Final answer:
When a sickle trait male and an Hgb C female produce offspring, there is a 25% chance that their offspring will be normal, with no symptomatic forms of anemia due to the possible genetic combinations of the beta-globin alleles.
Step-by-step explanation:
To determine the percentage of offspring that would be normal from a mating between a sickle trait male and an Hgb C female, we must understand the genetics behind sickle cell disease and Hgb C disease. In this case, both diseases are caused by mutations in the beta-globin gene. A sickle cell trait individual has one normal allele and one sickle cell allele (HbA/HbS), and an Hgb C individual has one normal allele and one Hgb C allele (HbA/HbC).
When these two individuals have children, there are four possible genetic combinations for their offspring: HbA/HbA (normal), HbA/HbS (sickle trait), HbA/HbC (Hgb C trait), and HbS/HbC (which will result in a form of anemia due to both abnormal hemoglobins). There is a 25 percent chance for each of these outcomes:
- 25% chance of being normal (HbA/HbA),
- 25% chance of inheriting the sickle cell trait (HbA/HbS),
- 25% chance of inheriting the Hgb C trait (HbA/HbC),
- 25% chance of having sickle cell disease and Hgb C disease (HbS/HbC).
Therefore, the percentage of offspring that would be normal (not showing symptoms of sickle cell disease or Hgb C disease) is 25%.