# BRS error?

Discussion in 'Step I' started by Erek94, May 19, 2014.

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1. ### Erek94 5+ Year Member

94
18
Mar 15, 2010
Is there an error in the BRS explaination or am I just missing the concept here. They say the answer is 1/64 but I got 1/16 .

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3. ### Enzymes 2+ Year Member

155
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Mar 13, 2012
I got 1/16 too. Stage IV has a 50% chance of being a carrier. And the only way to get disease expression is two carriers having a baby (1/4 chance). So I got 1/4 * 1/2 * 1/2 = 1/ 16.

4. ### dyspareunia

1,019
587
Mar 6, 2014
It's 1/64.

It's (1/2)*(1/2)*(1/2)*(1/2)*(1/4)

Gen II for patient's mom = 1/2
Gen III for patient's mom = 1/2

Gen II for patient's dad = 1/2
Gen III for patient's dad = 1/2

Then a child has a 1/4 chance of being homo-recessive from two hetero parents.

Both generation II and generation III are 1/2 because the people are already born and must be heterozygous or there is no chance the unborn child will be affected. In a heterozygous cross you have 1 AA, 2 Aa, 1 aa. So 2/4 are heterozygous. Since the parents are both heterozygous, their child has a 1/4 chance of being homorecessive.

Last edited: May 19, 2014
5. ### Erek94 5+ Year Member

94
18
Mar 15, 2010
I guess my confusion is trying to wrap my head around how gen II to III is 1/2? I woudl think its 100% because one of the parents has the disease.

6. ### dyspareunia

1,019
587
Mar 6, 2014
And the other parent is heterozygous. The only 100% chance is if both parents are homozygous recessive.

7. ### DrEnderW 2+ Year Member

1,598
1,201
Dec 5, 2012
Rishi Maze
If II-2 is affected (aa) and has a child with a homozygous WT (AA) that would be:

AA x aa = 100% chance offspring is a carrier

Why is that not the case?

8. ### dyspareunia

1,019
587
Mar 6, 2014
Oh I see what you're all saying now. I agree, 1/16 looks right.

Edit: Nope, nevermind.. back to 1/64, here's why.

Generation III is 100% carrier (both maternal and paternal sides).

For some reason the tendency to assume heterozygosity resonates with me. I remember my genetics teacher explained this, but I don't remember why or how.

So, assuming all 4 maternal grandparents are heterozygotes, you now have (1/2)*(1/2) for BOTH maternal and paternal odds and another (1/2)*(1/2) for baby

Thus, (1/2)^6.

You get (1/2)^4 if you assume that 2 grandparents are homozygous dominant and the other 2 are heterozygotes though.

Last edited: May 19, 2014
9. ### Erek94 5+ Year Member

94
18
Mar 15, 2010
I appreciate your time and effort.
Here is the explaination the book gives. I just dont understand how there is a 1/2 chance from II-2 to III-1. Your logic obviously got the question correct for you and I cant argue that. I would just continue to get this type of problem wrong over and over again if presented like this.

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