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266    Chapter 7    Anatomy and Function of a Gene: Dissection Through Mutation


                    detected  after growth in BrdU). These chromosomes   a.  Indicate what phenotypes and ratios you would
                  are caused by high levels of chromosome breakage         expect in the F 2  generation of this cross if there is
                  followed by repair through homologous recombina-       one pigment pathway, with orange and black being
                  tion. In some patients, every cell has many harlequin   different intermediates on the way to brown.
                  chromosomes. In other patients, the majority of cells   b. Indicate what phenotypes and ratios you would
                  have many harlequin chromosomes, but about 10% of        expect in the F 2  generation if orange pigment is a
                  the cells surprisingly have none.                      product of one pathway, black pigment is the prod-
                  a.  What kinds of events produce the 10% of the        uct of another pathway, and brown is the effect of
                    cells in certain Bloom syndrome patients with no     mixing the two pigments in the skin of the snake.
                    harlequin chromosomes? (Hint: Think about        39.  In a certain species of flowering plants with a  diploid
                      recombination.) What does the existence of these   genome, four enzymes are involved in the generation
                    cells lacking harlequin chromosomes say about      of flower color. The genes encoding these four en-
                    the  alleles of the Bloom syndrome gene carried    zymes are on different chromosomes. The biochemi-
                    by these patients?                                 cal pathway involved is as follows; the figure shows
                  b. In what way do Bloom syndrome patients of both    that either of two different enzymes is sufficient to
                    classes reflect the results of complementation tests?  convert a blue pigment into a purple pigment.
                  c.  Why does it make sense that the events you described                       →
                    in part (a) might occur in Bloom syndrome patients?         white → green → blue  →  purple
                  d. What is different about the events you described in       A true-breeding green-flowered plant is mated with a
                    part (a) from the events that give rise to harlequin   true-breeding blue-flowered plant. All of the plants in
                    chromosomes?                                       the resultant F 1  generation have purple flowers. F 1
                  e.  Could the events you described in part (a) occur   plants are allowed to self-fertilize, yielding an F 2  gen-
                    during G 1  of the cell cycle? During G 2 ?        eration. Show genotypes for P, F 1 , and F 2  plants, and
                  f.  The events that give rise to the cells without harle-  indicate which genes specify which biochemical
                    quin chromosomes are very rare, occurring in less   steps. Determine the fraction of F 2  plants with the
                    than one in a million cell divisions even in Bloom     following phenotypes: white flowers, green flowers,
                    syndrome patients. Surprisingly, however, roughly   blue flowers, and purple flowers. Assume the green-
                    10% of the cells in certain patients lack harlequin   flowered parent is mutant in only a single step of
                    chromosomes. How can these two statements be       the pathway.
                    true  simultaneously?                            40.  The intermediates A, B, C, D, E, and F all occur in
                                                                       the same biochemical pathway. G is the product of the
                                                                                                           −
                                                                       pathway, and mutants 1 through 7 are all G , meaning
              Section 7.5                                              that they cannot produce substance G. The following
                37.  The pathway for arginine biosynthesis in Neurospora   table shows which intermediates will promote growth
                  crassa involves several enzymes that produce a series   in each of the mutants. Arrange the intermediates in
                  of intermediates.                                    order of their occurrence in the pathway, and indicate
                                                                       the step in the pathway at which each mutant strain is
                        ARG-E    ARG-F   ARG-G        ARG-H            blocked. A + in the table indicates that the strain will
               N-acetylornithine → ornithine → citrulline → argininosuccinate → arginine
                                                                       grow if given that substance, an O means lack of growth.
                                                 −
                                                            −
                  a.  If you did a cross between ARG-E  and ARG-H
                    Neurospora strains, what would be the distribution                        Supplements
                          +
                                  −
                    of Arg  and Arg  spores within parental ditype and   Mutant  A     B    C     D     E    F     G
                    nonparental ditype asci? Give the spore types in
                    the order in which they would appear in the ascus.   1       +     +    +     +     +    O     +
                                                                         2       O     O    O     O     O    O     +
                  b. For each of the spores in your answer to part (a),   3      O     +    +     O     +    O     +
                    what nutrients could you supply in the media to get   4      O     +    O     O     +    O     +
                    spore growth?
                                                                         5       +     +    +     O     +    O     +
                38.  In corn snakes, the wild-type color is brown. One   6       +     +    +     +     +    +     +
                    autosomal recessive mutation causes the snake to be   7      O     O    O     O     +    O     +
                  orange, and another causes the snake to be black. An
                  orange snake was crossed to a black one, and the F 1      41.  In each of the following cross schemes, two true-
                  offspring were all brown. Assume that all relevant   breeding plant strains are crossed to make F 1  plants,
                  genes are unlinked.                                  all of which have purple flowers. The F 1  plants are
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