Homework Assignment #4 BIOS 2500 Name: Due Monday (June 11) at the beginning of

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Homework Assignment #4 BIOS 2500 Name: Due Monday (June 11) at the beginning of recitation Problem I: Metabolic pathway analysis (ala Beadle and Tatum) Background: A species of thrip (Pseudothysamoptas nauseatus) is able to repel potential predators by secretinga noxious chemical called mephitamine, or "MA". MA is known to be produced from a plant alkaloid (cryptofetidate, a.ka. "CFD") that is present in the diet of P. nanseaus, but the specific steps and enzymes in the overali CFD- MA pathway have not yet been characterized biochemically Data: Recently, a geneticist carried out a genetic screen and obtained four different mutant strains of P. nauseatus that predators), the mutations were designated as 1-dfl4 (in the order of discovery). Tests of the various mutants showed that, in certain cases, production of MA could be restored by feeding the mutant animals a diet supplemented with one of three different compounds (called pre-mephitamines or PMA's, and arbitrarily designated A, B, & C) that are produced as metabolic intermediates during MA synthesis. Results are summarized in the following table lementis) that could rescue (ie, restore MA production) dil mutant dfi-1 PMA-B (only) PMA-A; PMA-B: PMA-C (ie, allany one of the three) none PMA-B: PMA-C an- dil-2 dfl-3 dfi-4 1) Based on the data provided above, determine the simplest possible pathway for MA synthesis in wild-type P nauseatus and draw this pathway in the space below. Guidelines: a) Your pathway should start with "CFD" and end with "MA", b) Be sure to determine both the order of the metabolic intermediates and the gene/enzyme that is required to catalyze each step. c) Your pathway should include a total of four steps, with each represented by an arrow. d) Remember that each intermediate (ie, one of the 3 PMA's) should be placed at a position between the arrows, and that each gene/enzyme (din1-1,-2,-3, or -4) will normally be listed above one of the arrows. 2) When a fith defenseless mutant, d1-5, was isolated in a subsequent screen, its phenotype was found to be identical to di-4 (ie, it could be rescued by either PMA-B or PMA-C, but not with PMA-A). However, further genetic analysis (specifically, a complementation test) indicated that the 4i-4 and di-5 mutations affect two different genes. Which of the following is NOT a reasonable explanation that could account for this result? (i.c, for why mutations in different genes might appear to have the same effect on the pathway?) a) The step affected by the two mutations might normally be carried out by an enzyme with two different subunits, with one polypeptide encoded by dy1-4 and the other by d-3 b) There might be two different (but functionally-equivalent) enzymes that can catalyze the same step, with one enzyme encoded by 4 and the other by 4.5. c) One of the genes (e.g. dfA-5) might encode a transcription factor or other regulatory protein that is needed for expression of the other gene (c-g,dn-4) d) There might be an additional, fifth step in the pathway (and also a fourth, "unknown" PMA/intermediate) that is normally catalyzed by the dfi-5 gene prodact, and which occurs just before or just after the "df-4" step. e) None of the above (ie, all are reasonable explanations for the similar behavior of dt-4 and di-5) Answer- (continued on other side

Explanation / Answer

1. 4 different genes are affected by these seven mutations. The complementation groups are (1,5), (2,6), (3,4), and 7.

2. one. Strain 7

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