Feedback

The man is colorblind X^dY, and women no carrier X^DX^D.

Crossing:            X^dY             x           X^DX^D

Gametes:        X^d      Y                      X^D

F1:                    X^dX^D                     X^DY

Phenotype: All daughters will be carriers (with normal vision) and all sons will have normal vision.

Feedback

The color-blind man will be X^dY. The woman, being the daughter of a color-blind person who transmitted X^d to him, will be X^DX^d .

Crossing:                  X^dY            x            X^DX^d

Gametes:              X^d        Y                   X^D             X^d

Genotypes:            X^DX^d             X^dX^d            X^DY            X^dY

        Carrier woman    Colorblind woman   Healthy man      Colorblind man

Feedback

X^dX^D          x             X^dY

             X^dX^d       x         X^DY

X^dY The man is color blind.

Feedback

X^dX^D       x      X^DY

         X^dX       x       X^dY (color blind)

                    X^DY       X^dX^d

                 Normal    Colorblind

Feedback

a) A normal son (XY) receives the X chromosome from the mother, which depends on whether or not he is color blind. The father provides the Y chromosome and the mother will provide the X chromosome. In the case in question, the father would be X^dY (color blind) and the mother X^dX (carrier) or XX (not color blind)

The mother could not be colorblind X^dX^d because his son would transmit X^d, so that in all cases it would be colorblind. A colorblind woman cannot have children with normal vision.

b) A normal father (XY) and a normal mother (XX) or carrier (X^dX) could have the following offspring:

If both are normal: XY   x     XX All descendants are normal.

If the father is normal (XY) and the mother is a carrier (XdX):

Gametes          X^D     Y           X^d         X^D

F1                X^dX^D         X^DX^D        X^dY           X^DY

All the daughters would have normal vision (50% carriers).

50% of male children would have normal vision, and 50% would be color blind.

Feedback

The woman is normal vision (X^DX^D or X^dX^D), but as his father was colorblind (X^dY), it has to be X^dX^D.

The man is not color blind (X^DY). It does not influence anything that the grandfather also outside, and conveyed the Y chromosome, not the X^D.

Parents:                X^dX^D          x               X^DY

Gametes:            X^d       X^D                     X^D      Y

Descendants:     X^dX^D       X^dY         X^DX^D         X^DY

Phenotype: all daughters have normal vision (half being carriers). Half of the children are normal and the other half are color blind.

Feedback

For a woman to be color blind (X^dX^d) her father has to be color blind (X^dY) and her mother a carrier (X^DX^d), even if she has normal vision.

P                          X^dX^d                 x                     X^DY 

Gametes               X^d                                     X^D                Y                          

F1                        X^DX^d                                           X^dY

All daughters will have normal vision, although they will be carriers.

All children will be color blind.

Feedback

P                      X^DX^d               x              X^DY

Gametes        X^D         X^d               X^D               Y

F1                   X^DX^D          X^DX^d          X^DY            X^dY

25% of the offspring will be color blind (X^dY), 50% of the males born to the couple.

None of his daughters will be color blind, they will all be healthy, although 50% of them will be carriers (X^DX^d).

25% of the descendants will be carriers (X^DX^d). The rest, either healthy or color blind.

Feedback

a)

P                       X^DY             x         X^dX^d

F1                                      X^dY

A man with normal vision cannot be the child of a color blind, since color blindness is an X-linked recessive disease. If the mother is color blind, she will have both recessive alleles in her genotype (Xd Xd) and one of them would pass to son, so he would be color blind.

P                      X^dY          x           X^DX^D

F1                                   X^DY

A normal man X^DY can have a father with color blindness X^dY, if his mother is normal X^DX^D or a carrier X^DX^d.

b)  

P          X^DY          x         X^DX^d

F1                        X^dY

For a normal mother to have a color-blind child, she has to be a carrier of the X^DX^d gene, since the Y chromosome comes from the father and she is the one who provides the X^d chromosome.

Feedback

As stated in the problem statement, color blindness is a recessive character linked to sex. We know that it is linked to the X chromosome.

In this case, it is a couple, in which he is normal (X^DY) and she is heterozygous (X^DX^d), they have a color-blind male child (X^dY).

P                      X^DY        x            X^DX^d

Gametes      X^D      Y         X^D      X^d

F1        X^DX^D      X^DX^d      X^DY      X^dY

a) What is the probability that the couple has a color-blind daughter?

The probability of having a colorblind daughter will be 0%, since no girl is X^dX^d.

b) (1 point) What is the probability that the couple will have a child with normal vision? Argue your answer.

50% of male children will have normal vision (X^DY).

c) (1 point) Which of the woman's (male) grandparents could not be a source of transmission of the gene for color blindness? Argue your answer.

The female is X^DX^d heterozygous.

His father contributed X^D or X^d and his mother the other X^D or X^d.

The maternal grandfather gave the woman's mother an X chromosome and the maternal grandmother another X, X^D or X^d chromosome.

The paternal grandfather contributed the Y chromosome to the woman's father, the maternal grandmother the X chromosome.

Therefore, the paternal grandfather cannot be a source of transmission of color blindness.