4) Use ATP and Swi.Snf type enzymes to move the histones along the DNA

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1 GENE 603 Exam 2, October 27, 2017 NAME 1. What are histones and how are they involved in regulating gene expression? Describe 3 common modifications and the effect they would be expected to have on the histone-dna interaction. Histones are highly conserved proteins in eukaryotes that form an octomer (2 each H2A, H2b, H3 and H4) around which two wraps of DNA are held in place by H1 to form a nucleosome. Modification 1: add (histone methyl transferases) or remove (histone demtheylases) a methyl group(s) from lysines and arginines in the exposed tails of histones to up or down regulate transcription 2) Add (HAT) acetyl groups to histone + charged amino acids to decrease the binding of DNA and potentially increase transcription or remove (HDAC) for tighter binding and lowering transcription. 3. add or remove phosphates 4) Use ATP and Swi.Snf type enzymes to move the histones along the DNA 2. What is: a) genomic imprinting? Give an example. Genomic imprinting is inactivation of different specific chromosome regions in the formation of male and female gametes so that only one allele will be expressed in the embryo. Prader Willi is a human example in humans where the male-functional allele is defective. b) A Barr body and how does it figure in Lyons Law? Barr bodies X chromosomes that are inactivated and condensed on a random basis per cell early in embryology whenever more than 1 X is present. Lyons law uses the X-inactivation to explain dosage compensation (= expression of X-linked genes in males and females) and the fact that the same X remains inactive in mitotic daughter cells to explain patches of color for example in calico cats. c) the significance of a LOD3 score? LOD stands for log of the odds and infers how much more likely data from mapping experiments are likely to occur assuming that the map distance calculated is correct than if random assortment is involved. In most cases the odds that two random genes would be close together is about 1% so LOD3 actually implies more like a 95-99% likelihood rather than 1000 fold more likely

2 3. In yeast, Gal4 protein has two tethered domains that are involved in regulating transcription of 3 enzymes needed for using the sugar galactose as a carbon source. A) What is the normal role of the 2 domains? The binding domain recognizes and upstream activator domain in the Gal 4 promoter. The activation domain bends over to the TATA box region in the presence of Galactose to transcribe genes needed to break it down. B) What was changed to create yeast-two-hybrid screening? Another protein is fused to the binding domain, which is uncoupled from the activation domain, to which proteins made in an EST library are fused. If any of the proteins in the library interact with the binding domain, expression is activated. Inmost cases the Gal structural gene is replaced with something like LacZ so that blue-white screening can be used. C) What is the goal of yeast-two-hybrid screening? Identification of interacting proteins 4. A) List features that puc vectors make then highly useful in cloning DNA. Lac Z gene for easy screening for plasmids with inserts A Multiple Cloning site with several unique (only present once in the puc) restriction endonuclease cut sites in the beginning of the Lac Z sequence An origin of replication that allows > 60 copies per cell A gene for ampicillin resistance for easy screening of amp- cells that contain the plasmid B) Why would someone choose a vector that automatically added 6 histidine codons to the 5 or 3 ends of the mrna? This allows for direct collection of the protein made on a Nickel column C) Why would someone choose a vector that can function in both E. coli and Pichia pastoris? Pichia makes almost the same post-translational protein modifications that animals make and neither E. coli nor standard yast does not.

3 4. An ATPase in animals is expressed in many types of tissue. Three forms, all from the same gene, AS1, AS2 and AS3 of the enzyme are found: AS3 is only found in muscle and AS1 And AS2 are found in liver and kidney. Propose a mechanism to explain these observations, including a theoretical model for how the tissue specificity might be accomplished. Expected, though no the only accepted answer: Differential splicing of the pre mrna could lead to different forms and enzymes in different tissue. A model to explain this could be that different micrornas with tissue-specific expression for example may interfere affect splicing by interfering with SNRNPs or by substituting for different SNRP components (some evidence for this in brain tissue for example) 5. The amount of DNA per genome, also called the C value, led to something called the C- value paradox. What is the paradox and how is it explained. The C-value paradox is why huge difference in the amount of DNA per genome can occur in organisms with similar coding requirements. The explanation is that those with a great deal more DNA have a lot more repeated sequence DNA, most of which is non-coding. 6. Genes D, E and G on chromosomes 1, 2 and 3 in a Drosophila male (no crossing over) are heterozygous in an individual. Make a drawing showing early anaphase of 1) a cell in mitosis; 2) a cell in Meiosis 1. Predict the gamete genotypes from your meiotic cell. D e G and d E g

4 7. When a green plant was self pollinated three phenotypes were seen in the progeny, in the ratio of 70 white : 180 green and 70 yellow, making it clear the plant was not truebreeding. One hypothesis was a single gene with incomplete dominance might be present and the other hypothesis suggested two genes with epistasis involved. A) Show a one gene model to fit the data and test the expected segregation ratio using Chi square. Model: incomplete dominance with a 1 white: 2 green: 1 yellow Expected 80 white: 160 green : 80 yellow X 2 = 5 P(between 5 and 10 percent the data could represent chance distribution. What do the results say? Under the standard 95% level of confidence, the hypothesis cannot be rejected. B) Make a two gene epsistasis ratio and pathway to fit the data and test the expected ratio using Chi Square. Pathway: w! yellow! green Prediction 9/16 green =180 3/16 yellow = 60 X 2, 2df = /16 white = 80 What do the results say? There is over a 25% chance that deviation this large or larger would occur by chance; Hypothesis can be accepted. C. Is one hypothesis better than the other? Both are acceptable; perhaps looking at in a Bayes context, the data are more likely to have occurred from the 2 gene model. 8. In humans, T_ individuals taste PTC while tt individuals do not. A taster man whose parents were also tasters but who has a non-taster sister marries a non taster woman. Their first 3 children are tasters. What is the probability that he is Tt, first from his own family information and then after he has 3 taster children? A) 2/3 B) 2/3 (1/2) 3 //2/3 (1/2) 3 + 1/3(1) 3

5 9. Feather color in budgies, a small parakeet, results from 2 genes. An autosomal gene with a dominant Y allele produces yellow feathers and a dominant allele (B) of a Z-linked gene produces blue feathers. When both dominant alleles are present, the bird is green and if neither is expressed, it is white. Give the Parental genotypes, the F1 genotypes and F2 ratios predicted from sib matings for a cross between a true breeding Blue feather male and a true breeding yellow female. P1 male Z B Z B, yy P1 female Z b W, YY F1 male Z B Z b,yy F1 female Z B W, Yy F2 males 6 green : 2 blue F2 females 3 green : 3 yellow 1 blue : 1 white F2 overall 9 green: 3blue: 3 yellow : 1 white but need to include sex since it matters. 10. Eight different genes have been implicated in CLN disease in humans. Symptoms include odd fingerprints, progessive dementia and vision failure. Most cases are first detected in children, but for the latter traits, juvenile and adult onset has been verified. Very mild cases have been found in individuals following extreme exposure to reactive oxygen species. What terms associated with extensions that can confound simple Mendelian inheritance can be associated with this human condition (tell why for each). Genetic heterogeneity (8 genes cause CLN) Age of onset (juvenile and adult onset) Pleiotropy (multiple organs/phenotypes involved) Phenocopy (ROS cause similar problem) Variable expressivity, penetrance and epistasis are almost certainly involved, but no solid information in the writeup.

6 11. In maize: Wx_ waxy M _ male fertile B_ brittle stalk wx/wx starchy mm male sterile bb flexible stalk Pollen from a truebreeding starchy, flexible stalk plant is used to fertilize a truebreeding Waxy, Brittle, Male sterile plant. The F1 is then crossed to a female homozygous for all three recessives.(phenotypes are indicated by the allele from the male) The progeny of this cross gave the following phenotype s in 1000 progeny. m, Wx, B 377 m, wx, B 14 m, Wx, b 77 m, wx, b 3 M, Wx, B 4 M, wx, B 19 M, Wx, b 96 M, wx, b 410 a) Give the genotypes of the original parents and the F1. Parents: wx M b Wx m B F1 wx M b Wx M b Wx m B Wx m B b) Map the genes: include Interference and the Coefficient of Coincidence cm 10.3 cm Wx M B c) Does any gene/allele suggest a potential problem with viability? How did you decide? I found these data in another text and assumed that the 14 and 19 would be reciprocal and the same for 77 and 96. However, in working it out, the 14 and 96 are one class etc so checked to see if I had made an error in rewriting. Since not, I left as is. However if you check B and b for equal frequency as should occur, the X 2 P value is and Wx vs wx numbers are also do not fit a 1:1 ratio.

7 12. The data below from yeast show the 4 spores produced in meiosis from a cross involving 3 genes (a, b & c) where + represents the normal allele. Note that the 4 meiotic spores are not in linear order in yeast. The parents were + b c and a + +. Ascus Composition and Frequency Spore a + + a b + a + + a b + a + c a b c 2 a + + a b + a + c a b c a + c a b c 3 + b c + + c + b b b c + + c + b c + + c + b # observed Which of the asci (1-6) are PD, NPD and TT for pairs!! of the genes? For a and b 1, 3 & 5 are Pd, 2, 4 & 6 are NPD and none are TT For a and c 1 & 2 are PD, 5 & 6 are NPD and 3 & 4 are TT For a and b 1 & 6 are PD, 2 & 5 are NPD and 3 & 4 are TT Are any of the genes linked? How do you know? A and C since there are way more PDs than NPDs 13. Fawn and chartreuse are two spore color mutants in Aspergillus nidulans, a fungus that can grow as a diploid. Synteny mapping shows that both are on the same chromosome. A double heterozygote f+/f, c+/c that is green spores because each color mutant is recessive was used to search for mitotic recombination. Rarely, when colonies were grown on plates a fawn or chartreuse sector appeared in colonies, with fawn twice as frequent as chartreuse, but twin color spots were never seen. Make a map showing the genotypes associated with parental chromosomes that would explain these observations. f+ c f+ c f c+ f c+ this is one possibility since a mitotic crossover would almost never occur between both genes and the centromere. If both are on the same side, tin spots could occur 14. Rarely, females, as based on appearance and mental status, are born who have an XY chromosome composition. Give 2 examples to explain how this can happen genetically. Expected: Swyer syndrome where the Sry gene itselsf is defective Complete androgen insensitivity where the testosterone receptor gene on the X chromosome is defective