Answer:
1/2
Explanation:
This is a cross involving a single gene coding for for possession or not of sickle cell anemia in humans. The 'S' allele encodes a normal cell while the 's' allele encodes a sickle cell. The allele for normal cell (S) is dominant over the allele for sickle cell (s) i.e. S allele will mask the phenotypic expression of s allele in a heterozygous state.
An heterozygous individual contains both dominant (S) and recessive (s) alleles. Hence, will have a genotype: Ss.
Using a punnet square, a cross between two heterozygous individuals i.e. Ss × Ss will produce four possible offsprings with genotypes: SS, Ss, Ss, ss
SS (1/4) is homozygous dominant i.e. normal offspring with same dominant alleles
Ss (2/4) is heterozygous dominant i.e. normal offspring with different alleles.
ss (1/4) is homozygous recessive i.e. sickle cell offspring with same recessive alleles.
Therefore, the probability that an offspring of the cross between two heterozygous parents will also be heterozygous is 2/4 = 1/2.
S. s
S SS Ss
s Ss ss
