1 / 61

Lecture 9 One Gene One enzyme

Lecture 9 One Gene One enzyme. Genes. Is there a ribosome binding site upstream of the ATG Is there a promoter upstream of the ribosome binding site. Prokaryotic Genes. PROMOTER. 3 ’. 5 ’. antisense. --- TTGACAT ------ TATAAT ------- AT -/-AGGAGGT-/- ATG CCC CTT TTG TGA

chuong
Download Presentation

Lecture 9 One Gene One enzyme

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 9 One Gene One enzyme

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

  3. 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? One gene One enzyme hypothesis

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

  5. 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

  6. Precursor Int1 Int2 Product EnzA EnzB EnzC GeneA GeneB GeneC 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?

  7. 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.

  8. 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.

  9. 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

  10. 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

  11. 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 To identify mutants Transfer mutants to minimal media (water, sugar, salts) 1 2 3 4 5 6 7 8 9 10 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?

  12. 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.

  13. 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) + 20 amino acids Minimal media (salt+sugar) Minimal media (salt+sugar) + vitamins Conclusion: strain1 is defective in the production of Vitamins and the mutant is rescued by adding back vitamins

  14. 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

  15. Which amino acid Minimal media + vitamin + all 20 amino acid Growth Minimal media + vitamin + lysine No growth Minimal media + vitamin + glutamine No growth Minimal media + vitamin + 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

  16. 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

  17. Add back Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3 Arg1 Arg2 Arg3 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 - - +

  18. 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! This would affect phenotype ratios in a cross

  19. Analogy Walnut Laurel Bay UCSC

  20. Another example I get three 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?

  21. Another example Compound E B N A Mut1 - - + + Mut2 - - + - Mut3 + - + + 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

  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----> 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

  23. This rationale currently is being used in many laboratories to elucidate more complex pathways in multicellular organisms • キDevelopment- formation of the body axis • キBehavior- courtship and mating • キBiological clocks • キAging • キCell cycle and Cancer • Review • キBiochemical processes occur as a series of discrete stepwise reactions • キEach reaction is catalyzed by a single enzyme • キEach enzyme is specified by a unique gene • 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:

  24. xxxxxxxxxx

  25. 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.

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

  27. Ts mutant and Cat coat color 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 -

  28. 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 very 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.

  29. xxxxxxxxx

  30. 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

  31. Precursor -----> ornithine -----> citrulline -----> arginine enz1 enz2 enz3 Arg1 Arg2 Arg3 Arg1 mutant -- phenotype is unable to make arginine Arg2 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?

  32. Multiple Genes affect single phenotype - 9:7 Precursor----> intermediate----> product white white blue EnzA EnzB AaBb x AaBb aB Ab ab AB AB Ab AaBb AABB AABb AaBB aB AAbb AAbB Aabb AabB ab 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.

  33. 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 yellow white blue EnzA EnzB Parental cross white x yellow

  34. 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 AaBb AB AABB AABb AaBB 4:3:9 Y:W:B AAbb AAbB Aabb AabB Ab aaBb aABb aABB aaBB aB aabB ab aAbb aAbB aabb 9 A-B- blue 3A-bb white 3aaB- yellow 1aabb yellow

  35. Labradors recessive Epistasis give 9:4:3 ratio Parental Cross: black x yellow BBEE bbee BbEe (black) x BbEe (black) 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 EB EEBb EEbb EeBb Eebb Eb 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 eB Eebb eebb EeBb eeBb eb Epistasis= When the Allele of One Gene Mask the Expression of Allele of Second Gene

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

  37. Dominant Epistasis 12:3:1 Summer squash color 12:3:1 (Dominant Epistasis) GeneW determines pigment production (repressor of GeneY) W No pigment w Pigment (color? Depends on Y gene) GeneY determines color of pigment Y yellow y green Cross a heterozygous white squash to itself WwYy x WwYy 9 W-Y- white 3 W-yy white 3 wwY- yellow 1 wwyy green 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

  38. 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

  39. 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

  40. 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

  41. Multiple genes regulate a single phenotype • Pepper Color • Gene 1: • R=red • r=yellow • Gene 2: • Y=absence of chlorophyll (no green) • y=presence of chlorophyll (green) • Possible genotypes: • R-/Y- : red (red/white no chlorophyll) • R-/yy : browny orange (red/green chlorophyll) • rr/Y- : yellow (yellow/white no chlorophyll) • rr/yy : green (yellow/green chlorophyll)

  42. Two genes affect Chicken Combs • 4 different chicken comb phenotypes result: • Rose Combs (R-pp) • Walnut Combs (R-P-) • Pea Combs (rrP-) • Single Combs (rrpp)

  43. Multiple gene inheritance a. so far been discussing traits that are governed only by one gene b. far from the truth. 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. c. most phenotypes that we are aware of are distributed in a bell-shaped curve like human height d. often multiple genes affect such traits e. height in plants might be affected by three genes each possessing two alleles f. the dominant allele of each gene might add 1 cm to basic height of plant g. the recessive allele of each gene would not affect the basic 10 cm height h. aabbcc X AABBCC i. F1 generation selfs itself j. 1/64 6/64 15/64 20/64 15/64 6/64 1/64 k. the more genes affecting a trait, the smoother is the bell curve l. the environment also affects phenotype smoothing off the curve even more

  44. 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 always 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?

  45. Genes for hair color • Hair Color • Hair color is controlled by multiple genes on chromosomes 3, 6, 10, and 18. • The more dominant alleles that appear in the genotype, the darker the hair!

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

  47. Skin Pigmentation ABC, AbC, aBC, ABc, abC, Abc, aBc, abc

  48. Additive Gene Interaction Model for Continuous Variation Continuously varying traits are also called quantitative traits. The width/height of each sub-phenotype class indicates the number of variable Genes/alleles

  49. Multi-gene Disease Multigenic diseases result from less severe mutations in more than one gene. Any of these mutations alone might not affect a trait, but together, they can lead to significant phenotypic differences. Complex interactions between genes. Number of traits result from mutations in single genes- MONOgenic trait. Interaction among the phenotypic effects of different genes- epistasis, adds a layer of complexity to the study of genetic disease. 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. This implies that multiple genes may interact to increase or decrease disease susceptibility. 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.

More Related