AP Bio: Chpt. 18, 19, & 20

(polycistronic) mRNA.

RNA polymerase first binds to a set of DNA bases that serve as the point of attachment, the PROMOTER

a. another kind of gene – a regulatory gene. Note this gene is not transcribed.
b. genes that are transcribed are called STRUCTURAL GENES

1. Promoter always capable of binding to RNA polymerase and therefore the genes in question are always transcribed (“on”) >genes that are always on are called CONSTITUTIVE GENES

2. Promoter usually incapable of binding to RNA polymerase and therefore the genes are usually not transcribed (“off”) but can be turned on

3. Promoter usually capable of binding to RNA polymerase and therefore the genes are usually transcribed (“on”) but can be turned off

discovered Control of a prokaryote gene

the model they developed was the called the lac operon for E coli

1. E. coli can digest and utilize lactose if it is present in the culture
2. three proteins are essential for this process.
3. They are made only when lactose is present. Therefore the genes for these proteins must be capable of being turned on when lactose
present.

functional units called an Operons existed on the
chromosome of E. coli.

1. two regulatory genes consisting of the:
-operator gene
– promoter gene

2. one or more structural genes.

structural genes

In addition to the genes of the operon another gene that is usually located some distance from the operon codes for another structural gene that was always on – the regulatory gene – which produced a protein called the “repressor”

repressor can be made in one of two forms depending on the particular regulator gene

a. active repressor
b. inactive repressor

. In the lac operon the regulatory gene lies next to the operon (an unusual situation) and makes active repressor

The lac operon – an inducible operon

1. Active repressor binds to the operator and prevents RNA polymerase from binding to
the promoter

2. the active repressor molecule has a binding site to which an chemical called the inducer can bind. When “induced”, the repressor becomes inactive.

3. inactive repressor falls way from the operator and RNA polymerase can bind to the promoter causing the structural genes to be transcribed.

4. Remember we want this to happen when lactose is present. The inducer for the lac operon is lactose and certain of its analogs (collectively referred to as allolactose)

5. The inducible operon model accounts for the situation when the gene is usually off.

The trp operon – a repressible operon.

E. coli can synthesize the amino acid tryptophan. Certain enzymes are necessary for this and the genes for these enzymes must be “on” for this to occur. If tryptophan is placed in the culture medium, the genes for these enzymes turn off.

1. the regulator gene, therefore, makes inactive repressor.

2. inactive repressor cannot bind to the operator and, therefore, RNA polymerase binds to
the promoter and the structural genes are transcribed.

3. When tryptophan is present, it binds to the inactive repressor activating it. The
tryptophan is said to be a corepressor

4. the active repressor binds to the operator turning off the promoter.

negative gene regulation
transcription

1. The lac operon not only requires the presence of lactose but also the absence of glucose
-by itself, the lac operon is has a very inefficient promoter sequence.

2. In order to work, the lac operon requires a positive regulator called Catabolite Activator Protein (CAP) to be present
-CAP binds to a site immediately adjacent to the promoter sequence.
– makes it easier for RNA polymerase to bind to promoter. Without CAP very little transcription takes place even in the presence of lactose

3. CAP must be activated by first binding to cAMP. It is actually a CAP-cAMP complex that activates the promoter. This is the key to how E. coli “knows” if glucose is present.

4. If glucose present, very little cAMP present and, therefore, very little active CAP

1. each cell of a muticellular eukaryote contains the entire genetic code for that individual

2. only certain genes are active however resulting in the specialized functions of the particular cell.

3. the process by which a cell becomes specialized is called cell differentiation.
-begins during embryonic development.

4. requires a high degree of selective gene control.
-These controls can occur at several levels.

a. the eukaryote chromosome differs organizationally in several ways from the prokaryote chromosome

1. much more DNA

2. great deal of repetition and much of the DNA with no apparent function

3. a close association with structural proteins

4. much greater complexity in the encoding of information and the regulation of expression

1. DNA long and linear and associated with proteins

2. in humans averages 1.5 X 108 nucleotide pairs; 4 cm long if stretched out.

3. In the non reproducing chromosomes = chromatin

4. more than half the chromatin is protein

1. most abundant proteins in chromatin

2. large amount of positively charged arginine and lysine

3. + charged histones attracted to – charged DNA molecules

4. involved in folding and packaging DNA

5. 5 types – each type similar to types in other organisms

a. nucleosome – a core of 8 (2x 4 different types) histones with DNA wrapped twice about it them like thread around a spool
– histone tails – amino end of histone protein

b. nucleosomes, in turn are packed into “fiber”

c. during condensation,
– fibers loop into loop domains
-scaffolding proteins
– loop domains fold further to form condensed coils

gene activity

a. euchromatin – weakly stained more opened – can be transcribed

b. heterochromatin – darker stained more condensed – can’t be transcribed – some structural – telomere and centromere regions

a. some areas of heterochromatin always present – e.g. area of centromere

b. other areas vary from cell type to another as does areas of euchromatin

– Prokaryote gene expression is all about
regulating transcription.

-However, eukaryotes gene expression can be controlled at additional points that either proceed or follow transcription

a. addition or removal of acetyl group -COCH3

b. acetyl groups binds to lysines in the histone tails

c. acetylated histones lose their positive charge and tails no longer bind to neighboring nucleosomes

d. acetylated histones grip DNA less tightly, providing easier access for transcription proteins in this region

e. deacetylation has the opposite effect.

f. Some of the enzymes responsible for acetylation or deacetylation are associated with or are components of transcription factors that bind to promoters.

g. In addition, some DNA methylation proteins recruit histone deacetylation enzymes, providing a mechanism by which DNA methylation and histone deacetylation cooperate to repress transcription.

a. methylation of histone tails can promote condensation.

b. phosphorylation adjacent to a methylated amino acid reverses effect of methylation

c. The histone code hypothesis – specific combinations of modifications rather than overall level of histones acetylation help determine the chromatin configuration which in turn influences transcription

a. attachment by specific enzymes of methyl groups (-CH3) to DNA bases (usually cytosine) after DNA synthesis.

b. Inactive DNA is generally highly methylated compared to DNA that is actively transcribed.

c. Genes are usually more heavily methylated in cells where they are not expressed.

d. Once methylated, genes usually stay that way through successive cell divisions
– methylation patterns become important in establishing specialized tissues during embryogenesis.

e. methylation pattern may maintained in this way also accounts for genomic imprinting in mammal where methlation permanently regulates the expression of the allele of an either maternal or
paternal chromosome

f. proteins bind to methylated DNA and recruit enzymes that deacetylated histones (which as we have seen will also inhibit gene expression)

does not involve changes in DNA yet these modifications may be passed on to offspring! What!!!!! Does this sound like a very old and discredited biological principle (think Lamark)

-important point – unlike changes in DNA these modifications can be reversed and therefore are not necessarily permanent or predictable.

proteins

Eukaryotic RNA polymerase is dependent on transcription factors. Only when a complete initiation complex has been assembled can
transcription begin.

a. One type of transcription factor recognizes the TATA box.
b. Others in the initiation complex are involved in protein-protein interactions.

High transcription levels require additional transcription factors binding to other control elements.

a. Proximal control elements

b. Distant control elements, enhancers, may be thousands of nucleotides away from the promoter or even downstream of the gene or within an intron.

c. Bending of DNA through the use of DNA -bending protein enables transcription factors, activators, bound to enhancers to contact the protein initiation complex at the promoter. This helps position the initiation complex on the promoter.

Eukaryotic genes also have repressor proteins that bind to DNA control elements called silencers.

a. At the transcription level, activators are probably more important than repressors, because the main regulatory mode of eukaryotic cells seems to be activation of otherwise silent genes.

b. Repression may operate mostly at the level of chromatin modification.

Gene transcription controlled by a surprisingly small number of enhancer sequences

a. yet controlled exerted over huge numbers of genes
b. made possible by combinations of available activators (or repressors)

a. introns and exons

1. alternative RNA splicing, different mRNA molecules are produced from the same primary transcript, depending on which RNA segments are treated as exons and which as introns
– editing can be done differently and result in different polypeptides

b. rate of mRNA degredation

a. inactive proteins can be activated by environmental factors

b. Protein degradation rates can be altered.

play multiple roles in controlling gene expression

1. only a small fraction of DNA codes for proteins, rRNA, and tRNA

2. A significant amount of the genome may be transcribed into noncoding RNAs

3. Noncoding RNAs regulate gene expression at two points: mRNA translation and chromatin configuration

MicroRNAs (miRNAs) are small single-stranded RNA molecules that can bind to mRNA
– These can degrade mRNA or block its translation

The inhibition of gene expression by altering chromatin configuration by RNA molecules is called RNA interference (RNAi)
– RNAi is caused by small interfering RNAs (siRNAs)
– siRNAs play a role in heterochromatin formation and can block large regions of the chromosome
– siRNAs may also block transcription of specific genes

1. During embryonic development, a fertilized egg gives rise to many different cell types

2. Cell types are organized successively into tissues, organs, organ systems, and the whole organism

3. Gene expression orchestrates the developmental programs of animals

4. The transformation from zygote to adult results from cell division, cell differentiation, and morphogenesis
-Cell differentiation is the process by which cells become specialized in structure and function
– The physical processes that give an organism its shape constitute morphogenesis

5. Differential gene expression results from genes being regulated differently in each cell type

Materials in the egg can set up gene regulation that is carried out as cells divide
-An egg’s cytoplasm contains RNA, proteins, and other substances that are distributed unevenly in the unfertilized egg
– CYTOPLASMIC DETERMINANTS are maternal substances in the egg that influence early development
– As the zygote divides by mitosis, cells contain different cytoplasmic determinants, which lead to different gene expression

The other important source of developmental information is the environment around the cell, especially signals from nearby embryonic cells

– In the process called INDUCTION, signal molecules from embryonic cells cause transcriptional changes in nearby target cells

cells go from the original totipotent condition to some determinant level

a. Determination commits a cell to its final fate

once the fate of the cell is determined, the cell is bound to differentiate along certain lines.

1. Cell differentiation is marked by the production of tissue-specific proteins

Example – Myoblasts produce muscle-specific proteins and form skeletal muscle cells
– MyoD is one of several “master regulatory genes” that produce proteins that commit the cell to becoming skeletal muscle
-The MyoD protein is a transcription factor that binds to enhancers of various target genes

1. Pattern formation is the development of a spatial organization of tissues and organs

2. In animals, pattern formation begins with the establishment of the major axes

3. Positional information, the molecular cues that control pattern formation, tells a cell its location relative to the body axes and to neighboring cells

a. Combining anatomical, genetic, and biochemical approaches, researchers have discovered developmental principles common to many other species, including humans

b. In Drosophila, cytoplasmic determinants in the unfertilized egg determine the axes before fertilization

c. After fertilization, the embryo develops into a segmented larva with three larval stages

genes direct the developmental process

1. They created mutants, conducted breeding experiments, and looked for corresponding genes

2. Breeding experiments were complicated by embryonic lethals, embryos with lethal mutations

3. They found 120 genes essential for normal segmentation

encode for cytoplasmic determinants that initially
establish the axes of the body of Drosophila

a. These maternal effect genes are also called egg-polarity genes because they control orientation of the egg and consequently the fly

One maternal effect gene, the bicoid gene, affects the front half of the body

1. An embryo whose mother has a mutant bicoid gene lacks the front half of its body and has duplicate posterior structures at both ends

2. This phenotype suggests that the product of the mother’s bicoid gene is concentrated at the future anterior end

3. This hypothesis is an example of the gradient hypothesis, in which gradients of substances called morphogens establish an embryo’s axes and other features

– It identified a specific protein required for some early steps in pattern formation

– It increased understanding of the mother’s role in
embryo development

– It demonstrated a key developmental principle that
a gradient of molecules can determine polarity and
position in the embryo

a. Cancer results from genetic changes that affect cell cycle control

b. The gene regulation systems that go wrong during cancer are the same systems involved in embryonic development

c. Cancer can be caused by mutations to genes that regulate cell growth and division

d. Tumor viruses can cause cancer in animals including humans

cancer-causing genes

are the corresponding normal cellular genes that are
responsible for normal cell growth and division

cell cycle

a. Movement of DNA within the genome: if it ends up near an active promoter, transcription may increase

b. Amplification of a proto-oncogene: increases the number of copies of the gene

c. Point mutations in the proto-oncogene or its control elements: causes an increase in gene expression

uncontrolled cell growth

a. Repair damaged DNA
b. Control cell adhesion
c. Inhibit the cell cycle in the cell-signaling pathway

cancer onset

Mutations in the ras proto-oncogene and p53 tumor-suppressor gene are common in human cancers

a. Mutations in the ras gene can lead to production of a hyperactive Ras protein and increased cell division

b. mutations of the p53 tumor-suppressor gene can result in the in problems with a transcription factor (p53) and thus the gene that cannot be properly transcribed and the cell cycle inhibitory protein cannot be produced

Multiple mutations are generally needed for full-fledged cancer; thus the incidence increases with age

a cancerous cell is usually characterized by at least one active oncogene and the mutation of several tumor-suppressor genes

move within the genome – “jumping genes”

two types
a. Transposons
b. Retrotransposons

1. first proposed by Barbara McClintock in 1940s (Nobel Prize in 1983)

2. cut and paste vs copy and paste
– cut and paste DNA element cut out of sequence and spliced in elsewhere – becomes noticeable when jumps into functional gene.
– copy and paste DNA sequence copied then copy spliced leaving the original in place

mechanism always leaves a copy behind

1. Sequence transcribed into RNA

2. include in the sequence is a translatable element that gets translated into an enzyme called reverse transcriptase
-.enzyme first discovered in retroviruses
-some think retrovirus’ gene actually came originally from eukaryote retrotransposon

3. reverse transcriptase creates new copy of retrotransposon which is then spliced into the genome elsewhere.

Multiple copies of transposable elements ( 100s to 1,000s bases long) are scattered throughout the genome.

In humans and other primates a large proportion (almost 10% of human genome) of transposable element related-DNA consists of a family of similar sequences called Alu elements

1. probably arose by mistakes that occurred in replication or recombination.

2. large segment-duplication
– long stretches of DNA 10,000 to 300,000 bas pairs long
-seem to have been copied from one chromosome area to another on the or even another chromosome

Many copies of repeated short sequences:
ex. ….GTTACGTTACGTTACGTTAC….
-repeated units often fewer than 15 nucleotides but may consist of as many as 500 nucleotides
– number of repeated units may vary
-in our example, GTTAC may be repeated several hundred thousand times

Much of the genomes simple sequence DNA found in telomeres and centromeres.

exist as a collection of identical or very similar genes.

a. These likely evolved from a single ancestral gene.

b. The members of multigene families may be clustered or dispersed in the genome.

are multigene families that are clustered tandemly.

a. usually consist of the genes for RNA products or those for histone proteins.

-For example, the three largest rRNA molecules are encoded in a single transcription unit that is repeated tandemly hundreds to thousands of times.

-This transcript is cleaved to yield three rRNA molecules that combine with proteins and one other kind of rRNA to form ribosomal subunits.

the nucleolus

multigene families

diverged since their initial duplication event.

a. The classic example traces the duplication and diversification of the two related families of globin genes, a (alpha) and b (beta), of hemoglobin.

1. the a subunit family is on human chromosome 16
2. the b subunit family is on chromosome 11.

The different versions of each globin subunit are expressed at different times in development, fine-tuning function to changes in environment.

Within both the a and b families are sequences that are expressed during the embryonic, fetal, and/or adult stage of development.
– The embryonic and fetal hemoglobins have higher affinity for oxygen than do adult forms, ensuring transfer of oxygen from mother to developing fetus.

These gene families probably arise by repeated gene duplications that occur as errors during DNA replication and recombination.

The differences in genes arise from mutations that accumulate in the gene copies over generations.

The locations of the two globin families on different
chromosomes indicate that they (and certain other gene families) probably arose by transposition

These mutations may even lead to enough changes to form pseudogenes, DNA segments that have sequences similar to real genes but that do not yield functional proteins.

1. Duplication of chromosomal sets – nondisjunction – during meiosis or mitosis during early embryogenesis can produce polyploidy individuals with an extra set of chromosomes.
– one set of genes can carry out functions of organism other set can undergo changes (mutations that can give rise to novel genes.

2. Gene duplication due to crossing over errors – one chromosome ends up with extra copy of gene

3. Transposition

1. globulin genes arose from single ancestral gene.

2. This gene underwent a duplication about 450 to 500 million years ago.

3. each gene diverged into two families the α and β lines

4. at some point a transposition probably occurred that moved these genes to separate chromosomes

5. each of these lines underwent additional duplications

6. these duplicate genes, in turn, underwent divergent changes to become the current family of genes and pseudogenes

Protein coat (capsid)
Nucleic acid core
– DNA viruses
-RNA viruses

a. The smallest viruses have only four genes, while the largest have several hundred.

Nature of the capsid
a. shape of virus
– spiral – tobacco mosaic virus, influenza virus.
-triangular plates arranged in a polyhedron – adenovirus, bacteriophages

b. specificity – protein coat contains proteins that assist in recognition of host cell

c. lipoprotein or glycolipid envelope – actually a part of host cell’s membrane

Viruses are obligate intracellular parasites.

1. They can reproduce only within a host cell.

2. An isolated virus is unable to reproduce – or do anything else, except infect an appropriate host.

3. Viruses lack the enzymes for metabolism or ribosomes for protein synthesis.

1. double strands of DNA
2. single strand of DNA
3. double strands of RNA
4 single strand of RNA

b. may be linear or circular
c. smallest viruses have only four genes; largest, several hundred to a thousand

-attachment to receptor site on host
-entry of nucleic acid core into cell

1. may come in by itself leaving capsid outside – remember bacteriophage experiments of Hershey and Chase

2. may enter along with capsid – adenoviruses; once inside capsid releases core

– What happens next depends on type of nucleic acid in core (RNA or DNA)

– e.g., bacteriophages

a. viral DNA replicated by host cell.

b. viral DNA transcribed and translated by host cell to make more viral protein for capsids.

c. . new DNA and capsid proteins assembled into new viruses

d. viruses shed from cells – may or may not kill host cells

Research on phages led to the discovery that some double-stranded DNA viruses can reproduce by two alternative mechanisms: the lytic cycle and the lysogenic cycle.

the phage reproductive cycle culminates in the
death of the host.

1. In the last stage, the bacterium lyses (breaks open) and releases the phages produced within the cell to infect others.

– Virulent phages reproduce only by a lytic cycle.

the phage genome replicates without destroying the host cell.

like phage lambda, use both lytic and lysogenic
cycles.

1. During a lytic cycle, the viral genes immediately turn the host cell into a virus-producing factory, and the cell soon lyses and releases its viral products.

2. The viral DNA molecule, during the lysogenic cycle, is incorporated by genetic recombination into a specific site on the thost cell’s chromosome.
– In this prophage stage, one of its genes codes for
a protein that represses most other prophage genes.
– Every time the host divides, it also copies the
viral DNA and passes the copies to daughter cells.

3. Occasionally, the viral genome exits the bacterial
chromosome and initiates a lytic cycle.
-This switch from lysogenic to lytic may be
initiated by an environmental trigger.

4. The lambda phage which infects E. coli demonstrates the cycles of a temperate phage.

a. In some with single-stranded RNA (class IV), the genome acts as mRNA and is translated directly.

b. In others (class V), the RNA genome serves as a template for mRNA and for a complementary RNA.

1. This complementary strand is the template for the synthesis of additional copies of genome RNA.

2. All viruses that require RNA -> RNA synthesis to make mRNA use a viral enzyme, RNA replicase, that is packaged with the genome inside the capsid.

(class VI) have the most complicated life cycles.

1. reverse transcriptase carried among the capsids proteins

a. single strand of DNA (cDNA) made from RNA.

b. second strand of DNA then replicated to form double stranded cDNA

c. cDNA spliced into hosts DNA (this DNA of viral origin called a provirus [similar to a prophage])

d. viral DNA then transcribed as any other DNA can be transcribed.

1. can be translated to make viral protein for capsid
used directly for RNA core

the virus that causes AIDS (acquired immunodeficiency syndrome) is a retrovirus.

1. The viral particle includes an envelope with
glycoproteins for binding to specific types of blood
cells, a capsid containing two identical RNA strands
as its genome and two copies of reverse transcriptase.

2. The reproductive cycle of HIV illustrates the pattern of infection and replication in a retrovirus.

a. After HIV enters the host cell, reverse transcriptase synthesizes double stranded DNA
from the viral RNA.

b. Transcription produces more copies of the viral
RNA that are translated into viral proteins, which self-assemble into a virus particle and leave the
host.

smaller and simpler than even viruses, consist of tiny molecules of naked circular RNA that infect plants

a. Their several hundred nucleotides do not encode for proteins but can be replicated by the host’s cellular enzymes.

b. These RNA molecules can disrupt plant metabolism and stunt plant growth, perhaps by causing errors in the regulatory systems that control plant growth.

are infectious proteins that spread a disease.

a. They appear to cause several degenerative brain diseases including scrapie in sheep, “mad cow disease”, and Creutzfeldt-Jacob disease in humans.

b. Slow-acting; incubation up to 10 years.

c. Virtually indistructable.

d. spread by eating meat “contaminated” with nervous tissue from infected individual.

e. According to the leading hypothesis, a prion is a misfolded form of a normal brain protein.

f. It can then convert a normal protein into the prion version, creating a chain reaction that increases their numbers

In addition to single chromosome found in all prokaryotes, prokaryotes almost always have additional information contained in plasmids

a. much smaller than chromosomes
b. Plasmids, generally, benefit the bacterial cell.
c. They usually have only a few genes that are not required for normal survival and reproduction.
1. 2-30 genes
2. Plasmid genes are advantageous in stressful conditions.
d. circular and self-replicating
e. may replicate more often than chromosome – cells will have more than one copy of plasmid
f. may replicate less often than chromosomes – some cells will have no plasmids
g. may act as an episome

genetic elements that can undergo reversible incorporation into the cell’s chromosome.

temperate viruses also qualify as episomes.

Conjugation transfers genetic material between two bacterial cells that are temporarily joined.

1. One cell (“male”) donates DNA and its “mate” (“female”) receives the genes.

2. A sex pilus from the male initially joins the two cells and creates a cytoplasmic bridge between cells.

3. “Maleness”, the ability to form a sex pilus and donate DNA, results from an F factor as a section of the bacterial chromosome or as a plasmid.

4. The F plasmid facilitates genetic recombination when environmental conditions no longer favor existing strains.

5. The F factor or its F plasmid consists of about 25 genes, most required for the production of sex pili.

6. Cells with either the F factor or the F plasmid are called F+ and they pass this condition to their offspring.

7. Cells lacking either form of the F factor, are called F-, and they function as DNA recipients.

8. When an F+ and F- cell meet, the F+ cell passes a copy of the F plasmid to the F- cell, converting it.

9 The plasmid form of the F factor can become integrated into the bacterial chromosome.(i.e., act as a episome)

10. The resulting Hfr cell (high frequency of recombination) functions as a male during conjugation.

a.In the 1950s, Japanese physicians began to notice that some bacterial strains had evolved antibiotic resistance.

b. The genes conferring resistance are carried by plasmids, specifically the R plasmid (R for resistance).

c. Some of these genes code for enzymes that specifically destroy certain antibiotics, like tetracycline or ampicillin.

d. When a bacterial population is exposed to an antibiotic, individuals with the R plasmid will survive and increase in the overall population.

e. Because R plasmids also have genes that encode for sex pili, they can be transferred from one cell to another by conjugation.

f. since genes are on plasmids and plasmids can replicate rapidly a single cell may have many genes for resistance. It can then make large quantities of the resistance chemicals

g. resistance genes can be transferred from plasmid to plasmid resulting in plasmids that confer resistance to many antibiotics

h. When resistant bacteria introduced to colony of non-resistant bacteria, entire culture rapidly becomes resistant

a. conjugation (see information on F plasmid)

b. transformation -the alteration of a bacterial cell’s genotype by the uptake of naked, foreign DNA from the surrounding environment.
– The resulting cell is now recombinant with DNA derived from two different cells.

c transduction -occurs when a phage carries bacterial genes from one host cell to another.

a small piece of the host cell’s degraded
DNA is packaged within a capsid, rather than the phage genome.

a. When this phage attaches to another bacterium, it will inject this foreign DNA into its new host.

b. Some of this DNA can subsequently replace the homologous region of the second cell.

c. This type of transduction transfers bacterial genes at random.

occurs via a temperate phage.

a. When the prophage viral genome is excised from the chromosome, it sometimes takes with it a small region of adjacent bacterial DNA.

b. These bacterial genes are injected along with the phage’s genome into the next host cell.

c. Specialized transduction only transfers those genes near the prophage site on the bacterial chromosome.

d. Can pass from species to species!

small enough to work with

bacterial enzymes that cut foreign DNA into small pieces
1. bacterial defense mechanism designed to break up invading viral DNA
2. 200 + have been discovered

b. recognition sequences
1. short set of nucleotides that can be recognized by restriction enzymes

2. cut can be:
a. straight – Hpal
b. staggered cut – EcoRI (5′-GAATTC-3′);
Hind III (5′-AAGCTT-3′)

1. can be used to create “sticky ends” ideal for splicing

c. bacteria protect themselves by methylating at least one of the bases in any of their DNA that corresponds to a recognition sequence.

DNA segments created in this fashion (broken up by restriction enzymes) are called gDNA (genomic DNA)

a. different restriction enzymes produce different fragments
b. fragments can be separated by electrophoresis (from cathode to anode in gel, smaller portions go farther)

a. DNA produced this way will not necessarily be one particular gene
1. may include several genes
2. may include genes that have been fragmented

a. isolate mRNA
b. use reverse transcriptase to make cDNA
c. can attach sticky ends and splice into other DNA

1. hard to isolate mRNA – can only find in cells actively synthesizing protein in question

2. in eukaryotes the mRNA being translated has been edited; the actual gene contains info for introns as well as exons

a. polynucleotides can be synthesized in the lab

b. severe size limitations – only good for very small DNA molecules

Plasmids:

1. The source of the plasmid is typically E. coli.
– This plasmid carries two useful genes, ampR, conferring resistance to the antibiotic ampicillin and lacZ, encoding the enzyme beta-galactosidase which catalyzes the hydrolysis of sugar.
-The plasmid has a single recognition sequence, within the lacZ gene, for the restriction enzyme used.

2. splice, using DNA ligase, the DNA fragment into plasmid that has been cut open with the same restriction enzyme that was used to create the
fragment
-as bacteria reproduce, plasmids will be reproduced
– harvest DNA fragments by treating with restriction enzyme

newer technique takes tiny sample of DNA and within hours synthesizes millions of copies

1. DNA cloning is the best method for preparing large quantities of a particular gene or other DNA sequence.

2. When the source of DNA is scanty or impure, the polymerase chain reaction (PCR) is quicker and more selective.

3. This technique can quickly amplify any piece of DNA without using cells.

a. DNA added to solution containing
– DNA polymerase
-a large supply of DNA nucleotides in triphosphate form
-DNA primers complimentary to bases at ends of fragment molecule

b. solution heated to unzip DNA

c. DNA primers attach

d. DNA polymerase replicates both strands to produce 2 molecules of fragment

e. solution cooled

f. process repeated over and over again

g. 20 cycles produces over 1 million DNA molecules

1. If you want to sequence DNA, you first must get the sample molecule to a manageable size
a. to do this, you perform Restriction Fragment Analysis
– This technique involves cutting the DNA with a restriction enzyme.
– Then separating the fragments by gel electrophoresis

2. Once you have identified the fragments, you can remove the fragment from the gel (the electrophoretic procedure does not alter the DNA).
a. sample of fragments then must be amplified
– DNA cloning through bacterial recombination
-PCR

3. Once you have amplified the small DNA fragments the next step is to sequence it

-by using different restriction enzymes that recognize different recognition sequences,
you can produce different fragments

sequence fragments – technique example

a. take a sample of the amplified DNA and denature into a single strand

b. to the sample, add a small amount of four different dideoxynucleotides, each tagged with a
specific fluorescent dye
1. ddATP (dideoxyribose adenine triphosphate)
2. ddCTP (dideoxyribose cyosine triphosphate)
3. ddGTP (dideoxyribose guanine triphosphate)
4. ddTTP (dideoxyribose thymine triphosphate)
5. during replication if this type of nucleotide is inserted replication stops at that point.

c. into each group also add
-large amounts of normal nucleotides of all types
– radioactive primers – to prime 3′ end of the template stand
– DNA polymerase

d. Just as we saw for PCR, this mixture will result in the synthesis of a complementary strand

e. the difference is that whenever a ddnucleotide is added elongation of the complimentary strand is halted.

f. Since the amount of ddnucleotides in the sample is small the addition of these nucleotides will be less frequent and random for which of the possible nucleotide compliments it will attach and therefore, how long the complimentary strand will be.
– For example, ddTTP may compliment with first A (stopping replication after only one base addition), or the second A (stopping replication after two base additions) or the third A (stopping replication after four base additions), or the fourth A stopping
replication after 10 base additions.

g. as a result, the sample will contain fragments of different length.

labeled strands are separated through a polyacrylamide gel (shortest go farther), then a fluorescence detector detects the color of the tags, thereby distinguishing length

results, complimentary to the template strand) can be printed on a spectrogram and read from the bottom (shortest strand) to top (longest strand)

Once all fragments have been sequenced, fragments produced by different restriction enzymes can be analyzed for areas of overlap and sequences established for DNA molecule

a. the hummingbird genomic DNA and a bacterial plasmid are isolated
– the plasmid has been engineered to carry two genes:
– ampR which males E. coli cells resistant to the antibiotic ampicillin
– LacZ which you will recall is one of the enzymes in the lac – operon
— codes for enzyme β galactosidase
–normally used by E. Coli to break down lactose
— can also hydrolyze a similar synthetic molecule called X-gal to form a blue product

b. both are digested with the same restriction enzyme that makes a single cut in the LacZ gene
and many cuts within the hummingbird DNA

c. the fragments are mixed, and DNA ligase is added to bond the fragment sticky ends

d. Some recombinant plasmids now contain hummingbird DNA
-not all will have the β-globin gene however. Some will contain the other fragments.

e. the DNA mixture is added to bacteria that have a mutation in their own LacZ gene and have been genetically engineered to accept it the plasmid by transformation.

the bacteria are plated on a type of agar that contains ampicillin and X-gal

1. only bacteria containing the plasmid will reproduce because of the ampicillin

2. Bacteria who have intact LacZ genes (i.e., did not take up the Hummingbird DNA) will hydrozyze the X-gal and turn blue. Those that have plasmids with hummingbird DNA fragments will be white.

-This end results is the cloning of many hummingbird DNA fragments, including the β-globin gene

store collections of fragments from an entire genome

a. plasmid library – collection of recombinant vector clones produced by cloning DNA fragments from an entire genome

b. Phage library that is made using bacteriophages is stored as a collection of phage clones

c. bacterial artificial chromosome (BAC) library contains bacteria with large plasmids that have been trimmed down and can carry a large DNA insert. Each “book” in the “library” carries more information than those in the previous two. Note that the bacterial colonies are contained in multi well plates. Each well contains a colony with a specific BAC clones.

d. cDNA libraries – made by cloning DNA made in vitro by reverse transcription of all the mRNA produced by a particular cell.
– represents only part of the genome—only the subset of genes transcribed into mRNA in the original cells
-does not contain whole gene in euakaryotes because introns are not included.

1. A clone carrying the gene of interest can be identified with a nucleic acid probe having a sequence complementary to the gene

2. A probe can be synthesized that is complementary to the gene of interest
a. For example, if the desired gene is
5′ GGCTTAACTTAGC 3′,
then we would synthesize this probe
3′ CCGAATTGAATCG 5′

3. the DNA probe can be used to screen a large number of clones simultaneously for the gene of
interest
4. once identified, the clone carrying the gene of interest can be cultured

after a gene has been cloned, its protein product can be produced in larger amounts for research

1. Cloned genes can be expressed as protein in either bacterial or eukaryotic cells

2. Several technical difficulties hinder expression of cloned eukaryotic genes in bacterial host cells.
-to overcome differences in promoters and other DNA control sequences, scientists usually employ an expression vector, a cloning vector that contains a highly active prokaryotic promoter

1. uses gDNA obtained from restriction enzyme digestion of two or more sources DNA. We can tie together several molecular techniques to compare DNA samples from individuals.

a. start by adding the restriction enzyme to each of the samples to produce restriction fragments.

b. then separate the fragments by gel electrophoresis.

c. Southern blotting (Southern hybridization) allows us to transfer the DNA fragments from the gel to a sheet of nitrocellulose paper, still separated by size.

d. This also denatures the DNA fragments.

e. Bathing this sheet in a solution containing our probe allows the probe to attach by base-pairing (hybridize) to the DNA sequence of interest and we can visualize bands containing the label with autoradiography.

Southern blotting can be used to examine differences in noncoding DNA as well.

1. RFLPs (Restriction Fragment Length Polymorphism )* are detected and analyzed by Southern blotting, frequently using the entire genome as the DNA starting material.

2. These techniques will detect RFLPs in noncoding or coding DNA.

3. Because RFLP markers are inherited in a Mendelian fashion, they can serve as genetic markers for making linkage maps.

4. The frequency with which two RFPL markers – or a RFLP marker and a certain allele for a gene – are inherited together is a measure of the closeness of the two loci on a chromosome

base substitutions, additions, deletions or sequence rearrangements within restriction enzyme sequences recognition sequences.

“RFLP; the acronym is pronounced “riflip”. Restriction fragment length polymorphism is the identification of specific restriction enzymes that reveal a pattern difference between the DNA fragment sizes in individual organisms. To discover RFLPs, restriction enzymes (RE) are used to cut DNA at specific 4-6 bp recognition sites. Sample DNA is cut with one or more RE’s and resulting fragments are separated according to molecular size using gel electrophoresis. Differences in fragment length result from base substitutions, additions, deletions or sequence rearrangements within RE recognition sequences.

Splicing genes into bacteria to produce specific proteins- research, resistance, therapy

1. biomedical:
-human insulin
-growth hormone
-interferon
-clotting factors

2. agro-industrial:
-renin
-cellulose

a. common soil bacterium causes tumor called crown gall
b. research shows genes that cause tumor located on plasmid
c. a portion of this plasmid known as Ti is actually spliced into the hosts DNA
d. the genes of Ti have been worked out
e. Ti plasmid is being used as a vector to insert genes into plants

lucerferase experiment
a. genes for luciferase spliced into E coli and cloned
b. genes then inserted into plant virus to get regulatory sequence
c. genes then spliced into Ti plasmids
d. tobacco plant cell cultures then infected e. cells from culture grown into plant
f. plants given luciferin – plant glowed!
g. modifications of this technique being used to introduce favorable genes into plants – tomatoes for example – resistance to disease

virus RNA can help with bone marrow patients