Dental Admissions Test (DAT) Practice Test

What is the probability, based on classical Mendelian genetics, of a child being born with cystic fibrosis if one parent is a carrier and the other is homozygous dominant?

1. 0%

2. 25%

3. 50%

4. 100%

The correct answer is: 25%

When one parent is a carrier for cystic fibrosis (heterozygous) and the other parent is homozygous dominant (no cystic fibrosis allele), the possible genotypes for the offspring will be 50% heterozygous carriers like the first parent and 50% homozygous dominant like the second parent. Hence, the likelihood of a child being born with cystic fibrosis (homozygous recessive) in this scenario is 0%, making choice A incorrect. Choice C (50%) is incorrect because there is no chance of the child having cystic fibrosis with these parental genotypes. Choice D (100%) is incorrect as it implies that all children would have cystic fibrosis in this scenario, which is not possible based on Mendelian genetics. Thus, the correct answer is B (25%) because there is a 25% chance of the child being a carrier for cystic fibrosis.