Question 1, continued Your collegue has two other colonies of pure-breeding mice. Colony III mice have white fur and long tails, Colony IV mice have white fur and short tail You cross mice from each of these colonies white fur and white fur and long tails short tails Black fur and ils c)How can you explain this result? There are mutations in two different genes, F and W that result in the same recessive phenotype, white fur. The cross could be fwwLL (white fur with long tail) x FFwwoll(white fur with short tail) to give. fFWwLl= black fur with long tails Question 2 In a hypothetical type of yeast, the biosynthetic pathway for phenylalanine(shown below)has five reaction steps each catalyzed by a different enzyme(1-5). The pathway is known to begin with phosphoenolpyruvate(PEp)and proceed via 4 intermediates W, X, Y, and Z to phenylalanine PEP phenvlalanine You want to identify the enzymes(five total) involved in this pathway by isolating yeast that fail to synthesize phenylalanine( these yeast are referred to as"phe"). You know that mutant yeast that fail to synthesize phenylalanine(and thus can not grow without addition of phenylalanine) are likely to be defective in one of the enzymes involved in the phenylalanin synthesis pathway. You start with a population of wild-type("phet/) yeast, mutagenize it with UV light, and allow the treated cells to grow into isolated colonies on plate 1(see FIGURE 1). You then use the replica plating technique to transfer some yeast from each colony onto plates 2, 3, 4, and 5. The contents of the growth medium are listed below each plate. Rich medium contains all nutrients and allows growth of both wild type, and mutant cells Minimal medium contains nutrients sufficient to allow only wild-type, prototrophic yeast to grow. It will not support the growth of auxotrophic cells 012Fal20037.012 Fall 2003 Question 1, continued Your collegue has two other colonies of pure-breeding mice. Colony III mice have white fur and long tails, Colony IV mice have white fur and short tails. You cross mice from each of these colonies. white fur and long tails X white fur and short tails
Black fur and long tails c) How can you explain this result? There are mutations in two different genes, F and W that result in the same recessive phenotype, white fur. The cross could be: ffWWLL (white fur with long tail) X FFwwll (white fur with short tail) to give: fFWwLl = black fur with long tails Question 2 In a hypothetical type of yeast, the biosynthetic pathway for phenylalanine (shown below) has five reaction steps each catalyzed by a different enzyme (1 – 5). The pathway is known to begin with phosphoenolpyruvate (PEP) and proceed via 4 intermediates W, X, Y, and Z to phenylalanine. PEP W X Y Z phenylalanine 123 4 5 You want to identify the enzymes (five total) involved in this pathway by isolating yeast that fail to synthesize phenylalanine (these yeast are referred to as “phe-”). You know that mutant yeast that fail to synthesize phenylalanine (and thus can not grow without addition of phenylalanine) are likely to be defective in one of the enzymes involved in the phenylalanine synthesis pathway. You start with a population of wild-type (“phe+”) yeast, mutagenize it with UV light, and allow the treated cells to grow into isolated colonies on plate 1 (see FIGURE 1). You then use the replica plating technique to transfer some yeast from each colony onto plates 2, 3, 4, and 5. The contents of the growth medium are listed below each plate. Rich medium contains all nutrients and allows growth of both wild type, and mutant cells. Minimal medium contains nutrients sufficient to allow only wild-type, protoptrophic yeast to grow. It will not support the growth of auxotrophic cells