The results of a monohybrid cross can be given as a probability, a ratio or a percentage. For two heterozygous parents, Tt x Tt, the Punnett square has four equal boxes: TT, Tt, Tt and tt. The phenotype ratio is 3 dominant : 1 recessive, written 3:1. The probability of tt is 1 in 4, which is 0.25 or 25%.
A cross between a heterozygous parent and a homozygous recessive parent, Tt x tt, gives Tt, Tt, tt and tt. The ratio is 1:1, so the probability of a tt offspring is 50%.
Cystic fibrosis is caused by a recessive allele, f. Two carriers, Ff x Ff, have a 25% chance of a child with cystic fibrosis (ff) and a 50% chance of a child who is a carrier (Ff). The probability is the same for every child, because each fertilisation is a separate event. These are expected results, so real families, or small samples, may not match exactly. If 80 offspring are produced from Tt x Tt, the expected number with the dominant phenotype is 3/4 of 80, which is 60.
Pedigrees can be analysed in the same way. If two unaffected parents have an affected child, the allele is recessive. If two affected parents have an unaffected child, the allele is dominant.