BIO Cross Over Q~!

[joonkimdds]

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This is from Kaplan Blue Book, page 667 number 16

Q: A colorblind male XcY is crossed with a normal female who is a carrier of fragile X and colorblindness (XfXc). What is the probability that a male child will be phenotypically normal?

A: 0%

My questions
1. how can a normal female carry fragile x and colorblindness?
I say that woman is abnormal.

2. XfXc, XcXc, XfY and XcY are the 4 diff combinations.
How can i tell which is normal and which is not to find the answer is 0%?

 

[dat_student]

This is from Kaplan Blue Book, page 667 number 16

Q: A colorblind male XcY is crossed with a normal female who is a carrier of fragile X and colorblindness (XfXc). What is the probability that a male child will be phenotypically normal?

A: 0%

My questions
1. how can a normal female carry fragile x and colorblindness?
I say that woman is abnormal....

recessive genes 🙂 0% is correct 🙂

 

[allstardentist]

haha, i guess u have to assume that the fragile gene is recessive meaning that she has to have XfXf to express the phenotype. However her BOY child will have a Y from his dad and either Xf or Xc from her. Thus, the possible genotype is XfY or XcY and these two are both abnormal. So the answer is 0%.

 

[dat_student]

haha, i guess u have to assume that the fragile gene is recessive meaning that she has to have XfXf to express the phenotype. ....

"carrier of fragile X" << tells me it's recessive

 

[allstardentist]

good looking

 

[DMD to Be]

This is from Kaplan Blue Book, page 667 number 16

Q: A colorblind male XcY is crossed with a normal female who is a carrier of fragile X and colorblindness (XfXc). What is the probability that a male child will be phenotypically normal?

A: 0%

My questions
1. how can a normal female carry fragile x and colorblindness?
I say that woman is abnormal.

2. XfXc, XcXc, XfY and XcY are the 4 diff combinations.
How can i tell which is normal and which is not to find the answer is 0%?

This problem requires no work and no crosses. You don't even need the different combinations that you listed. This is just knowing your genetics. The question is what is the probability that a male child will be phenotypically normal?? Okay, simple. The dad is colorblind, so his genotype looks like this: Xc Y, since colorblindness is found only on the X chromosome. His genotype means nothing here, b/c his sons will only inherit his Y chromosome which is not affected by either disease. The mom's info is key in this problem. She has the genotype: Xf Xc. Her son will receive one of these two. Either way, he will not be normal. His possible genotypes are:
Xf Y = Son w/ fragile x
or
Xc Y = Son w/ colorblindness

Understand this?

 

[joonkimdds]

haha, i guess u have to assume that the fragile gene is recessive meaning that she has to have XfXf to express the phenotype. However her BOY child will have a Y from his dad and either Xf or Xc from her. Thus, the possible genotype is XfY or XcY and these two are both abnormal. So the answer is 0%.

hm, if fragile gene means recessive she has to have XfXf? but the problem states that she has XfXc?

and I thought the words Dominant and Recessive is used only when they have Capital letter and small case letter such as Zz and dominant(let's say Z) will own z and thus it has ZZ phenotype?

 

[Notoriousjae]

you dont even need to make the diff combinations to figure out its 0% given that the male is colorblind and can only transfer the Y to its male offspring

 

[allstardentist]

hm, if fragile gene means recessive she has to have XfXf? but the problem states that she has XfXc?

and I thought the words Dominant and Recessive is used only when they have Capital letter and small case letter such as Zz and dominant(let's say Z) will own z and thus it has ZZ phenotype?

Doesn't have to be that..

 

[tinman831]

Joon, it's really simple. In the case of color blindness and fragile x, you have to possess of both recessive alleles (XcXc or XfXf) in order to exhibit the phenotype. The fact that she only has one of each, XcXf, means she is only a carrier, and is "normal" (will not exhibit the phenotype). 🙂

 

[joonkimdds]

The mom's info is key in this problem. She has the genotype: Xf Xc. Her son will receive one of these two. Either way, he will not be normal. His possible genotypes are:
Xf Y = Son w/ fragile x
or
Xc Y = Son w/ colorblindness

Understand this?

I understand that son will have either fragile X or Colorblindness but then how can someone with Fragile X and Colorblindness be called normal?
it said normal woman with Fragile X and Colorblindness?? 😕

if carrier means carrying those genes but they don't show outside, can't their son also carry it but those symtoms don't show outside just like his mother?

 

[Notoriousjae]

I understand that son will have either fragile X or Colorblindness but then how can someone with Fragile X and Colorblindness be called normal?
it said normal woman with Fragile X and Colorblindness?? 😕

joon, its because she is only a carrier of the disease and is not affected by it, just like someone can carry STD's and not have the physical symptoms for it but can transmit it to someone else who can be physically affected. i dont know if this makes sense to you but its the best example i could come up with

 

[joonkimdds]

joon, its because she is only a carrier of the disease and is not affected by it, just like someone can carry STD's and not have the physical symptoms for it but can transmit it to someone else who can be physically affected. i dont know if this makes sense to you but its the best example i could come up with

then couldn't her son also be a carrier of the disease and is not affected by it just like his mother?

 

[aranjuez]

Joon,

To express these phenotypes, i.e. to be affected by the fragile X or colorblindness, you need two copies of the same recessive attribute if you're female. For example, XfXf would be affected, XcXc would be colorblind. She only has 1 of each, which means she doesn't show any symptoms for either (as far as the question is concerned).

Her son, though, will inherit a Y from his dad, and one of these recessives from the mom will be given to him, so either Xf or Xc. Since he doesn't have a normal one to combat either one, as the Y chromosome doesn't have these genes, then he'll be affected for sure.

Repost if it still doesn't make sense to you.

aranjuez

 

[joonkimdds]

Since he doesn't have a normal one to combat either one, as the Y chromosome doesn't have these genes, then he'll be affected for sure.

you said the son doesn't have a normal one to combat either Xf or Xc in XfY or XcY. Then could u show me the situation when he would have the normal one to fight either Xf or Xc so he can be a carrier but symtom doesn't show outside?

 

[Notoriousjae]

if the mother had XcXn where Xc=color blind and Xn=normal and the father had XcY
then XcY * XcXn=male would have XnY and XcY so the XnY male would be normal. get it? and the XcY would be color blind

 

[allstardentist]

Males cannot be a carrier for sex-linked traits. Males can only have one copy of a sex linked gene which means that it will express even if its recessive.

 

[Notoriousjae]

best option for you is to bust out a genetics textbook and look up sex linked genetics.

 

[dental#1]

Allstar you said exactly what I was thinking!!!!!!!!!!!!!!!!! I was just talking to my genetics prof this morning and he even showed an example of this same problem that would trip us up on the test if we didn't read carefully.

 

[joonkimdds]

If the problem doesn't state that they actually had sex, can we just say they adopted a child who has nothing to do with any of their genes ?

 

[dental#1]

Now that made me laugh!!!!!!!! I hope that statment doesn't pop into my head when taking the exam!

 

[DonExodus]

The question ONLY asks about male offspring: The only two possible genoypes for male offspring are XcY and XfY.
In either one of these cases, the child will be colorblind, or have fragile X.

There is therefore a 0% chance of having a male without either of these conditions, since the only two possibilities (XcY and XfY) = abnormal child.