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Announcements

Announcements. -First midterm exam will be in this room on Friday (4-25) from 10:30AM-12:20PM -Exam will cover material presented in lecture and quiz section through the end of last week, however…. -this weeks material will reinforce some of the previous concepts you have learned

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Announcements

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  1. Announcements -First midterm exam will be in this room on Friday (4-25) from 10:30AM-12:20PM -Exam will cover material presented in lecture and quiz section through the end of last week, however… -this weeks material will reinforce some of the previous concepts you have learned -this weeks material WILL be covered on the next midterm exam

  2. Leo Pallanck’s office hours: Friday afternoons by appointment (pallanck@u.washington.edu)

  3. how is genetic information read within cells? how are traits transmitted to progeny? how are unique physical traits determined by genes? Today how and why are model organisms used in genetics? how does that information apply to humans? What this course is about From Lecture 1 Inheritance More to come! Mutant analysis how are biological processes studied by analyzing mutants? Genomics what can we learn by studying whole genomes? Throughout the quarter . . .

  4. Mutant analysis (AKA Genetic Analysis) The use of mutants to understand how a biological process normally works* Very powerful - can be used to study metabolic pathways, animal development, neurobiology, cell division, etc. A simple analogy… *See the Salvation of Doug article at the following site: http://bio.research.ucsc.edu/people/sullivan/savedoug.html

  5. Analysis of pizza synthesis Analagous to genes

  6. Analagous to a mutation The mutant phenotype Analysis of pizza synthesis Analagous to genes No red sauce?!

  7. Mutant analysis involves model organisms What is a model organism? A species that one can experiment with to ask a biological question Why bother with model organisms? • All organisms are related at the molecular level • Not always possible to do experiments on the organism you want • If the basic biology is similar, it may make sense to study asimple organismrather than acomplexone

  8. Which of these cameras do you think would be easier to understand? IMAX 3-D camera Box camera

  9. • Mendel used a model organism—the garden pea • relatively short generation time—one per year • lots of progeny per cross • self-pollination and out-crossing possible • - true-breeding varieties readily available from local merchant Telomeres Features of a good model organism • Short generation time • Small, easy to maintain • Large numbers of progeny • Well-studied life cycle, biology • Appropriate for the question at hand 96 million telomeres per cell!

  10. Some commonly used model organisms • Bacteria — Escherichia coli • Budding yeast — Saccharomyces cerevisiae • Fruit fly — Drosophila melanogaster • Nematode — Caenorhabditis elegans • Mouse — Mus musculus

  11. Quiz Section this week: Complementation analysis of yeast mutants An introduction to yeast . . .

  12. mitosis cytokinesis mutation Yeast as a model “genetic” organism Budding yeast—a single-celled fungus that divides by budding Yeast cells can exist as haploids… Mutagenesis is easier in single-cell organisms with haploid lifestyles Haploid life cycle: The haploid life cycle (1n)

  13. mating The life cycle of “budding” yeast Yeast cells can also exist as diploids… 1n a haploid a haploid The haploid life cycle (1n) diploid zygote

  14. Mendelian segregation occurs here 2 a cells 1n 2 a cells The life cycle of “budding” yeast (cont) a/a diploid life cycle (2n) A tetrad with 4 haploid spores (“gametes”) meiosis

  15. Conducting a mutant analysis with yeast Case study: analyzing the adenine biosynthetic pathway by generating and studying “ade” mutants Wild-type yeast can survive on ammonia, a few vitamins, a few mineral salts, some trace elements and sugar… They synthesize everything else they need, including adenine What genes does yeast need to synthesize adenine? (Why might we care about adenine?)

  16. Adenine-requiring colonies (ade mutants) Treat wt haploid cells with a mutagen: plate cells -adenine plate m2 m1 sterile piece of velvet m3 Identifying yeast mutants that require adenine “complete” plate “Replica-plating”

  17. Genotypes: ade ADE diploids replica-plate using velvet Are the adenine-requiring mutants recessive? That is, are they LOF mutations? Why do we care? “a” mating type “” mating type m1 wild-type -adenine plate “complete” plate What do you conclude? ADE is dominant over ade

  18. Are all of the mutations in one gene? Are m1 and m2 alleles of the same gene? What would you predict if… • only one enzyme is needed for synthesis of adenine? • many enzymes are needed for synthesis of adenine? All mutants would be alleles of the same gene. Different genes might be mutant. How to find out how many different genes we have mutated? Do complementation test to ask: are the mutationsalleles of the same gene orof different genes?

  19. diploids -adenine 4 replica-plate 4 Performing a complementation test “a” mating type “” mating type m1 m2 “complete” Do m1 and m2 complement, or fail to complement? Are m1 and m2 alleles of the same gene, or alleles of different genes?

  20. Complementation tests with ade mutants What do you conclude from the pair-wise crosses shown below? Conclusion? m1 m2 m3 m4 m5 m6 m7 x m1, m5, m7 are mutations inone gene m1 + + + + o o o m2 o m3 o m4 o m5 o m6 o o=no growth on -ade += growth on -ade m7 o

  21. o + + + + + + + + + + + + o + Complementation tests with ade mutants What do you conclude from the pair-wise crosses shown below? Four complementation groups Usually means four genes Conclusion? m1 m2 m3 m4 m5 m6 m7 x m1, m5, m7 are mutations inone gene m1 + + + + o o o m2 o m2, m4 are inonegene m3 o m4 o m3 m5 o m6 m6 o o=no growth on -ade += growth on -ade m7 o

  22. Practice Question Yeast cells can normally grow on a sugar called galactose as the sole carbon source. Seven mutant “” haploid yeast strains have been isolated that are unable to grow on galactose (“gal”) plates. Six of these mutant strains were each cross-stamped on a gal plate with a wild type “a” strain. The resulting pattern of growth on the gal plates is depicted below (shading = growth). In all plates, the wild type strain is in the horizontal streak. On the leftmost plate, mark the location of the a/ diploid with a circle. What is the mode of inheritance of mutant phenotype in mutants 1-6? How can you tell? Diploids grow on gal plate… so, wild type is dominant

  23. Practice question (continued) Each of the seven “” mutant strains was cross-stamped on gal plates against “a” versions of the seven mutants. The results are depicted below: m1, m2, m5 m3, m6 m4 Looking just at mutants 1–6 for now… group these six mutants by complementation group.

  24. Practice question (continued) Now consider mutant 7. What is surprising about the result in the complementation table? Fails to complement any of the others… how could it be an allele of 3 different genes? Mutant 7 was cross-stamped on gal plate with wild type as you saw with the other six mutants earlier: What do you conclude about the mode of inheritance of mutant 7? How does that help you explain the complementation test result for mutant 7? Complementation test fails with a dominant mutation… heterozygote will always show the mutant phenotype What can you conclude about how many genes are represented in this collection of seven mutants? At least 3 genes (can’t tell about m7)

  25. Complementation is relevant to humans Family A Family B = deaf aaBB AAbb AaBb Within each family, does deafness look like it’s dominant or recessive? Assign genotypes (A, B, etc.) to the deaf individuals in these pedigrees. recessive

  26. normal NPC NPC1:NPC2 NPC1:NPC3 Complementation is relevant to humans Niemann Pick Type C disease (NPC): a recessive human neurodegenerative disease resulting in premature death Cellular cholesterol accumulation accompanies the disease (can be detected using a chemical called ‘filipin’ which fluoresces upon binding cholesterol) Do NPC1 & NPC2 complement? Do NPC1 & NPC3 complement?

  27. Practice Question Hearing mice… independently assorting genes A and B, both needed for hearing: a, b: complete LOF, recessive AaBb AaBb x WITHOUT drawing a Punnett square, predict the progeny phenotypes and proportions with respect to hearing ability.

  28. For example, this molecule accumulates in an ade13 mutant. -OOC N C CH C N H2N R N HC CH O COO- C N C H2N N CH2 O R C HN HN CH C HC N HC C C ADE13 NH N CH COO- encodes the enzyme that carries out the next step C N H2N R using mutants to order the steps in a pathway . . . (AIR) (SAICAR) (CAIR) . . . Adenine

  29. The Yeast Adenine Biosynthetic Pathway A B D E F Y G C H I J K

  30. A second phenotype of some ademutants… plate cells Mutagenize: complete plate Replica-plate Some of the adenine-requiring mutants arered! -adenine plate

  31. Genotypes: ade ADE replica-plate using velvet Are thered ademutations recessive? m8 wild-type -adenine plate complete plate White color is dominant Ability to make adenine is dominant

  32. LOF mutation ade1 X red pigment Hypothesis: Two phenotypes/one LOF mutation How can one LOF mutation generate two very different phenotypes? another intermediate “Y” some intermediate “X” ADE1 adenine X adenine Y UNK1* *not a real gene name! This gene has not yet been identified

  33. Suppose we isolate LOTS of independent red mutants: Are all red mutants defective in the SAME GENE? How to tell?

  34. m9 m10 m9 m11 diploids white diploids Complementation tests of red mutants mutations fail to complement = same gene mutations complement = different genes All pairwise combinations reveal two complementation groups. Must modify the hypothesis.

  35. Y adenine X “UNK1” red pigment Modified hypothesis for red phenotype ADE2 ADE1 Y adenine X ade2 ADE1 X But how do mutations in ADE1 result in a build-up of X?

  36. “UNK1” red pigment Modified hypothesis for red phenotype ADE2 ade1 X Y X Y adenine Mutations in either ADE1 or ADE2 lead to a defect in adenine biosynthesis and lead to the build-up of intermediates in the pathway. Excess “X” is converted to a red pigment.

  37. ADE13 ADE7 ade1 ade2 Same as ade7 single mutation! White and adenine-requiring. Whiteand able to grow on -ade plates. Test your understanding X Y X Y adenine “UNK1” red pigment 1. Phenotype of ade1 ade2 double mutation? Same as ade2 single mutation! Red and adenine-requiring. 2. Phenotype of ade2 ade7 double mutation? 3. Phenotype of ade2 ade13 double mutation? Same as ade2 single mutation! Red and adenine-requiring. 4. Phenotype of unk1?

  38. Practice Questions ADE1 ADE3 ADE2 X Y adenine X Y “UNK1” red pigment 1A. A MATa ade2 ADE3 mutant was mated to a MAT ADE2 ade3 mutant to create a diploid. What are the phenotypes of the three strains? Assume all other genes are wild type.

  39. no red no white white yes Practice Questions ADE1 ADE3 ADE2 X Y adenine X Y “UNK1” red pigment 1A. A MATa ade2 ADE3 mutant was mated to a MAT ADE2 ade3 mutant to create a diploid. What are the phenotypes of the three strains? Assume all other genes are wild type.

  40. 1B. ADE2 and ADE3 assort independently. Draw the chromosomes at metaphase of meiosis I such that the two WILD TYPE alleles face the same pole. Place a crossover on the other chromosome arm relative to the ADE2 and ADE3 genes. ADE2 ADE3 ADE2 ADE3 ade2 ade3 ade2 ade3 A B C D 1C. Recall that each tetrad contains the products of a single meiosis. Predict the genotypes and growth properties of each spore resulting from this meiosis. complete genotype? grow without adenine? Spore yes ADE2 ADE3 A yes ADE2 ADE3 B no ade2 ade3 C ade2 ade3 no D 1D. Analysis of many tetrads demonstrates that three types are found, depending on the behavior of the chromosomes in meiosis. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table. Tetrad on complete plates #1 #2 #3 A B C D red

  41. 1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. Place a crossover on the other chromosome arm relative to the Adenine genes. A ADE2 ade3 B ADE2 ade3 1F. Predict the genotypes and growth properties of each spore resulting from this meiosis. complete genotype? grow without adenine? Spore A B C D 1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table. Tetrad on complete plates #1 #2 #3 red

  42. 1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. Place a crossover on the other chromosome arm relative to the Adenine genes. A C B D ADE2 ade3 ade2 ADE3 ADE2 ade3 ade2 ADE3 A B C D ADE2 ade3 ADE2 ade3 ade2 ADE3 ade2 ADE3 1F. Predict the genotypes and growth properties of each spore resulting from this meiosis. complete genotype? grow without adenine? Spore no A no B no C no D 1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table. Tetrad on complete plates #1 #2 #3 red

  43. 1H. Now draw the chromosomes at metaphase of meiosis I such that one wild type and one mutant allele face each pole. On one chromosome, place a crossover on the other chromosome arm relative to the Adenine gene. On the other chromosome, place a crossover BETWEEN the centromere and the Adenine gene. 1I. Predict the genotypes and growth properties of each spore resulting from this meiosis. complete genotype? grow without adenine? Spore A B C D 1J. Which tetrad best fits the meiosis you just drew? Letter the spores below to match the genotypes in your table. Tetrad on complete plates #1 #2 #3 red

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