Question

COMMON MUD SNAIL (Ilyanassa obsoleta), ROUNDWORM (Caenorhabditis elegans), SEA SQUIRT (Ciona spp.), ZEBRAFISH (Danio rerio), MOUSE (Mus musculus)

a. One of these species has individuals who are either male, female, or hermaphroditic (hint = Phylum Mollusca). Identify the species, and then explain two ways in which the hermaphrodites participate in mating in this species.

b. One of these species has individuals who are either male or hermaphroditic, with the latter being most prevalent (hint = Phylum Nematoda). Identify the species, and then explain one advantage of having each type of individual in this species.

c. Two of these species exhibit holoblastic rotational cleavage. Identify the two species, and then describe how their cleavage is different from the cleavage associated with zebrafish development, making sure to describe how the processes are influenced by the amount and distribution of yolk. Limit of three sentences per type of cleavage.

d. Biologists have analyzed the effects of exogenous agents on zebrafish development. Briefly describe one study in which such an agent is critical for normal embryogenesis but disruptive at high concentrations, and another study in which such an agent is disruptive at all concentrations tested. Limit of three sentences per study.

Answer 1

a) Common mud snail (Ilyanassa obsoleta) belongs to phylum Mollusca and has either male or female or hermaphrodite individuals. Hermaphrodites contribute to mating in two ways that are simultaneous reciprocal mating and unilateral mating. In simultaneous reciprocal mating, both the snails involved in mating can act as male and female at the same time. For this to happen the genitalia should be placed exactly opposite before copulation. This is difficult for snails as they cannot hear and have very limited vision. So, chemoreception and sensing through contact come to their rescue. Usually, there is an equal exchange of sperms between the mating snails but sometimes sperm transfer can be one way. In unilateral mating, genders are decided beforehand for the mating snails, one will be male, and the other will be female. The male will transfer sperm to the female who receives it. For the second round of mating, they can reverse their roles as the male will become female and vice versa. Sperms will not get exchanged in this as was the case with reciprocal mating.

b) Roundworm (Caenorhabditis elegans) belongs to phylum Nematoda and has either male or hermaphrodite individuals (androdioecy). Hermaphrodites in nematodes resemble females of similar species and are considered females having functional sperms. The sperms are stored and used later for self-fertilization. So, they can either reproduce by self-fertilization by mating with males. Before their sperm gets depleted, hermaphrodites can produce around 300 offsprings by self-fertilization. But if a good number of virile males are available for the hermaphrodite, approx 1400 offsprings can be produced which is far more than the previous case. So, we can very well explain the advantage of two individuals in the phylum.

c) The answer here would be roundworm and mouse. In rotational cleavage, one daughter cell divides meridionally and the other, equatorially. In mammals (mouse is an example here), the yolk is less and is distributed evenly (isolecithal distribution). Because of less yolk, cells need to be implanted immediately in the uterus to start receiving nutrients as early as possible. Nematodes also undergo holoblastic rotational cleavage. Zebrafish eggs undergo discoidal cleavage (partial meroblastic). Fertilized egg contains a large amount of yolk as contrary to holoblastic cleavage. In this type of cleavage, the yolk is not penetrated and a disc of cells (blastodisc) forms above the yolk. In zebrafish eggs, the yolk is concentrated at one end (telolecithal) as opposed to fertilized eggs of nematodes or mammals that are isolecithal.

d) An agent that is critical for normal embryogenesis but disruptive at high concentrations. PTU (1-phenyl 2-thiourea) inhibits tyrosinase and aids in the development of visualization in zebrafish by blocking pigmentation. At a concentration of 0.003%, it inhibits melanogenesis (other processes are least affected). 0.003% PTU alters (has a beneficial effect) the regulation of retinoic acid and insulin-like growth factors during craniofacial development. On the other hand, at higher concentrations of 0.03%, PTU has an inhibitory effect on neural crest development.  

Endocrine-disrupting chemicals bisphenol A is disruptive at all concentrations. It is a nonsteroidal xenoestrogen.