Biology
4th Edition
ISBN: 9781259188121
Author: Peter Stiling, Robert Brooker, Linda Graham, Eric Widmaier
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 4.2, Problem 1BC
Summary Introduction
To explain: The effects of alternative splicing on protein structure and function.
Introduction: Precursor mRNA is converted into mRNA through a method called splicing. Alternative splicing is a method of RNA regulation in which splicing occurs in more than one way to produce two or more different polypeptides. Production of two or more polypeptides from a single gene results in a reduction of the size of genome and increase in the size of proteome which can be beneficial.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Topic: Splicing
a. What is the big picture of splicing and what is the importance of this concept in biology?b. What are the processes that are controlled or regulated or affected by this concept and why must these processes be controlled or regulated in the first place?
Instructions: Express your own gene! (1) Make up a DNA sequence of at least 18nucleotides and then (2) show the mRNA sequence that will be made via transcription,(3) show the tRNAs that will base pair and deliver the amino acids, and (4) the aminoacid sequence of the resulting protein. You can use the single letter abbreviations forDNA and RNA nucleotides and the three-letter abbreviations for the amino acids.
INSTRUCTION:
= IF BOTH STATEMENT ARE TRUE
= IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE
= IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE
= IF BOTH STATEMENTS ARE FALSE
STAMENT 1: Amino acyl tRNA synthase is the enzyme responsible for joining amino acid together
STAMENT 2: Nucleus is the part of the cell where translation takes place
ANSWER:
STAMENT 1: DNA sequences where RNA polymerase binds initially is called promoter sequences
STAMENT 2: UV light causes adenine to dimerize
ANSWER:
STAMENT 1: Guanosine is the name of the compound formed when guanine is bonded to ribose
STAMENT 2: DNA pairing is the term that refers to the process when two complementary and single stranded DNA combine
ANSWER:
Chapter 4 Solutions
Biology
Ch. 4.1 - Prob. 1CCCh. 4.1 - Prob. 2CCCh. 4.2 - Prob. 1BCCh. 4.2 - Prob. 1CCCh. 4.2 - Prob. 2BCCh. 4.3 - Prob. 1BCCh. 4.3 - Prob. 1CCCh. 4.3 - Prob. 2BCCh. 4.4 - Prob. 1CCCh. 4.4 - Prob. 1BC
Ch. 4.4 - Prob. 2CCCh. 4.4 - The Nucleus and Endomembrane System Experimental...Ch. 4.4 - Prob. 2EQCh. 4.4 - Prob. 3EQCh. 4.4 - Prob. 3CCCh. 4.5 - Prob. 1CCCh. 4.5 - Core Skill: Connections Look ahead to Figure...Ch. 4.5 - Prob. 2CCCh. 4.6 - Prob. 1CCCh. 4.6 - Prob. 2CCCh. 4 - The cell theory states that a. all living things...Ch. 4 - Prob. 2TYCh. 4 - Prob. 3TYCh. 4 - Prob. 4TYCh. 4 - Each of the following is part of the endomembrane...Ch. 4 - Prob. 6TYCh. 4 - Functions of the smooth endoplasmic reticulum...Ch. 4 - Prob. 8TYCh. 4 - Prob. 9TYCh. 4 - Which of the following observations would not be...Ch. 4 - Prob. 1CQCh. 4 - Explain how motor proteins and cytoskeletal...Ch. 4 - Prob. 3CQCh. 4 - Discuss the roles of the genome and proteome in...Ch. 4 - Prob. 2COQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Learning Task 3: TRACE THE CODE dentify the amino acids coded for by the MRNA codon using the Genetic Code Table below. Order of bases in mRNA (codon) AUC Order or bases Order of bases Amino acid Coded in DNA in tRNA into Proteins TAG CAT GC СА UAC Methionine Valine Procedures: Copy and fill in the table Refer to the Genetic Code table to identify the amino acid. To determine the order of bases in the first column (DNA), second column 1. 2. 3. (codon) and third column (anticodon), consider the complementary base pairs in DNA adenine pairs with thymine and guanine pairs with cytosine. 4 Example, AUG using the Genetic Code Table. Look for the first letter of the MRNA codon on the left side of the genetic code table (A). The second letter of the MRNA on the second letter column (U) and the third letter on the right-side column (G). AUG codes for the amino acid methionine. To identify the amino acid. Look at the bases in the MRNA codon. 5. Do the same with the other codons in the chart.…arrow_forwardtopic: translation how is the growing polypeptide released from the ribosomal assembly?arrow_forwardTranscription Know the central dogma Where transcription takes place What proteins are used for Know how to make a complementary MRNA strand from DNA Enzymes: RNA polymerase Vocabulary: promoter The steps of transcription (initiation, elongation, and termination) What is the outcome of transcription? Translation Where translation takes place Vocabulary: MRNA, TRNA, polypeptide, amino acid, codon, ribosomes Know the start and stop codons Be able to go from DNA->MRNA->codons->anticodons->amino acids Know the tools for translation The steps for translation What are the steps of protein synthesis?arrow_forward
- Transcribe the given DNA sequences to mRNA. DNA: 5'-GATCCGC-3' DNA: 5'-TTACGCTAA-3' DNA: 5'-ACGTCAATGGA-3' DNA: 5'-GCGCGGATTAGCGAT-3' mRNA: 3'- mRNA: 3'- mRNA: 3'- mRNA: 3'- -5' -5' -5' -5'arrow_forwardINSTRUCTION: = IF BOTH STATEMENT ARE TRUE = IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE = IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE = IF BOTH STATEMENTS ARE FALSE STAMENT 1: Transfer RNA is the RNA that contains anticodon STAMENT 2: The tail added to an mRNA to protect it from nuclease digestion is polyA tail ANSWER: STAMENT 1: Heterogenous RNA is a term that refers to mRNA that has not been processed STAMENT 2: If the %A of a bacteria is 20%, the amount of guanine is 30% ANSWER: STAMENT 1: A frameshift mutation involves a change in the reading frame used in the translation of an mRNA STAMENT 2: The genetic code is specific because each codon specifies only for one amino acid ANSWER:arrow_forwardPlease solve and answer in complete sentences. (thank you) Sequence divergence vs conservation for exon vs introns, and peptides A, B, and C of the CDS. for pig (Sus scrofa) vs. us (Homo sapiens). Comparing the (unprocessed) mRNAarrow_forward
- Date: Class: Name: RNA Modification Questions Answer the following questions. 1. What is the initial transcript called? 2. What is the final messenger MRNA called? 3. What two things are added to the messenger RNA? 4. What fragments are removed from the messenger RNA? 5. Why are the poly A tail and methyl G cap important? 6. Below, on the left, are the sequences of 3 pre-mRNAs. The exons are underlined and the introns are not underlined. Draw the mature mRNA's (ready to leave the nucleus) below each pre-mRNA. a. AUGGGGCCCAAACCCCAGUUUUAA b. AUGCAGUUGUUACGCCAAGGCCCGCGCGAUAG c. AUGUCAUGUAUCAUGUAUGUAUGUAUGUAUGUAGUAUGUAGUAUGUUUGUAUAAA (C) 2015 Bethany Lau.arrow_forwardTask #2: Examine the gene, transcript and polypeptide sequences Next, it's important to understand how the gene would be transcribed and translated. A 1 agagtctcct cagacgccga gatgctggtc atggcgcccc gaaccgtcct cctgctgctc 61 tcggcggccc tggccctgac cgagacctgg gccggtgagt gcgggtcggg agggaaatgg 121 cctctgccgg gaggagcgag gggaccgcag gcgggggcgc atgacctcag gagccgcgcc 181 gggaggaggg tcgggcgggt ctcagcccct cctcaccccc aggctcccac tccatgtggt 241 atttctacac ctccgtgtcc cggcccqgcc gcggggagcc ccgcttcatc tcagtgggct 301 acgtggacga cacccagttc gtgaggttcg acagcgacgc cgcgagtccg agagaggagc 361 cgcgggcgcc gtggatagag caggaggggc cggagtattg ggaccggaac acacagatct 421 acaaggccca ggcacagact gaccgagaga gcctgcggaa cctgcgcttc tactacaacc 481 agagcgaggc cgttgcgtga ccccggcccg gggcgcaggt cacgactccc catcccccac 541 gtacggcccg ggtcgccccg agtctccggg tccgagatcc gcctccctga ggccgcggga a) coding region i) What is the length of the coding region (in nucleotides)? ii) Explain how you arrived at your answer for (i). iii) Which nucleotides…arrow_forwardPlease answer fast You have been given the DNA sequence for a particular fragment of DNA. You then isolated the mRNA made from that DNA and amplified it by PCR. You then determined the sequence of the cDNA obtained from different cells. You notice a difference. In the sequence obtained from DNA sequencing you see that a codon is 5' CAG however on the cDNA sequence it is TAG. These results are confirmed by repeated DNA sequence analysis using DNA and cDNA from different cell cultures (same species and tissue samples). What can explain this? a.The DNA must have been mutated in all the cells that were used to isolate mRNA since the cDNA should always match the genomic sequence. b.Any cDNA made through RT-PCR will have T's substitued for genomic C's that are methyulated. c.The mRNA must have been deaminated at the cytosine. d.The cDNA generated most likely had a technical mistake caused by poor fidelity of the Taq enzyme.arrow_forward
- 5’ATCGCGCTAGGCGCATGCTACCTAGGCTATCTGCCTAGCTATCGACTAATCTGATCGAGTCAG3’ 3’TAGCGCGATCCGCGTACGATGGATCCGATAGACGGATCGATAGCTGATTAGACTAGCTCAGTC5’ Write out the pre-mRNA for this geneWrite out the mRNA for this geneHow many amino acids does this protein have? Translate the protein Label your 5’ and 3’ UTR’sarrow_forwardWhat would the sequence of the immature mRNA be? Place the sequence of ALL the transcript that would be synthesized from the gene. In each of the available line you should transcribe all nucleotides that form the immature mRNA. If the sequence indicated is not transcribed then write NA. WRITE THE NUCLEOTIDES IN CAPITAL LETTERS. Promoter: EXON 1: Intron 1: Exon 2: Intron 2: Exon 3 without terminator : Exon 3 terminator:arrow_forwardQuick activity Use the genetic code to translate the following 30 nt mRNA sequence: AUGAUCCUAGGGUGCAUGAUGCCAAAAUAA Prezi 1st letter UUU Phe UCU U UUC JUA UUG U CUA CUG Leu AUU A AUC lle AUA CUU CCU C CUC Leu CCC Pro CCA CCG UAU UCC Ser UAC UCA UAA UCG UAG ACU ACC Second Letter ACA AUG Met ACG C GUU GCU G GUC Val GCC GUA GCA GUG GCG AAU Thr AAC AAA AAG Ala CAU His CAC CAA Gin CAG GAU GAC A GAA GAG Tyr UGU Cys U UGC Stop UGA Stop Stop UGG Trp CGU CGC Arg CGA CGG Asn Lys Glu G AGA AGG Asp GGU Arg DCAG DCA AGU Ser U letter AGC GGC Gly GGA GGG A U VAG U JOJO 3rd G Albarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSON
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Anatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
QCE Biology: Introduction to Gene Expression; Author: Atomi;https://www.youtube.com/watch?v=a7hydUtCIJk;License: Standard YouTube License, CC-BY