Chapter 12 Molecular Genetics Section 1 DNA The

  • Slides: 142
Download presentation

Chapter 12 Molecular Genetics Section 1: DNA: The Genetic Material Section 2: Replication of

Chapter 12 Molecular Genetics Section 1: DNA: The Genetic Material Section 2: Replication of DNA Section 3: DNA, RNA, and Protein Section 4: Gene Regulation and Mutation Click on a lesson name to select.

Hint/Preview of first product assignment (due tomorrow)…so you can take better notes! Create an

Hint/Preview of first product assignment (due tomorrow)…so you can take better notes! Create an illustrated “timeline” of the events that led up to the discovery of DNA as the genetic material and its structure. Click on a lesson name to select.

Discovery of DNA as the genetic material Start: No knowledge of DNA (cavemen to

Discovery of DNA as the genetic material Start: No knowledge of DNA (cavemen to ~1928) -------------------------------------- 1. Griffith finds transforming factor 2. Avery says transforming factor is DNA (no one believes though) 3. Chargaff finds nucleotide ratios 4. Hershey / Chase prove to all DNA is genetic material 5. Watson / Crick find structure of DNA --------------------------------------Present: Understanding of DNA as the genetic material of cells Click on a lesson name to select.

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Griffith § Performed the

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Griffith § Performed the first major experiment that led to the discovery of DNA as the genetic material

 • The discovery of the genetic role of DNA began with research by

• The discovery of the genetic role of DNA began with research by Frederick Griffith in 1928. • He studied Streptococcus pneumoniae, a bacterium that causes pneumonia in mammals. – One strain, the R strain, was harmless. – The other strain, the S strain, was pathogenic. • In an experiment Griffith mixed heat-killed S strain with live R strain bacteria and injected this into a mouse. • The mouse died and he recovered the pathogenic strain from the mouse’s blood. Copyright © 2002 Pearson Education, Inc. , publishing as Benjamin Cummings

 • Griffith called this phenomenon transformation, a change phenotype due to the assimilation

• Griffith called this phenomenon transformation, a change phenotype due to the assimilation of a foreign substance into a cell. (His exp did not show it was DNA –just that something moved from dead pathogenic cells to normal cells- making them pathogenic)

 • For the next 14 years scientists tried to identify the transforming substance.

• For the next 14 years scientists tried to identify the transforming substance. • Finally in 1944, Oswald Avery, and others announced that Griffith’s transforming substance was DNA. Protein destroying enzymes used on Giffith’s exp = bact still transformed DNA destroying enzymes used on Griffith exp = bact not transformed No one believed them though. DNA thought to be too simple DNA has 4 different monomers (4 nucleotides) Protein has 20 different monomers (20 AA’s)

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Chargaff § Chargaff’s rule:

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Chargaff § Chargaff’s rule: C = G and T = A

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Hershey and Chase §

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Hershey and Chase § Used radioactive labeling to trace the DNA and protein § Concluded that the viral DNA was injected into the cell and provided the genetic information needed to produce new viruses

Watson and Crick -In 1953 Used the work of others and their own brain

Watson and Crick -In 1953 Used the work of others and their own brain power to deduce the structure of DNA Erwin Chargaff figured out that the number of A’s in an organism was equal to the number of T’s and also that C’s = G’s Rosalind Franklins took X-ray pictures of DNA and figured out it must be a helix

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Watson and Crick §

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Watson and Crick § Built a model of the double helix that conformed to the others’ research 1. two outside strands consist of alternating deoxyribose and phosphate 2. cytosine and guanine bases pair to each other by three hydrogen bonds 3. thymine and adenine bases pair to each other by two hydrogen bonds

James Watson and Francis Crick won Nobel Prize in 1962 for figuring out the

James Watson and Francis Crick won Nobel Prize in 1962 for figuring out the structure of DNA

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material DNA Structure § Nucleotides

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material DNA Structure § Nucleotides § Consist of a five-carbon sugar, a phosphate group, and a nitrogenous base

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material DNA Structure § DNA

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material DNA Structure § DNA often is compared to a twisted ladder. § Rails of the ladder are represented by the alternating deoxyribose and phosphate. § The pairs of bases (cytosine–guanine or thymine–adenine) form the steps.

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Orientation § On the

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Orientation § On the top rail, the strand is said to be oriented 5′ to 3′. § The strand on the bottom runs in the opposite direction and is oriented 3′ to 5′.

Chapt 12 test questions will be on final exam on thurs… there will be

Chapt 12 test questions will be on final exam on thurs… there will be no separate chapt 12 test.

ACTG Test: Draw a double stranded stretch of DNA if one strand has the

ACTG Test: Draw a double stranded stretch of DNA if one strand has the sequence ACGT

Describe how DNA is packaged into chromosomes What are histones and what is their

Describe how DNA is packaged into chromosomes What are histones and what is their purpose in the cell?

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Chromosome Structure § DNA

Chapter 12 Molecular Genetics 12. 1 DNA: The Genetic Material Chromosome Structure § DNA coils around histones to form nucleosomes, which coil to form chromatin fibers. § The chromatin fibers supercoil to form chromosomes that are visible in the metaphase stage of mitosis.

Chapter 12 Molecular Genetics 12. 2 Replication of DNA Semiconservative Replication § Parental strands

Chapter 12 Molecular Genetics 12. 2 Replication of DNA Semiconservative Replication § Parental strands of DNA separate, serve as templates, and produce DNA molecules that have one strand of parental DNA and one strand of new DNA.

 • The replication of a DNA molecule begins at special sites, origins of

• The replication of a DNA molecule begins at special sites, origins of replication. • In bacteria, this is a single specific sequence of nucleotides that is recognized by the replication enzymes. – These enzymes separate the strands, forming a replication “bubble”. – Replication proceeds in both directions until the entire molecule is copied.

 • In eukaryotes, there may be hundreds or thousands of origin sites per

• In eukaryotes, there may be hundreds or thousands of origin sites per chromosome. – At the origin sites, the DNA strands separate forming a replication “bubble” with replication forks at each end. – The replication bubbles elongate as the DNA is replicated and eventually fuse.

Define what each of these do: -DNA polymerase: -DNA helicase: -RNA primase: -DNA Ligase:

Define what each of these do: -DNA polymerase: -DNA helicase: -RNA primase: -DNA Ligase: Draw a sketch of DNA replication taking place at the replication fork. Label each enzyme.

 • The strands in the double helix are antiparallel. • The sugar-phosphate backbones

• The strands in the double helix are antiparallel. • The sugar-phosphate backbones run in opposite directions. -phosphate – Each DNA strand has a 3’ end with a free hydroxyl group attached to deoxyribose and a 5’ end with a free phosphate group attached to deoxyribose. – The 5’ -> 3’ direction of one strand runs counter to the 3’ -> 5’ direction of the other strand.

 • DNA polymerase catalyzes the elongation of new DNA at a replication fork.

• DNA polymerase catalyzes the elongation of new DNA at a replication fork. • As nucleotides align with complementary bases along the template strand, they are added to the growing end of the new strand by the polymerase. DNA polymerases can only add nucleotides to the free 3’ end of a growing DNA strand (the 3’ end of a nucleotide on the end of a strand).

 • The leading strand is created continuously towards to fork 3’ • The

• The leading strand is created continuously towards to fork 3’ • The lagging strand, is copied away from the fork in short segments (Okazaki fragments). 5’ 5’ 3’ 3’ 5’

 • DNA polymerases cannot initiate synthesis of a polynucleotide because they can only

• DNA polymerases cannot initiate synthesis of a polynucleotide because they can only add nucleotides to the end of an existing chain that is base-paired with the template strand. • To start a new chain requires a primer, a short segment of RNA. – The primer is about 10 nucleotides long in eukaryotes. • RNA Primase, links ribonucleotides(RNA) that are complementary to the DNA template into the primer. – RNA polymerases can start an RNA chain from a single template strand.

 • After formation of the primer, DNA polymerases can add DNA nucleotides to

• After formation of the primer, DNA polymerases can add DNA nucleotides to the 3’ end of the RNA. Later, a different DNA polymerase will come through, cut out the RNA nucleotides, and replace them with DNA Finally DNA Ligase comes through to make covalent phosphate sugar “backbone” bonds RNA Primase making an RNA Primer

 • Returning to the original problem at the replication fork, the leading strand

• Returning to the original problem at the replication fork, the leading strand requires the formation of only a single primer as the replication fork continues to separate. Primer was here

The lagging strand requires formation of many new primers as the replication fork progresses.

The lagging strand requires formation of many new primers as the replication fork progresses. After the primer is formed, DNA polymerase can add new nucleotides away from the fork until it runs into the previous Okazaki fragment

Helicase untwists and separates the template DNA strands at the replication fork. • Single-strand

Helicase untwists and separates the template DNA strands at the replication fork. • Single-strand binding proteins keep the unpaired template strands apart during replication.

Chapter 12 Molecular Genetics 12. 2 Replication of DNA Joining § DNA polymerase removes

Chapter 12 Molecular Genetics 12. 2 Replication of DNA Joining § DNA polymerase removes the RNA primer and fills in the place with DNA nucleotides. § DNA ligase links the two sections.

DNA Ligase: After Primer RNA replaced with DNA by DNA polymerase, There is a

DNA Ligase: After Primer RNA replaced with DNA by DNA polymerase, There is a space where the sugar of one nucleotide before the gap is not covalently bonded to the phosphate of the next. DNA Ligase makes this bond

 • To summarize, at the replication fork, the leading stand is copied continuously

• To summarize, at the replication fork, the leading stand is copied continuously into the fork from a single primer. • The lagging strand is copied away from the fork in short segments, each requiring a new primer. Primer was here. RNA later gets replaced with DNA by polymerase

Summary of enzymes in DNA Replication: -DNA Polymerase: Lays down new DNA nucleotides Replaces

Summary of enzymes in DNA Replication: -DNA Polymerase: Lays down new DNA nucleotides Replaces RNA nucleotides put down by primase Proofreads finished copy looking for errors -Primase: Lays down RNA primer so DNA polymerase can add DNA -Ligase: Seals up nicks in DNA “backbone” after RNA primers get replaced -Helicase: Unzipper of parent DNA pulls apart original double strand

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein § RNA § Contains

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein § RNA § Contains the sugar ribose and the base uracil § Usually is single stranded

 • The pentose joined to the nitrogen base is ribose in nucleotides of

• The pentose joined to the nitrogen base is ribose in nucleotides of RNA and deoxyribose in DNA. – The only difference between the sugars is the lack of an oxygen atom on carbon two in deoxyribose. – The combination of a pentose and nucleic acid is a nucleoside. Sugars Deoxyribose (in DNA) Ribose (in RNA) (c) Nucleoside components: sugars

What is the role of each of the 3 types of RNA? m. RNA

What is the role of each of the 3 types of RNA? m. RNA r. RNA t. RNA

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein Messenger RNA (m. RNA)

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein Messenger RNA (m. RNA) § Copy of gene that leaves nucleus to go to ribo. Ribosomal RNA (r. RNA) § RNA makes ribosomes Transfer RNA (t. RNA) § Brings correct AA to the m. RNA at the ribosome

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein

Explain the “Central Dogma” of Biology:

Explain the “Central Dogma” of Biology:

Transcription and translation are the two main processes linking gene to protein: an overview

Transcription and translation are the two main processes linking gene to protein: an overview • Genes provide the instructions for making specific proteins. • The bridge between DNA and protein synthesis is RNA.

 • In DNA or RNA, the four nucleotide monomers act like the letters

• In DNA or RNA, the four nucleotide monomers act like the letters of the alphabet to communicate information. • To get from DNA, written in one chemical language, to protein, written in another, requires two major stages: • Transcription (DNA gene copied into m. RNA) • and • Translation (m. RNA copy turned into AA seq).

 • During transcription, a DNA strand provides a template for the synthesis of

• During transcription, a DNA strand provides a template for the synthesis of a complementary RNA strand. – This process is used to synthesize any type of RNA from a DNA template. – Transcription of a gene produces a messenger RNA (m. RNA) molecule. • During translation, the information contained in the order of nucleotides in m. RNA is used to determine the amino acid sequence of a polypeptide. – Translation occurs at ribosomes.

Define Transcription & where does it take place? :

Define Transcription & where does it take place? :

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein Transcription § Through transcription,

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein Transcription § Through transcription, the DNA code is transferred to m. RNA in the nucleus. § DNA is unzipped in the nucleus and RNA polymerase binds to a specific section where an m. RNA will be synthesized.

 • Not all of the m. RNA that comes off the DNA is

• Not all of the m. RNA that comes off the DNA is part of the gene. These noncoding portions of the RNA molecule (INTRONS) must be cut out. • Most eukaryotic genes and their RNA transcripts have long noncoding stretches of nucleotides that must be removed – Noncoding segments = introns. These lie between coding regions. (intron = intervening sequence) – Coding regions = exons. These are translated into amino acid sequences. (exon = expressed sequence)

Explain the difference between Introns & Exons:

Explain the difference between Introns & Exons:

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein RNA Processing § The

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein RNA Processing § The code on the DNA is interrupted periodically by sequences that are not in the final m. RNA. § Intervening sequences are called introns. § Remaining pieces of DNA that serve as the coding sequences are called exons.

 • RNA splicing removes introns and joins exons to create an m. RNA

• RNA splicing removes introns and joins exons to create an m. RNA molecule with a continuous coding sequence.

In the genetic code, nucleotide triplets specify amino acids • If the genetic code

In the genetic code, nucleotide triplets specify amino acids • If the genetic code “word” consisted of a single nucleotide, each base would call for 1 unique amino acid in the protein (4 bases would = 4 different AA’s) • But there are 20 amino acids that the cell needs to be able to call for specifically • Code words of two nucleotides would give 24 = 16 “words”, Not enough • Triplets of nucleotide bases are the smallest units of “code word length that can code for all the amino acids. • In the triplet code, three consecutive bases specify an amino acid, creating 43 (64) possible code words.

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein The Code § Experiments

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein The Code § Experiments during the 1960 s demonstrated that the DNA code was a three-base code. § The three-base code in DNA or m. RNA is called a codon.

 • During transcription, one DNA strand, the template strand, provides a template for

• During transcription, one DNA strand, the template strand, provides a template for ordering the sequence of nucleotides in an RNA transcript. During translation, blocks of three nucleotides, codons, are decoded into a sequence of amino acids.

 • By the mid-1960 s the entire code was deciphered. – 61 of

• By the mid-1960 s the entire code was deciphered. – 61 of 64 triplets code for amino acids. – The codon AUG not only codes for the amino acid methionine but also indicates the start of translation. – Three codons do not indicate amino acids but signal the termination of translation.

What AA is called for by the m. RNA codon: CCA? A. B. C.

What AA is called for by the m. RNA codon: CCA? A. B. C. D. Gly Pro Ile Glu

What AA is called for by the DNA codon: CCC? A. B. C. D.

What AA is called for by the DNA codon: CCC? A. B. C. D. Pro Gly Glu Ala

What AA is called for by the m. RNA codon: UGA? A. B. C.

What AA is called for by the m. RNA codon: UGA? A. B. C. D. Thr Cys Leu None

Why is the genetic code table used as evidence supporting evolution?

Why is the genetic code table used as evidence supporting evolution?

The genetic code must have evolved very early in the history of life •

The genetic code must have evolved very early in the history of life • The genetic code is universal, shared by organisms from the simplest bacteria to the most complex plants and animals. The near universality of the genetic code indicates the code as it is must have been operating very early in the history of life. (evidence of evolution)

Define Translation & where does it take place?

Define Translation & where does it take place?

Translation is the RNA-directed synthesis of a polypeptide: a closer look • In the

Translation is the RNA-directed synthesis of a polypeptide: a closer look • In the process of translation, a cell interprets a series of codons along a m. RNA molecule. • Transfer RNA (t. RNA) transfers amino acids from the cytoplasm’s pool to a ribosome. • The ribosome adds each amino acid carried by t. RNA to the growing end of the polypeptide chain.

 • During translation, each type of t. RNA links a m. RNA codon

• During translation, each type of t. RNA links a m. RNA codon with the appropriate amino acid • Each t. RNA arriving at the ribosome carries a specific amino acid at one end and has a specific nucleotide triplet, an anticodon, at the other. • The anticodon base-pairs with a complementary codon on m. RNA. – If the codon on m. RNA is UUU, a t. RNA with an AAA anticodon will bind to it. • Codon by codon, t. RNAs deposit amino acids in the prescribed order and the ribosome joins them into a polypeptide chain.

 • A t. RNA molecule consists of a strand of about 80 nucleotides

• A t. RNA molecule consists of a strand of about 80 nucleotides that folds back on itself to form a three -dimensional structure. – It includes a loop containing the anticodon and an attachment site for an amino acid.

 • Codon by codon, amino acids are brought in until the polypeptide chain

• Codon by codon, amino acids are brought in until the polypeptide chain is completed.

What RNA carries the genetic code from the DNA to the ribosome? A. B.

What RNA carries the genetic code from the DNA to the ribosome? A. B. C. D. r. RNA ss. RNA t. RNA m. RNA

What RNA drags the correct Amino Acid to the ribosome/m. RNA to be inserted

What RNA drags the correct Amino Acid to the ribosome/m. RNA to be inserted into a growing protein A. B. C. D. r. RNA ss. RNA t. RNA m. RNA

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein One Gene— One Enzyme

Chapter 12 Molecular Genetics 12. 3 DNA, RNA, and Protein One Gene— One Enzyme § The Beadle and Tatum experiment showed that one gene codes for one enzyme. We now know that one gene codes for one polypeptide.

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Prokaryote Gene Regulation §

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Prokaryote Gene Regulation § Ability of an organism to control which genes are transcribed in response to the environment § An operon is a section of DNA that contains the genes for the proteins needed for a specific metabolic pathway. § Operator § Promoter § Regulatory gene § Genes coding for proteins

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation The Trp Operon Trp

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation The Trp Operon Trp operon is for making all proteins needed to make the AA tryptophan If Trp is not in environment…then bacteria cell needs to make it

Trp operon is for making all proteins needed to make the AA tryptophan If

Trp operon is for making all proteins needed to make the AA tryptophan If Trp is in environment…then doesn’t make sense to make more Trp activates the “off switch"

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation The Lac Operon Lac

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation The Lac Operon Lac operon is for making all proteins needed to utilize the sugar Lactose for energy If there is no Lactose in environment doesn’t make sense to make these proteins

If Lactose is present then Lactose deactivates the “off switch” and these genes are

If Lactose is present then Lactose deactivates the “off switch” and these genes are turned into m. RNA, then proteins

Operon summary -Operons are ways for prokaryotes (bact) to easily control a series of

Operon summary -Operons are ways for prokaryotes (bact) to easily control a series of related genes. By packaging the genes in a row with an “On/Off” switch at the start, the cell can quickly turn the whole series on or off with out having to regulate each gene individually

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Eukaryote Gene Regulation §

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Eukaryote Gene Regulation § Controlling transcription § Transcription factors ensure that a gene is used at the right time and that proteins are made in the right amounts § Once transcription factor binds in front of gene then RNA polymerase can make the m. RNA

2 genes needed for a related task (ex. grow new skin to heal a

2 genes needed for a related task (ex. grow new skin to heal a cut) are regulated together by both requiring the same unique transcription factor. Only these two genes will be active when this transcription factor is released. These two genes might be on separate chromosmes (the two related genes are not near each other like in operon model)

Gene A on chromosome 17 2 genes needed to start cell growth to heal

Gene A on chromosome 17 2 genes needed to start cell growth to heal cut require same transcription factor Gene B on chromosome 20

Transcription factors summary -Transcription factors are proteins that go out and bind to the

Transcription factors summary -Transcription factors are proteins that go out and bind to the “On switch” (promoter) for genes -Each transcription factor will bind to and activate the promoter for a unique series of related genes -These genes are all needed for a related function -The individual genes might be far apart in our chromosomes, but the unique Trans Factor will go out and find them -Transcription factors are how euk. Cells can control a series of related genes without having to worry about turning each one on and off individually

Match each point mutation to its description 24___Silent A. 25___missense B. 26___nonsense C. 27___base

Match each point mutation to its description 24___Silent A. 25___missense B. 26___nonsense C. 27___base pair deletion D. 28___base pair insertion E.

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutations § A permanent change

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutations § A permanent change that occurs in a cell’s DNA is called a mutation.

Point mutations • 3 types of point mutations: A. Base pair substitutions 1. Silent

Point mutations • 3 types of point mutations: A. Base pair substitutions 1. Silent 2. Missense 3. Nonsense B. Base pair deletion C. Base pair insertion

silent missense nonsense Change in codon, but still indicates the same amino acids because

silent missense nonsense Change in codon, but still indicates the same amino acids because of redundancy in the genetic code Change in codon that makes it now code for a different amino acid than what it called for originally. change an amino acid codon into a stop codon, nearly always leading to a nonfunctional protein

Explain why base pair deletions and insertions are considered to be much worse mutations

Explain why base pair deletions and insertions are considered to be much worse mutations in general than are the substitutions:

Insertions and deletions are additions or losses of nucleotide pairs in a gene. –

Insertions and deletions are additions or losses of nucleotide pairs in a gene. – These have a disastrous effect on the resulting protein more often than substitutions do. • Unless these mutations occur in multiples of three, they cause a frameshift mutation. – All the nucleotides downstream of the deletion or insertion will be improperly grouped into codons. – The result will be extensive missense, ending sooner or later in nonsense - premature termination.

 • Mutagens are chemical or physical agents that interact with DNA to cause

• Mutagens are chemical or physical agents that interact with DNA to cause mutations. • Physical agents include high-energy radiation like X-rays and ultraviolet light. • Chemical mutagens (cigarette anyone? ) may operate in several ways. – Some chemicals are base analogues that may be substituted into DNA, but that pair incorrectly during DNA replication. – Other mutagens interfere with DNA replication by inserting into DNA and distorting the double helix. – Still others cause chemical changes in bases that change their pairing properties.

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Body-cell v. Sex-cell Mutation

Chapter 12 Molecular Genetics 12. 4 Gene Regulation and Mutation Body-cell v. Sex-cell Mutation § Somatic cell mutations are not passed on to the next generation. § Mutations that occur in sex cells are passed on to the organism’s offspring and will be present in every cell of the offspring.

In class Prod = Poster titled “Beware… Mutagens in the greater KM area”

In class Prod = Poster titled “Beware… Mutagens in the greater KM area”

HW Assign: Eukaryotic vs Prokaryotic regulation of gene expression as “Album Art” One bands

HW Assign: Eukaryotic vs Prokaryotic regulation of gene expression as “Album Art” One bands makes album “Operon” Other band makes album “Transcription Factors”. How would they illustrate their albums themes?

Chapter 12 Molecular Genetics Chapter Resource Menu Chapter Diagnostic Questions Formative Test Questions Chapter

Chapter 12 Molecular Genetics Chapter Resource Menu Chapter Diagnostic Questions Formative Test Questions Chapter Assessment Questions Standardized Test Practice biologygmh. com Glencoe Biology Transparencies Image Bank Vocabulary Animation Click on a hyperlink to view the corresponding lesson.

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Which scientist(s) definitively proved that DNA transfers

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Which scientist(s) definitively proved that DNA transfers genetic material? A. Watson and Crick B. Mendel C. Hershey and Chase D. Avery

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Name the small segments of the lagging

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Name the small segments of the lagging DNA strand. A. ligase B. Okazaki fragments C. polymerase D. helicase

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Which is not true of RNA? A.

Chapter 12 Molecular Genetics Chapter Diagnostic Questions Which is not true of RNA? A. It contains the sugar ribose. B. It contains the base uracil. C. It is single-stranded. D. It contains a phosphate.

Chapter 12 Molecular Genetics 12. 1 Formative Questions The experiments of Avery, Hershey and

Chapter 12 Molecular Genetics 12. 1 Formative Questions The experiments of Avery, Hershey and Chase provided evidence that the carrier of genetic information is _______. A. carbohydrate B. DNA C. lipid D. protein

Chapter 12 Molecular Genetics 12. 1 Formative Questions What is the base-pairing rule for

Chapter 12 Molecular Genetics 12. 1 Formative Questions What is the base-pairing rule for purines and pyrimidines in the DNA molecule? A. A—G and C—T B. A—T and C—G C. C—A and G—T D. C—U and A—G

Chapter 12 Molecular Genetics 12. 1 Formative Questions What are chromosomes composed of? A.

Chapter 12 Molecular Genetics 12. 1 Formative Questions What are chromosomes composed of? A. chromatin and histones B. DNA and protein C. DNA and lipids D. protein and centromeres

Chapter 12 Molecular Genetics 12. 2 Formative Questions True or False The work of

Chapter 12 Molecular Genetics 12. 2 Formative Questions True or False The work of Watson and Crick solved the mystery of how DNA works as a genetic code.

Chapter 12 Molecular Genetics 12. 2 Formative Questions Which is not an enzyme involved

Chapter 12 Molecular Genetics 12. 2 Formative Questions Which is not an enzyme involved in DNA replication? A. DNA ligase B. DNA polymerase C. hilicase D. RNA primer

Chapter 12 Molecular Genetics 12. 2 Formative Questions During DNA replication, what nucleotide base

Chapter 12 Molecular Genetics 12. 2 Formative Questions During DNA replication, what nucleotide base sequence is synthesized along an original strand that has the sequence TCAAGC? A. AGTTCG B. ATGGCG C. CTGGAT D. GACCTA

Chapter 12 Molecular Genetics 12. 3 Formative Questions Which shows the basic chain of

Chapter 12 Molecular Genetics 12. 3 Formative Questions Which shows the basic chain of events in all organisms for reading and expressing genes? A. DNA RNA protein B. RNA DNA protein C. m. RNA r. RNA t. RNA D. RNA processing transcription translation

Chapter 12 Molecular Genetics 12. 3 Formative Questions In the RNA molecule, uracil replaces

Chapter 12 Molecular Genetics 12. 3 Formative Questions In the RNA molecule, uracil replaces _______. A. adenine B. cytosine C. purine D. thymine

Chapter 12 Molecular Genetics 12. 3 Formative Questions Which diagram shows messenger RNA (m.

Chapter 12 Molecular Genetics 12. 3 Formative Questions Which diagram shows messenger RNA (m. RNA)? A. C. B. D.

Chapter 12 Molecular Genetics 12. 3 Formative Questions What characteristic of the m. RNA

Chapter 12 Molecular Genetics 12. 3 Formative Questions What characteristic of the m. RNA molecule do scientists not yet understand?

Chapter 12 Molecular Genetics 12. 3 Formative Questions A. intervening sequences in the m.

Chapter 12 Molecular Genetics 12. 3 Formative Questions A. intervening sequences in the m. RNA molecule called introns B. the original m. RNA made in the nucleus called the pre-m. RNA C. how the sequence of bases in the m. RNA molecule codes for amino acids D. the function of many adenine nucleotides at the 5′ end called the poly-A tail

Chapter 12 Molecular Genetics 12. 4 Formative Questions Why do eukaryotic cells need a

Chapter 12 Molecular Genetics 12. 4 Formative Questions Why do eukaryotic cells need a complex control system to regulate the expression of genes?

Chapter 12 Molecular Genetics 12. 4 Formative Questions A. All of an organism’s cells

Chapter 12 Molecular Genetics 12. 4 Formative Questions A. All of an organism’s cells transcribe the same genes. B. Expression of incorrect genes can lead to mutations. C. Certain genes are expressed more frequently than others are. D. Different genes are expressed at different times in an organism’s lifetime.

Chapter 12 Molecular Genetics 12. 4 Formative Questions Which type of gene causes cells

Chapter 12 Molecular Genetics 12. 4 Formative Questions Which type of gene causes cells to become specialized in structure in function? A. exon B. Hox gene C. intron D. operon

Chapter 12 Molecular Genetics 12. 4 Formative Questions What is an immediate result of

Chapter 12 Molecular Genetics 12. 4 Formative Questions What is an immediate result of a mutation in a gene? A. cancer B. genetic disorder C. nonfunctional enzyme D. amino acid deficiency

Chapter 12 Molecular Genetics 12. 4 Formative Questions Which is the most highly mutagenic?

Chapter 12 Molecular Genetics 12. 4 Formative Questions Which is the most highly mutagenic? A. chemicals in food B. cigarette smoke C. ultraviolet radiation D. X rays

Chapter 12 Molecular Genetics Chapter Assessment Questions Look at the following figure. Identify the

Chapter 12 Molecular Genetics Chapter Assessment Questions Look at the following figure. Identify the proteins that DNA first coils around.

Chapter 12 Molecular Genetics Chapter Assessment Questions A. chromatin fibers B. chromosomes C. histones

Chapter 12 Molecular Genetics Chapter Assessment Questions A. chromatin fibers B. chromosomes C. histones D. nucleosome

Chapter 12 Molecular Genetics Chapter Assessment Questions Explain how Hox genes affect an organism.

Chapter 12 Molecular Genetics Chapter Assessment Questions Explain how Hox genes affect an organism. A. They determine size. B. They determine body plan. C. They determine sex. D. They determine of body segments. number

Chapter 12 Molecular Genetics Chapter Assessment Questions Explain the difference between body-cell and sex-cell

Chapter 12 Molecular Genetics Chapter Assessment Questions Explain the difference between body-cell and sex-cell mutation.

Chapter 12 Molecular Genetics Chapter Assessment Questions Answer: A mutagen in a body cell

Chapter 12 Molecular Genetics Chapter Assessment Questions Answer: A mutagen in a body cell becomes part of the genetic sequence in that cell and in future daughter cells. The cell may die or simply not perform its normal function. These mutations are not passed on to the next generation. When mutations occur in sex cells, they will be present in every cell of the offspring.

Chapter 12 Molecular Genetics Standardized Test Practice What does this diagram show about the

Chapter 12 Molecular Genetics Standardized Test Practice What does this diagram show about the replication of DNA in eukaryotic cells?

Chapter 12 Molecular Genetics Standardized Test Practice A. DNA is replicated only at certain

Chapter 12 Molecular Genetics Standardized Test Practice A. DNA is replicated only at certain places along the chromosome. B. DNA replication is both semicontinuous and conservative. C. Multiple areas of replication occur along the chromosome at the same time. D. The leading DNA strand is synthesized discontinuously.

Chapter 12 Molecular Genetics Standardized Test Practice What is this process called?

Chapter 12 Molecular Genetics Standardized Test Practice What is this process called?

Chapter 12 Molecular Genetics Standardized Test Practice A. m. RNA processing B. protein synthesis

Chapter 12 Molecular Genetics Standardized Test Practice A. m. RNA processing B. protein synthesis C. transcription D. translation

Chapter 12 Molecular Genetics Standardized Test Practice What type of mutation results in this

Chapter 12 Molecular Genetics Standardized Test Practice What type of mutation results in this change in the DNA sequence? TTCAGG TTCTGG A. deletion B. frameshift C. insertion D. substitution

Chapter 12 Molecular Genetics Standardized Test Practice How could RNA interference be used to

Chapter 12 Molecular Genetics Standardized Test Practice How could RNA interference be used to treat diseases such as cancer and diabetes?

Chapter 12 Molecular Genetics Standardized Test Practice A. by activating genes to produce proteins

Chapter 12 Molecular Genetics Standardized Test Practice A. by activating genes to produce proteins that can overcome the disease B. by interfering with DNA replication in cells affected by the disease C. by preventing the translation of m. RNA into the genes associated with the disease D. by shutting down protein synthesis in the cells of diseased tissues

Chapter 12 Molecular Genetics Standardized Test Practice True or False The structure of a

Chapter 12 Molecular Genetics Standardized Test Practice True or False The structure of a protein can be altered dramatically by the exchange of a single amino acid for another.

Chapter 12 Molecular Genetics Glencoe Biology Transparencies

Chapter 12 Molecular Genetics Glencoe Biology Transparencies

Chapter 12 Molecular Genetics Image Bank

Chapter 12 Molecular Genetics Image Bank

Chapter 12 Molecular Genetics Image Bank

Chapter 12 Molecular Genetics Image Bank

Chapter 12 Molecular Genetics Vocabulary Section 1 double helix nucleosome

Chapter 12 Molecular Genetics Vocabulary Section 1 double helix nucleosome

Chapter 12 Molecular Genetics Vocabulary Section 2 semiconservative replication DNA polymerase Okazaki fragment

Chapter 12 Molecular Genetics Vocabulary Section 2 semiconservative replication DNA polymerase Okazaki fragment

Chapter 12 Molecular Genetics Vocabulary Section 3 RNA intron messenger RNA ribosomal RNA transfer

Chapter 12 Molecular Genetics Vocabulary Section 3 RNA intron messenger RNA ribosomal RNA transfer RNA transcription RNA polymerase codon exon translation

Chapter 12 Molecular Genetics Vocabulary Section 4 gene regulation operon mutation mutagen

Chapter 12 Molecular Genetics Vocabulary Section 4 gene regulation operon mutation mutagen

Chapter 12 Molecular Genetics Animation § DNA Polymerase § Transcription § Visualizing Transcription and

Chapter 12 Molecular Genetics Animation § DNA Polymerase § Transcription § Visualizing Transcription and Translation § Lac-Trp Operon