Final answer:
The statement that all children of heterozygous parents for an autosomal recessive trait will show the trait is false. Genetics predict that there is a 25% chance for a child to exhibit the trait if both parents carry one recessive allele each.
Step-by-step explanation:
When both parents are heterozygous for an autosomal recessive trait, there's a 25% chance that any given child they have will exhibit the recessive trait. This is because when both parents carry one copy of a recessive allele (heterozygous), they can each pass on either the recessive allele or the dominant allele to their offspring. With both parents being heterozygous, the Punnett square shows that there is a 1 in 4 chance for the child to receive both recessive alleles (rr), and thus, show the trait. Therefore, the statement that all of their children will exhibit the trait if both parents are heterozygous for an autosomal recessive trait is false.
A correct genotype probability distribution for children of parents both heterozygous for a recessive trait is as follows: 25% chance to be homozygous recessive (rr) and show the trait, 50% chance to be heterozygous (Rr) and be carriers without showing the trait, and another 25% chance to be homozygous dominant (RR) and not show the trait nor be a carrier.
In accordance with Mendel's Laws, specifically the law of segregation, each parent contributes one allele, and these alleles segregate during gamete formation so that each gamete carries only one allele for each inherited trait. Following Mendel's monohybrid cross, a 3:1 phenotypic ratio is predicted in the second generation, though in real-life human family scenarios, chance can result in variations from this expected ratio.