Biological Science (7th Edition)
7th Edition
ISBN: 9780134678320
Author: Scott Freeman, Kim Quillin, Lizabeth Allison, Michael Black, Greg Podgorski, Emily Taylor, Jeff Carmichael
Publisher: PEARSON
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Chapter 19, Problem 6TYU
Summary Introduction
To review:
Consider discovering a loss-of-function mutation in a eukaryotic gene. The gene’s
The location where the mutation might occur and to study the behavior of the mutation from the given case.
Introduction:
A loss-of-function mutation occurs in a eukaryotic gene. The reason for mutation is determined by covering the entire length of the DNA (deoxyribonucleic acid) from the point of initiation to its termination.
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Check out a sample textbook solutionStudents have asked these similar questions
Here is a eukaryotic gene. The numbers given are base pairs of exon and intron.
How long in bases will the pre mRNA transcript be? Explain briefly.
What is the maximum number of amino acids that could make up the protein product from the final mRNA? Explain briefly.
The following four mutations have been discovered in a gene that has more than 60 exons and encodes a very large protein of 2532 amino acids.
Indicate which mutation would likely cause a detectable change in the size of the mRNA and/or the size of the protein product. Consider a detectable change to be >10% of the wild-type size. A table of the
genetic code is shown below.
First letter
0 00
U
O
A
บบบ
UUC
UUA
UUG
U
CUU
CUC
CUA
CUG
Phe
GUU
GUC
GUA
GUG
Leu
>Leu
AUU
AUC lle
AUA
AUG Met
>Val
UCU
UCC
UCA
UCG
CCU
CCC
CCA
CCG
ACU
ACC
ACA
ACG
GCU
GCC
GCA
GCG
Second letter
C
Ser
Pro
Thr
Ala
CAU
CAC
CAA
CAG
UAU
UGU
Tyr
UAC
UGC
UAA Stop UGA
UAG Stop UGG
AAU
AAC
AAA
AAG
A
GAU
GAC
GAA
GAG
His
Gin
Asn
Lys
Asp
G
Glu
CGU
CGC
CGA
CGGJ
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
O AAG576UAG (changes codon 576 from AAG to UAG)
Cys
Stop
Trp
O GUG326AUG (changes codon 326 from GUG to AUG)
Arg
Ser
Arg
Gly
DUAG DUA G
DCAG DO AG
deletion of codon 779
insertion of 1000 base pairs into the sixth intron
(this particular…
2a) Suppose you have a gene in which a single base substitution has created the nonsense mutation 5'TAA3' (which will be transcribed into 5'UAA3' in the mRNA - but recall that mutations are changes in the DNA sequence). Name all the amino acids that could have been coded for by the original, unmutated codon at that position in the gene.
Chapter 19 Solutions
Biological Science (7th Edition)
Ch. 19 - What is chromatin? a. the histone-containing...Ch. 19 - Prob. 4TYKCh. 19 - Compare and contrast the items in each pair: (a)...Ch. 19 - Prob. 6TYUCh. 19 - Prob. 8TYUCh. 19 - Prob. 9TYPSSCh. 19 - 10. QUANTITATIVE Imagine repeating the experiment...Ch. 19 - Prob. 12PIATCh. 19 - Prob. 13PIATCh. 19 - Prob. 14PIAT
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- For a specific type of mutation at a given location in a particular gene, identify whether it will affect the size of the mRNA, the protein, or both. How would the mutant appear on a gel in comparison to the originalarrow_forwardConsider the following original coding sequence of a gene that codes for a short 5- amino acid polypeptide: 5'-ATGGGCTCGAACTCATAA-3' Using the genetic code and the amino acid table below, which of the following sequences arises from a non-conservative missense mutation in the original sequence shown above? First base in codon U U A UUU UUC- UUA UUG- CUU CUC CUA CUG- U Phe (F) Leu (L) Leu (L) Second base in codon Val (V) UCU - UCC UCA UCG CCU CCC CCA CCG AUU ACU- AUC Ile (1) ACC AUA- ACA AUG Met (M) start ACG GUU GCU- GUC GCC GUA GCA GUG GCG- C Ser (S) Pro (P) Thr (T) Ala (A) UAU UAC UAAT UAG CAU CAC CAA CAG AAU AAC AAA AAG GAU GAC GAA GAG A Tyr (Y) STOP His (H) Gln (Q) Asn (N) Lys (K) Asp (D) Glu (E) G UGU UGC UGA STOP UGG Trp (W) Cys (C) CGU CGC CGA CGG AGU AGC AGA 1 AGG GGU- GGC GGA GGG Arg (R) Ser (S) Arg (R) Gly (G) U C A G U C A G U C A G U C A G Last base in codonarrow_forwardTay Sachs disease is an autosomal recessive disease in which a protein – Hex A - is abnormal. To make the Hex A protein: The promoter and transcription termination sites are 33,000 base pairs apart. The Hex A protein has 600 amino acids 5’ and 3’ UTR’s are each 500 bp long. a)How many base pairs would you expect in the final mRNA? Show your work b)How many bases were spliced out? Show your workarrow_forward
- ) 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).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_forwardConsider the tryptophan codon 5′ - UGG - 3′ in the standard genetic code . Can a single base change in this codon create a synonymous mutation? Can a single base change in this codon create a nonsense codon?arrow_forward
- Introns in protein-coding genes of some eukaryotes are rarely shorter than 65 nucleotides long. What might be a rationale for this limitation?arrow_forward2b) Prokaryotic cells can and do produce "polycistronic" mRNAs, which have multiple independent coding sequences coding for separate proteins on the same mRNA strand. Eukaryotic cells don't have polycistronic mRNAs, because only the first (most 5') coding sequence on such an mRNA would ever be translated in a eukaryotic cell. Explain why this is the case - why wouldn't a second, more 3' coding sequence on an mRNA be translated in a eukaryotic cell?arrow_forwardA mutation has occurred to a wild type mRNA sequence: Wild Type: 5’-AUG-UUG-CAA-GCG-3’ The new mutated sequence: 5’-AUG-UUG-UAA-GCG-3’ What kind of mutation has occurred? a)Silent mutation b)Missense mutation c)Nonsense mutation d)Frameshift mutationarrow_forward
- Given 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_forwardWhich of the following mutations in the protein-coding region of a gene is more likely to lead to complete loss of function of the encoded protein: an insertion of six nucleotides or a deletion of two nucleotides? Briefly explain your answer.arrow_forwardGiven the following Wild Type and Mutated DNA sequences: 1.) Identify where the base pair change occurs ( what letter changed?) 2.) For BOTH sequences, write the mRNA strands, define the codon regions and amino acid sequences. 3.) Describe what kind of mutation has occurred (missense, nonsense, or silent), and what effect this may have on the protein. Wild Type DNA Sequence: 3' - AGGCTCGCCTGT - 5' Mutated DNA Sequence: 3' - AGTCTCGCCTGT - 5'arrow_forward
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