# Assume you have freckles and you are heterozygous, and your partner does not have freckles; what are the chances your child will not have freckles?

If one has freckles and is heterozygous, but their partner does not have freckles, there is a 50% chance that their child will have freckles and a 50% chance they will not.

If you have freckles and you are heterozygous for the trait, it means that the trait itself is dominant, not recessive. This means that you only need one dominant allele in order to display the trait. Furthermore, if your partner does not have freckles, this means that their genotype is recessive. This is because, again, if the condition of freckles is a dominant trait, the only way for a person not to have the condition is to have two recessive alleles. Even having one dominant allele (being heterozygous) will mean that a person has the freckles phenotype, because the expression of that allele will overpower the expression of the recessive one.

We can easily represent this using a Punnett square. Let’s say that the big “A” allele codes for freckles, while the little “a” allele codes for non-freckles. You are heterozygous, so you have the Aa genotype. You also have freckles, which means that the A-allele (having freckles) overpowers the expression of the a-allele (not having freckles). Your partner must be homozygous recessive not to have freckles (aa), because both of these alleles code for the absence of the trait. Use a Punnett square to chart the probability of your child having freckles.

Aa x aa = 1/4 probability of each: Aa aa Aa aa.

You can see here that during fertilization, there is a 50% chance that your child will have the Aa genotype (and the freckles phenotype) and a 50% chance that they will have the aa genotype (and the no-freckles phenotype). The answer is 50%.