Bio Question Help. (1 Viewer)

[blackratpoo]

2005 HSC - 24 (b)

Nick's wife Maria has a history of red-green colour blindness in her family. Jack, their two-year old son, may be red-green colour blind. Maria's brothers, Vincent and Paul, are colour blind but her brother, James is not. Maria's mother Anne, is a carrier of red-green colour blindness. Her father, John, is unaffected.

(a) Construct a pedigree - This part is fine.
(b) Predict whether Jack will be colour blind. Justify your answer.

I have Success One and the answer reckons there is a 50% chance he is or isnt. But i think 75% isnt, 25% is. My response....

Anne is a carrier and hence has genotype XCXc. John must be unaffected and has genotype XcY. Therefore, Maria has either genotype XcXc or XCXc. Jack receives the Y chromosome from his father making his genotype irrelevant. If Maria is XcXc, then Jack is not colour blind. If Maria is XCXc, then there is a 50% chance Jack is or isnt colour blind. Therefore, in total, the chance Jack is colour blind is 25% and that he is not 75%.

Thanks for your time. We did Blueprint of Life first unlike most who did Maintaining a Balance so if 09' kids could help it'd be great.

 

[Schoey93]

2005 HSC - 24 (b)

Nick's wife Maria has a history of red-green colour blindness in her family. Jack, their two-year old son, may be red-green colour blind. Maria's brothers, Vincent and Paul, are colour blind but her brother, James is not. Maria's mother Anne, is a carrier of red-green colour blindness. Her father, John, is unaffected.

(a) Construct a pedigree - This part is fine.
(b) Predict whether Jack will be colour blind. Justify your answer.

I have Success One and the answer reckons there is a 50% chance he is or isnt. But i think 75% isnt, 25% is. My response....

Anne is a carrier and hence has genotype XCXc. John must be unaffected and has genotype XcY. Therefore, Maria has either genotype XcXc or XCXc. Jack receives the Y chromosome from his father making his genotype irrelevant. If Maria is XcXc, then Jack is not colour blind. If Maria is XCXc, then there is a 50% chance Jack is or isnt colour blind. Therefore, in total, the chance Jack is colour blind is 25% and that he is not 75%.

Thanks for your time. We did Blueprint of Life first unlike most who did Maintaining a Balance so if 09' kids could help it'd be great.

I'll have to muse over this question - genetics is definitely the hardest part of biology!

I'll edit this post to include the answer and my reason - once I've worked it out...

 

[blackratpoo]

ok, but your yr11. meh, any help is a bonus i guess.

 

[Schoey93]

Yes, I am a Year 11 student, but I studied genetics for the School Certificate. What I did was draw the pedigree like yourself and then I allocated a genotype to the relevant people, ie Anne, John, Maria and Maria's husband.

Your reasoning seems logic and you may very well be correct, but I doubt it. This is how I account for the likely fact that Jack has a 50% chance of being colour blind:

There is a 50% chance that Jack will be colour blind, hence there is also a 50% chance that he will not be. This is because Anne carries the trait, although she is not colour blind. From this one can safely infer that Anne's genotype for colour blindness is XCXc. As Jack receives the Y chromosome, Nick's (his father's) genotype for colour blindness is irrelevant. Maria's genotype must be either XCXc or XcXc. It is not stated whether or not Nick's father is colour blind himself. Therefore one cannot assume that he has normal vision, nor can one assume that he is colour blind. Hence the biologist must refer to the only available information, that is, Maria's family history. Jack's diagnosis will not be 100% accurate in this case, however, if his mother has the genotype XCXc, Jack will be colour blind and if she has the genotype XcXc, he will not be colour blind. This translates to a 50% chance that Jack will be colour blind.

 

[waller]

Yes, I am a Year 11 student, but I studied genetics for the School Certificate. What I did was draw the pedigree like yourself and then I allocated a genotype to the relevant people, ie Anne, John, Maria and Maria's husband.

Your reasoning seems logic and you may very well be correct, but I doubt it. This is how I account for the likely fact that Jack has a 50% chance of being colour blind:

There is a 50% chance that Jack will be colour blind, hence there is also a 50% chance that he will not be. This is because Anne carries the trait, although she is not colour blind. From this one can safely infer that Anne's genotype for colour blindness is XCXc. As Jack receives the Y chromosome, Nick's (his father's) genotype for colour blindness is irrelevant. Maria's genotype must be either XCXc or XcXc. It is not stated whether or not Nick's father is colour blind himself. Therefore one cannot assume that he has normal vision, nor can one assume that he is colour blind. Hence the biologist must refer to the only available information, that is, Maria's family history. Jack's diagnosis will not be 100% accurate in this case, however, if his mother has the genotype XCXc, Jack will be colour blind and if she has the genotype XcXc, he will not be colour blind. This translates to a 50% chance that Jack will be colour blind.

doesn't matter about Nick's genotype, as Jack gets the Y chromosome from him, and it's sex-linked

I reckon it's 75%/25% as well but I have a suspicion that I'm missing something glaringly obvious which annoys me

what do the Success One answers say?

 

[Schoey93]

Now that I think about it, I agree that there is a 25%/75% split chance of Jack being colour blind. The flaw was in my working, not BRP's.

I don't mean to be rude, but: have you considered that given textbook writers are human and make mistakes, Success One's answer is in all probability quite wrong?

 

[Essjaybee]

You dont know if the mum carries the gene or not. So you think about all possibilities. Either she's a carrier or she's not.

Use punnett squares. But rule out both female options, because the baby is a boy. You have two results for each square.

If she is a carrier
option 1: baby has it
option 2: baby doesnt have it

If not.
option 1: baby doesnt have it
option 2: baby doesnt have it

Therefore, 1/4 chance of getting it aka 25%.
Therefore textbook is wrong

Draw the punnet squares it helps a lot. XcX and XY for one and XX and XY for the other.

 

[waller]

that's what I got

 

[blackratpoo]

thanks everyone. wow, success one answers are crap.

 

[waller]

thanks everyone. wow, success one answers are crap.

what explanation did they offer?

 

[blackratpoo]

50%. Here is there answer....

Jack may or may not be colour blind. Maria has a 50% chance of being a carrier as her mother was a carrier; hence she has two colour blind brothers. Jack has a 50% chance of being colour blind, depending on whether Maria is a carrier or not.

 

[Essjaybee]

50%. Here is there answer....

Jack may or may not be colour blind. Maria has a 50% chance of being a carrier as her mother was a carrier; hence she has two colour blind brothers. Jack has a 50% chance of being colour blind, depending on whether Maria is a carrier or not.

Thats wrong. Carriers have a 50% chance of passing the gene, not 100%.

 

[Schoey93]

50%. Here is there answer....

Jack may or may not be colour blind. Maria has a 50% chance of being a carrier as her mother was a carrier; hence she has two colour blind brothers. Jack has a 50% chance of being colour blind, depending on whether Maria is a carrier or not.

Perhaps a monkey writes the Success One answers? I recommend you download the official 2005 BOS "Notes from the Marking Centre" as their answers are very thoroughly checked, standards-referenced (the HSC system is "a standards-referenced framework", which is why you don't get your raw marks). Um I'd give you a link but cbf (literally! ) and I'm tired and pissed off at my brothers ... so I just cbf right now to do much. But just go to the Board of Studies webpage, Home - Board of Studies NSW and follow the links to the 2005 HSC exams and marking notes.

I hope this helps and that you didn't already do this, cuz if you did, I would look quite the fool!

 

[blackratpoo]

Perhaps a monkey writes the Success One answers? I recommend you download the official 2005 BOS "Notes from the Marking Centre" as their answers are very thoroughly checked, standards-referenced (the HSC system is "a standards-referenced framework", which is why you don't get your raw marks). Um I'd give you a link but cbf (literally! ) and I'm tired and pissed off at my brothers ... so I just cbf right now to do much. But just go to the Board of Studies webpage, Home - Board of Studies NSW and follow the links to the 2005 HSC exams and marking notes.

I hope this helps and that you didn't already do this, cuz if you did, I would look quite the fool!

I did do this.
But for the 2005 papers they dont have sample answers. Just, marking criterias that say:

Makes a prediction based on pedigree in part (a)
• Supports prediction with logical argument 2
• Makes a correct statement from pedigree drawn 1

They dont give an answer though so its kinda useless.

One thing is for sure, i have lost all my faith in Success One Answers. DONT TRUST THEM!

 

[eviemarch]

Okay so the first part where you have gone wrong is that colour blindness is a recessive trait... so it would be a lower case letter not a capital letter. A capital letter in this case would mean an unaffected allele because being unaffected is the dominant trait. So if the dad is unaffected his genotype would be XC Y and the mothers would be XC Xc.

So because Maria is female she obviously has to get the X chromosome from her father with is a dominant capital C, but there is a 50% chance whether she gets the dominant or recessive allele for colour blindness from her mother. Regardless of being homozygous or heterozygous, she is still unaffected.

Just as you said yourself, it does not matter what Maria's partner Nicks genotype is, because they had a son (Jack), and the son must receive the Y chromosome from his father Nick. Whether Jack is affected or not simply depends on the allele he gets from his mother.
If his mother was homozygous unaffected (XC XC) which is a 50% chance, then jack would definitely be unaffected so it is also a 50% chance he unaffected.
If his mother is heterozygous unaffected (XC Xc) which in itself is a 50% chance, then there is a 50% of him receiving the unaffected dominant C allele so 50% of 50% is 25%.

In summary, there is a 75% chance jack is unaffected because 50% (if maria is homozygous) plus 25% (if she is heterozygous) equals 75% chance. There is only a 25% he is affected which would be if Maria was heterozygous and he then inherited her recessive allele.

Confusing for people just learning genes, alleles, punnet squares and sex linkage... but hopefully most understand!

So yes you are correct its a 75%:25% chance.