Question

Paragraph - Individual 5 The 2 alleles at the ALK locus are denoted ALK+ and ALK. The 3 alleles at the ABO blood group locus are A, B (co-dominant) and 0 (the 0 allele is recessive to A and B). Example: Individual 5 has a genotype of ALK-ALK- for alkaptonuria and 00 for blood type Genotype Individual 5 ALK-ALK-00 (both mutant ALK alleles and Type O blood) Questions 1. What is the genotype of individual 1 at the ALK and ABO loci? 2. What is the genotype of individual 3 at the ALK and ABO loci? 3. Individuals 3 and 4 are expecting their fifth child. A physician draws a prenatal blood sample. What is the probability that the child will have alkaptonuria AND type B blood? Show work.

Answer 1

1. Individual 1 is not affected by alkaptonuria. But two of its offsprings are affected. It means that one recessive allele is passed on from each parent. Therefore genotype of Individual 1 is ALK+/ALK-

As individual 1 has four offsprings of which 3 have blood group A and one has blood group O. As one of the offspring has O blood type, his genotype is OO. It means he recieved one recessive allele from each parent. Therefore as the blood group of individual 1 is A, the genotype is AO.

So the genotype of individual 1 is ALK+/ALK- and AO.

2. From the data provided, it can be inferred that individual 5 is affected as it has genotype ALK-/ALK- for alkaptonuria.

As individual 5 is affected, it has received one recessive allele ALK- from each of its parents 3 and 4. From the figure above, individual 4 is also affected. Therefore genotype of individual 4 is ALK-/ALK-

Also from the figure, we observe that the four offsprings of 3 and 4 have the blood groups A,B,AB and O. So the ABO genotype of 3 and 4 can be deduced. As O is recessive allele to A and B, and O is a blood group of one of the offsprings of 3 and 4, the recessive allele O is masked in the genotype of 3 and 4. So the genotype of 3 is BO and that of 4 is AO. This can be demonstrated as follows:

Genotype of 3 and 4 AO x BO

Gametes A O B O

Genotype of offsprings AB AO BO OO

Phenotype of offsprings AB A B O

Therefore individual 4 has genotype is ALK-/ALK- and AO.

As discussed above, the genotype of 3 is BO.

As individual 5 is affected, it has genotype ALK-/ALK-. This means that it received one recessive allele each from each of its parents 3 and 4. As individual 3 is not affected by alkaptonuria, the genotype of individual 3 is ALK+/ALK-

Therefore individual 3 has genotype is ALK+/ALK- and BO.

3. As described above, the offsprings of 3 and 4 can have the blood groups A,B,AB and O, as genotype of 3 is BO and genotype of 4 is AO.

Therefore the probability that the offspring of 3 and 4 will have a type B blood group is 1/4.

As individual 4 has genotype is ALK-/ALK- and individual 3 has genotype is ALK+/ALK-, the genotype of offspring can be either ALK+/ALK- or ALK-/ALK-

Genotype of 3 and 4   ALK+/ALK- x   ALK-/ALK-

Gametes ALK+ ALK-   ALK- ALK-

Genotype of offsprings ALK+/ALK- ALK+/ALK- ALK-/ALK- ALK-/ALK-

Therefore the probability that the offspring of 3 and 4 will have ALK-/ALK- is 1/2

Therefore the probability that the fifth child will have alkaptonuria and B type blood group is (1/2)*(1/4) = 1/8