The statement given is false.
When we analyze the genetic probabilities of two heterozyous parents having offspring, we use genetic principles including the Punnett square to understand the potential genotypes of their children. Heterozygous means that each parent has two different alleles for a particular gene: one dominant (which we can denote as 'A') and one recessive (which we can denote as 'a'). When such parents (Aa & Aa) have children, the possible genotypes of the offspring are as follows:
1. AA - where the offspring inherits the dominant allele from both parents.
2. Aa - where the offspring inherits the dominant allele from one parent and the recessive allele from the other.
3. aa - where the offspring inherits the recessive allele from both parents.
The Punnett square for this mating would be:
```
Parent 1
A a
P A AA Aa
a r aA aa
e
n
t
2
```
From the Punnett square, it is evident that there are four equally likely outcomes for the offspring's genotype:
- 1 out of 4 chance to be AA (25%)
- 2 out of 4 chance to be Aa (50%)
- 1 out of 4 chance to be aa (25%)
The offspring will only express the recessive trait if they inherit the recessive allele from both parents, which corresponds to the genotype 'aa'. This outcome occurs in one out of the four scenarios presented in the Punnett square.
Hence, the probability of the offspring expressing the recessive trait (having genotype 'aa') is 1 out of 4, or 1:4. Dividing both terms by 1 gives the same ratio 1:4 which means 25% of the time, on average, two heterozygous parents will have an offspring with the recessive trait. This is not equivalent to the ratio of 2:1 as stated in the given question, so the correct answer is "B False".
