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Modern Genetic Analysis Chapter 16. The Genetic Basis of Development
Problems

1. | In Drosophila, individual flies with two X chromosomes and three sets of autosomes (X:A ratio = 0.67) are intersexes. When examined more closely, it turns out that their intersexual phenotype is due to mosaicism. Some cells differentiate with an entirely male phenotype and other cells with an entirely female phenotype. Explain this observation in regard to how the Sxl onoff switch is established and maintained.

2. | In Drosophila, homozygotes for mutations in the tra gene transform XX flies into phenotypic males with regard to somatic secondary sexual characteristics. The gonad in Drosophila forms from somatic mesoderm tissue and germ-line cells. XX ; tra homozygotes are sterile and have rudimentary gonads. You suspect that the reason for this sterility is that the somatic tissue is sexually transformed into male by tra but the germ-line cells remain female. Design an experiment that tests this prediction.

3. | XYY humans are fertile males. XXX humans are fertile females. What do these observations reveal about the mechanisms of sex determination and dosage compensation? See answer

4. | Humans who are mosaics of XX and XY tissue occasionally occur. They generally exhibit a uniform sexual phenotype. Some of them are phenotypically female, others male. Explain these observations in regard to the mechanism of sex determination in mammals.

5. | There are dominant mutations of the Sxl gene, called SxlM alleles, that transform XY animals into females but do not affect XX animals. Reversions of these SxlM alleles can be readily induced with mutagen treatment. These reversions, called Sxlf alleles, yield normal individuals in XY animals, but, in homozygotes, they transform XX animals into phenotypic males. Provide a possible explanation for these observations, keeping in mind that Sxl is ordinarily dispensable for male development. See answer

6.

In C. elegans vulva development, one anchor cell in the gonad interacts with six equivalence group cells (cells with the potential to become parts of the vulva). The six equivalence group cells have three distinct phenotypic fates: primary, secondary, and tertiary. The equivalence group cell closest to the anchor cell develops the primary vulva phenotype. If the anchor cell is ablated, all six equivalence group lineages differentiate into the tertiary state.
a. Set up a model to explain these results.
b. The anchor cell and the six equivalence group cells can be isolated and grown in vitro; design an experiment to test your model.

7.

There are two types of muscle cell in C. elegans: pharyngeal muscles and body wall muscles. They can be distinguished from each other, even as single cells. In the following figure, letters designate particular muscle precursor cells, black triangles () are pharyngeal muscles, and white triangles () are body wall muscles.

a. It is possible to move these cells around during development. When the positions of ABa and ABp are physically interchanged, the cells develop according to their new position. In other words, the cell that was originally in the ABa position now gives rise to one body wall muscle cell, whereas the cell that was originally in the ABp position now gives rise to three pharyngeal muscle cells. What does that tell us about the developmental processes controlling the fates of ABa and ABp?
b. If EMS is ablated (by heat inactivation with a laser beam aimed through a microscope lens), no AB descendants make muscles. What does that suggest?
c. If P2 is ablated, all AB descendants turn into muscle cells. What does that suggest?

(Figure from J. Priess and N. Thomson, Cell 48, 1987, 241.)

See answer

8. | In humans, the male steroid hormone testosterone binds to the testosterone receptor. The hormonereceptor complex then activates transcription of male-specific genes. A mutation in the testosterone receptor prevents binding to testosterone. What would be the phenotype of a person who is XY and is homozygous for such a mutation? Explain your conclusion in relation to how steroid receptors act.

9. | For many of the mammalian Hox genes, it has been possible to determine that some of them are more similar to one of the insect HOM-C genes than to the others. Describe an experimental approach using the tools of molecular biology that would allow such a determination.

10.

a.When you remove the anterior 20 percent of the cytoplasm of a newly formed Drosophila embryo, you can cause a bicaudal phenotype, in which there is a mirror-image duplication of the abdominal segments. In other words, from the anterior tip of the embryo to the posterior tip, the order of segments is A8-A7-A6-A5-A4-A4-A5-A6-A7-A8. Explain this phenotype in regard to the action of the anterior and posterior determinants and how they affect gap gene expression.
b. Females homozygous for the maternally acting mutation nanos (nos) produce embryos in which the abdominal segments are absent and in which the head and thoracic segments are broader. In regard to the action of the anterior and posterior determinants and gap gene action, explain how nos produces this mutant phenotype. In your answer, explain why there is a loss of segments rather than a mirror-image duplication of anterior segments.

11. | The three homeodomain proteins ABD-B, ABD-A, and UBX are encoded by genes within the BX-C of Drosophila. In wild-type embryos, the Abd-B gene is expressed in the posterior abdominal segments, abd-A in the middle abdominal segments, and Ubx in the anterior abdominal and posterior thoracic segments. When the Abd-B gene is deleted, abd-A is expressed in both the middle and the posterior abdominal segments. When abd-A is deleted, Ubx is expressed in the posterior thorax and in the anterior and middle abdominal segments. When Ubx is deleted, the patterns of abd-A and Abd-B expression are unchanged from wild type. When both abd-A and Abd-B are deleted, Ubx is expressed in all segments from the posterior thorax to the posterior end of the embryo. Explain these observations, taking into consideration the fact that the gap genes control the initial expression patterns of the homeotic genes.

12. | When an embryo is homozygous mutant for the gap gene Kr, the fourth and fifth stripes of the pair-rule gene ftz (counting from the anterior end) do not form normally. When the gap gene kni is mutant, the fifth and sixth ftz stripes do not form normally. Explain these results in regard to how segment number is established in the embryo. See answer

13. | In considering the formation of the AP and DV axes in Drosophila, we noted that, for mutations such as bcd, homozygous mothers uniformly produce mutant offspring with segmentation defects. This outcome is true regardless of whether the offspring themselves are bcd+/bcd or bcd/bcd. Some other maternal-effect lethal mutations are different, in that the mutant phenotype can be "rescued" by introducing a wild-type allele of the gene from the father. In other words, for such rescuable maternal-effect lethals, mut+/mut animals are normal, whereas mut/mut animals have the mutant defect. Rationalize the difference between rescuable and nonrescuable maternal-effect lethal mutations.

14.

The anterior determinant in the Drosophila egg is bcd. A mother heterozygous for a bcd deletion has only one copy of the bcd+ gene. With the use of P elements to insert copies of the cloned bcd+ gene into the genome by transformation, it is possible to produce mothers with extra copies of the gene.

Shortly after the blastoderm has formed, the Drosophila embryo develops an indentation called the cephalic furrow that is more or less perpendicular to the longitudinal body axis. In the progeny of bcd+ monosomics, this furrow is very close to the anterior tip, lying at a position one-sixth of the distance from the anterior to the posterior tip. In the progeny of standard wild-type diploids (di-somic for bcd+), the cephalic furrow arises more posteriorly, at a position one-fifth of the distance from the anterior to the posterior tip of the embryo. In the progeny of bcd+ trisomics, it is even more posterior. As additional gene doses are added, it moves more and more posteriorly, until, in the progeny of hexasomics, the cephalic furrow is midway along the AP axis of the embryo.
a. Explain the gene dose effect of bcd+ on cephalic furrow formation in relation to bcd's contribution to AP pattern formation.
b. Diagram the relative expression patterns of mRNAs from the gap genes Kr and kni in blastoderm embryos derived from bcd monosomic, trisomic, and hexasomic mothers.

15. | In the Drosophila embryo, the 3 untranslated regions (3 UTR) of the mRNAs [the regions between the translation-termination codons and the poly(A) tails] are responsible for localizing bcd and nos to the anterior and posterior poles, respectively. Experiments have been done in which the 3 UTRs of bcd and nos have been swapped. Suppose that we make P element transformation constructs with both swaps (nos mRNA with bcd 3 UTR and bcd mRNA with nos 3 UTR) and transform them into the Drosophila genome. We then make a female that is homozygous mutant for bcd and nos and carries both swap constructs. What phenotype would you expect for her embryos in regard to AP axis development? See answer

© 1999 by W. H. Freeman and Company. All rights reserved.