Topic 3.5 & Some Topic 7: Biotechnology and Genetic Modification/Nucleic Acids Flashcards
DP2- U2- Molecular Genetics
A. Intro to Molecular Genetics
- State that DNA and RNA are nucleic acids and are polymers of nucleotides.
Define nucleotide/consists and nucelic acids
A nucleotide: a single unit of a nucleic acid
Nucleic acids: very large molecules– linking together nucleotides to form a polymer
Nucleotide consists of a phosphate group (acidic, negatively charged), sugar (5-C/pentose), nitrogenous base (ATGC or U). All lined by covalent bonds.
A. Intro to Molecular Genetics
2.Compare and contrast the structures of DNA and RNA
plus purine and pyrimidine
Purine= Adenine and Guanine
Pyrimidine= Thymine/Uracil and Cytosine
—
DNA:
* Adenine, Guanine, Thymine, Cytosine
* Deoxyribose sugar
* Two antiparallel complementary strands that form a double helix
RNA:
* Adenine, Guanine, Uracil, Cytosine
* Ribose sugar
* Single stranded and often linear in shape
A. Intro to Molecular Genetics
3.Outline the detailed structure of the DNA double helix including the antiparallel strands and the complimentary base pairs.
phopic&phylic regions/h-bond/complementary base pairing/antiparallel
- Each polynucleotide chain (strand) consists of a chain of nucleotides bonded covalently The sugar-phosphate is on the outside (hydrophilic) and the reactive nitrogenous bases are on the inside (hydrophobic)
- Two polynucleotide chains of DNA are held together by hydrogen bonds between complementary base pairs:
-(A=T) via 2 hydrogen bonds
-(G=C) via 3 hydrogen bonds - Complementary base pairing reduces the chance of error during DNA replication or transcription as only T goes with A and C with G.
- Antiparallel: In order for bases to be orientated correctly and thus able to pair, the two strands must run in opposite directions
-One strand at C5 at the top and the other has C3 at the top. 5’ and 3’ directions
A. Intro to Molecular Genetics
4.Draw a simple diagram of the structure of single nucleotides of DNA and RNA, using circles, pentagons and rectangles to represent phosphates, pentoses and bases.
how to draw one and 4 nucleotides
view sheet
A. Intro to Molecular Genetics
5.Analyze the results of the Hershey and Chase experiment that provided evidence that DNA is the genetic material.
aim/context/procedure/method/conclusion
Aim:
wanted to see if proteins or chromosome were the genetic material of cells
Context:
* Took T2 bacteriophage virus which can infect E. Coli bacterium. It attaches to the outside of E coli. and injects a substance into the bacteria to create many new copies of the virus.
* These newly cultured copies of the virus are then expelled when the bacteria bursts.
* Therefore, the substance that is initially injected into the bacteria, whether it is DNA or proteins, will have the instructions on how to make the new viruses. They wanted to discover if it was the DNA core or protein coat that was injected into the bacteria.
Procedure:
* They ran two experiments to test their aim
1. Radioactive sulfur 35 was labeled for the protein
2. Radioactive phosphorus 35 was labeled for the dna
-Elegant method because phosphorus is only found in the nucleotides that make up dna and not in proteins. Vise versa for phosphorus.
* After the viruses infected the bacteria they wanted to see if the radioactive signal was found in the bacteria or outside of it.
* They separated the viruses attached to the bacteria by using a waring blender to do so
-They then centrifuged the samples- since bacteria is bigger it is centrifuged to the bottom into a pellet and the virus was at the top of the tube in the supernatant
-They found that the sulfur labeled virus remained outside of the bacteria and none of the replicated copies had the radioactive virus included. Therefore the protein wasn’t being inherited
-On the other hand, the phosphorus labeled virus was injected into the bacteria and the cultured viruses all had the radioactivity from it. Therefore, DNA was in fact inherited.
Conclusion:
This experiment showed that DNA is in fact the genetic material and protein was not
A. Intro to Molecular Genetics
7.Outline what Crick and Watson were able to deduce regarding the structure of DNA using their model.
method/why/first model-faults/ what they realized/second model suggests
- Watson and Crick used a stick and ball modeling method to test their hypotheses on the possible structures of DNA.
- This allowed them to visualize the molecule and to quickly see how well it fitted the available evidence.
- Their first model, a triple helix, was rejected:
-The ratio of Adenine to Thymine was not 1:1 (as discovered by Chargaff)
-It required too much magnesium to be true (identified by Franklin) - Through this they realized:
-DNA must be a double helix.
-The complementary relationship between the bases and base pairing
-The strands must be anti-parallel to allow base pairing to happen - Because of the visual nature of their work the second and the correct model quickly suggested:
-Possible mechanisms for DNA replication
-That information was encoded in triplets of bases - Roslalind franklin whose x-ray diffraction of DNA was crucial to their discovery
B. DNA Replication
1.Outline Meselson and Stahl’s evidence for the semi-conservative replication of DNA.
state the three theories of dna replication/results from procedure
There were three theories to how DNA could have been replicated:
* Semi-conservative replication: each new DNA molecule is composed of one parent strand and one newly synthesized strand
* Conservative replication: the new DNA molecule consists entirely of newly synthesized DNA
* Dispersive replication: each new DNA molecule consists of segments of new and old DNA interspersed
Meselson and Stahl’s work shows that only semi-conservative replication was biologically significant.
Procedure:
* Gen 0: Grew e-coli in heavy nitrogen= N15 100% of the bacteria contained N15.
* Then they transferred them to e coli living in the lighter normal nitrogen N14.
* They spun the e coli in an ultracentrifuge, and DNA with the heavy nitrogen will be moved to the bottom of a test tube whereas those that contain lighter nitrogen (N14) will be moved to the top due to densities.
* Gen 1: The N15 DNA copied again but with the N14 added. This made one half of the DNA be heavy and the other half be lighter (hybrid strands)- due to the different densities of nitrogen- showed up in the middle of the test tube. Debunks conservative replication as it’s a hybrid.
* Gen 2: the strand from gen 1 stayed the same but another DNA molecule appeared above the hybrid location signifying that it consisted of 100% N14 DNA. Dispersive replication is falsified as this shows for there to be one 100% newly synthesized strand and one hybrid while dispersive would just show two that are partially segmented.
B. DNA Replication
2.Explain DNA replication.
both/leading/lagging– reference all 6 enzymes
REMEMBER: DNA replication occurs during (S phase of ) interphase, in preparation for mitosis and cell division
BOTH STRANDS:
* Helicase unwinds the double helix and separates the two polynucleotide strands of DNA. Binding proteins keep the strands apart.
* It uses ATP to break the hydrogen bonds between the two strands and move along the DNA strands (single stranded binding proteins keep the separated strands apart so that nucleotides can bind)
* DNA topoisomerase moves in advance of helicase to prevents the DNA from supercoiling.
* Each strand of parent DNA is used as template for the synthesis of the new strands
* synthesis always occurs in 5´ → 3´ direction on each new strand
* To synthesize a new strand an RNA primer is synthesized on the parent DNA using DNA primase.
LEADING STRAND:
* DNA primase synthesizes an RNA primer on the leading parent DNA for attachment and initiation of DNA polymerase III.
* DNA polymerase III adds the nucleotides (only in a 5’ to 3’ direction on the complementary strand- parent strand would be opposite for each side- refer to drawing on sheet) according to the complementary base pairing rules; adenine pairs with thymine and cytosine pairs with guanine; (names needed, letters alone not accepted).
* The free nucleotides which are added by the DNA polymerase III for complementary base pairing are in the form of deoxynucleoside triphosphates. Two phosphate groups are released from each nucleotide and the energy is used to form the covalent phosphodiester bonds between sugars and phosphate groups on the growing chain.
* Once the DNA polymerase III begins it work of adding the nucleotides, DNA polymerase I then removes the RNA primer and replaces them with DNA
* The new strand is made continuously towards the replication fork and as mentioned the bonds being made are phosphodiester covalent bonds.
LAGGING STRAND:
* DNA primase synthesizes a series of RNA primers at intervals leading away from the replication fork of the lagging parent DNA.
* Next the DNA polymerase III adds the nucleotides (to the 3´ end) added according to the complementary base pairing rules; between the primers.
* The free nucleotides which are added by the DNA polymerase III for complementary base pairing are in the form of deoxynucleoside triphosphates. Two phosphate groups are released from each nucleotide and the energy is used to form the covalent phosphodiester bonds between sugars and phosphate groups on the growing chain.
* This leads to the formation of Okazaki fragments on the lagging strand, in order to elongate it away from the replication fork.
* Once the DNA polymerase III begins it work of adding the nucleotides, DNA polymerase I then removes the RNA primer and replaces them with DNA
* DNA ligase next joins Okazaki fragments on the lagging strand to form a continuous strand
B. DNA Replication
3.Outline the important functions of non-coding regions of DNA.
define/role- 6 key non-coding +role
Non-coding DNA: all organisms have regions of DNA which are not expressed as polypeptides. They still do however have an important role to organisms, for example:
* Produces tRNA (t=transfer) and rRNA (r=ribosomal)
* Important role in regulating (controlling) gene expression.
Introns:
* Most eukaryotic DNA have introns and exons.
* Introns do not code for a protein so are removed when the mature mRNA is made (the introns and exons are both in DNA and translated to mRNA but spilced out for mature mRNA)
* Since mature mRNA is translated by ribosomes into polypeptides only exons code for polypeptides
* Introns however are needed because it enables multiple proteins to be made from a single gene.
Promoters:
* attachment point for RNA polymerase enabling transcription
Enhancers:
* sequences that increase the rate of transcription when bound to an activator protein
Silencers:
* inhibit transcription when bound to a repressor protein. Block or reduce transcription.
Telomeres:
* There are two types of repetitive sequences that are non-coding: moderately and highly repetitive sequences AKA satellite DNA
* Telomeres are non-coding highly repetitive sequences and they protect the DNA molecule from degradation during replication. They are located towards the end of chromosomes.
Tandem repeats (more below):
* used in DNA profiling because there is more variation in the repeats than in coding sections of genes.
B. DNA Replication
4.Outline how tandem repeats are used in DNA profiling.
define/how its used in profiling
5 is skipped- HL
- Tandem repeat sequences are short sequences of (non-coding) DNA, normally of length 2-5 base pairs, that are repeated numerous times.
- In a population, the repeating sequence may stay the same but varies greatly in the number of times it may repeat.
For profiling:
* dyes/markers are attached to the tandem repeats during PCR (DNA Replication).
* Restriction enzymes are used to cut DNA between the tandem repeats.
* Gel Electrophoresis enables scientists therefore to calculate the length of the tandem repeat sequence of an individual.
* If different tandem repeats at different loci are used, then a unique profile for an individual can be identified.
C. Transcription & Translation
1.Explain transcription.
4-1st, 4-2nd, 2-3rd
Occurs in the nucleus then moves to the cytoplasm for translation.
Initiation
* Transcription begins at a region upstream of the gene called a promoter (eukaryotes call it the TATA box because of the appearance before every gene)
* Proteins called transcription factors first bind to the promoter.
* The transcription factors then assist the enzyme RNA polymerase II to bind. (transcription factors + RNA polymerase II + promoter DNA = transcription initiation complex)
* RNA polymerase then binds to unwind the DNA.
Elongation
* RNA polymerase moves along the DNA and adds ribonucleoside triphosphates that are complementary to the exposed DNA nucleotides to the 3’ end of the growing mRNA molecule -> Transcription occurs in a 5’ to 3’ direction!
* RNA polymerase moves along the anti-sense (template) strand. Like in DNA replication, this complementary strand runs the opposite way ( 3’ to 5’)
* The energy from the cleavage of the additional phosphate groups is utilized to make the covalent bond between nucleotides
* The DNA double helix re-forms behind RNA polymerase and the new mRNA molecules peels away from the DNA template
Termination
* RNA polymerase continues to move along the DNA and synthesize mRNA until it reaches a terminator sequence. This sequence will be located downstream of a stop codon.
* It then detaches from the DNA strand and the newly formed pre-mRNA strand is released.
C. Transcription & Translation
2.Outline the role of the promoter.
Promoter: a region of DNA located close to the specific gene and the attachment point for RNA polymerase enabling transcription during initiation. This is where the majority of gene expression is controlled, by either accessing or blocking regions to this site from the RNA polymerase. Once the RNA polymerase is attached it then begins to create the mRNA strand by transcribing the gene.
C. Transcription & Translation
3.Outline how gene expression is regulated by proteins that bind to a specific base sequence in DNA.
two components and their functions
- Transcription is regulated by both transcription factors and regulatory proteins which mediate the binding of RNA polymerase to the promoter
- Transcription factors:
-Form a complex with the RNA polymerase at the promoter
-Since RNA polymerase cannot initiate transcription without these factors– it is crucial part of gene expression - Regulatory proteins
-Bind to DNA sequences outside of the promoter and interact with the transcription factors
-Activator proteins bind to enhancer sites and increase the rate of transcription (by mediating complex formation)
-Repressor proteins bind to silencer sequences and decrease the rate of transcription (by preventing complex formation)
C. Transcription & Translation
4.Use examples to outline how the environment has an impact on gene expression.
Human hair and skin color are impacted by sunlight and high temperatures.
The Himalayan rabbit produces a different fur pigment depending on the temperature (>35ºC = white fur ; <30ºC = black fur)
C. Transcription & Translation
6.Outline nucleosome role in DNA supercoiling and eukaryotic gene expression.
structure and why its done
Structure- formed by wrapping dna around histone proteins
* A nucleosome consists of an octamer of histones (2 copies of 4 different types of histone proteins)
* Histone #1 binds dna in a way to form a 30 nm fiber structure which facilitates further packing
* Linker DNA which holds it all together
Why
* Nucleosomes protect DNA and allow it to be packaged through supercoiling
* It’s essential to supercoil DNA because it helps with…
-organizing DNA in prep for cell division
-controlling DNA expression as supercoiled DNA cannot be transcribed
-allows for the specialization of cells by permanently supercoiling DNA like heterochromatin
-promoting or inhibiting transcription of active chromatin through the use of associated histones
