Presentation on theme: "PCB5065 Fall 2010 Name _key____________________________________ Exam 4 Total value = 70 points Question 1___________________________ Question 2___________________________."— Presentation transcript:
PCB5065 Fall 2010 Name _key____________________________________ Exam 4 Total value = 70 points Question 1___________________________ Question 2___________________________ Question 3 ___________________________ Question 4 ___________________________ Important – please keep your answers short; confine your answers to the space provided; do not write on the back of any pages !
PCB5065 Fall 2010 Name ____________________________________________ Exam 4 page 2 of 5 1. (15 pt, 5 pt each part) The questions below pertain to the genetic transformation of organelle genomes, which presents some special challenges. a) Transgenes are introduced into organelles via bombardment with DNA-coated particles and inserted into the organelle genomes via homologous recombination. Immediately after this occurs, the bombarded cells are heteroplasmic for the transgene. What genetic behavior of organelle genomes allows for the eventual selection of homoplasmic organelle genome transformants? Explain your answer. Somatic (mitotic) segregation Through this process organelle genomes having different haplotypes sort into different cells. Over multiple generations this leads to homoplasmic cells. In the absence of selection the each of the initial haplotypes should be represented within the cell population. At this point the even a recessive selectable marker, e.g. ribosome-based antibiotic resistance, could be selected to eliminate the antibiotic sensitive haplotype and recover the resistant haplotype. b) Because organelle genome transformation occurs by homologous recombination, mutant correction by homologous gene replacement is possible. Explain how you would select for yeast cells in which a mutant gene essential for mitochondrial respiration had been successfully replaced with a wild-type gene. wild type yeast cells can grow on glucose, via fermentation, or on glycerol, a carbon source that can must be respired. Yeast mutants that cannot respire will grow on glucose but not on glycerol, so cells could be tested for growth on glycerol or for large colony size on glucose + glycerol media. c) Because organelle genome transformation occurs by homologous recombination, mutant correction by homologous gene replacement is possible. Explain how you would select for a a chlamydomonas cell in which a mutant gene essential for chloroplast photosynthesis had been successfully replaced with a wild-type gene? wild-type chlamydomonas cells can grow in the presence of light without any supplied carbon source in the media. (They fix CO2 via photosynthesis.) Chlamydomonas mutants that cannot carry out photosynthesis will grow in the dark when provided acetate as a carbon source, but will not grow in the presence of light and absence of acetate. So cells could be tested for growth in the presence of light and absence of acetate. or Photosynthetic mutants of chlamydomonas are often light-sensitive due to damaging light reactions that result from partially assembled photosynthetic complexes. So you could select for cells that survive and grow in the presence of high light.
2) (15 pt, 5 pt each part) In mice, the imprinted insulin growth factor 2 (Igf2) gene is expressed from the paternally-inherited chromosome. The maternally contributed gene is silent. The complete loss of Igf2 function is viable, the mutant mice are just much smaller than wild-type mice. a) If a mouse is heterozygous for a loss-of-function mutation at the Igf2 locus (genotype Igf2 – / + ), will this mouse have a mutant or wild-type phenotype? Explain your answer. This depends upon whether the paternal or maternal parent contributed the mutant allele. If the paternal parent was mutant, the moue in question will have a mutant phenotype. If the maternal parent was mutant, the mouse will have a normal phenotype. b) If an Igf2 – / + male mouse is mated with a wild-type (Igf2 +/+ ) female mouse, what are the expected frequencies of Igf2 genotypes and resulting phenotypes in the offspring? Explain your answer. Transmission is Mendelian but only the paternal allele is expressed, so half are mutant and half wild-type PCB5065 Fall 2010 Name ____________________________________________ Exam 4 page 3 of 5 genotypesfrequenciesphenotypes Igf2 – / + 0.5small (mutant) Igf2 +/+ 0.5normal (wild-type) c) If an Igf2 +/+ male mouse is mated with an Igf2 – / + female mouse, what are the expected frequencies of Igf2 genotypes and resulting phenotypes in the offspring? Explain your answer. Transmission is Mendelian but only the paternal allele is expressed, so all are wild-type, the mutant allele derived from the female is not expressed genotypesfrequenciesphenotypes Igf2 – / + 0.5normal (wild-type) Igf2 +/+ 0.5normal (wild-type)
PCB5065 Fall 2010 Name ____________________________________________ Exam 4 page 4 of 5 3d) (2 pt)Which of the gene classes listed above are highly conserved in gene sequence and gene product function, between insects and mammals? Segment polarity and segment identity genes 3e) (3 pt) Which of the gene classes listed above have functions unique to drosophila and related insects, and what developmental feature or advantage is conferred by those gene functions? Maternal effect, gap and pair-rule genes. These allow for rapid establishment of embryo patterning and therefore a rapid developmental program Order of action in fly development Gene classGeneral function of protein products Loss-of-function mutant phenotype in fly larva or adult 1maternal effect genes transcription factorlethal larvae lacking polarity larvae with mirror imaged anterior- posterior features 2gap genestranscription factorlethal larvae missing central features mirror imaged remaining features 3pair rule genestranscription factorlethal failure to establish larval segments 4segment polarity genes secreted peptides receptors transcription factor (engrailed) in whole embryo: lethal loss of larval segment polarity loss of parasegment boundary in mutant sectors: missing adult structures mirror-imaged adult structures 5segment identity genes transcription factoraltered identity of adult segments homeotic transformation of adult organs 3. (20 pt.) 3a) (5 pt)The major gene classes that act in drosophila development are: homeotic (segment identity) genes, gap genes, maternal effect genes, pair-rule genes and segment polarity genes. In the table below, list these classes of genes in the order that they come into play during the drosophila developmental program (first =1, last=5). 3b) (5 pt) In the table below, indicate the general function of the proteins encoded by each gene class (e.g. transcription factor, receptor, etc.) 3c) ( 5pt)In the table below, indicate one phenotypic feature that is commonly observed in fly larvae or adults that are homozygous for a loss-of-function mutation within each gene class
5a) (2 pt)Based upon the ABC model diagrammed above, indicate which gene or combination of genes that acts to specify: Sepals A alone Petals A+B Stamens B+C Carpels C alone 5b) (1 pt) Which two gene classes in the diagram above mutually repress each other’s expression ? A and C 5c) Over x (above), use boxes to diagram the spatial distribution of remaining gene products if B function is entirely lost by mutation; underneath your boxes, indicate the flower structures that are predicted to develop. Just label them (e.g. sepal, petal, etc.); no need to draw them. 5d) Over y (above) use boxes to diagram the spatial distribution of gene products and then indicate the flower structures predicted to occur if C function is entirely lost by mutation. 5e) Over z (above) use boxes to diagram the spatial distribution of gene products and then indicate the flower structure predicted to occur if A function is entirely lost by mutation. 5f) (3 pt) In the space below, briefly compare and contrast the nature of the plant floral organ identity genes with the drosophila segment identity genes respect to their evolutionary origins and functions. 1 pt each, any three; note the question specifically asked about organ or segment identity, not about development in general Both encode transcription factors Drosophila, indeed all animals, use the homeobox transcription factors to specify segment or organ identity. Although plant genomes contain homeobox transcription factor genes, plants recruited a different class of transcription factor genes, the MADS box genes, to specify organ identity Evolved independently Mutual repression observed in both systems Mutants condition homeotic phenotypes in both systems In animals the homeobox genes are clustered, whereas the plant MADS box genes are not 5. (20 pt.) The diagram below shows the ABC model describing the specification of different floral organs in the four whorls of the arabidopsis flower. The spatial positioning of floral organ identity gene (A, B and C) expression is shown, along with the resulting pattern of floral organs. x (4 pt) y (5 pt) z (5 pt) sepal petal stamen carpel B B A A C C PCB5065 Fall 2010 Name ____________________________________________ Exam 4 page 5 of 5 A A C C B B A A A A B B C C C C sepal sepal carpel carpelsepal petal petal sepal carpel stamen stamen carpel