# Genetics Question, Probability GS-5

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#### boaz

##### shanah alef
10+ Year Member
15+ Year Member
This is from Gold Standard #5, Q 21:

[FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif][FONT=Verdana, Arial, Helvetica, sans-serif]21) Von Willebrand's disease is an autosomal dominant bleeding disorder. A man who does not have the disease has two children with a woman who is heterozygous for the condition. If the first child expresses the bleeding disorder, what is the probability that the second child will have the disease?
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1. 0.25
2. 0.50
3. 0.75
4. 1.00

Explanation:

BIO 15.3, 15.3.1

We are told that Von Willebrand's is an autosomal dominant disease. Let V be the allele for the disease while v is the absence of the allele for the disease. Thus the father is vv and the mother who is heterozygous is Vv. The Punnett square (BIO 15.3) reveals the following:
V v
v Vv vv
v Vv vv

Thus the frequency of Vv is 0.50 (= 2/4) and that of vv is also 0.50 (= 2/4). Since the gene is dominant all heterozygotes (Vv) will have the disease. The chance that the next child expresses the disease is not dependent on what happened to the parent's previous children (BIO 15.3.1).

Why not A? Don't we need to consider the fact that there is a 50% probability for each, so the probability that BOTH would have the allele would be (0.5)(0.5)=0.25?

The question stated that the first child already had the disease so you only have to consider the second child, which is 50%

So I guess technically the probability should be calculated as (1.0)(0.5), since the "probability" that the first child is just 1.0.

The percent chance of the second child getting the disorder is independent of the first child getting it. This is a common trick on genetics questions. Each child is an individual case.

The punnett square would be like this:

a a

A Aa Aa

a aa aa

There is a 50% chance to have the dominant allele and thus have the disorder. What the first child has doesn't affect the chances of the next child.

Edit: I didn't read the explanation, now that I read it, I basically said the exact same thing. Maybe you'll understand mine better though. Anyways, they're not asking the probability of both having the disorder, they're only asking for the probability of the second one having it.

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So I guess technically the probability should be calculated as (1.0)(0.5), since the "probability" that the first child is just 1.0.

Exactly.

Mendelans Law's of Independent assortment. The genes will separate randomly for each child. The first child is just put in there to throw you off. If they did not include the information about the first child the answer would still be the same and still solvable. You dont have to consider the first child at all.

There is a 2/4 chance he will be Vv Vv so 50% is the answer.