what is the probability that if each parent has one brown eye gene and one blue eye gene will have one chance in four that there children will get two blue eye genes?please help.
Your question is nonsense..you want probability, but you give it by saying one chance in four....
Bb x Bb= BB bb bB Bb
so bb is one chance in four.
To determine the probability of a child inheriting two blue eye genes when each parent has one brown eye gene and one blue eye gene, we need to understand how eye color genes are passed on.
Eye color is determined by the combination of genes inherited from parents. There are different variations of eye color genes, with blue eyes being recessive and brown eyes being dominant.
In this scenario, assuming that the blue eye gene is recessive, both parents would have the genotype Bb, where B represents the brown eye gene and b represents the blue eye gene.
When the parents have the genotype Bb, there are four possible combinations for their children's genotypes: BB, Bb, bB, and bb.
Out of these four combinations, only one combination (bb) has two blue eye genes.
Therefore, the probability of a child having two blue eye genes in this scenario is 1 out of 4, or 1/4.
To recap:
- Assume the blue eye gene is recessive (represented as b).
- Each parent has the genotype Bb (one brown eye gene and one blue eye gene).
- The possible genotypes for their children are: BB, Bb, bB, and bb.
- Out of the four possibilities, only one combination (bb) has two blue eye genes.
- Therefore, the probability of a child having two blue eye genes is 1/4 or 25%.