genetics

[Dencology]

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Color blindness (rg) and is a result of a recessive x-linked gene. If a father of a family is color blind and he happens to have a son that is color blind and a daughter that has a normal vision, what are the respective genes of the parents?
Father mother

a. rg/Y rg/rg
b. rg+/Y rg/rg+
c. rg/Y rg+/rg+
d. rg/Y rg+/rg
e. rg+/Y rg+/rg+

a can not be because all the kids would be colorblind. B can't be because the father is colorblind so we must have rg/Y. for D, then we have both guys and girls colorblind and it says that only boys are color blind. so how does this work?

 

[waystinthyme]

Color blindness (rg) and is a result of a recessive x-linked gene. If a father of a family is color blind and he happens to have a son that is color blind and a daughter that has a normal vision, what are the respective genes of the parents?
Father mother

a. rg/Y rg/rg
b. rg+/Y rg/rg+
c. rg/Y rg+/rg+
d. rg/Y rg+/rg
e. rg+/Y rg+/rg+

a can not be because all the kids would be colorblind. B can't be because the father is colorblind so we must have rg/Y. for D, then we have both guys and girls colorblind and it says that only boys are color blind. so how does this work?

d. rg/Y rg+/rg

immediately eliminate b & e, because the father is not colorblind.
for a: BOTH the son and daughter will have the trait 100% of the time
for c: NEITHER the son or daughter will have the trait 100% of the time

males: XY / females: XX

to have the trait...
males: must have the recessive allele on their x-gene
females: must have the recessive allele on BOTH x-genes...if the allele is only on one x-gene then they are only CARRIERS of the trait.

remember that males get their Y gene from the father and their X gene from their mother. this means that there is a 50% chance that the mother may pass the recessive or dominant allele if she is heterozygous. since the son is colorblind, the mother passed him the RECESSIVE allele, so he has the trait.

remember that females get one X gene from their father and the other X gene from their mother. again there is a 50% chance that a heterozygous mother will pass the RECESSIVE allele and a 50% chance that she will pass the DOMINANT allele to the child. in this case, the father passes the recessive allele and the mother passes the dominant allele, so the daughter is a CARRIER of the trait, but does not display characteristics of the trait.

note: if this is hard to see, you may want to write out punnent squares for the possible parent pairs, until you get the hang of how to do this sort of problem...

-waystinthyme

 

[Sea of ASH]

father is color-blind and mother is a carrier

 

[Sea of ASH]

immediately eliminate b & e, because the father is not colorblind.

in the question it stated that the father is color blind

 

[waystinthyme]

in the question it stated that the father is color blind

exactly, but in answer choices b & e the father is not colorblind, and therefore those are INCORRECT answers...i.e. you can eliminate them.

-waystinthyme

 

[Sea of ASH]

oooh i get what ur saying lol, sorry

 

[Dencology]

but when i do the punnet sq i also get Xrg Xrg which is daughter colorblind. so why this does not work.

 

[lhtran07]

rg/Y x rg+/rg

From their dad the daughters will get the rg gene no matter what. But from their mom, who's a carrier, the daughter may get a normal dominant rg+ gene (therefore, making the daughter a carrier with normal vision) or the rg gene (and with no normal rg+ gene to mask it...the daughter ends up colorblind)

Girls can have the following genotype: rg+/rg or rg/rg, so they have a 50/50 chance of being colorblind or a carrier (with normal vision)

 

[waystinthyme]

but when i do the punnet sq i also get Xrg Xrg which is daughter colorblind. so why this does not work.

with a heterozygous mother and a colorbind father, it IS possible for them to have a colorblind daughter, if the father passes the recessive allele AND the mother passes the recessive allele to the daughter, but the question states that the daughter is NOT colorblind.

since the mother is heterozygous, there is a 50/50 chance that she will pass either of her alleles to the daughter. since the daughter is not colorblind, we must assume that she passed the DOMINANT allele to the daughter.

in the other pairs, the father is either not colorblind, or the mother is heterozygous dominant or recessive. in these cases the mother will always pass the same gene to their offspring, and hence, it would not be possible to have a son with the trait and a daughter without the trait...since BOTH offspring receive atleast one x-gene from the mother...

-waystinthyme

 

[Dencology]

i got it now. thanks.