Fundamentals of genetics.pptx

Entire fundamentals including mendelian genetics upto linkage has been covered up....

1. GPB 121: Fundamentals of Genetics Krishnendu Chattopadhyay Principal Scientist (Plant Breeding) ICAR-NRRI, Cuttack 7/18/2022 Molecular Genetics-I

2. The First Piece of the Puzzle: Miescher Discovers DNA Although few people realize it, 1869 was a landmark year in genetic research, because it was the year in which Swiss physiological chemist Friedrich Miescher first identified what he called "nuclein" inside the nuclei of human white blood cells. (The term "nuclein" was later changed to "nucleic acid" and eventually to "deoxyribonucleic acid," or "DNA.") Laying the Groundwork: Levene Investigates the Structure of DNA other scientists continued to investigate the chemical nature of the molecule formerly known as nuclein. One of these other scientists was Russian biochemist Phoebus Levene . A physician turned chemist, Levene was a prolific researcher, publishing more than 700 papers on the chemistry of biological molecules over the course of his career. Levene is credited with many firsts. For instance, he was the first to discover the order of the three major components of a single nucleotide (phosphate-sugar-base ); the first to discover the carbohydrate component of RNA (ribose); the first to discover the carbohydrate component of DNA (deoxyribose); and the first to correctly identify the way RNA and DNA molecules are put together.

3. During the early years of Levene's career, neither Levene nor any other scientist of the time knew how the individual nucleotide components of DNA were arranged in space; discovery of the sugar-phosphate backbone of the DNA molecule was still years away. The large number of molecular groups made available for binding by each nucleotide component meant that there were numerous alternate ways that the components could combine. Several scientists put forth suggestions for how this might occur, but it was Levene's "polynucleotide" model that proved to be the correct one. Based upon years of work using hydrolysis to break down and analyze yeast nucleic acids, Levene proposed that nucleic acids were composed of a series of nucleotides, and that each nucleotide was in turn composed of just one of four nitrogen-containing bases, a sugar molecule, and a phosphate group. Levene made his initial proposal in 1919, discrediting other suggestions that had been put forth about the structure of nucleic acids. In Levene's own words, "New facts and new evidence may cause its alteration, but there is no doubt as to the polynucleotide structure of the yeast nucleic acid" (1919).

4. Indeed, many new facts and much new evidence soon emerged and caused alterations to Levene's proposal. One key discovery during this period involved the way in which nucleotides are ordered. Levene proposed what he called a tetranucleotide structure, in which the nucleotides were always linked in the same order (i.e., G-C-T-A-G-C-T-A and so on). However, scientists eventually realized that Levene's proposed tetranucleotide structure was overly simplistic and that the order of nucleotides along a stretch of DNA (or RNA) is, in fact, highly variable. DNA is in fact composed of a series of nucleotides and that each nucleotide has three components: a phosphate group; either a ribose (in the case of RNA) or a deoxyribose (in the case of DNA) sugar; and a single nitrogen-containing base. We also know that there are two basic categories of nitrogenous bases: the purines (adenine [A] and guanine [G]), each with two fused rings, and the pyrimidines (cytosine [C], thymine [T], and uracil [U]), each with a single ring. Furthermore, it is now widely accepted that RNA contains only A, G, C, and U (no T), whereas DNA contains only A, G, C, and T (no U)

5. Strengthening the Foundation: Chargaff Formulates His "Rules" Erwin Chargaff was one of a handful of scientists who expanded on Levene's work by uncovering additional details of the structure of DNA, thus further paving the way for Watson and Crick. Chargaff, an Austrian biochemist, had read the famous 1944 paper by Oswald Avery and his colleague s  at Rockefeller University, which demonstrated that hereditary units, or genes, are composed of DNA. This paper had a profound impact on Chargaff, inspiring him to launch a research program that revolved around the chemistry of nucleic acids. As his first step in this search, Chargaff set out to see whether there were any differences in DNA among different species. After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions (Chargaff, 1950).

6. . Figure 2: What is Chargaff's rule? All DNA follows Chargaff's Rule, which states that the total number of purines in a DNA molecule is equal to the total number of pyrimidines. First, he noted that the nucleotide composition of DNA varies among species. In other words, the same nucleotides do not repeat in the same order, as proposed by Levene. Second , Chargaff concluded that almost all DNA--no matter what organism or tissue type it comes from--maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is usually similar to the amount of thymine (T), and the amount of guanine (G) usually approximates the amount of cytosine (C). In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. (This second major conclusion is now known as "Chargaff's rule.") Chargaff's research was vital to the later work of Watson and Crick, but Chargaff himself could not imagine the explanation of these relationships--specifically, that A bound to T and C bound to G within the molecular structure of DNA.

7. Putting the Evidence Together: Watson and Crick Propose the Double Helix Chargaff's realization that A = T and C = G, combined with some crucially important X-ray crystallography work by English researchers Rosalind Franklin and Maurice Wilkins, contributed to Watson and Crick's derivation of the three-dimensional, double-helical model for the structure of DNA. Watson and Crick's discovery was also made possible by recent advances in model building, or the assembly of possible three-dimensional structures based upon known molecular distances and bond angles, a technique advanced by American biochemist Linus Pauling . In fact, Watson and Crick were worried that they would be "scooped" by Pauling, who proposed a different model for the three-dimensional structure of DNA just months before they did. In the end, however, Pauling's prediction was incorrect.

8. Using cardboard cutouts representing the individual chemical components of the four bases and other nucleotide subunits, Watson and Crick shifted molecules around on their desktops, as though putting together a puzzle. They were misled for a while by an erroneous understanding of how the different elements in thymine and guanine (specifically, the carbon, nitrogen, hydrogen, and oxygen rings) were configured. Only upon the suggestion of American scientist Jerry Donohue did Watson decide to make new cardboard cutouts of the two bases, to see if perhaps a different atomic configuration would make a difference. It did. Not only did the complementary bases now fit together perfectly (i.e., A with T and C with G), with each pair held together by hydrogen bonds, but the structure also reflected Chargaff's rule.

9. X-ray diffraction data collected by Rosalind Franklin and Maurice Wilkins showed that the B-form of DNA (which is more hydrated than the A-form) is a regular helix, making a complete turn every 34 Å (3.4 nm), with a diameter of about 20 Å (2 nm). The distance between adjacent nucleotides is 3.4 Å (0.34 nm); thus, there must be 10 nucleotides per turn. (In aqueous solution, the structure averages 10.4 nucleotides per turn.). The density of DNA suggests that the helix must contain two polynucleotide chains. The constant diameter of the helix can be explained if the bases in each chain face inward and are restricted so that a purine is always paired with a pyrimidine, avoiding partnerships of purine–purine (which would be too wide) or pyrimidine–pyrimidine (which would be too narrow). Chargaff also observed that regardless of the absolute amounts of each base, the proportion of G is always the same as the proportion of C in DNA, and the proportion of A is always the same as that of T. Consequently, the composition of any DNA can be described by its G-C content, or the sum of the proportions of G and C bases. (The proportions of A and T bases can be determined by subtracting the G-C content from 1.) G-C content ranges from 0.26 to 0.74 among different species.

10. Each base pair is rotated about 36° around the axis of the helix relative to the next base pair, so approximately 10 base pairs make a complete turn of 360°. The twisting of the two strands around each other forms a double helix with a minor groove that is about 12 Å (1.2 nm) across and a major groove that is about 22 Å (2.2 nm) across, as can be seen from the scale model presented in FIGURE 1.14 . In B-DNA, the double helix is said to be “righthanded”; the turns run clockwise as viewed along the helical axis. (The A-form of DNA, observed when DNA is dehydrated, is also a right-handed helix and is shorter and thicker than the B-form. A third DNA structure, Z-DNA (named for the “zig-zag” pattern of thebackbone), is longer and narrower than the B-form and is a lefthanded helix.

11. Although scientists have made some minor changes to the Watson and Crick model, or have elaborated upon it, since its inception in 1953, the model's four major features remain the same yet today. These features are as follows: DNA is a double-stranded helix, with the two strands connected by hydrogen bonds. A bases are always paired with Ts, and Cs are always paired with Gs, which is consistent with and accounts for Chargaff's rule. Most DNA double helices are right-handed; that is, if you were to hold your right hand out, with your thumb pointed up and your fingers curled around your thumb, your thumb would represent the axis of the helix and your fingers would represent the sugar-phosphate backbone. Only one type of DNA, called Z-DNA, is left-handed. The DNA double helix is anti-parallel, which means that the 5' end of one strand is paired with the 3' end of its complementary strand (and vice versa). As shown in Figure 4, nucleotides are linked to each other by their phosphate groups, which bind the 3' end of one sugar to the 5' end of the next sugar. Not only are the DNA base pairs connected via hydrogen bonding, but the outer edges of the nitrogen-containing bases are exposed and available for potential hydrogen bonding as well. These hydrogen bonds provide easy access to the DNA for other molecules, including the proteins that play vital roles in the replication and expression of DNA (Figure 4).

12. One of the ways that scientists have elaborated on Watson and Crick's model is through the identification of three different conformations of the DNA double helix. In other words, the precise geometries and dimensions of the double helix can vary. The most common conformation in most living cells (which is the one depicted in most diagrams of the double helix, and the one proposed by Watson and Crick) is known as B-DNA. There are also two other conformations: A-DNA, a shorter and wider form that has been found in dehydrated samples of DNA and rarely under normal physiological circumstances; and Z-DNA, a left-handed conformation. Z-DNA is a transient form of DNA, only occasionally existing in response to certain types of biological activity (Figure 5). Z-DNA was first discovered in 1979, but its existence was largely ignored until recently. Scientists have since discovered that certain proteins bind very strongly to Z-DNA, suggesting that Z-DNA plays an important biological role in protection against viral disease (Rich & Zhang, 2003).

13. Mutation – Molecular basis

14. DNA Polymerase III ONLY Responsible for spontaneous mutation ??

15. Tautomeric shift

16. Female with ClB chromosome are mated with irridiated male ClB daughter with ClB chro and irridiated X chro from male, crossed with wild male If recessive lethal mutation- only female will produce

17. Potent chemical mutagens

18. Evolution of Gene Concept

19. Gene: The basic unit of function

20. One factor – one character : Gregor J Mendel (1866)

21. One mutant gene one metabolic block: Sir Archibald Garrod (1909) Inborn errors of metabolism

22. Gene Control of Enzyme Structure Genes encode proteins including enzymes, which catalyze reactions Genes work in sets to accomplish biochemical pathways Genes often work in cooperation with other genes

23. Garrod and Bateson’s Hypothesis of Inborn Errors of Metabolism – work in Alkaptonuria Alkaptonuria is symptomized by blackened, oxidized urine and late onset arthritis The etiology of the disease (genetic) was elucidated by examining familial inheritance Pathologically, patients with alkaptonuria lack the necessary enzyme to metabolize homogenistic acid (HA) due to a recessive mutation on chromosome 3 (found in studies performed later on) Garrod’s work provided the first evidence of a specific relationship between genes and enzymes.

24. Notice the relationship of pathways within the metabolome

25. Next, George Beadle and Edward Tatum exposed a bread mold (haploid fungus Neurospora crassa) to X-rays, creating mutants unable to survive on minimal medium due to an inability to synthesize certain molecules Using crosses, they identified three classes of arginine-deficient mutants, each lacking a different enzyme necessary for synthesizing arginine They developed a one-gene one-enzyme hypothesis , stating that each gene dictates the production of a specific enzyme One-Gene One-Enzyme Hypothesis Animation: Meselson-Stahl Experiment

26. Life Cycle of Neurospora Asexual propogation occurs through the duplication of condidia or mycelium (haploid genome – mitotic) 1 Sexual reproduction occurs through the fusion of two mating types producing ascospores (diploid genome – meiotic) 2 Prototrophs (wild-type) grow on minimal medium (precursors only) Auxotrophs (mutants) need nutritional medium supplements (products) 1 1 2

27. Experimental Protocol One mating type of conidia were mutagenized with X-rays and crossed with wild-type conidia of the opposite mating type (Why?) Fruiting bodies were produced; microscopic ascospores were dissected and transferred to culture tubes containing complete or minimal medium No growth on minimal medium identified a nutritional mutant that was further investigated and specifically identified

28. One Gene One Enzyme: Beadle and Tautam (1942)

29. One Gene One Enzyme: Beadle and Tautam (1942) Beadle and Tatum received a Nobel Prize in 1958

30. Mutants affirm that methionine biosynthesis proceeds through a series of reactions catalyzed by enymes

31. But some proteins aren’t enzymes. So what then? Thus, the researchers revised the hypothesis to: one-gene one-protein But, But! Many proteins are composed of several polypeptides, each of which has its own gene Therefore, Beadle and Tatum’s hypothesis is now restated as the one-gene one- polypeptide hypothesis Note that even though it is not accurate (or sometimes correct) it is common to refer to gene products as proteins rather than polypeptides

32. One Gene One Polypeptide: Ingram (1957) Tryptophan synthetase

33. Gene Control of Protein Structure Genes also encode proteins that are not enzymes Structural proteins, such as hemoglobin, are often abundant, making them easier to isolate and purify (enzymes are generally produced in much smaller amounts and thus, are more difficult to purify to homogeneity)

34. Sickle Cell Anemia and Hemoglobin Hemoglobin is formed by four polypeptide chains and is responsible for O 2 and CO 2 distribution Two subunits of the protein contain the a polypeptide Two subunits of the protein contain the b polypeptide Each subunit associates with a heme group, which contains an iron center (reactive site)

35. Sickle Cell Hemoglobin Phenotypically, the mutant red blood cells change shape (sickle) under low O 2 tension Sickled cells are fragile, causing anemia. They are also less flexible, blocking up capillaries Effects are pleiotropic, including damage to the extremities, resulting in heart failure, pneumonia, paralysis, kidney failure, abdominal pain, and rheumatism Heterozygous individuals have the sickle-cell trait, a much milder form of the disease

36. Hemoglobin Form Electrophoresis of the protein showed the sickle cell form of Hb (Hb-S) has altered mobility compared with normal hemoglobin (Hb-A) Hemoglobin from individuals with the sickle-cell trait shows equal amounts of Hb-A and Hb-S, indicating that heterozygotes make both forms of Hb Thus, the sickle-cell mutation changes the form of its corresponding protein. Since protein structure is controlled by genes…

37. Hemoglobin Genetics The 6 th amino acid of the b chain in sickle-cell hemoglobin is valine (no electrical charge) rather than the negatively charged glutamic acid in the b chain of normal hemoglobin Over 200 types of hemoglobin mutants have been genetically characterized (most are point mutants) The mutant form is codominantly expressed

38. Cystic Fibrosis The affected gene is on the long arm of chromosome 7 and encodes a protein called Cystic Fibrosis Transmembrane Conductance Regulator Mutation results in an abnormal CFTR protein, preventing chlorine ion transport, resulting in mucus accumulation Cystic fibrosis (CF) affects the pancreas, lungs, and digestive system, and sometimes the vas deferens in males Treatment regimen includes antibiotics and percussion of the thoracic cavity

39. Genetic counseling Genetic counseling is advice based on genetic analysis, focusing either on the probability that an individual has a genetic defect or the probability that prospective parents will produce a child with a genetic defect Information for genetic counseling is obtained from two sources: Pedigree Analysis (prior history of disease in a family?) Carrier Detection (identify genotype of parents - Aa?) Fetal Analysis (fetal biochemical or genetic assay)

40. Fetal Analysis Amniocentosis Involves removal of a sample of amniotic fluid using a syringe needle inserted through the uterine wall (post 12 wk) Fetal cells are cultured Biochemical and genetic analyses are performed

41. Chorionic villus sampling Involves removal of chorionic villus tissue either through the abdomen or vagina (wk 8-12) No need for further culture of fetal cells Biochemical and genetic analyses are performed

42. Gene: The basic unit of structure

43. Beads on String Concept The gene is the fundamental unit of Structure : indivisible by crossing over 2. Change : Mutations change alleles from one form to another 3. Function : parts of genes cannot function alone in tests of complementation

44. Recombination within the Gene Lozenge eye locus in Drosophila – Oliver 1940 Two mutations, lz s and lz g , were considered alleles of the same gene because lz s / lz g heterozygotes have lozenge, not wild-type, eyes. But when lz s / lz g females are crossed to lz s or lz g males, about 0.2% of the progeny are wild-type

45. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 171 Determining the Number of Genes for Mutations with the Same Phenotype 1. Relationship between phenotype and gene can be studied through mutants identified by phenotype distinct from wild type. 2. Complementation test ( cis-trans test) determines whether independently isolated mutations for the same phenotype are in the same or different genes by crossing two mutants (Figure 13.1). a. If mutations are in different genes, phenotype will be wild type (complementation). b. If mutations are in the same gene, phenotype will be mutant (no complementation).

46. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 172 Fig. 12.1 Complementation test to determine whether two mutations resulting in the same phenotype are in the same or different genes

47. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 173 3 Drosophila provides an example. Wild-type body color is grey-yellow. If two true-breeding mutant black-bodied strains are crossed, all F1 are wild type (Figure 13.2). a. Genes are e ( ebony ) and b ( black ). Black parents are homozygous mutant but in different genes ( e / eb +/ b +) and ( e +/ e + b / b ). b. F1 are heterozygous at both loci ( e +/ eb +/ b ) and therefore wild type, showing complementation.

48. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 176 Multiple Alleles 1. Not all genes have only two forms (alleles); many have multiple alleles (figure 13.3). No matter how many alleles for the gene exist in the multiple allelic series, however, a diploid individual will have only two alleles, one on each homologous chromosome.

49. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 177 Peter J. Russell, iGenetics : Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig. 13.3 Allelic forms of a gene

50. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 178 Biochemical Genetics of the Human ABO Blood Group 1. Cellular antigens are important in blood transfusions, since recipient antibodies may respond to antigens on donor cells. 2. Karl Landsteiner discovered human ABO blood groups in the early 1900s, and received the 1930 Nobel Prize in Physiology or Medicine for this work. Properties of the human ABO blood group: a. There are three alleles at the ABO locus, I A, I B, and i . From these three alleles, four phenotypes are produced: i. Type A individuals have the A antigen on their RBCs, and anti-B antibodies in their blood. Their genotype is I A/ I A or  I A/ i . ii. Type B individuals have the B antigen on their RBCs, and anti-A antibodies in their blood. Their genotype is I B/ I B or  I B/ i . iii. Type AB individuals have both the A and the B antigen on their RBCs, and neither anti-A nor anti-B antibodies in their blood. Their genotype is I A/ I B. iv. Type O individuals have neither the A nor the B antigen on their RBCs, and both anti-A and anti-B antibodies in their blood. Their genotype is i / i . 3. Antigen–antibody relationships, and their impact in blood transfusions, are summarized in Figure 13.4.

51. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 179 4. Summary of the relationship between the ABO alleles and RBC antigens: a. The ABO locus produces RBC antigens by encoding glycosyltransferases, which add sugars to existing polysaccharides on membrane glycolipid molecules. These polysaccharides act as the antigen in the ABO system (Figure 13.5). b. In most people, the glycolipid is the H antigen. i. Activity of the I A gene product, a- N -acetylgalactosamyl transferase, converts the H antigen to the A antigen. ii. Activity of the I B gene product, a-D-galactosyltransferase, converts the H antigen to the B antigen. iii. Both enzymes are present in an I A/ I B individual, and some H antigens will be modified to the A antigen while others are modified to the B antigen. iv. Neither enzyme is present in an i / i individual, and so the H antigen remains unmodified. 5. Production of the H antigen is controlled by a different genetic locus from the ABO enzymes. Rarely, an individual lacks the dominant allele H needed for H antigen production. This h / h genotype results in the Bombay blood type, which is similar to type O except that Bombay blood type individuals produce anti-O antibodies that are not seen in true type O individuals.

52. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 180 Drosophila Eye Color 1. Drosophila has over 100 mutant alleles at the eye-color locus on the X chromosome. Example designations for alleles at this locus: a. The white-eyed variant allele is designated w . b. The wild-type (brick red) allele is w + . c. A recessive allele, w e , produces eosin (reddish-orange) eyes. 2. Soon after Morgan’s discovery of X-linkage, he found new genes for eye shape. and color, including a red one called vermilion ( v + ) . a. He experimentally crossed a white-eyed female with a vermilion eyed male. b. The F 1 females were all red-eyed (wild type), rather than either vermilion or white. c. He concluded two different genes were involved in Drosophila eye color (white and vermilion) rather than just alleles of a single locus. d. The original cross was: w v + / w v + (white-eyed female) with w + v/ Y (vermilion-eyed male). e. The F 1 females (wild-type red eyes) would be w v + / w + v, doubly heterozygous. 3. The eosin allele is recessive to wild-type. Morgan (1912) crossed an eosin-eyed female with a white-eyed male. All F 1 females had eosin eyes.

53. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 181 4. Sturtevant (1913) concluded that eosin and white are mutations of a single gene. The relationship between these multiple alleles is: a. The allele red (wild-type) is dominant to eosin and white. b. The eosin allele is recessive to red, but dominant to white. c. For example, in the cross of an eosin-eyed ( w e / w e ) female . with a white-eyed male ( w /Y), the F l females are all w e /w. They have eosin eyes, showing that w e is dominant over w. d. Next the eosin-eyed F l females ( w e /w) are crossed with red-eyed males w + / Y (Figure 13.6). (1) All female progeny are red-eyed (w + /w or w + /w e ). (2) Male progeny are 1/2 eosin-eyed ( w e / Y), and 1/2 white-eyed (w/Y). e. Many alleles of the white-eye gene exist, producing a wide range of colors depending on the deposition of pigment in the eye cells (Table 13.2).

54. 台大農藝系 遺傳學 601 20000 Chapter 12 slide 182 Peter J. Russell, iGenetics : Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig. 13.6 Results of crosses of Drosophila melanogaster involving two mutant alleles of the same locus

55. Complementation test Operational definition of gene Cis-trans position effect

56. Intragenic complementation

57. Fine structure analysis of gene Complementation test

58. Fine structure analysis of gene Recombination test

59. Deletion Mapping

60. Collinearity between a gene and its polypeptide

61. Genes within gene and Overlapping genes

62. Modern definition of gene

63. A gene is a genomic sequence on DNA (or RNA) It encodes (one or many) functional product molecules (RNA or protein) Functional products sharing overlapping genomic regions are united the union must be coherent i.e. union built separately for RNA and protein products does not require that all products necessarily share a common subsequence Example: Three functional protein products built from genomic elements A,B,C: A+B, A+C, C only belong to the same gene even though A+B and C only do not share a common subsequence. Notice: sharing of UTRs or regulatory regions is not sufficient (see D,E). “The gene is a union of genomic sequences encoding a coherent set of potentially overlapping functional products.” Modern definition of gene

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