AP Biology Review #3

DNA Replication DNA Tutorial

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

Enzyme Quiz Single Strand Binding Proteins DNA Polymerase Helicase RNA Primase DNA Ligase A. Unwinds DNA B. Links new nucleotides together C. Holds DNA Apart D. Initiates building of new DNA strand E. Joins DNA Fragments Together

DNA Replication VOCAB Replication fork Leading Strand Lagging Strand Okazaki Fragment RNA Primer Shorter pieces of DNA that are built in the 5’3’ on the antiparallel strand Name for A Beginning of both the leading and lagging strands Where the DNA is split Continuous strand of DNA build in the 5’  3’

DNA Replication VOCAB Replication fork Leading Strand Lagging Strand Okazaki Fragment RNA Primer Shorter pieces of DNA that are built in the 5’3’ on the antiparallel strand Name for A Beginning of both the leading and lagging strands Where the DNA is split Continuous strand of DNA build in the 5’  3’

DNA Replication VOCAB Replication fork Leading Strand Lagging Strand Okazaki Fragment RNA Primer Shorter pieces of DNA that are built in the 5’3’ on the antiparallel strand Name for A Beginning of both the leading and lagging strands Where the DNA is split Continuous strand of DNA build in the 5’  3’

DNA Replication VOCAB Replication fork Leading Strand Lagging Strand Okazaki Fragment RNA Primer Shorter pieces of DNA that are built in the 5’3’ on the antiparallel strand Name for A Beginning of both the leading and lagging strands Where the DNA is split Continuous strand of DNA build in the 5’  3’

DNA Replication VOCAB Replication fork Leading Strand Lagging Strand Okazaki Fragment RNA Primer Shorter pieces of DNA that are built in the 5’3’ on the antiparallel strand Name for A Beginning of both the leading and lagging strands Where the DNA is split Continuous strand of DNA build in the 5’  3’

Errors in DNA Replication Mismatch Substitution Deletion Insertion Frameshift Mutagens: - Carcinogens - UV  Thymine Dymers

Protein Synthesis Transcription RNA Processing Translation

Transcription RNA Polymerase unzips DNA and Build RNA (5’3’) Transcription factors bind to promoter region RNA Polymerase Binds to TATA box

Transcription Transcription factors bind to promoter region RNA Polymerase Binds to TATA box RNA Polymerase unzips DNA and Build RNA (5’3’)

RNA Processing GTP Cap added to ________ end Poly A tail added to ________ end ______ are removed by _______ ______ are spliced together This happens in the __________ but not in _____________ cells.

RNA Processing GTP Cap added to 5’ end Poly A tail added to 3’ end Introns are removed by SNRnPs Exons are spliced together This happens in the nucleus but not in prokaryotic cells.

Translation – place the following steps in order and label whether they are initiation, elongation or termination. 1.Ribosome reaches a termination sequence (UAA, UAG, UGA) 2. Ribozyme catalyzes the formation of peptide bond. 3. t-RNA brings in Start Methonine to match to m-RNA 4. New t-RNA brings in next amino acid to A site 5. Small subunit of ribosome attaches to 5’cap. 6. Translation Complex Breaks Apart 7. Large subunit lines up so Start Meth is in the P site 8. Release factor comes in instead of t-RNA. 9. t-RNAs translocate: A  P, P  E (exit) 10. Process Repeats

Translation – place the following steps in order and label whether they are initiation, elongation or termination. 5. Small subunit of ribosome attaches to 5’cap. 3. t-RNA brings in Start Methonine to match to m-RNA 7. Large subunit lines up so Start Meth is in the P site 4. New t-RNA brings in next amino acid to A site 2. Ribozyme catalyzes the formation of peptide bond. 9. t-RNAs translocate: A  P, P  E (exit) 10. Repeat 1. Ribosome reaches a termination sequence (UAA, UAG, UGA) 8. Release factor comes in instead of t-RNA. 6. Translation Complex Breaks Apart

Chromatin structure Chromatin structure is based on successive levels of DNA packing Eukaryotic DNA Is precisely combined with a large amount of protein Eukaryotic chromosomes Contain an enormous amount of DNA relative to their condensed length

Nucleosomes, or “Beads on a String” Proteins called histones Are responsible for the first level of DNA packing in chromatin Bind tightly to DNA The association of DNA and histones Seems to remain intact throughout the cell cycle

(a) Nucleosomes (10-nm fiber) Each “bead” is a nucleosome The basic unit of DNA packing 2 nm 10 nm DNA double helix Histone tails His- tones Linker DNA (“string”) Nucleosome (“bad”) Histone H1 (a) Nucleosomes (10-nm fiber) Figure 19.2 a

Higher Levels of DNA Packing The next level of packing Forms the 30-nm chromatin fiber Nucleosome 30 nm (b) 30-nm fiber Figure 19.2 b

(c) Looped domains (300-nm fiber) The 30-nm fiber, in turn Forms looped domains, making up a 300-nm fiber Protein scaffold 300 nm (c) Looped domains (300-nm fiber) Loops Scaffold Figure 19.2 c

(d) Metaphase chromosome In a mitotic chromosome The looped domains themselves coil and fold forming the characteristic metaphase chromosome 700 nm 1,400 nm (d) Metaphase chromosome Figure 19.2 d

DNA Expression Control Histone acetylation Seems to loosen chromatin structure and thereby enhance transcription Figure 19.4 b (b) Acetylation of histone tails promotes loose chromatin structure that permits transcription Unacetylated histones Acetylated histones

DNA Methylation Addition of methyl groups to certain bases in DNA Is associated with reduced transcription in some species

Epigenetic Inheritance Is the inheritance of traits transmitted by mechanisms not directly involving the nucleotide sequence

Organization of a Typical Eukaryotic Gene Segments of noncoding DNA that help regulate transcription by binding certain proteins Enhancer (distal control elements) Proximal control elements DNA Upstream Promoter Exon Intron Poly-A signal sequence Termination region Transcription Downstream Poly-A signal Primary RNA transcript (pre-mRNA) 5 Intron RNA RNA processing: Cap and tail added; introns excised and exons spliced together Coding segment P G mRNA 5 Cap 5 UTR (untranslated region) Start codon Stop 3 UTR tail Chromatin changes RNA processing degradation Translation Protein processing and degradation Cleared 3 end of primary transport Figure 19.5

Figure 19.6 Distal control element Activators Enhancer Promoter Gene TATA box General transcription factors DNA-bending protein Group of Mediator proteins RNA Polymerase II RNA synthesis Transcription Initiation complex Chromatin changes RNA processing mRNA degradation Translation Protein processing and degradation A DNA-bending protein brings the bound activators closer to the promoter. Other transcription factors, mediator proteins, and RNA polymerase are nearby. 2 Activator proteins bind to distal control elements grouped as an enhancer in the DNA. This enhancer has three binding sites. 1 The activators bind to certain general transcription factors and mediator proteins, helping them form an active transcription initiation complex on the promoter. 3 Figure 19.6

Movement of Transposons and Retrotransposons Eukaryotic transposable elements are of two types Transposons, which move within a genome by means of a DNA intermediate Retrotransposons, which move by means of an RNA intermediate Figure 19.16a, b

Genetics of Viruses and Bacteria

Figure 18.4 Viral structure 18  250 mm 70–90 nm (diameter) 80–200 nm (diameter) 80  225 nm 20 nm 50 nm (a) Tobacco mosaic virus (b) Adenoviruses (c) Influenza viruses (d) Bacteriophage T4 RNA Capsomere of capsid DNA Capsomere Glycoprotein Membranous envelope Capsid Head Tail fiber Tail sheath

Figure 18.5 A simplified viral reproductive cycle VIRUS Capsid proteins mRNA Viral DNA HOST CELL Entry into cell and uncoating of DNA Replication Transcription DNA Capsid Self-assembly of new virus particles and their exit from cell

Figure 18.6 The lytic cycle of phage T4, a virulent phage Attachment. The T4 phage uses its tail fibers to bind to specific receptor sites on the outer surface of an E. coli cell. Entry of phage DNA and degradation of host DNA. The sheath of the tail contracts, injecting the phage DNA into the cell and leaving an empty capsid outside. The cell’s DNA is hydrolyzed. Synthesis of viral genomes and proteins. The phage DNA directs production of phage proteins and copies of the phage genome by host enzymes, using components within the cell. Assembly. Three separate sets of proteins self-assemble to form phage heads, tails, and tail fibers. The phage genome is packaged inside the capsid as the head forms. Release. The phage directs production of an enzyme that damages the bacterial cell wall, allowing fluid to enter. The cell swells and finally bursts, releasing 100 to 200 phage particles. 1 2 4 3 5 Phage assembly Head Tails Tail fibers

The lytic and lysogenic cycles of phage , a temperate phage Many cell divisions produce a large population of bacteria infected with the prophage. The bacterium reproduces normally, copying the prophage and transmitting it to daughter cells. Phage DNA integrates into the bacterial chromosome, becoming a prophage. New phage DNA and proteins are synthesized and assembled into phages. Occasionally, a prophage exits the bacterial chromosome, initiating a lytic cycle. Certain factors determine whether The phage attaches to a host cell and injects its DNA. Phage DNA circularizes The cell lyses, releasing phages. Lytic cycle is induced Lysogenic cycle is entered or Prophage Bacterial chromosome Phage DNA

Figure 18.13 Model for how prions propagate Normal protein Original prion New prion Many prions

Figure 18.14 Replication of a bacterial chromosome Replication fork Origin of replication Termination of replication

Bacterial Genetics - Matching Transduction Transformation Conjugation Pick of small pieces of DNA Get Bacterial DNA from a faulty virus Transfer of plasmids via a pilus

Bacterial Genetics Transformation Transduction Conjugation Pick of small pieces of DNA Get Bacterial DNA from a faulty virus Transfer of plasmids via a pilus

Figure 18.16 Generalized transduction Phage DNA Donor cell A+ B+ Phage infects bacterial cell that has alleles A+ and B+ Host DNA (brown) is fragmented, and phage DNA and proteins are made. This is the donor cell. A bacterial DNA fragment (in this case a fragment with the A+ allele) may be packaged in a phage capsid. 1 2 3

Figure 18.16 Generalized transduction Phage DNA Donor cell Recipient cell A+ B+ B– A– Recombinant cell Crossing over Phage infects bacterial cell that has alleles A+ and B+ Host DNA (brown) is fragmented, and phage DNA and proteins are made. This is the donor cell. A bacterial DNA fragment (in this case a fragment with the A+ allele) may be packaged in a phage capsid. Phage with the A+ allele from the donor cell infects a recipient A–B– cell, and crossing over (recombination) between donor DNA (brown) and recipient DNA (green) occurs at two places (dotted lines). The genotype of the resulting recombinant cell (A+B–) differs from the genotypes of both the donor (A+B+) and the recipient (A–B–). 1 2 3 4 5

Figure 18.17 Bacterial conjugation Sex pilus 1 m

Figure 18.18 Conjugation and recombination in E. coli (layer 1) A cell carrying an F plasmid (an F+ cell) can form a mating bridge with an F– cell and transfer its F plasmid. A single strand of the F plasmid breaks at a specific point (tip of blue arrowhead) and begins to move into the recipient cell. As transfer continues, the donor plasmid rotates (red arrow). 2 DNA replication occurs in both donor and recipient cells, using the single parental strands of the F plasmid as templates to synthesize complementary strands. 3 The plasmid in the recipient cell circularizes. Transfer and replication result in a compete F plasmid in each cell. Thus, both cells are now F+. 4 F Plasmid Bacterial chromosome Bacterial chromosome F– cell F+ cell Hfr cell F factor The circular F plasmid in an F+ cell can be integrated into the circular chromosome by a single crossover event (dotted line). The resulting cell is called an Hfr cell (for High frequency of recombination). Since an Hfr cell has all the F-factor genes, it can form a mating bridge with an F– cell and transfer DNA. A single strand of the F factor breaks and begins to move through the bridge. DNA replication occurs in both donor and recipient cells, resulting in double-stranded DNA The location and orientation of the F factor in the donor chromosome determine the sequence of gene transfer during conjugation. In this example, the transfer sequence for four genes is A-B-C-D. 5 The mating bridge usually breaks well before the entire chromosome and the rest of the F factor are transferred. 6 Two crossovers can result in the exchange of similar (homologous) genes between the transferred chromosome fragment (brown) and the recipient cell’s chromosome (green). 7 The piece of DNA ending up outside the bacterial chromosome will eventually be degraded by the cell’s enzymes. The recipient cell now contains a new combination of genes but no F factor; it is a recombinant F– cell. 8 Temporary partial diploid Recombinant F– bacterium Conjugation and transfer of an F plasmid from an F+ donor to an F– recipient (a) Conjugation and transfer of part of the bacterial chromosome from an Hfr donor to an F– recipient, resulting in recombination (b) A+ B+ C+ D+ F– cell A– B– C– D– Mating bridge

Figure 18.18 Conjugation and recombination in E. coli (layer 2) A cell carrying an F plasmid (an F+ cell) can form a mating bridge with an F– cell and transfer its F plasmid. A single strand of the F plasmid breaks at a specific point (tip of blue arrowhead) and begins to move into the recipient cell. As transfer continues, the donor plasmid rotates (red arrow). 2 DNA replication occurs in both donor and recipient cells, using the single parental strands of the F plasmid as templates to synthesize complementary strands. 3 The plasmid in the recipient cell circularizes. Transfer and replication result in a compete F plasmid in each cell. Thus, both cells are now F+. 4 F Plasmid Bacterial chromosome Bacterial chromosome F– cell F+ cell Hfr cell F factor The circular F plasmid in an F+ cell can be integrated into the circular chromosome by a single crossover event (dotted line). The resulting cell is called an Hfr cell (for High frequency of recombination). Since an Hfr cell has all the F-factor genes, it can form a mating bridge with an F– cell and transfer DNA. A single strand of the F factor breaks and begins to move through the bridge. DNA replication occurs in both donor and recipient cells, resulting in double-stranded DNA The location and orientation of the F factor in the donor chromosome determine the sequence of gene transfer during conjugation. In this example, the transfer sequence for four genes is A-B-C-D. 5 The mating bridge usually breaks well before the entire chromosome and the rest of the F factor are transferred. 6 Two crossovers can result in the exchange of similar (homologous) genes between the transferred chromosome fragment (brown) and the recipient cell’s chromosome (green). 7 The piece of DNA ending up outside the bacterial chromosome will eventually be degraded by the cell’s enzymes. The recipient cell now contains a new combination of genes but no F factor; it is a recombinant F– cell. 8 Temporary partial diploid Recombinant F– bacterium Conjugation and transfer of an F plasmid from an F+ donor to an F– recipient (a) Conjugation and transfer of part of the bacterial chromosome from an Hfr donor to an F– recipient, resulting in recombination (b) A+ B+ C+ D+ F– cell A– B– C– D– Mating bridge

Figure 18.18 Conjugation and recombination in E. coli (layer 3) A cell carrying an F plasmid (an F+ cell) can form a mating bridge with an F– cell and transfer its F plasmid. A single strand of the F plasmid breaks at a specific point (tip of blue arrowhead) and begins to move into the recipient cell. As transfer continues, the donor plasmid rotates (red arrow). 2 DNA replication occurs in both donor and recipient cells, using the single parental strands of the F plasmid as templates to synthesize complementary strands. 3 The plasmid in the recipient cell circularizes. Transfer and replication result in a compete F plasmid in each cell. Thus, both cells are now F+. 4 F Plasmid Bacterial chromosome Bacterial chromosome F– cell F+ cell Hfr cell F factor The circular F plasmid in an F+ cell can be integrated into the circular chromosome by a single crossover event (dotted line). The resulting cell is called an Hfr cell (for High frequency of recombination). Since an Hfr cell has all the F-factor genes, it can form a mating bridge with an F– cell and transfer DNA. A single strand of the F factor breaks and begins to move through the bridge. DNA replication occurs in both donor and recipient cells, resulting in double-stranded DNA The location and orientation of the F factor in the donor chromosome determine the sequence of gene transfer during conjugation. In this example, the transfer sequence for four genes is A-B-C-D. 5 The mating bridge usually breaks well before the entire chromosome and the rest of the F factor are transferred. 6 Two crossovers can result in the exchange of similar (homologous) genes between the transferred chromosome fragment (brown) and the recipient cell’s chromosome (green). 7 The piece of DNA ending up outside the bacterial chromosome will eventually be degraded by the cell’s enzymes. The recipient cell now contains a new combination of genes but no F factor; it is a recombinant F– cell. 8 Temporary partial diploid Recombinant F– bacterium Conjugation and transfer of an F plasmid from an F+ donor to an F– recipient (a) Conjugation and transfer of part of the bacterial chromosome from an Hfr donor to an F– recipient, resulting in recombination (b) A+ B+ C+ D+ A– B– C– D– Mating bridge

Figure 18.18 Conjugation and recombination in E. coli (layer 4) A cell carrying an F plasmid (an F+ cell) can form a mating bridge with an F– cell and transfer its F plasmid. A single strand of the F plasmid breaks at a specific point (tip of blue arrowhead) and begins to move into the recipient cell. As transfer continues, the donor plasmid rotates (red arrow). 2 DNA replication occurs in both donor and recipient cells, using the single parental strands of the F plasmid as templates to synthesize complementary strands. 3 The plasmid in the recipient cell circularizes. Transfer and replication result in a compete F plasmid in each cell. Thus, both cells are now F+. 4 F Plasmid Bacterial chromosome Bacterial chromosome F– cell F+ cell Hfr cell F factor The circular F plasmid in an F+ cell can be integrated into the circular chromosome by a single crossover event (dotted line). The resulting cell is called an Hfr cell (for High frequency of recombination). Since an Hfr cell has all the F-factor genes, it can form a mating bridge with an F– cell and transfer DNA. A single strand of the F factor breaks and begins to move through the bridge. DNA replication occurs in both donor and recipient cells, resulting in double-stranded DNA The location and orientation of the F factor in the donor chromosome determine the sequence of gene transfer during conjugation. In this example, the transfer sequence for four genes is A-B-C-D. 5 The mating bridge usually breaks well before the entire chromosome and the rest of the F factor are transferred. 6 Two crossovers can result in the exchange of similar (homologous) genes between the transferred chromosome fragment (brown) and the recipient cell’s chromosome (green). 7 The piece of DNA ending up outside the bacterial chromosome will eventually be degraded by the cell’s enzymes. The recipient cell now contains a new combination of genes but no F factor; it is a recombinant F– cell. 8 Temporary partial diploid Recombinant F– bacterium Conjugation and transfer of an F plasmid from an F+ donor to an F– recipient (a) Conjugation and transfer of part of the bacterial chromosome from an Hfr donor to an F– recipient, resulting in recombination (b) A+ B+ C+ D+ A– B– C– D– Mating bridge

Figure 18.20 Regulation of a metabolic pathway (a) Regulation of enzyme activity Enzyme 1 Enzyme 2 Enzyme 3 Enzyme 4 Enzyme 5 Regulation of gene expression Feedback inhibition Tryptophan Precursor (b) Regulation of enzyme production Gene 2 Gene 1 Gene 3 Gene 4 Gene 5 –

 Trp Operon (a) Tryptophan absent, repressor inactive, operon on. RNA polymerase attaches to the DNA at the promoter and transcribes the operon’s genes. Genes of operon Inactive repressor Protein Operator Polypeptides that make up enzymes for tryptophan synthesis Promoter Regulatory gene RNA polymerase Start codon Stop codon trp operon 5 3 mRNA 5 trpD trpE trpC trpB trpA trpR DNA mRNA E D C B A

Lac Operon lacl mRNA 5' DNA mRNA Protein Allolactose (inducer) Inactive repressor lacl lacz lacY lacA RNA polymerase Permease Transacetylase -Galactosidase 5 3 (b) Lactose present, repressor inactive, operon on. Allolactose, an isomer of lactose, derepresses the operon by inactivating the repressor. In this way, the enzymes for lactose utilization are induced. mRNA 5 lac operon

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