Effect of Test Sensitivity on the Risk of CF Childbirth.

Genetic Screening for Cystic Fibrosis

(Table from Wilfond & Fost 1990)

		Risk of Cystic Fibrosis in Offspring
		One Tested	Both Partners Tested
% of CF Mutations Detectable	Carrier Risk for Person with Test Negative	Negative	Both Negative	One Positive, One Negative
0	1 in 25.5	1 in 2500	1 in 2500	1 in 2500
55	1 in 55	1 in 5660	1 in 12300	1 in 222
75	1 in 99	1 in 10100	1 in 39200	1 in 396
95	1 in 491	1 in 50100	1 in 964000	1 in 1960
96	1 in 613	1 in 62600	1 in 1500000	1 in 2450

The table shows risks associated with a variety of screening situations. It substantiates the claim made in the case that increasing test sensitivity reduces ambiguity about the risks to child-bearing couples, but does not eliminate it. Some ambiguity is inevitable. The probabilities are calculated in a straightforward manner. The 1 in 25.5 risk with no test is an empirical observation on our population.

A person with a negative test can still be a carrier if his or her allele is one of those not detected. For example, with "75%" and "95%" tests, 1 in 4, or 1 in 20, carriers will not be detected. After such a negative test, the chances of being a carrier are:

75%: 1/4 x 1/25 = 1/100, about the same as in the table

95%: 1/20 x 1/25 = 1/500

If neither parent is screened (0% test) then the chances of a CF child are those calculated in problem 5d. This is the product of each parents probability of being a carrier times 1/4, the chances of two carriers having CF child. For the 75% test:

one negative: 1/100 x 1/25 x 1/4 = 1/10000, (10100 in table)

two negative tests: 1/100 x 1/100 x 1/4 = 1/40000, (39200)

one negative, one positive: 1/1 x 1/100 x 1/4 = 1/400 (396)

Calculations here assume no new mutation, table values are thus slightly higher.