1. Evidence supporting/refuting a single-recessive-gene interpretation:
strong familial patterns of occurence suggests genetic base, 3:1 ratios among offspring
suggests monohybrid cross (if the "1" is CF it is recessive, if the
"3" is CF it is dominant.)
non-familial occurrence and ratios other than above suggest another mechanism
2. Childless couple, both diagnosed carriers.
chances first child normal: 1 - chance of CF child
1 - 0.52 = 1.00 - 0.25 = 0.75
chances of two CF children:
chance of one = 1/4, chances of two = (1/4)2 = 1/16
first child is CF, chances the second will have CF:
they already have the first one, prob = 1.00
so the chances of having another one are still ¼
3. Screening test probabilities
test for most common mutation, detect 71% of CF cases:
chance of detecting CF is 0.7 for each carrier parent
thus chance of detecting CF in both carrier parents is
0.72 = 0.49 or 49%, less than half.
test for 4 mutations, detect 85% of CF cases:
detection rate = 0.852 = 0.722, or 72.2%
In general, increasing test sensitivity reduces uncertainty. (See page 3.)
this would not be likely to turn up new CF carriers. Patients test negative because they
don't carry the specific mutations tested for, not because of errors in testing. Test is accurate,
but not sensitive.
4. Is a CF child possible
Both test negative for four common mutations, but can have one op the rarer, unscreened
mutations.
Husband has two normal alleles, wife is a carrier: 1) Husband has a CF mutation
somewhere else on chromosome 7, or elsewhere in the genome. A few such cases have been
found so far. 2) Husband had a CF mutation occur in the cell line leading to the
fertilizing sperm cell. 3) someone else may be the father (oops!).
5. Population genetics
You would expect about 4%, or 1/25, to be carriers. 0.04 x 2,000 = 80 carriers with
all-european ancestry; less than that in our more diverse population.
CF carriers in the US: literature suggests about 8 million, given ethnic mix
approximately: (4 x 10-2) x (2.5 x 108) = 1 x 107 = 10
million
chance randomly chosen pair both carriers = 1/252 = 1/625