I'm convinced that this Kaplan problem is wrong

[axp107]

Advertisement - Members don't see this ad

A condition called A-T is an autosomal recessive disease

What is the probability that an unaffected sibling of an affected individual will be heterozygous for A-T?

A) 0%
B) 25%
C) 50%
D) 67%

The affected individual can have parents: Aa x aa or Aa x Aa

Each of the above has a 50% chance of producing Aa per reproduction.
So the answer SHOULD be 50%

Kaplan's answer:

For some reason assume parents are Aa x Aa.. do a cross.. you get:

1 AA, 2 Aa, 1 aa

Since 1 aa was already created (the unaffected person), the probability of the sibling being heterozygous is 2/3, or 67%

This is obviously wrong. You can't just eliminate aa... each reproduction is SEPARATE

 

[barto123]

the question asks about an "unaffected sibling"

aa- affected
Aa(2) - unaffected
AA- unaffected

unaffected possibilities = 3
2 of them the heterozygous

correct answer = 67%

 

[smeagol]

A condition called A-T is an autosomal recessive disease

What is the probability that an unaffected sibling of an affected individual will be heterozygous for A-T?

A) 0%
B) 25%
C) 50%
D) 67%

The affected individual can have parents: Aa x aa or Aa x Aa

Each of the above has a 50% chance of producing Aa per reproduction.
So the answer SHOULD be 50%

Kaplan's answer:

For some reason assume parents are Aa x Aa.. do a cross.. you get:

1 AA, 2 Aa, 1 aa

Since 1 aa was already created (the unaffected person), the probability of the sibling being heterozygous is 2/3, or 67%

This is obviously wrong. You can't just eliminate aa... each reproduction is SEPARATE

Why do we assume parents are Aa x Aa? What happened to Aa x aa?



edit: oops just read the word sibling... thought it was referring to son/daughter.

 

[PreMD86]

the question asks about an "unaffected sibling"

aa- affected
Aa(2) - unaffected
AA- unaffected

unaffected possibilities = 3
2 of them the heterozygous

correct answer = 67%

Hmm I still have a problem with that answer...

There is an equal probability that the parents can be either Aa x aa, or Aa x Aa.

The first cross produces:
50% Aa
50% aa

The second cross produces:
50%Aa
25%AA
25%aa

Since there is an equal chance of the parents being (Aa, aa) and (Aa, Aa),
The phenotypes are:
50%Aa
12.5%AA
37.5%aa

The question is asking for the odds that UNAFFECTED sibling is heterozygous, which is a way of asking the chance that the child is Aa out of all Aa AND AA.
Which is (50% Aa)/(50%+12.5%AA)= 74%

Am I making this problem much harder than it is or something lol?

 

[RoadRunner17]

Tricky ain't it? Just goes to show that the sciences' are a more benign version of VR.

 

[xlr8]

going by that logic wouldn't it be
2/3 from Aa x Aa
and
2/2 from Aa x aa

1) (2+2)/(3+2) = 80%?
or
2) 2/3*2/2 = 2/3?

that first one gives me a feeling that im violating some law of probability

 

[PreMD86]

going by that logic wouldn't it be
1) 2/3 from Aa x Aa
and
2) 2/2 from Aa x aa

3) (2+2)/(3+2) = 80%?

that third step to me gives me a feeling that im violating some law of probability

You would infact be violating laws of probability if you had any laws to break in that calculation.

 

[smeagol]

I still don't understand what happened to Aa x aa

edit: What does it say about the parents being Aa x aa or Aa x Aa? Is it a 50/50 chance between the two?

 

[dill_punjabi]

I think in random population when one offspring is affected we can assume that parents are heterozygous.

 

[barto123]

Hmm I still have a problem with that answer...

There is an equal probability that the parents can be either Aa x aa, or Aa x Aa.

The first cross produces:
50% Aa
50% aa

The second cross produces:
50%Aa
25%AA
25%aa

Since there is an equal chance of the parents being (Aa, aa) and (Aa, Aa),
The phenotypes are:
50%Aa
12.5%AA
37.5%aa

The question is asking for the odds that UNAFFECTED sibling is heterozygous, which is a way of asking the chance that the child is Aa out of all Aa AND AA.
Which is (50% Aa)/(50%+12.5%AA)= 74%

Am I making this problem much harder than it is or something lol?

although i think we're making too much of this problem my logic is that the parents are much more likely to be Aa than aa, with aa being so uncommon it is negligible. carriers of a recessive disorder are far more common than the homozygous recessive, especially if the disorder is early acting or lethal.

 

[Kaydubz]

Typical of Kaplan to put a calculation intensive (comparably) problem on the Bio section.

Parents_____chance_____genotype of sibling
P1____P2
Aa___Aa_____0.33_____AA, Aa, Aa, aa
Aa___aa_____0.33_____Aa, Aa, aa, aa
aa___Aa_____0.33_____Aa, Aa, aa, aa

For the second and third sets of parents, the unaffected individual is 100% likely to be heterozygous. For the first, the unaffected individual is 67% likely to be heterozygous. Therefore:

(0.67) * 33 + 33 + 33 = 88%

I might've messed up my probability math somewhere along the way.

Or:

Parents_____chance_____genotype of sibling
P1___P2
Aa___Aa_____0.5_____AA, Aa, Aa, aa
Aa___aa_____0.5_____Aa, Aa, aa, aa

For the second set of parents, the unaffected individual is 100% likely to be heterozygous. For the first, the unaffected individual is 67% likely to be heterozygous. Therefore:

(0.67) * 0.5 + 50 = 83.5%

Oh, and you can even go crazier and modify the 33/33/33 or 50/50 probabilities of parental genotype if you want to claim that the "aa" of the original infected individual is more likely to have had parent sets 2 and 3 rather than 1. *sigh*

Yep, way too complicated. 🙂



In any case, Kaplan assumed Aa x Aa, which yields 67%. Throw in even the smallest possibility of Aa x aa parents, with all the unaffected inviduals being heterozygous, and that percentage rises. Wow, this problem was way too interesting to reason through.. I edited this post like 5 times.

 

[dill_punjabi]

How many different ways 46 chromosomes can arrange during meiosis I

 

[smeagol]

although i think we're making too much of this problem my logic is that the parents are much more likely to be Aa than aa, with aa being so uncommon it is negligible. carriers of a recessive disorder are far more common than the homozygous recessive, especially if the disorder is early acting or lethal.

That makes sense. Kind of frustrating though because I have no idea what A-T is. I hope they'd tell us that the disorder was likely lethal.

 

[PreMD86]

although i think we're making too much of this problem my logic is that the parents are much more likely to be Aa than aa, with aa being so uncommon it is negligible. carriers of a recessive disorder are far more common than the homozygous recessive, especially if the disorder is early acting or lethal.

If that is the case, the answer would indeed be 67%.
I'm not sure that's a valid assumption however, since Aa is only twice as likely to appear as aa.
I suppose the moral of the story is Kaplan sucks, I'm sure that's something we can all agree on.

 

[axp107]

Ah dammit.. I reread the question

they're asking.. what the probability of the unaffected sibling being Aa.. dammit ahhh. how did i miss that

not

what is the probability of the sibling being Aa.

Dammit

 

[Nikki2002]

Ah dammit.. I reread the question

they're asking.. what the probability of the unaffected sibling being Aa.. dammit ahhh. how did i miss that

not

what is the probability of the sibling being Aa.

Dammit

nice.....😉

 

[ssquared]

Ah dammit.. I reread the question

they're asking.. what the probability of the unaffected sibling being Aa.. dammit ahhh. how did i miss that

not

what is the probability of the sibling being Aa.

Dammit

This question is very typical of both Kaplan and AAMC type questions. You must read the question very carefully! Good thing you're learning this lesson while studying, unlike the hard way (when taking the real thing).

 

[Creightonite]

A condition called A-T is an autosomal recessive disease

What is the probability that an unaffected sibling of an affected individual will be heterozygous for A-T?

A) 0%
B) 25%
C) 50%
D) 67%

The affected individual can have parents: Aa x aa or Aa x Aa

Each of the above has a 50% chance of producing Aa per reproduction.
So the answer SHOULD be 50%

Kaplan's answer:

For some reason assume parents are Aa x Aa.. do a cross.. you get:

1 AA, 2 Aa, 1 aa

Since 1 aa was already created (the unaffected person), the probability of the sibling being heterozygous is 2/3, or 67%

This is obviously wrong. You can't just eliminate aa... each reproduction is SEPARATE

hey, which problem is it? There was one in lesson book, but i think it is from elsewhere.

 

[quiksilver87]

so who has the answer right?
if you take both crosses into consideration = Aa x aa and Aa x Aa


for the Aa x aa, all the siblings that are unaffected are hetero
for Aa x Aa, only 3 out of 4 are unaffected

thus totall number of possibilities of being unaffected = 2 + 3
Total number of unaffected heteros = 4

Thus, the prob should be 4/5 or 80%

FINALLY

 

[DiscoDoc]

I still don't understand what happened to Aa x aa

edit: What does it say about the parents being Aa x aa or Aa x Aa? Is it a 50/50 chance between the two?

Generally with autosomal recessive genes you make an "ass" out of "u" and "me" and say that the frequency of double recessives is rare. Thus parents are wildtype unless otherwise specified. So we assume AA for both. But we know one child is affected so both parents have to be Aa.
I haven't come across any AAMC questions that require this assumption though.

 

[ahmedqman]

OMG... that is a dirty question. I hope they aren't all like this. Oh well, I'll be sure to read carefully!

 

[mrmilad]

Well since its an autosomal recessive maybe they are assuming that it is a rare to be affected, so thus its more likely that the parents are heterozygous. Although that is still a terrible question.


Aa more likely to occur than aa

 

[BrokenGlass]

although i think we're making too much of this problem my logic is that the parents are much more likely to be Aa than aa, with aa being so uncommon it is negligible. carriers of a recessive disorder are far more common than the homozygous recessive, especially if the disorder is early acting or lethal.

I agree. Recessive mutations in homozygous form are very rare. The reason is probability and the assumption that the 2 alleles at the same locus would mutate independently of each other (so you MULTIPLY probabilities due to the independence assumption, thus making the overall probability a very small number). Since one of the children is affected (i.e has the disease), each of the parents has at least one recessive allele. And since it's extermently unlikely that either parent is homozygous recessive (for the reasons stated above0, both parents are heterozygous recessive.

I know that for an MIT graduate I have a very poor command of English, but I hope this made sense.

 

[DiscoDoc]

Correction: There actually was a question on AAMC CBT 3 that required this assumption.