This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Name_______________________________ ID #________________________________ FINAL EXAM June 16, 2004 Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total A. B. C. D. Points 15 15 15 6 12 8 20 9 11 15 12 10 20 14 18 200 Score MCB164 Spring 2004 Page 1 of 9 This examination is closed book. There are 9 pages to the exam. Please count them before you start to make sure all are present. Please write your name on each page of the exam. Answer each question in the space provided. If additional space is required use the back of the page and indicate clearly that you continued your answer on the back. Do not attach additional pages. E. GOOD LUCK! Student authorization: I authorize the instructor to return the graded exam to me in the main office or to distribute it in the bin for me to pick up. Signature _____________________________ Date_____________________ Name_______________________________ MCB164 Spring 2004 Page 2 of 9 1. (15 points) Consider the four levels of chromosome folding to achieve a packing ratio of 10000. Match the characteristics of each level. (choose one) A. Nucleosomes B. 30 nm fiber C. Loops/Scaffold D. Coiling of loops (condensation) E. None of the above __B___ It likely involves a simple solenoid of six nucleosomes per turn with linker DNA tucked in the middle and stabilized by histone H1 protein __E___ A/T-rich DNA forms R-bands and G/C-rich DNA forms G-bands __A___There are 146 bp of DNA wrapped around the histone octamer, and two of them are connected by about 19 bp of linker DNA __C___In meiotic chromosomes, the bases of these are also likely the site of sister chromatid cohesion ___D__This is seen for the metaphase chromosome of higher eukaryotes (+3 points each) 2. (15 points) When telomeres become too short, cells possess a series of safety mechanisms to ensure the arrest of cell division and apoptosis (cell death). A. Name the critical cellular component that activates apoptosis in order that late crisis and chromosome fusion leading to anaphase bridges is avoided. (one word answer is ok) p53 (+3 points) B. What are anaphase bridges, how do they arise and what is their fate in the cell? Anaphase bridges are caused by the fusion of chromosome ends due to the loss of telomeres. This leads to a single long chromosome arm with a centromere at each end. When the centromeres are pulled to opposite poles the connecting fused arms form a bridge between the poles, which will eventually break to allow completion of mitosis. (+4 points each) Name_______________________________ MCB164 Spring 2004 Page 3 of 9 3. (15 points) Write a short paragraph 2-4 sentences using the following terms correctly: small interfering RNA (siRNA), microRNA (miRNA), Dicer, RISC (RNA induced silencing complex, autosilencing. Autosilencing of a transgene and endogenous sequence is achieved through the use of siRNA which is derived from double stranded RNA through the cleavage action of Dicer. miRNA participates in heterosilencing and also formed from the cleavage of a dsRNA (encoded in the genome) by Dicer. Both forms of small RNA are bound by the RISC complex which targets the small RNA to the mRNA. (+3 points each) 4. (6 points) What experimental evidence can you cite from budding yeast that meiotic recombination occurs at the four-strand stage (i.e. after DNA replication)? (hint: what type of tetrad would you get if recombination were to occur prior to DNA replication?) The presence of tetratype (TT) tetrads shows that recombination happens at the 4-strand stage. Consider 2 markers Aa and Bb on 2 different chromosomes. Only PD (AB and ab) and NPD (Ab and aB) could arise from crossing over at the two-strand stage. A tetratype will result when a crossover occurs between one of these markers and its adjacent centromere. 5. (12 points) Barbara McClintock established that recombination involves the physical exchange of genetic material using maize. For her analysis she followed cytologically detectable regions of heterochromatin called "knobs" during meiosis using a strain that was heterozygous for such knobs. The drawing below shows a representation of the bivalent she might have seen during metaphase I. The bivalent as shown has one chiasma. Knobs are shown as dark circles and centromeres are shown as open circles. A. Draw the structures of the bivalent at metaphase I (i.e. the cross stage) and the two chromosomes immediately following the first meiotic division. A. (+ 6 points) B. B. Name two sources of tension that stabilize the bivalent at metaphase I when the centromeres of the two homologs are pulled apart by the spindle? Name_______________________________ MCB164 Spring 2004 Page 4 of 9 Possible answers: chiasma, crossovers between nonsister homologs; sister cohesion between the chiasma and the telomeres, the spindle, monoorientation of the kinetochore (+6 points) 6. (8 points) Position effect variegation can be suppressed in male flies containing an extra copy of the Y chromosome (i.e. XYY). Give one possible way such a strain might arise that involves a meiotic nondisjunction event. Diagram meiosis for the parent in which nondisjunction occurred, but be sure to show the genotype of both parental gametes for the XYY fly. Do not worry about how you would score the phenotype of such an animal. A YY gamete could arise from a MII nondisjunction event in the male. Fertilization by an Xcontaining gamete from the mother will result in a XYY male fly. They should draw a picture of MI and MII divisions- you can work out the best way to divide up the points based on what people tended to do. 7. (20 points) Dosage compensation equalizes expression of genes from the X chromosome in the two sexes of the same species. Mark the following statements related to dosage compensation as TRUE or FALSE and give a brief explanation in support of your answer. (5 points each; 2 pts for T/F and 3 for the written answer) a. Female identical twins share more phenotypic characteristics than male identical twins. False: X inactivation in females makes them each a different mosaic for X-linked polymorphisms. In contrast, male identical twins have the same maternal X chromosome which is not inactivated. Thus, more phenotypic characteristics that are X-linked will be shared by the male twins than the female twins b. A XXY human male has two active X chromosomes. This is why men with Klinefelter's syndrome exhibit some feminization of secondary sex characteristics. False: Only one X chromosome is active in any given cell in XXY males. Feminization occurs due to reduced testosterone arising by testicular atrophy due to death of meiotic cells trying to segregate the extra X chromosome. c. One mechanism for dosage compensation is to increase X chromosome expression in the heterogametic sex. True: This mechanism occurs in XY flies so that the level of X expression matches that of homogametic (XX) flies. Name_______________________________ MCB164 Spring 2004 Page 5 of 9 d. Expression of a gene positioned near the centromeric heterochromatin of chromosome 1 in humans is more likely to be variegated in males versus females. (hint: the answer might be true or false but you have to justify whichever answer you choose) TRUE Since there is extra heterochromatin in female cells due to X-inactivation there may be fewer heterochromatin factors available for heterochromatic spreading. Or FALSE Although there is extra heterochromatin in female cells due to X-inactivation and therefore more use of limiting heterochromatin factors, it is likely that the cell has evolved a mechanism to ensure equal amounts of heterochromatin in the cell, perhaps by upregulating the limiting components. 8. (9 points) You insert a reporter gene into the orange bread mold Neurospora crassa at two different (non-allelic) positions in the genome to produce two isogenic haploid N. crassa strains, which faithfully express the reporter. When you cross the strains (cross 1), you surprisingly find that none of the spores expresses the reporter. You then insert the reporter gene into allelic positions in two haploids, and now all spores from the cross (cross 2) do express the reporter. Finally, you insert the reporter gene into only one haploid and the other is left untransformed, and spores from the cross (cross 3) do not express the reporter. (3 points each) A. At what stage in Neurospora life cycle is the reporter gene apparently silenced? i. vegetative haploid stage ii. dikaryon stage iii. before karyogamy iv. meiosis I prophase v. After meiosis II iv. meiosis I prophase B. This appears to be an example of: i. repeat-induced point mutation (RIP) ii. quelling (RNAi) iii. transvection iv. Silencing of meiotic unpaired DNA (SMUD) v. gene conversion v. SMUD C. What pattern of expression would you expect of the spores from cross 1 if the diploid is unable to make meiotic double-strand breaks and undergo homologous chromosome pairing? i. 4+: 4ii. 3:1:1:3 aberrant segregation iii. 2+:6- or 6+:2iv. 3+: 5- or 3-:5+ v. 8+:0vi. 0+:8vi. 0+:8- Name_______________________________ MCB164 Spring 2004 Page 6 of 9 9. (11 points) You mate male flies homozygous for a recessive su(var) mutation to female flies homozygous for an independently isolated recessive su(var) mutation and score the F1 progeny for a variegated brown eye phenotype. (Watch out! This question is a little trickier than it first seems) A. Circle the answer that best describes this type analysis? i. Epistasis test ii. Complementation test iii. Segregation analysis iv. Linkage analysis ii. complementation test (3 points) B. You find that the eyes of F1 progeny are more variegated than in either parent. Give two possible interpretations of these results. i. That these mutations complement one another (due to the lack of the su(var) phenotype) and are likely in different genes. ii. This may be an example of intragenic complementation caused by transvection. (e.g. intragenic complementation in yellow) (+4 points each) 10. (15 points) A model for choosing the inactive X chromosome during mammalian dosage compensation was presented in class in which an autosome encoded binding factor (BF) binds to a counting element located downstream of the Xist gene one the X chromosomes in XX mammals. In this model, BF is a negative regulator of Xist expression. A. For the diagram below, indicate which of the two X chromosomes will be inactive. BF CE CE Maternal X Paternal X The paternal X (+3 points) Xist Xist B. What is the nature of the gene product encoded by Xist? (e.g. RNA or Protein?) RNA (+3 points) Name_______________________________ C. Which chromosome(s) does the Xist gene product binds: 1. All chromosomes 2. Autosomes only 3. Both X chromosomes 4. The inactive X chromosome 5. The active X chromosome 4. The inactive X chromosome (+3 points) D. List three consequences to chromatin structure when Xist is bound. MCB164 Spring 2004 Page 7 of 9 bound by MacroH2A variant; methylation of H3 Lys9; condensed; hypoacetylated H4; late replicating; located at nuclear periphery, high levels of DNA methylation not correct: repeated DNA sequences; low gene density; low meiotic recombination (+6 points; 3 points each) 11. (12 points) You have created a mouse knockout mutation for the BGFT gene and find that the homozygous mutant has very large feet compared to normal mice. How would you determine if the wild type gene is normally imprinted? Compare BGFTDp/BGFT+m and BGFTDm/BGFT+p phenotypes to the BGFTDp/BGFTDm phenotype. If the BGFT null male crossed to a wild type female gives mice with big feet then BGFT is imprinted (silenced) on the maternally-derived chromosome. Likewise, if the BGFT null female crossed to a wild type male gives mice with big feet then BGFT is imprinted on the paternally-derived chromosome). If none of the progeny has big feet then the gene probably is not imprinted. (I don't know how to assign points now. Take a look at some and see what you think.) 12. (10 points) Why is a chromatid suffering a double-strand break often referred to as the "acceptor" of genetic information while the unbroken nonsister strand is referred to as donor of information? Use diagrams if you think they will help. If there is a DNA polymorphism near the break (i.e. it is a heteroallele) then resection of the DNA to give 3'ss overhangs will result in loss of information for that particular strand. Since in meiosis the nonsister strand is used as a template for repair, new DNA synthesis will be complementary to the "donor" strand. The broken strand is thus the "acceptor" of this new information. Name_______________________________ MCB164 Spring 2004 Page 8 of 9 13. (20 points) The expression of a URA3 reporter gene inserted near a telomere (~2 kb away) in yeast exhibits variegated expression. Telomere position effect (TPE) can be measured by growth on media containing the drug 5FOA where silencing leads to growth of Ura- cells. Four genes in yeast are required for silencing: RAP1, SIR2, SIR3, and SIR4. Rap1 protein binds to a specific DNA sequence found only at telomeres. Binding of Rap1 recruits Sir3 and Sir4, which can also bind to unacetylated histone H3. SIR2 encodes a histone deacetylase. Spreading is thought to occur by the action of Sir2 clearing a binding site for Sir3 and Sir4 proteins. Sir3 is the limiting component of silencing. Write the relative levels of growth on 5FOA that you would expect in the following mutant backgrounds and in the presence of a boundary element positioned either on the telomere proximal or telomere distal side of URA3. Use a scale of -, +, ++ Genotype Wild type Over expression of SIR3 Over expression of RAP1 rap1D sir3D sir3D rap1D sir3D + over expression of RAP1 rap1D + over expression of SIR3 rap1D + over expression of SIR2 Wild type with boundary element telomere proximal to URA3 Wild type + boundary element telomere distal to URA3 Growth on 5FOA + ++ + + If a student has clearly written the opposite of all the answers (i.e. if he or she misinterpreted what 5FOA growth measured then give full credit but subtract 5 points for the error. 14. (14 points) Tor1 is an evolutionarily conserved protein kinase that controls cellular growth in response to nutrients. In fission yeast a tor1D has a temperature sensitive phenotype for growth at 36oC. Gad8 which encodes a Ser/Thr kinase was isolated as a high-copy suppressor that partially suppresses the tor1D ts phenotype. A gad8D strain also exhibits a temperature sensitive phenotype for growth at 36oC. A. Describe an experiment to test whether or not tor1 and gad8 act in the same or in parallel (i.e. redundant) pathways. Include how you would interpret possible results. Construct a double deletion strain of tor1D and gad8D and test to see if the phenotype is the same as both of the single deletion mutants. If the phenotype is the same then they are Name_______________________________ MCB164 Spring 2004 Page 9 of 9 probably in the same pathway. If the phenotype is enhanced then they are probably in parallel pathways. (+8 points) B. You find from your results in part A that gad8 and tor1 appear to act in the same pathway. Would you expect that over expression of tor+ would suppress a gad8D ts phenotype? Why or why not? If the two genes are in the same pathway then the fact that overexpression of gad8+ suppresses tor1D suggests that gad8 acts downstream of tor1. Therefore it is not expected that overexpression of tor1 should be able to compensate for the loss of gad8. (+6 points) 15. (18 points) RNAi an important genetic tool for genetic research in C. elegans. Interestingly only a small amount of RNA is required to elicit gene silencing throughout the animal and its progeny. It is often used to infer something about wild type gene function. A. What form is the RNA generally introduced into the animal? i) antisense to target gene ii) the sense strand of the target gene iii) double-stranded RNA B. How is RNA generally introduced into C. elegans? i) by microinjection into the animals ii) by soaking animals in a solution of RNA iii) by feeding animals E. coli expressing the RNA iv) any of the above means v) none of the above means C. What enzyme expressed by C. elegans allows for the amplification and spreading of silencing throughout the animal? i) Dicer ii) Argonoute iii) RNA-dependent RNA polymerase iv) DNA methyltransferase v) Histone methyltransferase D. Circle the following statements that are true concerning the use of RNAi in C. elegans: i) RNAi usually produces a mutation at the target DNA locus which is then inherited as a mendelian trait. ii) RNAi is generally used to create a phenocopy of a loss-of-function mutation for a given target gene iii) RNAi is generally used to create a phenocopy of a gain-of-function mutation for a given target gene. Name_______________________________ MCB164 Spring 2004 Page 10 of 9 iv) The use of RNAi to infer the function of a wild type gene product is an example of reverse genetics v) forward genetic screens can be carried out using RNAi ...
View Full Document
This note was uploaded on 04/04/2010 for the course MCB 164 taught by Professor Burgess during the Spring '08 term at UC Davis.
- Spring '08