BIO question!!

[supera]

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Suppose an animal has a gene for coat color that is sex-linked (X-linkage) and incompletely dominant, and in which females with an AA genotype have a black coat color; aa individuals have a yellow color; and those with Aa have a marble coloration. If a marble female was crossed with a yellow male, then each of the following might result under normal Mendelian conditions EXCEPT one. Which one is the EXCEPTION?
A. Black male
B. Black female
C. Yellow male
D. Yellow female
E. Marble female

So I thought this was going to be crossing XAXa XaY and comes out of
XAXa(Marble female), XaXa(Yellow female), XAY(Black male), XaY(Yellow female)

I chose B but I got it wrong and it seems like 2009 sample answer is telling me the right one in A. What is right?

 

[ScarletKnight24]

Suppose an animal has a gene for coat color that is sex-linked (X-linkage) and incompletely dominant, and in which females with an AA genotype have a black coat color; aa individuals have a yellow color; and those with Aa have a marble coloration. If a marble female was crossed with a yellow male, then each of the following might result under normal Mendelian conditions EXCEPT one. Which one is the EXCEPTION?
A. Black male
B. Black female
C. Yellow male
D. Yellow female
E. Marble female

So I thought this was going to be crossing XAXa XaY and comes out of
XAXa(Marble female), XaXa(Yellow female), XAY(Black male), XaY(Yellow female)

I chose B but I got it wrong and it seems like 2009 sample answer is telling me the right one in A. What is right?

I crossed Aa vs AA using the punnett square and got

Aa (marble), aa(yellow), Aa(marble), aa(yellow)

We know right away that we can eliminate choices C, D, and E since they are included in the punnett square. So it is now between A or B

I believe since it is sex-linked, answer choice A is an exception since sex-linked disease are more common in male. Males never carry the gene however.

Don't know if this makes any sense. I was flipping through my bio notes

 

[ugination87]

You're right. It is B. because you can get both black and yellow males, but only marble or yellow females.

 

[Golfguy]

I think the answer is B.

I dont get an XAXA genotype from the cross, which I think is needed for a black female. All others seem possible. Black male; XAY is produced in the cross....

 

[AlbinoPolarBear]

Suppose an animal has a gene for coat color that is sex-linked (X-linkage) and incompletely dominant, and in which females with an AA genotype have a black coat color; aa individuals have a yellow color; and those with Aa have a marble coloration. If a marble female was crossed with a yellow male, then each of the following might result under normal Mendelian conditions EXCEPT one. Which one is the EXCEPTION?
A. Black male
B. Black female
C. Yellow male
D. Yellow female
E. Marble female

So I thought this was going to be crossing XAXa XaY and comes out of
XAXa(Marble female), XaXa(Yellow female), XAY(Black male), XaY(Yellow female)

I chose B but I got it wrong and it seems like 2009 sample answer is telling me the right one in A. What is right?

Yeah, might be a mistake on the answer sheet.

I crossed XAXa with XaY and got the same answer as you.

Answer should be B.

 

[Jackson-Ave-Dental]

Instead of crossing this out... I think how they arrived at the conclusion was...

Since the genes are X-linked... and you need two genes to fully express the black color trait. It would be physically IMPOSSIBLE for a male to have a black color because they only have one X...

They can be yellow or marble (at most)! But never completely black.

 

[HannibalLecter]

Instead of crossing this out... I think how they arrived at the conclusion was...

Since the genes are X-linked... and you need two genes to fully express the black color trait. It would be physically IMPOSSIBLE for a male to have a black color because they only have one X...

They can be yellow or marble (at most)! But never completely black.

Nope.

OP this error is not even mentioned in the ADA 2009 Errata sheet and I am kinda surprised at the number of errors seen in this thing. You are absolutely right that when you cross an XAXa female with an XaY male you get the offsprings:

XAXa Marble female
XaXa Yellow female
XAY Black male (Watch out for this one and leaving him alone with tvs in the dark, lol)
XaY Yellow female

And for Jo's explanation you are talking about non-Mendelian inheritance, the question clearly states Mendelian inheritance.

 

[Jackson-Ave-Dental]

Nope.

OP this error is not even mentioned in the ADA 2009 Errata sheet and I am kinda surprised at the number of errors seen in this thing. You are absolutely right that when you cross an XAXa female with an XaY male you get the offsprings:

XAXa Marble female
XaXa Yellow female
XAY Black male (Watch out for this one and leaving him alone with tvs in the dark, lol)
XaY Yellow female

And for Jo's explanation you are talking about non-Mendelian inheritance, the question clearly states Mendelian inheritance.

The XA Y should not be a black male. Not enough pigmentation is produced for the male to be totally black. It needs a double dose of A for it to be black.

Similarly in the case to snapdragons... the aa = white, Aa = pink, AA = red.
The a allele produces no color while the A allele produces color but it needs a double dose of A for it to be completely red.

Thus, shouldn't the XA Y be a marbled male?


If XA Y is a black male, then we are under the assumption of complete dominance are we not?
I think these animals are all inherently yellow due to a different gene at a different locus and the recessive allele 'a' does not actually produce a color...

 

[AlbinoPolarBear]

The XA Y should not be a black male. Not enough pigmentation is produced for the male to be totally black. It needs a double dose of A for it to be black.

Similarly in the case to snapdragons... the aa = white, Aa = pink, AA = red.
The a allele produces no color while the A allele produces color but it needs a double dose of A for it to be completely red.

Thus, shouldn't the XA Y be a marbled male?


If XA Y is a black male, then we are under the assumption of complete dominance are we not?
I think these animals are all inherently yellow due to a different gene at a different locus and the recessive allele 'a' does not actually produce a color...

Yeah, this makes sense too, with XAY being a marbled male because of incomplete dominance.

But, the question still remains.. if the answer is not black male, then what is the answer then?

how can we get a black female from those gametes? XAXa crossed with XaY.

*head explodes*

 

[Jackson-Ave-Dental]

Yeah, this makes sense too, with XAY being a marbled male because of incomplete dominance.

But, the question still remains.. if the answer is not black male, then what is the answer then?

how can we get a black female from those gametes? XAXa crossed with XaY.

*head explodes*

Yes... *head explodes* lol

I think its just a sloppy question. To add onto my previous post...

If 'a' added a color and since we all know that 'A' adds a color...

Then Aa would both be adding color to the mix and that would enter the domain of codominance wouldn't it? So 'a' cannot add any color...

Is this totally incorrect and I'm crazy lol? Someone tell me!

 

[AlbinoPolarBear]

Yes... *head explodes* lol

I think its just a sloppy question. To add onto my previous post...

If 'a' added a color and since we all know that 'A' adds a color...

Then Aa would both be adding color to the mix and that would enter the domain of codominance wouldn't it? So 'a' cannot add any color...

Is this totally incorrect and I'm crazy lol? Someone tell me!

Alright, back to this question...
JO and the answer sheet is right.

The question does not state that that the observed phenotype has to be in the F1 generation.

Crossing marble female (XAXa) with yellow male (XaY) gets you:

XAXa Marble female
XAY Marble male
XaXa Yellow female
XaY Yellow male

So from the answer choices, we're left with:
A. Black male
B. Black female

The black female can be produced the in F2 generation, while, because of incomplete dominance, producing a black male phenotype would be impossible.

 

[LetsGo2DSchool]

Alright, back to this question...
JO and the answer sheet is right.

The question does not state that that the observed phenotype has to be in the F1 generation.

Crossing marble female (XAXa) with yellow male (XaY) gets you:

XAXa Marble female
XAY Marble male
XaXa Yellow female
XaY Yellow male

So from the answer choices, we're left with:
A. Black male
B. Black female

The black female can be produced the in F2 generation, while, because of incomplete dominance, producing a black male phenotype would be impossible.

I think you have to assume it is F1 generation. I mean the exact words are:

If a marble female was crossed with a yellow male, then each of the following might result under normal Mendelian conditions EXCEPT one.

This is plain English. At least to me.

 

[AlbinoPolarBear]

I think you have to assume it is F1 generation. I mean the exact words are:

If a marble female was crossed with a yellow male, then each of the following might result under normal Mendelian conditions EXCEPT one.

This is plain English. At least to me.

Yeah, these are the trickier questions where they trap you into an answer.

Either way, it's impossible to get a black male, as it's not in the F1 generation either. Under normal Mendelian conditions, a black female can be produced in the F2 generation, while black males will never be produced in any generation.

 

[supera]

Thank you for everyone's input!
I guess I shouldn't just worry too much for this one tricky/sloppy question.

 

[Troyvdg]

wasnt there a thread somewhere about the errors in that version of the sample test..hmmm

 

[FROGGBUSTER]

This was an error that has been corrected in later editions of the sample test, the answer is indeed B. ADA really, really blows.

 

[Troyvdg]

This was an error that has been corrected in later editions of the sample test, the answer is indeed B. ADA really, really blows.

thought so 👍 thanks