In mice, a recessive mutation causes them to waltz. Such mice are called waltzers.
The crosses in the problem have been shown below:
Waltzers x homozygous normal
Progeny: single waltzing female among normal.
Single waltzing female x waltzing male
Progeny: all waltzing.
Single waltzing female x homozygous normal
Progeny: all normal.
Some males and females of the above progeny were crossed
Progeny: all normal.
The single waltzing female, which was seen amidst many normal individuals, was due to expression of a recessive gene. This could be a new “waltzer” mutation inherited from the normal male.
But, based on the information a shortened chromosome was present, we can say that this female inherited a deletion of a wild-type allele. This allowed the expression of the recessive allele in the single waltzing female.
When the single female was crossed with a waltzing male, all progeny were waltzers. When she was crossed with a homozygous normal male, all progeny were normal. When some of these normal progeny were crossed, all the subsequent progeny were normal.
If she had inherited some new waltzing allele, the normal progeny should have been heterozygous. When two heterozygous progeny are crossed, they should produce 25% waltzing individuals.
If a deletion had occurred, half of progeny would be heterozygous and half would be heterozygous, for the deletion. When such progeny are intercrossed, 25% will be homozygous, for the deletion and the rest of the progeny would not possess the waltzer allele.
Thus, the theory is consistent with the data presented in the question and is the right answer.