1. 1 Lecture 9 One Gene One enzyme

2. 2 Genes ---TTGACAT------TATAAT-------AT-/-AGGAGGT-/-ATG CCC CTT TTG TGA ---AACTGTA------ATATTA-------TA-/-TCCTCCA-/-TAC GGG GAA AAC ATT (-10) (-35) PROMOTER 5’5’ 3’3’ 3’3’ 5’5’ antisense sense RIBOSOME BINDING SITE U-/-AGGAGGU-/-AUG CCC CUU UUG UGA 5’5’ 3’3’ 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. 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. 4 The work of biochemists showed that chemical compounds in the cell are synthesized through a series of intermediates-a biochemical pathway OrnithineCitrulineArginine Enzyme1Enzyme2 Glutamic acid-  How do you link genes to enzymes

5. 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: 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

6. 6 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?

7. 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. 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. 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. 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. 11 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!!! Vitamins and Amino acids are missing- CONCLUSION? Complete All mutants grow minimal 12345678910 Irradiate spores. Take mutant spores. Plate individual spores on complete media (sugar, salts, water, AND vitamins AND all 20 amino acids).

12. 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. 13 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

14. 14 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 Take strain 7

15. 15 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 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 :

16. 16 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????

17. 17 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

18. 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. 19 Analogy UCSC WalnutLaurelBay

20. 20 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?

21. 21 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

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---->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. 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:

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25. 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 26 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 K253E Cs Interacts very stably with RFC C752T Ts Mis folding PCNA

27. 27 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 - Ts mutant and Cat coat color

28. 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.

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30. 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. 31 Precursor ----->ornithine----->citrulline----->arginine enz1enz2enz3 Arg1Arg2Arg3 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. 32 Multiple Genes affect single phenotype - 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 AaBbx AaBb Altered 9:3:3:1 ratios are a hallmark of epistasis involving two genes.

33. 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 yellowwhiteblue EnzA EnzB Parental crosswhitexyellow

34. 34 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 Parental cross:AAbbxaaBB whiteyellow F1AaBb (blue) x AaBb (blue)

35. 35 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 E works upstream of B Epistasis= When the Allele of One Gene Mask the Expression of Allele of Second Gene

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

37. 37 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-yywhite 3 wwY-yellow 1 wwyygreen 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 Dominant Epistasis 12:3:1

38. 38 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 13:3

39. 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 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 Dominant Epistasis give 13:3 ratio

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41. 41 The 9:3:3:1 ratio in the F 2 suggests two genes control coat color You have to know the rules for each color Gene Interaction: Range of Phenotypes From Combined Action of Alleles of Two Genes

42. Multiple genes regulate a single phenotype 42 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 : browny orange (red/green chlorophyll) zrr/Y- : yellow (yellow/white no chlorophyll) zrr/yy : green (yellow/green chlorophyll)

43. Two genes affect Chicken Combs 43 z4 different chicken comb phenotypes result: zRose Combs (R-pp) zWalnut Combs (R-P-) zPea Combs (rrP-) zSingle Combs (rrpp)

44. 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. aabbccX AABBCC i. F1 generation selfs itself j. 1/646/6415/6420/6415/646/641/64 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

45. 45 zThe 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. zIn addition Gene L is always found in a dominant condition and always contributes 40 cm to the height. za) What would be the height of a plant with the genotype AABBCCLL? 3+3+3+3+3+3+40 zb) What would be the height of a plant with a genotype aabbccLL? 0+0+0+0+0+0+40 zc) 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 zd) What would be the heights of the offspring produced from a cross between AaBbCcLL and AaBbCcLL?

46. Genes for hair color 46 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!

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

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

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

50. Multi-gene Disease 50 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.

51. Alzheimers 51 Alzheimer's disease, is a progressive neurodegenerative disorder that causes memory loss and dementia. A mutation in a gene called apolipoprotein E4 was associated with a higher risk of developing Alzheimer's. While having one or two copies of mutant apolipoprotein E4 increase one's risk of Alzheimer's, not all carriers of apolipoprotein E4 develop the disease. This suggested that gene-gene interactions were involved. They confirmed 27 different genetic interactions in 4 different biochemical pathways: cholesterol metabolism, beta-amyloid production, inflammation, oxidative stress. Some interactions were synergistic, while others were antagonistic. The synergistic interactions indicate that the pair of involved genes together increase the risk of Alzheimer’s. The strongest synergistic interactions involved the pairing of apolipoprotein E4 mutation with mutations in three different genes: alpha(1)-antichymotrypsin, β-secretase, and butyrylcholinesterase K. These genes are not acting alone, but in a pathway that affect one another. The APP gene produces a transmembrane protein that is modified, cleaved by secretase and then inserted into membranes. Function of APP is not fully known- could be membrane or Ca+ trafficking in neurons. APP interacts with chymotrypisn and Apolipoprotein. Alterations in these interactions lead to formation of APP plaques (amyloid fibers) leading to Alzheimers. ApoE4 mutation leads to alterations in cholesterol in membranes. The ApoE4 allele causes greater aggregation of APP and greater amyloid deposition.

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53. 53 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

54. 54 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

55. 55 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.

56. 56 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.

57. 57 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.

58. 58 Muscular Dystrophy 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.

59. 59 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

60. 60 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. Formal genetics provides little information about these intermediary steps. Since the 1970’s, a set of techniques have been developed that enable us to elucidate each step in a pathway These techniques are generally placed under the rubric of Molecular Biology or Molecular Genetics

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