In humans, cystic fibrosis is caused by a recessive allele, f. A man and a woman are both heterozygous for the cystic fibrosis allele. What is the probability that they will produce a girl who has cystic fibrosis?

The genotype for someone heterozygous for the cystic fibrosis allele would be Ff. Thus the father’s genotype (including sex chromosomes) would be FfXY and the mother’s would be FfXX. In order to work out the probability of them producing a girl with cystic fibrosis, we must do a cross between FfXY and FfXX. The potential paternal gamete combinations are FX, fX, FY, and fY. The potential maternal gamete combinations are FX and fX.The cross yields these 16 combinations: FFXX, FfXX, FFXY, FFXY, FfXX, ffXX, FfXY, ffXY, FFXX, FfXX, FFXY, FfXY, FfXX, ffXX, FfXY, ffXY. Only 2 out of those 16 are ffXX, a female with cystic fibrosis. 2/16 * 100 = 12.5%. Thus there is a 12.5% chance of their child being a girl with cystic fibrosis.