Examine the table below for the pathway with precursors / products A, B, C, D, and E and mutant enzymes 1, 2, 3, and 4. Use the information given to draw the pathway. (+ indicates growth, 0 indicates no growth) A B C D E 1 + 0 0 0 + 2 + + 0 0 + 3 + + 0 + + 4 0 0 0 0 + Later a new mutation (5) is found that fails to complement mutant 2. Between what two compounds on the pathway is this new mutant located?
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Examine the table below for the pathway with precursors / products A, B, C, D, and E and mutant enzymes 1, 2, 3, and 4. Use the information given to draw the pathway. (+ indicates growth, 0 indicates no growth)
|
A |
B |
C |
D |
E |
1 |
+ |
0 |
0 |
0 |
+ |
2 |
+ |
+ |
0 |
0 |
+ |
3 |
+ |
+ |
0 |
+ |
+ |
4 |
0 |
0 |
0 |
0 |
+ |
Later a new mutation (5) is found that fails to complement mutant 2. Between what two compounds on the pathway is this new mutant located?
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- In this problem, just put the order of the intermediates on the pathway, starting with P and ending with Z and explain your reasoning. (So to get you started, notice the class 1 mutants—and there is only mutant in this class, which is mutant #5—won’t grow on minimal medium, but will grow if you give it substance Z. However, giving it substances, p, w, x, or y won’t help. This means that the gene that is mutated lies on the pathway in a place that the Z substances is to its right and all the other substances are to its left. The easiest next one to look at is the line with 2 pluses, and then the line with 3 pluses, and then the line with 4 pluses).In our feeding experiment with vermilion/brown and cinnabar/brown mutant flies, we were supposed to use kynurenine or hydroxykynurenine to overcome the enzymatic block in the ommochrome pathway, caused by the vermilion or cinnabar mutations. Which eye color would you expect for each of the two double mutants, if we had supplemented the medium with N-formylkynurenine? Use a diagram of the pathway to explain your answer.The intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.
- The intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +There is a group of six or so genes, called luxlux genes, whose gene products are necessary for light formation. Given that these bacterial genes are regulated together, propose a hypothesis for how the genes are organized and regulated. Match the terms in the left column to the appropriate blanks in the sentences on the right. Terms can be used once, more than once, or not at all. Terms: metaboliozed, density, quorum sensing, state, signal amplification, inhibit, activate, detected, secrete, transcrption initiation complex, consume, operon Bacterial cells blank molecules that can be blank by other bacterial cells. Sensing the concentration of such signalling molecules allows bacteria to monitor the local blank of cells, a phenomenon called blank. There is a group of genes in the genome of the described bacterium, called luxgenes, which are organized in an blank. Reaching a certain concentration, molecules involved in this process blank the luxgenes. As a result, the…W, X, and Y are the intermediates (in that order) in abiochemical pathway whose product is Z. Z− mutantsare found in five different complementation groups.Z1 mutants will grow on Y or Z, but not W or X. Z2mutants will grow on X, Y, or Z. Z3 mutants will onlygrow on Z. Z4 mutants will grow on Y or Z. Finally,Z5 mutants will grow on W, X, Y, or Z.a. Order the five complementation groups in terms ofthe steps they block.b. What does this genetic information reveal aboutthe nature of the enzyme that carries out theconversion of X to Y?
- Auxotrophic mutation 103 grows on minimal medium supplemented with A, B, or C; mutation 106 grows on medium supplemented with A or C, but not B; and mutation 102 grows only on medium supplemented with C. What is the order of A, B, and C in a biochemical pathway?Feather color in parakeets is produced by the blending of pigments produced from two biosynthetic pathways shown below. Four independently assorting genes (AA, BB, CC, and DD) produce enzymes that catalyze separate steps of the pathways. For the questions below, use an uppercase letter to indicate a dominant allele producing full enzymatic activity and a lowercase letter to indicate a recessive allele producing no functional enzyme. Feather colors produced by mixing pigments are green (yellow + blue) and purple (red + blue). Red, yellow, and blue feathers result from production of one colored pigment, and white results from absence of pigment production. What is the genotype of a pure-breeding purple parakeet strain? Express your answer as combination of allelic symbols. Example: AaBBccDd.The reason why Beadle and Tatum observed four different categories of mutants that could not grow on media without methionine is because a. the enzyme involved in methionine biosynthesis is composed of four different subunits. b. the enzyme involved in methionine biosynthesis is present in four copies in the Neurospora genome. c. four different enzymes are involved in a pathway for methionine biosynthesis. d. a lack of methionine biosynthesis can inhibit Neurospora growth in four different ways.
- . In a certain plant, the flower petals are normally purple.Two recessive mutations arise in separate plants and arefound to be on different chromosomes. Mutation 1 (m1)gives blue petals when homozygous (m1/m1). Mutation2 (m2) gives red petals when homozygous (m2/m2).Biochemists working on the synthesis of flower pigments in this species have already described the following pathway:colorless (white)compoundblue pigmentred pigmentenzyme Aenzyme Ba. Which mutant would you expect to be deficient inenzyme A activity?b. A plant has the genotype M1/m1 ; M2/m2. Whatwould you expect its phenotype to be?c. If the plant in part b is selfed, what colors of progenywould you expect and in what proportions?d. Why are these mutants recessive?A number of auxotrophic mutant strains were isolated from wild-type haploid Neurospora crassa. These strains responded to the addition of certain nutritional supplements to minimal culture medium either by growth (+) or no growth (0) The data from this experiment are presented in the table below. Diagram a biochemical pathway, complete with positions of intermediates, that is consistent with the data. Indicate where in the pathway each mutant strain is blocked.Suppose that you are hired by a biotechnology firm to produce a strain of giant fruit flies, by using recombinant DNA technology, so that genetics students will not be forced to strain their eyes when looking at tiny flies. You go to the library and learn that growth in fruit flies is normally inhibited by a hormone called shorty substance P (SSP). You decide that you can produce giant fruit flies if you can somehow turn off the production of SSP. SSP is synthesized from a compound called XSP in a single-step reaction catalyzed by the enzyme runtase: XSP———————>SSP runtase A researcher has already isolated cDNA for runtase and has sequenced it, but the location of the runtase gene in the Drosophila genome is unknown. In attempting to devise a strategy for turning off the production of SSP and producing giant flies by using standard recombinant DNA techniques, you discover that deleting, inactivating, or otherwise mutating this DNA sequence in Drosophila turns out to be extremely…