Essentials of Genetics, Books a la Carte Plus Mastering Genetics with eText -- Access Card Package (9th Edition)
9th Edition
ISBN: 9780134319070
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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Chapter 14 Solutions
Essentials of Genetics, Books a la Carte Plus Mastering Genetics with eText -- Access Card Package (9th Edition)
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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- In studies using repeating copolymers, AC . . . incorporates threonine and histidine, and CAACAA . . . incorporates glutamine, asparagine, and threonine. What triplet code can definitely be assigned to threonine?arrow_forwardIn the fungus Neurospora, some mutants were obtained that lacked activity for a certain enzyme. The mutations were found, by mapping, to be in either of two unlinked genes. Provide a possible explanation in reference to quaternary protein structure.arrow_forwardIf this molecule can inhibit toxic effects at 0.5mM against gastric carcinoma cells, does it mean that the dimer of this nucleoside analogue will exhibit an IC50 of about 0.25mM?arrow_forward
- . Seven E. coli mutants were isolated. The activity ofthe enzyme β-galactosidase produced by cells containing each mutation alone or in combination with othermutations was measured when the cells were grown inmedium with different carbon sources.Lactose +Glycerol Lactose GlucoseWild type 0 1000 10Mutant 1 0 10 10Mutant 2 0 10 10Mutant 3 0 0 0Mutant 4 0 0 0Mutant 5 1000 1000 10Mutant 6 1000 1000 10Mutant 7 0 1000 10F′ lac from mutant 0 1000 101/ mutant 3F′ lac from mutant 0 10 102/ mutant 3Mutants 3 + 7 0 1000 10Mutants 4 + 7 0 0 0Mutants 5 + 7 0 1000 10Mutants 6 + 7 1000 1000 10Assume that each of the seven mutations is one andonly one of the genetic lesions in the following list.Identify the type of alteration each mutation represents.a. superrepressorb. operator deletionc. nonsense (amber) suppressor tRNA gene (assumethat the suppressor tRNA is 100% efficient in suppressing amber mutations)d. defective CRP–cAMP binding sitee. nonsense (amber) mutation in the β-galactosidase genef.…arrow_forwardThe intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.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_forward
- . Give two reasons to explain why a proline residue in the middle of an ahelix is predicted to be destabilizing to the helical structurearrow_forwardIn site-directed mutagenesis experiments of an enzyme, scientists altered an aspartate residue to glutamate, lysine, phenylalanine, or valine. Which substitution is expected to have the least effect on enzymatic acitivity? Group of answer choices Glutamate Valine Lysine Phenylalaninearrow_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_forward
- An enzyme isolated from rat liver has 193 amino acids and is encoded by a 1440 base pair long gene. Explain the connection between the amino acid number of the enzyme and the number of nucleotide pairs in its gene.arrow_forwardWhich of the following statements comparing the synthesis of pyrimidines vs. purines is correct? (A) Both types of nucleotides are formed by constructing the nitrogenous base step-by-step on a phosphorylated ribose foundation.(B) Aspartate serves as a nitrogen donor only for purines, while pyrimidines use only glutamine as a nitrogen donor. (C) Both pathways require the synthesis of 5-phophoribosyl-1-pyrophosphate.(D) During the de novo synthesis of both types of nucleotides, nucleosides are formed first and then phosphates are added to the 5’ carbon of ribose.(E) Ribonucleotide reductase is used to form dNTPs from pyrimidine NTPs but not purines.arrow_forwardFor the anticodon sequences 5' IAA and 5' xm^3s^2UAA, considering the DNA sequences of the genes encoding the tRNAs(assuming both tRNAs exist even if that is not true), What is the sequence of the RNA-like strand of each tRNA gene that corresponds to the tRNA's anticodon? be sure to indicate polarities.arrow_forward
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