Molecular+Genetics


 * Lesson Date: Thursday, November 14th, 2013**
 * By: Lahina Sivanantharajah **


 * Structure of DNA**


 * DNA structure- the monomer of DNA is a nucleotide
 * The phosphate group and deoxyribose from the rails of the molecule, while the base can be either a purine or a pyrimidine nitrogenous base


 * Edwin Chargaff's Data (Early 1950's)**
 * Did not believe there were equal amounts of the 4 nitrogenous bases like everyone else
 * He looked at purine vs. pyrimidine (nitrogen base) composition in different samples of DNA
 * He measured molar ratios of the various nucleotide
 * Ratios of A:T and G:C were very similar in many species investigated
 * Thus suggesting the paring of these bases with each other for some reason
 * This lead to the notion of complementarity
 * Wilkins and Franklin** ( take notes on page 266-267)


 * History- Structure of DNA**
 * Structure of DNA- Watson and Crick**


 * Drawing on their own research, and that of others they sugg ested a structure for the DNA molecule
 * Consisting of a double helix
 * Purine, pyrimidine paring (A=T, CG) from one helix to another
 * Pairing was held together by two hydrogen bonds between A and T and three bonds between G and C
 * Helices were arranged anti-parallel to one another (one spiraled up to the right, the other down to the right)


 * Homework:**
 * Page 279: # 3,5,6,9
 * Read section 6.4

Helpful resources:
 * 1) [|DNA Animation]

=Monday, November 18th, 2013=

Sharyse Bell
**Meselson and Stahl (p.282-283) - DNA Replication**


 * Investigated whether DNA replications conservative or semi-conservative
 * Used e-coli
 * Grew it in heavy Nitrogen, 15N medium, so all nitrogenous bases of the DNA would be labeled
 * Transferred there bacteria to a 14N medium
 * Compared density of ones only in 15N meduim and those only in 14N medium
 * Density of nitrogen present reflected semi-conservative process

Current Model of DNA

 * DNA polymers consist of nucleotides linked by 3'-5' phosphodiester bonds
 * Chemically, the concentrations or purines equals that of pyrimidines in any sample of DNA
 * A=T and G=C
 * Heating DNA slightly changed its physical properties (without breaking the 3'-5' phosphodiester bonds)
 * DNA molecules have a doubled helical structure

(occurs during the interphase (section 5)

 * As shown by Meselson and Stahl, DNA replicates in a semi-conservative manner
 * This involves the unzipping off the DNA double helix by enzymes
 * DNA polymeres (enzymes) add new nucleotides to the now unzipped portion
 * New nucleotides are attached to the 3' end of the newly synthesized strand by a DNA polymerase
 * Bacteria has 3 types of DNA polymerase and eukaryotes have 5 types

Process of DNA Replication
Recall: DNA in prokaryotes = circular, DNA in eukaryotes = linear
 * 1) Hydrogen bonds hold the two strands of DNA together and it is twisted in the helical shape
 * 2) **DNA gyrase** unwinds DNA
 * 3) **DNA helicase** unzips DNa by breaking those hydrogen bonds
 * 4) Nowyou have two single stranded sections BUT they have a natural tendency to **re-anneal** (pair up again)
 * 5) To keep them separate, **single standed binding proteins (SSB's)** are bound *DNA cannot be fully unwound due to its large size, so replication occurs immediately when areas of the DNA are exposed as single strands* **Replication forks** and **replication bubbles** form
 * 6) Enzyme **primase** puts down RNA primers that will indicate tp DNA polymease 3 as starting point of replication
 * 7) **DNA polumerase 3 (5' to 3')** adds the complimentary nucleotides to the 3' end of the new strand - this is called the leading strand VS. the **lagging strand (Okazaki fragments),** that are built discontinuously from the replication fork.
 * 8) **DNA polymase 1** excises (cuts out) the RNA primers and replaces then with appropriate deoxyribonucleotides (DNA nucleotides)
 * 9) **DNA** **ligase** joins the gaps between the Okazaki fragments by creating phospodiester bonds
 * 10) **Finally** DNA polymerase 1 and 3 proofread the complimentary strands.

Homework:

 * Test on Photsynthesis Wednesday
 * Read pg 282-289


 * Lesson Date: November 19, 2013**
 * By: Randy Ketwaroo**

We reviewed for the test on Wednesday


 * Homework:**


 * Make sure to study for the test tomorrow
 * Bring colours for Thursday

November 21st, 2013 Viththakan Arunthavanathan

=DNA REPLICATION=

Overview:
In today’s class, the students were to bring colour pencils to outline the steps that are involved in DNA replication. DNA Replication:
 * Occurs during interphase
 * Replicates by using semi-conservative method (proven by Meselson and Stahl)
 * Involves unwinding and unzipping of DNA double helix (by enzymes)
 * DNA polymerases (enzymes) add new nucleotides to the now unzipped portion
 * New nucleotides are attached to the 3’ end of the newly synthesized strand by DNA polymerase (3 types in bacteria, 5 types in eukaryotes)

The 10 steps briefly...
// Remember that DNA in prokaryotes are circular and in eukaryotes they are linear //


 * 1) Hydrogen bonds hold the 2 strands of DNA together and it is twisted in the helical shape.
 * 2) DNA gyrase unwinds the DNA
 * 3) DNA helicase unzips the DNA by breaking those hydrogen bonds
 * 4) This results in 2 single stranded sections, however, they have a natural tendency to re-anneal or in other words, pair up again
 * 5) To keep them separated, single stranded binding proteins (SSB’s) are bound *DNA cannot be fully unwound due to its large size so replication occurs immediately when areas of DNA are exposed as single strands *Replication forks and replication bubbles form
 * 6) Enzyme primase puts down RNA primers that will indicate to polymerase III as starting point of replication
 * 7) DNA polymerase III (5’ to 3’) adds the complementary nucleotides to the 3’ end of the new strand – this is called the leading strand whereas the lagging strand also known as Okazaki fragments that are built discontinuously from the replication fork
 * 8) DNA polymerase I cuts out the RNA primers and replaces them with appropriate deoxyribonucleotides (DNA nucleotides)
 * 9) DNA ligase pins the gaps between the Okazaki fragments by creating phosphodiester bonds
 * 10) Finally DNA polymerase I and II proofread the complementary strands

Here, the enzymes gyrase and helicase unwind and unzip the double helix



The Single stranded binding proteins add onto the strands to keep them from reattaching



Primase adds on the RNA primers



DNA Polymerase III attaches and builds new complementary strands Leading strand: grows towards the replication fork Lagging strand: grows in pieces away from the replication fork



DNA polymerase I cuts out the primers and primers are joined with DNA. Lagging strands: DNA ligase joinds the Okazaki fragments via Phosphodiester bonds



DNA Polymerase I and II proofread and fix any errors

Homework:
Read Pages (282-289) and do #1-9 on 290 Prepare for the quiz Wednesday Bring Colours Monday!

If you are still having trouble understanding the process, check out these links: [] media type="custom" key="24513076" []

Friday November 22nd 2013 Luxena Sribaskaran

Today's class was a work period. Students were told to read sections 6.5 and 6.6 of the textbook and answer the following questions:
 * Page 294 #1, 3-6, 8-9
 * Page 298 #1,3,6

__** 6.5 DNA Organization in Eukaryotes and Prokaryotes **__


 * The Packing of Eukaryotic DNA**
 * DNA is wound around special proteins, called //Histones//, that act as spools
 * Histones are positively charged proteins and DNA strands are negatively charged. This attraction is what causes the DNA to wrap tightly around a cluster of eight histones, thus greatly reducing the amount of space that a DNA occupies
 * Each unit of eight histone proteins and the wrapped DNA is called a //nucleosome//
 * //Linkers// connect the nucleosomes together
 * Coiling strings of nucleosomes into cylindrical fibres ( //solenoids//) allows for further packing of the DNA
 * When a cell enters the reproductive stage of it's life cycle, the solenoids can be even further supercoiled to form the X-shaped chromosome


 * Prokaryotic DNA organization **
 * Bacterial DNA consists almost entirely of one chromosome that is commonly circular
 * Linear chromosomes sometimes occur in bacteria but are rare
 * Conjugation: smaller circular pieces of DNA called plasmids are able to exit one cell and enter another
 * The recipient bacterium incorporates the new plasmid into its genome
 * Coiling technique is used to package DNA
 * Supercoiling: The continuous twisting of prokaryotic DNA that reduces the volume of the DNA


 * Telomeres: A Key Difference **
 * The loss of DNA during every cycle of replication causes chromosomes to continually shorten and can lead to the loss or damage of important genes
 * To prevent this, telemeres are used
 * Telemeres are a repeating sequence of DNA at the end of a chromosome that protects coding regions from being lost during replication

__**6.6 DNA Replication and Aging**__

Telomeres have several functions including
 * They help prevent chromosome ends from fusing to other chromosomes
 * They prevent DNA degradation from enzymes called nucleases
 * They assist DNA repair in distinguishing DNA breaks from chromosomal ends
 * They may play a role in determining the number of times that a cell can divide, and therefore may play a critical role in determining the lifespan of an organism


 * Role of Telomeres during Replication **
 * DNA polymerase I cannot replace the final RNA primer on the lagging strand
 * Chromosomes are capped by telomeres thus portions of the telomere is lost instead of a small portion of a gene near the end of the chromosome
 * The DNA that is inherited by new cells after mitosis is fully functional with all gees intact however the only difference is that the telomere portions of the the chromosome are slightly shorter
 * After many DNA replications, the telomeres are completely lost and therefore provide no protection for the chromosomes
 * As a result of essential portions of DNA being lost during subsequent replications, the new cell may lose its ability to grow, metabolize or divide. This period of decline is known as cell senescence
 * This shows that cells have a total number of times they can divide limiting cell division. This is known as the Hayflick limit
 * Some cells have telomerase which is an enzyme that adds new telomerase sequences to the ends of chromosomes


 * Telomeres and Aging**
 * The remaining length of a telomere in a cell can indicate the age of the cell and how many more times it will be able to divide effectively
 * As an organism ages, more of its cells reach senescence making it difficult for healthy cells to function at their optimum level
 * Although research in this field is preliminary, it seems safe to say that the secret to a longer life may simply be a healthy, active lifestyle


 * Telomeres and Cancer: When Cells Do Not Age **
 * Cancer cells can be dangerous because they never stop dividing
 * healthy cells use up the telomeres in their DNA overtime and begin cell senescence but cancer cells never do
 * This is due to the fact cancer cells produce telomerase in great quantities
 * As a result, telomerase replaces the telemeres lost during cell division thus these cancer cells are able to continue dividing indefinitely



__**Additional Links**__
 * []

__ **Additional Reminders** __ :
 * Bring trip form for Monday
 * Bring colours for Monday
 * Quiz on chapter six is Wednesday November 27 2013

=Monday, November 25th, 2013=

**__Function of Genes__**
Beadle and Tatum - one gene one enzyme hypothesis
 * It states "each gene is unique and codes for the synthesis of a single enzyme.

**__Experiment__**
==
 * Neurospora (bread mold) can generally be grown on a medium ("original") of sugar, inorganic acids and salts, amonium compounds + biotin.
 * They reasoned that Neurospora used these simple chemicals to make complex ones needed for growth and reproduction.
 * They used x-rays to cause mutations in many samples of the mould cells (i.e. samples A,B, and C).
 * They took the x-rayed/"mutated" samples and saw some did not grow in the original medium anymore! They needed something else.
 * The radiation produced mutant Neurospora which required additional specific nutrient supplements in order to grow.
 * Their Hypothesis:** A mutation has blocked a metabolic step leading to the synthesis of a specific compound.
 * There is no product after B product.*

__**Protein Synthesis**__

 * DNA does not just replicate - it encodes to be decoded.
 * DNA is synthesized during replication, but it is **transcribed** during protein synthesis.
 * 1) DNA acts as an original tape, preserving information while temporary transcripts **make use** of the information.
 * 2) The transcripts are made in the molecule can consist of RNA molecules, aka. mRNA
 * 3) The mRNA is then **translated** into polypeptides in the cytoplasm, which eventually becomes active proteins.

**__3 Major classes of RNA__**

 * 1) mRNA - the info that is transcribes into the final protein
 * 2) tRNA - transfers the appropriate amino acid to the ribosome as determined by the mRNA
 * 3) rRNA - structural component of the ribosome - site of protein synethsis

**__PROTEIN SYNTHESIS is made up of two major processes:__**

 * 1) **__Transcription__**
 * 2) **__Translation__**

- mRNA contains the coding for a specific protein - mRNA is usually PROCESSED before it leaves the nucleus
 * __Transcription__ **
 * Information is transcribed onto another medium (DNA to RNA)
 * Enzyme called RNA polymerase transcribes the DNA
 * RNA polymerase READS 3' to 5'
 * The newly synthesized single strand of RNA is COMPLIMENTARY to the one strand of DNA to mRNA
 * __Post Transcriptional Changes__**
 * 1) Addition of 5' can (consisting of 7 G's) which protects the mRNA from digestion by enzymes
 * 2) Removal of **introns** aka non-coding regions using **spliceosomes** (p.322 - draw a labeled diagram of how this happens)
 * 3) Addition of poly-Adenine tail - this extends the life of the mRNA molecule.

__**Overall Process of Transcription**__ - The first bit of "writing" is the 5' UTR (untranslated region) followed by the exons and introns and ends with the 3' UTR - This newly transcribed RNA is the mRNA - It proceeds until it reaches the termination sequence single mRNA is released and processed. Transcription --> making mRNA + post transcribed changes
 * RNA polymerase does not require primer
 * RNA polymerase begins unzipping the DNA at an INITIATION SEQUENCE called **promotor region (TAC)**
 * RNA polymerase reads in the 3'-5' direction
 * RNA polymerase then compliments the single DNA strand

__**In nucleus:**__ __**Homework:**__


 * Read chapter 7.2
 * Do p. 324 #1-6
 * Bring the field trip from + money ($14) for tomorrow
 * Quiz on chapter 6 on Wednesday

=**Tuesday November 26, 2013** = =**By: Randy Ketwaroo** = =**Lesson of Study: Translation **=
 * At the ribosomes we translate & read the mRNA in the 5' to 3' direction
 * We also read mRNA in triplets aka codons
 * Begin at the start codon [AUG] found at the beginning of a exon
 * There are 3 stop codons [UGA, UAA, UAG]

**What is Involved?** a) Aminocyl site aka the A site is for reading the mRNA and retrieving the correct tRNA b) Peptidyl site aka the P site is for holding onto the amino acid and forming peptide bonda c) Exit site is where the empty tRNA is moved then released
 * 1) Ribosomes is a two unit structure - small subunit + large subunit
 * In the ribosome, there are three active sitesa

(A site --> P site --> E site)
 * As the ribosome move along the mRNA, it will read it and each codon on the mRNA will go into each site

2. Transfer RNA
 * tRNA is the smallest of the 3 types of RNA
 * on the bottom (opposite of the 3' end) you will find the __anticodon__ - a three nucleotide segment that pairs with a codon in the mRNA
 * tRNAs are active or non-active, which is determined by whether it has an amino acide or not
 * The process of adding an amino acid to a tRNA is called aminocylation
 * The a.a is added to the 3' end of the tRNA via an ester bond
 * It is done by an enzyme called synthetase
 * The final product is called an aminocyl-tRNA

__**Initiation**__
 * 1) Small ribosomal subunit binds to mRNA and scans it until it finds the start codon [AUG]
 * 2) [AUG] is sitting in the P site
 * 3) Active tRNA enters the P site ad matches its anticodon to the codon
 * 4) This matching signals for the large ribosomal subunits to bind

__**Elongation**__
 * The ribosome will read the mRNA codon by codon. This is also known as the reading frame

5. A site "reads" the second codon. The correct aminocyl-RNA binds in the A site 6. Peptidyl transferase breaks the ester bond of the 1st tRNA and creates the peptide bond that conncects the 1st amino acid to the 2nd amino acid (that is still connected to the 2nd tRNA) 7. Ribosomes shifts the reading frame a) non-active tRNA enters into the E site, then exits the ribosome b) 2nd tRNA enters into the P site c) 3rd codon enters into A site (repeat steps 5-7) until a stop codon enters

__**Termination**__

8. Stop codon enters into the A site 9. Instead of an aminocyl-tRNA entering, a protein release factor binds to the site 10. Polpeptide is released from the ribosome. Ribosome subunits separate and protein release factor and mRNA are released



__**Homework **__
 * pg 331 Q. 1,2,6,8
 * Bring in trip money

__**Additional Help**__

[]

//Independent Study Period// Things to review: __**Protein synthesis**__
 * Thursday November 28th, 2013**
 * By: Megan Lin**
 * DNA is //synthesized// during replication, but it is //transcribed// during protein synthesis
 * Recall the 3 major classes of RNA (All of which participate in protein synthesis):
 * mRNA - the info that is transcribed into the final protein
 * tRNA - transfers the appropriate amino acid to the ribosome, as determined by the mRNA
 * rRNA - structural component of the ribosome - site of protein synthesis
 * Two processes are involved:
 * TRANSCRIPTION [Pg. 319 in your textbook!]


 * TRANSLATION [Pg. 325 in your textbook!]

__**Homework:**__ - Read and take notes on 7.4 (Include: Diagrams of Lac and Trp Operon) - Do #1-5 on pages. 339 - Start build project

__**Additional Links:**__ More on Transcription: []

More on Translation: []

Friday November 29th 2013 Luxena Sribaskaran

Today's class was a work period. Students were told to read and make notes on section 7.5 of the textbook and answer the following questions:

>
 * Page 345 #1-8
 * Work on build assignment when this was completed


 * __7.5 Genetic Mutations __**


 * Small-Scale Mutations **


 * Small- scale mutations include mutations of an individual base pair, called point mutations, and of small groups of base pairs
 * There are several different types of point mutations including:
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">The substitution of one base for another
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">The insertion or deletion of a single base pair
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">The inversion of two adjoining base pairs
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">The differences in the DNA of individuals within a population that are caused by point mutations are referred to as single nucleotide polymorphisms (SNPs)
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">The effects of small-scale mutations can range from being positive, through having no effect, to being severe
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Functionally, small-scale mutations can be categorized into four groups: missense mutations, nonsense mutations, silent mutations and frameshift mutations


 * <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Large-Scale Mutations **


 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Can involve multiple nucleotides, entire genes, or whole regions of chromosomes
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Has various effects on the organism of a genome and the functioning of the organism
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Amplification also known as gene duplication occurs when a gene or group of genes is copied to multiple regions of chromosomes. This duplication leads to a larger number of copies of the gene or group of genes, which compounds it effects
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">In large-scale deletions, entire coding regions of DNA are removed
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Chromosomal translocation occurs when entire genes or groups of genes are moved from one chromosome to another
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Translocation between two non-homologous chromosomes usually occurs when portions of each chromosome break of and exchange places
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Inversion occurs when a portion if a DNA molecule, often containing one or more genes, reverses its direction in the genome
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">A trinucleotide is a triplet of nucleotides. This is normal in the genome however if a mutation occurs, these repeats become unstable and expand uncontrollably. This mutation is known as trinucleotide repeat expansion and increases in the number of repeats from one generation to the next


 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Causes of Genetic Mutations **

>> >>
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Mutations can be grouped into two categories: spontaneous mutations and induced mutations
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Spontaneous mutations is a mutation that is caused by an error in DNA replication
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Induced mutations is a mutation that is caused by an environmental agent known as mutagens
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">The two most common forms of mutagens are chemical mutagens and radiation
 * <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Chemical mutagens is any chemical agent that can enter the cell nucleus and chemically alter the structure of the DNA
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">Confuses replication machinery and results in inaccurate copying
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">Mimics a DNA nucleotide
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt;">Inaccuracies in replication and damage of future generations of cells
 * <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Radiation can also cause mutations
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Bonds that form between adjacent nucleotides along a DNA strand can cause a kink in the backbone of the DNA strand and complicate replication and transcription ( Lower energy radiation)
 * o <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">Can strip molecules of electrons and break bonds within the DNA molecule, causing the rearrangement or deletion of large portions of chromosomes ( Higher energy radiation)


 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Mutations: Positive or Negative? **


 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">Mutations are ultimately responsible for the variety of individuals and species of organisms
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">A mutation that is truly harmful is only harmful to the individual
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">A negative mutation renders the individual less fit than other members of its group and will be selected by nature
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">A beneficial mutation is advantageous to to the individual. The individual experiences greater survival and reproductive success, and passes the beneficial mutation on to future generations
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">The majority of mutations are neutral
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">A mutation is only beneficial or negative in a given time and situation- what may seem like a negative trait may become useful under certain circumstances


 * <span style="font-family: Arial,sans-serif; line-height: 1.5;">Additional Links **
 * <span style="font-family: Arial,sans-serif; font-size: 10pt; line-height: 1.5;">[]


 * December 2, 2013**
 * By: Randy Ketwaroo**
 * Lesson of Study: Control of Gene Expression**

Gene Expression is controlled at various levels There are four types of control in Eukaryotic Cells
 * 1) Transcriptional
 * Regulates which genes are transcribed
 * Controls the RATE of transcription

2. Post transcriptional
 * Remove Introns
 * 5' caps and 3' poly A tail added

3. Translational - Poly A tail: shorter tail = shorter lifespan
 * How often and how fast mRNA is translated
 * Length of time for mRNA to be active

4. Post translational
 * Modification to protein in Golgi: addition of funcitonal groups or cleavage of parts of the protein

- The gene for the enzyme are normally turned off - An activator needs to be present - Once lactose is broken down, the [lactose] decreases and the inhibitor is free to find to the operator again - This system helps the bacteria's survival by limiting the energy used for the manufacturing of enzyme
 * Example 1 Lac Operon**
 * Found in E.Coli: breaks down lactose into glucose and galactose as it source of energy
 * It uses the enzyme B - galactosidase
 * A negative control system is used to block the production of B-galactosidase until it is needed in the bacteria

- Lac Z - B-galactosidase - Lac Y - B-galactosidase parmease - Lac A - unknown functioning protein
 * In this Specific Example**
 * Lac Operon has 3 genes that code for proteins
 * Repressor Protein = Lac L binds to the protein
 * Inducer = Lactose

- Genes for enzyme are usually "on" - When tryptophan is present, trp binds to the repressor - This trp + repressor (trp-trp repressor complex) binds to the operator region and presents transcription - This system helps the bacteria survive via limiting energy used manufacturing enzymes only when needed (i.e. when tryp is not present)
 * Example 2 Trp Operon**
 * Found in E.Coli
 * It needs the a.a tryptophan to produce proteins
 * A positive control system


 * In this Example**
 * Trp operon has 5 genes that code for 5 polypeptides that make 3 enzymes used to synthesize tryptophan
 * Repressor proteins inactive trp repressor + tryptophan


 * Normal Situation - E.Coli need Trp **


 * Cell NEEDS tryptophan
 * Trp repressor protein is inactive
 * RNA polymerase will transcribe
 * [Trp] increase
 * When [Trp] is too high, Trp will bind with the environment and it will change shape so that it will be able to bind to the operator to inhibit transcription of Trp gene


 * Homework**
 * Pg 339 Q. 1-5

[]
 * Additional Help**


 * Dec 4 2013**
 * Alex Tam**
 * Plasmid Construction: How plasmids are made**


 * Plasmids**


 * Contains numerous sites for restriction enzymes to act.
 * Single, double-stranded, supercoiled circular DNA in bacteria.
 * Separate from chromosomal DNA, and usually codes for a gene giving specific characteristics.
 * Can be designated as a vehicle to transfer genetic information into a cell. In this case, the plasmid is called a vector.
 * A cell that can take up foreign DNA and express it is called a competent cell.


 * Plasmid Maps **


 * Shows restriction enzyme sites on a plasmid
 * Helps determine which plasmid is best suited for a give recombinant DNA
 * Distances between restriction sites are measured in base pairs (bp)


 * To create a plasmid map, cut plasmids with different restriction enzymes (alone and in combination) and determine the lengths of the fragments formed by cleaving (gel electrophoresis.)
 * Then use fragment lengths to create map.
 * Vectors are created by using restriction enzymes to insert the gene of interest.


 * Things to consider...**


 * Plasmids may have many restriction sites, but the one you use must only occur once (and have a sticky end if possible.)
 * Point of insertion should be downstream of an existing control sequence such as //lac// or //trp// operons
 * Allows for control of eventual gene product
 * Gene of interest must have no introns or control sequences
 * Genes must insert in the correct direction. (reads 3' - 5')

If the base pairs in a double digest do not add up to the same total base pairs as the single digest, it is likely that there are double bands Example: EcoRI: 7kb, 3kb = 10kb BamHI 7kb, 3kb = 10kb EcoRI + BamHI: 3kb,2kb = 5kb, The base pairs should add up to 10kb. Therefore, there must be 2 3kb fragments and 2 2kb fragments, otherwise, it would not add up to 10kb.


 * Help with Plasmid mapping:**

[] - Scroll down to example 2. []


 * Homework:**


 * Do**


 * Plasmid mapping worksheet.

__**Gel Electrophoresis**__
 * By: Megan Lin**


 * Gel electrophoresis is a technique used to separate pieces of DNA based on their size (kb) kilbases
 * Samples are inserted into each well
 * One sample is the control: It has fragments of known molecular sizes also known as "the standard" - aka standard ladder
 * Set up the power supply with the +'ve terminal at the bottom, -'ve value at the top
 * Recall: DNA is negatively charged on the outside due to the phosphate groups
 * When the current is turned on, the negatively charged fragments will travel towards the positive electrode
 * The shorter the fragment, the faster it will travel through the gel (made of agarose)
 * To see the DNA fragments, we use a loading dye
 * It is made up of a dye (Ethidium bromide) plus glycerol (heavy molecule)




 * Gel electrophoresis can be used for protein separation as well. It would use gels made up of polyacrylamide (smaller pores)
 * Gel electrophoresis involves the separation of charged molecules on the basis of size/mass
 * The medium used is some kind of gel, that acts as filter screen or sieve
 * DNA is negatively charged due to the phosphates, and may be combined with SDS (a type of detergent)
 * Once loaded into the gel, a change will attract the DNA towards the positive end with the smallest molecules moving the most quick

//Animation of gel electrophoresis: (Click Narrated)// []

__**Homework:**__
 * Plasmid Mapping worksheet
 * Build project

//-* Reminders!//
 * Plasmid Mapping quiz is on Monday!
 * Unit Test is on Wednesday!

Luxena Sribaskaran

Friday December 06, 2013 The purpose of this process is to amplify a gene fragment. The steps of this reaction are listed below. > - This allows the primers to anneal (stick to) the ssDNA > - Polymerase III cannot be used for reasons including temperature, no RNA primers to work from and will not work on reverse strand. > - DNA in a hair follide or saliva is ver little
 * __Polymerase Chain Reaction (PCR)__**
 * 1) . Given a short segment of DNA, heat it to 95°C until it dissociates into single stranded (ssDNA)(Replaces helicase and gyrase)
 * 2) Cool it to about 57°C. Incubate the single stranded DNA with a forward and reverse DNA primers. (5’-3’)(Replaces RNA primers and primase)
 * 1) Heat it to 72°C, then add Taq polymerase (heat insensitive) and allow new complementary strands to form.
 * 1) Heat the strands up again to 95°C to dissociate the new pieces and the templates
 * 2) Cool it to 57°C again and allow the primers to anneal, and heat to 72°C to allow Taq to replicate the DNA. (Repeat the heating/cooling, synthesizing)
 * 3) After the second cycle, the new pieces have gone from primer to primer exactly. They are shorter than either the original DNA or the first copy
 * 4) Continued cycling causes an exponential increase in the number of short but ‘consistent’/constant length pieces. Eventually the proportion of strands that are ‘odd’/variable lengths will be so small as to be insignificant.
 * __Application of PCR:__**
 * Forensic criminal investigations
 * Forensic criminal investigations
 * Use PCR to amplify the DNA so that various tests can be used on it (Ex. Paternity test)
 * Genetic testing


 * __Technique #4: Restriction Fragment Polymorphism (RFLP)__**
 * A technique in which organisms may be differentiated by analysis of patterns derived from digestion of their DNA
 * If two organisms differ in the distance between sites of digestion of a particular restriction endonuclease, __the length of the fragments will__ differ when the DNA is digested with a restriction enzyme
 * The similarity of the patterns generated can be used to differentiate species(and even strains) from one another

> - e.g.) TAGTAGTAGTAGTAG
 * __VNTRs (Variable number of tandem repeats) aka microsatellites__**
 * A noncoding region of DNA
 * Often used in RFLP
 * A sequence of base pairs that repeat over and over
 * Between individuals, VNTRs can vary in length and position within the genome


 * __DNA Sequencing__**
 * A sample of DNA is obtained and divided into 4 test tubes
 * Each test tube contains everything necessary for DNA replication
 * Also each test tube has a small percent of nucleotides made with ddDNA 9dioxyribonucleic acid)
 * Each tube has a different ddDNA nucleotide
 * As the DNA is synthesized, occasionally a ddDNA is used
 * This stops the synthesis at that point due to the absence of an –OH on carbon #3 of the deoxyribose
 * Each test tube has various sized pieces of DNA each stopped at the same base
 * By running each sample in parallel in a gel, each piece can be separated by size


 * __Additional Links__**
 * []


 * __Homework__**
 * Page 367-374 #1-7
 * Correction for plasmid mapping on question level 3 H&B, turn 500 into 800
 * Study for plasmid mapping quiz on Monday and molecular genetic test on Wednesday

December, 12, Thursday Alex Tam


 * Today was a work period.
 * Work on a flowchart for the pGLO lab tomorrow.
 * (If you didn't go to the musical,) the DNA build project is due tomorrow

Friday, December 13th, 2013 By: Viththakan Arunthavanathan

=Transformation Lab=
 * In class, the transformation lab was done all period.

Procedure:
<span style="font-family: Arial,Helvetica,sans-serif;">1. Label one closed micro test tube +pGLO and another –pGLO. Label both tubes with your group’s name. Place them in the foam tube rack.

<span style="font-family: Arial,Helvetica,sans-serif;">2. Open the tubes and using a sterile transfer pipet, transfer 250 micro liters of transformation solution (CaCl2) into each test tube.

<span style="font-family: Arial,Helvetica,sans-serif;">3. Place the tubes on ice.

<span style="font-family: Arial,Helvetica,sans-serif;">4. Use a sterile loop to pick up the 2-4 large colonies of bacteria from your starter plate. Select starter colonies that are “fat” (i.e 1-2 mm in diameter). It is important to take individual colonies (not a swab of bacteria from the dense portion of the plate), since the bacteria must be actively growing to achieve high transformation efficiency. Choose only bacterial colonies that are uniformly circular with smooth edges. Pick up the +pGLO tube and immerse the loop into the transformation solution at the bottom of the tube. Spin the loop between your index finger and thumb until the entire colony is dispersed in the transformation solution (with no floating chunks). Place the tube back in the tube rack in the ice. Using a new sterile loop, repeat for the –pGLO tube.

<span style="font-family: Arial,Helvetica,sans-serif;">5. Examine the pGLO DNA solution with the UV lamp. Note your observations. Immerse a new sterile loop into the pGLO plasmid DNA stock tube. Withdraw a loopful. There should be a film of plasmid solution across the ring. This is similar to seeing a soapy film across a ring for blowing soap bubbles. Mix the loopful into the cell suspension of the +pGLO tube. Optionally, pipet 10 micro liters of pGLO plasmid into the +pGLO tube and mix. Do not add plasmid DNA to the –pGLO tube. Close both the +pGLO and -pGLO tubes and return them to the rack on ice.

<span style="font-family: Arial,Helvetica,sans-serif;">6. Incubate the tubes on ice for 10min. Make sure to push the tubes all the way down in the rack so the bottom of tubes stick out and make contact with the ice.

<span style="font-family: Arial,Helvetica,sans-serif;">7. While the tubes are sitting on ice, label your four LB nutrient agar plates on the bottom (not the lid) as follows: <span style="font-family: Arial,Helvetica,sans-serif;">Label one LB/amp plate: +pGLO <span style="font-family: Arial,Helvetica,sans-serif;">Label the LB/amp/ara plate: +pGLO <span style="font-family: Arial,Helvetica,sans-serif;">Label the other LB/amp plate: -pGLO <span style="font-family: Arial,Helvetica,sans-serif;">Label the LB plate: -pGLO

<span style="font-family: Arial,Helvetica,sans-serif;">8. Heat shock. Using the foam rack as a holder, transfer both the (+)pGLO and (-) pGLO tubes into the water bath, set at 42 degrees Celsius, for exactly 50 seconds. Make sure to push the tubes all the way down in the rack so the bottom of the tubes stick out and make contact with the warm water. Double-check the temperature of the water bath with two thermometers to ensure accuracy. <span style="font-family: Arial,Helvetica,sans-serif;">When the 50 seconds are done, place both tubes back on ice. For the best transformation results, the transfer from the ice (0 degrees Celsius) to 42 degrees Celsius and then back to the ice must be rapid. Incubate tubes on ice for 2 minutes.

<span style="font-family: Arial,Helvetica,sans-serif;">9. Remove the rack containing the tubes from the ice and place on the bench top. Open a tube and, using a new sterile pipet, add 250 micro liters of LB nutrient broth to the tube and reclose it. Repeat with a new sterile pipet for the other tube. Incubate the tubes for 10 minutes at room temperature.

<span style="font-family: Arial,Helvetica,sans-serif;">10. Gently flick the closed tubes with your finger to mix and resuspend the bacteria. Using a new sterile pipet for each tube, pipet 100 micro liters of the transformation and control suspensions onto the appropriate nutrient agar plates.

<span style="font-family: Arial,Helvetica,sans-serif;">11. Use a new sterile loop for each plate. Spread the suspensions evenly around the surface of the LB nutrient agar by quickly skating the flat surface of a new sterile loop back and forth across the plate surface. DO NOT PRESS TOO DEEP INTO THE AGAR. Uncover one plate at a time and re-cover immediately after spreading the suspension cells. This will minimize contamination.

<span style="font-family: Arial,Helvetica,sans-serif;">12. Stack up your plates and tape them together. Put your group name and class period on the bottom of the stack and place the stack of plates upside down in the 37 degree Celsius incubator until the next day. The plates are inverted to prevent condensation on the lid which may drip onto the culture and interfere with your results.

<span style="font-family: Arial,Helvetica,sans-serif;">Homework:
Look over the lab and come in Tuesday prepared to see any observations DNA Builds are due Monday!