Colorblind man w/ blood type AB crosses w/ woman who has normal vision & blood type A. What are odds of ....?

Question

having a daughter that is colorblind and has blood type A?

a. 1 in 8

b. 1 in 16

c. 1 in 2

d. 1 in 4

Answer 1

the mom would have to be a carrier of the colorblind gene,as for blood type idk, and if the mom is a carrier, im pretty sure the chance of the daughter being colorblind is increased if shes a carrier from her mom or grandma

Answer 2

A little bit more info is needed to answer this question: is the mom a carrier of colorblindness gene? Because if she is not, then there is no chance of having a colorblind daughter (although all daughters can be carriers of the gene even if mom is not a carrier).

Lets assume that the mom is a carrier and the allele for colorblindness is "c". The allele for color blindness is on X chromosome (Xc) therefore a carrier mother is XcX and a colorblind father is XcY.

There are 2 ways that a person can have blood type A. they can either be AA or AO. Lets say that the mother is AA first.

Mom: XcXAA--------> Possible gametes: XcA and XA

Dad: XcYAB---------> Possible gametes: XcA, XcB, YA, YB

Kids: XcXcAA, XcXcAB, XcYAA, XcYAB, XXcAA, XXcAB, XYAA, XYAB

Therefore the kids will be:

1/8 colorblind daughter with A blood type (XcXcAA)

1/8 colorblind daughter with AB blood type (XcXcAB)

1/8 colorblind son with AB blood type (XcYAB)

1/8 carrier daughter with blood type A (XXcAA)

1/8 carrier daughter with blood type AB (XXcAB)

1/8 healthy son with blood type A (XYAA)

1/8 healthy son with blood type AB (XYAB)

Now lets assume that mom is a carrier with blood type A (genotype AO):

Mom: XcXAO---------> Possible gametes: XcA, XcO, XA, XO

Dad: XcYAB-----------> Possible gametes: XcA, XcB, YA, YB

Kids: XcXcAA, XcXcAB, XcYAA, XcYAB, XcXcAO, XcXcBO, XcYAO, XcYBO, XcXAA, XcXAB, XYAA, XYAB, XcXAO, XcXBO, XYAO, XYBO

Therefor the kids will be:

1/8 colorblind daughter with blood type (=BT) of A (XcXcAA and XcXcAO)

1/16 colorblind daughter with BT of  AB (XcXcAB)

1/8 colorblind son, BT of A (XcYAA and XcYAO)

1/16 colorblind son, BT of  AB (XcYAB)

1/16 color blind son, BT of B (XcYBO)

1/16 colorblind daughter, BT B (XcXcBO)

1/8 carrier daughter, BT of A (XcXAA and XcXAO)

1/16 carrier daughter, BT of AB (XcXAB)

1/8 healthy son, BT of A (XYAA and XYAO)

1/16 carrier daughte, BT B (XcXBO)

1/16 healthy son, BT of AB (XYAB)

1/16 healthy son, BT of B (XYBO)

Therefore, in both cases, chances of having a color blind daughter with blood type A is 1 in 8.

Answer 3

The man is AB and XcY (colorblind) and the woman is A- and XcX (since none of the choices are zero, the woman has to be a carrier or else the daughters would have no chance of being colorblind, just carriers) (the dash in blood type represents either an A or an O, as either will give type A and not affect what we are looking at)

Blood Type:

_ _ A _ _ -

A _ AA _ A-

B _ AB _ B-

50% chance of having blood type A

Colorblind:

_ _ _ Xc _ _ X

Xc_ XcXc _ XcX

Y _ _XcY _ _ XY

50% chance of having a daughter

50% chance that a daughter will be colorblind

To find the chance of having a daughter with blood type A who is colorblind you multiple the chances, so .5 x .5 x .5 = .125

So having a colorblind daughter with blood type A is a 1 in 8 chance, answer A