Final answer:
Parents with freckles, who are heterozygous, have a 25% chance to have a child without freckles. The probability that two more children will also have freckles is about 56.25%, as determined by multiplying the individual probabilities of each child having freckles.
Step-by-step explanation:
The scenario described by the student, in which parents with freckles have a child without freckles, is a classical example of Mendelian inheritance, specifically a monohybrid cross involving dominant and recessive alleles. Freckles are controlled by a dominant allele (let's say 'F'), so individuals with at least one 'F' allele will show freckles. Non-freckled individuals must have two recessive alleles ('f').
If Tim and Jan both have freckles, but their son does not, it indicates that both parents must be heterozygous (Ff), carrying one dominant allele for freckles and one recessive allele for no freckles. The Punnett square for Tim and Jan's offspring would show that there is a 25% chance for a child to be non-freckled (ff), 50% chance for a child to be heterozygous with freckles (Ff), and a 25% chance for a child to be homozygous with freckles (FF).
In terms of probability, if Tim and Jan have two more children, there is a 3/4 chance for each child to have freckles. To find the probability that both children will have freckles, we need to multiply the probabilities for each child having freckles: (3/4) × (3/4) = 9/16 or approximately a 56.25% chance that both children will have freckles.