1. Genetic Fine Structure Nature of the Gene at the Molecular Level

2. Bead Theory The gene is the fundamental unit of 1.Structure The gene is indivisible by crossing over. The gene is indivisible by crossing over. Crossing over occurs only between genes. Crossing over occurs only between genes.

3. Bead Theory The gene is the fundamental unit of 2. Change The whole gene must change from one allelic form to another, there are no smaller components within gene can change by mutation. The whole gene must change from one allelic form to another, there are no smaller components within gene can change by mutation.

4. Bead Theory The gene is the fundamental unit of 3. Function The gene functions as a unit, parts of a gene cannot function on their own. The gene functions as a unit, parts of a gene cannot function on their own.

5. Revised Bead Theory The nucleotide pair is the fundamental unit of 1.Structure 2.Change The gene is the fundamental unit of 3. Function

6. How Can the Expression of a Gene by Altered By: 1. Intragenic recombination? 2. Mutation? 3. Complementation?

7. Intragenic Recombination Recombination within a gene is shown by recombination between two mutants to give a wild type (non-mutant) form of the gene. Recombination within a gene is shown by recombination between two mutants to give a wild type (non-mutant) form of the gene. O O X Mutant 1 Mutant 2 OO Wild type Double Mutant

8. Application: Deletion Mapping Deletions prevent recombination. Deletions prevent recombination. 1.If no wild type recombinants can be produced in a cross between two deletion mutants, the deletions are overlapping. 2.Regions of a gene can be defined by deletion mutations, and point mutations can be located within those regions.

9. Application: Deletion Mapping X Non-overlapping deletions Overlapping deletions Unable to achieve recombination to restore wild type Wild Type Double Mutant

10. Application: Deletion Mapping Deletion and Point Mutation do not overlap Unable to achieve recombination to restore wild type XO Wild Type Double Mutant O Deletion and Point Mutation overlapO

11. Deletion mapping of the rII region of Bacteriophage T4.

12. Application: Deletion Mapping Problem 1, page 3-4 In a particular bacteriophage, four deletion mutants are crossed in pairwise combinations to test for their ability to produce wild-type recombinants. The results are given beside where + indicates that recombinants were found. Draw a deletion map for these mutations and divide it into subdivisions according to overlapping mutations. 1234 1----++ 2----+ 3---- 4-- Deletion Mutants

13. Application: Deletion Mapping Solution Problem 1, page 3-4 2 13 4

14. Application: Deletion Mapping Problem 1, page 3-4 There are several site-specific point mutations (A, B and C) that map in the region covered by the deletions. By coinfection of phage with one of the deletions and phage with each of the site-specific mutations, recombinant phage are observed in the following cases. Assign each site-specific mutation to one of the subdivisions of the deletion map. 1234 A+----+ B----++ C++---- Deletion Mutants Site-Specific Mutations

15. Application: Deletion Mapping Solution Problem 1, page 3-4 2 13 4 1 & 2 overlap 2 & 3 3 & 4 AB C

16. How Can the Expression of a Gene by Altered By: 1.Intragenic recombination? Recombination between two mutant forms gives a wild type version of the gene --- changes in both genotype and phenotype occur. Recombination between two mutant forms gives a wild type version of the gene --- changes in both genotype and phenotype occur.

17. Mutations Silent AGG  CGG Arg Arg Synonymous AAA  AGA Lys Arg Missense AAA  GAA Lys Glu Nonsense CAG  UAG Gln Stop FrameshiftAA(A)GACUUACCAA Lys-asp-leu-pro  Lys-thr-tyr-gln Change in a nucleotide can lead to change in amino acid found in the protein.

18. How Can the Expression of a Gene by Altered By: 2. Mutation? Change in DNA triplet can alter amino acid sequence of protein. Change in DNA triplet can alter amino acid sequence of protein.

19. Complementation Production of the wild type phenotype when two different mutations are combined in a diploid or heterokaryon. Production of the wild type phenotype when two different mutations are combined in a diploid or heterokaryon. PRECURSOR PRECURSORINTERMEDIATE PRODUCT PRODUCT Ab enzyme A enzyme A mutant 1 mutant 1 aB enzyme B enzyme B mutant 2 mutant 2

20. Application: Complementation Tests 1.If a wild type phenotype cannot be produced in a cross between two mutants, the mutations are in the same gene (cistron). 2.If wild type phenotype can be produced, the mutations are in different genes.

21. Application: Complementation Tests Problem 2, page 3-4 Problem 2, page 3-4 Five mutant strains of Neurospora give the following results in complementation tests where a plus signifies complementation and a minus shows no complementation. Determine how many cistrons are represented by these mutations and indicate which mutants belong to each cistron.

22. Application: Complementation Tests Problem 2, page 3-4 Problem 2, page 3-4 12345 1--+++-- 2--+++ 3----+ 4--+ 5-- Mutant Strain Mutant Strain Mutant Strain

23. Application: Complementation Tests Solution Problem 2, page 3-4 (1, 5) (2) (3, 4)

24. How Can the Expression of a Gene by Altered By: 3. Complementation? Production of the wild type phenotype when two different mutations are combined in a diploid or heterokaryon—genotypes are unchanged. Production of the wild type phenotype when two different mutations are combined in a diploid or heterokaryon—genotypes are unchanged.

25. Application: Determining the Order in a Biochemical Pathway

26. Application: Observing Complementation for Genes within the Same Pathway Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on minimal medium? Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on minimal medium? 1.ARG-E _, ARG-G _ combined with ARG-F _, ARG-H _ 2.ARG-E _, ARG-F _ combined with ARG-F _, ARG-H _

27. 1.ARG-E _, ARG-G _ combined with ARG-F _, ARG-H _ XX X X

28. 2. ARG-E _, ARG-F _ combined with ARG-F _, ARG-H _ X X X X X

29. Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on minimal medium? Suppose we have two different haploid cells, each with mutations in two of the genes in the pathway, and these haploid cells fuse to form a diploid cell. Which of the following diploid cells can grow on minimal medium? 1.ARG-E _, ARG-G _ combined with ARG-F _, ARG-H _ YES 2.ARG-E _, ARG-F _ combined with ARG-F _, ARG-H _ NO Ornithine will build up because the ARG-F product is missing. SOLUTION SUMMARY