CHAPTER 24
Application and Experimental Questions
E1.
Researchers have used the cloning methods described in Chapter 19 to clone the
bicoid gene and express large amounts of the Bicoid protein. The Bicoid protein was then
injected into the posterior end of a zygote immediately after fertilization. What
phenotypic results would you expect to occur? What do you think would happen if the
Bicoid protein was injected into a segment of a larva?
Answer: The expected result would be that the embryo would develop with two anterior
ends. It is difficult to predict what would happen at later stages of development. At that
point, the genetic hierarchy has already been established so its effects would be
diminished. Also, at later stages of development, the embryo is divided into many cells so
the injection would probably affect a smaller area.
E2.
Compare and contrast the experimental advantages of Drosophila and C. elegans
in the study of developmental genetics.
Answer: Drosophila has an advantage in that researchers have identified many more
mutant alleles that alter development in specific ways. The hierarchy of gene regulation is
particularly well understood in the fruit fly. C. elegans has the advantage of simplicity
and a complete knowledge of cell fate. This enables researchers to explore how the
timing of gene expression is critical to the developmental process.
E3.
What is meant by the term cell fate? What is a lineage diagram? Discuss the
experimental advantage of having a lineage diagram. What is a cell lineage?
Answer: The term cell fate refers to the final cell type that a cell will become. For
example, the fate of a cell may be a muscle cell. A lineage diagram depicts the cell
lineages and final cell fates for a group of cells. In C. elegans, an entire lineage diagram
has been established. A cell lineage is a description of the sequential division patterns
that particular cells progress through during the developmental stages of an organism.
E4.
Take a look at Solved problem S3 before answering this question. Drosophila
embryos carrying a ts mutation were exposed to the permissive (25°C) or nonpermissive
(30°C) temperature at different stages of development. Explain these results.
Time After
Fertilization (hours)
Group: 1
2
3
4
5
0–1
25°C 25°C 25°C 25°C 25°C
1–2
25°C 30°C 25°C 25°C 25°C
2–3
25°C 25°C 30°C 25°C 25°C
3–4
25°C 25°C 25°C 30°C 25°C
4–5
25°C 25°C 25°C 25°C 30°C
5–6
25°C 25°C 25°C 25°C 25°C
SURVIVAL:
Yes No
No
Yes Yes
Answer: These results indicate that the gene product is needed from 1 to 3 hours after
fertilization for the embryo to develop properly and survive. The gene product is not
needed at the other developmental stages that were examined in this experiment (0–1
hours, or 3–6 hours after fertilization).
E5.
All of the homeotic genes in Drosophila have been cloned. As discussed in
Chapter 19, cloned genes can be manipulated in vitro. They can be subjected to cutting
and pasting, site-directed mutagenesis, etc. After Drosophila genes have been altered in
vitro, they can be inserted into a Drosophila transposon vector (i.e., a P element vector)
and then the genetic construct containing the altered gene within a P element can be
injected into Drosophila embryos. The P element will then transpose into the
chromosomes and thereby introduce one or more copies of the altered gene into the
Drosophila genome. This method is termed P element transformation.
With these ideas in mind, how would you make a mutant gene with a “gain-offunction” in which the Antp gene would be expressed where the abd-A gene is normally
expressed? What phenotype would you expect for flies that carried this altered gene?
Answer: As discussed in Chapter 17, most eukaryotic genes have a core promoter that is
adjacent to the coding sequence; regulatory elements that control the transcription rate at
the promoter are typically upstream from the core promoter. Therefore, to get the Antp
gene product expressed where the abd-A gene product is normally expressed, you would
link the upstream genetic regulatory region of the abd-A gene to the coding sequence of
the Antp gene. This construct would be inserted into the middle of a P element (see next).
The construct shown here would then be introduced into an embryo by P element
transformation.
abd-A regulatory
P element
region
Antp coding
/
sequence
P element
The Antp gene product is normally expressed in the thoracic region and produces
segments with legs, as illustrated in Figure 24.13. Therefore, because the abd-A gene
product is normally expressed in the anterior abdominal segments, one might predict that
the genetic construct shown above would produce a fly with legs attached to the segments
that are supposed to be the anterior abdominal segments. In other words, the anterior
abdominal segments might resemble thoracic segments with legs.
E6.
You will need to understand Solved problem S4 before answering this question. If
the artificial gene containing the stripe 2 enhancer and the β-galactosidase gene was
found within an embryo that also contained the following loss-of-function mutations,
what results would you expect? In other words, would there be a stripe or not? Explain
why.
A.
Krüppel
B.
bicoid
C.
hunchback
D.
giant
Answer:
A.
Yes, because Krüppel protein acts as a transcriptional repressor, and its
concentration is low in this region anyway.
B.
Probably not, because Bicoid protein acts as a transcriptional activator.
C.
Probably not, because Hunchback protein acts as a transcriptional
activator.
D.
Yes, because giant protein acts as a repressor, and its concentration is low
in this region anyway.
E7.
Two techniques that are commonly used to study the expression patterns of genes
that play a role in development are Northern blotting and in situ hybridization. As
described in Chapter 19, Northern blotting can be used to detect RNA that is transcribed
from a particular gene. In this method, a specific RNA is detected by using a short
segment of cloned DNA as a probe. The DNA probe, which is radioactive, is
complementary to the RNA that the researcher wishes to detect. After the radioactive
probe DNA binds to the RNA within a blot of a gel, the RNA is visualized as a dark
(radioactive) band on an X-ray film. For example, a DNA probe that is complementary to
the bicoid mRNA could be used to specifically detect the amount and size of the mRNA
in a blot.
A second technique, termed fluorescence in situ hybridization (FISH), can be
used to identify the locations of genes on chromosomes. This technique can also be used
to locate gene products within oocytes, embryos, and larvae. For this reason, it has been
commonly used by developmental geneticists to understand the expression patterns of
genes during development. The photograph in Figure 24.8b is derived from the
application of the FISH technique. In this case, the probe was complementary to bicoid
mRNA.
Now here is the question. Suppose a researcher has three different Drosophila
strains that have loss-of-function mutations in the bicoid gene. We will call them bicoidA, bicoid-B, and bicoid-C; the wild type is designated bicoid+. To study these mutations,
phenotypically normal female flies that are homozygous for the bicoid mutation were
obtained, and their oocytes were analyzed using these two techniques. A wild-type strain
was also analyzed as a control. In other words, RNA was isolated from some of the
oocytes and analyzed by Northern blotting, and some oocytes were subjected to in situ
hybridization. In both cases, the probe was complementary to the bicoid mRNA. The
results are shown here.
[Insert Text Art 24.4]
A.
How can phenotypically normal female flies be homozygous for a loss-offunction allele in the bicoid gene?
B.
Explain the type of mutation (e.g., deletion, point mutation, etc.) in each of
the three strains. Explain how the mutation may cause a loss of normal function for the
bicoid gene product.
C.
Discuss how the use of both techniques provides more definitive
information compared to the application of just one of the two techniques.
Answer:
A.
The female flies must have had mothers that were heterozygous for a
(dominant) normal allele and the mutant allele. Their fathers were either homozygous for
the mutant allele or heterozygous. The female flies inherited a mutant allele from both
their father and mother. Nevertheless, because their mother was heterozygous for the
normal (dominant) allele and mutant allele, and because this is a maternal effect gene,
their phenotype is based on the genotype of their mother. The normal allele is dominant,
so they have a normal phenotype.
B.
Bicoid-A appears to have a deletion that removes part of the sequence of the gene
and thereby results in a shorter mRNA. Bicoid-B could also have a deletion that removes
all of the sequence of the bicoid gene or it could have a promoter mutation that prevents
the expression of the bicoid gene. Bicoid-C seems to have a point mutation that does not
affect the amount of the bicoid mRNA.
With regard to function, all three mutations are known to be loss-of-function
mutations. Bicoid-A probably eliminates function by truncating the Bicoid protein. The
Bicoid protein is a transcription factor. The bicoid-A mutation probably shortens this
protein and thereby inhibits its function. The bicoid-B mutation prevents expression of
the bicoid mRNA. Therefore, none of the Bicoid protein would be made, and this would
explain the loss of function. The bicoid-C mutation seems to prevent the proper
localization of the bicoid mRNA in the oocyte. There must be proteins within the oocyte
that recognize specific sequences in the bicoid mRNA and trap it in the anterior end of
the oocyte. This mutation must change these sequences and prevent these proteins from
recognizing the bicoid mRNA.
E8.
Explain one experimental strategy you could follow to determine the functional
role of the mouse HoxD-3 gene.
Answer: You could follow the strategy of reverse genetics. Basically, you would create a
HoxD-3 gene knockout. This inactivated HoxD-3 gene would be introduced into a mouse
by the technique of gene replacement described in Chapter 20. By making the appropriate
crosses, homozygous mice would be obtained that carry the loss-of-function allele in
place of the wild-type HoxD-3 gene. The phenotypic characteristics of normal mice
would then be compared to mice that were homozygous for a defective HoxD-3 gene.
This would involve an examination of the skeletal anatomies of mice at various stages of
development. If the HoxD-3 plays a role in development, you might see changes in
morphology suggesting anterior transformations. In other words, a certain region of the
mouse may have characteristics that are appropriate for more anterior segments.
E9.
Another way to study the role of proteins (e.g., transcription factors) that function
in development is to microinject the mRNA that encodes a protein, or the purified protein
itself, into an oocyte or embryo, and then determine how this affects the subsequent
development of the embryo, larva, and adult. For example, if Bicoid protein is injected
into the posterior region of an oocyte, the resulting embryo will develop into a larva that
has anterior structures at both ends. Based on your understanding of the function of these
developmental genes, what would be the predicted phenotype if the following proteins or
mRNAs were injected into normal oocytes?
A.
Nanos mRNA injected into the anterior end of an oocyte
B.
Antp protein injected into the posterior end of an embryo
C.
Toll mRNA injected the dorsal side of an early embryo
Answer:
A. The larva would develop with two posterior ends. The larva would not survive
to the adult stage.
B.
The posterior end of the larva and adult fly would develop structures that
were appropriate for thoracic segments. The adult fly may not survive.
C.
The dorsal side of the larva and adult fly would develop structures that
were appropriate for the ventral side. The adult fly may not survive.
E10. Why have geneticists been forced to use reverse genetics to study the genes
involved in vertebrate development? Explain how this strategy differs from traditional
genetic analyses like those done by Mendel.
Answer: Geneticists who are interested in mammalian development have used reverse
genetics because it has been difficult for them to identify mutations in developmental
genes based on phenotypic effects in the embryo. This is because it is difficult to screen a
large number of mammalian embryos in search of abnormal ones that carry mutant genes.
It is easy to have thousands of flies in a laboratory, but it is not easy to have thousands of
mice. Instead, it is easier to clone the normal gene based on its homology to invertebrate
genes and then make mutations in vitro. These mutations can be introduced into a mouse
to create a gene knockout. This strategy is opposite to that of Mendel, who characterized
genes by first identifying phenotypic variants (e.g., tall versus dwarf, green seeds versus
yellow seeds, etc.).