A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{16}\]
D) \[\frac{1}{64}\]
Correct Answer: C
Solution :
| It is already mentioned that the couple has a child with blood group O and the child is diseased. Thus, the parents are heterozygotes for the diseased allele (as parents are normal and the disease is recessively inherited) and for the blood group. The probability of O blood group in progeny is |
| \[\begin{matrix} {{\text{I}}^{\text{A}}}\text{i }\!\!\times\!\!\text{ }{{\text{I}}^{\text{B}}}\text{i} \\ \downarrow \\ {{\text{I}}^{\text{A}}}{{\text{I}}^{\text{B}}}{{\text{I}}^{\text{A}}}\text{i}{{\text{I}}^{\text{B}}}\text{i}\,\text{ii} \\ \end{matrix}\] |
| i.e.\[\frac{1}{4}\] individual is blood group O. |
| Similarly, for the disease to be inherited, the individuals should have both the recessive alleles, thus \[Dd\times Dd=dd\](probability to get the disease is\[\frac{1}{4}).\]Therefore, probability that the second child has the disease = P (the child is blood group O) \[\times \,P\](the child has both the recessive alleles). Thus, the P (diseased child) \[=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}.\] |
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