Manx cats are heterozygous for a dominant mutation that results in no tails t. The mating of two Manx cats yield three Manx Kittens for each normal, long-tailed and medium tail length. Explain using genetic cross.
That result shouldn't be possible. Manx cats must be heterozygous in order to be Manx. So their genotype would be written as Tt. A long tail would be tt and a medium tail would TT.
The question said that two Manx cats are crossed. That would be Tt x Tt. The expected result of that cross is one TT, two Tt, and one tt. That cross doesn't often result that way, since the Manx gene is semi-lethal. A homozygous dominant (medium tail) cat typically dies because of other genetic issues. So instead of the expected 1:2:1 ratio, Manx crosses result in a 2:1 ratio. That means it's expected that two Manx are born for every one normal, long tail cat.
The only way for a 3:1 ratio to develop in Manx cats is if the gene is no longer lethal and completely a regular dominant gene. If that is the case, then the 1:2:1 genotypic ratio does lead to a 3:1 phenotypic ratio. Any cat born with the dominant T gene would be Manx. So TT is Manx and Tt is Manx. tt is long tail. That gives you the three to one ratio in your question, but a medium tail would never develop.
The result is entirely possible given that the mutation causing the Manx cat phenotype would most likely be a lethal allele. Using Mendelian genetics, the expected genotype ratio from crossing two heterozygotes would be 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive. A phenotypic ratio of 2:1 hints that there is most likely a lethal allele (tt in this case). That lethal allele means that those offspring that are tt would not survive embryonic development.
The predicted segregation pattern is 1/4 homozygous dominant, 1/2 heterozygous, and 1/4 embryonic lethal.