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ONE GENE – ONE ENZYME HYPOTHESIS
In 1942, Beadle and Tatum were studying the Neurospora crassa, a type of bread mold (fungus) which is used as a model organism because it's easy to grow and has a haploid life cycle that makes genetic analysis simple, since the recessive traits are shown in the offspring. They discovered that there's a direct correlation between genes and enzymes and they led to the formulation of the hypothesis "one gene - one enzyme".
The Neurospora crassa mycelium produces asexual orange spores. Neurospora can be reproduced asexually by inoculating fragments of mycelium or asexual spores (conidia) in a medium to grow a new mycelium. It can also reproduce sexually: there're 2 sexes, A and a, which are identical in appearance, but crosses between individuals of the same sex aren't possible.
The sexual cycle starts when A and a sexes are mixed in a medium poor in nitrogen. Cells of the 2 sexes fuse together to produce a diploid A/a nucleus.
which undergoes a meiosis that produces 4 haploid cells (2 A and 2a) in a long sack called ascus. A mitotic division produces 8 haploid cells, around which the spore's walls form to produce 8 sexual spores (4 A and 4 a). Several asci develop inside of the fruiting body.
When the ascus is mature, the sexual spores are expelled from the ascus and the fruiting body. The germination of a spore produces a new haploid mycelium.
The wild-type can grow in a minimal medium (prototroph), which only contains inorganic salts (nitrogen source), simple sugars (organic carbon source) and biotin (vitamin H or B7). Neurospora synthesizes the other molecules necessary for the growth by using the simple chemical compounds present in the minimal medium.
Beadle and Tatum believed that it was possible to isolate nutritional mutants, which couldn't grow in a minimal medium because they needed other nutritional components. They isolated the nutritional mutants by treating conidia with X rays, which are
Mutagens (they cause mutations). Then, they crossed the mutants with the wild-types.
Sexual spores were grown in a complete medium, so that mutants unable to synthetize molecules by using the simple compounds of a minimal medium could grow by using those that were present in the complete one.
Each culture was analyzed during its growth in a minimal medium. The strains that didn't grow were mutants and they were individually analyzed to see if they were able to grow in minimal mediums with the addition of amino acids or vitamins.
A mutant that has lost the ability to synthetize a certain amino acid grows in a minimal medium which contains amino acids, but not vitamins.
To know which of the 20 amino acids is needed for the growth of a certain mutant strain, the strain is analyzed in 20 test tubes, each of which contains a minimal medium to which one of the 20 amino acids was added.
The strains that required arginine to grow were further studied. 4 genes involved in the arginine
formatted as follows:Four genes involved in the biosynthetic pathway of arginine were discovered: argE, argF, argG, and argH. A mutation in one of these genes causes the loss of the ability to synthesize arginine. The growth of the mutants was analyzed in different mediums containing different arginine precursors, and the sequence of the reactions occurring in the arginine biosynthetic pathway could be deduced.
argH mutants grow when arginine is added to the medium, but not when its precursors are added. argG mutants grow when arginine and argininosuccinate are added, but not when citrulline and ornithine are added. argF mutants don't grow when ornithine only isn't added to the medium. This means that ornithine is produced before citrulline, which is produced before argininosuccinate, which is produced before arginine.
If the metabolic chain is blocked between compound B and compound C, the product D isn't produced, B accumulates, and the mutant can grow if D is provided to it. It can also grow if C is provided to it.
The same experiment can be...
Conducted on E. coli. Nutritional mutants are isolated with the replica plate technique: E. coli cultures are grown on a complete medium; a plastic block covered with a sterilized velvet surface is pressed on the complete medium (master plate); the colonies removed from the master plate and that are now on the velvet surface are imprinted in a minimal medium (replica plate), where nutritional mutants can't grow; nutritional mutants are those bacteria that were present in the master plate, but aren't in the replica plate.
Complementation is a genetic process during which 2 strains of organisms with different homozygous recessive mutations that produce the same mutant phenotype are crossed to restore the wild-type phenotype, which is shown in their offspring. Complementation occurs if the 2 mutations are in different genes.
A complementation test is used to test whether the mutations in the 2 strains are in different genes. In a complementation test, 2 mutants with the same phenotype are crossed.
If the 2 mutations are located indifferent genes, all the progeny is composed of wild-type/mutant heterozygotes for each of the 2 genes and it shows the wild-type phenotype.
If the 2 mutations are in the same gene, the progeny inherits a mutant version of the gene in each of the 2 homologous and its phenotype is mutant.
Another example of complementation involves the anthocyanin pathway in flowers. There’re 3 types of mutants: $ mutants have a mutation in gene w1; £ mutants have a mutation in a different position in gene w1; ¥ mutants have a mutation in gene w2.
When a $ mutant is crossed with a £ mutant, there’s no complementation and their progeny has a mutated $ allele and a mutated £ allele. Therefore they produce the first precursor of anthocyanin, but not the second one.
When a £ mutant is crossed with a ¥ mutant, there’s complementation, because the progeny has only 1 mutated £ allele and 1 mutated ¥ allele, whose effects are
dominated by the ones of the wild-type alleles. Therefore they produce both precursors and anthocyanin, which makes the flowers' petals blue.
GENE INTERACTION
Epistasis is the interaction between alleles of 2 or more genes, which control a single phenotype, in which the expression of a gene is affected by the expression of one or more genes. This phenomenon often involves genes which contribute to different passages of the same pathway. The gene that masks the expression of the other gene is called epistatic, while the gene whose expression is masked is called hypostatic.
Given 1 non-allelic loci A and B, with A epistatic on B, epistasis can be caused by the recessive homozygosis of allele A (a/a, recessive epistasis) or by the presence of a dominant allele in locus A (A/-, dominant epistasis). In both cases, the expression of the genes in locus B is masked, regardless of the alleles it contains (B/- or b/b). All these possibilities produce different modifications of the 9:3:3:1 ratio in a
If you consider the genotypes of a F2 offspring in a dihybrid cross (A/- B/-, A/-b/b, a/a B/-, a/a b/b), in a RECESSIVE EPISTASIS of locus A and locus B, the individuals a/a B/- and a/a b/b have the same phenotype. This determines a phenotypic ratio of 9:3:4, instead of 9:3:3:1.
A recessive epistasis takes place when recessive alleles at one locus hide the expression of both (dominant and recessive) alleles at another locus.
Es. Gene w encodes for enzyme 1, which is needed to produce the pink pigment (1 passage of the pathway), while gene m encodes for enzyme 2, which is needed to produce the blue pigment (2 passage of the pathway).
When a double heterozygote (w /w m /m) is selfed, an offspring with the following phenotypic ratio is obtained: 9 blue w /- m /- (both enzymes are active) : 3 pink w /- m/m (only the 1 enzyme is active) : 4 white w/w m /m or w/w m/m (both enzymes are inactive).
The recessive allele w is epistatic on m and m.
In the DOMINANT EPISTASIS, A
is epistatic on B and A/- B/- and A/- b/b individuals show the same phenotype.
Dominant epistasis occurs when the expression of both alleles (dominant and recessive) at another locus is masked by a dominant allele at one locus. Therefore, the expression of one dominant or recessive allele is masked by another dominant gene. This determines a phenotypic ratio of 12:3:1.
Es. Pumpkins generated by crosses between white and yellow pumpkins or white and green pumpkins are always white, while those generated by crosses between yellow and green pumpkins are always yellow. This means that yellow is recessive with respect to white, but dominant over green.
Considering 2 genes each with an allelic couple: W/w Y/y. The pumpkin's yellow if there's at least one Y allele, while it's green if there're 2 y alleles in the Y locus. This happens in absence of the product of the W locus, which inhibits the phenotypic expression of yellow and green and pumpkins are white. Therefore, W/- Y/-
and W/- y/y pumpkins are white, w/w Y/- pumpkins are yellow and w/w y/y pumpkins are green.
The progeny of selfed F1 pumpkins show a phenotypic ratio of 12:3:1 of white : yellow : green pumpkins.
In the DUPLICATED RECESSIVE EPISTASIS, a/a is epistatic on B and b and b/b is epistatic on A and a, as for the color of pea flowers, where purple is dominant over white.
Duplicate recessive epistasis occurs when recessive alleles at either of the two loci can hide the expression of dominant alleles at the two loci. This determines a phenotypic ratio of 9:7.
Es. A recessive mutation in gene C or P, if in homozygosis, produces white flowers. Purple flowers are produced only by genotypes with at least one normal C allele and one normal P allele.
Crosses between white pure lines with genotype C/C p/p and white pure lines with genotype c/c P/P produce only purple flowers with genotype C/c P/p in the F1.
When F1 individuals are selfed, they produce a F2 offspring with the following relations: 9 C/- P/- (purple
flowers) : 3 C/- p/p(white flowers) : 3 c/c P/- (white flowers) : 1 c/c p/p (whiteflowers). The resulting phenotypic ratio is 9 purple flowers : 7white flowers, because c/c is epistatic on P and p, and p/p isepistatic on C and c.
The following pathway can be imagined to explain the production of thepurple pigment:
Genes C and P code for products that control different passages of the same pathway and the productof both is necessary for the production of the final purple product. Therefore, if the product of locus Cor locus P is missed or both products are missed, the conversion of the white intermediate into thefinal purple product is blocked (if the genotype is C/- p/p) or the conversion of the white precursorinto the white intermediate is blocked (if the genotype is c/c P/- or c/c p/p).
In the DUPLICATED DOMINANT EPISTASIS, A and B are epistatic on a and b.Duplicate dominant epistasis occurs when a dominant allele at either of two locican hide the expression of
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