Answer:
The chances of a child being non - tongue roller is ![0.5\\](https://tex.z-dn.net/?f=0.5%5C%5C)
Explanation:
Let the allele for dominant "tongue rolling trait" be represented by "T"
The allele for recessive "non-tongue rolling trait" be represented by "t"
The genotype of a heterozygous tongue rolling parent will be "Tt"
The genotype of a homozygous non- tongue rolling parent will be "tt"
If a cross is carried out between these two individuals , the offspring produced are as shown in the punnet square below-
T t
t Tt tt
t Tt tt
So the number of offsrpings which are non- tongue roller are two in numbers represented by "tt"
So the chances that a child will be a non-tongue roller are
![\frac{2}{4} \\= 0.5\\](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B4%7D%20%5C%5C%3D%200.5%5C%5C)
Where , "4" represents the total number of offsprings produced