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 17, Problem 10TYPSS
Summary Introduction
To review:
The maximum rate of translation by a ribosome in a bacterial cell. To calculate the time period taken by the bacterial cell to translate an mRNA (messenger ribonucleic acid) containing 1800 codons.
Introduction:
The transcriptional and translational rates should proceed at a controlled rate to avoid collisions between the ribosomes and the RNA polymerases. The transcription rate of RNA polymerase into mRNA is 60
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Chapter 17 Solutions
Biological Science (7th Edition)
Ch. 17 - Prob. 1TYKCh. 17 - Prob. 3TYKCh. 17 - Prob. 4TYKCh. 17 - 5. RNases and proteases are enzymes that destroy...Ch. 17 - Prob. 6TYUCh. 17 - The nucleotide shown below is called cordycepin...Ch. 17 - Prob. 10TYPSSCh. 17 - What better not be for dinner? Eating even a...Ch. 17 - 12. α-Amanitin inhibits transcription by binding...Ch. 17 - Prob. 13PIAT
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- Which is the coding strand? Which is the template strand? What is the direction of mRNA transcription by RNA-Polymerase for this gene?arrow_forwardTranscription occurs at a rate of about 30 nucleotides per second. is it possible to calculate the time required to synthesize a titin mRNA from the information given here?arrow_forwardGive Detailed Solution (no need Handwritten)arrow_forward
- The figure below shows the stage in translation when an incoming aminoacyl-tRNA has bound to the A site on the ribosome. Using the components shown in A as a guide, show on B and C what happens in the next two stages to complete the addition of the new amino acid to the growing polypeptide chain.arrow_forwardWhat is the total size of the mature i.e. fully processed mRNA in nucleotides?arrow_forwardGive typing answer with explanation and conclusionarrow_forward
- I've attached the table of transcription ans translation for a DNA and Bees work, Genes A and B are exons while C is an intron. Gene A has a silent mutation and Gene B has a nonsense mutation. Please answer the below for me The 3 genes code for different proteins: • Gene A = protein essential for stinger • Gene B = DNA replication enzyme • Gene C = fuzzy hair protein Do you think it matters which protein is mutated? Is one protein more important than another? How would you try to help the bees stay healthy using the information from the mutations?arrow_forwardThe flu virus maximizes the use of its limited (13.5 kb) genome by using alternative translation initiation sites, overlapping reading frames, and ribosomal frameshifting. For example, part of the viral PA gene includes a rarely used CGU codon. When the ribosome pauses to translate this codon, it may slip ahead by one nucleotide and produce a polypeptide with a diff erent C-terminal sequence. From the partial mRNA sequence shown here, determine the normal polypeptide sequence and the sequence with the frameshift.arrow_forwardThe sequence below shows the non-coding strand from the whole of the transcribed region of a very short gene. 5’-GGCTTCTTTAGTACTGGCCAGTGGGATCCAAGTAGGCTGCCATTTCGT-3’ Write out the sequence of the mRNA from this gene in the orientation 5′ → 3′ and, using the genetic code (see Fig. 1. overleaf) deduce the amino acid sequence of the peptide it encodes (NB you should read about the operation of the genetic code prior to attempting this question).arrow_forward
- Determine whether each event occurs during initiation, elongation, or termination. Initiation Peptidyl transferase transfers the peptidyl group to water. In E. coli, EF-Tu hydrolyzes GTP. Elongation In E. coli, mRNA binds to the 30S ribosomal subunit. Answer Bank In prokaryotes, the Shine-Dalgarno sequence pairs with rRNA. In E. coli, EF-Tu delivers an aminoacyl-tRNA to the ribosome. Termination Translocation occurs. Initiator tRNA enters the P site.arrow_forwardE22. The method of Northern blotting is used to determine the amount and size of a particular RNA transcribed in a given cell type. Alternative splicing (discussed in Chapter 12) produces mRNAs of different lengths from the same gene. The Northern blot shown here was made using a DNA probe that is complementary to the MRNA encoded by a particular gene. The mRNA in lanes 1 through 4 was isolated from different cell types, and equal amounts of total cellular MRNA were added to each lane. 2 3 4 Lane 1: MRNA isolated from nerve cells Lane 2: MRNA isolated from kidney cells Lane 3: MRNA isolated from spleen cells Lane 4: MRNA isolated from muscle cells Explain these results. | |arrow_forwardPlease asaparrow_forward
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