So since one child is affected in family with aut. recessive disease and parents are normal it means that both of them (parents) are Aa and Aa.chance that II-3 has trait is 50 %.

 

So since one child is affected in family with aut. recessive disease and parents are normal it means that both of them (parents) are Aa and Aa.chance that II-3 has trait is 50 %.

So Aa Aa parents could have 25% AA 50% Aa and 25 aa.
since the II-3 isn't affected (not black) it leaves only first two variants. so 2 of 3. so appr. 66% ....

Didn't pay attention in the beginning.

 

Becuase one child is already affected for sure you can state that the chances of a sibling to be affected is 2/3. One has to be healthy and 2 should be carriers. 50% is only true if they would have asked you to calculate the chances without knowing that anyone is sick or healthy.

 

I think is D 66%
2/3 carrier
1/3 sick
right?

So Aa Aa parents could have 25% AA 50% Aa and 25 aa.
since the II-3 isn't affected (not black) it leaves only first two variants. so 2 of 3. so appr. 66% ....
Didn't pay attention in the beginning.

That's conceptually incorrect. What happened to one child has no bearing on what'll happen to the next one. 25% and 50% describe the probability for each child or the overall picture you can expect to see in a very very large sample size(ideally infinity).

To give a simple and oft quoted example, if the odds of winning a mini-lotto is 1/100, and I've been buying lotto tickets without winning for the past 99 weeks, what are my chances of winning the 100th time?? That's right, 1/100! And if I've been winning the lottery for the last 99 weeks, what are the chances that I'll win the next one? Again its 1/100. Previous outcomes do not influence the next event.

Sent from my Desire HD

 

I think is D 66%

2/3 carrier
1/3 sick

right?

 

I think it's A 25%.

Every child has an equal chance of getting the disease and since cystic fibrosis is autosomal recessive, it would be 1/4 = 25%

 

2/3

I always miss this question somehow.
Thank you for an extra drill!

 

Fcuk i misread the question ... It's asking if the offspring is a carrier ... I'd go with 50% since every time they hav a child, each one has the same risk.

 

D) 2/3

Since we already know individual II-3 doesn't have the disease it only leaves the possibility they they're either a heterozygous carrier or homozygous normal. The probability of being homozgous affected (1/4) is obsolete since we already know they're normal.

So the probability that this individual is a carrier is 2/3 and the probability that she's homozgous normal is 1/3.

 

Guys, how can it be 2/3 ?

All but one (II-2) of their offspring's is affected. It doesn't mean the other's have AA,Aa,Aa genotype. They can all be carriers (Aa) or normal (AA) as far as anyone knows...

Think about it this way - They had their first child (lets say II-1)- He had 1/4 risk of being affected, 1/2 risk of being a carrier and 1/4 risk of being homozygous. Then they had their second child (let's say II-2) and again, she had the same risk! Just because they had one normal child before her doesn't mean, the affected offspring had a higher chance of being affected !!

IF they were quadruplets or some how the couple had all four children at one time .. then you would have 2/3 possibility of II-3 being a carrier. That probability would also be the same for II-1 and II-4.

I hope that's the right answer or I just made a fool of myself .. either way, please discuss and tell me if what I said makes sense or not?

 

Guys, how can it be 2/3 ?

All but one (II-2) of their offspring's is affected. It doesn't mean the other's have AA,Aa,Aa genotype. They can all be carriers (Aa) or normal (AA) as far as anyone knows...

Think about it this way - They had their first child (lets say II-1)- He had 1/4 risk of being affected, 1/2 risk of being a carrier and 1/4 risk of being homozygous. Then they had their second child (let's say II-2) and again, she had the same risk! Just because they had one normal child before her doesn't mean, the affected offspring had a higher chance of being affected !!

IF they were quadruplets or some how the couple had all four children at one time .. then you would have 2/3 possibility of II-3 being a carrier. That probability would also be the same for II-1 and II-4.

I hope that's the right answer or I just made a fool of myself .. either way, please discuss and tell me if what I said makes sense or not?

In AR disease, one have to eliminate the rr (homozygous) in order to make the calculation. From the possible combination:

2/3 is carrier
1/3 is non sick

 

Guys, how can it be 2/3 ?

All but one (II-2) of their offspring's is affected. It doesn't mean the other's have AA,Aa,Aa genotype. They can all be carriers (Aa) or normal (AA) as far as anyone knows...

Exactly, except individual II-2 who's affected, all the other offspring can be either carriers (Aa) or normal (AA), but they cannot be (aa). So that eliminates the possibility of (aa) for all the unaffected offspring, meaning now you only have 3 choices in the punnet square, the choice (aa) can be removed. Normally the punnet square has four choices: AA, Aa, Aa, aa... So with aa removed as an option, the probability of being a carrier is 2/3.

 

AHHH I get it ... So when they ask the risk of being a carrier, that eliminates homozygous affected because if the genotype was aa then the person would be affected (not carrier or normal). Damn, no wonder I used to get these wrong! Well, now I won't get these wrong lol ... Good discussion fellas!

 

read it!!

the answer is D 66% here how i get it:

1. first thing the disease is recessive because she has the disease and non of here parents are affected.

2.draw a punnett square of here parents:

D d

D DD Dd

d Dd dd

notice that 25% affected and 50% are carrier and 25% are normal
we see form the pedigree that II-3 is carrier or normal so we cancel the chance of being affected

3. II-3 is Dd or Dd or DD so the probability of Dd is 33% so its 66% since we got 2 Dd 66% are carrier and 33% are normal

 

3. II-3 is Dd or Dd or DD so each probability is 1/3 and thus the answer is 33%

And Dd or Dd equals to 2 Dd = 33 *2 = 66 ?

Isn't it right ?

 

correct ans is D

correct ans is D 66%

i was scratchin my head over dis question & thought of skippin it

the explanation given in the source was pathetic as it was far more confusing than the question...

thnx for the explanation

 

Where is this Q from ?

 

If the question was asking about an unborn child, then the chance of being carrier is 50%. But since the offspring the are asking about is already not affected, then the chance of being affected should be excluded, and hence the chance for being carrier is 2 out of 3.

 

Lets extend the question.

thanks for such a nice discussion. I got it! What you guys meant to say.
But i was thinking Just imagine if the couple had 5 children then question was asked about the II 5. Then what should be the provability.
Will it be 1/4 for AA and 2/4 for Aa.
May be a idiotic question for some fella's but still wanna polish my concept. Thanks