Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
3rd Edition
ISBN: 9780135564172
Author: Mark Sanders, John Bowman
Publisher: PEARSON+
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Chapter 18, Problem 14P
Given that maternal Bicoid activates the expression of hunchback (see Figure
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In Drosophila subobscura, the presence of a recessive gene called grandchildless (gs) causes the offspring of homozygous females, but not those of homozygous males, to be sterile. Can you offer an explanation as to why females and not males are affected by the mutant gene?
In Drosophila sex determination
(Check all that apply.)
A) XY individuals transcribe the Sxl gene early but not late.
B
XX individuals carry the Sxl gene, XY individuals do not.
XX individuals transcribe the Sxl gene, XY individuals do not.
XX individuals express the Sxl protein, XY individuals do not.
E) The Sxl transcript (Pre-MRNA) is spliced differently in XX compared to XY individuals.
Why is it that mutations in bicoid and nanos exhibit genetic maternal effect in Drosophila (a mutation in the maternal parent produces a phenotype that shows up in the offspring), but mutations in runt and gooseberry do not?
Chapter 18 Solutions
Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
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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- Suppose a researcher has three different Drosophila strains that have mutations in the bicoid gene called bicoid-A, bicoid-B, and bicoid-C; the wild type is designated bicoid +. To study these mutations, phenotypically normal female flies that are homozygous for the given bicoid mutation were obtained, and their oocytes were analyzed using a Northern blot to determine the size and/or amount of the bicoid mRNA and in situ hybridization to determine the bicoid mRNA location within the oocyte. A wild-type strain was also analyzed as a control. In both cases, the probe was complementary to the bicoid mRNA and the results are shown below. (Anterior is on the left; posterior is on the right.) Northern blot 1 2 - 3 4 In situ hybridization Wild type Lane 1. Wild type (bicoid*) Lane 2. bicoid-A Lane 3. bicoid-B Lane 4. bicoid-C bicoid-B bicoid-A bicoid-C Which mutation is likely to cause the embryo to develop two "anterior" ends? bicoid-B Obicoid-A bicoid-Carrow_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_forwardYou 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 sectorsarrow_forward
- Human females who are heterozygous for an X-linked recessive allele sometimes exhibit mild expression of the trait. However, such mild expression of X-linked traits in females who are heterozygous for Xlinked alleles is not seen in Drosophila. What might cause this difference in the expression of X-linked genes between human females and female Drosophila? (Hint: In Drosophila, dosage compensation is accomplished by doubling the activity of genes on the X chromosome of males.)arrow_forwardThe locations of six deletions have been mapped to a Drosophila chromosome, as shown in the following deletion map. Recessive mutations a, b, c, d, e, and f are known to be located in the same region as the deletions, but the order of the mutations on the chromosome is not known. (refer image for contination )arrow_forwardHas the DNA sequence of the eye color gene been changed in part (b) compared with part (a)? How do we explain the phenotypic difference?arrow_forward
- Researchers have exploited Minute mutations in orderto study the phenotypes associated with recessive lethal mutations (l−) that decrease the rate of cell divisionand thus make only very tiny homozygous mutant clones that are difficult to analyze. Many differentstrains of Drosophila carry dominant loss-of-functionMinute (M) mutations in a variety of genes encodingribosomal protein subunits. The M genes are haploinsufficient; flies with only one wild-type M+ gene copyhave a slower pace of cell division, and thus prolongeddevelopment and subtle morphological abnormalities.To circumvent the tiny clone problem, researchersgenerate GFP-marked homozygous l−/ l− clones thatare also M+/ M+, in flies that are l−/ l+ and M−/ M+.The loss of the Minute mutation only in cells withinthe clone gives the l−/ l− cells a growth advantageover their neighbors, enabling the mutant clone togrow large enough to study. Diagram chromosomesthat could be used to generate such clonesarrow_forwardExplain the Retrotransposons—the Copia –White-Apricot System in Drosophila ?arrow_forwardHow do we know that the mutant Bar-eye phenotype in Drosophila is due to a duplicated gene region rather than to a change in the nucleotide sequence of a gene?arrow_forward
- The locations of six deletions have been mapped to a Drosophila chromosome, as shown in the following deletion map. Recessive mutations a, b, c, d, e, and f are known to be located in the same region as the deletions, but the order of the mutations on the chromosome is not known.arrow_forwardOur understanding of maternal effect genes has been greatly aided by their identification in experimental organisms such as Drosophila melanogaster and Caenorhabditis elegans. In experimental organisms with a short generation time, geneticists have successfully searched for mutant alleles that prevent the normal process of embryonic development. In many cases, the offspring die at early embryonic or larval stages. These are called maternal effect lethal alleles. How would a researcher identify a mutation that produced a recessive maternal effect lethal allele?arrow_forwardIn Drosophila, one of the genes controlling wing length is located on the X chromosome. A recessive mutant allele of this gene makes the wings miniature—hence, its symbol m; the wild-type allele of this gene, m_, makes the wings long. One of the genes controlling eye color is located on an autosome. A recessive mutant allele of this gene makes the eyes brown—hence, its symbol bw; the wildtype allele of this gene, bw_, makes the eyes red. Miniature-winged, red-eyed females from one true-breeding strain were crossed to normal-winged, brown-eyed males from another true-breeding strain. 1. Predict the phenotypes of the F1 flies. 2. If these flies are intercrossed with one another, what phenotypes will appear in the F2, and in what proportions?arrow_forward
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