Chapter 22: Population Genetics and Evolution
● Quantitative Trait Loci (QTL)
○ Quantitative trait loci: genes that contribute to quantitative traits.
○ QTL are not fundamentally different to the genes that control discrete phenotypes, but they are more difficult to detect. There will be several genes that contribute to the same phenotype and the phenotype is more variable, so the analysis of genotypes is more difficult than saying if a genotype does/does not have a specific phenotype.
○ Mapping of a discontinuous trait relies on our ability to determine the genotype of a particular phenotype.
○ QTL mapping is a method that detects linkage between a trait of interest and markers at known locations throughout the genome.
■ Step 1: Interbreed two divergent parental lines to generate an F1 generation (all heterozygous) If you want to learn more check out What does unpitched mean in music?
■ Step 2: Backcross the F1 offspring to either of the two parental lines and
genotype the offspring for the markers.
Don't forget about the age old question of How is exocrine different than endocrine?
● GWAS -QTL mapping in humans
○ QTL mapping requires controlled crosses → not possible in humans
○ Genome wide association studies (GWAS) is a method used to identify the vast majority of human disease genes.
○ GWAS: analyze associations between a trait and genetic variation present in a population. Variants associated with a trait are near genes that contribute to that trait.
■ Considers each marker that has a different version in the two parents as an allele, and tests for linkage between the marker and the phenotype of interest.
■ The closer a marker is to the gene affecting fruit size that differs in effect
between the two parents, the more tightly linked the marker and the gene will be. They will tend to inherit together the backcross offspring.
If you want to learn more check out What improves the quality of life for the community?
○ Distinct distributions for genotypic classes at a marker locus signal the location of a QTL near the marker.
Don't forget about the age old question of Reactive relationship refers to what?
If you want to learn more check out What will happen if the stalk of the pituitary gland is cut or destroyed in any way?
We also discuss several other topics like What is the satisfaction a consumer obtains from the consumption of a good or service?
Odds = Prob(Data|QTL)
■ “LOD” = Log (10) of the Odds
● How does genetic/phenotypic variation arise and get maintained?
○ Natural selection
○ Genetic drift
● Population genetics
○ Population: a group of interbreeding individuals
○ Gene Pool: the collection of alleles found in a population → this is the source of genetic information from which the next generation is produced. There will almost always be more alleles in a population than in any one individual
○ Hardy-Weinberg (H-W) equilibrium hypothesis predicts the frequency of alleles and genotypes in the next generation of an idealized population with a known gene pool.
■ The population is infinite.
■ Mating is random.
■ No difference in fitness of different genotypes.
*In practice, these assumptions are not met: H-W serves as a NULL model to assess the influence of deviations from assumptions*
● H-W equilibrium predicts genotype frequencies from allele frequencies
○ Prediction 1: if a population has a known frequency of two alleles, the frequency of genotypes in the next generation can be calculated using:
p2 + 2pq + q2 = 1 → 2 alleles
p2 + 2pq + q2 + 2pr + r2 + 2qr = 1 →3 alleles
○ Prediction 2: H-W equilibrium predicts that the genotype frequencies will not change between generations. Even if a population starts away from H-W equilibrium, a single round of mating is sufficient for equilibrium to be reached.
Example: A1A1 36 30
A1A2 48 50
A2A2 16 20
Chi-Square Test (X2):
X2 = ∑(observed-expected)2
= 1 + 0.08 + 1
P = 0.15 (df = 1)
● H-W is used as a null hypothesis. If we follow genotype frequencies in a population and find that they are changing, it suggests that the population is violating one of the H-W assumptions.
● How can we determine if genotype frequencies are changing?
○ Allele/genotype counting can be determined directly by observing phenotypes of individuals for codominant/incompletely dominant traits where each genotype has a distinct phenotype.
○ Square root method can be used if one allele is known to be recessive to the other.
● How do allele frequencies change?
○ Natural Selection
■ Describes the differential reproductive success of individuals with different phenotypes. This causes a change in alleles that determine these phenotypes ■ The fitness of individuals with the best phenotype in the population is defined as: w = 1
■ Relative to this best phenotype, individuals with other genotypes have a fitness reduced by a proportion called the selection coefficient, s, and the relative fitness is defined as: w = 1 – s
● If there are several reduced fitness genotypes, we can distinguish them as s1, s2, etc.
○ Example: 2 co-dominant alleles, giving 3 genotypes, each with different fitness:
Genotype: B1B1 B1B2 B2B2
Frequency: 0.36 0.48 0.16
Number born: 360 480 160
Relative fitness (w): 1 0.8 0.4
Number that mate: 360 x 1 = 360 480 x 0.8 = 384 160 x 0.4 = 64
New allele frequencies: B1 = 360/808 + (384/2)/808 = 0.68
B2 = 64/808 + (384/2)/808 = 0.32 (or 1 – 0.68)
B1 = 720/1616 + 384/1616 = 0.68
B2 = 128/1616 + 384/1616 = 0.32 (or 1 – 0.68)
○ Over successive generations you can see that B1 will continue to increase, causing B2 to decrease –
eventually B2 will fix, or reach a frequency of 1, in the population.
○ The speed with which B1 increases depends on its fitness relative to B2. The larger the difference, the more quickly B1 will fix.
○ Called directional selection because selection drives the frequency of B1 in a consistent direction → up. ○ Mutation is the ultimate source of genetic variation, but by itself, it changes allele frequencies slowly. Variation acted on by natural selection can cause rapid changes.
● Random sampling causes a deviation from expected allele frequencies.
○ An assumption of H-W with important, but not intuitive, consequences is that the population has infinite size.
○ This assumption cannot be met, since all populations are finite. Violating the assumption of infinite population size can lead to big deviations from H-W predictions. The random sampling of events in finite populations will cause genotype, and therefore, allele, frequencies to change each generation. This is known as genetic drift.
Infinite population size Finite population size
Over successive generations, genetic drift continually alters allele frequencies. Eventually,
an allele will go extinct. Here, 4 replicate populations each started with two alleles at equal frequency.
○ In small populations, a sampling effect can cause allele frequencies to differ substantially from expectations.
■ Founder effect: where a small sample of large population establishes a new population.
■ Genetic bottleneck: describes a dramatic decrease in population size (can be due to migration or ecological event)
● Bottlenecks are one reason why some alleles are common in some
○ Inbreeding, also known as consanguineous mating, is mating between individuals that share a greater proportion of alleles than would two random members of a population.
○ An effect of inbreeding is an increase in the frequency of homozygous genotypes.
○ Related individuals are more likely to share alleles, so they are more likely to produce homozygotes.
○ Sewall Wright investigated the
consequences of inbreeding and devised
the coefficient of inbreeding (F) as an
arithmetic measure of the probability of
homozygosity for an allele obtained in
identical copies from an ancestor.
○ The coefficient quantifies the probability
that two alleles in the homozygous
individual are identical by descent (IBD),
having descended from the same copy of
the allele carried by a common ancestor.
● Non-random mating reduces variation
○ Non-random mating tends to reduce the
amount of variation in a population.
○ Example: if there were two alleles in a
population and three genotypes, A/A; A/a;
a/a, each only mating with other individuals of the same genotype. The frequency of the A/a genotype would quickly decline.
○ The reason for this is that offspring of A/A x A/A and a/a x a/a parents have the parental genotype whereas only half the offspring of A/a x A/a parents have the parental genotype. The fraction of this genotype declines by half each generation.
Chapter 7: DNA Structure and Replication
● 7.1: DNA is the hereditary molecule of life
○ Before DNA was known to be the hereditary molecule, there were five characteristics of hereditary molecules identified:
■ Localized to the nucleus, component of chromosomes.
■ Present in stable form in cells.
■ Sufficiently complex to have information needed for structure, function, development,, and reproduction of an organism.
■ Able to accurately replicate itself so that the daughter cells have the same information as the parent cells.
■ Mutable, undergoing a low rate of mutations that introduces genetic variation and serves as a foundation for evolutionary change.
○ In 1923, DNA was localized to chromosomes and made a candidate for the hereditary material.
○ Proteins and RNA are also found in chromosomes.
○ Lipids and carbohydrates were also considered as candidates of being the hereditary molecule.
● Transformation factor
○ Frederick Griffith identified two strains of Pneumococcus:
■ S caused fatal pneumonia in mice.
■ R did not.
○ A single nucleotide change can convert the R (rough) strain into the S (smooth) strain.
○ These strains occur in four different antigenic types (I, II, III, IV) that cannot be altered by mutation alone.
■ Mice infected with strain SIII
had pneumonia and died.
■ Mice infected with strain RII or with heat-killed strain SIII
■ Mice infected with heat-killedstrain SIII and live strain RII
had pneumonia and died →
live SIII bacteria were
recovered from the mice.
○ DNA is the transformation factor
■ Biochemical tests of the
heat-killed SIII extract
showed that it contained
mainly DNA, with small
amounts of RNA, protein,
lipids, and polysaccharides.
■ However, more tests were needed to identify the
■ Avery, MacLeod, and McCarty used heat-skilled SIII bacteria, live RII bacteria and infected mice.
■ The extract of heat-killed SIII bacteria was divided into aliquots and treated to destroy either DNA, RNA, proteins, or lipids and polysaccharides.
■ All aliquots killed the mice except the one with the DNA destroyed.
● Bacteriophage: Viruses which infect Bacteria
○ Alfred Hershey and Martha Chase decided to study viruses that infect bacteria
○ They conducted experiments with the Waring
○ Proteins contain large amounts of sulfur but almost
○ DNA contains large amounts of phosphorus but no
○ Hershey and Chase separately labeled either phage proteins (with 35S) or DNA (with 32P) and then traced
each radioactive label in the course of infection.
○ After infection, in both experiments, agitation by a
blender separated the empty phage particles from the
○ In the protein labeling experiment, the radioactivity
was detected in the empty phage particles (ghosts).
○ In the DNA labeling experiment, the radioactivity was detected inside the infected bacteria.
● 7.2: The DNA double helix consists of two complementary and antiparallel strands
○ DNA is a polymer composed of four kinds of nucleotides joined by covalent phosphodiester bonds with two polynucleotide chains that join to form a double helix.
POSITION IN THE
○ Each polymer/strand is made of nucleotides connected by phosphodiester bonds. ○ Strands have directionality.
○ Two strands are antiparallel.
○ Complementation of bases and hydrogen bonds hold strands together.
● Chargaff’s Rule
● Rosalind Franklin
○ X-ray diffraction pattern of
● Watson and Crick
○ DNA model
● 7.3: DNA replication is semiconservative and bidirectional
○ Three possible models of DNA replication:
■ Semiconservative: each daughter duplex has one parental and one daughter strand.
■ Conservative: one daughter duplex has both parental strands and the other has both daughter strands.
■ Dispersive: each daughter duplex has mixed parental and daughter segments.
● The Meselson-Stahl Experiment
○ In 1958, Meselson and Stahl used cesium chloride (CsCl) centrifugation to test the models of DNA replication.
○ This method is capable of separating molecules with slightly different molecular weights.
○ They began by growing E. coli in a medium containing heavy nitrogen (15N) for many generations.
○ Once all of the bacterial cells in the culture had DNA containing only the heavy nitrogen, they transferred the bacteria to the medium containing 14N.
○ After one round of replication, the DNA of a aliquot of cells was isolated and centrifuged to determine its density.
○ The same was done after
successive replication cycles.
○ After one round of replication, the DNA molecules all had the densities expected from 14N/15N molecules.
○ After two rounds of replication, half of the molecules had densities expected of 14N/15N molecules and the other half had densities expected for 14N/14N molecules.
○ These results are consistent only with the semiconservative model of replication. ● Origin of replication in bacterial DNA
○ DNA replication is most often bidirectional, proceeding in both directions from a single origin of replication in bacterial chromosomes.
○ Eukaryotic chromosomes have multiple origins of replication.
○ John Cairns reported the first evidence of bacterial origins of replication in 1963. ● Multiple replication of origins in eukaryotes
○ Autoradiograph analysis shows multiple origins of replication in eukaryotic chromosomes.
○ Large eukaryotic genomes contain thousands of origins of replication separated by 40,000 to 50,000 base pairs.
○ The human genome contains more than 10,000 origins.
○ DNA replication rate varies among different types of cells.
● DNA replication precisely duplicates the genetic material
○ Replication is best studied in bacteria.
○ Though replication is very similar among bacteria, archaea, and eukarya, the processes are not identical.
○ The enzymes and proteins involved are parts of large complex aggregations of proteins and enzymes called replisomes.
○ These assemble at the replication fork.
● DNA sequences at replication origins
○ Replication origins have sequences that attract replication enzymes.
○ The origin of replication sequence of E. coli is called oriC, and it contains about 245 base pairs of A-T rich DNA.
○ The origin is divided into three 13 base pair sequences followed by four 9 base pair sequences Saccharomyces cerevisiae (yeast) has the most fully characterized origin-of-replication sequences.
○ The multiple origins of replication are called autonomously replicating sequences (ARS). ○ ARS organizations and sequences is similar throughout the yeast genome. ○ Replication origins of other eukaryotes are less well characterized.
○ Consensus sequence:
● Initiation of replication
○ DnaA first binds the 9-mer sequences, bends the
DNA, and breaks hydrogen bonds in the A-T rich
sequences of the 13-mer region.
○ DnaB is a helicase that uses ATP energy to
break hydrogen bonds of complementary bases
to separate the strands and unwind the helix.
○ DnaB is carried to the DNA helix to DnaC.
○ The unwound DNA strands are kept from
reannealing by single-stranded binding protein
○ Unwinding of circular chromosomes will create
torsional stress, potentially leading to supercoiled
○ Enzymes called topoisomerases catalyze
controlled cleavage and rejoining of DNA that
● RNA primers are needed for DNA replication
○ DNA polymerase can only elongate DNA strands by adding nucleotides to the 3’ end of a pre-existing strand → direction of synthesis is 5’ to 3’.
○ They cannot initiate DNA strand synthesis on their own.
○ RNA primers are needed; they synthesized by an RNA polymerase called primase. ○ Primase and some additional proteins join DnaA at oriC to form the primosome. ○ In E. coli, daughter cells DNA strands are synthesized by the DNA polymerase III (pol III) holoenzyme.
○ Holoenzyme refers to a multiprotein complex in which a core enzyme is associated with the additional components needed for full function.
○ The replisome is found at each replication fork and contains two copies of pol III. ● Leading and lagging strand synthesis
○ Leading strand: synthesized by pol III continuously in the same direction as fork progression.
○ Lagging strand: elongated by pol III discontinuously in the opposite direction of the fork progression via short segments (Okazaki fragments).
● RNA primer removal and Okazaki fragment ligation
○ DNA polymerase I (pol I) used two activities to complete replication.
■ Its 5’ to 3’ exonuclease activity removes the RNA primers.
■ Its 5’ to 3’ polymerase activity add DNA nucleotides to the 3’ end of the DNA segment preceding the primer.
○ DNA ligase seals the gap between the resulting DNA segments.
PUTTING IT ALL TOGETHER
→ → → → → → → →→ → → →
● Simultaneous synthesis of leading and lagging strands
○ Each replisome complex carries out replication of the leading and lagging strand simultaneously → this is a challenge/problem
○ The DNA pol III holoenzyme contains 11 protein subunits, with the two pol III core polymerase each tethered to a different copy of the tau (��) protein.
○ The tau proteins are joined to a protein complex called the clamp loader; two additional proteins form the sliding clamp.
● DNA proofreading
○ DNA replication is very accurate, mainly because DNA polymerases undertake DNA proofreading, to correct occasional errors.
○ Errors in replication occur about one every billion nucleotides in E. coli.
○ Proofreading ability of DNA polymerase enzymes is due to a 3’ to 5’ exonuclease (cutting the ends off) activity.
○ Replication errors produce a DNA mismatch and inability of the mismatched bases to form the appropriate H-bonds.
○ This leads to displacement of the 3’-OH into the 3’ to 5’ exonuclease “site” of the enzyme.
○ Several nucleotides (including the incorrect one) are removed and new nucleotides are
○ The leading strand of linear chromosomes can be replicated to the end.
○ The lagging strand requirement for a primer means that lagging strands cannot be completelyreplicated.
○ This problem is resolved by repetitive sequences at the ends of chromosomes, called telomeres.
○ These repeats ensure that incomplete
chromosome replication does not affect vital
○ Telomeres are synthesized by the
○ The RNA in telomerase is complementary to the
telomere repeat sequence and acts as a template for addition of DNA.
○ The template RNA of telomerase allows new DNA replication to lengthen the telomere sequences.
○ Once telomeres are sufficiently elongated, the α polymerase synthesizes additional RNA primers.
○ New DNA replication then fills out the
○ Telomere sequences in most organisms are quite similar.
● Importance of telomerase activity
○ Mice that are homozygous for loss-of-function mutations of the TERT (telomerase reverse
transcriptase) gene give rise to developmental defects.
○ These defects are first observed in the fourth and fifth generations, due to loss of telomere length with each generation.
○ By the fourth and fifth generations, shortening of the chromosomes is critical and apoptosis is
● Telomeres, aging, and cancer
○ Telomere length is important for chromosome stability, cell longevity, and reproductive success.
○ Telomerase is active in germline cells and some stem cells in eukaryotes.
○ Differentiated somatic cells and cells in culture have virtually no telomerase activity. They have limited life spans (30 to 50 cell divisions).
● Abnormal reactivation of telomerase activity
○ Telomerase is normally turned off in somatic cells.
○ Reactivation of telomerase can lead to aging cells that continue to proliferate, a feature of many types of cancer.
○ TERT reactivation is one of the most common mutations in cancers of all types. ● Molecular genetic analytical methods make use of DNA replication processes
○ Molecular biologists have used their understanding of DNA replication to develop new methods of molecular analysis.
○ Two widely used methods include: polymerase chain reaction (PCR) and dideoxynucleotide DNA sequencing.
● Polymerase chain reaction (PCR)
○ An automated version of DNA replication that produces millions of copies of a short target DNA segment.
○ There are numerous application of PCR.
○ PCR reactions are carried out in small volumes (less than 100 ��l)
■ A double-stranded DNA template containing the target sequence to be amplified.
■ A supply of four DNA nucleotides. ■ A heat-stable DNA polymerase.
■ Two different single-stranded DNA primers.
■ A buffer solution.
○ The most commonly used DNA polymerase, Taq, is isolated from Thermus aquaticus, which occurs naturally in hot springs.
○ Process of PCR:
■ PCR uses two DNA sequences called PCR primers that provide a starting
point for Taq polymerase to add
■ The PCR primers define the 5’ to 3’ boundaries of the replication products.
■ PCR is composed of three steps that result in exponential amplification of large numbers of the target DNA.
○ Steps of PCR:
■ Denaturation: the reaction is heated to ~95°C to denature the DNA into single strands.
■ Primer annealing: the reaction temperature is reduced to ~45-68°C to allow primers to hybridize to their complementary sequences in the target DNA.
■ Primer extension: the reaction’s temperature is raised to 72°C to allow Taq polymerase to synthesize DNA.
○ Limitations of PCR:
■ Some knowledge of the target DNA sequences is required in order to determine primer sequences.
■ Amplification products longer than 10 to 15 kb are difficult to produce.
○ Despite these limitations, PCR is a practical way to obtain large quantities of DNA from a particular gene for molecular analysis.
○ Separation of PCR products:
■ Amplified DNA fragments are separated from the rest of the reaction mixture by gel electrophoresis and visualized by ethidium bromide staining.
■ PCR product sizes are measured in base pairs (bp)
■ Differences in the size of DNA amplified by a pair of primers are related to the amount of DNA between the primers.
● Dideoxynucleotide DNA sequencing (dideoxy sequencing)
○ The ultimate description of a DNA molecule is its precise sequence of bases.
○ The Sanger (dideoxynucleotide) method wasthe most open to automation and is the method of choice today.
○ This method uses DNA polymerase to replicate new DNA from a single-stranded template.
○ The four standard deoxynucleotide bases (dNTPs) are present in large amounts.
○ Each reaction contains a small amount of dideoxynucleotide (ddNTP), which lacks a 3’-OH group.
○ The principal of Sanger sequencing
■ Whenever a ddNTP is incorporated into the product DNA molecule,
■ A separate reaction is carried out for A, T, G, and C using a corresponding amount of ddNTP.
■ Each reaction tube produced a series of partial DNA molecules, each of which ends with that nucleotide.
■ All four reactions must run side by side on a gel in order to determine complete sequence.
○ Visualization of DNA sequence
■ After the reactions are complete, the reactions are run side by side on a gel.
■ The bands shown in the autoradiograph are visible because the primers that begin each fragment were labeled with radioactive isotopes (end-labeling).
■ The shortest bands are the DNA products closest to the primer and these travel fastest on the gel. The gel is read from the bottom up, all four lanes together.
○ Automated DNA sequencing
■ Automated DNA sequencers use a single reaction for each DNA sequence, in which all four ddNTPs are included.
■ Each ddNTP is labeled with a unique fluorescent marker.
■ The DNA is synthesized and a mixture of fragments is produced and run on a DNA gel.
■ The fluorescent label on each ddNTP has a different wavelength, and a laser light excites the fluorescent tag on each
fragment as it passes.
■ The wavelength of the fluorescence is read as the fragment passes, and the
information is recorded by computer.
■ The fluorescence pattern produced shows the sequence of the DNA.
Chapter 8: Molecular Biology of Transcription and RNA Processing ● RNA transcripts carry the messages of genes
○ RNA ribonucleotides are composed of a sugar, nucleotide base, and one or more phosphate groups, with two critical differences compared to DNA nucleotides:
■ Thymine is replaced by uracil.
■ The sugar ribose is used instead of deoxyribose.
● RNA synthesis
○ RNA polymerase catalyzes the addition of each ribonucleotide to the 3’ end of the growing strand.
○ Two phosphates are eliminated in the process, as in DNA synthesis.
● RNA classification
○ Messenger RNA (mRNA): produced by protein-encoding genes and is a short-lived intermediary between DNA and protein.
■ The only type of RNA that undergoes translation.
■ Transcription of mRNA is often followed by post-transcriptional processing.
○ Transfer RNA (tRNA): encoded in dozens of forms and is responsible for binding an amino acid and depositing it for inclusion into a growing protein chain.
○ Ribosomal RNA (rRNA): combines with numerous proteins to form ribosomes.
○ Small nuclear RNA (snRNA): found in the nucleus of eukaryotes and plays a role in mRNA processing.
○ Micro RNA (miRNA): active in plant and animal cells and is involved with post-transcriptional regulation of mRNA.
○ Long non-coding RNA (lncRNA): belongs to a class of genes that is nearly as numerous as protein coding genes.
● Gene structure
○ The gene contains several segments with distinct functions.
○ The promoter is immediately upstream (5’) to the start of the transcription, referred to as the +1 nucleotide.
○ The promoter controls the access of RNA polymerase to the gene.
○ The coding region of the gene is the portion that contains the information needed to synthesize the protein product.
○ The termination region of the gene regulates the end of transcription.
● Essential steps of transcription in E. coli
○ Promoter recognition
○ Transcription initiation
○ Chain elongation
○ Chain termination
● Bacterial RNA polymerase
○ A single type of RNA polymerase catalyzes transcription of all RNAs in E. coli.
○ The bacterial RNA polymerase holoenzyme is composed of pentameric core enzyme that binds a sixth subunits, called the sigma (σ) subunit.
○ The large core enzyme is composed of two α subunits, one �� and one ��’, and an �� subunit.
○ The core enzyme can transcribe RNA from a DNA template but cannot bind the promoter or initiate RNA synthesis without the σ subunit.
● Bacterial promoters
○ A promoter is a double-stranded DNA sequence that is the RNA polymerase binding site and is required for the initiation of transcription.
○ The promoter is located a short distance upstream of the coding sequence, within a few nucleotides of +1.
○ RNA polymerase is attracted to promoters by the presence of consensus sequences. ● Promoter consensus sequences
○ Consensus sequences are written in single-stranded shorthand form, 5’ to 3’ on the coding strand.
○ At the -10 position is the Pribnow box, or -10 consensus sequence, 5’ - TATAAT - 3’ ○ At -35 is a 6-bp region, the -35 consensus sequence, 5’ - TTGACA - 3’ ○ RNA polymerase binds to -10 and -35 sequences and occupies the space between and
● Transcription Initiation
○ RNA polymerase initiates through a two-step process
■ First, the holoenzyme makes a loose attachment to the promoter sequence to form the closed promoter complex.
■ The holoenzyme next unwinds about 18 bp of DNA around the -10 position to form the open promoter complex.
● Transcription termination mechanisms
○ Termination of transcription in bacteria is signaled by a DNA termination sequence that usually has a repeating sequence.
○ In intrinsic termination, a mechanism dependent only on the presence of the repeat, induces secondary structure needed for termination.
○ Rho-dependent termination requires a different termination sequence and the rho protein.
● Intrinsic termination
○ Most bacterial termination occurs via intrinsic termination.
○ Termination sequences include an inverted repeat followed by a string of adenines.
○ mRNA containing the inverted repeats form into a short stem-loop structure, called a hairpin.
○ The hairpin followed by a series of “U”s in the mRNA causes the RNA polymerase to slow down and destabilize.
○ The instability caused by the slowing polymerase and the U-A base pairs induces the polymerase to release the transcript and separate from the DNA.
● Rho-dependent termination
○ Certain bacterial genes require the action of rho protein for termination.
○ Rho-dependent termination sequences do not have a string of uracils. Instead, they have a rho utilization (rut) site, a stretch of about 50 nucleotides rich in cytosines.
○ Rho-dependent termination also uses a terminator site which forms a hairpin structure when transcribed.
● Eukaryotic transcription uses multiple RNA polymerases
○ Eukaryotic promoters and consensus sequences are more diverse than those of bacteria.
○ Eukaryotes have three different RNA polymerases that recognize different types of RNAs.
○ The complex that assembles to initiate and elongate transcription is more complex in eukaryotes than in bacteria.
○ Eukaryotic genes carry introns and exons, and require processing to remove introns.
○ Eukaryotic DNA is associated with proteins to form chromatin; the chromatin composition of a gene affects its transcription.
○ Chromatin thus plays an important role in gene regulation of eukaryotes. ● Eukaryotic polymerases
○ RNA polymerase I (RNA pol I): transcribes three ribosomal RNA genes.
○ RNA polymerase II (RNA pol II): transcribes protein coding genes and most small nuclear RNA genes.
○ RNA polymerase III (RNA pol III): transcribes tRNA, one small nuclear RNA, and one ribosomal RNA.
● Promoter elements
○ The most common eukaryotic promoter consensus sequence is the TATA box, or the Goldberg-Hogness box, located at about position -25.
○ The consensus sequence is 5’-TATAAA-3’
○ A CAAT box is often found near the -80 position.
○ A GC-rich box (consensus 5’-GGGCGG-3’) is located at -90, or further upstream. ● Promoter recognition
○ RNA pol II recognizes and binds to promoter sequences with the aid of proteins called transcription factors (TFs).
○ TFs bind to regulatory sequences and interact directly, or indirectly, with RNA polymerase; those interacting with pol II are called TFII factors.
○ The TATA box is the principal binding site during promoter recognition. ● Detecting promoter consensus elements
○ Research to verify that a segment of DNA is functionally important component of a promoter has two components:
■ Discovering the presence and location of DNA sequences that transcription factors will bind to.
■ Mutational analysis to confirm that functionality of each sequence.
● Mutational analysis of promoters
○ Researchers produce many different point mutations and compare the level of transcription generated by the mutant sequence relative to wild type.
○ Mutations inside the consensus region significantly reduce levels of transcription. ○ Mutations outside the consensus region have nonsignificant effects on transcription.
● Enhancer sequences
○ Enhancer sequences: increase the level of transcription of specific genes.
○ They bind proteins that interact with the proteins that are bound to gene promoters, and together the promoters and enhancers drive gene expression.
○ Enhancers may be variable distances from the gene they affect and may be upstream or downstream of the gene.
● Silencer sequences
○ Silencer sequences: DNA elements that act at a distance to repress transcription of their target genes.
○ Silencers bind transcription factors called repressor proteins that induce bends in DNA. ○ These bends reduce transcription of the target gene.
○ Silencers may be located variable distances from their target genes, either upstream or downstream.
● Post-transcriptional processing modifies RNA molecules
○ Eukaryotic transcripts are more stable than bacterial transcripts.
○ In eukaryotes, transcription and translation are separated in time and location. ○ Eukaryotic transcripts have introns, which are not found in bacterial transcripts.
○ These features are all related to post-transcriptional modification of eukaryotic transcripts.
○ The initial eukaryotic gene mRNA is called pre-mRNA whereas the fully processed mRNA is called the mature mRNA. Modifications include:
■ 5’ capping
■ 3’ polyadenylation
■ Intron splicing
● Functions of 5’ Capping
○ Protection of mRNA from rapid degradation.
○ Facilitating transport of mRNA out of the nucleus.
○ Facilitating subsequent intron splicing.
○ Enhancing translation efficiently by orientation the ribosome on the mRNA.
● Functions of 3’ polyadenylation
○ Facilitating transport of mature mRNA across the nuclear membrane to the cytoplasm. ○ Protecting the mRNA from degradation.
○ Enhancing translation by enabling the ribosomal recognition of mRNA. **Some eukaryotic transcripts (e.g. histone genes) do not undergo polyadenylation** ● Pre-mRNA intron splicing
○ Intron splicing requires great precision to remove intron nucleotides accurately. ○ Errors in intron removal would lead to incorrect protein sequences.
● Splicing signal sequences
○ Specific short sequences define the junctions
between introns and exons.
○ The 5’ splice site is at the 5’ intron end and contains a
consensus sequence with an invariant GU dinucleotide at the 5’- most end of the intron.
○ The 3’ splice site at the opposite end of the intron has an 11 nucleotide consensus with a pyrimidine rich region and a nearly invariant AG at the 3’- most end.
○ A third consensus region, called the branch site, is 20 to 40 nucleotides upstream of the 3’ end of the intron.
○ It is pyrimidine-rich and contains an invariant adenine called the branch point adenine near the 3’ end of the consensus.
○ Mutation analysis shows that all three consensus sequences are required for accurate splicing.
○ Introns are removed from the pre-mRNA by an snRNA-protein complex called spliceosome.
○ The 5’ splice site is cleaved first and a lariat intron structure is formed when the 5’ intron end binds to the branch point adenine.
○ Then the 3’ splice site is cleaved and the exon ends are ligated together.
● Spliceosome composition
○ The spliceosome is a large complex made up of many snRNPs (U1 through U6).
○ The composition is dynamic, changing through the steps of splicing.
○ Spliceosome components are recruited to 5’ to 3’ splice sites by SR proteins; these bind to sequences in exon called exonic splicing
enhancers (ESEs) and ensure accurate
● Intron self-splicing
○ RNAs can contain introns that catalyze their own removal.
○ There are three categories: group I, group II, and group III.
○ Group I introns are large, self-splicing ribozymes that catalyze their own excision from mRNAs, and from tRNA and rRNA precursors of bacteria, simple
eukaryotes, and plants.
○ Intron self-splicing takes place via two
transesterification reactions that excise the intron and ligate the exon ends.
● Alternative transcripts of single genes
PUTTING IT ALL TOGETHER ← ← ← ← ← ← ← ← ← ← ←
○ It is common for large eukaryotic genomes to express more proteins than there are genes in the genome.
■ Example: human cells can produce over 100,000 distinct polypeptides but contain ~22,000 genes.
○ Three transcription-associated mechanisms can explain this.
● Mechanisms for producing alternative transcript
○ Pre-mRNA can be spliced in alternative patterns in different cell types.
○ Alternative promoters can initiate transcription at distinct start points.
○ Alternative localizations of polyadenylation can produce different mature mRNAs. **TOGETHER, THESE COMPRISE ALTERNATIVE PRE-mRNA PROCESSING** ● Alternative intron splicing
○ Alternative intron splicing: processing of identical transcripts in different cells can lead to mature mRNAs with different combinations of exons and thus different polypeptides.
○ Approximately 70% of human genes are thought to undergo alternative splicing. ○ It is less common in other animals and rare in plants.
○ The Drosophila Dscam gene has one of the most complex patterns of alternative splicing.
○ Of the 24 exons, numbers 4, 6, 9, and 17 have numerous alternative sequences. ○ More than 38,000 different polypeptides can be produced through alternative splicing. ○ Not all of the possible arrangements are observed, however.
● Alternative processing
○ Alternative splicing is mainly controlled by variation in SR proteins in different cell types.
○ Use of alternative promoters can occur when more than one sequence upstream of a gene can initiate transcription.
○ Alternative polyadenylation requires more than one polyadenylation signal in a gene. ○ Alternative promoters of polyadenylation are controlled by variable expression of
regulatory proteins in specific cell types.
● Post-transcriptional RNA editing
○ In the mid-1980s, RNA editing was uncovered; thatis responsible for post-transcriptional modifications to the nucleotide sequence (and the protein
produced) of some mRNA.
○ In one kind of RNA editing, uracils are added with the assistance of a guide RNA (gRNA), which contains a sequence complementary to the mRNA that it edits.
○ Editing may sometimes involve deletion of uracils, too.
● Base substitution
○ A second type of RNA editing is base substitution, frequently replacement of cytosine with uracil.
○ This has been identified in mammals, most land plants, and some single-celled
Chapter 9: The Molecular Biology of Translation ● 1 base code = 41 = 4 amino acids
● 2 base code = 42 = 16 amino acids
● 3 base code = 43 = 64 amino acids
● The triplet code
○ Groups of three consecutive nucleotides
(codons) in an mRNA each correspond to one amino acid.
○ The genetic code contains 64 different
codons; with only 20 common amino acids,
this leads to redundancy–some amino acids
are specified by more than one codon.
● The genetic code displays third-base wobble
○ The triplet genetic code, with 64
combinations, provides enough variety to
code 20 amino acids.
○ 61 codons specify amino acids, and 3 are STOP codons.
○ All amino acids except methionine and
tryptophan are specified by at least two
codons, called synonymous codons.
○ Though there are 61 codons that specify amino acids, most genomes have 30-50 different tRNA genes.
○ A relaxation of the strict complementary base-pairing rules at the third base of the codon is called third-base wobble.
○ tRNA molecules with different anticodons for the same amino acids are called iso-accepting tRNAs.
● How third-base wobble works
○ Most synonymous codons can be grouped into
pairs that differ only in the third base; the pairs
either both carry a purine (A or G) or both carry
a pyrimidine (C or U).
○ Third-base wobble occurs through flexible
pairing at the 3’-most nucleotide of the codon
and the 5’-most nucleotide of the anticodon.
○ However, a pyrimidine must still base-pair with a purine.
● Charging tRNA Molecules
○ tRNA molecules are transcribed from tRNA genes.
○ Correct charging of each tRNA molecule is crucial for theintegrity of the genetic code.
○ Enzymes called aminoacyl-tRNA synthetases or tRNA synthetases catalyze the addition of the correct amino acid to tRNAs.
○ Recognition of the iso-accepting tRNAs by the enzyme is complex with no single set of rules.
● Experiments Deciphered the Genetic Code
○ A remarkable set of experiments in the 1960s deciphered the genetic code and answered the following questions:
■ Do neighboring codons overlap one another?
■ How many nucleotides make up an mRNA codon?
■ Is the polypeptide-coding information of mRNA continuous or does it contain
● No Overlap in the Genetic Code
● Conclusive Evidence of Nonoverlap
○ In 1960, a study of single nucleotide substitutions by Fraenkel-Conrat and colleagues showed that single nucleotide changes led to single amino acid changes.
○ An overlapping code would have led to multiple amino acid changes as a result of altering one nucleotide.
○ The results of the study are consistent with a nonoverlapping genetic code. ● A Triplet Genetic Code
○ Proof of a triplet genetic code came in 1961 when researchers (Crick, Barnett, Brenner, and Watts-Tobin) created mutations by insertion or deletion of single nucleotides.
○ This leads to a change in reading frame of the mRNA.
○ Reading frame refers to the specific codon sequence as determined by the start codon ● Frameshift Mutations
○ Mutations that alter reading frame are called frameshift mutations and garble the sense of the translated message.
■ Wild type: YOU/MAY/NOW/SIP/THE/TEA
■ Mutant addition: YOU/MAC/YNO/WSI/PTH/ETE/A
■ Mutant deletion: YOU/MAY/NOS/IPT/HET/EA
○ All the codons after the addition or deletion will specify the wrong amino acids ● Reversion of Frameshift Mutations
○ Frameshift mutations may be restored if a second mutation in a different location in the gene restores part of the reading frame.
■ Mutant addition: YOU/MAC/YNO/WSI/PTH/ETE/A
■ Reversion mutant deletion: YOU/MAC/YNO/SIP/THE/TEA
○ The frameshift is now confined to just a small area between the original mutation and the reversion mutation – the rest of the protein (sentence) is normal.
● Interpretations of the Experiments Confirm That the Genetic Code Is a Triplet Code
○ Single additions or deletions of nucleotides into the rII gene of T4 bacteriophage caused frameshift mutations.
○ Addition of two nucleotides or deletion of two nucleotides produced frameshifts as well.
○ Addition of three nucleotides or deletion of three nucleotides produced a “mutant” region between the first and last of the added or deleted nucleotides but a non-frameshifted protein outside of the altered region.
● No Gaps in the Genetic Code
○ Crick and colleagues suggested that the genetic code is read continuously, with no gaps, spaces, or pauses between codons.
○ For example, if a spacer were present a transcript might read:
YOUxMAYxNOWxSIPxTHExTEAx; if the spacer existed, an inserted or deleted nucleotide would affect only one codon.
● Deciphering the Genetic Code
○ The genetic code was deciphered between 1961 and 1965.
○ It was a milestone in establishing the central dogma of biology: DNA → RNA → protein.
● First Steps in Deciphering the Genetic Code
○ Nirenberg and Matthai performed an experiment in 1961 that laid the groundwork for later work.
○ Strings of repeating nucleotides were translated in vitro and the resulting polypeptide identified.
○ For example, an artificial mRNA containing only uracils, poly(U), resulted in polypeptides
containing only phenylalanine, so the codon UUUcorresponds to the amino acid Phe.
● Khorana Extended the Analysis of the Genetic Code
○ Khorana synthesized mRNA molecules with repeating di-, tri- and tetranucleotides, and
translated them in vitro to define more codons.
○ For example, a dinucleotide repeat (UC)n produces an mRNA with the sequence
5-UCUCUCUCUCUCUCUC-3 and two possible codons, UCU and CUC.
○ The resulting polypeptides had alternating amino acids, serine (Ser) and leucine (Leu). ● Nirenberg and Leder’s Results
○ Nirenberg and Leder tested all 64 possible codons.
○ They identified all 61 of the codon-amino acid associations.
○ They also identified the three stop codons, UAA, UAG, and UGA.
● The (Almost) Universal Genetic Code
○ In all organisms, the processes of transcription and translation are similar.
○ Because the genetic code is universal, bacteria can be used to produce important proteins from plants and animals.
○ However, there are a few exceptions to the universality of the genetic code, found principally in mitochondria, though there are two exceptions in living organisms.
● Ribosomes Are Translation Machines
○ The main function of ribosomes is to read the message in the mRNA and catalyze the formation of peptide bonds between appropriate amino acids.
● Ribosome Composition
● Important Regions of Ribosomes
○ The peptidyl site (P-site) holds the tRNA to which the polypeptide is attached.
○ The aminoacyl site (A-site) binds a new tRNA molecule containing an amino acid to be added to the growing polypeptide chain.
○ The exit site (E-site) provides an avenue for exit of the tRNA after its amino acid has been added to the chain.
○ Ribosomes also form a channel from which the polypeptide chain emerges. ● Translation Occurs in Three Phases
○ Translation can be divided into three phases: initiation, elongation, and termination. ○ The phases are similar in bacteria and eukaryotes, though there are several differences. ● Bacterial Translational Initiation
○ rRNA aligns with the consensus sequence in mRNA and positions A site over the start codon.
● The Shine-Dalgarno Sequence
○ The preinitiation complex forms when the 16S rRNA and the Shine-Dalgarno sequence on the mRNA base pair.
○ The Shine-Dalgarno sequence is a purine-rich sequence of about six nucleotides three to nine nucleotides upstream of the start codon.
○ A complementary pyrimidine-rich sequence is found near the 3 end of the 16S rRNA. ● The Second Step of Initiation
○ The initiator tRNA binds to the start codon where the P-site will be once the ribosome is fully assembled.
○ The amino acid on the initiator tRNA is a modified amino acid, N-formylmethionine (fMet); the charged initiator tRNA is called tRNAfMet.
○ Initiation factor, IF-2, and a GTP molecule are bound to the tRNAfMet and IF-1 joins the complex; together these form the 30S initiation complex.
● The Final Step of Initiation
○ The 50S subunit joins the 30S subunit to form the intact ribosome.
○ The union of the two subunits is driven by hydrolysis of GTP to GDP.
○ The dissociation of IF1, IF2, and IF3 accompanies the joining of the subunits to create the 70S initiation complex.
● Polypeptide Elongation
○ Elongation begins with recruitment of elongation factor (EF) proteins that use energy of GTP hydrolysis to:
■ Recruit charged tRNAs to the A-site.
■ Form peptide bonds between sequential amino acids.
■ Translocate the ribosome in the 3 direction along the mRNA.
● Polypeptide Elongation in Bacteria
● Translation Termination
→ → → → → → → → → → → →
● Translation and Transcription
○ In bacteria, the coupling of transcription and translation allows ribosomes to begin translating mRNAs that have not yet been completed.
○ In eukaryotes, mRNAs are produced in the nucleus.
○ They are processed to form mature mRNAs, then sent to the cytoplasm for translation. ● Translation of Bacterial Polycistronic mRNA
● Transfer RNAs and Genetic Code Specificity
○ Chaperville and others determined that the specificity of the codon-amino acid correspondence was due to the mRNA-tRNA interactions.
○ They prepared normal Cys-tRNACys, then converted the attached cysteine to alanine to produce a tRNA that recognizes the codon for Cys but carried Ala.
○ In in vitro translation, the resulting polypeptide had Ala instead of Cys.
● Two Important Conclusions:
○ The genetic code derives specificity from the interaction between tRNA & mRNA.
○ The fidelity of the aminoacyl-tRNA synthetases in recognizing each of their tRNAs and correctly charging them, is extremely important.
● Translation Is Followed by Polypeptide Processing and Protein Sorting
○ Posttranslational Processing: modifies polypeptides into functional protein by removal or chemical alteration of amino acids.
■ Example: fMet is not found in functional bacterial proteins, and methionine is not always the first amino acid in eukaryotic proteins.
■ The absence of these from the N-terminus of proteins is due to their removal after translation.
○ Protein sorting uses signal sequences, also called leader sequences, to direct proteins to their cellular destinations.
● Modification of Amino Acids
○ Phosphorylation, carried out by kinases, can activate or inactivate a protein.
○ Other enzymes may add methyl, hydroxyl, or acetyl groups to amino acids.
○ Carbohydrate side chains are added to some proteins. ● Cleavage of Polypeptides
○ Polypeptides may be cleaved into multiple segments that have separate functions or that aggregate to form a functional protein.
○ The hormone insulin is first produced as preproinsulin, from which the “pre-amino” segment at the N-terminus is cleaved to produce proinsulin.
○ Proinsulin forms disulfide bonds and is cleaved again to produce insulin, a functional protein consisting of two separate chains held together by the disulfide bonds.
● Protein Sorting
○ The signal hypothesis suggests a mechanism by which proteins are transported to their correct locations.
○ It proposes that the first 15-20 amino acids of many polypeptides contain an “address label” that directs proteins to their correct locations.
○ Blobel suggested that the signal sequence directs proteins to the ER, where they are sorted for their specific destinations.
● Destruction of Incorrectly Folded Proteins
○ Mutations cause protein defects by changing the amino acid sequence, usually by altering protein folding and stability.
○ Incorrectly folded proteins will not function normally.
○ Within the ER, proteins that are incorrectly folded are bound by molecules called chaperones that help proteins fold correctly.
● Chaperones and Protein Folding
○ Once a chaperone has assisted a protein to fold correctly, it releases the protein. ○ Proteins that cannot fold properly are irreversibly bound to chaperones.
○ These chaperone-protein complexes are sequestered and then destroyed. ○ There are some diseases that result from a failure of this process.
Chapter 11: Chromosome Structure
● Bacterial Chromosomes Are Simple in Organization
○ Bacteria have single chromosomes that are almost always circular.
○ Some species have linear chromosomes.
○ A few carry more than one chromosome.
● Bacterial and Archaeal Chromosomes
○ Most bacterial and archaeal species contain a single closed circular chromosome. ○ These are variable in size.
○ The single or largest (if more than one) chromosome carries essential genes required for survival and reproduction.
○ Bacteria also carry multiple copies of one or more plasmids.
○ These are extrachromosomal DNA molecules.
○ They carry nonessential genes, unless the bacteria are exposed to certain antibiotics. ● Chromosome Organization
○ Bacterial and archaeal chromosomes are densely packed to form the nucleoid. ○ The chromosome is organized into a series of tight loops.
○ These allow for efficient packaging of relatively long DNA molecules into small spaces.
● Bacterial Chromosome Compaction
○ Bacterial chromosomes are compacted in two ways:
■ Proteins help put DNA into loops that pack the chromosome into the nucleoid.
■ The circular DNA undergoes supercoiling.
● Proteins Associated with Chromosomes
○ Small nucleoid-associated proteins participate in the DNA bending that contributes to folding and
condensation of chromosomes.
○ Structural maintenance of chromosome (SMC) proteins attach directly to the DNA, holding it in coils or V-shapes to form large nucleoprotein complexes.
○ Covalently closed circular chromosomes exist in several forms.
○ The relaxed circle is the least twisted.
○ The highly supercoiled form is the most tightly twisted; the chromosome is compacted so that it occupies less space in the nucleoid.
● Eukaryotic Chromosomes Are Organized as Chromatin
○ A eukaryotic chromosome has one DNA double helix, with a diverse array of proteins. ○ The DNA and associated proteins of a chromosome are called chromatin.
○ Proteins that organize chromosomes are essential and provide a mechanism for condensation, segregation, and organization of chromosomes.
● Chromatin Composition
○ Each chromosome is approximately half DNA and half protein.
○ About half of the proteins are histone proteins, small proteins that tightly bind DNA.
○ The remaining proteins, the nonhistone proteins, are very diverse and perform a variety of tasks in the nucleus.
○ There are five types of histone proteins: H1, H2A, H2B, H3, and H4; they are highly conserved among eukaryotes.
○ Two molecules each of histones H2A, H2B, H3, and H4 form an octamer. ○ A span of DNA ~ 146 bp long wraps around each octamer to form a nucleosome. ● Nucleosome Assembly
○ Histones H2A and H2B assemble into dimers; H3 and H4 also form dimers.
○ Two H3-H4 dimers form a tetramer, after which two H2A-H2B dimers associate with it to form the octamer.
○ The wrapping of DNA around the nucleosome is the first level of DNA condensation, and compacts the DNA about sevenfold.
● Higher Levels of Chromatin Compaction
● Higher-Order Chromatin
● Higher-Order Chromatin Condensation
○ Chromatin loops of 20 to 100 kb are anchored to the chromosome scaffold by nonhistone proteins at sites called MARs (matrix attachment regions).
○ The radial loop-scaffold model suggests that the loops gather into “rosettes” and are further compressed by nonhistone proteins.
○ Metaphase chromatin is compacted 250-fold compared to the 300-nm fiber. ● Roles of Higher-Order Chromatin Condensation
○ Chromosome compaction allows for efficient separation of chromosomes at anaphase.
○ The chromatin loops formed during condensation play a role in the regulation of gene expression.
○ Active transcription takes place in segments of loops distant from MARs; thus larger loops have more active transcription than small loops.
● Nucleosome Distribution and Synthesis During Replication
○ The presence of nucleosomes on chromosomes raises several questions about DNA replication.
○ Experimental evidence suggests that old histones are retained as single molecules, dimers, or tetramers.
○ Most nucleosomes present after replication are assembled partially from old nucleosome components and partially from new histones.
*This is the basis for epigenetic inheritance.*
● Nucleosomes and Replication
○ As the replication fork passes, nucleosomes break down into component parts.
○ H3-H4 tetramers immediately reassociate randomly with one of the sister chromatids. ○ H2A-H2B dimers disassemble and are reassembled from both old and new histones. ● Chromosome Regions Are Differentiated by Banding
○ Chromosome condensation reaches a maximum at metaphase.
○ Cytogeneticists can distinguish such chromosomes microscopically based on size, shape, and banding pattern.
○ Chromosome bands appear light or dark when chromosomes are treated with specific dyes and stains.
● Chromosome Structures and Banding Patterns
○ Centromeres divide chromosomes into chromosome arms, segments of unequal length. ○ The short arm is called the p arm and the long arm is the q arm.
○ Chromosome shapes are named based on centromere position, which determines the relative sizes of the arms.
● Chromosome Shapes
○ Metacentric: the centromere is near the middle of the chromosome.
○ Submetacentric: the centromere is between the center and the tip.
○ Acrocentric: the centromere is close to one end.
○ Telocentric: the centromere is at the tip of the chromosome and there is no p arm.
○ A karyotype is an ordered photographic display of a complete set of chromosomes for a species.
○ The chromosomes are grouped into homologous pairs in descending order of size.
○ The sex chromosomes are identified separately.
○ Chromosomes may be stained with dyes to show distinctive banding patterns for each chromosome.
● Chromosome Banding Techniques
○ Chromosome banding allows cytogeneticists to identify each chromosome in a karyotype.
○ Different stains and dyes are used to produce banding patterns.
○ The standard for human chromosome banding is G (Giemsa) banding; the patterns are distinct and reproducible.
● Uses of Karyotypes
○ Karyotypes allow for recognition of abnormalities in chromosome number or structure.
○ Extra or missing chromosomes can be easily identified, as can chromosome rearrangements such as insertions, deletions, or others.
○ Comparison between species allows for tracing of evolutionary history of related species.
● In Situ Hybridization
○ In situ hybridization uses molecular probes, labeled with fluorescence or radioactivity, to detect their target sequences.
○ First-generation methods used nucleotide probes labeled with 32P.
○ New generation methods utilize fluorescent labels with a greatly improved resolution, so that each labeled chromosome can be identified.
● Fluorescent In Situ Hybridization
○ Currently, fluorescent in situ hybridization uses
molecular probes labeled with compounds that emit
fluorescent light when excited by UV or visible light.
○ Various labels that emit light of different wavelengths
can be used simultaneously.
○ For human chromosomes, there are 24 different
fluorophores available, unique to each chromosome.
● Chromosome Territory During Interphase
○ Chromosomes are not uniformly distributed within a nucleus.
○ Boveri, who first observed this, suggested that the variation in position might be related to gene activity.
○ Cremer and Cremer showed that chromosomes are partitioned into specific regions, chromosome territories, during interphase.
● Dynamic Chromosomes
○ Chromosomes do not occupy the same territory in each nucleus, but once confined to a territory, a chromosome does not leave until the M phase is initiated.
○ However, chromosomes are active within their territories and move, twist, and turn during
transcription and DNA replication.
○ Chromosomes appear to be anchored in their territories by their centromeres. ○ Interchromosomal domains are regions between territories.
○ These are channels for movement of proteins, enzymes, and RNA molecules.
○ Early replicating parts of chromosomes are generally near the center of the nucleus and late replicating parts are near the periphery.
● Chromosome Position and Transcriptional Activity
○ Transcriptionally active portions of chromosomes are found nearer to interchromosomal domains, probably due to:
■ Greater access to needed proteins and enzymes.
■ Faster dispersal of RNA transcripts, once they are completed.
● Dynamic Chromatin Structure
○ Changes in level of compaction regulate access to DNA by proteins for replication, transcription, recombination, and repair.
○ Position effect variegation (PEV) in Drosophila illustrates the effect of chromatin compaction on gene expression.
○ Muller identified x-ray-induced mutations in fruit flies that resulted in variegated eye color.
● Analysis of Position Effect Variegation
○ Muller noticed that the X chromosomes of flies with variegated eye color had undergone inversion.
○ The white gene had been moved from its normal position near the telomere to a region near the centromeric heterochromatin.
○ The extent to which heterochromatin spreads from the centromere outward varies from one chromosome to the next.
● Heterochromatin and PEV
○ The stopping point of
determines whether or not the relocated white gene will be expressed.
○ If the w gene remains
euchromatic, it is expressed; if the gene is compacted into heterochromatin, it is not
○ Some eye cells will have an active white gene, others an inactive one, giving a speckled appearance.
● The Occurrence of PEV Shows
○ Gene expression can be silenced by the gene’s chromosomal position.
○ Silencing is a feature of chromatin structure that can be transmitted from one cell generation to the next.
● PEV Mutations
○ Genetic analysis of eukaryotic genomes shows that PEV is widespread.
○ Mutations modifying PEV led to identification of proteins that play a role in establishing and maintaining chromatin structures.
○ Two types of mutations have been identified:
■ E(var) Mutations
● E(var) mutations, enhancers of position-effect variegation, increase or enhance the appearance of the mutant white phenotype in flies with a
variegating allele of the white gene.
● E(var)s encourage the spread of heterochromatin beyond its normal
● These mutations produce a greater number of eye cells lacking pigment. ■ Su(var) Mutations
● Su(var) mutations, suppressors of position-effect variegation, increase or enhance the appearance of the mutant white phenotype in flies with a
variegating allele of the white gene.
● Su(var)s restrict the spread of heterochromatin or interfere with its
● These mutations produce a greater number of pigmented eye cells.
Chapter 13: Chromosome Aberrations and Transportation
● Nondisjunction Leads to Changes in Chromosome Number
○ Nondisjunction is the process of failed chromosome and sister chromatid segregation. ○ It can result in abnormalities in chromosome number.
○ The number of chromosomes in a nucleus and the relative size and shape of each chromosome are species-specific characteristics.
○ The euploid number of chromosomes of an individual is the number of complete sets (e.g., n, 2n, 3n).
○ If cells contain a number of chromosomes that is not euploid, chromosome number is aneuploid.
● Chromosome Nondisjunction
○ Chromosome nondisjunction is the failure of homologous chromosomes or sister chromatids to separate as they normally do during cell division.
○ In somatic cells, it can result in one daughter cell with an extra chromosome (2n+1) and the other missing one chromosome (2n-1).
○ The relatively poor survival of these cells normally limits their number in animals. ● Nondisjunction in Germ-Line Cells
● Nondisjunction in Meiosis II
● Gene Dosage Alteration
○ Blakeslee and Belling (1913) studied aneuploidy in the jimson weed (Datura stramonium; 2n=24).
○ They identified 12 phenotypically distinct lines of trisomic Datura, one for each chromosome.
○ Phenotypic effects resulted from changes in gene dosage, the number of copies of a gene; aneuploidy alters dosage of all the genes on the affected chromosome
● Gene Balance
○ Changes in gene dosage lead to an imbalance of gene products from the affected chromosome relative to the unaffected chromosomes.
○ Most animals are highly sensitive to changes in gene dosage.
○ In contrast, plants tolerate gene dosage changes more readily.
● Aneuploidy in Humans
○ Humans are enormously sensitive to gene dosage changes and aneuploids usually do not survive.
○ Only trisomies of chromosomes 13, 18, and 21 are seen in newborn infants, and no autosomal monosomies are observed.
○ Multiple forms of sex-chromosome trisomies, and one type of sex-chromosome monosomy occur.
○ Trisomies and monosomies other than those found in newborn infants are known to occur in humans.
○ Studies that monitor human pregnancies indicate that about half of all conceptions spontaneously abort during the first trimester.
○ About half of these pregnancies carry abnormalities of chromosome number or structure.
● Trisomy 21, or Down Syndrome
○ Research has identified a link between the risk of trisomy 21 and maternal age.
○ A small number of genes on chromosome 21 are responsible for the cognitive disabilities and heart abnormalities that are principal symptoms.
○ Critical Region on Chromosome 21
■ A portion of the chromosome called DSCR (Down syndrome critical region) can be correlated with the majority of the Down syndrome symptoms.
■ A candidate gene called DYRK, known to produce dosage-sensitive learning defects in mice and flies, makes a major contribution to Down syndrome.
● Turner Syndrome
○ Turner syndrome is a monosomy of the X chromosome, with no second sex chromosome (XO).
○ In XO embryos, the single copy of the gene SHOX, which is not inactivated by dosage compensation, is insufficient to direct normal development.
○ The haploinsufficiency of this gene plays a central role in producing the symptoms of the syndrome.
● Reduced Fertility in Aneuploidy
○ In trisomics, chromosome segregation during meiosis is disturbed because of failure to properly pair and
○ Two patterns of synapsis are possible: a trivalent synapsisor a bivalent and univalent arrangement.
○ Neither mechanism can segregate three chromosomes equally at anaphase I.
○ In trisomics, meiosis results in two chromosomes moving to one pole and one chromosome moving to the other.
○ Thus half the gametes contain two copies of the chromosome; these will produce trisomic offspring that areunlikely to survive.
○ This results in a form of semisterility, in which only some of the offspring produced are viable.
○ Mosaicism can develop as a result of mitotic
nondisjunction early in embryogenesis.
○ For example, 25-30% of Turner syndrome cases occur in females that are mosaic, with some 45, XO cells and some 46, XX.
○ Some Turner syndrome individuals carry 47, XXX cells too.
○ In some insects such as butterflies and fruit flies, sex-chromosome mosaicism produces
○ These are individuals with some female and some male cells.
○ These occur due to sex-chromosome nondisjunction in early development.
● Changes in Euploidy Result in Various Kinds of Polyploidy
○ Polyploidy is the presence of three or more sets of chromosomes in the nucleus of an organism.
○ It can result from duplication of chromosome sets within a species (autopolyploidy).
○ It can also occur from combining the chromosome sets of different species (allopolyploidy).
○ Many types of polyploidy are possible.
○ Three mechanisms lead to autopolyploidy:
■ Multiple fertilizations of one egg by
multiple pollen grains.
■ Mitotic nondisjunctions in sex stem cells.
■ Meiotic nondisjunction leading to a
diploid rather than haploid gamete.
○ Allopolyploids carry multiple sets of
chromosomes that originate in different species, and result in a hybrid offspring that is infertile because the chromosome sets are not
○ Chromosome nondisjunction in these hybrids leads to cells with double the number of
chromosomes so that now each chromosome has a homolog for pairing, and the hybrid is
● Consequences of Polyploidy
○ Allopolyploids occur naturally or via human manipulation.
○ Fruit and flower size is increased in polyploids.
○ Fertility is decreased, particularly in odd-numbered polyploids (3n, 5n, etc.).
○ Hybrid vigor, more rapid growth, increased fruit and flower production, and improved resistance to disease occur in polyploids.
○ Polyploids have more than two alleles of each gene.
○ Recessive phenotypes are produced only in individuals that are homozygous for that allele.
○ This occurs much less frequently in polyploids than in normal diploids.
● Chromosome Breakage Causes Mutation by Loss, Gain, and Rearrangement of Chromosomes
○ Mutations that result in loss or gain of chromosome segments can produce severe abnormalities due to gene dosage imbalances.
○ Changes may be large enough to detect microscopically.
○ Others may be too small for microscopy but may be detectable with molecular methods. ● Partial Chromosome Deletion
○ When a chromosome breaks, both DNA strands are severed at a location called a chromosome break point.
○ The broken chromosome ends can adhere to other broken ends or the termini of intact chromosomes.
○ Or, since part of the broken chromosome is acentric (lacks a centromere), it may be lost during cell division.
● Breakage Can Result in Partial Chromosome Deletion
○ Partial chromosome deletion: loss of a portion of a chromosome.
○ Small deletions, or microdeletions, are too small to be seen microscopically, and remove one or a few genes.
○ These can behave as recessive mutations if the genes involved are haplosufficient; if one or more of the genes are dose sensitive, the mutation will act as a dominant allele.
● Larger Chromosome Deletions
○ Larger chromosome deletions can be detected microscopically and affect more genes.
○ Detachment of one chromosome arm leads to a terminal deletion; the broken fragment lacks a centromere and is lost during cell division.
○ Organisms with one normal and one
terminally deleted chromosome are called partial deletion heterozygotes.
○ An interstitial deletion is the loss of an internal portion of a chromosome, and results from two chromosome breaks.
○ These are observed in many
organisms, including humans.
○ For example, WAGR syndrome is a series of conditions caused by deletion of multiple genes on chromosome
11–patients may have all or just some symptoms depending on exactly which genes are deleted.
● Unequal Crossover
○ Occasionally an unequal crossover takes place between two homologs.
○ This can result in partial duplication on one homolog and partial deletion on
○ An organism with one normal and one duplication homolog is a partial duplication heterozygote; an organism with one normal and one deleted homolog is a partial deletion heterozygote.
● Detecting Duplication and Deletion
○ Large deletions or duplications can be detected by microscopy that reveals
altered chromosome banding patterns.
○ Microdeletions and microduplications are too small to detect this way.
○ Instead molecular techniques such as FISH (fluorescent in situ hybridization) can be used to detect presence or
absence of DNA sequence.
● Chromosome Breakage Leads to Inversion and Translocation of Chromosomes
○ Sometimes chromosome breakage leads to reattachment of the wrong
○ Reattachment in the wrong orientation leads to chromosome inversion; reattachment to a nonhomologous chromosome leads to chromosome translocation.
○ If no critical genes are mutated, and dose-sensitive genes remain in balance, there may be no phenotypic consequences
● Chromosome Inversion
○ Two types of chromosome inversion can occur, depending on the relative position of the centromere.
○ In a paracentric inversion, the centromere is outside of the inverted region.
○ In a pericentric inversion, the centromere is within the inverted region.
○ Inversion heterozygotes have one normal and one inverted homolog.
● Chromosome Translocation
○ Translocations occur when broken ends of nonhomologous chromosomes are reattached.
○ Translocation heterozygotes, with one normal copy and one translocated copy of each chromosome, may be normal in phenotype if no genes are disrupted by the breakage and reattachment events.
○ Certain translocation heterozygotes may experience semisterility due to segregation abnormalities.
● Three Types of Translocations
○ Unbalanced translocations arise when a piece of one chromosome is translocated to a nonhomolog and there is no reciprocal event.
○ Reciprocal balanced translocations occur when pieces of two non-homologs switch places.
○ Robertsonian translocations, also called chromosome fusions, involve fusion of two nonhomologous chromosomes.
● Transposable Genetic Elements Move Throughout the Genome
○ Transposable genetic elements are DNA sequences that can move within the genome by an enzyme-driven process, transposition.
○ They exist in many forms, lengths, and copy numbers.
○ A transposable element can cause a mutation if it inserts into a wild-type allele and disrupts its function; this is called insertional inactivation.
● Discovery of Transposition
○ Barbara McClintock discovered transposition using crosses involving three linked genes, C, Sh and Wx, in maize.
○ C = purple kernels, c1 = colorless; Sh = plump kernels, sh = shrunken; Wx = shiny kernels, wx = waxy.
○ In experiments with trihybrid C Sh Wx/c1 sh w, she observed some kernels that were mostly purple but had sectors that lacked color in various patterns; the colorless sectors were all shrunken and waxy.
● McClintock’s Results
○ McClintock observed the nuclei of cells from the colorless sectors and noticed a terminal deletion of one chromosome 9 homolog.
○ The chromosome 9 homologs in the purple sectors were both intact.
○ McClintock observed that the break-points of chromosome 9 all occurred in the same
position of the affected chromosome.
● Interpretation of McClintock’s Results
○ McClintock concluded that a genetic element, later called a dissociation (Ds) element, was located at the site of chromosome breakage.
○ She suggested that Ds could not cause
chromosome breakage without a second
element, called the activator (Ac) element.
○ She was able to use the same explanation for a second observation of an unstable mutant
● Unstable Mutants
○ McClintock’s second observation was colorless kernels with varied patterns of purple
○ She concluded that the unstable mutant alleles were caused by insertion of Ds into the C locus to produce a kernel lacking
pigmentation (the mutation is called c1DS).
○ Occasional, random transposition of Ds out of the gene caused reversion to wild type and the array of purple spots on the kernels.
● Characteristics of Transposable Genetic Elements
○ Transposition requires the enzyme transposase; the gene for transposase is carried by some transposable elements.
○ Some transposable elements also carry genes that impart specific characteristics to cells or organisms.
○ Some elements contain only repetitive sequences.
● Types of Transposable Genetic Elements
○ Autonomous transposable elements carry a transposase gene and all DNA sequences needed to carry out transposition (e.g., Ac).
○ Non-autonomous transposable elements have no transposase gene and may lack the sequences needed for transposition (e.g., Ds).
○ These elements are unable to move unless transposase is provided by an autonomous element elsewhere in the genome.
● Transposition Modifies Bacterial Genomes
○ Bacterial genomes contain two categories of transposable elements.
○ IS (insertion sequence) elements are simple transposable elements containing only the genes and sequences needed for autonomous transposition.
○ Bacteria also have two types of transposons, that carry multiple genes and confer new traits on bacteria that contain them.
● Insertion Sequences
○ Each IS element contains about 1000 bp of DNA.
○ They have a transposase gene that is bracketed by a short, inverted repeat (IR) sequence.
○ Different IS elements have different IR sequences.
○ The inverted repeats are integral to transposase-mediated transposition.
○ Transposons (Tn) are composed of two types, composite and simple.
○ These are distinguished by the types of sequences flanking the transposon.
○ The genes most commonly carried by transposons confer antibiotic-resistance to bacteria that carry them.
● Composite Transposons
○ Composite transposons have a central region of several kb containing one or more functional genes.
○ The central region is flanked by complete IS elements in opposite orientation; at least one of these (sometimes both) contains a copy of the transposase gene.
○ The transposon Tn10 has a typical composite transposon structure
● Simple Transposons
○ Simple transposons are flanked by very short IR sequences of less than 50 bp.
○ The IR sequences do not encode transposase; this enzyme is encoded by the simple transposon itself.
○ There are additional genes in the central element.
● Transposition Mechanisms
○ A common outcome of transposition is the duplication of the target site and creation of direct repeats to either side of the inserted element.
○ There are two mechanisms of transposition that lead to target site duplication. ○ These are conservative transposition and replicative transposition.
■ Conservative Transposition
● Conservative transposition excises a transposable element from one
position and inserts it into a new location.
● It is similar to a cut-and-paste process used to edit text.
● The process moves transposable elements around the genome but results in no increase in the number of transposable elements.
■ Replicative Transposition
● Replicative transposition involves the copying of the transposable element and an increase in the number of elements per genome, similar to
● Following initiation of replication of the element (e.g., in plasmids),
transposase facilitates formation of a cointegrate – a temporary fusion of
● A recombination-like process resolves the cointegrate, and leaves both plasmids with a copy of the element.
● Transposition Modifies Eukaryotic Genomes
○ Eukaryotic transposable elements are divided into two groups:
■ DNA transposons are transposed through conservative or replicative
■ Retrotransposons are transcribed, then reverse transcriptase produces a double-stranded DNA copy of the element, which is then inserted into the
○ Retroviruses infect eukaryotic cells, and have genomes composed of single-stranded RNA.
○ On infection the RNA is transcribed into double-stranded DNA by reverse transcriptase, allowing the DNA to integrate into the host’s genome.
○ The viral genes gag and env, encoded by the integrated virus, are needed to produce new retroviral particles; pol encodes reverse transcriptase.
○ Retrotransposons are related to retroviruses; all carry pol and some contain gag.
○ None encode env, thus they can be reverse transcribed and inserted into host DNA, but are unable to produce viral particles.
○ The gene(s) carried on retrotransposons are flanked by long terminal repeats (LTRs).
**MAKE SURE TO READ THE TEXTBOOK AND DO THE END OF CHAPTER QUESTIONS** GOOD LUCK!