w is the diagram of a eukaryotic gene that encodes a protein. The promoter and n start sites are indicated on the map. The numbers shown correlate with nucleotide on along the DNA sequence. Transcriptional Transcription start site occurs at base position 1 terminator 5100 ter Exon 1 Intron 1 Exon 2 Intron 2 Exon 3 100 1100 2000 3000 4000 Positions 200-202 Start codon positions 4801-4803 Stop codon he above sequence where RNA polymerase will bind? in bases) will the primary mRNA transcript of this gene be? in bases) is the processed mRNA transcript?
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- Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. Question 6: How many amino acids are present in the wild-type human B-globin protein? = exons = introns Transcription termination site (also poly A site) Promoter Start of transcription 3' 5' ATG 50 TẠC TAA 126 132 |ATT 90 130 222 850 3 5' Start codon Stop codon А. 438 В. 146 C. 620 D. 206 © 2013 John Wiley & Sons, Inc. All rights reserved.Shown below are different regions of an eukaryotic gene. Which of the above regions of a gene will be transcribed? Or which regions will be part of the new RNA molecule that is synthesized during transcription? Select all that apply. Promoter Intron Exon Transcription stop sile Transcription start site 5- 31 ATG 75 TAC TAA ATT 3' 5' 50 100 300 200 228 50 3' 5' Start Splice donor codon Splice аcсeptor Splice acceptor Splice donor Stop codon Exons Ribosomal binding site Promoter IntronsThe diagram below shows an imaginary eukaryotic structural gene containing two exons. The exon nucleotides are numbered beginning at the transcription start site and a portion of the intron is not shown to save space: Help Center? transcription start site promoter U STACAGTATAAATGAATTAATTGACGTATGTCAATCGGTAAGT...TCAGGTACT U UUU} Phe UUG} Leu exon 1 3 ATGTCATATTTACTTAATTAACTGCATACAGTTAGCCATTCA...AGTCCATGAATGACTTATGTGCGGTTATTTACTGAT... Second letter C Predict the amino acid sequence of the polypeptide encoded by this structural gene. The genetic code is provided below.
- Figure 2 is a schematic drawing of ABC gene, which encodes ABC protein. Transeriptional terminator Promoter Intron 1 Intron 2 1 100 s100 base pairs 1100 2100 3100 4100 Positions 200-203 = Start codon Positions 4800-4802 = Stop codon Figure 2. (i) The transcript first produced by ABC gene would be approximately how many nucleotides long?Please match the gene element with the the correct definition. control elements 5' UTR 3' UTR Promoter coding region +1 site [Choose ] stretch of DNA that contains the sequence that will encode the protein the active form of RNA polymerase region of DNA upstream of transcription start site that is recognized by RNA polymerase region of mRNA on the 5' end that goes untranslated RNA polymerase and the sigma factor regions of DNA that increase or decrease transcription rate the first nucleotide to be transcribed region of mRNA on the 3' end that goes untranslated a transcription initiation factor that helps RNA polymerase recognize the promoter the inactive form of RNA polymerase [Choose ] [Choose ]Shown below is the genomic structure of the human B-globin gene. The numbers within the boxes indicate the length in nucleotides of each region. = exons Transcription termination site (also poly A site) = introns Promoter Start of transcription 3' 5'. TAA ATG 50 TAC 130 222 850 126 132 90 ATT 5' 3 QUESTION 3: What is the length in nucleotides of the mature, processed B-globin mRNA? A.620 B.980 C.438 D.1600
- 5 5 S 6 5 5 5 6 U 6 U 6 5:14 PM | 0.2KB/s HHHHH R R U RUUR ARU AP AP R U U R R AP R R R AP MOLECULAR...GENETICS. Describe gene regulation at transcription level. Explain the role of antsense RNA in control mechanism. Describe translational control mechanisms. Describe common DNA damages. Distinguish excision and mismatch repair. Describe the role of recA protein in recombination repair Elaborate on SOS repair mechanism. Define thymine dimer. How are they formed and repaired? Describe the molecular basis of mutation. 11 Leu+ Met+ Arg+ Write a detailed note on spontaneous mutation. Explain about mutant detection methods. Define reverse mutation. Describe the mechanism underlying Intragenic and intergenic suppressor mutations Describe the transposition mechanisms. 13 Vo LTE UNIT IV Time (Min) Describe the process of generalised transformation occurring in bacterial chromosome and plasmid. Elaborate on molecular mechanism and significance of transformation 22 Describe the process of…Identify the DNA elements and protein factors, #1-7, in the figure below that are involved in the initiation and elongation of transcription at a eukaryotic promoter of a gene TRANSCRIPTION ON RNA PROCESSING PefA A eukaryotic promoter commonly includes a TATA box (a nucleotide sequence TRANSLATION Ritoom containing TATA) about 25 nucleotides upstream from Nontemplate (coding) strand of DNA 7 Pupeptide 1 the transcriptional start point. Nontemplate (coding or sense) strand DNA TATAAAA ATATILTI 5 2 3 Start point 2 Several transcription TCCAA 3' 5' factors, one recognizing the TATA box, must bind 3' end to the DNA before RNA 4 polymerase Il can bind in the correct position and orientation. UCCA 3 5' 3' 3 Additional transcription GGTT factors (purple) bind to the DNA along with RNA 5' Direction of transcription polymerase II, forming the transcription initiation 3 complex. RNA polymerase Il then unwinds the DNA double helix, and RNA synthesis begins at the start point on the temple strand.…The following double-stranded DNA sequence is part of a hypothetical yeast genome which contains a very small gene. Transcription starts at the Transcription Start Site (TSS), proceeds in the direction of the arrow and stops at the end of the Transcription Terminator (green box). 5' 3' TSS CTATAAAAATGCCATGCATTATCTAGATAGTAGGCTCTGAGAAATTTATCTCACT | | | | | | | | | | GATATTTTTACGGTACGTAATAGATCTATCATCCGAGACTCTTTAAATAGAGTGA - 5' PROMOTER TERMINATOR 3' a) Which strand (top or bottom) is the template strand? Explain why. b) What is the sequence of the mRNA produced from this gene? Label the 5' and 3' ends. c) What is the sequence of the protein produced from the mRNA? d) If a mutation (an insertion) were found where a T/A (top/bottom) base pair were added immediately after the T/A base pair shown in red, what would be the sequence of the mRNA? What would be the sequence of the protein?
- Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices 1.Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds. 2.Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. 3.Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. 4.Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination.BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation LeadpleA bacterial species has a hypothetical sigma promoter that has the following sequence: TTGGCA - 18 bases - TATAAT What change in the level of transcription would there be if the sequence was mutated to: TTCGCA -18 bases -TATAAT Group of answer choices 1.The mutation would inhibit the promoter thereby inhibiting transcription 2.No change the consensus TATAAT sequence in the same. 3.The mutation would move the promoter away from consensus and reduce the level of transcription 4.The mutation would bind the promoter to the consensus and produce normal levels of transcription