Question

Color blindness is X-linked recessive and blood type is autosomal. If two parents who are both Type A and have normal vision produce a son who is color blind and type O, what is the probability that their next child will be a female who has normal vision and is type O? Please explain step by step and simplified.

Answer 1

there are three alleles at the ABO lcous which determines the blood type of the individual.

IA, IB and i.

IAIA, IAi- have A type blood,

IBIB, IBi- have B type blood.

IAIB- AB type blood

ii- O type blood.

both parents have A type blood and the child has O type blood so the genotype of both parents are IAi, so that the child inherits i allele from both parents.

colourblindness is a sex-linked recessive trait when the mother is a heterozygous carrier it is possible to have a son with colourblindness because the sons inherit the X chromosome from their mother.

let the alleles be XC- normal and Xc-colourblind

both parents have normal vision but they have a child with colourblindness, means the mother is a carrier. genotype of mother is XCXc and genotype of father is XCY.

IAi * IAi

	IA	i
IA	IAIA ( A type blood)	IAi ( A type blood)
i	IAi ( A type blood)	ii ( O type blood)

probability of getting child with O type blood= number of children with O type blood/total number=1/4

XCY * XCXc

	XC	Xc
XC	XCXC (normal female)	XCXc ( carrier female, normal vision)
Y	XCY ( male, normal vision)	XcY ( male with colourblindness)

probability of getting female with normal vision= number of the female with normal vision/total number=2/4=1/2

having is a child is an independent event so it does not depend on the genotype or phenotype of the first child.

so probabiltiy that the child is a female with normal vision and O type blood= probabiltiy of getting child with O type blood * probabiltiy of getting female with normal vision= 1/4*1/2=1/8

so the probabiltiy that the child is a female with normal vision and O type blood = 1/8