MKSAP Quiz: Intermittent rectal bleeding of 3 months' duration

A 46-year-old man is evaluated for intermittent rectal bleeding of 3 months' duration. Colonoscopy finds 12 polyps, and genetic testing reveals biallelic mutations in the MYH gene. What is the most cost-effective approach to determining if the patient's children have inherited MYH-associated polyposis?


A 46-year-old man is evaluated for intermittent rectal bleeding of 3 months' duration. He is otherwise well and takes no medications. His father had a few polyps removed from the colon when he was 71 years old, but no other details are known about his father's medical history. The patient and his wife have three children aged 8, 12, and 18 years.

On physical examination, vital signs and the remainder of the examination are normal.

Colonoscopy reveals 12 polyps ranging in size from 2 to 7 mm, all of which are removed from the colon. He undergoes genetic testing, which reveals biallelic mutations in the MYH gene and confirms a diagnosis of MYH-associated polyposis (MAP).

Genetic testing of which of the following persons will provide the most cost-effective approach to determining if the patient's children have inherited MAP?

A. The patient's father
B. The patient's mother
C. The patient's 18-year-old child
D. The patient's wife


MKSAP Answer and Critique

The correct answer is D. The patient's wife. This item is available to MKSAP 17 subscribers as item 63 in the Gastroenterology & Hepatology section. More information on MKSAP 17 is available online.

The most important person to receive genetic testing is the patient's wife. MYH-associated polyposis (MAP) is the only known autosomal recessive hereditary colorectal cancer syndrome. MAP should be considered as a cause for multiple colorectal adenomas (>10) in patients with an apparent autosomal recessive inheritance of colorectal polyposis or cancer. MAP is caused by biallelic mutations in the base excision repair gene MYH. Because MYH-associated polyposis is an autosomal recessive disorder, the affected patient must have two abnormal copies of the gene (that is, he is homozygous for the abnormal gene), having received an abnormal copy of the gene from each parent. The status of the patient's wife for the abnormal gene will provide information to help assess risk and guide the need for genetic testing of the couple's children. If the patient's wife is negative for the MYH mutation, all of the children will be heterozygous for the mutation (carriers), having received a normal copy of the gene from their mother and an abnormal copy of the gene from their father. In this case, the children would not be affected, but their carrier status would be known and no further genetic testing would be indicated. If the mother is a carrier of the mutation, 50% of their children would be affected with the syndrome and 50% would be carriers. Therefore, testing of each child would be indicated to identify whether the child is homozygous for the mutation and is affected, or whether the child is a heterozygous carrier of the mutation.

Performing genetic testing on the patient's father or mother would not provide any useful information at this time. The patient has already been diagnosed with the two mutations; therefore, both his mother and father must be positive for the mutation.

If the patient's wife was unavailable for genetic testing, it would be appropriate to perform genetic testing on the children. Genetic testing is usually done at the age at which high-risk colorectal cancer screening would begin, which is 18 years in MAP and attenuated familial adenomatous polyposis and 12 to 15 years in classic FAP.

Key Point

  • Because MYH-associated polyposis is an autosomal recessive disorder, both parents must be carriers of an MYH mutation in order for a child to inherit the syndrome.