When a disease is present in a family, knowledge of defective gene transmission patterns can be used to compute the probability of prospective parents having a child with the disorder. Genetic counselor is the one who calculates genetic probability. Calculation of such genetic probability is important for prospective parents because they will be able to plan their family and the child with healthy mind and body. In this article I will be discussing a simple example of a pedigree and how to calculate the probability of prospective parents having a child with the disease.
Pedigree Analysis Examples
There is a couple named Jack and Martina. They’re planning for a baby. When they are discussing about their child, Jack said that his uncle died of Tay-Sachs disease. Suddenly, Martina also says that her uncle also died of Tay-Sachs disease. After sharing this information, both got nervous because Tay-Sachs disease is present in both the families. Without wasting any time, they visit a genetic counselor named Hudson. After hearing both Jack and Martina, the genetic counselor says, he needs to draw pedigree of both the families first to calculate the probability of the child with the disorder. Hudson forms Jack’s family pedigree and Martina’s family pedigree separately.
If you look at Jack’s pedigree, Jack’s parent is normal, but Jack’s uncle has Tay-Sachs disease and he died of the same. Jack’s grandparent is also normal. On the other hand, if you see Martina’s pedigree, her parent is normal, but her uncle died of Tay-Sachs disease. Martina’s grandparent is normal as well. Now, Jack and Martina marry and as a prospective parent, they plan for a baby. The probability of having Tay-Sachs disease needs to be calculated.
It is important to mention about Tay-Sachs disease before I calculate the genetic probability. Tay Sachs is an autosomal recessive disorder caused by a genetic mutation in the HEXA gene present on chromosome 15 responsible for coding an enzyme Hexosaminidase A. Mutation in the gene disrupts the activity of the enzyme and thus results in the accumulation of the molecule GM2 ganglioside within cells, leading to toxicity. This causes destruction of the nerve cells in the brain and spinal cord. The disorder becomes apparent when a child reaches three to six months of age. The baby loses the ability to sit or crawl. The condition becomes severe with time and eventually leads to death of the child.
The gene has two alleles T and t. T is dominant over t. A child with normal condition will have the genotype as either T/T or T/t. And a child with the disorder will have a genotype of t/t (homozygous recessive condition).
Calculation of the Pedigree Analysis Probability of the Child with the Disorder
1. The grandparent’s genotype must have been T/t because they have a son with Tay-Sachs disorder (t/t) – Jack’s uncle. Jack’s father’s genotype could either be T/T or T/t, but we know from the Punnet square cross that the relative probability of having T/t is more than T/T. Now, by crossing Jack’s grandparent’s genotype, you can get Jack’s father’s genotype probability of T/t.
From the above calculation, we can conclude that the probability of having T/t for Jack’s father is 2 /3.
2. Jack’s mother must be T/T since she was introduced into the family by marrying Jack’s father and disease alleles generally are rare. So, we can conclude that Jack’s father’s genotype is T/t and Jack’s mother’s genotype is T/T. If we draw a cross between Jack’s parents genotype, the expected progeny proportion would be 1 / 2 (T/T) and 1 / 2 (T/t) as shown below:
3. The overall probability of Jack being heterozygote (T/t) can be calculated using a product rule. According to the product rule, the joint probability of two independent events is the product of their individual probabilities – this is the probability of one event and another event occurring. For instance, the probability of rolling a 5 with a single throw of a single six sided die is 1/6 and the probability of rolling 5 in each of three successive rolls is 1/6 x 1/6 x 1/6 = 1/216. Now, in our discussion, the probability of Jack’s being heterozygous is the product of the probability of his father being a heterozygote and the probability of the father having a heterozygote son, which is 2/3 x ½ = 1/3.
4. The probability of Martina being heterozygous is also 1/3.
5. The prospective parents (Jack and Martina) of the child are both heterozygous (T/t). So, you can calculate the probability of the child being (t/t) as ¼. So, the overall probability of the couple having an affected child would be 1/3 x 1/3 x ¼ = 1 / 36; in other words, a 1 in 36 chance.
Conclusion:
From the above probability calculation, it is clear that the probability of having Tay-Sachs disease to the child is 1/36 meaning, 1 in 36 child. In other words, we can say that there is a very low chance of child to carry the disease. So, Jack and Martina can definitely plan for a healthy baby. So, from the discussion, probability calculation gives a value that we can rely on and plan accordingly.