You're on a roll — Here is the working for the sickle cell anaemia question: a) Proportion of offspring with sickle cell anaemia: Step 1: Identify the genotypes of the parents. The problem states that both parents have sickle cell trait. This means they are heterozygous for the gene. Let H represent the gene for normal haemoglobin and S represent the gene for abnormal haemoglobin. Parent 1 genotype: HS Parent 2 genotype: HS Step 2: Construct a Punnett square to determine the possible genotypes of the offspring. $$ \begin{array}{|c|c|c|} \hline \text{Gametes} & \text{H} & \text{S} \\ \hline \text{H} & \text{HH} & \text{HS} \\ \hline \text{S} & \text{HS} & \text{SS} \\ \hline \end{array} $$ Step 3: Determine which genotype results in sickle cell anaemia. Sickle cell anaemia occurs in individuals who are homozygous for the abnormal haemoglobin gene, which is the SS genotype. Step 4: Calculate the proportion of offspring with sickle cell anaemia. From the Punnett square, there are 4 possible outcomes for the offspring's genotypes: HH, HS, HS, SS. One out of these four outcomes is SS. Therefore, the proportion of offspring with sickle cell anaemia is $\frac{1}{4}$. The proportion of their offspring that have sickle cell anaemia is $\boxed{\frac{1}{4}}$. b) Explanation for prevalence of sickle cell trait in tropical countries: Sickle cell trait (HS genotype) is more prevalent in tropical countries because it offers a survival advantage against malaria. These regions often have a high incidence of malaria, a parasitic disease. Individuals with the sickle cell trait are resistant to malaria because the presence of some abnormal haemoglobin makes their red blood cells less hospitable for the malaria parasite. This resistance increases their chances of survival and reproduction in malaria-prone areas, leading to a higher frequency of the sickle cell allele in the population through natural selection. What's next?