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University of Toronto St. George
Human Biology
Stephen Wright

Pages 219-233 Genes act by controlling cellular chemistry – several human diseases have defects in metabolism (the general set of chemical reactions taking place in an organism) - Archibald Garrod – made this discovery, came up with the notion that such genetic diseases are “inborn errors of metabolism” - Observations of alkaptonuria (AKU) led to the possibility that the cell’s chemical pathways were under control of a large set of interacting genes George Beadle and Edward Tatum used Neurospora fungus to demonstrate this control - Irradiated Neurospora cells to produce mutations, then tested cultures grown from ascospores for unusual mutant phenotypes relevant to biochemical function - Found a number of mutations with defective nutrition, specifically auxotrophic mutants o Wild-type Neurospora can use cellular biochemistry to synthesize almost all its cellular components from inorganic nutrients and a carbon source in the medium, but auxotrophic Neurospora can’t o To grow, mutants require a nutrient to be supplied, suggesting a defect in some normal synthetic step (wild-types can synthesize their own nutrients) - Each mutation that generated a nutrient requirement was inherited as a single-gene mutation b/c each gave a 1:1 ratio when crossed with a wild type o + x aux  progeny: ½ + and ½ aux - Classified the specific nutritional requirement of each auxotroph – nutrients needed include proline, arginine, methionine, pyridoxine... o Concentrating on arginine, they found that the genes that mutated to give arginine auxotrophs mapped on 3 different loci on 3 separate chromosomes (arg-1, arg-2, arg-3) o Discovered that the auxotrophs for each locus differed in response to structurally related cpds ornithine and citrulline - Biochemical pathway for interconversions of related compounds o precursor (enzyme X)  ornithine (enzyme Y)  citrulline (enzyme Z)  arginine o arg-1 lacks X, arg-2 lacks Y, arg-3 lacks Z; proves that a mutation of a particular gene is assumed to interfere with the production of one single enzyme o Defective enzyme creates a block in some biosynthetic pathway, circumvented by giving the cells any cpd that normally comes after the block in the pathway; creates a more complete biochemical model (see page 220) - One-gene-one-enzyme hypothesis – source of the first exciting insight into the functions of genes; genes were somehow responsible for enzyme function, and each gene controlled one specific enzyme in a series of interconnected steps - All proteins, enzymes or not, were found to be encoded by genes, so the phrase was redefined to become the one-gene-one-polypeptide hypothesis - A gene encodes the physical structure of a protein, which in turn dictates its function Neurospora arginine pathway – synthetic pathway – chain of enzymatic conversions that synthesizes essential nutrients Signal-transduction pathway – chain of complicated signals from the environment to the genome and from one gene to another; these are crucial to the proper function of an organism - Yeast – 2 mating types are necessary for yeast mating to occur; when a cell is near an opposite mating type, it undergoes a series of changes in shape and behaviour to prepare for mating - Mating response is triggered by a signal-transduction pathway requiring the sequential action of a set of genes (discovered through a standard interaction analysis of mutants with aberrant mating response, most sterile) - Signal that gets everything going – is a mating pheromone (hormone) released by the opposite mating type, binds to membrane receptor, coupled to G protein inside membrane, activates protein o G protein sets in motion a series of sequential protein phosphorylations (a kinase cascade), which activates the transcription of a set of mating-specific genes that enable the cell to mate - A mutation at any of these steps will disrupt the mating process Developmental pathways – comprise the steps by which a zygote becomes an adult organism - Involves many genetically controlled steps, e.g. laying down body plans for organs, tissue differentiation/movement... - All these steps require gene regulation and signal transduction To find out the interacting genes for a certain biological property, we have to take 3 steps - Obtain many single-gene mutants and test for dominance - Test the mutants for allelism – one or several loci? - Combine mutants in pairs to form double mutants to see if the genes interact Gene interaction is inferred from the phenotype of the double mutant – if the genes interact, then the phenotype is different from the simple combination of both single-gene mutant phenotypes - If mutant anlleles from different genes interact, we infer that the wild-type genes also interact normally as well - In these case, a modified 9:3:3:1 Mendelian ratio will usually result Complementation test – a way of deciding whether 2 mutations belong to the same gene - Diploid – intercross 2 individuals that are homozygous for different recessive mutations, observer whether the progeny have wild-type phenotype o If the progeny are wild-type, the recessive mutations must be in different genes – recessive wild-type alleles provide wild-type function (the 2 mutations have complemented) o If the progeny are not wild-type, the recessive mutations must be alleles of the same gene – both alleles are mutants so there’s no wild type to distinguish btw 2 different mutant alleles o Alleles could have different mutant sites on the same allele, but they would both be non- functional - Complementation – the production of a wild-type phenotype when 2 haploid genomes bearing different recessive mutations are united in the same cell - Using harebell plants as an example (see pages 223-225), it can be deduced that complementation is actually a result of the cooperative interactions if the wild-type alleles of the 2 genes - Haploid – complementation can’t be performed by intercrossing; in fungi, an alternative way brings mutant alleles together to test complementation – fusing to form a heterokaryon o When 2 different strains fuse, the haploid nuclei from the different strains occupy one cell, which is the heterokaryon o The nuclei don’t fuse, so this is sometimes known as a “mimic diploid” - To learn whether genes interact, we have to analyze the phenotype of the double mutant, to see if it’s different from the combination of both single mutations - Double mutant obtained by intercrossing - F obt1ined through complementation test, F 2 obtained by selfing the F 1rogeny and should contain the double mutant - Double mutant may then be identified by looking for Mendelian ratios o Standard 9:3:3:1 – phenotype present in only 1/16 of the progeny will represent the double mutant (the “1” in 9:3:3:1) o In cases of gene interaction, the phenotype of the double mutant may not be distinct but will match that of one of the single mutants; a modified ratio will result – 9:3:4 or 9:7 Standard 9:3:3:1 ratio is the simplest case, expected when there’s no gene interaction and if the 2 mutations under test are on different chromosomes (the null hypothesis – any modified Mendelian ratio will be informative) - Inheritance of skin coloration in corn snakes – natural colour is a repeating black-and-orange camouflage pattern; the 2 colours are produced by 2 separate pigments, both under genetic control - Orange pigment – o (presence) and o (absence); black pigment – b (presence) and b (absence) + + o These two genes are unlinked – natural pattern produced by genotype o /-; b /- o Genotype for a black snake is o/o; b /- (lacks orange pigment) o Genotype for an orange snake is o /o ; b/b (lacks black pigment) o Double homozygous recessive o/o; b/b is albino – pink colour is from the haemoglobin of the blood that is visible through the snake’s skin o The albino snake also shows that there is another element to skin pigmentation – the repeating motif in/around the places where pigment is deposited Homozygous orange female o /o ; b/b x homozygous black male o/o; b /b  F o /o; b /b1 + + (camouflaged) - The F2 generation shows a 9:3:3:1 ratio – camouflaged female o /o; b /b x camouflaged male + + o /o; b /b – will produce progeny with the following phenotypes o 9 camouflaged – o /-; b /- o 3 orange – o /-; b/b + o 3 black – o/o; b /- o 1 albino – o/o; b/b - 9:3:3:1 ratio is produced because the 2 pigment genes act independently at the cell level - If the orange pathway fails, the black pathway is still functional, and vice versa; no pigment is produced only when both mutants are present and neither colour pigment is produced 9:7 ratio – genes in the same pathway; modification of the 9:3:3:1 ratio where the 3:3:1 combines to make 7, so some kind of interaction must be happening + + + + + + - w1/w1; w2 /w2 (white) x w1 /w1 ; w2/w2 (white)  F generation1of w1 /w1; w2 /w2 (blue) + + + + - w1 /w1; w2 /w2 x w1 /w1; w2 /w2  F generati2n with the following phenotypes o 9 blue – w1 /-; w2 /- o 3 white – w1 /-; w2/w2 + o 3 white – w1/w1; w2 /- o 1 white – w1/w1; w2/w2 o The only way in which a 9:7 ratio is possible is if the double mutant has the same phenotypes as the 2 single mutants - The modified ratio constitutes a way of identifying the double mutant’s phenotype;
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