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.
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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.
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