Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Textbook Question
Chapter 9, Problem 32P
A eukaryotic
Locate the start codon and stop codon in this sequence.
Determine the amino acid sequence of the polypeptide produced from this
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Using the genetic code table provided below, identify the open reading frame in this mRNA
sequence, and write out the encoded 9 amino acid long peptide sequence:
5'- CGACAUGCCUAAAAUCAUGCCAUGGAGGGGGUAACCUUUU
C
A
G
U
UUU Phe UCU Ser
UUC Phe UCC Ser UAC
UCA Ser UAA
UCG Ser
UAG
UUA Leu
Leu
G
C
CUU Leu
CUC Leu CCC
CUA Leu
CUG Leu
AUU lle
AUC lle
AUA lle
AUG Met ACG
ACU
Thr
ACC
Thr
ACA Thr
Thr
A
UAU Tyr
UGU Cys
Tyr UGC Cys
CCU Pro CAU His CGU Arg
Pro CAC His
Pro CAA Gln
CGC Arg
CGA
Arg
CCA
CCG Pro
CAG Gln CGG
Arg
GUU Val GCU
Ala
GAU
GUC Val GCC
Ala
GAC
GUA Val GCA Ala
GAA
GUG Val GCG Ala GAG
Stop UGA
Stop UGG
AAU Asn
AAC
AAA
AAG
AGU
Asn AGC
G
Lys
Lys
Asp
Asp
Glu
Glu
Stop A
Trp
Ser
Ser
AGA Arg
AGG Arg
GGU Gly
GGC Gly
UCAG
GGA Gly
GGG Gly
с
U
C
A
G
U
C
A
G
U
C
A
G
For the messenger RNA sequence below, find the beginning of the amino acid coding sequence and translate the sequence using the
genetic code provided below.
5' - AAUUAUGGGCAAUAUGCCGGGCcGGUUAAGCG - 3'
Second Letter
A
UGU cys u
UGC
Phe UCU
UU
U UUC
UUA
UAU
Tyr
Ser UAC
UAA
UAG
Leu
UCA
Stop UGA Stop
UUG
UCG
Stop UGG
Trp
CUU
CU
CAU
His
CGU
c cuc
Leu ccc
ССА
CCG
Pro
CÁC
CAA
CAG
CGC
CGA
CGG
Arg
CUA
CUG
Gin
1st
3rd
letter
Ser u letter
AUU
ACU
AAU
AAC
AAA
AAG
Asn
A AUC
AUA
AUG
lle ACC
ACA
AGU
AGC
AGA
AGG
Thr
Lys
Arg
Met ACG
GUU
G GUC
GUA
GUG
GCU
GAU
GAC
GAA
Asp
GGU
Val
GCC
Ala
GGC
Gly
GCA
GGA
Glu
GCG
GAG
GGG
G
Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start):
5’- GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’
By in vitro translating the mRNA, you determined that the translated peptide is 15 amino acids long. What is the expected peptide sequence in single letter abbreviations?
Chapter 9 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 9 - 9.1 Some proteins are composed of two or more...Ch. 9 - In the experiments that deciphered the genetic...Ch. 9 - 9.3 Several lines of experimental evidence pointed...Ch. 9 - Outline the events that occur during initiation of...Ch. 9 - 9.5 A portion of a DNA template strand has the...Ch. 9 - Describe three features of tRNA molecules that...Ch. 9 - Prob. 7PCh. 9 - For each of the anticodon sequences given in the...Ch. 9 - What is the role of codons UAA, UGA and UAG in...Ch. 9 - Compare and contrast the composition and structure...
Ch. 9 - Consider translation of the following mRNA...Ch. 9 - Prob. 12PCh. 9 - Third-base wobble allows some tRNAs to recognize...Ch. 9 - The genetic code contains 61 codons to specify the...Ch. 9 - 9.15 The three major forms of (,, and ) interact...Ch. 9 - The accompanying figure contains sufficient...Ch. 9 - 9.17 The line below represents a mature eukaryotic...Ch. 9 - 9.18. After completing Problem, carefully draw a...Ch. 9 - 9.19 Define and describe the differences in the...Ch. 9 - 9.20. Describe the roles and relationships...Ch. 9 - 9.21 In an experiment to decipher the genetic...Ch. 9 - Identify and describe the steps that lead to the...Ch. 9 - Prob. 23PCh. 9 - Har Gobind Khorana and his colleagues performed...Ch. 9 - 9.25 An experiment by Khorana and his colleagues...Ch. 9 - Prob. 26PCh. 9 - 9.27 The mature transcribed from the human gene is...Ch. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - 9.31 A portion of the coding strand of for a gene...Ch. 9 - A eukaryotic mRNA has the following sequence. The...Ch. 9 - Diagram a eukaryotic gene containing three exons...Ch. 9 - Prob. 34PCh. 9 - 9.35 Table lists and gene sequences for or ...Ch. 9 - Prob. 36PCh. 9 - In terms of the polycistronic composition of mRNAs...Ch. 9 - Prob. 38PCh. 9 - 9.39 Answer the following questions about the...Ch. 9 - 9.40 for each of the following anticodon...Ch. 9 - Prob. 41PCh. 9 - Prob. 42P
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- 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…arrow_forwardThe amino acids, in one-letter symbols and no spaces, coded by the following mRNA sequence is 5’ AAUGGAACGUCGGUACUGCCAUCGCAUUAGUACCAUGGCAAGCUGAAGC 3’arrow_forwardAssume the following portion of an mRNA. Find a start signal, and write the amino acid sequence that is coded for. 5'-GCCAUGUUUCCGAGUUAUCCCAAAGAUAAAAAAGAG 3'arrow_forward
- The bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt-5’ Answer the following questions: Q1. Which is the coding strand? Which is the template strand? [10%] Top-bottom. Bottom-Top. Both can be used as either coding or template for this gene.arrow_forwardThe bacterial gene shorty (sh) encodes for a small protein. The DNA sequence of the sh gene is shown below. The ORF is in CAPITAL LETTERS 5’-tataatgggcttaacaATGAGTAAAAGAGGTCCTTTACTCCGGTATCACCAATAGaaatattatttaa-3’ 3’-atattacccgaattgtTACTCATTTTCTCCAGGAAATGAGGCCATAGTGGTTATCtttataataaatt What is the length in AA’s of the Sh protein? Assume fMet is NOT CLEAVED [10%] 39 AA 13 AA 12 AA 21 AAarrow_forwardTranslate the following mRNA nucleotide sequence into an amino acid sequence, starting at the first base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’arrow_forward
- For the following mRNA sequence (reading from left to right) what will be the amino acid sequence following translation? (Use chart provided) AUGCCAGUUGAAUAA First position U C A UUU UUC UUA UUG CUU CUC CUA CUG U >Phe GUU GUC GUA GUG >Leu >Leu AUU ACU AUC lle ACC AUA ACA AUG Met/start ACG >Val UCU UCC UCA UCG ©2019 Pearson Education, Inc. CCU CCC CCA CCG GCU GCC GCA GCG Second position A >Ser >Pro Thr >Ala UAU UGU U UAC UGC C UAA Stop UGA Stop A UAG Stop UGG Trp G CAU CAC CAA CAG AAU AAC AAA AAG GAU GAC GAA GAG Tyr >His >Gin >Asn >Lys >Asp >Glu CGU CGC CGA CGG AGU AGC AGA AGG G GGU GGC GGA GGG >Cys Arg >Ser >Arg >Gly DOAG SCAG U с А UCA с А G Third position Correct answer not given Met-Val-Tyr-Pro Met-His-Phe-Ala-Arg Pro-Val-Met-Leu-His Met-Pro-Val-Gluarrow_forwardIndicate the amino acid sequence of the protein encoded by the following mRNA molecule. Use the genetic code table and assume that the very first “AUG” the ribosome encounters will serve as the start codon and specify methionine. 5’-AAUUCAUGCCCAAAUUUGGGGCACGAAGCUUCUUAGGCUAGUCCUAAAAAA-3’arrow_forwardIf an antisense RNA is designed to silence the following mRNA sequence, which of the following antisense oligos (a-d) could be used for this purpose? mRNA sequence: 5' UAGGACUAUUAAGGUACACCCAUU 3' O 5' AUCCUGAUAAUUCCAUGUAAAUAA 3' O 5' AAUGGGUGUACCUUAAUAGUCCUA 3' O 5' UAGGACUAUUAAGGUACACCCAUU 3' O 5' UUACCCACAUGGAAUUAUCAGGAU 3¹arrow_forward
- An mRNA transcript is listed below and contains both start and termination codons. Assume that the initial methionine will stay on the polypeptide in this case. What amino acid sequence will be specified during translation? List the amino acids. The start codon is highlighted. 5’ – CAGCCAAGCAUGCUCGCAAAUGGACGUUGAUAUUUUGUC – 3’arrow_forwardTranslate the following mRNA nucleotide sequence into an amino acid sequence, starting at the second base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’arrow_forwardUsing the genetic code table provided below, write out the sequence of three different possible mRNA sequences that could encode the following sequence of amino acids: Met-Phe-Cys-Trp-Glu C A G U C UUU Phe UCU Ser UUC Phe UCC Ser UCA Ser UUA Leu UUG Leu UCG Ser CUU Leu CCU Pro CUC Leu CUA Leu CUG Leu CCG A CAU His CGU Arg CCC Pro CAC His CGC Arg UAU Tyr UGU Cys U UAC Tyr UGC Cys C Stop UGA Stop Stop UGG Trp UAA A UAG G CCA Pro CAA Gln Pro CAG AUU lle AUC lle AUA lle AUG Met ACG G 등등 Gln CGA Arg CGG Arg ACU Thr AAU Asn AGU Ser ACC AAC Asn AGC Ser Thr ACA Thr Thr AAA Lys AGA Arg AAG Lys AGG Arg GUU Val GCU Ala GAU Asp GGU Gly GGC Gly GUC Val GCC Ala GAC Asp GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly U C A G U C A G SUAUarrow_forward
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