Biology = genetics
please do Challenge problem 3 Now, figure out the complementation groups. This shows how many genes are involved in this pathway. Some of the mutant classes in the previous problem must contain more than one complementation group (gene) For example class 2 is defined by mutants 1,3,4,6, and 7. These mutants may, or may not, all be in the same gene. Complementation testing will help you know if they are, or are not, in the same gene. So there are actually additional steps between some of the letters. For example, maybe instead of XàY, it is really X->F->G->Y.
Extracted text: CHALLENGE PROBLEM 3 The 10 mutants in Challenge Problem 2 were tested for complementation in all pairwise combinations using heterokaryons. The results are shown in the matrix, in which + indicates the ability of the hetero- karyon to grow in minimal medium and - indicates inabil- ity to grow in minimal medium. 4 6 7 8 9. 10 2 Mutant 1 + 3 4 + 7 8. Assume that each complementation group defines a different gene, and assume further that each gene encodes an enzyme that catalyzes a single step in the metabolic pathway, which converts one molecule of substrate into one molecule of product. (a) Redraw the metabolic pathway deduced from Chal- lenge Problem 2. Use a right arrow to indicate each enzymatic step in the pathway, and label each arrow with the mutant number 1-10 that blocks the enzy- matic step. In some cases, you will not be able to specify the order in which the enzymes occur in the pathway, so you may write them in any order you wish. (b) In the metabolic pathway that you have deduced from the data, how many unknown intermediates are there between the precursor P and the vitamin Z? 9 + + + + + + + | + + + + + + + LO 3. + + LO
Extracted text: Challenge problem 2 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) P- No Class mutants grows X - Class 2 mutants ie (1,3,4,6,7) grows on this medium w- Class 2 (1,3,4,6,7) and Class 3(2) mutants grow in this medium Y- Class 2 (1,3,4,6,7) Class 3(2) and Class 4(8,9,10) mutants grow in this medium Z- All the Class mutants grow in this medium. -P -> X -> W -> Y -> P W Y Class I (mutant 5) Class II (mutants 1, 3, 4, 6, 7) Class III (mutant 2) Class IV (mutants 8, 9, 10) Draw a linear metabolic pathway with P on the left and z on the right, in which each of the intermediates W, X, and Y is shown in the order in which it occurs in the metabolic pathway in the synthesis of Z from P. 区 N + +