Individuals 3 and 4 are expecting their fifth child. A physician draws a prenata
ID: 255993 • Letter: I
Question
Individuals 3 and 4 are expecting their fifth child. A
physician draws a prenatal blood sample and determines
that the child has blood type B. What is the probability that
the child will have alkaptonuria? Explain your answer.
Please incorporate and explain 11% recombinant frequency .
Alkaptonuria is an extremely rare disease. The gene for Alkaptonuria (ALK) has been shown to lie on human chromosome 9 and is linked to the gene encoding the ABO blood group, with a recombination frequency of 11% between the loci. A pedigree of a family with the disease is shown below, with affected individuals indicated in black. In addition, the blood type of family members is given. AB O 3 ABExplanation / Answer
According to the information, the individuals 3 and 4 have childrent of blood types A, B, AB and O. This gives that both the parents are heterozygous for their blood types i.e. have genotypes AO and BO. Thus, there are 25% chances to have blood types A, B, AB or O. On the other hand, the individual 4 is alkaptonuric in nature. This genetic disorder is inherited in autosomal recessive manner. This suggests that the individual must carry two copies for being affected by the disease.
Now carefully consider that two offsprings of this couple are affected from the disease suggesting that the parent 4 must be homozygous whereas parent 3 must be heterozygous for the condition. Thus, there are 75% chances everytime that this couple would bear a child with alkaptonuria. Considering the fact that the recombination frequency is 11%, the chances of having the disease would be given by:
Probability of having the disease = 75 % / 11% = 6.81 or nearly 68%.
This gives that although free recombination would give 75% chances of a child with alkaptonuria, 11% recombination rate would reduce this probability to 68% in this couple.