Campbell Biology in Focus, Books a la Carte Edition; Modified Mastering Biology with Pearson eText - ValuePack Access Card - for Campbell Biology in Focus (2nd Edition)
2nd Edition
ISBN: 9780134433769
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman
Publisher: PEARSON
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Chapter 18.5, Problem 2CC
Summary Introduction
To explain:
How multiple exons might have arisen in the ancestral EGF and fibronectin genes, with reference to the data provided in the “Figure 18.14”.
Introduction:
The part of the gene that encodes for RNA is known as exon. During RNA splicing the non-coding part introns are removed and the exons are covalently bond to generate mature mRNA (messenger RNA). Just as the genome is the entire set of gene, exome is the entire set of exons.
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Explain how the following mutations would affect transcription of the yeast GAL1 gene in the presence of galactose. (a) A deletion within the GAL4 gene that removes the region encoding amino acids 1 to 100. (b) A deletion of the entire GAL3 gene. (c) A mutation within the GAL80 gene that blocks the ability of Gal80 protein to interact with Gal3p. (d) A deletion of one of the four UASG elements upstream from the GAL1 gene. (e) A point mutation in the GAL1 core promoter that alters the sequence of the TATA box.
In a disorder called gyrate atrophy, cells in the retina begin to degenerate in late adolescence, causing night blindness that progresses to total blindness. The cause is a mutation in the gene that encodes an enzyme, ornithine aminotransferase (OAT). Researchers sequenced the OAT gene for 5 patients with the following results:
Patient A: A change in codon 209 of UAU to UAA
Patient B: A change in codon 299 of UAC to UAG
Patient C: A change in codon 426 of CGA to UGA
Patient D: A two-nucleotide deletion at codons 64 and 65 that results in a UGA codon at position 79.
Patient E: Exon 6, including 1071 nucleotides is entirely deleted.
Which patient(s) have a frameshift mutation?
How many amino acids is patient E missing?
Which patient(s) will produce a shortened protein?
Chapter 18 Solutions
Campbell Biology in Focus, Books a la Carte Edition; Modified Mastering Biology with Pearson eText - ValuePack Access Card - for Campbell Biology in Focus (2nd Edition)
Ch. 18.1 - Prob. 1CCCh. 18.2 - Prob. 1CCCh. 18.2 - Explain the advantage of the systems biology...Ch. 18.2 - Prob. 3CCCh. 18.3 - The best estimate is that the human genome...Ch. 18.3 - Prob. 2CCCh. 18.3 - Prob. 3CCCh. 18.4 - Discuss the characteristics of mammalian genomes...Ch. 18.4 - Which of the three mechanisms described in Figures...Ch. 18.4 - Prob. 3CC
Ch. 18.5 - Describe three examples of errors in cellular...Ch. 18.5 - Prob. 2CCCh. 18.5 - Prob. 3CCCh. 18.6 - Would you expect the genome of the macaque (a...Ch. 18.6 - Prob. 2CCCh. 18 - Prob. 1TYUCh. 18 - Prob. 2TYUCh. 18 - Two eukaryotic proteins have one domain in common...Ch. 18 - SCIENTIFIC INQUIRY The scientists mapping human...Ch. 18 - FOCUS ON EVOLUTION Genes important in the...Ch. 18 - FOCUS ON INFORMATION The continuity of life is...Ch. 18 - SYNTHESIZE YOUR KNOWLEDGE Insects have three...
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- a) hours: 0 2 567 DMC1 SPS1 hours: 0 2 5 6 7 9 11 DIT1 SPS100 Oxygen level (% normal) b) 120 100 FIGURE 4.6. Comparison of Northern blots with DNA microarray data. a) Results from four individual Northern blots examining four different genes and measuring mRNA production over time, as indicated. b) Results from a series of microarrays for the same four genes of interest. Note the color scale on the bottom of b), where bright green indicates a 20-fold repression and bright red indicates a 20-fold induction. Black indicates no change in transcription (i.e., the merged microarray spot would have appeared yellow). 80 60 40 20 0 1 2 gene X gene Y gene Z 3 1 hour 1.0 1.0 b) 1.0 hours: 0.5 2 5 7 9 11 DMC1 SPS1 DIT1 SPS100 fold repressed >20 10x 3x3x10x >20 1:1 fold induced 4 5 6 7 8 9 10 11 Time (hours) 3 hour 2.2 4.5 1.5 5 9 hour hour 1.0 0.15 0.95 0.05 2.0 2.0 FIGURE 4.7 Transcriptional response of three genes to the gradual loss of oxygen. a) Graph of oxygen con- sumption over time by…arrow_forwardSearching the yeast Saccharomyces cerevisiae genome, researchers found approximately 4,000 DNA sites with a sequence which could potentially bind the yeast transcription factor GAL4. GAL4 activates the transcription of galactose genes. Yet there are only 10 GAL4-binding sites which control the genes necessary for galactose metabolism. The GAL4 binding sequence is CGGAT#AGAAGC*GCCG, where # is T, C or G, and * is C or T. In one chromatin immunoprecipitation experiment (ChIP), yeast growing on galactose were lysed, and subjected to cross-linking reagents which cross-linked transcription factors and activators to DNA. Next the DNA was sheared into small fragments, and antibodies to GAL4 were added. These antibodies coprecipitated the GAL4 and the DNA it was cross-linked to. The cross-linking was then chemically reversed, and the DNA was isolated, cloned into a library of plasmids and sequenced. Results showed that only 10 different DNA sequences had GAL4 bound. Since the…arrow_forwardIn the module, you have learned about P-element mediated transgenesis in Drosophila and the concept of using transgenes to rescue mutant phenotypes. In the figure below, you will see a wild type fly with its natural eye colour and three mutants with their eye colours changed to vermillion, white and rosy, respectively. A schematic of P-element mediated transgenesis (as shown in the lectures) is also included in the figure. Please inspect the schematic carefully and choose which of the following statements is true: I. Injection of the white experimental transgene into the vermillion mutant embryo will not change the vermillion mutant phenotype II. Injection of the white experimental transgene in the rosy mutant embryo will change rosy eye colour to red (wild type) III. Injection of the white experimental transgene in the white mutant embryo will not change the white mutant phenotype IV. Injection of the white experimental transgene in the rosy mutant…arrow_forward
- The figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature MRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 в в 11 E2 12 E4 Exon (E) Intron (1)arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides.arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 457 nucleotides in length, and the other is 439 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 457 and 439 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 120 nucleotides. 99 62 120 84 102 27 117 Gene X E1 11 E2 12 ЕЗ 13 E4 Exon (E) Intron (I)arrow_forward
- The diagram below shows a closeup of regulatory proteins binding to one of the UASG elements near the GAL7, GALI0, and GALI genes, which code for the protein products needed for yeast to use the sugar galactose. The red triangle symbolizes an "effector" molecule that binds to Gal80p. In this hypothesis (which has since been shown to be incorrect), what could be happening to Gal80p when it is bound to the effector molecule that causes it to change its position and uncover the Gal4p transcriptional activation domain. Hint: think about what effector molecules do upon binding to proteins such as the the Lac repressor protein or the CAP protein. Galactose absent, glucose absent Gal80p. _Activation domain Gal4p dimer -Binding domain UASG Galactose present, glucose absent Activation domain Gal80p- Binding domain UASG For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).arrow_forwardThe figure below shows the introns and exons found in gene X. The size of each exon and intron is shown as well. A study on this organism found that two mature mRNA molecules are produced for this gene. One is 867 nucleotides in length, and the other is 685 nucleotides in length. Name the process responsible for producing this variation. Also explain how these 867 and 685 nucleotide fragments were produced by referring to the information provided. Hint: This organism produces a poly-A tail of 150 nucleotides.arrow_forwardBased on Figure 9-19, can you predict the position of amutation that would affect the synthesis of one isoformbut not the other?arrow_forward
- Describe how transcription would be affected in the Galactose metabolizing pathway in Yeast in the presence of the following mutations. 1. A mutation that resulted in an inability of Gal80 to enter the nucleus. 2. A mutation that resulted in a lack of ability of Gal3 to bind galactose.arrow_forwardYou take DNA samples from a family with a history of the genetic disease spinal muscular atrophy, which results from many different mutations in the SMA gene. One mutation in the SMA gene is a 240-bp in-frame deletion mutation in the middle of an exon. Data from this family shows you that several individuals from the family have this mutation. What do you predict that you would find when comparing the mRNA and protein products of the mutated and unmutated SMA gene? Select all that apply. 1. The mRNA from the SMA is more stable than unmutated SMA mRNA. 2. The pre-mRNA from the SMA mutant is shorter than the unmutated SMA pre-mRNA. 3. The mature mRNA from the SMA mutant is longer than the unmutated SMA mRNA. 4. The protein from the mutated SMA is shorter than the unmutated SMA protein.arrow_forwardDirect mutagenesis of Ca2+ ATPase gene resulted in the replacement of two amino acid residues - Asn111 and Asn114 to Ala. These substitutions led to the reduction in Ca2+ transport activity by 10% and 50%, respectively. On the other hand, directed mutagenesis that resulted in the alteration of four Glu residues in the lumenal loop of this transport protein to Ala, did not affect the Ca2+ transport. Provide the possible explanation for the observed differences in the Ca2+ transport activity between the protein with Asn->Ala substitution and the protein with Glu->Ala substitution.arrow_forward
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