What does the genome of mammalian mitochondria contain A total of 37 genes Genes for proteins used in oxidative phosphorylation Genes for rRNA and tRNA Two of the above All of the above
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What does the genome of mammalian mitochondria contain
A total of 37 genes
Genes for proteins used in oxidative phosphorylation
Genes for rRNA and tRNA
Two of the above
All of the above
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- During the biosynthesis of ribosomes, rRNA synthesized in the (fill in) Nuclear membrane nucleolus Cytoplasm Nuclear pore complex And Ribosomal proteins are assembled together into the rirbosomal subunits which are then shipped to the cytoplasm where mature ribosomes are finally assembled. Free ribosomes are localized in the cytoplasm and synthesize cytoplasmic proteins such as [ans2]. While no protein synthesis takes place in the sER, it is an important site for the synthesis of (fill in) DNA tyrosine kinase receptors alcohol integrins tubulin proton pumps RNA phospholipids The sER plays also an important role in detoxification of different types of drugs, which is why ( fill in) Liver Bone Brain skin cells display a large sER. In addition, the sER functions as a reservoir for the storage of (fill in) Iron Iodine Calcium Sodium Ions in many type of cellsCytosine deaminationoccurs ~100 cytosinesper genome per day in a human genome. Eukaryotic cells also contain residues of 5’-methylcytosine, which is involved in regulating gene transcription rates. Mutation of 5’-methylcytosine by deaminationconverts it to thymine. This presents the cell with a much more severe problem than normal cytosine deaminationof cytosine to uracil. Why?Given the following genomic sequence which contains 2 exons and CDNA sequence predidt the amino acid sequence of the exons which are part of the coding region of a protein. How many nucleotides make up the intron? Second letter UAU TV UACJ UAA Stop|UGA Stop A UAG Stop UG Trp UUU1 Phe UUCJ UGU1, UGCJ UCU) Cys UCC Ser UCA ULA UCG CUUT CUC CUA CUG CCU CAUT C4C His CAAT CAGJ OGU CCC Leu CA CGC FArg CGA Gln CGG Pro CCG AUU AUC lle AUA ACU AGU AAU AACJ AAA 1 AAGYS Asn Ser ACC Thr ACA AGC. AGA Arg AUG Mer ACG, AGG GCU GAUT GACJ GGU GGC GUU Asp GUC Val GUA Gly Ala GAA Glu GAGJ GGA GGG GUG GCG G. Genomic antisense sequence: 5'-TAACTGATTCTAGGCTACTCTTTCACCTAA-3" CDNA sequence: 5'-TAACTGATTTCTTTCACCTAA-3" I Use the editor to format your answer First letter 皇 Third letter
- The primary complexes for protein synthesis and very long chain fatty acid degradation in eukaryotes are A O ribosomes and proteasomes O ribosomes and peroxisomes O proteases and lysosomes proteases and proteasomes O ribosomes and mRNA and respectivelyMicrotubules in an animal cell: O a. undergo assembly if the concentration of GTP-bound tubulin dimers is high and a cell is exposed to vinblastine O b. undergo disassembly if the concentration of GDP-bound tubulin dimers is high and a cell is exposed to taxol O c. undergo disassembly if the concentration of GTP-bound tubulin dimers is high and a cell is exposed to colchicine O d. undergo assembly if the concentration of GTP-bound tubulin dimers is high and a cell is exposed to colchicineHow do sigma factors help regulate gene expression during the transition to stationary phase? O Sigma factors create extra nutrients by fixing CO2 and N2 in the air, thereby preventing the onset of stationary phase. O Different sigma factors recognize different promoters. Sigma factors that recognize stress response and catabolic metabolism promoters are expressed during stationary phase and the housekeeping sigma factor is downregulated, Thus, transcription shifts toward stress response and away from growth, O Sigma factors bind the ribosome and block it from translating certain MRNAS that no longer need to be translated in stationary phase O Signa factors are transcriptional inhibitors, they stop transcription entirely so the bacteria don't waste energy on transcription during starvation O Stationary phase is caused by rifampin treatment, and sigma factors block rifampin activity. thereby preventing the onset of stationary phase.
- Searching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4. GAL4 activates the transcription of galactose genes. Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism. The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T. In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added. These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to. The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced. Results showed that only 10 different DNA sequences had GAL4 bound. Since the…Mutations in the nuclear laminproteins cause a large number ofdiseases called laminopathies. What dowe not understand about the nuclearlamina that could account for this fact?A Leu →Ala mutation at a site buried in the core of the enzyme lysozymeis found to be destabilizing. Explain the observed effect of this mutationon lysozyme stability by predicting how enthalpy (ΔH°), conformationalentropy (ΔS°peptide), and the hydrophobic effect (ΔS°solvent) are expected to change for the mutant compared to wild-type lysozyme. Explain how ΔG°for unfolding is affected by your predicted changes in enthalpy or entropy.
- HbS results from the substitution of valine forglutamic acid at the number 6 position in the b chainof human hemoglobin. HbC is the result of a change atthe same position in the β chain, but in this case lysinereplaces glutamic acid. Return to the genetic code table and determine whether single-nucleotide changes can account for these mutations. Then view and examine the R groups in the amino acidsglutamic acid, valine, and lysine. Describe the chemicaldifferences between the three amino acids. Predict how thechanges might alter the structure of the molecule and leadto altered hemoglobin function.Genes participating in informational processessuch as replication, transcription, and translation aretransferred between species much less often than aregenes involved in metabolism. The basis for this inequalityis unclear at present, but one suggestion is that it relatesto the underlying complexity of the two types of processes.Informational processes tend to involve large aggregatesof different gene products, whereas metabolic reactionsare usually catalyzed by enzymes composed of a singleprotein. Why would the complexity of the underlying pro-cess—informational or metabolic—have any effect on therate of horizontal gene transfer?A eukaryotic ribosome contains four diff erent rRNA molecules and ∼82 diff erent proteins. Why does a cell contain many more copies of the rRNA genes than the ribosomal protein genes?
