Draw a genetic diagram to show how unaffected parents could have a son with Haemophilia but also other children who do not suffer from the disease.

First figure out the phenotype (what is visible/expressed) of parents: Normal male x Normal female Next write down their gametes: X^H y and X^H X^h The disease is sex linked (X linked) so can only be carried on the X chromosome. The male only has one X and is phenotypically normal so he must carry the normal (H) allele. the female has 2 X chromosomes and is also phenotypically normal so must carry one normal allele (h) on one X chromosome and one disease allele (H) on the other X chromosome.Next draw a punett square and put in the games of the parents: Female X^H X^h Male X^H X^H X^H X^H X^h Y X^H Y X^h YNext transfer results of punett square into genotypes and phenotypesoffspring genotypes: X^HX^H ,X^HX^h ,X^H Y ,X^hYoffspring phenotypes: Normal female, Normal (carrier)female, Normal male, Affected male