Topic A:Molecular biology, biochemistry, cell biology Flashcards
(124 cards)
(2014) Cdc25 was identified in Lee Hartwell’s screen for cell division cycle mutants in budding yeast. Assuming the mutation ihibits the function of the protein, what effect would you expect it to have on cell cycle progression? Explain your reasoning
The wee1 kinase inhibits cyclin which regulates progression of the cell cycle. Cdc25 phosphotase reverses wee1 inhibition and allows cell cycle progression. Thereby a mutation in cdc25 could result in an arrest of the cell cycle.
(2014) If a diploid cell in G1 phase of the cell cycle contains 4 picograms (4 x 10−15 grams) of DNA, how much DNA would you expect to find in a gamete?
2 x 10−15 grams DNA
Consider a G-protein linked receptor in a signaling pathway that leads to the activation of adenylyl cyclase. What happens to cAMP levels in the cell after ligand binding to the receptor?
cAMP levles increase and cAMP is synthesized from ATP by adenylyl cyclase
Leptin is a circulating hormone important for the regulation of body weight. This protein is synthesized and secreted by adipose (fat) cells. Although many cells are exposed to leptin, only a subset of cells seem to respond to the hormone. What is the simplest explanation for why some cells respond to this hormone and others do not?
some cells have receptors that bind the hormone allowing it ot respond to the signal whereas other cells do not have receptors for this hormone and thus the hormone has no effect
The first human oncogene identified encodes a mutant form of the Ras protein in which the 12th amino acid is glycine instead of valine. This mutation decreases the GTPase activity of Ras. When the cell expressing the mutant Ras protein is exposed to EFG, what will happen to the MAP kinase activi
n MAPk/ERK pathway, ligand EGF binds to EGF receptor that activates the receptor to phos- phorylate itself, the phosphorylated receptor binds to the adaptor protein, which couples the signal downstream and activates the MAP kinase. MAPK then attaches phosphates to target proteins like Myc transcription factor and alters gene transcription and cell cycle preoduction. In the cell with mutatnt Ras, MAPK activity will increase, decreasing GTPase activity of Ras.
What type of cell-cell signalling involves direct interactions between cells? Name one class of proteins required for this type of signalling.
juxtacrine signaling; extracellular matrix proteins are required
A) Which of the following mutations would you expect might be associated with cancer? Briefly explain your reasoning.
a. A mutation that causes Ras to hydrolyze GTP less efficiently
b. A mutation that casues Ras to hydrolyze GTP more efficiently
c. A mutation that prevents Ras from binding GTP
a
Mutations in the Ras GTPase protein.Do you expect this mutation would be dominant or recessive? Briefly explain your reasoning.
dominant because loss of function is leading to uncontrolled prolifera
A Ras mutation that occurs spontaneously in an epithelial cell in the colon would increase the risk of colon cancer. Would this increased risk be inherited? Why or why not?
No, only a germline mutation is inherited
Which of the following proteins couple a chemical cycle of ATP binding, hydrolysis, and release to a mechanical cycle?
a. Adenylyl cyclase
b. cAMP-dependent protein kinase (PKA)
c. Dynein
d. Ras GTPase
e. a potassium channel
f. kinesin
g. MAP kinase
a. catalyse the conversion of adenosine triphosphate (ATP) to 3’,5’-cyclic AMP (cAMP) and pyrophosphate?
In which of the following polymers are the monomers connected by non-covalent bonds?
a. DNA
b. RNA
c. Proteins
d. Microtubules
d
Protein kinases typically remove phosphates from:
a. adenosine
b. serine
c. threonine
d. tyrosine
e. none of the above
e
Describe the core splicing sequencing elements that are required for the splicing of a mammalian intron.
5’ splice consensus - (AC)AG/GT(AG)AGT
3’ splice site consensus-YYYYYYYYYYNCAG/G Branchpoint consensus- YNCURAC
What is the predominant RNA polymerase that generates mammalian miRNAs? What se- quence features do miRNAs and mRNAs share?
RNA pol II generates mammalian miRNAs
hairpin, stem loops, 3’ poly A tail and 5’ cap, 5’ splice consensus sequence
Describe the main RNA cleavage steps in the nucleus and cytoplasm that are involved in the processing of mammalian microRNAs and at least one protein component involved in each cleavage step.
A gene is transcribed into pri-miRNA in the nucleus. Drosha and Pasha cleave the 5’ cap and 3’ poly-A tail . The resulting product is pre-miRNA. Ran-GTP, Exportin-5 exports pre-miRNA into the cytosol. Dicer cleaves the hairpin loop, and an miRNA/miRNA* duplex is formed.Argo recognizes the miRNA and form the miRNP silencing complex(RISC).
Describe one epigenetic modification that is associated with active gene transcription and one that is associated with repression of transcription.
histone methylation is typically associated with repression and histone acetylation is associated with activation
Provide 4 examples of histone modifications that can influence gene expression.
H3K27me1 - mono-methylation = activation H3K27ac - acetylation = activation H3K27me2 - di-methylation = repression H3K27me3 - tri-methylation = repression H3K4me3 - tri-methylation = activation H3K4me1 - mono-methylation = activation H3K9me1 - mono-methylation = activation H3K9me2 - di-methlaytion = repression H3K9me2 - tri-methylation = repression H3K9ac - acetylation = activation H3K14ac - acetylation = activation H3K36me3- activation
Describe the difference between a eukaryotic gene promoter and enhancer.
Both are needed for transcription. A promoter is near the gene (at TSS), and general TF involved in pre-intiation complex bind to the promoter. An enhancer is upstream of the gene(up to kb away) and specific TF bind here. Enhancers can interact w/ promoter by bringing sites into direct contact via loop-like structure.
Give two examples of genome-wide approaches that can be carried out to identify regions of active transcription in a eukaryotic cell
- Nuclear run on experiments like GRO-seq (GRO=genomic run on) -isolate cell nuclei, label with nucleotides -genes in the process of transcription are detected by hybridization of extracted RNA to gene specific probes on a blot -GRO allows for calculation of transcription rates
- DNase I hypersensitive sites sequencing (DNase seq)– cuts open chromatin, if it cuts, you see a peak, indicating active transcription
- ChIP-Seq for acetylated/active histone markers
How do the mechanisms of translation initiation differ between prokaryotes and eukaryotes?
Shine-Delgarno sequence in prokaryotes helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning it with the start codon In eukaryotes, initiation of translation usually involves the interaction of certain key proteins with a special tag bound to the 5’-end of an mRNA molecule, the 5’ cap, as well as with the 5’ UTR (Kozak sequence which occurs on eukaryotic mRNA and has the consensus (gcc)gccRccAUGG)
What is the typical nucleosome positioning in the promoter of a housekeeping gene
This canonical promoter chromatin structure = ( 1 nucleosome/Nucleosome Free Region/+ 1 nucleosome) .NFR width correlates somewhat with transcription levels in yeast.
Name two similarities and two differences between histone methylation and acetylation
Similarities: 1) both modify lysine 2) both are associated with activation 3) act on histone tails
Differences: 1) gene regulation (acetylation–activation; methylation–mostly repression) 2) can add 1,2,or 3 methyl groups to lysine, but only one acetyl group 3) arginine can by methylated (only 1 or 2 times) but not acetylated 4) electrical charge: methylation is neutral, acetylation is negatively charged
For each question provide a molecular/cellular explanation for the described phenomena?
A loss-of-function mutation that has a dominant phenotype.
If the remaining wild-type allele cannot compensate for the loss-of-function in the mutated allele, then the dominant phenotype will be the loss-of-function mutant (haploinsufficiency)
Explain why A null mutation that has no obvious phenotype.
If heterozygous, the null mutant is compensated for by the wild-type allele. If homozygous, there may be redundant pathway or a second copy of the gene may exist elsewhere.