Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
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Chapter 14, Problem 26PDQ
The initial discovery of IS elements in bacteria revealed the presence of an element upstream (5') of three genes controlling galactose
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The authors state: "Properties of the single mutants K557R, D591V and K617A, a double mutant,
D591V/K617A, and a triple mutant, K557R/D591V/K617A, were evaluated to find an allosteric
binding site for citrate."
The numbers between the letters represent the residue number altered. The letter in front was the
wild type and the letter afterwards is the mutant.
How would the the triple mutations K557R and D591V and K617A alter the overall protein?
O lower the # of charges, increase the pl
O increase the # of charges, increase the pl
O lower the # of charges, lower the pl
O lower the # of charges, increase the pl
O no change in the # of charges, increase the pl
O no change in the # of charges, lower the pl
Genes involved in DNA to protein route are highly conserved in comparison to genes of enzymes that catalyze steps in metabolic pathways individually. What could be the reason for this?
In the procedure shown, why was it necessary to link thecoding sequence for the A or B chains to the sequence forβ-galactosidase? How were the A or B chains separated fromβ-galactosidase after the fusion protein was synthesized in E. coli?
Chapter 14 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 14 - CASE STUDY| Genetic dwarfism Seven months...Ch. 14 -
CASE STUDY | Genetic dwarfism
Seven months...Ch. 14 -
CASE STUDY| Genetic dwarfism
Seven months...Ch. 14 - CASE STUDY | Genetic dwarfism Seven months...Ch. 14 -
HOW DO WE KNOW?
1. In this chapter, we focused on...Ch. 14 - Review the Chapter Concepts list on page 257....Ch. 14 - What is a spontaneous mutation, and why are...Ch. 14 -
4. Why would a mutation in a somatic cell of a...Ch. 14 - Why is a random mutation more likely to be...Ch. 14 - Most mutations in a diploid organism are...
Ch. 14 - What is meant by a conditional mutation?Ch. 14 -
8. Describe a tautomeric shift and how it may...Ch. 14 - Contrast and compare the mutagenic effects of...Ch. 14 - Why are frameshift mutations likely to be more...Ch. 14 - Why are X rays more potent mutagens than UV...Ch. 14 -
12. DNA damage brought on by a variety of natural...Ch. 14 - Contrast the various types of DNA repair...Ch. 14 -
14. Mammography is an accurate screening...Ch. 14 - Describe how the Ames test screens for potential...Ch. 14 - What genetic defects result in the disorder...Ch. 14 - In a bacterial culture in which all cells are...Ch. 14 - Human equivalents of bacterial DNA mismatch repair...Ch. 14 - A number of different types of mutations in the...Ch. 14 -
20. Some mutations that lead to diseases such as...Ch. 14 - In maize, a Ds or Ac transposon can cause...Ch. 14 -
22. Presented here are hypothetical findings from...Ch. 14 -
23. Cystic fibrosis (CF) is a severe autosomal...Ch. 14 -
24. Electrophilic oxidants are known to create...Ch. 14 - Skin cancer carries a lifetime risk nearly equal...Ch. 14 -
26. The initial discovery of IS elements in...Ch. 14 -
27. It is estimated that about 0.2 percent of...Ch. 14 -
28. It has been noted that most transposons in...Ch. 14 - Two related forms of muscular dystrophy–Duchenne...
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- Eukaryotes modify mRNA with an m7G cap: an N7-methylated guanosine linked via a 5′ to 5′ triphosphate linkage, and by methylation at the 2'-hydroxyl of the penultimate nucleoside. Provide a summary of the chemical steps and cofactors involved in the capping reaction.arrow_forwardIn E. coli, a methyltransferase enzyme encoded by the dam generecognizes the sequence 5′–GATC–3′ and attaches a methyl groupto the nitrogen at position 6 of adenine. E. coli strains that have thedam gene deleted are known to have a higher spontaneous mutationrate than normal strains. Explain why.arrow_forwardIn a mixed heteropolymer experiment, messages were createdwith either 4/5C:1/5A or 4/5A:1/5C. These messages yielded proteinswith the amino acid compositions shown in the followingtable. Using these data, predict the most specific coding compositionfor each amino acid.4/5C:1/5A 4/5A:1/5CPro 63.0% Pro 3.5%His 13.0% His 3.0%Thr 16.0% Thr 16.6%Glu 3.0% Glu 13.0%Asp 3.0% Asp 13.0%Lys 0.5% Lys 50.0% 98.5% 99.1%arrow_forward
- An enzyme isolated from rat liver has 193 amino acids and is and encoded by a 1440 base pair long gene. What is the connection between the amino acid number of the enzyme and the number of nucleotide pairs in its gene. Explain clearly pleasearrow_forwardTransposable elements, “jumping genes”, were first described by Barbara McClintock in her Nobel prize winning studies with corn. The genome of every multicellular eukaryote that has been sequenced is composed of a large percentage of transposable elements (often 30% to 50%), but they are nearly absent from the genomes of eubacteria and archaea. What are these things and what do they do? Describe and draw the structural organization and mechanism for movement of transposons and retro- transposable elements. What effect do these elements have on the stability of genomes? can you help me answer the bolded question?arrow_forwardThe antibiotic paromomycin binds to a ribosome and induces the same conformational changes in 16S rRNA residues A1492 and A1493 as are induced by codon–anticodon pairing (Fig.). Propose an explanation for the antibiotic eff ect of paromomycin.arrow_forward
- Wild-type E. coli cells can synthesize all 20 common amino acids, but some mutants, called amino acid auxotrophs, are unable to synthesize a specific amino acid and require its addition to the culture medium for optimal growth. Besides their role in protein synthesis, some amino acids are also precursors for other nitrogenous cell products. Consider the three amino acid auxotrophs that are unable to synthesize glycine, glutamine, and aspartate, respectively. Match each nitrogenous product with the mutant or mutants that would fail to synthesize it. Each mutant may fail to synthesize more than one nitrogenous product, and some nitrogenous products may fail to be synthesized by more than one mutant. glycine auxotrophs glutamine auxotrophs aspartate auxotrophs Answer Bank adenine nucleotides cytosine nucleotides guanine nucleotides uridine nucleotidesarrow_forwardHbS results from the substitution of valine forglutamic acid at the number 6 position in the b chainof human hemoglobin. HbC is the result of a change atthe same position in the β chain, but in this case lysinereplaces glutamic acid. Return to the genetic code table and determine whether single-nucleotide changes can account for these mutations. Then view and examine the R groups in the amino acidsglutamic acid, valine, and lysine. Describe the chemicaldifferences between the three amino acids. Predict how thechanges might alter the structure of the molecule and leadto altered hemoglobin function.arrow_forwardIn HbS, the human hemoglobin found in individuals with sickle-cell anemia, glutamic acid at position 6 in the beta chain is replaced by valine. Q.) Show that one of the glutamic acid codons can be converted to a valine codon by a single substitution mutation (i.e., by changing one letter in one codon).arrow_forward
- Consider the structure of Cro repressor protein from bacteriophage lambda E. It is a DNA binding protein, and like many sequence- specific DNA binding proteins, it must function as a homodimer Ex. Notice the mutual docking of a phenylalanine residue from each subunit into a hydrophobic pocket of the partner subunit. These hydrophobic interactions are required for dimerization. The noncovalent interactions highlighted in yellow are also required for dimerization. These interactions represent examples of: Osecondary structure O tertiary structure O quaternary structure O secondary AND quaternary structure Ⓒ tertiary AND quaternary structurearrow_forwardCTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mamma- lian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbal- anced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.arrow_forwardShown below is a schematic diagram illustrating a very short gene with 5000 bp region of an unknown Schizosaccharomyces pombe genome. (Note: Transcription starts at Transcription Start Site (TSS).) TSS 5. 3' 3 +1 (i) Name the specific regions that can be recognized by Transcription Factor IID (TF ID) and indicate the locations in the diagram above. (ii) List the mechanistic steps that can trigger the initiation of transcription by Transcription Factor IIH (TF IIH).arrow_forward
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