Exercise 11: Development and Evolution
Objectives:
-Students will identify and explain the stages of development and be able to apply this knowledge
to the development of the starfish embryo.
-Students will create and conduct an experiment investigating the regenerative properties of
Dugesia
-Students will understand the influence mutation and mutagens have on the evolution of bacteria.
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Introduction:
Fertilization of an egg by a sperm cell produces a zygote. Embryonic development begins at
this point and ultimately produces a free-living organism. All the instructions required for
development are packed into the fertilized egg. Once the zygote formed when egg and sperm nuclei
fuse; mitotic division begin the developmental activity. The initiation of development depends
primarily on the DNA in the zygote nucleus, and on cytoplasmic determinants, that is, mRNA and
protein molecules stored in the cytoplasm. The mRNAs and proteins direct the first stages of
development up until the genes of the zygote become active.
In addition to cytoplasmic determinants, the egg cytoplasm also contains ribosomes and
other cytoplasmic components required for protein synthesis and the early cell divisions of
embryonic development. It also contains mitochondria, nutrients stored in granules in the yolk and
in lipid droplets. It is the distribution of the yolk that influences the rate and location of cell division
during embryonic cell development. Cell division, typically, proceeds more slowly in the region of
the egg containing the yolk. Unequal distribution of yolk and other components in a mature egg is
termed polarity. For example, in most species the egg nucleus is located toward one end of the egg.
This end of the egg, the animal pole, typically gives rise to surface structures and the anterior end of
the embryo. The opposite end of the egg, the vegetal pole, typically gives rise to internal structures
such as the gut and the posterior end of the embryo.
Soon after fertilization, the zygote beings cleavage, a series of mitotic division in which
cycles of DNA replication and division occur without the production of new cytoplasm. As a result,
the cytoplasm of the egg is partitioned into successively smaller cells without increasing the overall
size or mass of the embryo. These cells are called blastomeres. Cleavage is the first of three major
developmental processes that, with modifications, are common to the early development of most
animals. Following cleavage, the second major process, gastrulation, produces an embryo with
three distinct primary tissue layers. The third process, organogenesis, accomplishes the
development of the major organ systems. At the end of organogenesis, the embryo has the body
organization characteristic of its species. Gastrulation and organogenesis both involve the
processes of cell division, cell movements, and cell rearrangements.
The cleavage divisions lead to three successive developmental stages that are common to the early
development of most animals. The first stage, morula, is a solid ball or layers of blastomeres. As a
cleavage divisions continue, the ball or layer hollows out to form the second stage, the blastula, in
which the blastomeres enclose a fluid-filled cavity, the blastocoel.
Once cleavage is complete, the blastomeres undergo extensive cellular rearrangements.
This morphogenetic process is called gastrulation and during this process the embryo is termed a
gastrula. The three germ layers are formed during this stage; the outer ectoderm, the inner
endoderm, and the mesoderm between the ectoderm and the endoderm. Gastrulation establishes
body pattern; that is, each tissue and organ of the adult animal originates in one of the three
primary cell layers of the gastrula.
As gastrulation proceeds, embryonic cells begin to differentiate: they become recognizably
different in biochemistry, structure, and function. The developmental potential of the cells also
becomes more limited than that of the fertilized egg from which they originated. That is, although a
fertilized egg is totipotent, meaning that it is capable of producing all the various types of cells of
the adult, progressively the cells produced become more specialized. That is, totipotent cells give
rise to pluripotent cells, which can give rise to multipotent cells which give rise to cells with
particular functions. Thus, for example, a multipotent mesoderm cell may develop into muscle or
bone but not normally into outside skin or brain.
Morphogenesis is the generation of the body form as differentiated cells and in their
appropriate positions. In animals, morphogenesis occurs by changes in cell shape, cell position, and
cell adhesion. For embryo morphogenesis, the orientation and rate of my mitotic cell division have
special significance. Regulation of these two features of mitotic cell division occurs at all stages of
development.
The orientation of cell division refers to the angles at which daughter cells are added to
older cells as development proceeds. Orientation is determined by the location of a furrow that
separates the cytoplasm after mitotic division of the nucleus. The furrow forms in alignment with
the spindle midpoint. Therefore when the spindle is centrally positioned in the cell, the furrow
leads to symmetric division of the cell. However, when the spindle is displaced to one end of the
cell the furrow leads to asymmetric division of the cell into a smaller and the larger cell.
The rate of cell division primarily reflects the time spent in the G1 period of interphase;
once DNA replication begins, the rest of the cell cycle is usually of uniform length in all cells of the
same species. As an embryo develops and cells differentiate, the time spent in interphase increases
and varies in length in different cell types. Therefore, different cell types proliferate at various rates,
giving rise to tissues and organs with different cell numbers. Ultimately, the rate of cell division is
under genetic control.
Changes in both cell shape and cell movement play important roles cleavage, gastrulation,
and organogenesis. Changes since Cell shape typically results from reorganization of the
cytoskeleton. For example, during the development of the plate in frogs, the ectoderm flattens and
thickens; that is, microtubules in cells in the ectoderm layer lengthen and slide farther apart,
causing the cells to transition from a cube-like to a columnar shape. Among the most striking
examples of whole-cell movements in embryonic development are the cell movements during
gastrulation and the often long distance migrations of neural crest cells.
Typically, cells migrate over the surface of stationary cells in one of the embryos layers. In
many developmental systems, migrating cells follow tracks formed by molecules of the extracellular
matrix, secreted by the cells among the route over which they travel. For example, migrating cells
recognize and adhere to the fibronectin; in response, internal changes in the cells trigger movement
in a direction based on the alignment of the fibronectin molecules. Some migrating cells follow
concentration gradients instead of molecular tracks. The gradients are created by the diffusion of
molecules (often proteins) released by cells in one part of an embryo. Cells with receptors for the
diffusing molecule follow the gradient toward its source, or move away from the source.
Selective cell adhesion, the ability of an embryonic cell to make and break specific
connections to other cells, is closely related to cell movement. As development proceeds, many cells
break their initial adhesions and move, forming new adhesions in different locations. Final cell
adhesions hold the embryo in its correct shape and form. Junctions of various kinds, including tight,
anchoring, and gap junctions, reinforce the final adhesions.
Task 1: Starfish Development
Examine the prepared Starfish development composite slide. Draw the image in the space below.
Locate the different developmental stages of the starfish and label each of these cells on your
drawing.
Examine the prepared Starfish early cleavage slide. Locate and label and defining features on your
drawing.
Examine the prepared Starfish blastula slide. Locate and label and defining features on your
drawing.
Examine the prepared Starfish gastrula slide. Locate and label and defining features on your
drawing. Label the three dermal layers (endoderm, mesoderm, and ectoderm).
Examine the prepared Starfish young slide. Locate and label and defining features on your
drawing.
Keep in mind: The next two slides are not of Starfish, but are representative of important
developmental stages.
Examine the prepared Frog neural fold slide. Locate and label and defining features on your
drawing.
Examine the prepared Pig embryo slide. Locate and label and defining features on your drawing.
Questions:
1. You have observed many stages of starfish embryo development. How does cell size change
from the zygote to the larval forms?
2. The blastula is also essentially the same size as the unfertilized egg. How are the cells
arranged at this stage? How many cell layers are there?
3. Observe the drawing of a blastula below. What might happen to the adult starfish that
grows from this blastula if the cell marked A was destroyed at the blastula stage?
4. How is the mesoderm formed in starfish?
5. Do you observe evidence of the beginning mesoderm development in any slide? Explain.
6. What stage of development is characterized by an increase in the number of cells but no
overall increase in the mass of the embryo?
7. Which of the starfish stages that you drew marks the beginning of morphogenesis?
8. Yolk in animal eggs is used for food energy for growth of the embryo until the embryo
hatches or becomes a large larva. The starfish egg has almost no yolk and develops into a
larva with a complete digestive tract fort eight hours after fertilization. A frog egg has a
moderate amount of yolk and develops into a larva (tadpole) 10 days after fertilization. A
bird egg has a very large amount of yolk and doesn’t hatch for three weeks. Does there seen
to be any connection between the amount of yolk present in a starfish egg and the rate at
which the larval stage and its digestive system develops in the starfish embryo? Explain.
9. How do cells know where to migrate during gastrulation and organogenesis?
10. How do cell size, number and shape differ as the starfish develops?
11. How does the larval starfish differ in terms of symmetry from the adult?
12. All animals begin life as a single fertilized egg, yet they grow to become a complex organism
consisting of billions of cells. Yet, even though every cell in an organism contains the same
DNA, not all cells are the same. Keeping in mind that all cell activities are controlled by
DNA, suggest a reason that different types of cells are different, even though they have the
same DNA.
13. Observing the neural fold. Predict, at the cellular level, how the developing embryo was able
to create the fold seen in the slide.
Task 2: Regeneration of Planarians
Planarians (Figure 1) are bilaterally symmetric metazoans of the phylum Platyhelminthes
commonly found in freshwater streams and ponds where they prey predominantly upon insects,
insect larvae, and other invertebrates. Planarians have the capacity to replace large regions of
missing structures through regeneration. Planarians lack a coelom, i.e., an organ-containing internal
cavity, and possess derivatives of all three germ layers (ectoderm, mesoderm, and endoderm).
The nervous system is organized into bi-lobed cephalic ganglia connected to two ventral
longitudinal nerve cords. Sensory structures, such as photoreceptors and chemoreceptors are
found at the anterior end of the animal and send projections to the cephalic ganglia. The body wall
musculature contains longitudinal, diagonal, and circular muscle fibers not used for locomotion, but
rather for negotiating obstacles. Ventral ciliated epithelial cells accomplish locomotion. Food is
ingested through a muscular, extensible pharynx that serves as both the mouth and the anus of the
animal; the pharynx connects to the three-branched digestive system, consisting of one anterior
and two posterior branches. Freshwater planarians reproduce either asexually by fission or
sexually as cross-fertilizing hermaphrodites. The reproductive system consists of paired ovaries
situated behind the cephalic ganglia, with numerous testes located dorsolaterally.
For over 100 years, scientists have known that planaria can regenerate. Early pioneers of
planaria biology noticed different regeneration abilities in different species. Moreover, they also
noticed that in some species, different parts of the body had different regeneration capacities. In
this experiment, we will investigate the capability of different body sections to regenerate.
When a planaria’s head is cut off, the remaining tail section will first regenerate a head.
Even if the cut is made very close to the tail, the small tail section first regenerates the head and
then continues to regenerate the rest of the tissue between the head and the tail. We will use this
property and compare how long it takes for worms cut in different places to regenerate a head. If
different parts of the planaria body have equal ability to regenerate, they should all regenerate the
head in the same amount of time. If not, they should regenerate the head in different amounts of
time. The regenerative capacity of different body sections may be an indicator of the location of
stem cells called neoblasts. For instance, if one body segment has a low capacity to regenerate,
perhaps only a few neoblasts exist in the area around the cut. Additional neoblasts may need to
migrate to the area or be created by cell division, slowing down the rate of regeneration.
Figure 1. Planarian, Dugesia sp.
Using the space provided below design an experiment where the regeneration capacity of
the planarian is tested. KEEP IN MIND ALL OF THE STEPS OF THE SCIENTIFIC METHOD!!
Procedure:
1. Obtain one petri dish and label the dish with your group, section, and instructor name.
2. Using a transfer pipette, transfer a planaria into the empty plastic petri dish. Make sure your
only transfer a minimal amount of water with the planaria to minimize the movement of the
planaria.
3. Using the dissecting microscope observe the planaria. Let the planarian acclimate for a few
minutes.
4. Using a plastic coverslip, cut the planaria into two pieces. Using the diagram below indicate
where the planarian was cut.
Planaria, Digesia
Transfer a very small pinch of egg albumin into the water. After a few minutes remove the
remaining water using a transfer pipette and fill the petri dish with spring water (just
enough to fill the bottom of the plate). Make sure to write down an initial observations!
6. Place the petri dish on the shelf to store until the following lab.
5.
Questions:
14. What environmental variables do you think may influence regeneration?
15. After you cut the planaria, how does the mobility of the tail fragments compare with the
mobility of the head fragments? Do they move the same or differently? If they move
differently, why do you think that is?
16. Do you think the head fragments prefer light or shade? How could you test this?
Task 3: Evolution
In nature, mutation is the process that creates new alleles. Mutations are mistakes that
happen during DNA replication before cell division, or abnormalities that develop in chromosomes
during cell division. Whatever the source, mutations contribute to genetic variability in populations.
However, their effect is usually much less than that of genetic recombination that occurs as a result
of crossing over. Mutations can be neutral, having no effect, or be detrimental or even beneficial.
Usually, but not always, mutations create recessive alleles. This means that the trait would not be
expressed until it appeared in a homozygous individual, a process that could take several
generations.
Natural selection is the agent in nature that determines whether a mutation is “good” or
“bad.” A detrimental mutation (allele) would be selected against; for example, organisms with the
mutation would not function as well in the environment and would leave fewer offspring. Over
time, the frequency of genotypes carrying a detrimental mutation should decrease in the
population. The opposite would be true for a beneficial mutation.
In this task, you will experimentally induce mutations in a two bacteria, Escherichia coli and
Serratia marcescens. Bacteria are good organisms to use in mutation studies because they are
haploid. If a mutation occurs, it is expressed because there is not a second allele present to mask it.
You will study mutations affecting viability in these two bacteria. Such mutations are called lethal
mutations because the mutant gene fails to produce a needed product and the cell dies.
Mutations can be induced by several means. Chemicals, called mutagens can change an
organism’s DNA, causing changes in hereditary information. Ultraviolet radiation has similar effects
and will be used in the experiment, Because mutations caused by UV exposures are random, many
different mutations will be induced in the bacteria. Some may affect pigment synthesis in Serratia.
This bacterium is normally red, but when a mutation occurs in the genes producing the enzymes
involved in pigment synthesis, no pigment is made and the bacteria are white. Other mutations will
kill the bacteria because the UV exposure damages genes that are essential to life.
Based on what you know about UV exposure, formulate hypotheses (Scientific ,Ho and Ha)
for what you expect to occur to the number of colonies at the different exposure times. Write the
hypotheses in the space provided and explain your reasoning for each.
Scientific:
Ho:
Ha:
Based on what you know about UV exposure, formulate hypotheses (Scientific, Ho and Ha) for what
you expect to occur to the number of Serratia marcescens colonies in comparison to the number of
Escherichia coli colonies. Write the hypotheses in the space provided and explain your reasoning for
each.
Scientific:
Ho:
Ha:
Make Predictions:
For each trial write down whether you expect the resulting bacterial populations to have a larger
amount of colonies or smaller amount of colonies than the other bacterial population.
Table 1. Predictions
Exposure Time
0 seconds
10 seconds
30 seconds
1 minute
2 minutes
3 minutes
5 minutes
Serratia marcescens
Escherichia coli
Procedure:
1. Using a plastic sterile loop pick up a very small amount of E. coli and place the bacteria into
a test tube with 10 mL of LB nutrient broth. Allow the bacteria about 10 minutes to
acclimate.
2. Make sure that the bacteria is evenly distributed in the broth before performing the next
step.
3. Add 0.1mL of the culture to 9.9mL of LB Broth.
4. Take 0.1mL of the dilution culture and add it to 9.9mL of LB Broth in the final test tube.
5. Plate the volumes of bacteria according to Table 2.
6. Repeat the steps for S. marcescens.
7. Spread the bacteria on the plates using proper aseptic techniques.
8. Expose the bacterial plates to UV light. Make sure the plates are dry before placing them
open and upside down on the UV light box. WIPE DOWN THE BOX ONCE THE EXPERIMENT
IS OVER.
Table 2. Exposure and Volumes
Exposure Time
0 seconds
10 seconds
30 seconds
1 minute
2 minutes
3 minutes
5 minutes
Sample Volume (µL)
10
20
20
40
40
40
80
Questions:
17. Which of the variables is (are) the independent variable(s)?
18. Which of the variables is (are) the dependent variable(s)?
19. What variables serve as controls and what do they control for?