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: 12345
0–125°C25°C 25°C 25°C 25°C
1–225°C30°C 25°C 25°C 25°C
2–325°C25°C 30°C 25°C 25°C
3–425°C25°C25°C 30°C 25°C
4–525°C25°C 25°C 25°C 30°C
5–625°C25°C 25°C 25°C 25°C
SURVIVAL:YesNo 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-of-function” 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 regulatoryAntp coding
P elementregion/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 bicoid-A, 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-of-function 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.).