11 Homework 1. (3 pts) In Drosophila, mutations at four genes known as the male-specific lethals (msl) cause male-specific lethality during the larval stages. Analysis has shown that malespecific lethality is caused by defects in the process of dosage compensation. Dosage compensation in Drosophila is a process by which the single X chromosome of a male is transcribed at twice the rate of either of a female's two X chromosomes). Examination of chromosomes in normal males and females shows that: 1. The X chromosome is less tightly condensed in males than in females. 2. The X chromosome of a male is enriched in an acetylated form of histone H4. 3. The protein products of each of the four msl loci are primarily associated with the X chromosome of the male but not that of the female. Examination of chromosomes in each of the four types of mutant msl males reveals that their X chromosome remains about as tightly condensed as an X chromosome in females, and that the absence of any single MSL-protein in a male results in the other three MSLproteins not associating with the X chromosome. a. Based on your understanding of the relationship between chromatin structure and gene transcription, how might a decrease in chromatin condensation be associated with dosage compensation? The acetylation of histone can destabilize chromatin packaging. As the DNA associated with nucleosomes becomes more accessible to transcription factors, the generally repressive action of histone on transcription is overcome and the rate of transcription is increased. In this case, the four MSL-proteins and an acetylated histone H4 appear to be necessary to decondensed the chromatin structure of the X chromosome of the male. This leads to an increase in the rate of transcription of genes on the single X chromosome of the male to a level equal to the rate of transcription of genes on the two X chromosomes of a female. b. Based on the data presented above, what potential roles might the protein products of the msl loci play in dosage compensation? The protein products of the msl loci may be involved in maintaining the looser chromatin structure and/or in increasing the rate of transcription once the chromatin has been decondensed. c. What evidence is there to suggest that the protein products of the msl loci act as a complex? Since mutations at any of the msl genes cause the remaining MSL-proteins not to bind the X chromosome of the male, it would appear that each is required for the binding of the others. This suggests that the four genes produce proteins that act together as a complex. 2. (3 pts) Both fragile X syndrome and Huntington disease are caused by trinucleotide repeat expression. Individuals with fragile X syndrome have at least 200 CGG repeats at the 5' end of the FMR-1 (fragile X mental retardation-1) gene. Individuals with Huntington disease have at least 36 CAG repeats within the protein-coding region of the huntingtin gene. In fragile X syndrome, the expanded CGG repeat results in DNA hypermethylation of the FMR-1 gene. Fragile X syndrome is the second leading genetic cause of mental retardation in the United States, although the function of the protein is
unknown. In Huntington disease, the expanded CAG repeat results in the inclusion of a polyglutamine stretch within the huntingtin protein, which causes it to have an abnormal function that overrides the function of a normal huntingtin protein. Huntington disease is characterized by involuntary movements and progressive central nervous system degeneration. a. How does a trinucleotide repeat effect the expression of these two genes differently? The trinucleotide in repeat the fmr-1 gene leads to transcriptional silencing (due to hypermethylation). The transcriptional silencing does not occur in the case of the hd gene. Instead, expression produces a protein with the abnormal polyglutamine stretch. b. Based on your answer to (a), why is the fragile X syndrome recessive, while Huntington disease is dominant? A heterozygote with a CGG repeat in the FMR-1 gene will still have one normal copy of the gene. The normal gene can produce a normal product, even if the other is silenced. In contrast, a novel, abnormal protein is produced by the CAG expansion in the disease allele in Huntington disease. Since the disease phenotype is due to the presence of the abnormal protein, the disease trait is dominant. c. Why is the number of trinucleotide repeats needed to cause the phenotype different for each disease? Transcriptional silencing may require significant amounts of hypermethylation, and so requires more CGG repeats for an effect to be seen. In contrast, protein function may be altered by a relatively smaller sized polyglutamine expansion. 3. (1 pt) If Actinomycin D, an inhibitor of transcription, is added to newly fertilized frog eggs, there is no significant effect on protein synthesis in the eggs. Similar experiments have shown that Actinomycin D has little effect on protein synthesis in embryos up until a particular developmental stage called the gastrula stage. After the gastrula stage, however, protein synthesis is significantly inhibited by Actinomycin D, and the embryo does not develop any further. How can you explain the lack of effect of Actinomycin D up until the gastrula stage? Preexisting mRNA that was made by the mother and packaged in the egg prior to fertilization is translated up to the gastrula stage. 4. (1 pt) Distinguish between maternal effect and non-Mendelian inheritance. Maternal effect is the determination of gene-controlled characters by the maternal genotype prior to the fertilization of the egg cell. A maternal effect is seen when nuclear genes of the mother function to specify some characteristic of the zygote. In contrast to nuclear genes that can function in the mother, the genes involved in nonMendelian inheritance are extranuclear and are found in mitochondria and chloroplasts. If the mitochondria are inherited with transmission of the mother's cytoplasm, mitochondrial genes will show maternal inheritance, a type of extranuclear inheritance. 5. (2 pts) A series of crosses performed with a recessive mutation in Drosophila called tudor. Homozygous tudor animals appear normal and can be generated from the cross of two heterozygotes, but a true-breeding tudor strain cannot be maintained. When homozygous tudor males are crossed to homozygous tudor females, both of which appear to be phenotypically normal, a normal-appearing F1 is produced. However, when F1
males are crossed to wild-type females, or when F1 females are crossed to wild-type males, no progeny are produced. The same results are seen in the F1 progeny of homozygous tudor females crossed to wild-type males. The F1 progeny of homozygous tudor males crossed to wild-type females appear normal, and they are capable of issuing progeny when mated either with each other or with wild-type flies. a. How would you classify the tudor mutation? Why? The tudor mutation is a maternal effect mutation. Homozygous tudor mothers give rise to sterile progeny, regardless of their mate. b. What might cause the tudor phenotype? The tudor phenotype results from the absence of some maternally packaged component in the egg needed for the development of the F1's germ line.