A person with a phenotype of HbSS has sickle cell disease. A person with a phenotype of HbAS allele carries the sickle cell trait. A person with a phenotype of HbAA does not have sickle cell disease and does not carry the trait. The Punnett square shows the phenotype of two people carrying the sickle cell trait.
Answers
A person with a phenotype of HbSS has sickle cell disease. A person with a phenotype of HbAS allele carries the sickle cell trait. A person with a phenotype of HbAA does not have sickle cell disease and does not carry the trait. The Punnett square in the image shows the resulting progeny after a cross between the two individuals of the genotype, HbAS, who are carrying the sickle cell gene. The probability of offspring getting sickle cell disease turns out to be, 25%.
Answer:
The punnet square that results in phenotype of two carriers is as a result of parents who both have a phenotype of HbAS
Explanation:
The punnet square that results from the reproduction of HbAS × HbAS is as follows:
HbAS × HbAS
HbAS ⇒ 50% - 2 carriers out of the the 4 offsprings in the punnet square
HbSS ⇒ 25% - 1 affected offspring out of the 4 offsprings in the punnet square
HbAA ⇒ 25% - 1 non affected and non - carrier offspring of the sickle cell among the 4 offsprings in the punnet square.
Interpretation
When two people who are carries of the sickle cell disease (HbAS) reproduce the chances of:
1) Having a child with sickel cell ( HbSS) is 25%
2) Having a child without sickle cell and who is not a carrier (HbAA) is also 25%
3) Having a child without sickle cell but who is a carrier like the parents (HbAS) is 50%
Attached is the image of the actual punnet square:
