MCDB 4650 Developmental Biology Problem Set 2 Answers 1) (1 ...

MCDB 4650 Developmental Biology
Problem Set 2 Answers
1) (1) In two successive embryonic cleavages, which factor(s) among the following can
determine spindle orientation in the second cleavage relative to the first cleavage?
a) Whether the first cleavage was meroblastic or holoblastic.
b) Whether the two blastomeres resulting from the first cleavage are unequal in size.
c) The plane in which the duplicated centrosomes migrate around the envelopes of the
nuclei that form following the first cleavage.
d)Whether microtubules pull on the spindle poles to change spindle orientation prior to
the second cleavage.
2) (1) Developmentally important proteins in the very early cleavage-stage embryo (e.g.
8-cell stage) in C. elegans and Xenopus are
a) all synthesized in the oocyte from transcripts made during oogenesis
b) partly synthesized in the oocyte and partly after fertilization from transcripts made
during oogenesis.
c) partly synthesized in the oocyte from transcripts made during oogenesis, and partly
after fertilization from transcripts made after fertilization from the embryo’s genome.
d) mostly synthesized in the early cleavage stage cells from transcripts made after
fertilization from the embyro’s genome.
3. (1) In C. elegans, when you separate the two cells present at the two-cell stage (called
AB and P1) , neither can form a complete larva. In humans, identical twins can form up
to the 16-cell stage by splitting of the embryo into two separate cell clusters. Moreover,
individual human embryonic cells (blastomeres) up until about the same stage can be
separated and cultured individually in a dish and still make a complete embryo. What do
these results suggest about differences in the characteristics of early blastomeres between
humans and C. elegans?
These results suggest that early cell specification by control of cleavage plane
orientations and segregation of determinants are much less important in human embryos
than in C. elegans. Most cells in the human embryo are still totipotent at the 16-cell
stage, so that split embryos can still regulate cell commitments and develop normally.
4. (1.5) In experiments on Xenopus cleavage, you treat 1-cell embryos with a drug that
completely blocks RNA polymerase II, which is required for all mRNA synthesis.
Explain briefly if and how you would expect this treatment to affect the processes of
cleavage and gastrualtion
All the machinery for cell division, cell cycle maintenance, DNA synthesis,
etc. during cleavage is either already present in the egg as proteins or maternal mRNAs
ready to be translated, so cleavage will not be affected. However, the onset of
gastrulation accompanied (possibly triggered) by the synthesis of new proteins from the
embryonic genome. Thus, gastrulation will not proceed.

MCDB 4650 Developmental Biology
5. (1)
Almost all animals form a blastocoel (a fluid-filled cavity in the middle of the embryo) of
some sort in the middle of the developing embryo, usually quite early in
development. The diagram below shows the progression of 8 cells in a
tight ball-shape to a ring of cells around the blastocoel. Describe what must
be happening at each of the two steps in order to generate these changes
within the embryo.
The first change must require a change in adhesivity, as the cells are no
longer adherent to each other for their entire extent. The second change is
impacted by the plane of division. In this case, all the cells divided the
same way (radially), so that no cells were invading the blastocoel space,
but rather were creating a larger space. If the cells had divided
perpendicularly to the way shown, some cells would be obscuring the
blastocoel. Lastly,
6. (.5) Which of the following are possible reasons for an embryo to have a
blastocoel? Select all that are correct.
a. space for future movement of cells
b. keeps cells separated that will signal each other later
c. pre-formed endodermal cavity
d. to allow for storage of nutrients
7. (1) Imagine that gastrulation is blocked in a Xenopus embryo. Predict one
consequence for the embryo regarding the differentiation of cell types.
Cells that normally come into contact for important signaling events will not do so.
Thus, not only will the embryo be disorganized, but specific cell types will be missing.—
Neurulation will not take place because the neural cells will not organize into a flat sheet
bounded by epithelial cells, because the signaling required for those two cell types to
form will not be present. Without these junction points, and the underlying notochord,
the neural tube cannot fold. There are certainly other possibilities.
8. (1) A cell that migrates through the dorsal lip of the blastopore at the very end of
gastrulation in Xenopus is likely to differentiate into what kind of cell type (endoderm,
ectoderm or mesoderm) and be located in what general region of the embryo? Mesoderm,
because after endoderm initially migrates in, that is the only cell type that migrates
during gastrulation, and at the end of gastrulation, cells are populating what will be the
posterior end of the embryo (specifically, ventral posterior mesoderm).
9. (1) In the figure below (A, B) are the results of an experiment showing the effects of
levels of Sonic hedgehog (Shh), secreted from notochord cells, on the differentiation of
neurons in the intact neural tube. In panel C, cells from chick neural tube were cultured
in vitro in the presence of different concentrations of Shh, then immunostained for

MCDB 4650 Developmental Biology
proteins characteristic of the different types of neurons shown in the picture (e.g. floor
plate=FP, Motor, etc.). The left diagram in (C) shows the positions of cells that stain for
these proteins in the normal intact neural tube.
Based on the data shown in the figure, describe the role of Shh. How is it functioning in
specification of neural tube cell fates?
a. Shh works as an autocrine signaling molecule
b. Shh works as a juxtacrine signaling molecule
c. Shh works as a paracrine signaling molecule
d. Shh works as a morphogen
e. Shh works to determine only the floorplate cells
10. (1) The diagram below shows a cartoon drawing of the developing neural tube (b) and
the overlying non-neuronal epithelium (a). If these cells are dissociated with trypsin into
single cells and put into a cell culture system called a “drop culture”, they first associate
loosely together as shown in (c). A drop culture is literally a small drop of culture
solution that allows the cells to associate with each other. After some time, the two cell
populations separate from each other (d and e).
Explain what is most likely responsible for the segregation of these two cell types.
The cells lose their external receptors when they are dissociated, thus they will
loosely associate together initially when in this kind of a culture system. However, over
the course of a few hours, the neural cells will make n-cadherin, while the epidermal
cells will make E-cadherin; thus the two types of cells will segregate.
During closure of the neural tube the two epithelia must segregate. The neural
tube expresses N-cadherin and the overlying epithelium on the dorsal surface expressess
E-cadherin. These are homophilic cell adhesion receptors and thus, the two tissues will
segregate and only interact with like receptors, allowing the neural tissue to segregate
from the non-neural tissue.