Page 179 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1
P. 179

Problems   171


                             the cross of the F 1  females with the scabrous, jave-  Section 5.2
                             lin males, and indicate for each class whether it is a     5.  In mice, the dominant allele Gs of the X-linked gene
                             recombinant or parental type.                     Greasy produces shiny fur, while the recessive wild-
                                                                                     +
                          b. The cross in part (a) yielded the following progeny:   type Gs  allele determines normal fur. The dominant
                             77 scabrous eyes and normal bristles; 76 wild type   allele Bhd of the X-linked Broadhead gene causes
                             (normal eyes and bristles); 74 normal eyes and    skeletal abnormalities including broad heads and
                                                                                                                  +
                             javelin bristles; and 73 scabrous eyes and javelin   snouts, while the recessive wild-type Bhd  allele
                             bristles. Are the genes governing these traits likely   yields normal skeletons. Female mice heterozygous
                             to be linked, or do they instead assort indepen-  for the two alleles of both genes were mated with
                             dently? Why?                                      wild-type males. Among 100 male progeny of this
                          c.  Suppose you mated the F 1  females from the cross   cross, 49 had shiny fur, 48 had skeletal abnormalities,
                             in part (a) to wild-type males. Why would this    2 had shiny fur and skeletal abnormalities, and 1 was
                             cross fail to inform you whether the two genes are   wild type.
                             linked?                                           a.  Diagram the cross described and calculate the
                          d. Suppose you mated females from the true-breeding    distance between the two genes.
                             stock with javelin bristles to males with scabrous   b. What would have been the results if you had
                             eyes and javelin bristles. Why would this cross     counted 100 female progeny of the cross?
                             fail to inform you whether the two genes are     6.  In Drosophila, males from a true-breeding stock with
                             linked?                                           raspberry-colored eyes were mated to females from a
                         3.  With modern molecular methods it is now possible   true-breeding stock with sable-colored bodies. In the
                          to examine variants in DNA sequence from a very      F 1  generation, all the females had wild-type eye and
                          small amount of tissue like a hair follicle or even a   body color, while all the males had wild-type eye
                          single sperm. (See the Fast Forward Box Mapping      color but sable-colored bodies. When F 1  males and
                          the Crossovers that Generate the Chromosomes of        females were mated, the F 2  generation was composed
                          Individual Human Sperm.) You can consider these      of 216 females with wild-type eyes and bodies, 223
                          variants to be alleles of a particular site on a chro-  females with wild-type eyes and sable bodies, 191
                          mosome (a locus; loci in plural). For example,       males with wild-type eyes and sable bodies, 188 males
                          AAAAAAA, AAACAAA, AAAGAAA, and                       with raspberry eyes and wild-type bodies, 23 males
                          AAATAAA at the same location (call it B) on ho-      with wild-type eyes and bodies, and 27 males with
                          mologous autosomes in different sperm might be       raspberry eyes and sable bodies. Explain these results
                                                                2
                                                             1
                          called alleles 1, 2, 3, and 4 of locus B (B , B , etc.).   by diagramming the crosses and calculating any rele-
                                                              1 3
                          John’s genotype for two loci B and D is B B  and     vant map distances.
                                                           4
                                                1 2
                              3
                                                         1
                            1
                          D D . John’s father was B B  and D D , while his     7.  If the a and b loci are 20 m.u. apart in humans and an
                                              2
                                      3 3
                                                 3
                          mother was B B  and D D .                            A B / a b woman mates with an a b / a b man, what is
                          a.  What is (are) the genotype(s) of the parental type   the probability that their first child will be A b / a b?
                             sperm John could produce?                       8.  CC DD and cc dd individuals were crossed to each
                          b. What is (are) the genotype(s) of the recombinant   other, and the F 1  generation was backcrossed to the
                             type sperm John could produce?                    cc dd parent. 997 Cc Dd, 999 cc dd, 1 Cc dd, and 3 cc
                          c.  In a sample of 100 sperm, 51 of John’s sperm were   Dd offspring resulted.
                                       1
                                              1
                             found to be B  and D , while the remaining 49 sperm   a.  How far apart are the c and d loci?
                                  3
                                     3
                             were B  D . Can you conclude whether the B and D   b. What progeny and in what frequencies would you
                             loci are linked, or whether they instead assort     expect to result from testcrossing the F 1  generation
                               independently?                                    from a CC dd × cc DD cross to cc dd?
                         4.  The Punnett square in Fig. 5.4 shows how Mendel’s   c.  In a typical meiosis, how many crossovers occur
                          dihybrid cross results would have been altered had the   between genes C and D?
                          two genes (A and B) been linked and had the P gener-  d. Assume that the C and D loci are on the same
                          ation cross been A B / A B × a b / a b.                chromosome, but the offspring from the testcross
                          a.  What would be the frequency of each F 2  pheno-    described in part (b) were 498 Cc Dd, 502 cc dd,
                             typic class if 80% of the gametes produced by the   504 Cc dd, and 496 cc Dd. How would your an-
                             F 1 s were parentals?                               swer to part (c) change?
                          b. Answer part (a) assuming that the original, P gen-    9.  In mice, the autosomal locus coding for the β-globin
                             eration cross was A b / A b × a B / a B.          chain of hemoglobin is 1 m.u. from the albino locus.
   174   175   176   177   178   179   180   181   182   183   184