Chapter 13

TRANSCRIPTION

The enzyme RNA polymerase catalyzes synthesis of RNA from one strand of the DNA double helix (template strand) by complementary base pairing analogous to DNA replication (5’--->3’).
(Fig. 13-11)

Ribose (2’ OH) replaces deoxyribose.

A, C, G, U (Uracil)--->U replaces T

RNA is considered single-stranded but it does base pair.

G-C (3 H-bonds)
A-U (2 H-bonds) (G:C>A:U>G:U)
G-U (2 H-bonds)

3 Classes of RNA

1. rRNA (ribosomal RNA)--->Component of ribosomes (16S, 23S, 5S).

2. tRNA (transfer RNA)--->Brings AAs to ribosome during translation.

3. mRNA (messenger RNA)--->Intermediate that transfers the genetic information from DNA to proteins. Directs protein synthesis.

The DNA template strand for a given mRNA is the sense strand.

The complementary non-template strand is the antisense strand.

(i.e.) RNA is the same sequence as the non-template strand except U replaces T.

RNA Polymerase--->In prokaryotes a single RNA pol synthesizes rRNA, tRNA, mRNA.

RNA can be monocistronic (one gene) or polycistronic (more than one gene).

Holoenzyme--->a₂bb’ws (6 subunits)
Core enzyme---> a₂bb’ (4 subunits)

bb’-->polymerization activity.
a₂-->interacts with transcription factors (chapter 18).
w--->no known function.
s--->provides specificity at promoters.

Different s factors allow recognition of different classes of promoters.

E. coli

Transcription (3 stages)

1. Initiation
2. Elongation
3. Termination

INITIATION

Promoter-->Nucleotide sequence that s recognizes.
 
 

TGTTGACA------------------TATAAT------ +1 (start)

-35                 11-15 nt        -10      5-8 nt

RNA pol holoenzyme binds to promoter, unwinds DNA, catalyzes condensation of incoming NTPs via DNA complementarity.

DNA G-C-A-T
RNA C-G-U-A

Energy provided by cleavage of high-energy triphosphate as in DNA replication.
(See Fig. 13-10)

ELONGATION--->Shortly after initiation, s is released, transcription continues with core enzyme.
(See Fig. 13-9)

TERMINATION--->RNA pol recognizes signals to terminate transcription. Eukaryotes different???

A. Rho-independent termination--->G-C rich RNA stem-loop followed by several U residues.

RNA pol dissociates from DNA template and releases the transcript when it encounters this signal.
(See Fig. 13-12)

B. Rho-dependent termination--->

Rho termination factor binds to rut sequence in the nascent transcript (C rich), rho catches RNA pol and causes the RNA to dissociate from RNA pol.
(See Fig. 13-13)

Eukaryotic RNA synthesis

Transcription in nucleus, translation in cytoplasm.
(i.e.) RNA transported from nucleus to cytoplasm.

1. RNA pol I--->rRNA
2. RNA pol II--->mRNA ("all" monocistronic)
3. RNA pol III--->tRNA, snRNA

Transcription similar to prokaryotes except different promoter sequences and more RNA pol subunits.

 
Eukaryotic RNA processing (Occurs in nucleus)
(Fig. 13-37)

1. 5’ cap added (7-methyl G).
2. Poly (A) tail added to 3’ end of transcript.
(Also occurs in prokaryotes but less frequently)
3. Splicing removes introns. (Fig. 13-39)

(exon-intron-exon--->exon-exon + intron)

SPLICING

1. Small nuclear ribonucleoprotein complexes (snRNPs) catalyze removal of introns via splicing. (See Fig. 13-40)

2. Some introns are self-splicing (Fig. 13-43)
RNA enzyme called a ribozyme.
(Tom Cech--->Nobel Prize-1989)
Also occurs in a few phage genes.

Introns may have evolved to allow rapid generation of new genes by exon shuffling, thus new proteins.

rRNA Processing (E. coli) (Not in book)
30S primary rRNA transcript--->16S, 23S, 5S rRNAs
 

5’           16S       tRNA       23S         5S    tRNA 3’

 

TRANSLATION--->Conversion of genetic information encoded in mRNA into proteins.

Genetic Code--->Nucleotides are "letters" in the code. (Fig. 13-18)

3 nts form the words (codons) representing different AAs. (i.e.) 3 nt triplet codon. (non-overlapping)

20 AAs and only 4 different nts.

4³=64 words (codons)

64 triplet codons but only 20 amino acids--->
genetic code is redundant (degeneracy).

(i.e.) most AAs represented by more than one codon.

Reading Frame--->The frame in which the triplet codons specify the AAs in a protein.

(e.g.)     5’ GCCAUACGCCUACUUGG 3’
Frame 1
Frame 2
Frame 3

Genetic code is universal (identical in all organisms) EXCEPT for a couple differences in mtDNA (mitochondrial DNA) and in some protozoans nuclear DNA.

Thus, the genetic code is NOT universal.
tRNA Recognition of Codons--->Anticodon in tRNA recognizes the codons in mRNA by base pairing. (Fig. 13-17)

tRNA and rRNA genes are genes that do not code for proteins.

Wobble--->Not a different tRNA for each codon.

In some cases the 3rd position in the anticodon (5’) can pair with more bases than its normal complementary base. (e.g.) G:U
(See Fig. 13-19; tables 13-5, 13-6)

Stop Codons--->UAA, UAG, UGA (Fig. 13-18)

No corresponding tRNA. Signals the termination of protein synthesis.

tRNA Charging
 
 

ATP ADP

AA1 + tRNA1 ------------> AA1-tRNA (charged tRNA)

tRNA                             synthetase

Ribosomes--->Huge complexes consisting of about 50 proteins and 3 rRNAs (23S, 16S, 5S) that are sites of protein synthesis (translation). (Fig. 13-22)

A. Prokaryotes

50S subunit + 30S subunit--->70S ribosome
50S subunit contains 23S and 5S rRNA
30S subunit contains 16S rRNA

B. Eukaryotes

60S subunit + 40S subunit--->80S ribosome

Translation--->initiation, elongation, termination

INITIATION--->Requires mRNA, ribosomes, charged tRNAs, initiation factors.

In prokaryotes the first AA is N-formylmethionine and is inserted by initiator tRNA (tRNA^fMet).
(Uses normal met codon and anticodon).

Ribosome binding site (rbs)--->Sequence on mRNA (Shine-Dalgarno sequence) is complementary to the 3’ end of 16S rRNA.

1. mRNA binds to 30S ribosomal subunit by base pairing with the 3’ end of the 16S rRNA.

2. fMet-tRNA binds to P (peptidyl) site on 30S.

3. 50S ribosomal subunit binds forming the 70S ribosome.

ELONGATION--->Requires elongation factors.

1. Aminoacyl-tRNA binds to the A site.
2. Peptide bond is formed by peptidyltransferase by transfer of growing peptide chain to AA at A site.
Peptidyltransferase enzyme is part of 23S rRNA.
3. Ribosome translocates by moving one codon along mRNA. tRNA at P site is removed and newly formed peptidyl-tRNA moves to P site.
4. Process is repeated..............

**GTP hydrolysis drives ribosome assembly,

AA-tRNA binding, and ribosome translocation.

TERMINATION (Fig. 13-28)

1. Release factors (proteins) recognize the stop codons. (Not tRNA)
2. Polypeptide is released from ribosome by cleavage from tRNA.
3. Ribosome dissociates into 2 subunits, mRNA is released.

In prokaryotes transcription and translation are coupled.

Not coupled in eukaryotes. Transcription in nucleus and translation in cytoplasm. (Fig. 13-48)

 
Chapter 14/15

Recombinant DNA technology allows us to isolate a specific gene from the genome so that we can study the function of that particular gene.

Recombinant DNA molecules can be made from any organism by inserting DNA fragments into a "small genome" or vector (gene cloning). (Fig. 14-1)

VECTOR--->The vehicle that DNA fragments are cloned into (plasmid or virus). Contain an origin of replication for gene amplification and usually an antibiotic resistance gene for selection.

RESTRICTION ENZYMES--->Make sequence specific cuts in DNA by cleaving each strand.

(e.g.) EcoRI--->sticky/cohesive ends
 
 

5’...GAATTC...3’ ...G        +      AATTC...

3’...CTTAAG...5’ ...CTTAA                    G...

(e.g.) HindII---> blunt ends
 

5’...GTPyPuAC...3’        ...GTPy     +     PuAC...

3’...CAPuPyTG...5’        ...CAPu     +     PyTG...

180° axis of symmetry, usually a 4 (1/256) or 6 nt sequence (1/4096). (See Table 4-1)

GENE CLONING

(Figs. 14-4, 14-6)

1. Digest chromosomal DNA and vector DNA with the same enzyme.

2. Mix together, sticky ends anneal due to base complementarity. Can also use blunt ends but less efficient.

3. Seal the nicks with DNA Ligase.

4. Transform host (usually E. coli).

5. Select for drug resistance or by complementation of a mutant defect.

6. Amplify and purify recombinant DNA.

Herb Boyer and Paul Berg--->Nobel prize
 

VECTORS

A. Plasmids--->small, circular, origin of replication, antibiotic resistance gene.
(Can clone several Kb) (Fig. 14-7)

B. Viruses or Phages--->remove unnecessary phage DNA, replace with DNA of interest.
(Can clone larger fragments than plasmids)
(Fig. 14-8)

C. EXPRESSION VECTORS--->contain transcription and translation signals to allow overexpression of the protein encoded by the gene. Express eukaryotic genes in bacteria.

D. SHUTTLE VECTORS--->contains origins for two organisms (e.g.) E. coli and SV40 virus.

Clone in E. coli, purify DNA, transform mammalian cell line.

E. Yeast Artificial Chromosomes (YACs)---> contains telomeres, a centromere and an ARS.
1000 kb can be cloned.

F. Bacterial Artificial Chromosomes (BACs)
Based on the F factor. 100-300 kb can be cloned.
( See Fig. 14-10)

DNA LIBRARY--->Random chromosomal or cDNA fragments cloned into one of the above vectors. A random population of clones should contain every gene. Used to isolate a specific gene.

DNA PROBES--->Radioactive DNA fragment complementary to the gene you want to clone. Often from a related organism. (Fig. 14-12)

1. Transfer bacterial colonies or phage plaques (independent clones) from a plate to nitrocellulose membrane.

2. Lyse cells and denature DNA. DNA sticks to nitrocellulose.

3. Add denatured probe to seal-a-meal bag containing membrane.

4. Visualize positive clones by autoradiography.

5. Pick corresponding colony from plate. Amplify/purify DNA containing cloned gene.

Cloning by complementation

1. Isolate a mutant giving the desired phenotype.

2. Transform mutant strain with DNA library and directly select for the positive clone by its ability to complement the mutant defect.

ELECTROPHORESIS--->Used to fractionate DNA based on its size. (RNA, Proteins)

Digest DNA, run on gel, isolate or probe for specific fragment if desired.

SOUTHERN BLOT--->Probing for DNA fragment using a DNA probe. (Fig. 14-20)

NORTHERN BLOT--->Probing for RNA fragment using a DNA or RNA probe.

WESTERN BLOT--->Probing for a protein using antibodies.

DNA SEQUENCING

Used to determine the specific nt sequence of any gene. Can resolve DNA fragments differing by 1 nt.

A. Maxam and Gilbert method (Gilbert Nobel Prize)

Base destruction method using chemicals specific for each base (Not used much anymore).

B. Dideoxy Sequencing (Sanger Nobel Prize)

Dideoxy nts lacking 3’ OH group can’t be extended by DNA pol once incorporated. Random incorporation.

(ddATP, ddCTP, ddGTP, ddTTP)

 
Polymerase Chain Reaction (PCR)

(Mullis Nobel Prize)

Used to amplify specific regions of DNA using two complementary primers. Uses a thermostable DNA pol. (Taq, Vent, Deep Vent) (Fig.14-29)

Can amplify DNA from a single cell.

RESTRICTION MAPPING---> Restriction sites in a DNA fragment can be used as markers or reference points to subclone fragments within the fragment.

1. Digest DNA with one of several enzymes.

2. Run digested DNA on an agarose or polyacrylamide gel to separate fragments.

3. Stain DNA with Ethidium bromide (EtBr) which intercalates between bases.

4. View under UV light (EtBr fluoresces).

5. Single, double or partial digests.

(figs. 14-30, 14-31)

Genetic Engineering

Site-Directed Mutagenesis--->Directing point mutations, insertions or deletions into previously cloned DNA fragments. (Fig. 15-3)

Used in structure-function studies of enzymes, regulatory proteins, etc.

Requires using a mutagenic DNA oligonucleotide of defined sequence (buy from a company).

1. Isolate ssDNA or use denatured dsDNA.

2. Hybridize DNA and mutagenic oligo.

3. Extend with DNA pol.

4. Ligate to seal nick.

5. Transform mutS E. coli (mismatch deficient).

6. Screen for change by DNA sequencing.

7. Analyze affect of the change.
 

Eukaryotic Gene Expression in Bacteria

Use specialized vectors to specifically overexpress biologically important human proteins in bacteria (usually E. coli).
(e.g.) Insulin, growth hormone, clotting factor VIII

Eukaryotic Transgenic Technology

E. coli--->4.2 million bp

Human--->3 billion bp

Plants--->Some even larger

Thus, specialized techniques developed to handle large genomes.

Transgenic Technology--->Methods used to transfect eukaryotic cells.

A. Plants & Fungi--->Remove cell wall and transfect resulting protoplasts.

B. Animal cells--->Inject DNA with microsyringe or use a defective virus.

C. Plants--->Shoot particles coated with DNA into plants w/ gene gun.

Transgenic Organism--->Organism that develops from transfected cell.

Transgenic Yeast

A. 2m (micron) plasmid

Naturally occurring yeast plasmid

B-E all contain a selectable marker & gene of interest. (Fig. 15-15)

B. Yeast Integrative Plasmid (YIp)

Must integrate in yeast genome with 1 or 2
X-overs (i.e.) suicide vector.

C. Yeast Episomal Plasmid (YEp)

Autonomously replicating because of 2m plasmid.

D. Yeast Centromere Plasmid (YCp)

Plasmid w/ yeast ARS & centromere

(ensures proper segregation during cell division)

E. Yeast Artificial Chromosome (YAC)

1. ARS
2. Centromere
3. Telomere ends

(Behaves like a linear chromosome.)

 
Gene Inactivation (Fig. 15-19)

1. Clone selectable marker in middle of gene.

2. Linearize with restriction enzyme.

3. Transform yeast.

4. Double X-over results in replacement of WT gene with disrupted gene.

5. Study affect of mutation.

Studying Gene Regulation

1. Clone regulatory (5') region adjacent to reporter gene (gene whose protein is easy to assay).

Expression of reporter gene depends on cloned regulatory elements.

2. Study regulation.

3. Repeat with deletions or point mutations in regulatory region.

Transgenic Plants

Ti Plasmid--->from Agrobacterium tumefaciens

Causes plant tumors. (Fig. 15-22)

Bacteria infects plant--->injects part of plasmid called T DNA (T=tumor)

Clone gene in middle of T DNA so that gene is inserted into plants with T DNA.

(e.g.) firefly luciferase gene--->resulted in plants that could glow in the dark

(e.g.) Flav-R-Savor tomato

TRANSGENIC ANIMALS

Applying similar techniques to study function of animal genes.

Can be used for gene therapy in humans.

Gene Therapy--->Try to correct genetic defect by transferring WT gene into germ line (egg cells) of animals or other actively dividing tissue.

Problem--->Have to target to actively dividing cells of a living person.

HUMAN GENETIC DISORDERS

Recessive disorders cause over 500 genetic diseases.

Would like to determine if individual carries mutant gene(s).

Restriction Fragment Length Polymorphism (RFLP) Mapping

(e.g.) sickle cell anemia

Affects 0.25% of U.S. African-Americans.
GAG--->GTG mutation eliminates MstII site.

Change detected by Southern blotting.

Change in banding pattern diagnostic for sickle allele.

REVERSE GENETICS

1. Purify a protein of known or unknown function.

2. Determine partial AA sequence.

3. Use genetic code to design a probe to clone gene encoding the protein by techniques already discussed.

4. Study gene.

 
Chapter 16

CHROMOSOME STRUCTURE

E. coli---> 4,200 kb--->4.2 million bp
Human cell--->6 million kb--->6 billion bp
2m DNA in 0.006mm=6um nucleus.

Each chromosme is one DNA duplex.

Chromatin--->Chromosomal DNA & protein.

Nucleosomes--->10nm fiber (diameter) (Fig. 16-5)

DNA and chromosomal proteins called histones.

First level of packaging (beads on a string).

Each nucleosome contains an octamer of 2 subunits each of the histones H2A, H2B, H3, and H4.

DNA wrapped » twice around histone octamer.

Solenoid--->(30 nm fiber) A coil of nucleosomes.

Stabilized by histone H1 that runs down center of the structure.

Supercoil--->(700 nm fiber) Seen during Mitosis and Meiosis (Fig. 16-8)

Held together by topoisomerase scaffold at scaffold attachment regions (SARs).

Euchromatin--->Most of the active genes reside in euchromatin.

Less tightly packaged--->compatible with transcription (gene activity).

Heterochromatin---> more compact, generally inactive DNA.

Position-Effect Variegation--->when a gene is placed next to an area of heterochromatin it can become inactivated in some cells.

Replication & transcription of Chromatin---> Nucleosomes do not dissociate. Probably loosen up to prevent steric hindrance of transcription and replication machinery.

CENTROMERES---> Attachment point for nuclear spindle fibers.

Constriction visible during Mitosis and Meiosis.

Replication is not completed through the centromere until Anaphase II of meiosis.

TELOMERES--->ends of chromosomes that consist of tandem arrays of simple DNA sequences that do not code for RNA or protein.

During replication of linear DNA the leading strand is completed but no priming can occur at the end of the lagging strand.

Telomerase adds the simple repeats to end of lagging strand to complete replication.

SEQUENCE ORGANIZATION

1. Single copy protein coding genes.

2. DNA present in more than 1 copy.

    A. Functional sequences

        1.gene families (e.g.) globin, actin, rRNA

        2.noncoding (e.g.) regulatory regions

    B. Sequences with no known function (20%)

3. Spacer DNA

Unique single copy genes are embedded in a diverse assembly of repetitive DNA.

VNTRs--->Variable Number Tandem Repeats

Humans--->1-5 kb sequences consisting of repeats 15-100 nt long.

DNA Fingerprinting--->Used in forensic medicine. (Fig. 16-29)

1. Digest DNA with restriction enzyme that does not cut within VNTRs.

2. Run DNA on gel.

3. Southern blot with VNTR probe.

4. Pattern on autoradiograph is highly individualistic.

DNA samples can be amplified by PCR using trace amounts of blood, semen, hair.

Chapter 18

Regulation of gene expression involves

protein-nucleic acid and protein-protein interactions.

Regulation occurs at several levels:

1. Initiation of transcription (Repression,Activation)
2. Transcript elongation (Attenuation)
3. mRNA stability (Steady-state level of mRNA)
4. Efficiency of translation (rbs, Codon usage)

OPERON--->Genetic unit of coordinate expression.
                    (e.g.) lacPOZYA

PROMOTER (P)--->Specific nucleotide sequence recognized by the sigma subunit (s) of RNA polymerase. (e.g.) -10 & -35 sequences

Different classes of promoters are recognized by alternative s factors.

OPERATOR (O)--->Specific nucleotide sequence recognized by the repressor. Often overlaps the promoter sequence.

STRUCTURAL GENES (ZYA)--->Encode mRNAs (proteins), rRNAs, tRNAs.

INDUCTION--->Relief of repression--->requires inducer to inactivate repressor.

Inducer binds to repressor--->repressor dissociates from DNA--->RNA pol binds to promoter---> transcription--->translation of mRNA (enzymes, structural and regulatory proteins)--> enzymes allow synthesis of compounds or utilization of compounds as carbon, nitrogen and/or energy source.

lac operon (lacPOZYA) (>1000 fold regulation) Utilization of lactose as carbon and energy source.

lac operon negatively regulated by LacI (repressor) because it blocks expression of lac genes in the absence of inducer. (Fig. 18-1)

Lactose(allolactose)-->inducer of lac operon.

IPTG--->gratuitous inducer--->can bind to LacI and inactivate repressor but is not a suitable substrate for b-galactosidase (lacZ).

lacI- --->mutations in gene encoding repressor results in constitutive expression (always on)--->caused by AA changes in DNA binding domain of protein.

lacI+is trans-dominant with respect to lacI-.

(i.e.) Wild type LacI can be supplied in trans (diffusible product) on a plasmid which complements the defect. (Fig. 18-8)

lacIs--->mutated inducer binding domain of repressor. (S=super repressor)

No induction by lactose or IPTG (always off).

lacI+ is trans-recessive with respect to lacIs .

(i.e.) Wild type LacI is not able to bind to the operator if LacIs is already bound to operator DNA.
(Fig. 18-9)

lacOc--->operator constitutive mutations---> mutation in operator DNA sequence prevents repressor binding (always on).

lacOc mutations are cis-dominant.

(i.e.) Wild type lacO supplied in trans has no effect on the operon containing the lacOc mutation.
(Fig. 18-10)

POSITIVE CONTROL

Positive control mechanisms require protein factor binding to allow expression of the operon in question.

(e.g.) cAMP-CAP control of lac operon.

Superimposed on the repression/induction system.

Catabolite Repression--->glucose is used for carbon and energy source before using other sugars.

Regulation is mediated by the catabolite activator protein (CAP) and cyclic adenosine monophosphate (cAMP).
(Fig. 18-17)

High [glucose]--->low [cAMP]
Low [glucose]--->high [cAMP]

cAMP binds to and activates CAP.

cAMP-CAP complex activates expression of lac operon by binding near the promoter which facilitates RNA pol binding by protein-protein interactions with the a subunits of RNA pol.

DUAL POSITIVE AND NEGATIVE CONTROL

(e.g.) arabinose operon (CaraBAD) (Fig. 18-20)

Positive control mechanism--->

1. Arabinose binds to araC protein (AraC).

2. Arabinose-AraC complex binds to initiator region near the promoter.

3. cAMP-CAP complex binding is also required.

4. Binding of both complexes facilitates RNA pol binding to the promoter.

5. Transcription of araBAD--->synthesis of
enzymes--->utilization of arabinose as carbon and energy source.

Negative control mechanism--->

In the absence of arabinose, AraC binds to initiator and to the operator, resulting in the formation of a DNA loop.

DNA loop prevents transcription by blocking

cAMP-CAP complex and RNA pol binding.

NEGATIVE CONTROL and ATTENUATION

tryptophan biosynthesis in E. coli. (trpEDCBA)

(Fig. 18-24)

FEEDBACK INHIBITION--->High [tryptophan]

Tryptophan binds to enzyme (TrpED complex) and prevents enzyme activity (first enzyme in biosynthetic pathway) which prevents tryptophan production.

NEGATIVE CONTROL (REPRESSION)
trp repressor (TrpR) encoded by trpR.

High [tryptophan]
1. TrpR activated by tryptophan binding.
2. Activated TrpR binds to operator.
3. Prevents RNA pol binding (100-fold regulation)

TRANSCRIPTION ATTENUATION--->any mechanism that results in the premature termination of transcription.

trpR- --->mRNA synthesized in the presence of tryptophan.

But: removal of tryptophan from medium resulted in an additional 10-fold increase in trp mRNA.

A deletion identified the attenuator located in the untranslated leader segment of the trp mRNA.
(Fig. 18-25)

Requires coupled transcription and translation.
Leader peptide--->encoded by a "mini" gene
(14 AA).

Two tandem trp codons are present in the middle of the mini gene. (Fig. 18-26)

If high [tryptophan], high [charged tRNAtrp].

Thus, tandem trp codons are efficiently translated.

If low [tryptophan], low [charged tRNAtrp].

Thus, ribosome stalls at tandem trp codons.

1. Initiation of transcription.

2. RNA pol transiently pauses at stem-loop 1:2 (pause structure).

3. Ribosome initiates translation of leader peptide, thereby releasing the paused RNA pol complex.

Results in coupled transcription-translation.

If tryptophan excess:

1. Ribosome translates tandem trp codons and continues to stop codon (in stem part 1).

Thus, ribosome interacts with stem part 2.

2. Transcription continues.

Stem-loop 2:3 (antiterminator structure) does not form because the ribosome blocks its formation.

3. Stem-loop 3:4 forms (rho-independent terminator) promoting transcription termination.

If limiting tryptophan:

1. Ribosome stalls at tandem trp codons.

No interaction with stem part 2.

2. As transcription continues stem-loop 2:3 (antiterminator) forms, thereby preventing terminator (3:4) formation.

3. Transcription continues into the trp structural genes (transcriptional read-through).

***The decision to terminate transcription or to allow transcription to proceed into the structural genes depends on whether the antiterminator or terminator forms which, in turn, depends on whether the ribosome can translate the tandem trp codons which, in turn, depends on the availability of charged-tRNAtrp which, in turn, depends on the [tryptophan] in the cell.

(i.e.) everything depends on which of the mutually exclusive RNA II° structures form.

TRANSLATIONAL REGULATION

A. Codon usage--->rare codons slow up translation resulting in lower expression levels.

B. Protein binding to the ribosome binding site (rbs) blocks initiation of translation by blocking ribosome access to the rbs.

C. If the rbs is sequestered in an RNA II° structure the mRNA is unable to base pair with the 3’ end of 16S rRNA which blocks translation initiation.

Often controlled by alternative II° structures and a specific RNA-binding protein
 

GENE REGULATION IN EUKARYOTES

All mRNAs transcribed by RNA pol II.

Requires promoters and enhancers. (Fig. 18-38)

trans-acting factors (regulatory proteins) interact with promoters and enhancers.

Eukaryotic Promoters--->Required for initiation.
 
 

GGGCGG           CCATT           TATA           +1

-110                    -40                 -10

TATA box->TATA binding protein (TBP). Directs RNA pol to initiate transcription downstream.
CAT box--->(Not present in all cases)
G-C box--->(Not present in all cases)

All 3 elements are recognized by protein factors to enable RNA pol to bind and initiate transcription.

Enhancers--->Greatly increase transcription initiation at promoters.

1. Can be several kb away from promoter.
2. Can be in either orientation.
3. Can be 5' or 3' of promoter.
4. Can be tissue specific. (e.g.) DNA-binding protein only present in some cell types.

(i.e.) Differential gene expression

Steroid hormones---> somewhat analogous to bacterial inducers.
(i.e.) Effect gene expression of a subset of genes.
(e.g.) Estrogen--->activates transcription by binding to an activator protein that binds to a tissue specific enhancer.

TBP associated factors (TAFs)

TBP, TAFs, trans-acting factors, RNA pol form a preinitiation complex via protein-nucleic acid and protein-protein interactions.

Many TAFs and trans-acting factors have 2 domains:

1. DNA-binding domain
2. Transcription activation domain.

(i.e.) interact with other protein factors.

The two domains can be swapped with domains from other factors and change the specificity of gene activation.

INITIATION COMPLEX--->(Fig. 18-47)

Ordered assembly of an enormous complex consisting of several protein factors and RNA pol.

Forms a DNA loop via protein-protein interactions bringing together the active complex.

DNA-BINDING MOTIFS

1. Helix-Turn-Helix---> The two a-helices interact with two successive major grooves on the same face of the double helix. (i.e.) 10 bp apart
(Fig. 18-40)

2. Zinc Finger--->Zinc atom complexed by histidine and cysteine residues. (Fig. 18-41)

3. Leucine Zipper--->Protein dimers form by hydrophobic interactions between leucine residues spaced 7 AAs apart present in each subunit. (Fig. 18-42)

4. Helix-Loop-Helix--->Form dimers without a zipper. Interface between dimers formed by interactions between the 2 helices and the loop.
(Fig. 18-45)

3 and 4 can be homodimers or heterodimers

Heterodimers increases the number of usable DNA sequences and the variety of protein combinations that can be used to turn genes on and off.
(Fig. 18-44) (See Fig. 18-43)

1.          Fill in the blanks of the following statement.

__________is responsible for processing the 23S rRNA from the E. coli 30S primary rRNA transcript, whereas__________ is responsible for processing the 5S rRNA from the E. coli 30S primary rRNA transcript.

A. RNase E; RNase P
B. RNase III; RNase E
C. RNase E; RNase III
D. RNase E; RNase E
E. RNase III; RNase III

            The E. coli RNA polymerase__________is involved in transcription initiation; whereas the __________ of RNA polymerase is involved in elongation.

A. holoenzyme; core enzyme
B. core enzyme; holoenzyme
C. holoenzyme; holoenzyme
D. core enzyme; core enzyme
E. none of the above

A. 3’ poly(A) tail addition
B. snRNP intron splicing
C. 5’ CAP addition
D. B and C
E. A, B, and C

4.     rut sequence            A. Rho-independent terminator
                                            B. Rho binding site
                                            C. Release factor binding site
5.     stop codon               D. Terminator tRNA recognition site
                                            E. AUG

6.     Fill in the blanks of the following statement.

        __________ adds a tandem array of simple DNA repeats to the end of the __________ strand to complete DNA replication.

 A. Centromerase; lagging
 B. Centromerase; leading
 C. DNA polymerase; lagging
 D. Telomerase; lagging
 E. Telomerase, leading

A.The peptidyltransferase enzyme is part of 23S rRNA.
B. GTP hydrolysis is involved in ribosome assembly.
C. GTP hydrolysis is involved in binding of amino acyl-tRNA (AA-tRNA).
D. GTP hydrolysis is involved in ribosome translocation.
E. All of the statments are true.

A. No protein at all.
B. A wild type protein.
C. A protein that terminates at or near the point of the insertion.
D. A protein in which several amino acids downstream from the insertion are changed.
E. Both C and D are possible.

A. 5’ AGCCUGUACGAUC 3’
B. 5’ GAUCGUACAGGCU 3’
C. 5’ CUAGCAUGUCCGA 3’
D. 5’ UCGGACAUGCUAG 3’
E. 5’ CTAGCATGTCCGA 3’

10.     Which of the following statements is false?

A. Topoisomerase is invovled in DNA replication.
B. Topoisomerase is involved in eukaryotic chromosome packaging.
C. Nucleosomes consist of DNA and the histones H1, H2A, H2B, H3 and H4.
D. Nucleosomes probably loosen up to allow transcription and replication.
E. All of the above are true.

A. 5’ GAATTC 3’             B. 5’ CGCG 3’
     3’ CTTAAG 5’                  3’ GCGC 5’

C. 5’ GCAACG 3’              D. 3’ AAATTT 5’
     3’ CGTTGC 5’                   5’ TTTAAA 3’

E. 3’ T(Py)CG(Pu)A 5’
    5’ A(Pu)GC(Py)T 3’

12.     Which of the following statements is false?

A. Suicide vectors cause cell death.
B. Ti plasmids are used to genetically engineer animals.
C. Yeast artificial chromosomes contain an origin of replication, a centromere and two telomeres.
D. A and B are false.
E. A, B and C are false.

A. Reverse genetics utilizes a partial amino acid sequence of a protein to design a probe in order to clone the gene encoding the protein.
B. RFLP mapping can be used to determine if an individual is likely to be a carrier of certain genetic disorders.
C. The polymerase chain reaction (PCR) is used in DNA fingerprinting.
D. Restriction enzymes are used in DNA fingerprinting.
E. Northern blots are used in DNA fingerprinting.

enzyme(s)                 resulting fragments (kb)
BamHI                                  4 and 7
EcoRI                                   3 and 8
HindIII                                 1 and 10
EcoRI and HindIII             1, 3 and 7
BamHI and EcoRI             1, 3 and 7

    What fragment sizes are expected if the 11 kb fragment is cut with BamHI and HindIII?

A. 1 kb, 3 kb and 7 kb fragments
B. 1 kb, 4 kb and 6 kb fragments
C. 1 kb, 1 kb and 9 kb fragments
D. 2 kb, 6 kb and 3 kb fragments
E. 1 kb, 1 kb, 3 kb and 6 kb fragments

15.     Produce a human protein in E. coli.         A. integrative plasmid
                                                                                  B. YAC
                                                                                  C. shuttle plasmid
16.     Inactivate a yeast gene.                              D. expression plasmid
                                                                                  E. episomal plasmid
 

17.     The following question involves regulation of the E. coli lac operon in a strain that is unable to synthesize cAMP. A partial diploid (merodiploid) of geneotype I- P- O- Z+ Y- A+/I- P+ O+ Z+ Y- A- will show...

A. high level inducible production of b-galactosidase.
B. low level inducible production of b-galactosidase.
C. no expression of b-galactosidase.
D. high level constitutive production of b-galactosidase.
E. low level constitutive production of b-galactosidase.

A. The level of histidine in the growth medium.
B. The level of charged tRNA^his in the cell.
C. The level of glucose in the growth medium.
D. RNA polymerase pausing.
E. Formation of alternative RNA secondary structures.

A. Mutations in the DNA sequence of the lac promoter.
B. Mutations in the DNA sequence of the lac operator.
C. This class of lacI mutation results in repressor proteins with a higher affinity for lactose such that lactose binding to the wild type repressor is diminished.
D. Some of the tetrameric repressor proteins would contain mutant as well as wild type subunits resulting in the inability of these repressor proteins to bind to the operator.
E. The mutant repressor proteins help to facilitate RNA polymerase binding to the promoter, thereby partially overcoming the ability of the wild type repressor to prevent RNA polymerase binding.

20.     Which of the following is not involved in regulating tryptophan biosynthesis in E. coli?

A. attenuation
B. activation
C. repression
D. feedback inhibition
E. Rho-independent termination

A. Binding of the cAMP-CAP complex to a region of the DNA upstream (5’) from the start of transcription.
B. Binding of the arabinose-AraC complex to the initiator sequence.
C. Binding of the arabinose-AraC complex to the operator sequence.
D. DNA loop formation.
E. All of the above influence expression of the araBAD operon.

22.     Which of the following statements is false?

A. Regulation of gene expression is involved in maintaining homeostasis.
B. An operon is defined as a genetic unit of coordinate expression.
C. Gene expression is not regulated at the level of translation.
D. Repressor binding interferes with binding of RNA polymerase.
E. Binding of an activator protein aids RNA polymerase binding to the promoter.

          __________ binds to the __________ in eukaryotic promoters.

A. TBP; CAT box
B. TBP; TATA box
C. RNA polymerase; TATA box
D. TAFs; CAT box
E. RNA polymease; G-C box

A. Dimerization occurs from interactions between leucine residues present in each subunit.
B. They can be homodimers or heterodimers.
C. The transcription activation domain consists of the leucine zipper itself.
D. Heterodimers increase the number of usable DNA sequences that can be used to turn genes on and off.
E. They bind to specific DNA sequences.

A. They can exert their effect on transcription when located several kb from promoters.
B. They can exert their effect on transcription in either orientation.
C. They can be tissue specific.
D. They can be upstream (5’) or downstream (3’) from promoters.
E. All of the above are characteristics of enhancers.