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?
*
4 points
75%
50%
25%
0%
The trait for having an extra finger is determined by a dominant gene. This means that if a person has one copy of the gene, they will have an extra finger. If they have two copies of the gene (homozygous), they will also have an extra finger.
In this scenario, one parent is normal and does not have the gene for the extra finger. This means that they do not have any copies of the gene (homozygous recessive). The other parent has an extra finger but is heterozygous for the trait, meaning they have one copy of the gene (heterozygous dominant).
When two parents have different genotypes for a trait, there are four possible combinations of their genes that can be passed on to their children: dominant-dominant, dominant-recessive, recessive-dominant, and recessive-recessive.
In this case, the dominant-dominant combination (DD) would result in the child having an extra finger. The dominant-recessive combination (Dr) would result in the child being normal but carrying the gene for the extra finger. The recessive-dominant combination (rD) would also result in the child being normal but carrying the gene for the extra finger. Lastly, the recessive-recessive combination (rr) would result in the child being normal and not carrying the gene for the extra finger.
Out of these four possible combinations, only one (rr) results in the child being normal. Therefore, the probability that the first child will be normal is 25%.