An extra finger in humans is rare but due to a dominant gene. When one parent is normal and the other parent has an extra finger but is heterozygous for the trait, what is the probability that the first child will be normal?
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4 points
25%
50%
75%
0%
The first parent is normal, so they have two normal copies of the gene (NN). The second parent is heterozygous, meaning they have one normal copy (N) and one extra finger copy (E).
When the two parents have a child, they will each randomly contribute one of their copies of the gene. There are four possible combinations:
1. The first parent contributes an N and the second parent contributes an N. This would result in a normal child (NN).
2. The first parent contributes an N and the second parent contributes an E. This would result in a child with an extra finger (NE).
3. The first parent contributes an E and the second parent contributes an N. This would result in a child with an extra finger (EN).
4. The first parent contributes an E and the second parent contributes an E. This would result in a child with an extra finger (EE).
Out of these four possibilities, only one (NN) results in a normal child. Therefore, the probability that the first child will be normal is 1 out of 4, or 25%.
So the correct answer is: 25%