Lecture 9 One Gene One enzyme

What is a Gene RIBOSOME BINDING SITE Prokaryotic Genes PROMOTER 3’ 5’ antisense -TTGACAT------TATAAT-------AT—AGGAGGT--ATG CCC CTT TTG TGA---- -AACTGTA------ATATTA-------TA—TCCTCCA--TAC GGG GAA AAC ATT---- sense 3’ (-35) (-10) RIBOSOME BINDING SITE 5’ 5’ 3’ U—AGGAGGU--AUG CCC CUU UUG UGA--- Met Pro leu leu stp When ALL OF THE RULES ARE SATISFIED THEN AND ONLY THEN WILL A PIECE OF DNA GENERATE A PROTEIN. EUKARYOTES ARE EVEN MORE COMPLICATED.

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?

Mendel analyzed several mutant Pea phenotypes. Genes produce enzymes. Green peas--------------------- yellow peas Sgr enzyme converts chlorophyl (green) to colorless (yellow) This shows that a specific gene produces a specific enzyme whose activity affects the phenotype. Dominant allele produces functional enzyme while recessive does not produce an enzyme. This is common BUT not always the case.

The work of biochemists showed that chemical compounds in the cell are synthesized through a series of intermediates-a biochemical pathway (E) Ornithine Citruline Arginine (R) Enzyme1 Enzyme2 How do you link genes to enzymes

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: Products are synthesized as a series of steps Each step is catalyzed by an unique enzyme Each enzyme is specified by a unique gene The logic: Precursor Int1 Int2 Product EnzA EnzB EnzC GeneA GeneB GeneC

Consequences of mutations Precursor Int1 Int2 Product EnzA EnzB EnzC GeneA GeneB GeneC 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?

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.

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.

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

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

The method Irradiate spores. Take mutant spores. Plate individual spores on complete media (sugar, salts, water, AND vitamins AND all 20 amino acids). Complete All mutants grow 1 2 3 4 5 6 7 8 9 10 To identify mutants Transfer mutants to minimal media (water, sugar, salts) minimal 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!!! Vitamins and Amino acids are missing- CONCLUSION?

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.

Strain1 and 7 are defective in either amino acid Conclusion- strain1 Strain1 and 7 are defective in either amino acid production or Vitamin production Take Strain 1 Complete media (salt+sugar+ Vitamin + amino acids) Minimal media (salt+sugar) Minimal media (salt+sugar) + 20 amino acids Minimal media (salt+sugar) + vitamins Conclusion: strain1 is defective in the production of Vitamins and the mutant is rescued by adding back vitamins

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

Which amino acid Minimal media + vitamin + all 20 amino acid Growth Minimal media + vitamin + only lysine No growth Minimal media + vitamin + only glutamine No growth Minimal media + vitamin + only arginine Growth Mutant7 is in a gene required for the production of Arginine. Beadle and Tatum found that three mutants could not produce arginine Arg1 Arg2 Arg3 The biochemical pathway for arginine synthesis was kind of known. Ornithine and citrulline are closely related to arginine and were thought to be precursors The pathway for arginine biosynthesis is : Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3

Add back Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3 There are three different enzymes required for arginine synthesis Enz1, enz2 and enz3 Beadle and Tatum isolated three different mutations in genes (three genes) Arg1 Arg2 Arg3 ?????Which mutant gene codes for which enzyme???? Instead of arginine, if they added ornithine or citrulline to the media, some mutants were rescued and others were not Ornithine Citrulline Arginine Mutant1 Mutant2 Mutant3

Add back Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3 Arg2 Arg1 Arg3 Instead of arginine, if they added ornithine or citrulline to the media, what happens? Ornithine Citrulline Arginine Mut1 (Arg1) - + + Mut2 (Arg2) + + + Mut3 (Arg3) - - +

Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3 Arg2 Arg1 Arg3 Mutant in Arg2- only precursor made Add ornithine or citrulline to media, downstream enzymes are functional and pathway continues---> arginine synthesized Mutant in Arg1- 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! This would affect phenotype ratios in a cross

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

+++ ++ + +++ ++ + Another example Compound E B N A Mut1 - - + + Rearrange the mutants Compound E B N A Mut3 + - + + Mut1 - - + + Mut2 - - + - +++ ++ + Rearrange the compounds Compound B E A N Mut3 - + + + Mut1 - - + + Mut2 - - - + +++ ++ + B----> E----> A----> N mut3 mut1 mut2

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 This process might also identify multiple mutants for the steps in the pathway! B----> E----> A----> N Mut3 mut5 Mut1 mut4 Mut2 Compound E B N A Mut1 - - + + Mut2 - - + - Mut3 + - + + Mut4 - - + + Mut5 + - + +

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Multiple genes and pathways Beadle/Tatum Results Also show that THREE different Genes/enzymes are necessary for ONE phenotype- synthesis of ARG! Precursor (E) -----> ornithine----->citrulline -----> arginine enz1 enz2 enz3 Arg2 Arg1 Arg3 Arg2 mutant -- phenotype is unable to make arginine Arg1 mutant-- phenotype is unable to make arginine Arg3 mutant-- phenotype is unable to make arginine What would happen if you crossed different Arg mutants with one another such as arg1 with arg2?

Mutants and Genetic pathways Altered PHENOTYPE RATIOS! The one gene/one enzyme helps explain altered phenotype ratios observed in a standard Mendelian dihybrid cross: (2 genes segregating independently) If the Two genes being analyzed affect the same genetic pathway Precursor----> intermediate----> product yellow white blue EnzA EnzB Parental cross white x yellow (True breeding) Genotype: AAbb x aaBB

Multiple genes affecting a single phenotype Precursor----> intermediate----> product yellow white blue EnzA EnzB Parental cross: AAbb x aaBB white yellow F1 AaBb (blue) x AaBb (blue) F2 AB Ab aB ab AB Ab aB ab AABB AABb AaBB AaBb 4:3:9 Y:W:B AAbB AAbb AabB Aabb aABB aABb aaBB aaBb aabB aAbB aAbb aabb 9 A-B- blue 3A-bb white 3aaB- yellow 1aabb yellow

Labradors recessive Epistasis give 9:4:3 ratio Parental Cross: black x yellow BBEE bbee BbEe (black) x Yellow-------> brown--------> black E B Given the pathway show above, what phenotypic ratios would be produced in progeny from the dihybrid cross: BbEe x BbEe EB Eb eB eb EEBB EEBb EeBB EeBb 4:3:9 Y:Br:Bl EEBb EEbb EeBb Eebb Recessive epistasis Homozygous ee gene alleles mask effect of B gene alleles e is epistatic to B E works upstream of B EeBB EeBb eeBB eeBb EeBb Eebb eeBb eebb Epistasis= When the Allele of One Gene Mask the Expression of Allele of Second Gene

9:4:3 Labrador coat color Black eumelanin deposited densely Enzyme E Enzyme B Colorless 1 Eumelanin Brown eumelanin lightly deposited Enzyme A Enzyme b- Colorless 1 Eumelanin Enzyme a- Enzyme B Yellow No Eumelanin Colorless 1 No eumelanin Yellow No eumelanin deposition Enzyme a- Enzyme b- Colorless 1 No eumelanin

Multiple Genes affect single phenotype - 9:7 Precursor----> intermediate----> product white white blue EnzA EnzB AaBb x AaBb AB Ab aB ab AABB AABb AaBB AaBb AAbB AAbb AabB Aabb aaBb aABB aABb aaBB aabB aAbB aAbb aabb 9 A-B- blue 3A-bb white 3aaB- white 1aabb white Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.

9:7 Enzyme A Enzyme B Colorless 1 Colorless Purple Enzyme A Enzyme b- White Enzyme a- Enzyme B Colorless 1 Colorless White Enzyme a- Enzyme b- Colorless 1 Colorless White

Dominant Epistasis 12:3:1 Summer squash color 12:3:1 (Dominant Epistasis) GeneY determines color of pigment Y yellow y green GeneW determines pigment deposition W No pigment deposition (white) w Pigment deposited (color? Depends on Y allele) The dominant allele W blocks the function of all of the alleles of the Y gene Cross a heterozygous white squash to itself WwYy x WwYy 9 W-Y- white 3 W-yy white 3 wwY- yellow 1 wwyy green Dominant allele of one gene hides effects of both alleles of second gene= 12:3:1 ratio Dominant allele of one gene hides effect of dominant allele of second gene = 13:3 ratio Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.

12:3:1 Summer Squash coat color Dominant Epistasis Enz Y Enz W No Pigment deposited Green Yellow WHITE Enz y- Enz W No Yellow Pigment No Pigment deposited Green WHITE Enz Y Enz w- Yellow Pigment deposited YELLOW Green Yellow Enz y- Enz w- Green Pigment deposited No Yellow Pigment GREEN Green

Dominant Epistasis 13:3 In Leghorn chickens Colored feathers are due to a dominant gene, C; White feathers are due to its recessive allele, c. CC= color Cc= color cc= white Another dominant gene, I, inhibits expression of color while ii allows expression of color In birds with genotypes CC or Cc or cc if there is II or Ii the birds are white! Therefore both CCII, CcII, CCIi, CcIi and cc– are ALL white. Only birds that are colored are C-ii

Dominant Epistasis give 13:3 ratio B/b is the epistatic gene. Any chicken with a dominant B in their genome will have white feathers. Being homozygous recessive bb at this locus enables the expression of genes coded for at the hypostatic locus (A). At the hypostatic locus A/a the dominant allele A codes for colored feathers while the recessive a codes for no color. Hence, a chicken that is homozygous recessive aa will also be white giving you a 13:3 ratio In white leghorn and white wyandotte chickens, a dominant B allele masks color production associated with the dominant A allele of a second gene. Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes. Dominant allele of one gene hides effects of both alleles of second gene= 12:3:1 ratio Dominant allele of one gene hides effect of dominant allele of second gene = 13:3 ratio

13:3 Chicken coat color Dominant Epistasis Enz A Enz B Brown Pigment No Pigment deposited Colorless WHITE Enz a- Enz B No Pigment No Pigment deposited Colorless WHITE Enz A Enz b- Brown Pigment Pigment deposited BROWN Colorless Enz a- Enz b- No Pigment No Pigment deposited Colorless WHITE

Gene Interaction: Range of Phenotypes From Combined Action of Alleles of Two Genes The 9:3:3:1 ratio in the F2 suggests two genes control coat color You have to know the rules for each color

Two Genes interact and give coat color in lentils (Brown x Green=9:3:3:1) EnzymeA Colorless 1 Tan BROWN COAT Colorless 2 Gray EnzymeB EnzymeA Colorless 1 Tan TAN COAT Colorless 2 Enzyme b- Enzyme a- Colorless 1 GRAY COAT Colorless 2 Gray Enzyme B Enzyme a- Colorless 1 COLORLESS COAT GREEN COTYLEDON Colorless 2 Enzyme b-

9:3:3:1 and variation Brown x Green = 9 Brown:3Tan:3Gray:1Green Tan x Green = 3 Tan:1Green Gray x Green = 3Gray: 1Green Brown x Gray = 3Brown: 1 Gray Brown x Tan = 3 Brown:1 Tan

Multiple Genes affect single phenotype – 15:1 Leaf Precursor1--------> product <-------- Leaf Precursor2 Leaf EnzA EnzB AaBb x AaBb AB Ab aB ab AABB AABb AaBB AaBb AAbB AabB aABB aAbB aABb AAbb Aabb aaBB aaBb aAbb aabB aabb 9 A-B- Leaf 3A-bb Leaf 3aaB- Leaf 1aabb No leaf Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.

15:1 Maize leaf shape - Redundant Genes EnzA Leaf Cell Precursor Leaf Leaf Cell Precursor Leaf EnzB EnzA Leaf Cell Precursor Leaf Leaf Cell Precursor Enz b- Enz a- Leaf Cell Precursor Leaf Cell Precursor Leaf Enz B Enz a- Leaf Cell Precursor No Leaf Leaf Cell Precursor Enz b-

Multiple gene inheritance Been discussing traits that are governed only by one or two genes Many human traits, such as height, skin color etc are determined by multiple genes. Multigenic ("many gene") traits exhibit a mode of inheritance that would have surprised Gregor Mendel. Most phenotypes that we are aware of are distributed in a bell-shaped curve like human height Often multiple genes affect such traits Basic Premise: The Quantitative Trait Values vary with Genotype. A gene that influences a quantitative trait’s value is called a quantitative trait locus Genotypic variance of the trait can be due to a single major QTL; few QTLs with large effects; many QTLs with small effects or a combination of these Individual genes may act on the variance in an additive manner or there may be genetic interaction- epistatic relationships

Additive effects The height of plants is controlled by 4 genes. 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 dominant condition and contributes 40 cm to the height. a) What would be the height of a plant with the genotype AABBCCLL? 3+3+3+3+3+3+40 b) What would be the height of a plant with a genotype aabbccLL? 0+0+0+0+0+0+40 c) What would be the height of the offspring produced from a cross between the plants in a) and b)? AaBbCcLL 3+0+3+0+3+0+40 d) What would be the heights of the offspring produced from a cross between AaBbCcLL and AaBbCcLL?

Additive Gene Interaction-Model for Continuous Variation For each gene there are two alleles Assume alleles exhibit Incomplete dominance One allele (a) adds 0 to phenotype. The other allele (A) increases phenotype by a fixed value 2 Genes 2 alleles each 5 phenotype class ab AB Ab aB 1 Gene 2 Alleles each 3 Phenotype class a A 1/4 1/2 1/16 6/16 Continuously varying traits are also called quantitative traits.

Additive Gene Interaction Model for Continuous Variation 2 Alleles each 3 Phenotype class 2 Genes 2 alleles each 5 phenotype class 3 Genes 2 alleles each 7 phenotype class Continuously varying traits are also called quantitative traits. The width/height of the curve indicates the number of variable Genes/alleles

Skin Pigmentation- 3 genes with 2 alleles aabbcc X AABBCC F1 generation selfs itself- 8 types of gametes- ABC, ABc, AbC, aBC, Abc, aBc, abC, abc 64 progeny classes 1/64 6/64 15/64 20/64 1/64 6/64 15/64 20/64 15/64 6/64 1/64 (6 white) (5 white) (4white) (3white) (2white) (1white) (0white) The more genes affect a trait, the smoother the bell curve

Multi-gene Disease Some traits result from mutations in single genes- MONOgenic trait. Multigenic diseases arise from less severe mutations in more than one gene. Any of these mutations alone might not lead to disease, but together, they can lead to significant phenotypic differences. There are Complex interactions between these genes. ----------- Interaction among the phenotypic effects of different genes- epistasis, adds a layer of complexit. Genes don't function alone; rather, they constantly interact with one another. These gene-gene interactions result in an output phenotype. Certain genes are known to modify the phenotype of other genes- modifiers- enhancers and suppressors. For some traits the environment might affect phenotype smoothing the curve even more If the effect of the disease-bearing gene is masked or altered by the effects of a second gene (by say altering expression level of the disease bearing gene), then identifying the first gene can be complicated. In addition, if more than one genetic interaction occurs to cause a disease, then identifying the multiple genes involved and defining their relationships becomes even more difficult.