Q.In a family, a woman is found to be colour blind. What must be true of the genotypes of both her mother and her father for this to be possible? Explain your reasoning using the X-linked recessive inheritance pattern of colour blindness.

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 …

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. …

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).

…

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 col …