The heterozygous probabilities will be 1/2; 1/4; 1/4
- We can infer that since this is a rare characteristic, individual 1's mother is heterozygous and her father is homozygous for the wild type allele.
- The mother passed on an X to Individual 1, giving him or her a 50% chance of inheriting a mutant allele.
- Individual 3 has a 50% probability of inheriting the affected X chromosome from her mother if 1 is heterozygous, and a 100% chance of inheriting an affected X from her father.
- Individual 3 has a total probability of being impacted of 1/2, 1/2, and 1. There is a 50% possibility that the mother of person 4 carries an afflicted X and that she will pass it on to her son.
- The son receives only a Y chromosome from his father. The total probability in this case is 1/2 ×1/2.
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