F 11/23 Happy Thanksgiving! 8 M 11/26 Gene identification in the genomic era Bamshad et al. Nature Reviews Genetics 12: , 2011

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1 3 rd Edition 4 th Edition Lecture Day Date Topic Reading Problems Reading Problems 1 M 11/5 Complementation testing reveals that genes are distinct entities Ch W 11/7 One gene makes one protein Ch F 11/9 The genetic code Ch M 11/12 Administrative holiday 4 W 11/14 Hardy-Weinberg populations Ch Ch. 7 #16-21, 23 Ch Ch. 7 #25, 26, 28; IG Biochemical Genetics #1-3 Ch Ch. 8 #1-9, 15 Ch Ch. 21 #2-5, 7, 9, 10, 12, 14 5 F 11/16 Evolution alters allele distribution Ch. 21 #20, 21; IG Population Genetics #1-5 6 M 11/19 Positional cloning Ch W 11/21 Individual identification Ch Ch. 11 #20-23, 25, 27; IG Molecular Markers #1-3, 5 Ch. 11 #1-5, 12, 13, 17; IG Molecular Markers #6 Ch Ch Read 3 rd edition Ch Ch. 7 #18-23, 25 Ch. 7 #27, 28, 30; IG Biochemical Genetics #1-3 Ch. 8 #1-9, 15 Ch. 19 #2-5, 7, 9, 10, 12, 14 Ch. 19 #20, 21; IG Population Genetics #1-5 Ch. 11 #20-23, 25, 27; IG Molecular Markers #1-3, 5 Ch. 11 #1-5, 12, 13, 17; IG Molecular Markers #6 Quiz 1 this week Quiz 2 next week Quiz 3 11/27-30 F 11/23 Happy Thanksgiving! 8 M 11/26 Gene identification in the genomic era Bamshad et al. Nature Reviews Genetics 12: , 2011 Quiz 4 12/4-7 9 W 11/28 Transgenic animals Ch Ch. 11 #26 Read 3 rd 10 F 11/30 Gene replacement in mice Ch M 12/3 Model organisms and mutagenesis screens 12 W 12/5 Human genetics from disease to treatment F 12/7 Problem solving clinics/tas edition Ch. 20 #12, 13 Ch Hafter et al. Development 123:1-36, 1996 Rock et al. Nature Genetics 40: , 2008 Ch. 11 #26 Ch. 18 #12, 13

2 Complementation vs. recombination Complementation Are two mutations are located in the same gene? Can two mutants provide each other with the required proteins? Recombination Used to calculate the distance between two mutations Can recombination restore a functional gene product?

3 Complementation vs. recombination What if a deletion spans adjacent genes in 2 complementation groups? COMPLEMENTATION RECOMBINATION complementation groups: Group A: 1, 2, 5 Group B: 3,4 2 does not complement 1,5 (gene A) or 3, 4 (gene B) => large deletion 2 recombines with 5 (gene A), but does not recombine with 1 (gene A) or 3, 4 (gene B) => 5 is outside of the deletion; deletion spans both genes A B 5 1 (3,4) 2

4 3 Complementation Practice problem Recombination Mutant Complementation groups A: 1,2,3 B: 4,5 3 must be a deletion spanning both genes because it does not complement any other mutants Gene A Mutant Recombination mapping 1 recombines with all others 2 does not recombine with 3 so it is within the deletion in gene A 4 does not recombine with 3 so it is within the deletion in gene B 5 recombines with all others Gene B

5 Summary - A gene can be thought of as a line, a contiguous string of nucleotides - Mutations affect the nucleotides along the line (chromosome) - Recombination occurs between the nucleotides at single base resolution - Nucleotides are the fundamental units of change (evolution) - Genes are the functional units of heredity

6 Beadle and Tatum: One gene encodes one enzyme N. Crassa life cycle Nobel Prize, 1958 haploid haploid diploid (transient) Arginine biosynthesis in Neurospora crassa

7 Advantages of N. crassa as a model Can dissect and grow Individual ascospores Can mutagenize haploid conidia Haploid hyphae fuse to form a heterokaryon

8 Beadle and Tatum: Isolated arginine auxotrophs Experiment: 1. Mutagenize (haploid) conidia 2. Cross with WT and sporulate to generate ascospores 3. Dissect out single spores and germinate on complete medium 4. Transfer to minimal medium to see which don t grow (auxotrophs) 5. Supplement with arginine and see that the auxotrophs now grow X-rays complete medium growth minimal medium no growth minimal medium + arginine growth In this way many (arginine) auxotrophs were isolated

9 How many genes can be mutated to produce arginine auxotrophy? Experiment: Complementation testing 1. Cross two (haploid) auxotrophs 2. See if the heterokaryon can grow without arginine supplementation Mut A Mut B same complementation group (gene) = no growth different complementation groups (genes) = growth 4 complementation groups were found

10 Do the different genes constitute a pathway for arginine biosynthesis? Experiment: Test growth on intermediates in the arginine pathway Minimal medium ornithine citrulline argininosuccinate arginine Mutant Mutant Mutant Mutant N-acetyl ornithine Mutant 1 ornithine Mutant 2 citrulline Mutant 3 argininosuccinate Mutant 4 arginine ARG-E ARG-F ARG-G ARG-H Mutants affect genes which encode enzymes in the arginine biosynthetic pathway

11 Conclusions from Beadle and Tatum s work Specific: Each mutant has a defect in one step in the biosynthetic pathway Each mutant affects the function of an enzyme in the pathway One gene, one enzyme General: Each gene encodes a single protein

12 Question: How does the linear DNA sequence specify the protein; what is the code? The Central Dogma What we knew replication DNA transcription RNA translation 1. DNA is the genetic material 2. There are 4 nucleotides, A, C, G & T 3. The gene is the functional unit 4. RNA is an intermediate molecule (U) 5. The protein is encoded in the DNA/RNA 6. There are 20 amino acids Protein

13 What is the code? 1. Mathematical prediction 2. Genetic evidence 3. Biochemical evidence

14 1. Mathematical prediction: How many bases are needed to encode 20 amino acids? Single base code: 4 amino acids maximum (4 1 ) = 4 N-N-N-N-N-N aa Doublet code: 16 amino acids maximum (4 2 ) = 16 NN-NN-NN-NN-NN-NN aa Triplet code: 64 amino acids maximum (4 3 ) = 64 NNN-NNN-NNN-NNN-NNN-NNN aa At least 3 nucleotides are needed to encode the 20 amino acids - the code must be degenerate (multiple codons per amino acid)

15 2. Genetic evidence: Frameshift rii mutations in T4 bacteriophage The cryptographers Francis Crick Nobel Prize, 1962 Leslie Barnett Sydney Brenner Nobel Prize, 2002 Crick FH, Barnett L, Brenner S, Watts-Tobin RJ (December 1961). "General nature of the genetic code for proteins"

16 rii mutants grow on E. coli B but not on E. coli K Phage E. coli strain B K(λ) r + Wild type plaques Wild type plaques rii Large plaques No growth r + rii r +

17 Mutations made with intercalating agents can add or delete a base Proflavin Acridine orange Experiment #1: Treat T4 phage and isolate rii mutants no growth on E. coli K..NNNNNNNNNNNNNNNNNNNNNNNNNNNNNN.. * delete 1 nucleotide..nnnnnnnnnnnnnnnnnnnnnnnnnnnnn.. Mutation results in loss of 1 base and an rii phenotype (FC0)

18 Experiment #2: Treat T4 FC0 and isolate suppressor mutants growth on E. coli K..NNNNNNNNNNNNNNNNNNNNNNNNNNNNN.. * insert 1 nucleotide..nnnnnnnnnnnnnnnnnnnnnnnnnnnnnn.. Mutation could just reverse the original mutation or could insert a nucleotide Distinguish by recombining with WT to see if rii results Hypothesis: Intragenic suppressor results in compensatory gain of 1 base (FC7)

19 It didn t matter whether FC0 or FC7 were known to be 1 base deletions or insertions, just that they compensated for each other

20 Experiment #3: Treat FC7 and isolate suppressors of suppressors have rii again! Isolated many mutations, rii, rii suppressors, rii suppressors of suppressors etc. Classified all mutations into one of two groups, (+) or (-) relative to F0 relative to each other Evaluated which combinations of mutations resulted in rii or WT phenotype

21 Experiment #4: Test combinations of mutations for their phenotype growth on E. coli K All mutations individually resulted in rii phenotype no plaques Any (+) with any (-) mutation resulted in WT phenotype Two (+) [or two (-)] mutations resulted in rii phenotype no plaques Drumroll please!

22 Any 3 (+) [or 3 (-)] mutations resulted in WT phenotype Interpretation: The genetic code is a triplet code, read three bases at a time Everything was in order. I looked across at Leslie. 'Do you realise,' I said, 'that you and I are the only people in the world who know it's a triplet code?'" - Francis Crick

23 Midterm results have been posted along with the key n = 533 Range 36 to 150 (there were three perfect scores) Mean (73.1%) Median 114 SD 24.7 From prior class final grades Estimate from this exam A s about 25% Roughly 128 and above B s about 41% Roughly 103 to 127 C s about 34% Roughly below 102 D s about 1%??? F s 0%???

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