Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 16, Problem 24P
A hunchback gene, a gene necessary for proper patterning of Drosophila embryo, is translationally regulated. The position of the coding region within the transcript is known. How could you determine if the sequences within the 5' UTR or 3' UTR', or both, are necessary for the proper regulation of the mRNA's translation?
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The hunchback gene, a gene necessary for proper patterning of the Drosophila embryo, is translationallyregulated. The position of the coding region withinthe transcript is known. How could you determine ifthe sequences within the 5′ UTR or 3′ UTR, or both,are necessary for proper regulation of the mRNA’stranslation?
In eukaryotes there is not a consistent relationship between the length of the coding sequence of a gene and the length of the mature mRNA it encodes, even though one nucleotide in DNA = one nucleotide in pre-mRNA or primary transcript. Explain why this is so.
Even though the lac Z, Y, and A structural genes are transcribed as a single polycistronic mRNA, each gene contains the initiation and termination signals essential for translation. Predict what will happen when a cell growing in the presence of lactose contains a deletion of one nucleotide (a) early in the Z gene and (b) early in the A gene.
Chapter 16 Solutions
Genetics: From Genes to Genomes, 5th edition
Ch. 16 - For each of the terms in the left column, choose...Ch. 16 - Does each of the following types of gene...Ch. 16 - List five events other than transcription...Ch. 16 - Which eukaryotic RNA polymerase RNA pol I, pol II,...Ch. 16 - You have synthesized an enhancerless GFP reporter...Ch. 16 - Prob. 6PCh. 16 - Yeast genes have cis-acting elements upstream of...Ch. 16 - A single UASG regulates the expression of three...Ch. 16 - Prob. 9PCh. 16 - a. Assume that two transcription factors are...
Ch. 16 - a. You want to create a genetic construct that...Ch. 16 - In Problem 12, you identified a genomic region...Ch. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Genes in both prokaryotes and eukaryotes are...Ch. 16 - Prob. 16PCh. 16 - Prader-Willi syndrome is caused by a mutation in...Ch. 16 - The human IGF-2R gene is autosomal and maternally...Ch. 16 - Follow the expression of a paternally imprinted...Ch. 16 - Reciprocal crosses were performed using two inbred...Ch. 16 - Interestingly, imprinting can be tissue-specific....Ch. 16 - Prob. 22PCh. 16 - a. How can a single eukaryotic gene give rise to...Ch. 16 - A hunchback gene, a gene necessary for proper...Ch. 16 - You know that the mRNA and protein produced by a...Ch. 16 - You are studying a transgenic mouse strain that...Ch. 16 - Prob. 27PCh. 16 - Scientists have exploited the siRNA pathway to...Ch. 16 - Researchers know that Fru-M controls male sexual...Ch. 16 - The Drosophila gene Sex lethal Sxl is deserving of...Ch. 16 - Prob. 31P
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- 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 domainarrow_forwardGenes can be transcribed into mRNA, in the case of protein coding genes, or into RNA, in the case of genes such as those that encode ribosomal or transfer RNAs. Define a gene. For the following characteristics, state whether they apply to (a) continuous, (b) simple, or (c) complex transcription units.i. Found in eukaryotesii. Contain intronsiii. Capable of making only a single protein from a given genearrow_forward"Upstream" "Downstream" Exons Start of transcription Termination codon 5 3' Promoter initiator codon Introns Polyadenylation signal (intervening sequences) 5' untranslated region 3' untranslated region Direction of transcription Please study the diagram above on eukaryotic gene expression. In order to provide instructions for gene expression, a eukaryotic gene should have the following sequences except for O A. Promoter B. Start codon also known as initiator codon C. Splicing signals (dinucleotide sequence in the intron) O D. 5' CAP sequencearrow_forward
- You are teaching a class on the regulation of eukaryotic gene expression. In order to demonstrate this complex process, you decide to draw for the class a typical eukaryotic gene/transcription unit with its major regions, such as the promoter regions, where the RNA polymerase II and transcription factors would bind From the list given - choose all components that you think are part of a typical eukaryotic gene From the list given - choose all the regulatory sequences that you think would control the expression of this eukaryotic gene From the list given - choose all of the regulatory proteins that would bind the eukaryotic gene to control its expressionarrow_forwardDraw the SMN2 pre-mRNA (10 exons and 9 introns), indicate the 5’ and 3’ SS location. Draw the SMN2 mature mRNA before treatment with Spinraza. What type of alternative splicing is occurring? Draw the SMN2 mature mRNA after treatment with Spinrazaarrow_forwardThe asterisk (*) in the diagram below indicates a single base mutation in the 5' splice site of the second intron of a eukaryotic gene. Due to this mutation, the second intron is now not ‘spliced out’ during the splicing process. What are the most likely consequences of this mutation with respect to the size of the pre-mRNA and the size of the mature mRNA? a. The pre-mRNA will be longer and the mature mRNA will be longer. b. The pre-mRNA will be longer and the size of the mature mRNA will not be affected c. The size of the pre-mRNA will not be affected and the mature mRNA will be longer d. The size of the pre-mRNA will not be affected and the size of the mature mRNA will not be affectedarrow_forward
- The interphase nucleus is a highly structured organelle with chromosome territories, interchromatin compartments, and transcription factories. In cultured human cells, researchers have identified approximately 8000 transcription factories per cell, each containing an average of eight tightly associated RNAP II molecules actively transcribing RNA. If each RNAP II molecule is transcribing a different gene, how might such a transcription factory appear? Provide a simple diagram that shows eight different genes being transcribed in a transcription factory and include the promoters, structural genes, and nascent transcripts in your presentation.arrow_forwardWhat proportion (in %) of the CFTR gene/DNA sequence is represented in the CFTR mRNA? The mRNA contains a 5’ UTR, ORF, and 3’ UTR. What proportion (in %) of the mRNA is represented in the ORF?arrow_forwardDefine both transcription and translation. In addition, describe the role(s) of each of the following in the processes of gene expression and protein synthesis: DNA, mRNA, tRNA, rRNA, ribosome(s), RNA polymerase, codon, anticodon, amino acid(s) and polypeptide(s). Be detailed in your answer.arrow_forward
- A common feature of many eukaryotic mRNAs is the presence of a rather long 3′ UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an important enzyme in cellular metabolism. Certain cells—termed U937D cells—have lots of CK-B mRNA, but no CK-B enzyme is present. In these cells, the 5′ end of the CK-B mRNA is bound to ribosomes, but the mRNA is apparently not translated. Something inhibits the translation of the CK-B mRNA in these cells. Researchers introduced numerous short segments of RNA containing only 3′ UTR sequences into U937D cells. As a result, the U937D cells began to synthesize the CK-B enzyme, but the total amount of CK-B mRNA did not increase. The introduction of short segments of other RNA sequences did not stimulate the synthesis of CK-B; only the 3′ UTR sequences turned on the translation of the enzyme. On the basis of these results, propose a mechanism for the inhibition of CK-B translation in the U937D cells. Explain how the introduction of short…arrow_forwarda) What is a mutation in molecular terms? b) a mutation deletes a base in the genomic DNA discuss how that will affect the reading frame and expression product production. Using the following list of codons describe, using diagrams etc., how information stored in the DNA is translated into a peptide. Be sure to discuss all steps. In other words, use a diagram and give me sequences, transcription and translation steps. Show the sequences of the sense and the other DNA strand, the mRNA and the tRNA’s. UUU -phenylalanine UCU -serine AUG –initiation/methionine CUU -leucine ACU -threonine GUU -valine UAA -Terminationarrow_forwardThe following double-stranded DNA sequence is part of a hypothetical yeast genome which contains a very small gene. Transcription starts at the Transcription Start Site (TSS), proceeds in the direction of the arrow and stops at the end of the Transcription Terminator (green box). 5' 3' TSS CTATAAAAATGCCATGCATTATCTAGATAGTAGGCTCTGAGAAATTTATCTCACT | | | | | | | | | | GATATTTTTACGGTACGTAATAGATCTATCATCCGAGACTCTTTAAATAGAGTGA - 5' PROMOTER TERMINATOR 3' a) Which strand (top or bottom) is the template strand? Explain why. b) What is the sequence of the mRNA produced from this gene? Label the 5' and 3' ends. c) What is the sequence of the protein produced from the mRNA? d) If a mutation (an insertion) were found where a T/A (top/bottom) base pair were added immediately after the T/A base pair shown in red, what would be the sequence of the mRNA? What would be the sequence of the protein?arrow_forward
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