Becker's World of the Cell (9th Edition)
9th Edition
ISBN: 9780321934925
Author: Jeff Hardin, Gregory Paul Bertoni
Publisher: PEARSON
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Chapter 24, Problem 24.5CC
A “knockout” mouse has been produced using the genetic techniques described in Chapter 21 that is unable to make caspase-9. What sorts of defects do you predict might be present in the brains of knockout embryos?
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The ability to selectively modify the genome in the mouse has revolutionized mouse genetics. Outline the procedure for generating a knockout mouse at a specific genetic locus. How can the loxP-Cre system be used to conditionally knock out a gene? What is an important medical application of knockout mice?
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Using baker's yeast it is possible to generate point mutations that destroy the kinase activity of CDK9/P-TEFB protein kinase. When genome-wide ChIP studies using anti-phosphorylated CTD antibodies and RNA-seq analysis of the mRNA purified from these cells were performed it was found that RNA Polymerase II was present in the 5' region of the coding regions of most protein coding genes, but mRNA levels were reduced for all genes tested. The cells that had this mutation were very sick and the mutation caused many of the cells to die. The abundance of mature mRNAs was lower than in non-mutant cells, and often lower than the levels of their corresponding pre-mRNA. Some of the mature mRNAs had short poly A tails and some were detected with none at all. Based on this information: Why are the poly A tails affected in this mutant?
Chapter 24 Solutions
Becker's World of the Cell (9th Edition)
Ch. 24 - Prob. 1QCh. 24 - Which DNA sequences are more alike: a pair of...Ch. 24 - Suppose you treat rapidly dividing tissue culture...Ch. 24 - Cyclin D is part of the G1 Cdk-cyclin complex....Ch. 24 - Both EGF (a mitogen) and TGF (when acting as an...Ch. 24 - A knockout mouse has been produced using the...Ch. 24 - Cell Cycle Phases. Indicate whether each of the...Ch. 24 - QUANTITATIVE The Mitotic Index and the Cell Cycle....Ch. 24 - Chromosome Movement in Mitosis. It is possible to...Ch. 24 - Cytokinesis. Predict what will happen in each of...
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- Using baker's yeast it is possible to generate point mutations that destroy the kinase activity of CDK9/P-TEFB protein kinase. When genome-wide ChIP studies using anti-phosphorylated CTD antibodies and RNA-seq analysis of the mRNA purified from these cells were performed it was found that RNA Polymerase II was present in the 5' region of the coding regions of most protein coding genes, but mRNA levels were reduced for all genes tested. The cells that had this mutation were very sick and the mutation caused many of the cells to die. The abundance of mature mRNAs was lower than in non-mutant cells, and often lower than the levels of their corresponding pre-mRNA. Some of the mature mRNAs had short poly A tails and some were detected with none at all. Based on this information: Provide one reason why this mutation in CDK9 affects the levels of mature mRNAs?arrow_forwardConsider the now dominant variant of the SARS-CoV-2 called the D614G mutation: a) The mutation changes an Aspartate (D, Asp) to a Glycine (G, Gly) at nucleotide position 614 (that’s why it’s called the D614G mutant) in the S1 subunit of the Spike protein. Using only the information above and a codon table, what are the mRNA codon sequences of the 2019-dominant and 2020-dominant Spike proteins? Note the figure above is not needed toanswer the question. Report in 5’ to 3' orientation __________________________________________ b) What type of substitution is this? In your answer, address the following: • The expected substitution in base sequence (e.g., A à C)• If the mutation is synonymous, nonsynonymous, or a frameshift• If the mutation is a transition or a transversion c) The mutation increases infectivity by reducing the stability of the Spike protein such that it can remain in the open conformation more often. The open conformation increases the chances of binding to the host ACE2…arrow_forwardThe codon change (Gly-12 to Val-12) in human rasH that convertsit to oncogenic rasH has been associated with many types ofcancers. For this reason, researchers would like to develop drugs toinhibit oncogenic rasH. Based on your understanding of the Rasprotein, what types of drugs might you develop? In other words,what would be the structure of the drugs, and how would theyinhibit Ras protein? How would you test the efficacy of the drugs?What might be some side effects?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_forwardIn this chapter you were introduced to nonsense suppressor mutations in tRNA genes. However, suppressormutations also occur in protein-coding genes. Using thetertiary structure of the β subunit of hemoglobin shownin Figure 9-3(c), explain in structural terms how a mutation could cause the loss of globin protein function. Nowexplain how a mutation at a second site in the same protein could suppress this mutation and lead to a normalor near-normal protein.arrow_forwardBriefly discuss (referring to the images provided) why mutant 2 fails to produce functional protein. Note that none of the mRNA transcribed from this gene is of the expected size; some of the mRNA molecules produced are 223 nucleotides shorter than expected, whilst others are 47 nucleotides longer than expected.arrow_forward
- Cx is a member of the family of connexin genes that encode the proteins of gap junction intercellular channels. Cx proteins in one cell form hemi-channels in the plasma membrane. Hemi-channels in adjacent cells dock to form complete intercellular channels through which ions and small molecules diffuse from cell to cell. Distinct Cx mutations were identified in three different families, F1, F2 and F3, affected by the same disease. To study their functional properties, normal (wild type, wt) and mutant (m) Cx proteins were expressed in cultured cells. Translation of the proteins was checked (Fig. 3). A extracellular EC 1 SM TM 1 membrane 2 3 4 F10 intracellular N F2 EC 2 F3 oricand c B 42 kDa C 42 kDa 35 kDa Control wt m-F1 m-F2 PM C PM C PM C PM C Western blot anti-Cx Control wt PM C m-F1 PM C PM C = Metabolically labelled m-F2 PM C m-F3 PM C m-F3 PM C WSEY Fig. 3 mont (A). Membrane topology of Cx protein indicating positions of mutations. Cx is an integral membrane protein with 4…arrow_forwardWhat would be the most likely effect of inhibiting the translation of hunchback mRNA throughout a Drosophila embryo?arrow_forwardAs we have learned in this chapter, the Nanos protein inhibits the translation of hunchback mRNA, lowering the concentration of Hunchback protein at the posterior end of a fruit-fly embryo and stimulating the differentiation of posterior characteristics. The results of experiments have demonstrated that the action of Nanos on hunchback mRNA depends on the presence of an 11-base sequence that is located in the 3′ untranslated region (3′ UTR) of hunchback mRNA. This sequence has been termed the Nanos response element (NRE). There are two copies of NRE in the 3′ UTR of hunchback mRNA. If a copy of NRE is added to the 3′ UTR of another mRNA produced by a different gene, that mRNA is repressed by Nanos. The repression is greater if several NREs are added. On the basis of these observations, propose a mechanism for how Nanos inhibits Hunchback translation.arrow_forward
- Explain one experimental strategy for determining the functional role of the mouse HoxD-3 gene.arrow_forwardThe following question refers to the transcriptional control of the Gal7locus in Saccramyces cerevisiae. What is the predicted Gal7 expression phenotype (inducible,uninducible, constitutive) of the following mutant strains?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
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