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 18, Problem 15P
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You are interested in studying position effect variegation in Drosophila using the chromosome depicted below:
Deactivation of the w+ gene gives a white eye phenotype and deactivation of the rst+ gene gives a rough eye
phenotype; the normal phenotypes are red and smooth. Because the rst+ and w+ genes have now been placed
close to a heterochromatic domain, some sections (or sectors) of the eye display mutant phenotypes due to gene
deactivation while others display the normal, wild type phenotype. Which phenotype would you not expect to see
rst w
Inverted X chromosome
white smooth eye sectors
white rough eye sectors
red smooth eye sectors
red rough eye sectors
You have identified a Drosophila gene that is expressed exclusively in the odd-numbered "stripes" in the cellular blastoderm. Assuming that this gene is not redundant, what
would be the most likely phenotype cause by a loss-of-function mutation in this gene?
an embryo missing odd numbered segments
an adult fly with a second pair of wings instead of halteres
an embryo with two anterior ends
an embryo missing even-numbered segments
an embryo missing larval segments 3-10
Here are schematic diagrams of mutant Drosophila larvae.
The left side of each pair shows a wild-type larva, with gray boxes showing the sections that are missing in the mutant larva. Which type of gene is defective in each larva: a gap gene, a pair-rule gene, or a segment-polarity gene?
Chapter 18 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 18 - 18.1 Explain why many developmental genes encode...Ch. 18 - Bird beaks develop from an embryonic group of...Ch. 18 - 18.3 How is positional information provided along...Ch. 18 - Early development in Drosophila is atypical in...Ch. 18 - 18.5 Consider the evenskipped regulatory sequences...Ch. 18 - What is the difference between a parasegment and...Ch. 18 - Why do loss-of-function mutations in Hox genes...Ch. 18 - 18.8 Compare and contrast the specification of...Ch. 18 - Prob. 9PCh. 18 - Ablation of the anchor cell in wild type C....
Ch. 18 - 18.11 In gain-of-function and. elegans mutants,...Ch. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - 18.14 Given that maternal Bicoid activates the...Ch. 18 - What phenotypes do you expect in flies homozygous...Ch. 18 - The pair rule gene fushitarazu is expressed in...Ch. 18 - 18.17 In contrast to Drosophila, some insects...Ch. 18 - Prob. 18PCh. 18 - 18.19 You are traveling in the Netherlands and...Ch. 18 - 19.20 A powerful approach to identifying genes of...Ch. 18 - The Hoxd 913 genes are thought to specify digit...Ch. 18 - Three-spined stickleback fish live in lakes formed...Ch. 18 - The flowering jungle plant Lacandoniaschismatica,...Ch. 18 - 18.24 Homeotic genes are thought to regulate each...Ch. 18 - Prob. 25PCh. 18 - Basidiomycota is a monophyletic group of fungi...Ch. 18 - Prob. 27PCh. 18 - In C. elegans there are two sexes: hermaphrodite...Ch. 18 - In Drosophila, recessive mutations in the...
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- In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…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_forwardTo understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5arrow_forward
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