Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Textbook Question
Chapter 15, Problem 29CTQ
A normal mRNA that reads 5’ -
UGCCAUGGUAAUAACACAUGAGGCCUGAAC- 3’ has an insertion mutation that changes the sequence to 5'
-UGCCAUGGUUAAUAACACAUGAGGCCUGAAC- 3’. Translate the original mRNA and the mutated mRNA, and explain how insertion mutations can have dramatic effects on proteins. (Hint: Be sure to find the initiation site.)
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) A normal mRNA that reads 5'- UGCCAUGGUAAUAACACAUGAAGGCCUGAAC-3' was an insertion mutation that changes the sequence to 5'- UGCCAUGGUUAAUAACACAUGAGGCGUGAAC-3'. Translate the original mRNA and the mutated mRNA and explain how insertion mutations can have dramatic effects on proteins. ( Hint; Be sure to find the initiation site).
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…
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 15 Solutions
Biology 2e
Ch. 15 - Figure 15.11 A scientist splices a eukaryotic...Ch. 15 - Figure 15.13 Errors in splicing are implicated in...Ch. 15 - Figure 15.16 Many antibiotics inhibit bacterial...Ch. 15 - The AUC and AUA codons in mRNA both specify...Ch. 15 - How many nucleotides are in 12 mRNA codons? 12 24...Ch. 15 - Which event contradicts the central dogma of...Ch. 15 - Which subunit of the E. coli polymerase confers...Ch. 15 - The -10 and -35 regions of prokaryotic promoters...Ch. 15 - Three different bacteria species have the...Ch. 15 - Which feature of promoters can be found in both...
Ch. 15 - What transcripts will be most affected by low...Ch. 15 - How do enhancers and promoters differ? Enhancers...Ch. 15 - Which pre-mRNA processing step is important for...Ch. 15 - What processing step enhances the stability of...Ch. 15 - A scientist identifies a pre-mRNA with the...Ch. 15 - The RNA components of ribosomes are synthesized in...Ch. 15 - In any given species, there are at least how many...Ch. 15 - A scientist introduces a mutation that makes the...Ch. 15 - Imagine if there were 200 commonly occurring amino...Ch. 15 - Discuss how degeneracy of the genetic code makes...Ch. 15 - A scientist sequencing itiRNA identifies the...Ch. 15 - If mRNA is complementary to the DNA template...Ch. 15 - In your own words, describe the difference between...Ch. 15 - A fragment of bacterial DNA reads: 3’...Ch. 15 - A scientist observes that a cell has an RNA...Ch. 15 - Chronic lymphocytic leukemia patients often harbor...Ch. 15 - Transcribe and translate the following DNA...Ch. 15 - Explain how single nucleotide changes can have...Ch. 15 - A normal mRNA that reads 5’ -...
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- Given the following mRNA transcript: 5’-UUUGGCAUGGGUAUCGUAGAGAUGGAAUUCAUAGUGGAGUAA-3’ What is the one-letter abbreviation of the protein product of the mRNA transcript?arrow_forwardWhat will be the overall anti-codon sequence in tRNA for this mRNA? 5’-GUAGCCUUAUCUAGCGAUCACCGUCCGUAUUACUAGUGGCCAGACUCUUUUCACCAUGUAUAGUUG-3’arrow_forwardUsing 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 Garrow_forward
- Template strand of DNA is: 3’ TACATAACCGGGCCCATATCGGCCATTTGC5’. 2a). Following transcription, what is the total number of codons in the mRNA transcript? 2 b). Where is the start codon located in this mRNA transcript? 2c). Following translation of this mRNA transcript, how many amino acids will the proteincontain and identify the amino acids sequence of this gene from a genetic code table*.*Note= using a genetic code tablearrow_forwardAnother thalassemic patient had a mutation leading to the production of an mRNA for the β chain of hemoglobin that was 900 nucleotides longer than the normal one. The poly(A) tail of this mutant mRNA was located a few nucleotides after the only AAUAAA sequence in the additional sequence. Propose a mutation that would lead to the production of this altered mRNA.arrow_forwardGiven the following DNA sequence of the template strand for a given gene: 5' TTTCCGTCTCAGGGCTGAAAATGTTTGCTCATCGAACGC3' Part A ) Write the mRNA that will be transcribed from the DNA sequence above (be sure to label the 5' and 3' ends). Part B ) Use the genetic code to write the peptide sequence translated in a cell from the mRNA in part A. Please use the 3 letter abbreviation for each amino acid. Part C: How would the peptide synthesized in a cell be different if the mRNA was translated in vitro (i.e. not in the cell)?arrow_forward
- The mRNA formed from the repeating tetranucleotide UUACincorporates only three amino acids, but the use of UAUC incorporates four amino acids. Why?arrow_forwardDetermine the amino acid sequence for a polypeptide coded for by the following mRNA transcript (written 5'-> 3'): AUGCCUGACUUUAAGUAGarrow_forwardReferring to the genetic code presented in Figure , give the aminoacids specified by the following bacterial mRNA sequences. Q. 5′ –GUACUAAGGAGGUUGUAUGGGUUAGGGG ACAUCAUUUUGA–3′arrow_forward
- Referring to the genetic code presented in Figure , give the aminoacids specified by the following bacterial mRNA sequences. Q. 5′ –UUUGGAUUGAGUGAAACGAUGGAUGAAAG AUUUCUCGCUUGA–3′arrow_forwardAs described earlier, DNA damage can cause deletion or insertion of base pairs. If a nucleotide base sequence of a coding region changes by any number of bases other than three base pairs, or multiples of 3, a frameshift mutation occurs. Depending on the location of the sequence change, such mutations can have serious effects. The following synthetic mRNA sequence codes for the beginning of a polypeptide: 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCUUAUC-AUGUUU-3′ First, determine the amino acid sequence of the polypeptide. Then determine the types of mutation that have occurred in the following altered mRNA segments. What effect do these mutations have on the polypeptide products? a. 5′-AUGUCUCCUACUUGCUGACGAGGGAAGGAGGUGGCUUAUCA-UGUUU-3′ b. 5′-AUGUCUCCUACUGCUGACGAGGGAGGAGGUGGCUUAUCAU-GUUU-3′ c. 5′-AUGUCUCCUACUGCUGACGAGGGAAGGAGGUGGCCCUUAUC-AUGUUU-3′ d. 5′-AUGUCUCCUACUGCUGACGGAAGGAGGUGGCUUAUCAU-GUUU-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
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