The problem states that normal haemoglobin is Hb^A and abnormal haemoglobin (sickle-cell) is Hb^S. Sickle-cell anaemia is due to a recessive gene, meaning Hb^S is recessive to Hb^A.

You're on a roll — Step 1: Define the alleles and genotypes. The problem states that normal haemoglobin is Hb^A and abnormal haemoglobin (sickle-cell) is Hb^S. Sickle-cell anaemia is due to a recessive gene, meaning Hb^S is recessive to Hb^A. Genotypes and their corresponding phenotypes: • Hb^AHb^A: Normal (no sickle-cell trait) • Hb^AHb^S: Sickle-cell trait (carrier) • Hb^SHb^S: Sickle-cell anaemia (affected) The cross is between a man who is heterozygous and a woman who is also heterozygous. Parental genotypes: Man (Hb^AHb^S) × Woman (Hb^AHb^S) a) What are the possible phenotypes of the offspring of a man who is heterozygous and a woman who is also heterozygous? Step 1: Perform the cross using a Punnett square. Gametes from each parent are Hb^A and Hb^S. |c|c|c| Gametes & Hb^A & Hb^S \\ Hb^A & Hb^AHb^A & Hb^AHb^S \\ Hb^S & Hb^AHb^S & Hb^SHb^S \\ Step 2: Determine the phenotypes from the genotypes. • Hb^AHb^A: Normal • Hb^AHb^S: Sickle-cell trait • Hb^SHb^S: Sickle-cell anaemia The possible phenotypes of the offspring are: Normal, Sickle-cell trait, Sickle-cell anaemia b) What proportion of the offspring would have been: b) (i) Sickle-cell anaemia? Sickle-cell anaemia corresponds to the genotype Hb^SHb^S. From the Punnett square, 1 out of 4 offspring has the genotype Hb^SHb^S. The proportion of offspring with sickle-cell anaemia is: (1)/(4) b) (ii) Sickle-cell trait? Sickle-cell trait corresponds to the genotype Hb^AHb^S. From the Punnett square, 2 out of 4 offspring have the genotype Hb^AHb^S. The proportion of offspring with sickle-cell trait is: (2)/(4) or (1)/(2) 4 down, 1 left today. What's next?