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What is the central dogma?
- DNA to RNA to Protein
- Conserved across almost all known organisms
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What are the two influential experiments that established the fact that DNA is the genetic material?
- Avery, Macleod & McCarty experiment: Pathogenic smooth cells broken down, DNA added to rough cells. Visible appearance of rough cells turned to smooth cells and became pathogenic. The DNA recombined into non-pathogenic bacteria and changed their behavior in a heritable manner. None of this happened when DNAase was added. Strong evidence, but not full confirmation.
- Hershey & Chase experiment: Using radioactive S-labeled coat protein and p-labeled DNA. They allow the phage to bind to the bacteria and then put the mixture in a blender. They observed that p-labeled DNA was inside the cell, S-labeled protein coat left the cell. The viral chromosomes multiplied and new phage particles were released by the infected bacteria. This was definitive proof.
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Explain the Avery, Macleod & McCarty experiment
- Start with pathogenic S (smooth) cells which were broken down. DNA from the cells was isolated. The DNA alone was added to R (rough) cells.
- The visible appearance of the rough cells turned to smooth cell walls. The cells also became pathogenic.
- The DNA isolated was able to recombine into non-pathogenic bacteria and change the behavior of these cells in a heritable manner.
- No transformation happened when DNAase was added.
- The purified DNA was not absolutely pure, so there was some concern that the mixture was not pure enough. This was very strong evidence, but did not fully confirm that DNA was the genetic material.
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Explain the Hershey & Chase experiment
- Working with bacteria and bacteriophage (a virus that infects a bacterium). Phage have DNA and protein and are very simple. The researchers used phage with radioactive S-labeled coat protein and p-labeled DNA.
- They allowed the phage to bind to the bacteria and then put the mixture into a blender.
- After violent agitation, the authors observed that the P-labeled DNA was inside the cell, and the S-labeled protein coat left the cell.
- The multiplication of viral chromosomes and the production and release of new phage particles showed some of the radio-labeled DNA.
- This was considered definitive proof that DNA was the genetic material.
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Explain nucleotides
- They’re the building blocks of nucleic acids, which include DNA (deoxyribonucleic acid) and RNA (ribonucleic acid).
- Nucleotides are in ATP (energy), cAMP (signaling), NAD (redox reactions)
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What is NMP (nucleoside monophosphate)
- Purine or pyrimidine base (attached to sugar with glycosidic bond)
- Ribose or deoxyribose
- phosphate (attached to sugar with an ester bond)
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Difference between RNA and DNA
RNA features a hydroxyl group on the 2’ carbon of the sugar ring
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Explain ATP
- Three phosphates bonded via ester bonds.
- Gamma and beta bonds are high energy bonds that are broken and formed to release/store energy for cellular processes.
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Explain key bonds within nucleotides
- Ester bond between phosphate group and pentose sugar.
- Glycosidic bond between pentose sugar and purine/pyrimidine base.
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Explain purines and pyrimidines
- Both planar
- Purines: Adenine and guanine (double ring)
- Pyrimidines: Cytosine and Thymine (Uracil)
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Explain phosphodiester backbone
- DNA and RNA are polymers are held together by phosphodiester bonds
- The covalent bond formed between 5’ phosphate of one nucleotide and 3’ hydroxyl of a second.
- Nucleic acid sequences are written 5’ to 3’
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Explain the base composition of DNA
- Chargaff’s rules: Studies from the 1940s exploring base composition in DNA. Purine content = Pyrimidine content. A = T and G = C. Base compositions differ between different species. Within a species, all tissues have the same base composition, as every cell has the same genome.
- Bases were thought to rapidly interconvert between their tautomer states. It was figured out that in physiological state, the bases were stable in one form.
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Explain how the structure of DNA was found
- Franklin used X-ray diffraction of DNA fibers. Crystalline DNA showed that DNA is a helical molecule. Found its diameter and the two different types of repeats within.
- Used Chargaff’s rule, tautomeric forms, X-ray diffraction data to inform physical models of DNA molecules.
- Antiparallel, right-handed double helix with major and minor grooves.
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Explain non-covalent interactions in DNA
- Hydrogen bonds between bases
- Repulsion between phosphate groups. Phosphate groups are negatively charged, negative charges repulse one another.
- Base stacking (profoundly important for the stability of DNA, you only get this in a structure like the double helix) When the planar hydrophobic rings are arranged vertically one above the other, the “stacked” hydrophobic interactions keep the DNA together.
- A single stranded DNA would be energetically more favorable, as the phosphate groups which are negative and repel each other would be further away. Hydrogen bonding is not as important, as the bases would also happily form hydrogen bonds with water molecules. The most important is base stacking.
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Explain the implications of the Watson-Crick model
- Base pairs underlie the replication mechanism of DNA
- Structure is independent of sequence, but sequence specific recognition of DNA is essential.
- Sequence-specific recognition of DNA is done through the chemical patterns in major and minor grooves.
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Three features of Watson-Crick base pairing:
- A binds to T. G binds to C.
- H bonds are possible bc of stable tautomeric forms in physiological conditions.
- AT and GC pairs are equivalent in size and shape.
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Explain how sequence-specific protein-DNA binding occurs if the structure of DNA is independent of the sequence of bases.
- The hydrogen patterns (acceptor, donor, hydrogen) in side-chains of the base pairings allows for sequence-specific binding.
- Minor grooves have less chemical information (as multiple pairings have the same hydrogen patterns), so most of the sequence-specific binding occurs in the major grooves.
- Proteins nestle into the major and minor grooves though non-covalent interactions with the side-chains of the bases.
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Explain nucleic acid double-helix forms
- There are different DNA forms, because bonds within monomers can rotate.
- A-form: Dehydrated DNA samples and RNA-double helix. More compact
- B-form: Almost all DNA in nature. Medium compact.
- Z-form; CG-repeat sequences (some evidence for Z-form in nature). Long form.
- Proteins that bind to one form can’t recognize sequences in the other.
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Explain protein translation in simple words
- mRNA directs protein synthesis, ribosome is the machinery.
- Pieces of the mRNA where each piece codes an aminoacid are codons.
- tRNAs bring individual amino acids to the ribosome.
- The code is collinear - nucleotide order corresponds to amino acid order.
- The code is non-overlapping. The proof of this is that a single base change will kill the function of the gene.
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Explain experiments to distinguish between overlapping code and non-overlapping code
- The researchers induced indel mutations in an mRNA.
- In a non-overlapping code, the reading frame shifts. A single indel shifts the reading frame. A combination of insertions and deletions would sometimes restore the reading frame.
- In overlapping code, the reading frame is not affected. The indel only changes the codons in a local environment.
- The gene function was killed, which mean that the code is non-overlapping.
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Explain how the genetic code was solved: Nirenberg and Khorana
- Cell-free translation system involved cell extract, ATP + one amino acid. Poly(U) + phenylalanine = polypeptide. Poly(U) + other amino acids did not result in a polypeptide. This shows that the UUU codon encodes phenylalanine. They showed that the CCC codon encoded proline and AAA encoded lysine. However, GGG did not encode anything, as it spontaneously formed stable tetraplexes.
- They then formed random polymers. They used polymers with different ratios of bases. For example, in a polymer with five times more A than C, we would expect AAA to be the most common, with AAC etc. being common. They then observed the frequency of incorporation of each amino acid and found the corresponding codon based on its expected frequency. This kind of approach does not give you an unambiguous match, it tells us for example that histidine is a codon with 2Cs and 1A, but not the order. This is where Khorana came in.
- Dinucleotides: For example, a polymer with an AC repeat could form Threonine or Histidine. Through this and Nirenberg’s experiments, they inferred that CAC must code for His and ACA must code for Thr.
- Through this approach, many codons were solved - but not all.
- Nirenberg worked out ways to purify ribosomes + synthetic RNA of 3nt (an mRNA analog) and bind specific aminoacylated tRNA. He tested different defined sequence RNA with exhaustively purified aminoacylated tRNA to test which trinucleotides could bind stably to specific aminoacylated tRNA. Nuremberg worked out 61 of the 64 possible codons - the remaining three were the stop codons.
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Explain codon use bias
- One AA can correspond to multiple codons, but different codons are not used equally.
- Codon usage differs among species, although there is redundancy. Codons used frequently correspond to tRNAs present at higher levels.
- Codon bias creates problems for recombinant protein expression: if rare codons are used it is difficult to efficiently translate the mRNA.
- In an experiment, we might want to translate a human mRNA in an E.coli system - this may not be efficiently translated due to different levels of tRNA availability.
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Degeneracy in the code
- Some mutations will not change the encoded amino acid sequence. Mutations that do not change the amino acid sequence can rarely still have an effect.
- Synonymous mutations - no change in encoding.
- Non-synonymous - change in encoding.
- Some non-synonymous mutations, such as Ser to Thr don’t have an impact, as the AA are similar chemically.
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Explain the structure of tRNA
- AA arm, attached to AA
- Anticodon arm, recognizes codons.
- 61 codons, but fewer tRNAs due to wobble pairing.
- All tRNA have CCA at their 3’ terminus (which is added after transcription)
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Explain wobble pairing
- Subtle differences in chemistry of RNA and DNA have profound structural consequences.
- Thymine replaced by Uracil in RNA. This allows Uracil to pair to Guanine as well as Adenine.
- This additional ability to form base pairs allows for wobble pairing, and 61 codons to bond with less than 61 tRNAs.
- Wobble pairing only occurs at the first position of the anticodon (third position of the codon).
- G can pair with C or U, U can pair with A or G
- Moreover, a new base called inosine (which is only found in tRNA) will bond with cytosine, uracil and adenine.
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What are the three key features of a gene
- Transcription of at least a part of the sequence.
- Regulated synthesis (via regulatory sequences or other mechanisms)
- Resulting transcripts are functional.
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Explain bacterial gene structure
- Promoter that immediately precede transcribed sequence
- Transcription unit (mRNA) which has 5’ untranslated region and 3’ untranslated region
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Explain bacterial operons
- Genes with related functions are often clustered in operons.
- Clustering facilitates coordinate regulation. Only a single promoter is responsible for the coding regions.
- This results in a single polycistronic transcript (mRNA).
- The cell then synthesizes independent proteins from each of the coding units.
- Operons are incredibly rare in eukaryotes.
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What is a polycistronic transcript?
- Transcript (mRNA) resulting from an operon that contains multiple coding sequences.
- Each coding region results in translation of a separate protein.
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Explain bacterial genome structure
- No nucleus.
- Genes are densely packed. Genomes are very efficiently organized.
- One circular DNA molecule of 5MB, ~5k genes.
- No transposons
- Plasmids are also typical in bacteria, made up of a few kb with <10 genes.
- Plasmids are autonomous mobile genetic elements. They can be transferred between bacteria. They can be edited and transferred to express genes as desired.
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Explain mammalian genome structure
- Multiple linear molecules.
- Nucleus.
- Packaged into chromatin.
- 3 billion base pairs.
- 25k genes.
- Not densely packed.
- Has transposons and repetitive DNA.
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Explain eukaryotic gene structure
- Distal Enhancers (multiple), which function with the promoter to regulate transcription
- Promoter + transcription unit
- The mRNA is initially the precursor with introns and exons, spliced to form mature mRNA with introns removed.
- Result is 5’UTR + cds + 3’UTR
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Exceptions to the genetic code: 21st amino acid
- Certain rare proteins require selenocysteine for function
- Rare mRNAs code selenoproteins.
- The 3’ end of the mRNAs have special structures (SECIS element) that recruit selenocysteine tRNA.
- The UGA stop codon then encodes selenocysteine
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Explain eukaryotic genome structure
- Telomeres on ends of linear DNA molecules
- Linear DNA molecules called chromosomes.
- Centromeres in each chromosome partition.
- Multiple replication origins
- Eukaryotic DNA is packaged and condensed to euchromatin and heterochromatin.
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Explain chromatin
- Euchromatin: less packed, gene rich regions of chromosomes
- Heterochromatin: densely packed, gene poor regions
- Chromatin has two core features: regulation and compactness
- DNA + histone = nucleosome.
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Explain human genome
- 3 Gb, 23 linear DNA molecules, ~25,000 genes with 20k coding
- 30% gene (1% exons), 45% transposons, 5% short-sequence repeats, segmental duplications, 20% miscellaneous.
- Less than 1% of our genome encodes proteins.
- Most transcribed DNA sequence is introns, 30x more introns than exons.
- LINEs: Retrotransposon, up to 9kb long, Many copies are no longer functional
- SINEs: ~1,5mil copies up to 300 nt long. Rely on other transposons for amplification and accumulations.
- Retrovirus-derived: 450,000 copies, non-functional, diversity of sizes and types
- Simple-sequence repeats and segmental duplications
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What is the human genome composition?
- 30% gene (1% exons), 45% transposons, 5% short-sequence repeats, segmental duplications, 20% miscellaneous.
- So, 50% is repetitive. 30% is gene. 1% is exon.
- 23 linear DNA molecules.
- 25k genes.
- Less than 1% of our genome encodes proteins.
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Explain transposons
- Mobile elements - segments of DNA that move with in a genome
- Many different types and mechanisms
- Examples in both eukaryotes and bacteria
- Contribute to insertions, deletions and translocations (mutagenic)
- Often present in many copies per genome. Not all copies are functional
- Some evidence that transposons contribute to gene regulation
- Genome size is dependent on how much repetitive DNA is present. The rat genome is 3 times bigger than our genome. The sea cucumber also has a larger genome than humans. This has nothing to do with organism complexity or gene number, but repetitive DNA accumulations.
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Explain organelle genomes
- Eukaryotic cells have genetic material in some organelles.
- Mitochondria - present in all eukaryotes, chloroplast - present in plants.
- Due to endosymbiotic theory: Eukaryotic mitochondria originated as bacteria, which were assimilated by early-Eukaryotes.
- Variations in genetic code (Exceptions to the genetic code) occur in mitochondria. This is because there are 13 proteins encoded, so the deleterious consequences of changes to the genetic code are very limited.
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Explain changes to the genetic code
- One change in a codon affects every protein that uses the codon.
- Mitochondria features variant codon assignments, which are only possible because of the tiny genome size.
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