- Joined
- Dec 31, 2007
- Messages
- 1,373
- Reaction score
- 25
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?
...........
1. 0.25
2. 0.50
3. 0.75
4. 1.00
Correct Answer: B.
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?
[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?
...........
1. 0.25
2. 0.50
3. 0.75
4. 1.00
Correct Answer: B.
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?