Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 11, Problem 12P
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Explain how Ac and Ds elements produce variegated corn kernels.
A yeast strain with a mutant spo11- allele has been isolated. The mutant allele is nonfunctional; it makes no spo11 protein. What do you suppose is the phenotype of this mutant strain?
In maize, a Ds or Ac transposon can alter the function of genes at or near the site of transposon insertion. It is possible for these elements to transpose away from their original insertion site, causing a reversion of the mutant phenotype. In some cases, however, even more severe phenotypes appear, due to events at or near the mutant allele. What might be happening to the transposon or the nearby gene to create more severe mutations?
Chapter 11 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 11 - 11.1 Identify two general ways chemical mutagens...Ch. 11 - 11.2 Nitrous acid and (BU) alter DNA by different...Ch. 11 - 11.3 What is the difference between a transition...Ch. 11 - What is the difference between a synonymous...Ch. 11 - 11.5 UV irradiation causes damage to bacterial...Ch. 11 - Ultraviolet (UV) radiation is mutagenic.
What...Ch. 11 - Researchers interested in studying mutation and...Ch. 11 - The effect of base - pair substitution mutations...Ch. 11 - Describe the purpose of the Ames test. How are...Ch. 11 - 11.10 In numerous population studies of...
Ch. 11 - 11.11 Two different mutations are identified in a...Ch. 11 - What is the phenotype effect of inserting a Ds...Ch. 11 - 11.13 Answer the following questions concerning...Ch. 11 - Several types of mutation are identified and...Ch. 11 - 11.15 A sample of the bacterium is exposed to...Ch. 11 - 11.16 A strain of is identified as having a null...Ch. 11 - Describe the difference between DNA transposons...Ch. 11 - 11.18 How are flanking direct repeat sequences...Ch. 11 - 11.19 Using the adeninethymine base pair in this...Ch. 11 - The partial amino acid sequence of a wild-type...Ch. 11 - Prob. 21PCh. 11 - 11.22 Many human genes are known to have homologs...Ch. 11 - The fluctuation test performed by Luria and...Ch. 11 - In this chapter, three features of genes or of DNA...Ch. 11 - Briefly compare the production of DNA double -...Ch. 11 - During mismatch repair, why is it necessary to...Ch. 11 - 11.27 Following the spill of a mixture of...Ch. 11 - 11.28 In an Ames test using Salmonella bacteria a...Ch. 11 - A wild - type culture of haploid yeast is exposed...Ch. 11 - A fragment of a wild - type polypeptide is...Ch. 11 - Prob. 31PCh. 11 - Alkaptonuria is a human autosomal recessive...Ch. 11 - 11.33 In an experiment employing the methods of...Ch. 11 - Using your knowledge of DNA repair pathways choose...Ch. 11 - 11.35 Ataxia telangiectasia is a human inherited...Ch. 11 - A geneticist searching for mutations uses the...Ch. 11 - 11.37 In a mousebreeding experiment a new mutation...Ch. 11 - 11.38 Considering the Dumbo mutation in a Problem,...Ch. 11 - 11.39 Thinking back to the discussion of...Ch. 11 - 11.40 Common baker’s yeast () is normally grown at...Ch. 11 - 11.41 The two gels illustrated below contain...
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- What is the S-gene and what role does it play in plants? Be specificarrow_forwardTwo morphotypes of the newly discovered plant in Mt. Banahaw de Lucban were tested for linkage of three genes such as the presence of tendril (t+), dense trichomes (d+), and the presence of secretory cells (sc+). The loci for the mutant genes have been mapped and are separated by the following map distances: t and d = 20CM; d and sc = 12CM. The interference between these genes is 0.4. The first morphotype is located at the lower elevation and is characterized by the absence of tendril, sparse trichome, and absence of secretory cells. On the other hand, the second morphotype is located at the higher elevation and is characterized by the presence of tendril, dense trichome, and presence of secretory cells. Further genetic analysis showed that the morphotypes were true breeding. The two morphotypes were intercrossed and the resulting F1 is testcrossed with the first morphotype. The cross resulted in 1800 progeny. Give the genotypes, phenotypes, and the expected numbers of phenotypes in…arrow_forwardThe normal sequence of markers on a certain Drosophila chromosome is ABCDE*FGHIJK, where the asterisk represents the centromere. Some flies were isolated with a chromosome aberration that has the following structure: ABCDE*FIJK . This represents a O a) deletion of GH segment O b) inversion of GH segment O c) deletion O d) deletion of centromerearrow_forward
- Why are mutations in the INK4 locus so dangerous?arrow_forwardAs shown in Figure 13-14, what is the fundamental distinction between a pair-rule gene and a segment-polaritygene?arrow_forwardThree haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typearrow_forward
- In yeast, LYS5, ADE1, and URA2 are genes required to synthesize lysine, adenine and uracil, respectively. A MATa lys5 ADE1 URA2 haploid was mated to a MATA LYS5 ade1 ura2 haploid to make a triply heterozygous diploid. This diploid was put through meiosis, 100 tetrads were dissected and the spore colonies transferred to medium lacking either lysine, adenine or uracil to determine which spores were prototrophic or auxotrophic for those nutrients. The following results were obtained: LYS5-ADE1 ADE1-URA2 LYS5-URA2 PD NPD Ι 20 22 58 18 20 62 50 8 42 Based on this information, select the statements below that are TRUE. Select 4 correct answer(s) A) The LYS5 and URA2 genes are linked. B) The genotypes of the four spores in the ADE1-URA2 tetratypes is ade1 ura2 ade1 URA2 ADE1 ura2 ADE1 URA2 C) The map distance between ADE1 and URA2 is 51 CM. D) ADE1 and URA2 are likely located on the same chromosome. E) The map distance between LYS5 and URA2 is 29 CM. ☐ F) LYS5 and ADE1 independently assort…arrow_forwardIn a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygousarrow_forwardExplain how (a) the absence of class B gene expression produces the flower structures seen in class B mutants (see Figure 22.15c) and (b) the absence of class C gene expression produces the structures seen in class C mutants (see Figure 22.15d).arrow_forward
- Consider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardThe oncogenic protein BETA promotes entry into the S phase of the cell cycle. Phosphorylation of BETA at the amino acid Tyr98 causes BETA to be degraded by the proteasome, thus limiting its abundance. A mutation in the codon encoding Tyr98 changes this residue to Cys, which cannot be phosphorylated. What is the best description of this mutant allele?a) antimorphb) hypermorphc) hypomorphd) amorphe) neomorpharrow_forwardIn roses, the synthesis of red pigment is produced by two steps in a pathway. gene O magenta intermediate - gene P colorless intermediate- red pigment What would the phenotype be of a plant homozygous for a null mutation of gene P? What would the phenotype be of a plant homozygous for a null mutation of gene Q? What would the phenotype be of a plant homozygous for null mutations of genes P and Q? magenta red Match a genotype to each strain. colorless Strain P locus Q locus homozygous null mutation of gene P homozygous null mutation of gene Q homozygous null mutations of genes P and Q Answer Bank plp PIP What F2 ratio is expected from crossing a plant that is homozygous for a null mutation of gene P with a plant that is homozygous for a null mutation of gene Q? Assume independent assortment. 9 colorless : 4 magenta : 3 red 9 red : 4 colorless : 3 magenta O 9 red : 4 magenta : 3 colorlessarrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY