Final answer:
The probability that the first child of IV-1 and IV-2, who are both heterozygous carriers of an autosomal recessive trait, will have the trait is 1/4 or 25%. This follows the Mendelian genetics pattern of a 3:1 ratio for two carrier parents, meaning the correct answer is option (a).
Step-by-step explanation:
If IV-1 and IV-2 in a pedigree that shows an autosomal recessive trait wish to have a child, we must assume both are heterozygous carriers if they are phenotypically normal but have an affected parent. This implies that each of them has one normal allele and one recessive allele for the trait in question. According to traditional Mendelian genetics for an autosomal recessive trait, when two carriers have a child, the child can either be affected, a carrier, or completely unaffected. This is visually represented through a Punnett square, showing the offspring's genotypes based on the parental alleles.
The possible genotypes for their offspring are: normal (homozygous dominant), two carrier genotypes (heterozygous), and affected (homozygous recessive). The classical Mendelian ratio for such a cross is 3:1, where for every four offspring, on average, three would be expected to be unaffected (either normal or carriers), and one would be affected by the autosomal recessive disorder. Therefore, the probability that their first child will have the trait is 1 out of 4, or 25%. This follows the inheritance pattern of an autosomal recessive disorder with two carrier parents. So the correct answer for the probability is 1/4, option (a).