HackMD - Collaborative Markdown Knowledge Base

1. **Describe the DNA packaging problem and the biological solutions to this problem.** - The total amount of DNA necessary to encode a human’s diploid genome is very large compared to the size it needs to fit into inside the nucleus of each cell. - Biology's solution is to compacting DNA by wrapping it around histones / nucleosomes - where "wrapping it" = 1 ( net ) negative supercoil 2. **Deduce the most abundant amino acids in the histone proteins,** - Lysine , it is basic and has a positively charged side chain - It balances negative charged DNA - https://en.wikipedia.org/wiki/Histone#Lysine_methylation 3. **Summarize the composition and structure of a nucleosome,** - Composed of 200 base pairs of DNA wound around a core of 8 histone molecules - Eukaryotic Histone Molecules : - H2A , H2B , H3 , H4 - There are 2 of each histone molecule , creating an octamer 4. **Predict the effect of displacement of the nucleosome histone octamer on DNA structure.** - Remove the negative supercoil by turning it clockwise 5. **Question : Why might histones ( especially histone H1 ) run with anomalously slow mobility on SDS-PAGE?** - Histone 1 has the highest molecular weight , almost double that of some of the other histones - SDS causes denaturation , increasing surface area and likely hood of getting trapped in the polyacrylamide "matrix" 6. **Problem : Based on your knowledge of chromatin packing in vivo, explain the nonrandom pattern of DNA sizes ( 200 bp , 400 bp , 600 bp , 800 bp , etc ) obtained after digestion of nuclear DNA with a randomly cutting nuclease ( micrococcal nuclease ).** - Nuclease will most likely attack the nucleosome linker DNA - So each fragment contains a proportional amount of nucleosomes all packed with roughly the same amount of base pairs - Depending on where the nuclease cuts , determines the proportional amount of base pairs to fragment length 7. **Describe the common types of epigenetic modifications to chromatin** - Methylation of DNA = addition of $CH_3$ group to carbon-5 of cytosine in [CpG](https://en.wikipedia.org/wiki/CpG_site) dinucleotides - Absence of Methylation at CpG sites = Increase in gene activity / transcription - Acetylation of lysine residues 8. **Relate chromatin packaging to heterochromatin and euchromatin** - Heterochromatin = inactive chromatine : - genes are not transcribed - the DNA encoding the genes remains tightly bound to histones and compacted in nucleosomes - Euchromatin = active chromatin : - genes are transcribed - the DNA encoding the genes loosens from the histones , allowing transcription machinery access to DNA 9. **Predict the effect of chromatin packaging on gene expression** - it's the same thing as the previous objective. - no matter what , if DNA is going to be transcribed then the polymerase and associated proteins need room to function - DNA will need to be unwound / loosened from histones - gene expression , or transcription of DNA into RNA also needs to be unwound from histones 10. **Problem : In order for epigenetic modifications of histones or DNA to act at specific genomic sites , something must target the modification(s) to those sites in chromatin. How many mechanisms can you hypothesize that might target epigenetic modifications to specific DNA sequences?** - 4 ? - acetylation , phosphorylation , methylation , ubiquitination - 1. Methylation : - There are sequence specific binding proteins that are going to target a specific cytosine - Then you have other enzymes that read where those sequence specific binding proteins are located - During transcription , histone H3 is methylated ( by specific histone methylases ) at Lys 4 in nucleosomes near the 5' end of the coding region and at Lys 36 within the coding region. - These methylations enable the binding of histone acetyltransferases ( HATs ) , enzymes that acetylate particular Lys residues. - Cytosolic ( type B ) HATs acetylate newly synthesized histones before the histones are imported into the nucleus. 2. Acetylation of Lysine Residues - sequence-specific binding transcription factors ( cAMP response element binding protein , p53 ) are used to guide histone acetyltransferases ( p300 / CBP SAGA ) 3. Phosphorylation of Serine or Threonine Residues : - asdf 4. Sumoylation ? : - asdf 5. Ubiquitination : - asdf 6. ATP-Dependent Chromatin Remodeling : - asdf - https://en.wikipedia.org/wiki/Epigenetics#Covalent_modifications - Chromatin regions where active gene expression (transcription) is occurring tend to be partially decondensed and are called euchromatin. In these regions, histones H3 and H2A are often replaced by the histone variants H3.3 and H2AZ, respectively - Incorporation of H2AZ stabilizes the nucleosome octamer, but impedes some cooperative interactions between nucleosomes that are needed to compact the chromosome. This leads to a more open chromosome structure that enables the expression of genes in the region where H2AZ is located. 11. **Detail the enzymatic activities of RNA polymerase** - About 17 bp are unwound at any given time. - RNA polymerase and the transcription bubble move from left to right along the DNA as shown , facilitating RNA synthesis. - The DNA is unwound ahead and rewound behind as RNA is transcribed. - As the DNA is rewound , the RNA-DNA hybrid is displaced and the RNA strand extruded. - The RNA polymerase is in close contact with the DNA ahead of the transcription bubble , as well as with the separated DNA strands and the RNA within and immediately behind the bubble. - A channel in the protein funnels new nucleoside triphosphates ( NTPs ) to the polymerase active site. - The polymerase footprint encompasses about 35 bp of DNA during elongation. - Two magnesium ions coordinated to the phosphate groups of the incoming nucleoside triphosphates ( NTPs ) and to 3 Aspartate residues. - One of the magnesium ions facilitates attack by the 3' hydroxyl group on the $\alpha$ phosphate on the NTP - The other magnesium facilitates displacement of the pyrophosphate - Both magnesium ions stabilize the pentacovalent transition state - Changes in the supercoiling of DNA brought about by transcription. - Movement of an RNA polymerase along DNA tends to create positive supercoils ( overwound DNA ) ahead of the transcription bubble and negative supercoils ( underwound DNA ) behind it. - In a cell , topoisomerases rapidly eliminate the positive supercoils and regulate the level of negative supercoiling 12. **Identify the enzymes responsible for RNA synthesis** - RNA polymerase holoenzyme = comprised of 6 subunits , including the $\sigma$ subunit in bacteria - Ribosomes - tRNA - transcription factors / promotors , the whole lot ? 13. **Describe the most abundant types of RNA in eukaryotes** - tRNA , rRNA , pre-mRNA , mRNA , miRNA 14. **Explain the function of each type of RNA in protein synthesis** - tRNA = transfer RNA = hairpin formation , used in translation to bring amino acid to ribosome - rRNA = ribosomal rRNA = used for translation - pre-mRNA = mRNA before processing ( 5' methyl guanosine cap , 3' poly A tail , intron splicing ) - mRNA = mature RNA codes for protein 15. **Distinguish between the modifications to tRNA , pre-mRNA , and rRNA** - All start out as something larger , and they get processed down to a smaller final product - tRNA = spliced and their 5' and 3' ends undergo maturation in the nucleus before export to the cytosol. - CCA is added to 3' end - Aminoacyl tRNA synthetase adds amino acid - pre-mRNA = 5' methyl guanosine cap , 3' poly A tail , intron splicing - rRNA = Small nucleolar ribonucleoproteins ( snoRNPs ) guide the chemical modification and RNase cleavages during rRNA maturation - 90S ➡️ : - pre-40S ➡️ 40S - pre-60S ➡️ 60S 16. **Explain how alternative patterns of pre-mRNA splicing can produce multiple proteins from a single gene** - alternative splicing removes different combination of introns - results in different combination of joined exons - different amino acid sequence = different protein 17. **Problem : What would be the effect of mutating nucleotide #5 from C to G?** - A. No effect on tRNA structure - B. No effect on translation - C. Loss of 5’-CCA-3’ addition - D. Change of tRNA structure - E. Loss of tRNA function - We change the structure , and therefore loose function of the tRNA molecule - D and E = correct 18. **Problem : In the mutant tRNA above, what would be the effect of an additional mutation of the apposed G to C, restoring C:G base pairing in the stem?** - A. No effect on tRNA structure - B. No effect on translation - C. Loss of 5’-CCA-3’ addition - D. Restore tRNA structure - E. Restore tRNA function - This change restores the tRNA back to normal - D and E = correct 19. **Question: Which DNA groove is most ‘informative’ for sequence-specific DNA recognition by proteins?** - Major groove - The alpha helix of some sequence-specific protein has the same width as the major groove of DNA = 1.2 nm 20. **The Lac operon: Expand on the general principles illustrated by the regulation of transcription by nutrients or metabolic products through LacR** - lac I makes repressor tetramer - repressor binds to promotor region - if lactose is present , it binds to repressor causing it to dissociate from promotor region - lac repressor = negative regulation - lac gene transcription needs another signal though before transcription can begin - when bacteria see lactose in their environment , an enzyme makes cAMP - cAMP = positive regulator - eat high sugar and lactose diet - in gut bacteria : - $[\ \text{high glucose}\ ]$ inhibits cAMP production - lac operon is not transcribed - The Lac I gene normally always produces the lac repressor - The lac repressor binds to the lac operator = negative regulator - If allolactose is present , it binds to the repressor and makes it incapable of binding to the lac operator - See Figure 28-18 from 7th edition 21. **The Trp operon: Expand on the general principles illustrated by the regulation of transcription by nutrients or metabolic products through attenuation** - If tryptophan concentration is high : - tryptophan will bind to the the Trp repressor - this blocks the Trp repressor from being able to bind to the promotor - A second "attenuation" signal is needed though , similar to cAMP in lac transcription - a leader sequence trp L needs to bind to the 5' end of the ribosome in order to begin transcription of the Trp operon - this requires forming the "attenuation" / hairpin structure - 2-3 hairpin = easier = continues transcription - 3-4 hairpin = stop signal - If tryptophan concentration is low : - there is no excess tryptophan to bind to the Trp repressor - therefore , the Trp repressor binds to the promotor of the Trp operon , and blocks transcription of the operon 22. **Generalize the type of factors that enhance transcription vs. those that suppress transcription.** - G-C rich ( 3-4 ) region of hairpins suppresses transcription - 2-3 base paired RNA doesn't stall out RNA polymerase 23. **Recognize similarities between the regulation of transcription in bacteria and eukaryotes** - Pages 38 - 56 on Lecture Slides 004.pdf - Initiation : - Bacteria = use sigma factor to recognize start sequence - Eukaryotes = TATA Binding Protein ( TBP ) - Positive / Feed-Forward : - Bacteria = asdf - Eukaryotes = - Negative / Feed-Back Inhibition : - Bacteria = asdf - Eukaryotes = asdf 24. **Explain the mechanisms by which miRNAs / siRNAs block protein expression** - they both bind to different sequences / targets and down-regulate gene expression 25. **Explain the role of CRISPR-Cas9 in gene disruption in vivo** - asdf 26. **Problem :** In procaryotes multiple genes involved in sequential steps of nutrient transport and metabolism are often arranged in tandem along the bacterial chromosome, and are transcribed from a single, regulated promoter. In eucaryotes, however, multiple genes involved in successive steps of transport and metabolism of a single nutrient are located at separate sites on different chromosomes with different promoters. Using the general models of transcriptional regulation, propose how multiple eucaryotic genes (located on different chromosomes) might be coordinately regulated (i.e. regulated together) by a single nutrient. (Try to draw analogies between the regulation of transcription in bacteria and eucaryotes). - asdf 27. **Problem : you have constructed four DNA libraries :** - i.) = a mouse liver cell genomic library - ii.) = a mouse muscle cell genomic library - iii.) = a mouse liver cell cDNA library - iv.) = a mouse muscle cell DNA library - A = how will the liver cDNA library differ from the muscle cDNA library ? - liver cDNA library will have special genes only used in the liver - muscle cDNA library will have special genes only used in the muscle - there might be some overlap of genes used ( housekeeping genes ) - B = How will liver genomic library differ from the muscle genomic library ? - They will be identical , both are cells with the same DNA - C = How will the muscle cDNA library differ form the muscle genomic library ? - Introns will have been removed in the cDNA library