One Gene, One Polypeptide • In the 1930’s, Beadle and Tatum did a series of experiments that went a long ways towards showing what genes actually do. The catchphrase that comes from their work is: “One gene, one polypeptide”. That is, each gene codes for a polypeptide. • Polypeptides are chains of amino acids. Proteins consist of one or more polypeptides, plus (in some cases), additional co-factors. For example, the oxygen-carrying protein hemoglobin consists of 4 polypeptides: 2 alpha chains plus 2 beta chains, and 4 heme co-factor molecules. • Proteins do most of the work in the cell. Many proteins are enzymes that catalyze chemical reactions in the cell. Other proteins are structural: collagen in skin and keratin in hair, for example. Others carry useful molecules around, such as hemoglobin. • The DNA base sequence of a gene codes for the sequence of amino acids in the polypeptide. Each group of 3 DNA bases is translated into a single amino acid.
Beadle and Tatum Experiments • B+T worked with Neurospora, bread mold, which is a eukaryote, a fungus. Neurospora normally grows as a haploid, but it can be mated to form a diploid-like organism (a dikaryon, actually). Neurospora can be grown from haploid spores • Wild type (prototrophic) Neurospora can be grown on a minimal defined medium: a set of chemicals that contains all that is necessary for life. Neurospora can synthesize all other necessary compounds from these raw materials. • It is possible to isolate mutant Neurospora that can’t make certain necessary compounds. These mutants are auxotrophs, and they need to have the chemical compounds supplied in the growth medium.
Selection of Auxotrophs • B+T mutated spores with UV light, then grew each individual spore in a separate tube on rich medium. Rich medium contains every possible nutrient: the Neurospora don’t need to make anything from scratch. Dead spores (inevitable with mutant generation) didn’t grow. • Next, some cells from each tube were transferred to minimal medium. Some that grow on rich medium did not grow on minimal: these are auxotrophs, they lack the ability to make some chemical compound that is found in rich medium but not in minimal. • To determine which particular compound the auxotrophs couldn’t make, each auxotroph was grown on minimal medium supplemented with a series of specific amino acids and vitamins. Each auxotroph proved to need a single additional compound. • The auxotrophs were thus grouped into categories such as Arg- (needed arginine) and Lys- (needed lysine), etc.
Complementation • B+T showed that each mutant was unable to synthesize a single chemical compound. • It is possible to use their system to generate hundreds or thousands of mutants of the same type: Arg-, for example. Clearly not all of these mutants affects a different gene. How to sort the mutants out? • The complementation test: make diploids of all combinations of the mutants. Two possible results: • 1. the diploid heterozygote is an auxotroph (same as each of the haploids by itself). This implies that the two mutations are in the same gene. We say the two mutations “fail to complement”. • 2. the diploid heterozygote is a prototroph (wild type). This implies that the two mutations are in different genes. We say that the two mutations “complement”.
Results • Each number represents an Arg- auxotroph. The squares of the table are the diploid heterozygotes. A “+” means that the heterozygote was a prototroph, and a “-” means that the heterozygote was an auxotroph. • There are 4 genes represented here, called (arbitrarily) A, B, C, and D. The mutants that affect each gene are: • A: 1, 3, 4 • B: 2, 6 • C: 5 • D: 7
Biochemical Pathways • The work of biochemists showed that chemical compounds in the cell are synthesized through a series of intermediates, a biochemical pathway. • For example, the amino acid arginine is synthesized from glutamic acid (another amino acid) through the following steps: • glutamic acid -> N-acetyl glutamate -> ornithine -> citruline -> arginine. • GLU -> NAG -> ORN -> CIT -> ARG • Each step is catalyzed by an enzyme, a protein consisting of one or more polypeptides plus co-factors (in some cases). Each step can be blocked by destroying its enzyme. The enzymes are destroyed by mutations in the genes that synthesize them. • If ANY of the steps is blocked by mutation, the organism becomes an Arg- auxotroph.
Linking Genes to Pathway Steps • By feeding the auxotroph various chemical intermediates on the pathway it is possible to determine which genes make which enzymes. • GLU -> NAG -> ORN -> CIT -> ARG • For example, imagine a mutation that destroys the enzyme that converts ornithine to citruline. Note that all other enzymes in the pathway are intact. Specifically, the CIT->ARG enzyme is still functional. Thus, if the Arg- auxotrophs were fed citruline, they would be able to convert it to arginine. • Thus a mutant blocked in the ORN-> CIT step can grow on either arginine or citruline. However, it cannot grow on GLU, NAG, or ORN: any intermediate in the pathway above the blocked enzyme. • A mutant that blocks a particular step in a biochemical pathway grows on any intermediate downstream from the block, but not on any intermediate upstream from the block.
Explanation • In this example, W, X, Y, and Z represent 4 chemical intermediates on the histidine biosynthesis pathway. The order of these intermediates is unknown: you have to figure it out. The numbers 1 through 6 are mutant genes. Each mutant was grown on the different intermediates. A “+” means that it grew and a “-” means that it did not grow. • Remember the basic rule: A mutant that blocks a particular step in a biochemical pathway grows on any intermediate downstream from the block, but not on any intermediate upstream from the block. • Start from the end of the pathway, the last step that leads directly to the synthesis of histidine. A mutation that blocks this step will grow on His but nothing else. Mutants 1 and 6 do this: they block the final step. • Next, determine the intermediate that is just before histidine. All mutants EXCEPT 1 and 6 will grow on it. Intermediate W fits this (look at the W row). Thus, the pathway so far is W -> His, with 1 and 6 affecting this step. • Continue to work up the pathway, alternating between determining the mutants that affect a step and the determining the intermediate leading to that step.