Myotonic Dystrophy Steinert disease









The nucleus of human cells contains 23 pairs of chromosomes, each pair composed of one chromosome from the mother and one chromosome from the father. Among these 23 pairs of chromosomes, only one pair is involved in determining the sex. The female sex is determined by the presence of two X chromosomes (the female pair of sex chromosomes is formed by 2 X chromosomes) and the male sex is determined by the Y chromosome (the male pair of sex chromosomes is formed by an X chromosome and a Y chromosome). Genes occupy the same positions on the two chromosomes from each pair and are called alleles. It is the sum of the genes that determines the hereditary characteristics of an individual: skin color, eye color, hair color, etc.

One allele can have more influence than the other. Thus, the allele for brown eyes usually wins over the allele for blue eyes. The winning allele is said to be dominant (allele for brown eyes in this example) and the other is said to be recessive (the allele for blue eyes). In order for the recessive trait (blue eyes) to occur, the recessive allele must be present on both chromosomes. When a gene is located on a non sexual chromosome, whether the disease is dominant or recessive, boys and girls will be affected at the same rate. For a dominant disease, the probability of having a child that is a carrier of the anomaly is 50% or one out of two.

Myotonic dystrophy transmission

Type 1 and 2 Myotonic Dystrophy are dominant forms of the disease that do not involve the sexual chromosome pairs. The anomaly is located on chromosome 19 for Type 1 Myotonic Dystrophy and on chromosome 3 for Type 2 Myotonic Dystrophy. This means that both girls and boys will be affected with the same frequency and that the risk of transmission of the disease will be one out of two for each child regardless of the sex.


Figure 1: Gene transmission. Each of our genes exists in duplicate (the scientific term for this is allele). One comes from our mother and the other is from our father. In the example shown, the Myotonic Dystrophy gene (black) is from the father who also has a normal gene (grey). The father is affected by the disease. His spouse has 2 normal genes (light and dark blue). The father can transmit the defective (black) gene or the normal (grey) gene in a 50-50 probability. If the father transmits the defective (black) gene, the child will have a 50-50 chance of receiving either the normal light blue or the normal dark blue gene from the mother. Since the defective (black) gene is dominant, the child will be affected by the disease. The other possibility is that the father transmits his normal (grey) gene. The child will receive one of the mother’s normal genes (light or dark blue). In all cases of a child without the defective (black) gene, the disease will not be present. Also, because the child does not carry the defective (black) gene, he/she will not be able to transmit the disease to his/her children.