Test 1 Review 1 Applied molecular Biology PDF

Preview

Summary

Explain genetic recombination and genetic mapping. Review the steps involved in transcription. Review the process of translation and protein production....

Description

Unit 1 Outcomes

APPLIED MOLECULAR BIOLOGY BIOT2027 Test 1 Review

Genetic Recombination • Recombination: crossing over between two homologous chromosomes. • Occurs in meiosis I • All DNA is recombinant DNA Recombination provides: • Genetic variation • Replacing or repairing damaged DNA • Restarting stalled or damaged replication forks • Regulation of expression of some genes

• Explain genetic recombination and genetic mapping. • Review the steps involved in transcription. • Review the process of translation and protein production.

Genetic Recombination • The frequency of crossing over between two genes on the same chromosome depends on the physical distance between these genes. Long distances give the highest frequencies of exchange. Therefore, the farther apart two genes are on a chromosome, the more likely they are to recombine © 2014 Oxford University Press

Create a Genetic Map Problem If five genes (A-E) gave the following frequencies of recombination, map the arrangement of the genes on a chromosome. Genes

Frequency of recombination

AB

2.5

AC

3.5

AD

0.5

AE

7.0

BE

9.5

CD

3.0

DE

6.5

Gene Expression 2 major steps involved: • Transcription = Synthesis of messenger RNA (mRNA) that is complementary to one of the strands of DNA (coding strand) representing the gene • Translation = Assembly of a protein (polypeptide) based on information encoded in the mRNA

B

Transcription

Step 1: Initiation of Transcription • RNA polymerase recognizes a region upstream of the gene called the promoter • RNA polymerase binds the promoter tightly, causing localized separation (~12 bp) of the DNA strand à transcription bubble

Three key stages: 1. Initiation 2. Elongation 3. Termination

RNA polymerase

Initiation in Prokaryotes • Single RNA polymerase • Polymerase requires an initiation factor at the promoter site: sigma (s) • Two conserved sequences of 6-7 bp in length upstream of transcription start site • –35 box • –10 box

Step 2: Elongation • RNA polymerase moves along the DNA template locally separating the DNA template, adding to the growing RNA strand in the 5’ à 3’ direction. • As RNA polymerase passes, the two DNA strands anneal. • RNA polymerase has proof reading functions. • Elongation occurs with very similar mechanisms in prokaryotes and eukaryotes.

Initiation in Eukaryotes • RNA Pol II à transcribes all protein-encoding genes into mRNA • RNA Pol requires several general transcription factors (GTFs) to initiate transcription Sigma

Promoter Structure Two parts: • Core promoter à TATA box (–25) • Upstream promoter elements

Step 3: Termination • The region at the end of the gene (downstream of the coding region) is called a terminator. • Essentially, opposite of promoter • RNA polymerase, RNA, and DNA dissociate

Termination in Eukaryotes

Termination in Prokaryotes Two Types: 1. (r) Rho-Independent (intrinsic) – A short inverted repeat (20 nucleotides) results in a stem-loop structure – Stem-loop interacts with RNAP causing it to pause – Followed by a stretch of adenine nucleotides, which are weakly bound and cause the RNA Polymerase to dissociate from RNA and DNA

•

Termination protein – RNAP I (rRNA) – A termination protein binds to the DNA at a specific site and physically blocks RNAP I from further transcription – RNAP I, rRNA, and DNA dissociate from each other and transcription stops

•

“Torpedo model” – RNAP II (mRNA) – RNAP continues past termination site, but endonucleolytic cleavage occurs in poly(A) site – Second (downstream) RNA strand is uncapped – Recognized by an RNase which quickly degrades remaining RNA

•

Polyadenylation of 3’ end – RNAP III (tRNA) – Lesser known model, linked to termination of transcription – Series of weak A=U binding causes dissociation (no stem-loop/pausing required)

2. (r) Rho-dependent – Ring shaped protein – Binds to mRNA and follows along behind RNA Pol – When hairpin forms and DNA pauses, rho acts as a helicase and causes the DNA and RNA to unwind and dissociate, releasing the mRNA – When RNAP pauses because of the hairpin, Rho is able to ‘catch up’ with the RNAP and dissociate the RNA from the DNA

RNA Processing in Eukaryotes 1. Capping of the 5’ end with a methylated guanine 2. Splicing – introns removed, exons (coding sequences) ligated together. Alternative splicing may occur. 3. Polyadenylation of 3’ end

Translation – Overview mRNA translated to protein (nucleic acids to amino acids) • Involves: 1. Ribosomes 2. Transfer RNA (tRNA) 3. mRNA

Ribosomes Large subunit • Peptidyl transferase center – formation of peptide bonds

Small subunit • Decoding center

Transfer RNA (tRNA) • Translates the nucleic acid language (mRNA) into protein language • tRNA is the “bifunctional molecule” • tRNA recognizes both mRNA and amino acids

Three binding sites for tRNA 1. Amino-acyl (A) site 2. Peptidyl (P) site 3. Exit (E) site

Transfer RNA (tRNA) • Anticodon found at the “bottom” of the tRNA • The 3 mRNA nucleotides are “codons” and the complement tRNA nucleotides are “anticodons” • For example, mRNA codon for amino acid alanine (Ala) attaches to anticodon on tRNA. This tells the ribosome to add alanine.

Amino acid attachment at 3’ OH group Acceptor stem

Codons • Codons are tri-nucleotide sequences of mRNA • The “reading frame” is the grouping, or order, in which the three-nucleotides are read and translated into amino acids • Every triplet of nucleotides specifies an amino acid • “Start” and “Stop” codons signify the beginning and end of amino acid (polypeptide) chain

Codons

Codon Chart

For example, the following mRNA sequence illustrates a start and stop codon, reading frame (read 5’ à 3’) and translated protein of 4 amino acids in length Start codon

Stop codon

5’ – AAUGAAAUUGAAUUAAAUU – 3’ Met - Lys - Leu - Asn

mRNA translation

Note: Stop codon does not code for an amino acid like the start codon does Each sequence could potentially be read in 3 reading frames

Initiating Protein Synthesis Three events must occur: 1. Ribosome recruited to mRNA 2. Charged tRNA must be placed in P site 3. Ribosome must be precisely positioned over start codon (sets reading frame)

Initiating Protein Synthesis – Prokaryotes 5’ of start codon à ribosome binding site (RBS) • Also called the “Shine-Delgarno sequence”

Initiator tRNA (fMet-tRNA) – located at P site base pairs with AUG start codon (AUG = Methionine) • fMet-tRNA is an N-formylated methionine Prokaryotic

rRNA from the small subunit of the ribosome recognizes RBS sequence on the mRNA, resulting in mRNA being recruited to the ribosome to begin translation.

Initiating Protein Synthesis – Eukaryotes • 5’ methylated guanine cap recruits small ribosome subunit – tRNA is already bound • Bound ribosome “scans” for AUG start codon • Once bound to start codon, the large subunit binds • Poly-A tail enhances the level of translation – promotes efficient recycling of ribosomes Eukaryotic

5’ methylated cap recruits the mRNA to the ribosome. The tRNA with the UAC anticodon for MET is already bound to the ribosome, recognizes the mRNA start codon, and begins translation.

Translation Elongation 3 Steps Per Cycle: 1. Aminoacyl-tRNA binding: initially fMet/Met-tRNA bound to the P site, then second AA enters the A site. 2. Peptide bond formation between two AA: formed in the A site via peptidyl transferase. 3. Translocation as the empty tRNA exits and the tRNA carrying the polypeptide chain enters the P site. This allows for a new tRNA to enter the newly emptied A site.

Termination of Translation 3 stop codons (UAG, UAA, and UGA) • Recognized by release factors (RFs) à activate hydrolysis of polypeptide from pepidyl-tRNA

UNIT 2 – PART 1 GENE CLONING AND EDITING TECHNIQUES

Molecular Cloning 1. A fragment of DNA that contains the gene to be cloned, is inserted into a vector. This produces a recombinant DNA molecule. 2. The vector transports the gene into a host cell. 3. Within the host cell the vector multiplies, producing numerous identical copies, not only of the vector, but also of the gene that it carries.

Why is Molecular Cloning Important? To obtain a pure sample of an individual gene, separated from all the other genes in a cell •

Recombinant protein expression Therapeutic drug research and development Food and nutritional enhancement Vaccine research and development Protein structure/function studies

•

Genome sequencing

•

Metabolic studies, gene knock out, expression patterns

•

Other: Gene therapy, tissue/organ growth

Molecular Cloning 4. When the host cell divides, copies of the recombinant DNA molecule are passed to the progeny and further vector replication takes place. 5. After a few cell divisions, a colony, or clone, of identical host cells is produced. Each cell in the clone contains one or more copies of the recombinant DNA molecule; the gene carried by the recombinant molecule is now said to be cloned.

Vectors for Gene Cloning Features: • The vector must be able to replicate within the host cell • Cloning vectors need to be relatively small. • Possess at least one origin of replication.

Two kinds of bacterial DNA molecules satisfy the above criteria: plasmids and bacteriophage chromosomes.

Plasmids

Plasmid Vectors

• Self-replicating, double stranded, circular DNA found in bacteria. • Plasmid DNA is separate from chromosomal DNA. • Virtually all bacteria contain plasmids and carry one or more genes responsible for useful characteristics displayed by the host bacterium, such as antibiotic resistance. • Antibiotics resistance is used as a selectable marker

Typically have the following characteristics: 1. Small size 2. Origin of replication (ori site) 3. One or more selectable marker (i.e., an antibiotic resistance gene for selection) 4. A Multiple Cloning Site (MCS) with one recognition site for each of the most common restriction enzyme (RE) sites Although plasmid vectors work well with bacteria and yeast, they may not work well for higher organisms.

Bacteriophage λ Phage Vectors • The Phage DNA molecule is 49 kb in size MCS – Multiple cloning site

• To use phage as a vector, some DNA sequences are removed from the phage DNA, but the genes needed for phage replication are retained. • The missing phage genes are replaced with foreign DNA, thus when a phage infects a cell it carries the cloned gene as well.

Bacteriophage M13 Phage Vector • Genome is less than 10 kb in size. • Behaves like a plasmid due to the double-stranded replicative form. • Used in phage display – a technique for identifying pairs of genes whose protein products interact with one another.

Total Cell DNA Total cell DNA consists of the genomic DNA along with any additional DNA molecules present in the cell. The procedure for total DNA preparation can be divided into four stages: 1. A culture of bacteria is grown and then harvested. 2. The cells are broken open to release their contents. 3. The cell extract is treated to remove all components except DNA. •

Ex. Organic extraction and RNase treatment, or ion-exchange chromatography and RNase treatment.

4. The resulting DNA solution is concentrated. •

Ex. Precipitation with ethanol and resuspension of dried pellet in a smaller volume of liquid.

Isolation and Purification of Nucleic Acids There are at least three distinct kinds of DNA which a genetic engineer will need to prepare: 1. Total cell DNA 2. Plasmid DNA 3. Phage DNA

Measurement of DNA Samples • DNA concentrations can be accurately measured by UV absorbance spectrophotometry. – The amount of UV radiation absorbed by a solution of DNA is directly proportional to the amount of DNA in the sample – A260 of 1.0 corresponds to 50 mg/ml – A260/A280 checks purity of DNA. Pure samples give a ratio of 1.8.

Preparation of Plasmid DNA

Plasmid DNA

Involves the same general strategy as preparation of total cell DNA, however the plasmid DNA must be separated from the bacterial chromosomal DNA. Methods used are based on physical differences between plasmids and bacterial DNA: • Size • Conformation

Supercoiled

linear

double-digested

Next step - excise plasmid from gel

Preparation of bacteriophage DNA • To prepare significant quantities of phage DNA, the infected culture must have a sufficiently high extracellular phage titer. • Titer – the number of phage particles per ml of culture • Practically, the maximum titer that can be expected for l is 1010, which yields ~500 ng of DNA; therefore culture volumes must be large.

RNA Isolation • Precautions must be taken to ensure no Rnase is present in any solution or on any tool/tube/material that may come in contact with the sample • Precautions must be taken when preparing solutions • Treat most solutions with diethyl pyrocarbonate (DEPC) à inactivates RNases, and then autoclaved to eliminate the DEPC • DEPC reacts with amino groups (example: cannot be used on Tris buffered solutions)

RNA Isolation • Isolate total RNA can be isolated using GIT or CsCl (guanidinium isothiocyanate or cesium chloride) along with the phenol:chloroform extraction procedure • Recover RNA by ultracentrifugation • RNA is purified by two ethanol precipitation steps • RNA is quantified using UV spectrophotometry

Back to Making a Recombinant Vector: Nucleases Nucleases degrade DNA molecules by breaking the phosphodiester bonds that link one nucleotide to the next in a DNA strand. Two different kinds of nucleases • Exonucleases: Remove nucleotides one at a time from the end of a DNA molecule. • Endonucleases: Break internal phosphodiester bonds within a DNA molecule. • Restriction endonucleases (REs) are used in cloning, as they cut at very precise locations – recognize a specific sequence

What sequences do REs cut? • The REs that recognize 4 base pair sequences will cut much more frequently than REs that recognize 6 bp sequences • This is because a 4 bp sequence will occur about every 44 = 256 bp whereas a 6 bp will occur about every 46 = 4096 bp • Restriction enzymes always cut DNA at the same site. This is a fundamentally important characteristic to carry out routine DNA digestion (cutting) for cloning, sequencing, and restriction mapping purposes

Nomenclature Example: EcoRI Derivation of the EcoRI name Abbreviation

Meaning

Description

E

Escherichia

Genus

co

coli

Species

R

RY13

Strain

I

First identified

Order of identification in the bacterium

How does a host cell’s DNA not get digested by REs? • All restriction enzymes are paired with enzymes called methylases that recognize and add –CH3 (methyl) groups to the restriction sites • This prevents the host cell from digesting its own DNA • Collectively, this is called the “restriction-modification system” (modification enzyme) . After methylation of DNA occurs, DNA sites are protected against digestion. • Foreign DNA does not become methylated and is therefore exposed to digestion by RE’s.

1. Identify restriction sites on either side of the gene of interest in the host cell DNA. 2. Identify a plasmid vector suitable to receive the cleaved DNA insert by choosing a plasmid with the same restriction sites as the host. 3. Expose both the plasmid and the host DNA to the necessary restriction enzymes. 4. Use the enzyme ligase to link the plasmid vector DNA and the DNA insert containing the gene of interest.

Restriction Mapping Example Single and double digestions performed on a linear λ DNA fragment Enzyme

REs and Cloning

Number of fragments

Sizes (kb)

Xba I

2

24.0, 24.5

Xho I

2

15.0, 33.5

Kpn I

3

1.5, 17.0, 30.0

Xba I + XhoI

3

9.0, 15.0, 24.5

Xba I + KpnI

4

1.5, 6.0, 17.0, 24.0

References Image References:

Brown, T.A. (2010). Gene Cloning and DNA Analysis (6th ed.). West Sussex, UK: Wiley-Blackwell.

• • • • •

https://www.mun.ca/biology/desmid/brian/BIOL2060/BIOL2060-20/CB20.html http://www.nature.com/scitable/topicpage/each-organism-s-traits-are-inherited-from-6524917 http://bioweb.wku.edu/courses/biol114/vfly1.asp https://www.biologycorner.com/worksheets/genetic_maps.html http://karimedalla.wordpress.com/2013/02/21/10-3-gene-linkage-polygenic-inheritance/

All Oxford University Press images are from: Craig, N.L., Cohen-Fix, O., Green, R., Greider, C., Storz, G., & Wolberger, C. (2014). Molecular Biology: Principle of Genome Function. Oxford University Press.

Unit Outcomes UNIT 2 – PART 2 Gene Cloning and Editing Techniques

• Describe and demonstrate the process of transformation. • Explain the use of inducible expression vectors. • Describe and discuss gene editing methodologies. • Discuss the innovations and applications of molecular cloning in biotechnology.

Ligation The vector molecule and the gene of interest are joined together via ligation, catalyzed by the enzyme DNA ligase. The function of DNA ligase in the cell is to: 1. Repair single-stranded breaks (discontinuities) that can occur from DNA replication. 2. Join together two individual fragments of double stranded DNA.

3 Possible Products of Ligation Blunt end ligation – less efficient Sticky end ligation – more efficient Double sticky end ligation with different REs – the most efficient.

Types of Transformation

Transformation Transformation – process used to introduce recombinant DNA molecules into a host cell Goal à adaptation: introduce specific genes of interest into a host cell Two elements required: – Culture of suitable host organism – Recombinant DNA molecule

Types: 1. Natural competence (bacteria) 2. Artificial competence and heat shock (CaCl2 exposure) (bacteria) 3. Transduction/transfection (DNA transfer by a viral vector) 4. Electroporation (pore formation using a high voltage charge) 5. Protoplast formation (cell wall removal) 6. Microprojectiles (DNA coated pellets that are shot into the cell using a particle gun)

Cloning Vectors of E.coli pGEM-3Z – Carries the ampR and lacZ’ genes – lacZ’ contains a cluster of restriction sites – Two phage promoter sequences that will synthesize 1–2 mg of RNA per minute

Cloning Vectors of E.coli

Cloning in Mammals

Screening tool: β– galactosidase gene expression β-galactosidase expression (no insert)

No β-galactosidase expression (disrupted by insert)

Grow on media with the indicator X-gal

Grow on media with the indicator X-gal

X-gal is cleaved, a blue dye released

X-gal is not cleaved, no color is released

Blue colonies (negative)

White colonies (positive for insert)

Three main reasons for gene cloning in mammals: 1. Gene knockout – used to determine the function of an unidentified gene. 2. Production of recombinant protein in a mammalian ce...