AP Biology Unit 6: Gene Expression and Regulation

Free Unit Notes

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I. DNA and RNA Structure (Topic 6.1)

A. Hereditary information and chromosomes

1. DNA stores hereditary information in its base sequence. Some viruses use RNA as their genetic material.

2. Prokaryotes typically have a circular chromosome in the nucleoid, a region without a surrounding membrane.

3. Eukaryotes typically have multiple linear chromosomes in a nucleus. DNA wraps around histones and associates with other proteins to form chromatin.

4. Plasmids are small, separate DNA molecules, usually circular. They occur in prokaryotes and some eukaryotes.

B. Nucleotides and complementary pairing

1. A nucleotide contains a phosphate group, a five-carbon sugar, and a nitrogenous base. Covalent bonds join nucleotides into a sugar-phosphate backbone.

2. DNA contains deoxyribose and the bases adenine (A), thymine (T), guanine (G), and cytosine (C). RNA contains ribose and uracil (U) instead of thymine.

3. Purines, A and G, have two rings. Pyrimidines, C, T, and U, have one ring. Complementary pairing matches a purine with a pyrimidine.

4. In DNA, A pairs with T through two hydrogen bonds, and G pairs with C through three. In RNA pairing, A pairs with U.

5. DNA usually forms an antiparallel double helix. One strand runs 5′ to 3′, while its partner runs 3′ to 5′. RNA is usually single-stranded and can fold through internal base pairing.

6. In double-stranded DNA, %A = %T and %G = %C. If A is 30%, T is 30%, and G and C are each 20%.

II. DNA Replication (Topic 6.2)

A. Copying the DNA molecule

1. DNA replication copies genetic information before cell division. In eukaryotes, it occurs during S phase of interphase.

2. Replication is semiconservative: each daughter DNA molecule contains one parental strand and one newly made complementary strand.

3. Replication begins at origins and proceeds in both directions. Eukaryotic chromosomes generally have many origins.

B. Opening and copying the strands

1. Helicase separates the DNA strands by disrupting hydrogen bonds between complementary bases.

2. Topoisomerase relieves twisting ahead of the replication fork, where the strands separate.

3. Short RNA primers provide a free 3′ end. DNA polymerase extends an existing strand; it cannot start one without a primer.

4. DNA polymerase reads the template 3′ to 5′ and adds complementary nucleotides to the new strand’s 3′ end. New DNA always grows 5′ to 3′.

5. Many DNA polymerases proofread newly added bases. Additional repair systems help correct errors that remain.

C. Leading and lagging strands

1. The leading strand grows continuously toward the advancing replication fork.

2. The lagging strand grows in short Okazaki fragments away from the fork. Each fragment begins with a new primer.

3. RNA primers are removed and replaced with DNA. DNA ligase seals remaining breaks in the sugar-phosphate backbone.

4. Both new strands grow 5′ to 3′. Their different assembly patterns result from the opposite directions of their templates.

III. Transcription and RNA Processing (Topic 6.3)

A. Making RNA from DNA

1. Transcription uses a DNA template to make RNA. For protein-coding genes, this is the first step in gene expression.

2. Transcription machinery assembles at a promoter, a DNA region associated with the start of transcription.

3. RNA polymerase reads one DNA template strand 3′ to 5′ and makes complementary RNA 5′ to 3′. It does not require a primer.

4. RNA matches the coding strand of DNA except that U replaces T. It is complementary and antiparallel to the template strand.
   a. Template DNA 3′-TAC GGA CTT-5′ produces RNA 5′-AUG CCU GAA-3′.

5. Transcription ends when termination signals cause the RNA transcript to be released.

B. Major RNA types

1. Messenger RNA (mRNA) carries the codon sequence used to build a polypeptide.

2. Transfer RNA (tRNA) carries an amino acid and has an anticodon that pairs with a complementary mRNA codon.

3. Ribosomal RNA (rRNA) forms the structural and catalytic core of ribosomes, including the site of peptide-bond formation.

C. Eukaryotic RNA processing

1. Eukaryotic transcription occurs in the nucleus. The initial pre-mRNA is processed before mature mRNA enters the cytoplasm.

2. A 5′ cap, containing a modified guanine nucleotide, helps protect mRNA and supports ribosome recognition.

3. A poly-A tail, a series of adenine nucleotides at the 3′ end, increases mRNA stability.

4. Splicing removes introns and joins retained exons. Exons remain in mature RNA; not every exon sequence necessarily codes for amino acids.

5. Alternative splicing joins exons in different combinations. One gene can therefore produce different mRNAs and protein forms.

6. Prokaryotes lack a nuclear envelope, so ribosomes can begin translating mRNA while it is still being transcribed.

IV. Translation (Topic 6.4)

A. The genetic code

1. Translation uses mRNA information to assemble a polypeptide, a chain of amino acids. It occurs on ribosomes in the cytoplasm or on the cytoplasmic surface of rough ER.

2. A codon is a group of three mRNA bases that specifies an amino acid or a stop signal. Read codons 5′ to 3′ when using a genetic code chart.

3. The code is redundant: several codons can specify the same amino acid. Each codon has only one meaning in a given genetic code.

4. The genetic code is nearly universal, which supports common ancestry and allows many organisms to express genes from other species.

B. Building the polypeptide

1. During initiation, ribosomal subunits and an initiator tRNA assemble on mRNA at the start codon, AUG, which codes for methionine.

2. During elongation, tRNA anticodons pair with successive mRNA codons. Each tRNA delivers the corresponding amino acid.

3. The ribosome catalyzes peptide-bond formation and moves along mRNA 5′ to 3′. The polypeptide grows from its amino end toward its carboxyl end.

4. During termination, a stop codon, UAA, UAG, or UGA, recruits a release factor. The completed chain is released; a stop codon adds no amino acid.

5. The start codon establishes the reading frame, the way bases are grouped into nonoverlapping triplets.
   a. An in-frame coding sequence with 300 nucleotides, including one terminal stop codon, specifies 99 amino acids before any protein processing.

6. The amino-acid sequence affects protein folding and function. Protein type and amount help connect genotype, an organism’s genetic makeup, to phenotype, its observable traits.

C. Reverse transcription in retroviruses

1. Retroviruses use reverse transcriptase to copy their RNA genome into DNA.

2. Viral DNA integrates into host DNA. The host then transcribes viral genes and translates viral mRNAs, supplying components for new viruses.

V. Regulation of Gene Expression (Topic 6.5)

A. Controlling transcription

1. Gene regulation controls when, where, and how much a gene is expressed. Regulatory proteins bind particular DNA sequences to affect transcription.

2. Constitutive expression occurs routinely under the relevant conditions. Inducible expression increases in response to a signal.

3. Activators increase transcription, often by helping transcription machinery assemble. Repressors decrease transcription, often by blocking its initiation or progress.

B. Coordinated regulation in prokaryotes

1. An operon contains genes controlled together by shared regulatory DNA. These genes can be transcribed into one mRNA.

2. A promoter recruits RNA polymerase. An operator is a regulatory DNA site where a repressor can bind.

3. In an inducible operon, an inducer can inactivate a repressor, allowing transcription to increase.
   a. In the lac operon, a lactose-derived inducer relieves repression. Strong expression also depends on low glucose availability.

4. In a repressible operon, a corepressor helps activate a repressor, reducing transcription.
   a. In the trp operon, abundant tryptophan helps repress genes needed to make more tryptophan.

C. Eukaryotic regulation and chromatin

1. Transcription factors bind regulatory DNA and influence transcription. Genes at different locations can respond together if they share regulatory sequences recognized by the same factors.

2. Epigenetic changes modify DNA or associated proteins without changing the DNA base sequence. These changes can be reversible and can affect DNA accessibility.

3. DNA methylation near promoters is often associated with reduced transcription. Histone acetylation often loosens chromatin and increases access to DNA.

4. Changes in RNA processing, mRNA stability, translation, protein modification, or protein breakdown can also change the amount of functional gene product.

VI. Gene Expression and Cell Specialization (Topic 6.6)

A. Regulatory DNA and transcription factors

1. Transcription factors help RNA polymerase assemble at promoters. Enhancers are regulatory DNA regions where activators bind to increase transcription.

2. Enhancers may lie far from a gene, upstream or downstream of its transcription start site. DNA looping can bring enhancer-bound activators near promoter-associated machinery.

3. Negative regulators can bind DNA and inhibit transcription. Changing a regulatory sequence can alter protein production without changing the protein’s amino-acid sequence.

B. Differential gene expression

1. Most cells in a multicellular organism contain essentially the same genome. Differential gene expression means that cells express different genes or express them at different levels.

2. Cell differentiation produces specialized structures and functions through these differences in gene expression.
   a. Muscle cells produce contraction proteins, while pancreatic beta cells produce insulin.

3. During development, signals activate transcription factors in sequence. These factors regulate other genes and help establish specialized cell types.

4. Environmental signals can change gene expression and phenotype without changing DNA sequence. Such changes are not necessarily inherited by offspring.

C. Small regulatory RNAs

1. Small RNAs can guide protein complexes to complementary mRNAs. This can block translation or promote mRNA breakdown, a process called RNA interference.

2. Reducing the available mRNA generally reduces production of its protein, which can change cell function.

VII. Mutations (Topic 6.7)

A. Changes in DNA sequence

1. A mutation is a change in DNA sequence. A substitution replaces one nucleotide with another.

2. In a coding region, a silent mutation leaves the amino-acid sequence unchanged. A missense mutation changes an amino acid. A nonsense mutation creates a premature stop codon.

3. Insertions add nucleotides, and deletions remove them. Within a coding sequence, changes not divisible by three usually cause a frameshift, altering downstream codons.

4. Insertions or deletions of complete codons preserve the downstream reading frame but can still change protein function. A substitution alone does not shift the reading frame.

5. Mutations in promoters, enhancers, or splice sites can change the amount or form of gene product even when an exon’s coding sequence is unchanged.

B. Causes, effects, and inheritance

1. Mutations arise from replication errors, failed repair, or mutagens, such as radiation and certain reactive chemicals. They do not arise because an organism needs a particular trait.

2. A mutation may be harmful, neutral, or beneficial depending on its location, its effect on gene products, and the environment. It does not automatically denature a protein.

3. Somatic mutations occur in body cells and may pass to daughter cells during mitosis. They generally do not pass to offspring through sexual reproduction.

4. Germ-line mutations can enter gametes and be inherited by offspring.

C. Chromosome changes

1. Chromosome segments can be deleted, duplicated, reversed, or moved to another location. These changes may disrupt genes or alter gene dosage and regulation.

2. Nondisjunction is the failure of homologous chromosomes or sister chromatids to separate properly during meiosis or mitosis.

3. Aneuploidy is the gain or loss of individual chromosomes, such as 2n + 1. Polyploidy involves extra complete chromosome sets; triploidy is 3n.

4. Abnormal chromosome number can change gene dosage and disrupt development. Meiotic nondisjunction can produce abnormal gametes; mitotic nondisjunction can produce different chromosome numbers among body cells.

D. Genetic variation and selection

1. Mutations create new genetic variation. Recombination and sexual reproduction reshuffle existing variants.

2. Prokaryotes can acquire genetic material through horizontal gene transfer, which occurs outside parent-to-offspring inheritance.
   a. Transformation is the uptake of DNA from the environment.
   b. Transduction transfers DNA through a virus.
   c. Conjugation transfers DNA through direct contact between cells.

3. Transposition moves DNA segments within or between DNA molecules and can alter gene function or regulation.

4. Related viruses can exchange genetic information when they infect the same host cell.

5. A heritable variant that improves survival and reproductive success in a particular environment can become more common through natural selection.

VIII. Biotechnology (Topic 6.8)

A. Gel electrophoresis

1. Gel electrophoresis uses an electric field to separate DNA fragments. Negatively charged DNA moves toward the positive electrode.

2. For linear DNA fragments in the same gel, smaller fragments generally move faster and farther from the wells. Compare bands with a DNA size standard to estimate fragment length.

3. Bands at the same position represent fragments of similar size. Matching band patterns can support relatedness or sample identification but do not prove that entire genomes are identical.

B. Polymerase chain reaction

1. PCR amplifies a selected DNA region. It requires template DNA, primers, free nucleotides, and a heat-stable DNA polymerase.

2. Denaturation separates the DNA strands by heating. Annealing allows primers to bind complementary sequences as the mixture cools. Extension builds new DNA from the primers.

3. Repeated cycles produce many copies. Under ideal doubling, N = N₀ × 2ⁿ, where N₀ is the starting number of target copies and n is the number of cycles.

C. Transformation and recombinant DNA

1. Recombinant DNA combines DNA from different sources. Restriction enzymes cut DNA at specific sequences, and ligase can join DNA fragments.

2. A vector, such as a plasmid, carries inserted DNA into a host. Bacterial transformation introduces foreign DNA into bacterial cells.

3. A selectable marker, such as antibiotic resistance, helps identify cells that acquired the vector. Survival alone does not prove that the desired gene was inserted correctly or expressed.

4. A no-plasmid comparison helps test selection. Growth without antibiotic helps confirm that the cells were viable.

D. DNA sequencing and applications

1. DNA sequencing determines the order of nucleotide bases. Sequence comparisons can identify variants and provide evidence of genetic relationships.

2. Biotechnology can investigate inheritance, identify samples, copy genes, and engineer organisms. Interpret results according to what each technique measures or changes.

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