This bio question doesn't make sense. frustrated

[adamrose]

From bootcamp:

A color blind man without hemophilia (both X-linked traits) marries a woman who is a carrier for both traits. What is the probability they will have a son with both color blindness and hemophilia? Hemophilia and color blindness are unlinked genes.

Answer: 1/8

"The man’s genotype is XcY and the woman’s genotype is XcX / XhX. The traits of hemophilia and color blindness are unlinked, meaning they are on either different X chromosomes or far enough apart on the same X chromosome so that they assort independently. In this case, since the mother can only donate one X chromosome (since the father will donate his Y chromosome for a male child), the genes must be far enough apart on the same X chromosome so that they assort independently, and are thus unlinked.

We want three things to happen, the man must donate the Y chromosome (1/2 probability), the woman must donate Xh (1/2 probability), and she must also donate Xc (1/2 probability). We just multiply these chances together to find (1/2)(1/2)(1/2) = 1/8"

Okay.. If she can only donate one chromosome, How is she donating 2 in the answer choice (see bold)
If she is a carrier for both, Why wouldn't she be XchX? if she was XcXh there would be no possible way for her to donate both disorders
My impression to solve this problem would be to set up a punnet square like this

Xch X
Xc XcXch XcX
Y XchY XcY

Answer: 1/4

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[bodmon]

now I'm frustrated

 

[questbar]

The man donates a Y chromosome, thus (1/2) probability.
We know that the woman is a carrier, but has only one chromosome to give (X). There are only two alleles that exist C/c for color vision and H/h for hemophilia. The question mentions they are unlinked so we can assume they can both assort independently.

The single X chromosome donated:
has a 1/2 chance of being normal for blood clotting and a 1/2 chance of being recessive for expressing hemophilia.
Likewise, the probability that that single X chromosome will be normal for color vision is 1/2 and recessive for expressing colorblindness is 1/2.

--> select your phenotypes and multiply them together. Y*c*h = (1/2)(1/2)(1/2) gives you 1/8 chance.

 

[Futuredentista62]

Deleted

 

[bodmon]

The man donates a Y chromosome, thus (1/2) probability.
We know that the woman is a carrier, but has only one chromosome to give (X). There are only two alleles that exist C/c for color vision and H/h for hemophilia. The question mentions they are unlinked so we can assume they can both assort independently.

The single X chromosome donated:
has a 1/2 chance of being normal for blood clotting and a 1/2 chance of being recessive for expressing hemophilia.
Likewise, the probability that that single X chromosome will be normal for color vision is 1/2 and recessive for expressing colorblindness is 1/2.

--> select your phenotypes and multiply them together. Y*c*h = (1/2)(1/2)(1/2) gives you 1/8 chance.

I understand 1/2 for Y, but I don't understand the logic of multiplying by the probability of both color blindness and hemophilia; if only one can be given.

 

[bco99]

I see what you are saying. Since the genes are not linked, they cannot be on the same X chromosome and a son can only accept one X chromosome, so he cannot get the colorblind gene and the hemo gene.

 

[adamrose]

Yes he can.

I see what you are saying. Since the genes are not linked, they cannot be on the same X chromosome and a son can only accept one X chromosome, so he cannot get the colorblind gene and the hemo gene.

Yes he can. the chance of this happening is 1/8 this is why i am confused

 

[adamrose]

I understand 1/2 for Y, but I don't understand the logic of multiplying by the probability of both color blindness and hemophilia; if only one can be given.

Same here. Don't understand multiplying by both color blindness and hemophilia either

 

[adamrose]

I see what you are saying. Since the genes are not linked, they cannot be on the same X chromosome and a son can only accept one X chromosome, so he cannot get the colorblind gene and the hemo gene.

the son CAN. according to the answer 1/8

 

[Typical Average Student]

I think it's just applying the principle of inheritance and probability on this problem... if one of the answers choices was zero, it would be zero. Nice catch though, I didn't notice that when I went through BC lol.

Edit: UNLESSSSSSS the son is an XXY which is difficult to assess.

But I think @Ari Rezaei should be able to answer this...

 

[cavitybrah]

The genes are on the same chromosome (x-linked).

Your possibilities are:

XCH XCh XcH Xch

The Y chromosome carries no trait so whatever the mother passes on is expressed.

1/2 Y * 1/4 Xch

You can essentially treat this as a dihybrid cross since the genes are unlinked therefore have no influence on one another

 

[bodmon]

The genes are on the same chromosome (x-linked).

Your possibilities are:

XCH XCh XcH Xch

The Y chromosome carries no trait so whatever the mother passes on is expressed.

1/2 Y * 1/4 Xch

You can essentially treat this as a dihybrid cross since the genes are unlinked therefore have no influence on one another

OP was saying that this is the only way to solve it that makes sense, but the problem was with how the solution was described. They solved it through 1/2*1/2*1/2, using the probability of each circumstance and multiplying it together. With that in mind, it would imply that both X chromosomes are donated from the mother

 

[bodmon]

OP was saying that this is the only way to solve it that makes sense, but the problem was with how the solution was described. They solved it through 1/2*1/2*1/2, using the probability of each circumstance and multiplying it together. With that in mind, it would imply that both X chromosomes are donated from the mother
I think.

 

[BWA1991]

So unlinked genes can still be on the same chromosome, just not close to each other. Linked genes are close to each other and when crossing over occurs would almost always stay together.

The question tell you unlinked so you know that they are two separate events that may occur if the unlinked genes are on the same x chromosome that is being passed on. Separate events mean you have to multiply them as two separate entities (1/2 * 1/2 for X and then *1/2 for Y), whereas if they are linked you would always have them together so it would be (1/2 for x multiplied by 1/2 for y).

When unlinked and on the same chromosome, you want to count them independently because of the fact crossing over can put them on different chromosomes or keep them on the same one. It all depends on the recombination.


If my explanation doesn't help... try this website: http://learn.genetics.utah.edu/content/pigeons/geneticlinkage/

The picture near the top illustrates what I was trying to describe.

Hope this has helped.

 

[cavitybrah]

OP was saying that this is the only way to solve it that makes sense, but the problem was with how the solution was described. They solved it through 1/2*1/2*1/2, using the probability of each circumstance and multiplying it together. With that in mind, it would imply that both X chromosomes are donated from the mother

1/2 Y * 1/4 Xch = 1/8 chance

 

[adamrose]

So unlinked genes can still be on the same chromosome, just not close to each other. Linked genes are close to each other and when crossing over occurs would almost always stay together.

The question tell you unlinked so you know that they are two separate events that may occur if the unlinked genes are on the same x chromosome that is being passed on. Separate events mean you have to multiply them as two separate entities (1/2 * 1/2 for X and then *1/2 for Y), whereas if they are linked you would always have them together so it would be (1/2 for x multiplied by 1/2 for y).

When unlinked and on the same chromosome, you want to count them independently because of the fact crossing over can put them on different chromosomes or keep them on the same one. It all depends on the recombination.


If my explanation doesn't help... try this website: http://learn.genetics.utah.edu/content/pigeons/geneticlinkage/

The picture near the top illustrates what I was trying to describe.

Hope this has helped.

Please tell me if I followed your logic:

So this woman must have a normal X chrom and a carrier x chrome with far apart unlinked genes
making her X X'ch

what you are saying is "h" is on the upper end of the X chrome, and "c" is on the lower end of the X chrome, making crossing over likely.

So in order for a son to be hemophilic and color blind,
(.5 chance of getting Y from dad) ((.5 chance of carrier x chrome from mother)(.5 chance that this carrier chromosome did NOT cross over the h or c and they are both still intact on the same chromosome)

is it rule of thumb that crossing over is a .5 chance?

 

[adamrose]

But I think @Ari Rezaei should be able to answer this...[/QUOTE]

I think it's just applying the principle of inheritance and probability on this problem... if one of the answers choices was zero, it would be zero. Nice catch though, I didn't notice that when I went through BC lol.

Edit: UNLESSSSSSS the son is an XXY which is difficult to assess.

But I think @Ari Rezaei should be able to answer this...

"Klinefelter syndrome is one of the most common chromosomal disorders, occurring in 1:500 to 1:1000 live male births"
choose hypothetical answer choice 1/5000 --> profit

 

[BWA1991]

Please tell me if I followed your logic:

So this woman must have a normal X chrom and a carrier x chrome with far apart unlinked genes
making her X X'ch

what you are saying is "h" is on the upper end of the X chrome, and "c" is on the lower end of the X chrome, making crossing over likely.

So in order for a son to be hemophilic and color blind,
(.5 chance of getting Y from dad) ((.5 chance of carrier x chrome from mother)(.5 chance that this carrier chromosome did NOT cross over the h or c and they are both still intact on the same chromosome)

is it rule of thumb that crossing over is a .5 chance?

I think of it as .5 chance of hemophilia, .5 chance of color blindness, and .5 of y from dad. Or .25 chance of both being on X from mom and .5 chance of getting y from dad.

Whereas if they were linked it would be .5 chance of colorblindness and hemophilia and .5 for y.

 

[mw18]

The way I see it, it is like this:

Mom could have both on the same X that goes to son, Dad gives Y
Mom could have both on same X that does not go to son, Dad gives Y
Mom could have Color Blindness trait on X that goes to son and hemophilia on the other, Dad gives Y
Mom could have Hemophilia trait on X that goes to son and Color Blindness on the other, Dad Gives Y

Mom could have both on the same X that goes to daughter, Dad gives X
Mom could have both on same X that does not go to daughter, Dad gives X
Mom could have Color Blindness trait on X that goes to daughter and hemophilia on the other, Dad gives X
Mom could have Hemophilia trait on X that goes to daughter and Color Blindness on the other, Dad Gives X



Only first one gives son who has both. 1/8.

 

[Jay20155555]

This is what I did

 

[Typical Average Student]

The answer is right, what OP is saying is the explanation for the problem is wrong, by saying Xc and Xh are on DIFFERENT chromosomes when in reality it would only work if Xc and Xh were on the same chromosome.