Q.Explain the inheritance of haemophilia in humans, which is caused by an X-linked recessive allele. If a carrier (heterozygous) woman marries a normal (non-haemophilic) man, use a Punnett square to work out what proportion of their sons is expected to be haemophilic, and what proportion of their daughters is expected to be carriers.

The Y-chromosome-linked (holandric) genes are inherited directly from father to son, since only males possess the Y chromosome and it is passed from father to all his sons but never to daughters.

Holandric genes, located on the Y chromosome, pass exclusively from father to son.

In humans, males are heterogametic (XY) and females are homogametic (XX). Genes located on the non-homologous (male-specific) region of the Y chromosome — called holandric genes — are transmitted only from father to son, generation after generation, and are never found in females, since daughters never receive a Y chromosome from their father. A classic example of a holandric trait is hypertrichosis (excessive hair growth on the pinna/ear), which shows this strict father-to-son pattern of inheritance, unlike autosomal or X-linked genes which can be inherited by both sexes.

Criss-cross inheritance is the pattern where an X-linked recessive trait passes from an affected father to his daughters (carriers) and then appears in his grandsons — seen in both haemophilia and colour blindness.

Criss-cross inheritance describes the transmission pattern of X-linked recessive genes: a trait in the father is transmitted to his grandsons through his (carrier, unaffected) daughters, rather than directly to his sons. This "criss-cross" pattern arises because:

• A son inherits his X chromosome only from his mother (his Y comes from his father), so he cannot inherit an X-linked trait directly from his father.
• A daughter inherits one X from her father (carrying the recessive allele) and one from her mother; if she is heterozygous she becomes a carrier and may pass the trait on to her own sons.

Both haemophilia (a blood-clotting disorder) and colour blindness (red-green colour vision defect) are caused by recessive alleles located on the X chromosome, so both show this criss-cross pattern. Thalassaemia, by contrast, is caused by autosomal recessive genes (on chromosome 11 or 16) and does not follow criss-cross inheritance.

Colour-blindness is an X-linked recessive trait; a colour-blind father crossed with a homozygous normal mother produces NO colour-blind children in the first generation.

Colour-blindness is caused by a recessive allele (Xc) located on the X chromosome. Let the colour-blind father be XcY and the normal (non-carrier) mother be XX (homozygous dominant, since she is stated to be a 'normal woman' with no indication of being a carrier).

Cross: XcY (father) x XX (mother)

Gametes: father gives Xc or Y; mother gives only X.

Offspring genotypes: XXc (daughters — carriers, phenotypically normal since the dominant X masks Xc) and XY (sons — normal, having received the mother's normal X and father's Y).

Since none of the offspring genotypes is XcXc (needs an Xc from BOTH parents) or XcY (needs an X-bearing egg carrying Xc, which the mother cannot supply), no child in this generation is colour-blind — though all daughters become carriers who may pass the trait to their own sons in the next generation.

Correct answer: (c) X^C X^c

Colour blindness is a recessive X-linked trait, so a heterozygous female (one normal allele, one colour-blind allele) is phenotypically normal — a carrier.

Colour blindness in humans is inherited as an X-linked recessive trait, with the normal-vision allele X^C dominant over the colour-blind allele X^c. A heterozygous female with genotype X^C X^c carries one copy of each allele; since X^C is dominant, she has normal vision phenotypically (though she is a carrier who can pass the X^c allele to her offspring). By contrast, X^c X^c (homozygous recessive female) would be colour-blind, and X^c Y (a male with the single recessive allele on his only X chromosome) would also be colour-blind, since males have no second X allele to mask it. X^c Y^c is not a valid genotype, as the Y chromosome does not carry this locus.