Answered: The synthesis of multiple different…

The synthesis of multiple different protein isoforms from just one gene in eukaryotic cells would most likely result from: polycistronic mRNAs exon shuffling alternative splicing intron shuffling

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter16: Gene Expression

Section: Chapter Questions

Problem 18RQ: IS. Alternative splicing has been estimated to occur in more than 95% of multi-exon genes. Which of...

See similar textbooks

Similar questions

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…
Imagine you are going to label a gene associated with apoptosis in Symbiodiniaceae with a Yellow Fluorescent Protein (YFP). To generate the YFP, you know the pre-MRNA looks as follows: Unspliced YFP premature mRNA Сap 5' UTR Exon 1 Intron Exon 2 Intron Exon 3 3' UTR Poly-A tail If Exon 2 is also required for mRNA stability, what can be predicted from the possible spliced alternative isoforms formed? One of the isoforms will not have a poly-A tail O The alternative splicing of YFP pre-MRNA prevents 5'-capping The MRNA isoform without Exon 2 will be degraded faster than the other isoform Exon 2 will be added to isoform B later to correct the mistake in splicing The protein translated from one of the mRNA isoforms will possess an additional functional domain
NASA has identified a new microbe present on Mars and requests that you determine the genetic code of this organism. To accomplish this goal, you isolate an extract from this microbe that contains all the components necessary for protein synthesis except mRNA. Synthetic mRNAs are added to this extract and the resulting polypeptides are analyzed: Synthetic mRNA Resulting Polypeptides AAAAAAAAAAAAAAAA Lysine-Lysine-Lysine etc. CACACACACACACACA Threonine-Histidine-Threonine-Histidine etc. AACAACAACAACAACA Threonine-Threonine-Threonine etc. Glutamine-Glutamine-Glutamine etc. Asparagine-Asparagine-Asparagine etc. From these data, what specifics can you conclude about the microbe’s genetic code? What is the sequence of the anticodon loop of a tRNA carrying a threonine? If you found that this microbe contained 61 different tRNAs, what could you speculate about the fidelity of translation in this organism?
The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post-transcriptional processing which is essentially way too different from the original code in the DNA template. Given: GUC-CAC-UUA-ACC-CCU-GAG-GAG-AAA-UCG-GCC (Protein with known amino acid sequence) Requirement: Original DNA code. Itemize the steps you would take to get to know the original DNA code of the protein in focus.
Matching type Choices are in the picture 1. simultaneous and rapid process producing mRNA and polypeptide 2. cleaving the polypeptide by adding water 3. three initiation factors are required to commence the process 4. removal of gene segment disrupting the message 5. single mRNA codes for the proteome
Which of the followings indicate the order of procaryotic mRNA degreadation? cleavage of the triphosphate 5′ terminus to yield a monophosphate- 3′ to 5′exonuclease digestion- The endonucleolytic cleavages occur in a 5′ to 3′ direction on the mRNA following the passage of the last ribosme cleavage of the triphosphate 5′ terminus to yield a monophosphate- The endonucleolytic cleavages occur in a 5′ to 3′ direction on the mRNA following the passage of the last ribosme- 3′ to 5′exonuclease digestion The endonucleolytic cleavages occur in a 5′ to 3′ direction on the mRNA following the passage of the last ribosme- cleavage of the triphosphate 5′ terminus to yield a monophosphate- 3′ to 5′exonuclease digestion
Using 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 SUAU
The primary RNA transcript of the chicken ovalbumin gene is 7700 nucleotides long, but the mature mRNA that is translated on the ribosome is 1872 nucleotides long. This size difference occurs primarily as a result of: capping cleavage of polycistronic mRNA removal of poly A tails reverse transcription splicing
A protein has the following amino acid sequence: Met-Tyr-Asn-Val-Arg-Val-Tyr-Lys-Ala-Lys-Trp-Leu-Ile-His-Thr-Pro You wish to make a set of probes to screen a cDNA library for the sequence that encodes this protein. Your probes should be at least 18 nucleotides in length. Q. Which amino acids in the protein should be used to construct the probes so that the least degeneracy results?
Consider the following mature mRNA from a human cell: 5' UAAUGUCGCAAUAACC 3¹ What is the sequence of amino acids in the translated protein? Second letter A First letter С U d A G บบบ UUC P U UUA UUG Leu CUU CUC CUA CUG GUU GUC GUA GUG -Phe Met-Ser-Gln Met-Arg-Lys-Ser Leu Stop-Cys-Arg-Asn-Asn C AUU AUC lle AUA ACA AUG Met ACG Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC GCU GCC GCA GCGJ Ser Pro Thr Ala UAU UAC Tyr UGU UGC. UAA Stop UGA Stop UAG Stop UGG Trp CAC CAGGin CAU His Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. AAU Asn AAA Lys AAG. There is no start codon, resulting in no translated protein. GAU ASP GAC Glu GAA GAG G CGU CGC CGA CGG Cys AGU AGC AGA AGG Arg GGU GGC GGA GGGJ Arg ser Gly MCAG U UCAG с А SCAG U SCAG U Third letter
The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post-transcriptional processing which is essentially way too different from the original code in the DNA template. Given: Val-His-Leu-Thr-Pro-Glu-Glu (Protein with known amino acid sequence) Requirement: Original DNA code. Itemize the steps you would take to get to know the original DNA code of the protein in focus.
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…