Question

Study the given pedigree chart for sickle-cell anaemia and select the most appropriate option for the genotypes.

• A. $H{b}^{A}H{b}^A,H{b}^{A}H{b}^S$
• B. $H{b}^{A}H{b}^A,H{b}^{A}H{b}^A$
• C. $H{b}^{A}H{b}^A,H{b}^{S}H{b}^S$
• D. $H{b}^{A}H{b}^S,H{b}^{S}H{b}^S$

Answer 1

Answer: (D)

Sickle cell anaemia is an autosomal recessive disorder in which if they want to express itself the gene should be present in homozygous condition. If we see the gene responsible for sickle cell anaemia is $H{b}^S$ . Those individuals in which this gene is present in homozygous condition only that individual will be affected. $H{b}^S$ is recessive so if in any individual $H{b}^{A}H{b}^S$ is present then it will be unaffected/carrier. If $H{b}^{A}H{b}^A$ is found then it will be normal. If we see the given pedigree chart in the question then in parents nobody is affected but there some progeny is affected which means male and female present in heterozygous condition.
$ParentH{b}^{A}H{b}^S\times H{b}^{A}H{b}^S$

	$H{b}^A$	$H{b}^A$
$H{b}^S$	$H{b}^{A}H{b}^S$ (Carrier)	$H{b}^{A}H{b}^S$ (Carrier)
$H{b}^A$	$H{b}^{A}H{b}^A$ (Normal)	$H{b}^{A}H{b}^A$ (Normal)

The carrier one is unaffected but they pass on this disease to the next generation.
So, the genotype of the parents is $H{b}^{A}H{b}^S$ , the genotype of the first child is $H{b}^{A}H{b}^S$ , the genotype of the third child will be $H{b}^{S}H{b}^S$ , because it is affected.