Final answer:
The probability of two heterozygous parents having two children with six fingers (recessive phenotype) is 6.25%, as calculated by multiplying the chances of each child inheriting the recessive allele independently, which is 1/4 for each child, resulting in a combined probability of 1/16.
Step-by-step explanation:
The probability that two parents, both heterozygous for the six-fingered phenotype (having one dominant and one recessive allele), will have two six-fingered offspring can be calculated using a Punnett square and the principles of Mendelian genetics. In this case, each parent has a genotype of Aa, where 'A' represents the dominant five-finger allele and 'a' represents the recessive six-finger allele. The chances of each parent passing on the 'a' allele is 1/2. When we use the product rule to calculate the probability of two independent events occurring together (both children inheriting the recessive allele from both parents), we would multiply the individual probabilities of each child being aa.
For the first child to have six fingers (aa), the probability is 1/4 (1/2 chance from each parent). For the second child to also have six fingers, the probability is again 1/4. To find the probability of both events happening, we multiply the individual probabilities: 1/4 × 1/4 = 1/16. The answer, therefore, is that there is a 6.25% chance of both children having six fingers.