Question

A pedigree affected by a rare human autosomal recessive trait is shown below. ODO What is the probability that III-2 is a carrier? (enter your answer as a percentage; if the answer is 83% enter 83) Answer:

Answer 1

Autosomal recessive disorder is a type of disorder that occurs due to a mutation in the allele of a gene present on the autosomes. For the disease to occur, two mutated alleles (in recessive condition) should appear together in a person. A single recessive allele will not cause the disease. Therefore, a homozygous recessive genotype is required for the disease to appear. A homozygous dominant genotype will not cause the disease. Heterozygous genotype with a dominant and a recessive allele will act as a carrier, as parents with heterozygous genotype (for example: Aa) will produce a child with autosomal recessive condition (aa), two carriers (Aa), and a normal child (AA).

Pedigree analysis is done to study a genetic disorder in a family for few generations. This helps in determining the type of the disorder by the appearance of the disease in the generations.

In the given question, the disease shown dies not appear in every generation, hence is a recessive disorder. The parents appear normal but the child is shown to be affected, which shows that the parents are carrier, with heterozygous genotype that produce on of the child with autosomal recessive condition.

Individual II-1(white circle representing a normal phenotype female) and individual II-2 ( white square representing a normal phenotype male) are married and produce two children. Out of these, the girl child is affected with the disease (III-1, dark circle), while the son appears, normal.

As told earlier, a child with autosomal recessive condition will be produced when both the parents have heterozygous genotype and are carriers of the condition.

Parents: Aa (Mother) X Aa (Father)

Children:  

AA (normal) Aa (carrier) Aa (carrier) aa (diseased)

Hence, out of the four children produced, two will be carrier of the disease (2/4=0.5), hence have heterozygous genotype and normal phenotype. Same us the case with individual III-2, whose sister has the disease, but himself appears normal. Therefore the probability of the individual III-2 to be the carrier of the disease will be 2/4=0.5.

0.5*100= 50%.

The answer is 50%.