1 Lecture 9 One Gene One enzyme

2 Genes DNA sequences that code for RNA- prm, RBS, ORF, Term ---TTGACAT------TATAAT AT-/-AGGAGGT-/-ATG CCC CTT TTG TGA ---AACTGTA------ATATTA TA-/-TCCTCCA-/-TAC GGG GAA AAC ATT (-10) (-35) PROMOTER 5’ 3’ 5’ antisense sense RIBOSOME BINDING SITE U-/-AGGAGGU-/-AUG CCC CUU UUG UGA 5’ 3’ Met Pro leu leu stp Prokaryotic Genes When ALL OF THESE RULES ARE SATISFIED THEN AND ONLY THEN WILL A PIECE OF DNA GENERATE A PROTEIN. EUKARYOTES ARE EVEN MORE COMPLICATED. Is there a ribosome binding site upstream of the ATG Is there a promoter upstream of the ribosome binding site

3 One gene One enzyme hypothesis In the next few lectures, the following questions will be Addressed: What is the structure of a gene? How does a gene function? How is information stored on the gene? What is the relationship between genotype and phenotype?

4 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 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? DNA Geneticist Bioinformatics RNA Molecular biologist Protein Biochemist Protein complex Biochemist Cellular phenotype Cell biologists Organism phenotype Physicians

5 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. 1909Garrod termed the defect- inborn error (genetic) 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. Garrods results and his explanation were ignored 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

6 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: 1Products are synthesized as a series of steps 2Each step is catalyzed by an unique enzyme 3Each enzyme is specified by a unique gene The logic: PrecursorInt1Int2Product EnzAEnzBEnzC GeneAGeneBGeneC

7 Consequences of mutations PrecursorInt1Int2Product EnzAEnzBEnzC GeneAGeneBGeneC 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?

8 Neurospora Beadle and Tatum analyzed biosynthetic mutations in the haploid fungus Neurospora (Red bread mold) It had the advantage in that it could be grown on a defined growth medium. Given salts like Na3 citrate, KH2PO4, NH4NO3, MgSO4, CaCl2 and sugars like sucrose Neurospora can synthesize the amino acids, vitamins etc required and grow to form colonies on agar plates.

9 Prototroph: a strain that utilizes sugar, salt and water to grow. Auxotroph: Mutant strain that needs a specific amino acid or vitamin along with sugar, salt and water to grow.

10 Arginine biosynthetic mutants 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

11 Logic of experiment 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. Conclusion: Enzyme for making arginine was missing

12 The method To identify mutants Transfer mutants to minimal media (water, sugar, salts) Strain1 and 7 can grow on complete media but not minimal media. They have a mutation in a gene required for growth on minimal media!!! Complete All mutants grow minimal Irradiate spores. Take mutant spores. Plate individual spores on complete media (sugar, salts and water, AND vitamins AND all 20 amino acids).

13 Analogy In the class: There are two kinds of students: Students who can climb trees Students that cannot climb trees. Under normal growth conditions when supermarkets are open, both kinds of students live happily When supermarkets are closed Students who can climb trees grow happily because they can climb trees and eat fruit Students who cannot climb trees do not grow. They cannot climb trees, and go hungry.

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: strain1 is defective in the production of Vitamins and the mutant is rescued by adding back vitamins Take Strain 1

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 Take strain 7

16 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: strain7 is defective in the production of Amino acids and the mutant is rescued by adding back amino acids Which of the 20 amino acids does strain7 fail to produce complete media (salt+sugar) Vitamin + amino acids

17 Which amino acid Minimal media + vitamin + all 20 amino acidGrowth Minimal media + vitamin + lysineNo growth Minimal media + vitamin + glutamineNo growth -  Minimal media + vitamin + arginineGrowth Mutant7 is in a gene required for the production of Arginine. Beadle and Tatum found that three mutants could not produce arginine Arg1Arg2Arg3 Precursor ----->ornithine----->citrulline----->arginine enz1enz2enz3 The pathway for arginine biosynthesis is :

18 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

19 Add back Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not OrnithineCitrulline Arginine 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 mutant gene codes for which enzyme????

20 Add back Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not OrnithineCitrulline Arginine Mutant1+++ Mutant2-++ Mutant3 --+ Precursor ----->ornithine----->citrulline----->arginine enz1enz2enz3 Arg1Arg2Arg3

21 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. The Results Also show that THREE different Genes/enzymes are necessary for ONE phenotype- synthesis of ARG! Let’s see how this would affect phenotype ratios in a cross

22 Temperature-sensitive mutations The one gene/one enzyme concept explains a number of genetic phenomena Temperature-sensitive mutations Some mutations exhibit a phenotype at one temperatures (the restrictive temperature) but function normally at another temperature (permissive temperature). Reasons: Slight destabilization/alteration of the 3D conformation of the enzyme or its ability to interact with other proteins 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.

Heat sensitive mutants 23 Cold sensitive Protein is functional at high temp and inactive at low temp Active at 30C but inactive at 15C Temperature sensitive Protein is functional at low temperature but inactive at high temperature Active at 23C but inactive at 32C K253E Cs Interacts very stably with RFC C752T Ts Mis folding PCNA

24 An example of a Ts mutation: Dogs and cats that are white with black feet or vice versa The genes for coat color are normal at one temperatures but are inactive at another temperatures One of the genes for coat color is Albino - in cats This gene affects melanin production. The normal or dominant form, C, is 'full color'. Various mutant alleles. These mutants are temperature sensitive -

25 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 is temperature sensitive. c is the most recessive form, also known as albino. In the homozygote cc this causes complete absence of any pigment and white fur. Cb is incompletely dominant over Cs; the heterozygote (Cb/Cs) gives a phenotype intermediate between Burmese and Siamese, known as Tonkinese.

26 Biosynthetic pathways at the grocery store Most of the red and blue colors found in higher plants are a result of pigments synthesized from one of two metabolic pathways, the carotenoid or the anthocyanin pathway. The biosynthetic pathway for corn kernel color is as follows: Precursor----->Chalcone ---->Flavanone ---->Anthocyanins (white)(yellow)(white) (blue) Grocery store corn is usually yellow. Which step in the pathway must be mutated to produce yellow corn? Beadle/Tatum Results Also show that THREE different Genes/enzymes are necessary for ONE phenotype- synthesis of ARG! Similarly for blue corn multiple genes/enzymes are required. Let’s see how this would affect phenotype ratios in a cross

27 Mutants and Genetic pathways 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

28 Multiple genes affecting a single phenotype Precursor---->intermediate---->product yellowwhiteblue ｷ A, B = normal alleles ｷ a, b= nonfunctional mutant alleles Parental cross:AAbbxaaBB whiteyellow F1AaBb (blue) x AaBb (blue) F2 EnzA EnzB

29 Multiple genes affecting a single phenotype AB Ab aB ab AB Ab aB ab 9 A-B-blue 3A-bbwhite 3aaB-yellow 1aabbyellow Precursor---->intermediate---->product yellowwhiteblue EnzA EnzB F2

30 Multiple genes affecting a single phenotype AB Ab aB ab AB Ab aB ab AABBAABb AaBB AaBb AAbB AabB aABB aAbB aABb 9 A-B-blue 3A-bbwhite 3aaB-yellow 1aabbyellow Precursor---->intermediate---->product yellowwhiteblue EnzA EnzB F2 AAbb Aabb aaBB aaBb aAbb aabB aabb 4:3:9 Y:W:B

31 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

32 Labradors recessive Epistasis give 9:4:3 ratio 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 EEBBEEBb EeBB EeBb EEBb EeBb EeBB EeBb EEbb Eebb eebb eeBB eeBb 4:3:9 Y:Br:Bl Recessive epistasis Homozygous ee gene alleles mask effect of B gene alleles e is epistatic to B Epistasis= When the Alleles of One Gene Mask the Expression of Alleles of a Second Gene

33 Gene interactions give 9:7 Precursor---->intermediate---->product whitewhiteblue EnzA EnzB AB Ab aB ab AB Ab aB ab AABBAABb AaBB AaBb AAbB AabB aABB aAbB aABb 9 A-B-blue 3A-bbwhite 3aaB-white 1aabbwhite AAbb Aabb aaBB aaBb aAbb aabB aabb

34 Enz V+ Precursor  Brown pigment \ (white) \ transporter W  Red / Precursor  Vermilion pigment / (white) Enz B+ WT -- Brown WT -- Vermilion WT -- White

35 Gene Interaction: A range of Phenotypes Arise From Combined Action of Alleles of Two Genes The 9:3:3:1 ratio in the F 2 suggests two genes control coat color.

Multiple genes regulate a single phenotype 36 zPepper Color zGene 1: zR=red zr=yellow zGene 2: zY=absence of chlorophyll (no green) zy=presence of chlorophyll (green) zPossible genotypes: zR-/Y- : red (red/white no chlorophyll) zR-/yy : brown/orange (red/green chlorophyll) zrr/Y- : yellow (yellow/white no chlorophyll) zrr/yy : green (yellow/green chlorophyll)

Two genes affect Chicken Combs 37 z4 different chicken comb phenotypes result: zRose Combs (R-pp) zWalnut Combs (R-P-) zPea Combs (rrP-) zSingle Combs (rrpp)

Multiple genes for hair color 38 zHair Color yHair color is controlled by multiple genes on chromosomes 3, 6, 10, and 18. yThe more dominant alleles that appear in the genotype, the darker the hair!

Multiple genes affect a single phenotype- additive effects 39 Height

Additive Gene Interaction for Continuous Variation Continuously varying traits are also called quantitative traits.

Additive Gene Interaction Model for Continuous Variation Continuously varying traits are also called quantitative traits.

42 zThe height of plants is controlled by 4 pairs of alleles. Alleles A, B, and C contribute 3 cm to the plant's height. Alleles that are recessive do not contribute to the height. In addition Gene L is always found in a homozygous dominant condition and always contributes 40 cm to the height. za) What would be the height of a plant with the genotype AABBCCLL? zb) What would be the height of a plant with a genotype aabbccLL? zc) What would be the height of the offspring produced from a cross between the plants in a) and b)? AaBbCcLL zd) What would be the heights of the offspring produced from a cross between AaBbCcLL and AaBbCcLL?

43 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

44 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

45 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.

46 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.

47 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: val his leu thr pro *glu *glu ｷ Hb S: val his leu thr pro *val* glu ｷ Hb C: val his leu thr pro *lys* glu 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.

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49 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?

50 Compound EBNA Mut3+-++ Mut1--++ Mut2--+- B---->E---->A---->N mut3mut1mut2 Compound BEAN Mut3-+++ Mut1--++ Mut2---+ Compound EBNA Mut1--++ Mut2--+- Mut3+-++ Rearrange the mutants Rearrange the compounds Another example

51 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---->S----->A---->N