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Announcements

Announcements. You should be working on chapter 6 problems: 10, 14, 15, 28 (not to turn in).

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Announcements

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  1. Announcements You should be working on chapter 6 problems: 10, 14, 15, 28 (not to turn in). Reminder- papers on “Monk in the garden” due in lab section 10/1, 10/2 (next week). I encourage you to see me to discuss your outlines. I will look over first drafts if received on or before Monday 9/30. Grading of Exam 1 should be completed Friday. Correct answers will be posted in the lab.

  2. Review of Last Lecture I. Genes linked on the same chromosome segregate together - linkage affects types of gametes that are formed; contrast to independent assortment when genes are not linked - linkage affects phenotypic ratios in the F2 and in testcrosses II. Crossing-over - Recombination is caused by linear arrangement of genes and crossing over. - Frequency of recombination is determined by distance between genes

  3. Outline of Lecture 12 Single crossovers Three-point mapping Determining the order of genes Linkage and mapping in haploid organisms Somatic cell hybridization - human chromosome maps

  4. I. Single Crossovers: Non-crossover (Parental) and Crossover (Recombinant) Gametes What is the maximum % recombination?

  5. Genetic Map of Drosophila melanogaster

  6. II. Three-Point Mapping • You can add % recombination between two genes to find the order of genes pretty well. • But the only way to be sure of the order of three genes is by Three-Point Mapping, which considers 3 genes at once. • You look for rare double-crossover events, and that is the clue to the gene order.

  7. Double Crossovers

  8. Probability of Double Crossovers • Equals product of each of their individual probabilities: • if PAxB = 0.20 and PBxC = 0.30 then • PAxBxC = (0.20)(0.30) = 0.06 = 6 % • Criteria for 3-point mapping cross: • Crossover gametes heterozygous at all loci • Genotypes can be determined from phenotypes • Sufficient numbers for representative sample

  9. 3-Point Mapping in Drosophila • Cross a y ec w female with wildtype male to get triply heterozygous mutant female and triply hemizygous mutant male. • Cross the F1 and examine the F2 phenotypes: • NCO: noncrossover • SCO: single crossover (2 types) • DCO: double crossover • NCO: y ec w 4685 + + + 4759 94.44% • SCO: y + + 80 + ec w 70 1.50% y + w 193 + ec + 207 4.00% • DCO: y ec + 3 + + w 3 0.06% • Total: 1000 100%

  10. 3-point Mapping Explanation

  11. III. To Deduce the Order from a 3-Point Cross: Method 1 1. Group the 8 phenotypic groups into 4 reciprocal pairs. 2. The Non-crossover (NCO) pair is the largest group. The Double crossover (DCO) pair is the smallest group. 3a. Note which gene “switches” from the parental arrangement in DCO (present on its own) - that one is in the middle.

  12. Possible Orders of 3 Genes • If yellow were in the middle, yellow phenotype would show up in DCO. • If echinus were in the middle, echinus phenotype would show up in DCO. • white is actually in the middle since white phenotype shows up in actual DCO data.

  13. To Deduce the Order from a 3-Point Cross: Method 2 3b. Assume one of the 3 possible gene orders and work the problem. If you later find a contradiction, try one of the other orders. 4b. Determine whether a DCO with your arrangement will produce the observed DCO phenotypes. • You will encounter a contradiction unless you have chosen the correct gene order. Keep trying until you get the right one.

  14. To calculate recombination %: • Total crossovers between y and w (SCO1 + DCO) : • Total crossovers between w and ec (SCO2 + DCO) :

  15. Types of Double Exchanges:Not All are Detectable

  16. Creighton and McClintock Experiment Proved Crossing Over was a Physical Event • In maize, colorless (c)/colored (C), starchy (Wx)/waxy (wx) linked on chromosome 9. • Cytological markers on one parental homolog (knob on one end and translocated segment on the other end) allowed direct observation.

  17. Crossovers Between Sister Chromatids (SCEs) • Revealed by “Harlequin” chromosomes labeled during DNA replication • Occurs between mitotic sister chromatids. • No recombination • Significance unknown, but increased incidence correlated with some human diseases.

  18. 3-point Mapping in Maize bm (brown midrib), v (virescent seedling), and pr (purple aleurone)

  19. Results of Cross

  20. Questions • What is the heterozygous arrangement of alleles in the female parent? • What is the gene order? • What are the map distances between each pair of genes?

  21. IV. Linkage and mapping in haploid organismsChlamydomonas and the Benefits of Sex Sex Reproduction

  22. Life Cycle of Fungi Neurospora and Sordaria Note different meaning of “tetrad”

  23. Tetrad Analysis

  24. Mapping the Centromere • Essentially like 2-point mapping problem between one gene locus and the centromere. • Identify first-division segregation (may or may not be most common group) from second-division segregation. • D = 1/2(second-division segregant asci)/total. • For example, if there are 65 first-division asci and 70 second-division asci, then D = 1/2(70/135) = 0.259 or 26 map units.

  25. V. Human Chromosomes have been Mapped by Somatic-cell Hybridization • Two cells from mouse and human fused to form heterokaryon (two nuclei in common cytoplasm). • Nuclei fuse to form synkaryon and lose human chromosomes over time. • Gene products are assayed and correlated with remaining human chromosomes. • Genes also mapped by pedigree analysis and recombinant DNA techniques.

  26. Example • Gene A: • Gene B: • Gene C: • Gene D:

  27. Human Chromosome Maps

  28. Why didn’t Mendel Observe Linkage? • There are 7 chromosomes and 7 genes • Did he get one gene per chromosome? • Genes are located on four chromosomes, but far enough apart to seem unlinked (frequent crossing over creates independent assortment). • He should have seen linkage if he had mated dwarf plants with wrinkled pea, but he apparently didn’t do this experiment.

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