1. (45) As stated in Solved Problem 2, certain mice called waltzers have a recessive mutation that causes them to execute bizarre steps. W. H. Gates crossed waltzers with homozygous normals and found, among several hundred normal progeny, a single waltzing female mouse. When mated with a waltzing male, she produced all waltzing offspring. When mated with a homozygous normal male, she produced all normal progeny. Some males and females of this normal progeny were intercrossed, and there were no waltzing offspring among their progeny. T. S. Painter examined the chromosomes of waltzing mice that were derived from some of Gates’s crosses and that showed a breeding behavior similar to that of the original, unusual waltzing female. He found that these mice had 40 chromosomes, just as in normal mice or the usual waltzing mice. In the unusual waltzers, however, one member of a chromosome pair was abnormally short. Interpret these observations as completely as possible, both genetically and cytologically.
2. (46)
Six bands in a salivary-gland chromosome of Drosophila are shown in the following illustration, along with the extent of five deletions (Del 1 to Del 5): Recessive alleles a, b, c, d, e, and f are known to be in the region, but their order is unknown. When the deletions are combined with each allele, the following results are obtained:
Recessive alleles a, b, c, d, e, and f are known to be in the region, but their orderis unknown. When the deletions are combined with each allele, the followingresults are obtained:
| | a | b | c | d | e | f |
| Del 1 | — | — | — | + | + | + |
| Del 2 | — | + | — | + | + | + |
| Del 3 | — | + | — | + | — | + |
| Del 4 | + | + | — | — | — | + |
| Del 5 | + | + | + | — | — | — |
In this table, a minus sign means that the deletion is missing the correspondingwild-type allele (the deletion uncovers the recessive), and a plus sign means thatthe corresponding wild-type allele is still present. Use these data to infer whichsalivary band contains each gene.
3.
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