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Announcements 1. Survey results: 87% like powerpoint 85% print notes before class 93% thought exam 1 covered appropriate material 43% thought exam 1 was.

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Presentation on theme: "Announcements 1. Survey results: 87% like powerpoint 85% print notes before class 93% thought exam 1 covered appropriate material 43% thought exam 1 was."— Presentation transcript:

1 Announcements 1. Survey results: 87% like powerpoint 85% print notes before class 93% thought exam 1 covered appropriate material 43% thought exam 1 was appropriate length Suggestions I will consider: posting lecture notes earlier, making exam 2 a bit shorter, more practice problems, continue doing problems during lecture. 2. Consider whether you prefer class to meet Wed. and not Fri., and no in- class review on Wed. before exam 2 OR in-class review Wed. and class meets Friday (day of exam). We’ll vote Friday. 3. Average on quiz 2 = 6.83/12 4. Lab this week: go over quiz and go over more linkage practice problems 5. Practice problems ch. 7: 9, 19.

2 Review of Last Lecture I. Determining the order of genes, continued - example in maize 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? II. Linkage and mapping in haploid organisms - ordered tetrad analysis D = 1/2(second-division segregant asci)/total

3 Outline of Lecture 14 I. Somatic cell hybridization - human chromosome maps II. Overview of Bacterial and Phage Genetics Conjugation Integration General Recombination Transformation Transduction

4 I. 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.

5 Example Gene A: Gene B: Gene C: Gene D:

6 Human Chromosome Maps

7 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. Why didn’t Mendel Observe Linkage?

8 II. Escherichia coli A model organism: useful for discovering general principles common to all organisms. The focus of genetic research from the 1940’s to 1960’s: What is a gene and how does it work? Advantages: short doubling time (30 min), simple culture media, pure cultures, haploid, lots of mutations. The advantage of being haploid is that a mutation in the parent is always seen in the offspring. In diploid organisms, mutations can be covered up if they are recessive. Bacteria are haploid Sordaria are haploid

9 Growth E. coli can grow on carbon source (e.g. glucose) + minimal inorganic salts. –Prototrophs: Grow well, are wildtype. –Auxotrophs: Require some other organic molecule that it cannot make, due to a mutation (e.g. amino acid leucine - leu - ). Grow in liquid culture flask or petri dish.

10 Genetic Recombination Revealed by Selective Media Colonies of prototrophs on minimal media met - bio - thr + leu + thi + met + bio + thr - leu - thi - AB A + B

11 Cells Must Contact Each Other for Mating: the Davis U tube How does genetic recombination occur? Cells that donate = F+ Cells that receive DNA = F- No growth!

12 Conjugation: process by which genetic information is transferred, recombined Discovered by Lederberg and Tatum (1946) Genetic info is transferred; basis for mapping Sex without reproduction Sex pilus is tube through which DNA is passed

13 Requirements for conjugation: F + X F - Bacteria Two mating types exist: donor F + (fertility) cells and recipient F - cells. Physical contact through F pilus on F + cells is required for conjugation. F + cells contain a fertility factor (F factor): - any cells grown with F+ become F+, F factor appears to be a mobile element - a plasmid (circular, extrachromosomal DNA) containing: (1) genes to allow transfer of plasmid (RTF) and (2) antibiotic resistance genes (r-determinants).

14 (tetracycline, kanamycin, streptomycin, sulfonamide, ampicillin, mercury) Typical Bacterial Plasmid Resistance transfer fragment Origin of Replication

15 Mechanism of Conjugation: F + X F - Pilus often breaks before complete transfer! two F+ cells result 1 F+ cell 1 F- cell

16 Hfr bacteria and chromosome mapping Hfr = high frequency of recombination This is a special type of F+, acts as donor of chromosome F+ x F- F+ Hfr x F- F- Some genes recombined more often than others???

17 Mapping by Interrupted Mating in Hfr Chromosome transferred linearly Gene order and distance between genes could be measured in minutes

18 Time Map of Experiment You can infer the order of the genes on the bacterial chromosome. “Minutes” = map units

19 Overlapping Time Maps The plasmid can insert randomly into the bacterial chromosome, allowing the complete chromosome to be mapped.

20 F + to Hfr by Integration into Bacterial Chromosome, Followed by General Recombination Recombination like crossing over F factor is last to transfer; F- stays F- Conjugation F factor integrates Chromosome transfer Replication

21 Circular Map of E. coli ~2000 genes Scaled in minutes One minute = ~ 20% recombination frequency

22 Transformation: a different process of recombination, can be used to map genes

23 Bacteriophages are viruses that use bacteria as hosts

24 Transduction: virus-mediated bacterial DNA transfer

25 T4 bacteriophage

26 T4 Phage Self-assembly: Development of a Simple Entity Head is an Icosahedron (20 faces)

27 recombinants Lawn of bacteria Larger, darker Smaller, lighter Smaller, darker parental Larger, lighter Recombination in Phage Strains with different plaque morphologies “crossed” by coinfection of bacteria: h r + X h + r –h mutant plaques are darker than h + –r mutant plaques are larger than r + Results: parental (h r + and h + r) and recombinant (h + r + and h r) plaques. # recombinants/total X 100% = recomb. frequency

28 T4 Map rII locus From Recombination Analysis

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