Figure 2 is a schematic drawing of ABC gene, which encodes ABC protein. Transeriptional terminator Promoter Iatron 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?

Figure 2 is a schematic drawing of ABC gene, which encodes ABC protein. Transeriptional terminator Promoter Iatron 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?

Name: Figure 2 is a schematic drawing of ABC gene, which encodes ABC protei
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Description: Figure 2 is a schematic drawing of ABC gene, which encodes ABC protein.Transeriptional terminatorPromoterIntron 1Intron 21 100s100 base pairs1100210031004100Positions 200-203 = Start codonPositions 4800-4802 = Stop codonFigure 2.(i)The transcript fi

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 16QP: Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also,...

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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 Leadple
Below is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'
What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'
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 BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINY
c) A gene in a bacteria has the following DNA sequences (the promoter sequence is positioned to the left but is not shown): 5'-CAATCATGGAATGCCATGCTTCATATGAATAGTTGACAT-3' 3'-GTTAGTACCT TACGGTACGAAGTATACTTATCAACTGTA-5' i) By referring to the codon table below, write the corresponding mRNA transcript and polypeptide translated from this DNA strand. 2 Second letter с A UUUPhe UAU Tyr UAC. UGU UGCJ UCU) UCC UCA UUG Leu UCG Cys UUC UUA Ser UAA Stop UGA Stop A UAG Stop UGG Trp G CUU CÚC CCU ССС CAU CGU His САC Pro CC CỦA Leu ССА CAA Arg CGA CUG J CCG) CAG Gin CGG AUU ACU AAU Asn AGU Ser AUC le АСC АCА AAC AAA AGC. Thr JArg AUA AGA AUG Met ACG AAG Lys AGG. GAU Asp GUU) GCU GCC GCA GCG GGU" GGC GGA GGG GUC Val GUA GAC Ala Gly GAA Glu GAGJ GUG ii) If the nucleotide indicated by the highlighted bold letter undergoes a mutation that resulted in deletion of the C:G base pair, what will be the resulting amino acid sequence following transcription and translation? Third letter DUAG DUAG DUAG A. First…
b) Shown below is a very short gene of an unknown bacteria genome (Figure 2). Transcription starts at Transcription Start Site (TSS) and terminates at the Terminator site. TSS 5'TATTATTAACGCATGACGAGCCATGCATTATCGGTATATGCACTGACCCGGRAAGGCTCCTTTTGGAGCCTTTTTT-3' 3' ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTCCGAGGAAAACCTCGGAAAAA-5' Promoter Terminator Figure 2 Based on the double-stranded DNA sequence of terminator, draw the structure of hairpin loop that will be formed during transcription. Illustrate how the hairpin loop structure initiates the termination of transcription.
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.
Telomerase supplies its own template RNA molecule as shown in Figure 3 below: B AAUCCCAAU TTAGGGTTAGGGTTAGGGTTAGGGTTAGGGTTAGGG-W' JAATCCCAATCCCAATCCCAA-X' Figure 3 (i) Label the ends (5' and 3') on the DNA and RNA strand at position X, Y and Z, in the Figure 3. (ii) Draw and explain the two loop structures at the end of telomere.
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?
The coding sequences of Gene F and Gene G are shown by the double-stranded DNA below: Gene F 5' ATGGGAGCACCAGGACAAGATGGATATCATTAG 3' 3' AGTTACCCTC GT GG TCCTGTTCTACCTATAGTAS Gene G Questions: 1. Write down the messenger RNA sequence when Gene F is transcribed. 2. Write down the polypeptide chain when Gene F is completely expressed. 3. Write down the messenger RNA sequence when Gene G is transcribed. 4. Write down the polypeptide chain when Gene G is completely expressed.
What will be the overall anti-codon sequence in tRNA for this mRNA? 5’-GUAGCCUUAUCUAGCGAUCACCGUCCGUAUUACUAGUGGCCAGACUCUUUUCACCAUGUAUAGUUG-3’