A dominat trait that is not sex-linked?

tim and christine have freckles (a dominant trait that is not sex-linked), but their son michael does not. show with a punnet square how this is possible. tim and christine are expecting a baby. what is the probability of freckles in that child?

A dominat trait that is not sex-linked?
Since freckles are a dominant trait, individuals that are heterozygous for the gene will have freckels. If both parents are heterozygous for the trait, then 1/4 of their children will have the recessive phenotype (i.e. no freckles).





Here's my best attempt at typing a punnet square:


.....F..| f


F |FF|Ff


f |Ff..|ff





Basically, if F is the gene for freckles, and f is the gene for no freckles, the FF and Ff individuals have freckles and ff do not (since F is dominant to f). If the parents are both Ff then 1/4 of their offspring will be FF, 1/2 will be Ff, and 1/4 will be ff. That means that 1/4 of their children will have the recessive (no freckles) phenotype. The remaining 3/4 will have freckles. So for any one child, there's a 3/4 chance they will have freckles.
Reply:3 out of 4
Reply:If freckles are a dominant trait and their son isn't have freckles, then both Tim and Christine must have one dominant and one recessive gene for this trait (Ff). Michael must be an ff since he doesn't have freckles. The possible outcomes are FF, Ff, Ff, and ff. In the first three combos result in a child with freckles; the last does not. Therefore the probability of a child with freckes is 75%.

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