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- Transcription and translation are separate processes in gene expression; however, they have similarities. The following terms all relate to translation. Which of these has a role that is most similar to that of the transcription start site during transcription? A)Start codon B)Stop codon C)tRNA D)Amino acidMutations in the ribosomal protein gene RPS19 have been found to be associated with Diamond-Blackfan anemia (DBA), a rare syndrome that disrupts erythropoiesis, the process by which new red blood cells are produced. Which of the following provides the most likely explanation for the connection between a mutation in RPS19 and a disruption in erythropoiesis? A B с D The ribosomes in the cell are unable to package and transfer the components that are required for erythropoiesis. The ribosomes in the cell are unable to facilitate the movement of nutrients that are re- quired for erythropoiesis across the cell membrane. The ribosomes in the cell are unable to correctly synthesize the proteins that are re- sponsible for erythropoiesis. The ribosomes in the cell are unable to effectively produce the ATP that is required for erythropoiesis.The following DNA sequence has been determined from DNA isolated from a bit of prehistoric amber material (picture). It corresponds to a complete transcriptional unit without introns. Use the Genetic Code to predict the primary sequence of the polypeptide encoded by this preserved DNA. (Show relevant molecular intermediates, and provide detailed and appropriate labels)
- Which of the following regulatory sequence allows transcription to continue? a) Sequence 4 b) Sequence 1 c) Sequence 2 d) Sequence 3a) Define the term gene expression b) State 4 difference between prokaryotes and eukaryotes gene expression c) state the importance of regulating gene expressiona) Examine the nucleotide sequence below, and determine the amino acid sequence encoded by this mRNA. (2) 5' CCUCCGGACCGGAUGCCCGCGGCAGCUGCUGAACCAUGGCCCGCGGGUGAGCCAAGGAGGAGGGC 3' b) What would be the consequence of a mutation that resulted in changing the underlined nucleotide to a G? (2) Second base U G. Consensus sequences functioning in transcription or translation (5-3): UGU UAU UCU Phe UCC Ser UCA Leu UCG UUU Tyr Cys TATA box (-25) TATAAA UUC UAC UGC UAA Stop UGA Stop A UAG Stop UGG Trp G UUA TFIIB recognition element /c/c/¢CGCC UUG TATAAT CGU CAU His CAC Pro CAA Gln CAG -10 (Pribnow) sequence CUU CCU CC Leu CCA CGC Arg CGA CUC TTGACA -35 sequence CỦA CUG CCG CGG Shine-Dalgarno sequence (Ribosome binding site) UAAGGAGGU YYANT/AYY AGU Asn AGC AUU ACU AAU Ser Initiator element AUC lle ACC Thr AAC AGA Lys AGG AUA ACA AAA lA AGLGU ^/G AGU Arg Intron 5' splice site AUG Met ACG AAG CAGIG GGU GAU Asp GAC Intron 3' splice site GCU GUU GCC Val GCA GGC Gly GGA GUC AAUAAA Ala Cleavage site…
- Why nonhomologous end-joining (NHEJ) is helpful ?1. (a) By binding one L-tryptophan molecule/monomer, the trp repressor binds to DNA to sup- press synthesis of L-tryptophan in E. coli. Below is the amino acid sequence of the helix - reverse turn - helix region of the trp repressor that binds to DNA compared to the sequence of the corresponding DNA binding motif of the Prl protein. A diagram of the trp repressor dimer is also shown. Trp Prl Trp Prl 80 -Gly-Glu-Met-Ser-Gln-Arg-Glu-Leu-Lys-Asn-Glu-Leu-Gly-Ala-Gly-Ile- -Ser-Glu-Glu-Ala-Lys-Glu-Glu-Leu-Ala-Lys-Lys-Cys-Gly-Ile-Thr-Val- trp helix 5 70 trp helix 4 Prl helix 80 Prl helix Ala-Thr-Ile-Thr-Arg-Gly-Ser-Asn-Ser-Leu-Lys-Ala-Ala- Ser-Gln-Val-Ser-Asn-Trp-Phe-Gly-Asn-Lys-Arg-Ile-Arg- reverse turn 90 Comparing the two protein sequences above, identify all amino acid pairs that differ in electrostatic charge due to proton dissociable groups (assume pH 7). Indicate the charge of both residues for each such pair. (b) Circle the pair of residues for which the electrostatic charge due to…RNA polymerases generally require a primer to begin transcription. (T) (F) The Death Cap Mushroom Amanita phalloides is toxic because of its ability to produce alpha-amanitin, which is an inhibitor of RNA Polymerases I and III. (T) (F) In bacteria, transcription and translation can occur simultaneously. (T) (F) In eukaryotes, transcription and translation can occur simultaneously (T) (F) RNA polymerase II has no form of proofreading activity. (T) (F) Sigma factors specify binding of bacterial RNA Polymerases to specific promoters (T) (F) An E. coli strain with mutations in genes encoding both the dam methylase and the RecA protein would likely be inviable (dead) (T) (F) An E. coli culture grown in a pure (100%) N2 atmosphere would likely have a lower rate of mutations than a culture grown under normal conditions (~30% O2 and 70% N2) (T) (F) Non-homologous end joining repairs double strand DNA breaks with no loss of information, restoring the original…
- If the sequence of a coding strand of a gene is 5' ATGGCAT 3', the sequence of the MRNA would be: 5’AUGGCAU 3' ОЗ ТАССGTA 5' 3' UACGGUA 5' 5' ATGGCAT 3' O 5' UACGGUA 3'Which of the following among A-Cis not needed for bacterial transcription Y A) O core RNA polymerase function B) Oa promoter C) O signma factor D) OA-C are all required for transcriptionThe following represent deoxyribonucleotidesequences in the template strand of DNA:Sequence 1: 5'@CTTTTTTGCCAT@3'Sequence 2: 5'@ACATCAATAACT@3'Sequence 3: 5'@TACAAGGGTTCT@3'(a) For each strand, determine the mRNA sequencethat would be derived from transcription.(b) determine the amino acidsequence that is encoded by these mRNAs.(c) For Sequence 1, what is the sequence of the codingDNA strand?