Answers, problem set E

42) a) BICOID, NANOS and DORSAL are expressed in nurse cells during oogenesis.

SNAIL is expressed in ventral cells at the syncytial blastoderm stage.

HUNCHBACK is expressed in a broad band in the anterior part of the embryo while it is a syncytium of dividing nuclei.

b) bicoid mutants lack anterior structures. nanos mutants lack posterior structures. dorsal mutants lack ventral structures. snail mutants lack ventral structures. hunchback mutants lack anterior structures.

c) bicoid, nanos, dorsal: maternal effects. snail, hunchback: zygotic effects. (Actually, there may be a partial maternal effect of hunchback mutations.)

43) 2) expression of HOX genes

1) expression of gap genes

4) metamorphosis (larva to adult transition)

3) determination of imaginal disc cells

44) a) The bicoid mutation eliminates anterior structures of Drosophila embryos.

b) BICOID is expressed in nurse cells during oogenesis.

c) Bicoid protein is found in a gradient with its highest concentration at the anterior end of the newly fertilized egg.

d) Injecting extra BICOID mRNA into the anterior end of a Drosophila egg would add more where it already is, and would therefore cause production of more Bicoid protein at the normal place. This might result in a larger overall concentration of Bicoid protein along the embryo, and a larger "anterior" domain.

e) Injecting extra BICOID mRNA into the posterior end of a Drosophila egg might cause a second anterior end to form there, although the presence of nanos mRNA at the posterior end would lessen the effect so that the posterior might also look fairly normal (the actual result might depend on the amount of BICOID mRNA injected.)

45) If the ULTRABITHORAX gene is inactivated by mutation, the third thoracic segment will appear like the second normally does (it will develop a pair of wings and the fly will have four wings). If a regulatory mutation causes misexpression of ULTRABITHORAX in the second thoracic segment, that segment will take on a "third thoracic segment" identity, and the fly will lack wings.

46) a) The Dorsal protein will be present throughout the embryo, but it will be in the cytoplasm on the dorsal side and in the nuclei on the ventral side. In between (on the flanks), it will be in both the nucleus and the cytoplasm.

b) Dorsal protein activates transcription of SNAIL and RHOMBOID, and represses transcription of DPP.

c) RHOMBOID

d) SNAIL

e) The RHOMBOID gene would be expressed throughout the entire ventral part of a snail twist homozygous double mutant embryo, because Snail would not repress it in the ventral-most part.

47) a) i) mother and embryo both homozygous mutant for dorsal - no ventral side (dorsalized embryo)

ii) mother homozygous mutant for dorsal, embryo heterozygous (+/-) - no ventral side (dorsalized embryo)

iii) mother and embryo both homozygous mutant for cactus - no dorsal side (ventralized embryo)

iv) mother and embryo both homozygous mutant for snail - ventral-most structures would be abnormal, probably having partial neurectoderm fate where mesoderm should be

v) mother homozygous mutant for snail, embryo heterozygous (+/-) - normal embryo

b) i) It would seem impossible for a mother to be homozygous mutant for snail (iv and v) because such a fly would have lacked mesoderm.

ii) A sector of homozygous mutant snail cells could be induced in a female fly heterozygous for a snail mutation. If this mosaic sector included the germline that gives rise to the nurse cells and oocytes, then the relevant part of the mother would be homozygous for the mutation whereas most of the rest of the fly (including the mesoderm, where a wild-type SNAIL gene is required), would be heterozygous for the mutation and hence viable. The mitotic recombination that could produce the mutant sector could be induced in these flies by irradiation with X-rays.