1. 1 Lecture 9 One gene One enzyme hypothesis In the next few lectures, the following questions will be Addressed:

2. 2 DMD Duchene muscular dystrophy is a human disease caused by an X-linked recessive mutation. DMD affects one in 3500 males. Age of onset is between one and six years. Affected individuals are often initially identified because they rise from the prone position in an unusual manner.

3. 3 Pathways Biologists and clinicians want to address the question of how altering a particular set of base pairs that make up the 3 billion base pairs in the human genome led to this phenotype.

4. 4 DMD Through these techniques, it was found that DMD patients have a mutation in a single gene. The normal function of the gene is to enable muscle fibers to make a protein called dystrophin. Dystrophin localizes to the plasma membrane in muscle cells. The normal dystrophin protein stabilizes the muscles during muscle contractions. Muscle fibers in people affected with DMD are extremely deficient in dystrophin. Without this protein, the plasma membrane ruptures during muscle contraction and degeneration of the muscle tissue occurs.

5. 5 Huntington's Disease Huntington's disease (HD) results from degeneration of neurons, in certain areas of the brain. This degeneration causes uncontrolled movements, loss of intellectual faculties, emotional disturbance and early death This disease is caused by a single dominant mutation on the forth chromosome. Each child of an HD parent has a 50-50 chance of inheriting the mutation. A person who inherits the mutation will sooner or later develop the disease! To understand this disease we need an interdisciplinary Approach. What is the normal function of the Huntington gene? What happens in the mutant? Can it be blocked?

6. 6 Alkaptonuria Degenerative disease. Darkening of connective tissue, arthritis Darkening of urine 1902Garrod characterized the disorder- using Mendels rules- Autosomal recessive. Affected individuals had normal parents and normal offspring. 1908Garrod termed the defect- inborn error of metabolism Homogentisic acid is secreted in urine of these patients. This is an aromatic compound and so Garrod suggested that it was an intermediate that was accumulating in mutant individuals and was caused by lack of enzyme that splits aromatic rings of amino acids. 1958La Du showed that accumulation of homogentistic acid is due to absence of enzyme in liver extracts 1994Seidman mapped gene to chromosome 3 in human 1996Gene cloned and mutant identified P230S &V300G 2000Enzyme principally expressed in liver and kidneys

7. 7 How does a gene generate a phenotype? The experiments of Beadle and Tatum in the 1940’s provided the first insight into gene function. They developed the one gene/one enzyme hypothesis This hypothesis has three tenets:

8. 8 Consequences of mutations Lets say we know the biochemical pathway. With this pathway, what are the consequences of a mutation in geneB? Would the final product be produced? Would intermediate2 be produced? Would intermediate1 be produced? What happens if we add intermediate1 to the media? What happens if we add intermediate2 to the media?

9. 9 Neurospora Beadle and Tatum analyzed biosynthetic mutations in the haploid fungus Neurospora. It had the advantage in that it could be grown on a defined growth medium.

10. 10 Prototroph: wildtype strain Utilizes sugar, salt and water to grow. Auxotroph: Mutant strain Needs a specific amino acid or vitamin along with sugar, salt and water to grow.

11. 11 Beadle and Tatum set out to identify genes involved in the biosynthetic pathway that led to the production of the amino acid arginine. Neurospora has approximately 15,000 genes and only 4-5 of these genes are involved in synthesizing arginine. How do you identify five genes from 15,000? The POWER OF GENETICS!!!!!! Typically the organism is exposed to a strong mutagen. This randomly mutagenizes genes. Then you look for a mutant in the pathway of interest.

12. 12 ARGININE BIOSYNTHESIS PATHWAY Irradiate (mutagenize) spores. Grow on medium containing arginine Transfer to medium lacking arginine DO THEY GROW OR NOT? If the cells cannot grow on medium lacking arg, then they must have a mutation in a gene required for making ARGININE Mutant needed arginine to grow. Enzyme for making arginine was missing

13. 13 The method 12345678910 complete minimal

14. 14 Conclusion- strain1 Strain1 and 7 are defective in either amino acid production or Vitamin production Complete media (salt+sugar+ Vitamin + amino acids) Minimal media (salt+sugar) + 20 amino acids Minimal media (salt+sugar) + vitamins Minimal media (salt+sugar) Conclusion: Complete media (salt+sugar) Vitamin + amino acids

15. 15 Conclusion- strain7 Strain1 and 7 are defective in either amino acid production or Vitamin production Complete media (salt+sugar+ Vitamin + amino acids) Minimal media (salt+sugar) + 20 amino acids Minimal media (salt+sugar) + vitamins Minimal media (salt+sugar) Conclusion: complete media (salt+sugar) Vitamin + amino acids

16. 16 Which amino acid? Mutant7 is in a gene required for the production of Arginine.

17. 17 Beadle and Tatum found that three mutants could not produce arginine Arg1Arg2Arg3 The biochemical pathway for arginine synthesis was kind of known. Ornithine and citrulline are closely related to arginine and were thought to be precursors Precursor ----->ornithine----->citrulline----->arginine enz1enz2enz3

18. 18 Add back Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not OrnithineCitrullineArginine Mutant1 Mutant2 Mutant3 Precursor ----->ornithine----->citrulline----->arginine enz1enz2enz3 There are three different enzymes required for arginine synthesis Enz1, enz2 and enz3 Beadle and Tatum isolated three different mutations in genes (three genes) Arg1Arg2Arg3 ?????Which mutation codes for which enzyme????

19. 19 Mutant in Arg1- only precursor made Add ornithine or citrulline to media, downstream enzymes are functional and pathway continues---> arginine synthesized Mutant in Arg2- You need to supplement media with citrulline for the pathway to continue. Adding the precursor or ornithine does not help. Mutant in Arg3- You need to supplement media with arginine. Adding the precursor, ornithine or citrulline does not help. These experiments demonstrated that a single gene (mutation) coded for a single enzyme. In addition, the combination of appropriate mutations and intermediates enabled Beadle and Tatum to define the biochemical pathway leading to Arginine synthesis.

20. 20 This rationale currently is being used in many laboratories to elucidate more complex pathways in multicellular organisms ｷ Review Solving biochemical pathways: The more mutations that a compound rescues, the later in the pathway the compound is located Conversely, the later a mutation is in a pathway, the fewer compounds will rescue it:

21. 21 Another example I get three mutants for a particular pathway I add back various intermediates in this pathway and determine the results Compound EBNA Mut1--++ Mut2--+- Mut3+-++ What is the order of the compounds and mutations in the pathway?

22. 22 The steps in a biochemical pathway identified by this procedure are dependent on the available intermediates and mutations. This procedure does not identify every step in the pathway This process does not identify every step in the pathway! B---->E---->A---->N B---->E---->X----->A---->N

23. 23 Temperature-sensitive mutations The one gene/one enzyme concept explains a number of genetic phenomena A) Temperature-sensitive mutations Some mutations exhibit a phenotype at high temperatures (the restrictive temperature) but function normally at lower temperatures (permissive temperature). The mutation results in a slight destabilization and alteration of the 3D conformation of the enzyme Low temp- structure of enzyme- normal- activity normal High temp- structure of enzyme-altered- No activity These kinds of conditional mutants allow you to turn on and off a function of a protein.

24. 24 An example of a Ts mutation: Dogs and cats that are white with black feet or vice versa The gene for coat color is normal at cold temperatures but is inactive at higher temperatures or Vice Versa Albino - C gene in cats This gene affects the intensity of melanin production. The normal or dominant form, C, is 'full color'. Various incompletely dominant mutant alleles. These mutants are temperature sensitive - The higher the temperature, the more effective they are at producing melanin In order of decreasing dominance we have C, Cb, Cs and c. C is wild-type or full color. It is dominant to all other alleles. Cb- 'Burmese' factor- it causes a slight lightening of color and is slightly temperature sensitive. Cs- 'Siamese' factor; it has a much greater lightening effect and appears more temperature sensitive. Cb is incompletely dominant over Cs; the heterozygote (Cb/Cs) gives a phenotype intermediate between Burmese and Siamese, known as Tonkinese. The most recessive form is c, also known as albino. In the homozygote cc this causes complete absence of any pigment and white fur.

25. 25 Dominance versus Recessive The one gene/one enzyme concept explains Dominance versus Recessive Genotypeenzyme activityphenotype A/A2Xnormal A/a1Xnormal a/a0Xmutant Usually substrate is limiting, enzyme is in excess By saying that a mutation is recessive, we are saying that 1 unit of enzyme (or 50% of the normal activity) is enough to produce a normal phenotype Precursor------> product = phenotype enzymeA geneA ^|^| ^|^| ｷ A allele produces functional enzyme ｷ a allele produces nonfunctional enzyme

26. 26 Genetic Ratios Altered PHENOTYPE RATIOS The one gene/one enzyme helps explain altered phenotype ratios observed in a standard dihybrid cross: (2 genes segregating independently) If the Two genes being analyzed affect the same genetic pathway Precursor---->intermediate---->product yellowwhiteblue EnzA EnzB Parental crosswhitexyellow

27. 27 AB Ab aB ab AB Ab aB ab 9 A-B-blue 3A-bbwhite 3aaB-yellow 1aabbyellow Precursor---->intermediate---->product yellowwhiteblue EnzA EnzB F2

28. 28 Labradors Parental Cross:blackxyellow BBEEbbee BbEe (black) x BbEe (black) Given the pathway show above, what phenotypic ratios would be produced in progeny from the dihybrid cross: BbEe x BbEe Yellow------->brown-------->black EB Eb eB eb EB Eb eB eb 9:3:4

29. 29 Precursor---->intermediate---->product whitewhiteblue EnzA EnzB AB Ab aB ab AB Ab aB ab 9 A-B-blue 3A-bbwhite 3aaB-white 1aabbwhite

30. 30 Biochemical Pathways and Linked Genes The F1 is testcrossed The following F2 progeny are produced: 50 yellow 40 blue 10 white Precursor---->intermediate---->product yellowwhiteblue EnzC EnzD GeneCGeneD ParentalC-Dxc-d C-Dc-d F1

31. 31 Biochemical Pathways and Linked Genes The following F2 progeny are produced: 50 yellow 40 blue 10 white Precursor---->intermediate---->product yellowwhiteblue EnzC EnzD GeneCGeneD ParentalC-Dxc-d C-Dc-d F1C-Dc-d c-dxc-d ParentalC-Dblue40 c-d c-dyellow40 c-d RecombC-dwhite10 c-d c-Dyellow10 c-d What is the map distance between these two genes? Map Distance+#Recombinants/Total Progeny x 100% 2(10)/100= 20 Map Units

32. 32 One gene: one polypeptide The concept of 1 gene/enzyme was modified to the concept of: 1 gene/ 1 protein Almost all enzymes are proteins but not all proteins are enzymes. Many proteins provide structural rather than enzymatic roles. For example polymers of the protein actin provide structural integrity to the eukaryotic cell. Perhaps the most notable example of this comes from studies of Hemoglobin. Hemoglobin is an iron carrying protein found in the red blood cells and is responsible for transporting oxygen from the lungs to the cells of the body.

33. 33 Hb Hemoglobin consists of four polypeptides (proteins) each associated with a specific Heme group (Heme is a small iron containing molecule to which oxygen can attach) Adults contain 2 alpha polypeptides and 2 beta polypeptides Alpha polypeptide = 141 amino acids Beta polypeptide= 146 amino acids Over 300 known hemoglobin variants are known and each is the result of a specific mutation Most of these are the result of a single amino acid substitution ｷ Hb A: ｷ Hb S: ｷ Hb C: These results demonstrate that: 1. Genes specify proteins that are not enzymes 2. Mutations can disrupt a single amino acid out of the many that make up the protein.