A woman whose mother is colourblind and father is haemophilic marries a man whose mother is haemophilic and father is colourblind. Which of the following is correct about its progeny?

Question

A woman whose mother is colourblind and father is haemophilic marries a man whose mother is haemophilic and father is colourblind. Which of the following is correct about its progeny?, They won't have any child who is free from both these diseases., Half of their children are haemophilic while the other half are colourblind., Half of their children will be colourblind while one fourth will be hameophilic, Half of their children will be haemophilic while one fourth will be colourblind

MARKS App · Accepted Answer

X h = X chromosome carrying colourblindness gene. X h = X chromosome carrying the haemophilia gene. Mother of the woman is colourblind, hence her genotype will be X c X c . Father of the woman is haemophilic, hence his genotype will be X h Y. The woman receives one X chromosome from her father (it has to be X h ) and the other X chromosome from her mother (it has to be X c ). Thus the genotype of the woman is X c X h (She is a carrier for colourblindness as well as haemophilia) Mother of the man is haemophilic, hence her genotype will be X h X h . Father of the man is colourblind, hence his genotype will be X c Y. The man receives Y chromosome from his father and X chromosome from his mother (it has to be X h ). Thus the genotype of the man is X h Y. The genotypes of their children can be represented as follows X c X h X h X c X h daughter: carrier for both diseases X h X h daughter: haemophilic Y X c Y son: colourblind X h Y son: haemophilic

	X^c	X^h
X^h	X^cX^h daughter: carrier for both diseases	X^hX^h daughter: haemophilic
Y	X^cY son: colourblind	X^hY son: haemophilic

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