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The exons and introns of a gene are shown below. Introns are shown as black bars. Exons are numbered white boxes. "nt" is abbreviation for nucleotides. For exon 1, the number of nucleotides labeled in the figure only include the sequence from the start codon to the 3' end of the exon. For exon 5, the number of nucleotides labeled in the figure only include the sequence from the 5' end of exon 5 to the stop codon. In each case, the number of nucleotides does include the start or the stop codon itself. Stop codon Start codon 4. 50 nt 150 nt 180 nt I Alternative splicing of this gene produces removes introns and joining different exons. Four different mRNA transcripts are formed by alternative splicing. All four transcripts utilize the same start codon (in Exon 1) and the same stop codon (in Exon 5) for translation. Based on this information, can you predict the exon combinations of the four transcripts and the length (l.e. number of amino acids) of each of the four proteins translated from the four transcripts? Please select all the correct answer options. 120 nt 40 nt Exons 1/4/5, length of the coding protein: 149 amino acids Exons 1/5, length of the coding protein: 99 amino acids Exons 1/2/3/5, length of the coding protein: 129 amino acids Exons 1/2/3/4/5, length of the coding protein: 179 amino acids Exons 1/2/4/5, length of the coding protein: 163 amino acids

The exons and introns of a gene are shown below. Introns are shown as black bars. Exons are numbered white boxes. "nt" is abbreviation for nucleotides. For exon 1, the number of nucleotides labeled in the figure only include the sequence from the start codon to the 3' end of the exon. For exon 5, the number of nucleotides labeled in the figure only include the sequence from the 5' end of exon 5 to the stop codon. In each case, the number of nucleotides does include the start or the stop codon itself. Stop codon Start codon 4. 50 nt 150 nt 180 nt I Alternative splicing of this gene produces removes introns and joining different exons. Four different mRNA transcripts are formed by alternative splicing. All four transcripts utilize the same start codon (in Exon 1) and the same stop codon (in Exon 5) for translation. Based on this information, can you predict the exon combinations of the four transcripts and the length (l.e. number of amino acids) of each of the four proteins translated from the four transcripts? Please select all the correct answer options. 120 nt 40 nt Exons 1/4/5, length of the coding protein: 149 amino acids Exons 1/5, length of the coding protein: 99 amino acids Exons 1/2/3/5, length of the coding protein: 129 amino acids Exons 1/2/3/4/5, length of the coding protein: 179 amino acids Exons 1/2/4/5, length of the coding protein: 163 amino acids

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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The exons and introns of a gene are shown below. Introns are shown as black bars. Exons are numbered white boxes. "nt" is abbreviation for nucleotides. For exon 1, the number of nucleotides labeled in the figure only include the sequence from the start codon to the 3' end of the exon. For exon 5, the number of nucleotides labeled in the figure only include the sequence from the 5' end of exon 5 to the stop codon. In each case, the number of nucleotides does include the start or the stop codon itself. Stop codon Start codon 120 nt 40 nt 50 nt 150 nt 180 nt Alternative splicing of this gene produces removes introns and joining different exons. Four different mRNA transcripts are formed by alternative splicing. All four transcripts utilize the same start codon (in Exon 1) and the same stop codon (in Exon 5) for translation. Based on this information, can you predict the exon combinations of the four transcripts and the length (I.e. number of amino acids) of each of the four proteins translated from the four transcripts? Please select all the correct answer options. Exons 1/4/5, length of the coding protein: 149 amino acids Exons 1/5, length of the coding protein: 99 amino acids Exons 1/2/3/5, length of the coding protein: 129 amino acids Exons 1/2/3/4/5, length of the coding protein: 179 amino acids Exons 1/2/4/5, length of the coding protein: 163 amino acids
Transcribed Image Text:The exons and introns of a gene are shown below. Introns are shown as black bars. Exons are numbered white boxes. "nt" is abbreviation for nucleotides. For exon 1, the number of nucleotides labeled in the figure only include the sequence from the start codon to the 3' end of the exon. For exon 5, the number of nucleotides labeled in the figure only include the sequence from the 5' end of exon 5 to the stop codon. In each case, the number of nucleotides does include the start or the stop codon itself. Stop codon Start codon 120 nt 40 nt 50 nt 150 nt 180 nt Alternative splicing of this gene produces removes introns and joining different exons. Four different mRNA transcripts are formed by alternative splicing. All four transcripts utilize the same start codon (in Exon 1) and the same stop codon (in Exon 5) for translation. Based on this information, can you predict the exon combinations of the four transcripts and the length (I.e. number of amino acids) of each of the four proteins translated from the four transcripts? Please select all the correct answer options. Exons 1/4/5, length of the coding protein: 149 amino acids Exons 1/5, length of the coding protein: 99 amino acids Exons 1/2/3/5, length of the coding protein: 129 amino acids Exons 1/2/3/4/5, length of the coding protein: 179 amino acids Exons 1/2/4/5, length of the coding protein: 163 amino acids
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