Explain how the head-tail axis is established during early embryonal development in the fruit fly Drosophila.

Explain how the head-tail axis is established during early embryonal development in the fruit fly Drosophila.

Expert Solution

The embryo is shown in side view and in cross section.Above is the relationship between the dorsoventral subdivision into future major tissue types and the anteroposterior pattern of future segments. A heavy line covers the region that will form segmental structures. During gastrulation the cells align the ventral midline invaginate to form mesoderm, while the cells fated to forms the gut invaginate near each end of the embryo.

As gastrulation nears completion, a series of indentations and bulges appear in the surface of the embryo, indicating the subdivision of the body into segments along its anteroposterior axis. Soon a fully segmented larva emerges, ready to start eating and growing. Within the body of the larva, small groups of cells remain dormant and undifferentiated, forming structures called imaginal discs. These will grow as the larva grows, and naturally they will give rise to most of the structures of the adult body, as we shall see later.

A head end and a tail end, a ventral (belly) side and a dorsal (back) side, a gut, a nervous system, a series of body segments.These are all features of the basic body plan that Drosophila shares with many other animals, including ourselves.

The embryos are seen in side view in drawings (A-C) and corresponding scanning electron micrographs (D-F).

(A and D) .At 2 hours the embryo is at the syncytial blastoderm stage and no segmentation is visible, although a fate map can be drawn showing the future segmented regions (color in A).
(B and E) At 5–8 hours the embryo is at the extended germ band stage: gastrulation has occurred, segmentation has begun to be visible, and the segmented axis of the body has lengthened, curving back on itself at the tail end so as to fit into the egg shell.
(C and F) At 10 hours the body axis has contracted and become straight again, and all the segments are clearly defined. The head structures, visible externally at this stage, will subsequently become tucked into the interior of the larva, to emerge again only when the larva goes through pupation to become an adult.

gladiator answered 2 years ago

How are the gradients of maternal and zygotic proteins established during early embryogenesis in Drosophila? Explain...

How are the gradients of maternal and zygotic proteins established during early embryogenesis in Drosophila? Explain how these genes control the expression of GAP genes along the AP axis of Drosophila.

A) How are the gradients of maternal and zygotic proteins established during early embryogenesis in Drosophila?...

A) How are the gradients of maternal and zygotic proteins established during early embryogenesis in Drosophila? B) Explain how these genes control the expression of GAP genes along the AP axis of Drosophila.

In the fruit fly, Drosophila melanogaster, a wingless female fly is mated to a male that...

In the fruit fly, Drosophila melanogaster, a wingless female fly is mated to a male that is white-bodied and legless. Assume that all alleles causing these phenotypes are recessive. Phenotypically wild-type F1 female progeny were mated to fully homozygous (mutant) males, and the following progeny were observed: Phenotypes Number Observed wingless 324 wild-type 34 white, wingless 135 white 8 white, legless 319 hairless, white, legless 32 hairless 140 hairless, legless 9 (a) With respect to the three genes mentioned in the problem, what are the genotypes...

In the fruit fly, Drosophila melanogaster, a spineless (sp, no wing bristles) female fly is mated...

In the fruit fly, Drosophila melanogaster, a spineless (sp, no wing bristles) female fly is mated to a male that is claret (cl, dark eyes) and hairless (h, no thoracic bristles). Phenotypically wild type F1 female progeny were mated to fully homozygous (mutant) males and the following progeny (1000 total) were observed. PHENOTYPES NUMBER OBSERVED spineless 316 wild 8 claret, spineless 136 claret 37 claret, hairless 304 hairless, claret, spineless 12 hairless 144 hairless, spineless 43 What is the correct gene map for these genes?...

Use the preamble below to answer Questions 27 and 28 In the fruit fly Drosophila melanogaster,...

Use the preamble below to answer Questions 27 and 28 In the fruit fly Drosophila melanogaster, red eye colour (R) is dominant to white eye colour (r). A student wishes to perform a test cross to determine whether a female red eyed fruit fly is homozygous or heterozygous for the eye colour. The student mates the female fruit fly with a red eye coloured male and studies the offspring, which are 100% red eye. The student concludes that the genotype...

In the fruit fly Drosophila melanogaster, wild type eye color is a brick red color and...

In the fruit fly Drosophila melanogaster, wild type eye color is a brick red color and is produced by the action of two genes, one producing red color, the other brown. Vermilion is an X-linked recessive mutation that results in a bright red eye. Brown is an autosomal recessive mutation that results in a brown eye. Flies carrying both the vermilion and brown mutations produce no pigment and have white eyes. True breeding Vermilion eyed females are crossed to brown...

4) In fruit fly (Drosophila melanogaster), grayish eye color and wingless bodies are caused by recessive...

4) In fruit fly (Drosophila melanogaster), grayish eye color and wingless bodies are caused by recessive genes located on different chromosomes. Red eyes (gr+) and normal wings (wl+) are dominant to gray eyes (gr) and wingless (wl), respectively. Suppose that a true-breeding male with gray eyes and normal wings is crossed with a true-breeding female with red eyes and wingless. a) What is the genotype and phenotype of the F1? b) If these F1 individuals were allowed to intermate, what...

The entire genome of the fruit fly Drosophila melanogaster consists of 1.65 x 108 base pairs...

The entire genome of the fruit fly Drosophila melanogaster consists of 1.65 x 108 base pairs (bp). If this organism’s DNA polymerase can copy DNA at the rate of 30 bp/sec calculate the minimum time (in days) required to replicate the entire genome. Remember there are 2 strands, and each are copied at the same time. 9,900,000,000 days 4,950,000,000 days 31.83 days 1909.7 days

I need a paragraph regarding Thomas hunt and his discover with the fruit fly drosophila melanogaster...

I need a paragraph regarding Thomas hunt and his discover with the fruit fly drosophila melanogaster and how those fruit flies are good model organisms in genetics. Thank you & I’ll rate!

Look up the RPL4 gene in Drosophila melanogaster (fruit fly) in Entrez Gene. (a) (2 marks)...

Look up the RPL4 gene in Drosophila melanogaster (fruit fly) in Entrez Gene. (a) What is RefSeq accession number of the genomic sequence of the gene? What is the RefSeq accession number of the corresponding mRNA sequence of the gene? (b) Identify the same gene in Bos taurus (cattle). What is the RefSeq accession number of the genomic DNA sequence of the RPL4 gene in this animal? What is the RefSeq accession number of the corresponding mRNA sequence? (c) Using...

Question