role in Heredity 18

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.18 Date 07-22-10

    A spontaneous or induced mutation of C occurs.

    This C cannot hydrogen-bond with G but instead pairs with A.

    This base cannot pair with G but instead pairs with A.

    1

    The mutated C pairs with A instead of G.

    2

    Although the mutated C usually reverts to normal C, either spontaneously or by DNA repair mechanisms…

    3

    …the "mispaired" A remains, propagating a mutated sequence.

    4

    H

    O

    H

    H

    H

    Cytosine (common tautomer)

    Original sequence

    Newly replicated strands

    Template strand

    Template strand

    Replication is normal

    Mutated sequence

    Cytosine (rare tautomer)

    Deaminated formof cytosine (= uracil)

    Deamination byN N

    CN

    O

    H

    C*N N

    N

    O

    H

    H

    CN N

    N

    O

    UN N

    O

    HNO2

    (A) A spontaneous mutation

    (C) The consequences of either mutation

    (B) An induced mutation

    AATGCTGTTACGAC

    AATGCTGTTACGAC

    AATGCTGTTACAAC AATGTTG

    TTACAAC

    AATGCTGTTACGAC

    AATGCTGTTACGAC

    FIGURE 9.18 Spontaneous and Induced Mutations (Page 182)

    RGreeneTypewritten Text- happen due to imperfect cellular processes

    RGreeneTypewritten Textp.182

    RGreeneTypewritten Text- an agent outside the cellcause the mutation-environment

    RGreeneTypewritten TextMutagens can be natural or artificial.* Plants have molecules that provide a defense against pathogens. Some are mutagenic and possibly carcinogenic. Ex. Aflatoxin produced by mold, Aspergillus. The aflatoxin bonds to guanine and cause mutations in the DNA.*Radiation made from man or natural radiation from space. Natural UV radiation can also cause mutations. In about 3 billion base pairs, there are about 16,000 DNA-damaging events per cell per day. About 80 percent are repaired.

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    Chapter 9 | DNA and Its Role in Heredity 19

    POL Sadava Sinauer AssociatesMorales Studio Figure 09.19 Date 07-22-10

    When 5-methylcytosine loses its amino group, thymine results. Since thymine is a normal DNA base, it is not removed.

    When DNA replicates, half the daughter DNA is mutant and half is normal.

    5-Methylcytosine Thymine

    Replication

    50%

    50%

    GGATmCGCTCCCTA GCGAG

    GGAT ACTCCCTA TGAG

    GGATTACTCCCTAATGAG

    GGATCACTCCCTAGTGAG

    O

    H

    H

    N N

    N

    CH3

    O

    O

    HN N

    CH3

    G

    T

    FIGURE 9.19 5-Methylcytosine in DNA Is a “Hotspot” for Mutations (Page 183)

    FIGURE 9.20 A Neanderthal Child (Page 185)