To answer this question, you must have a basic knowledge of sex-linked disease genetics. Men are much more likely to suffer from X-linked disorders; unlike women, they have only one X chromosome, and therefore do not have a second copy that can compensate for the affected one. A man cannot, however, inherit a defective X chromosome from his father, because fathers pass on their Y chromosome to their sons. So, Jacob must inherit one of his mother's healthy X chromosomes, and there is no chance that he will be colorblind.

This trait is X-linked, which can be determined by considering individual 4. This individual has a 50% chance of receiving an affected maternal chromosome (she is a carrier) and a 0% chance of receiving an affected paternal chromosome (he is unaffected), and yet he has the defect. We know that the son inherits the Y-chromosome from his father, and a singular X-chromosome from his mother; thus, we can assume he inherits the affected chromosome from his mother. He has no dominant X-chromosome to veil the trail, thus he expresses the trait despite it being recessive. This pattern indicates that the trait resides on the X-chromosome.

Individual 22 is male, and the trait it X-linked recessive. We know he will inherit the Y-chromosome from the unknown father, and a singular X-chromosome from the affected mother. Because the mother is affected, we know she must have two affected X-chromosomes. No matter which chromosome is passed to individual 22, he will inherit the trait.

We can tell from the pedigree that the trait is X-linked recessive. Individual 18 is female; her mother is a carrier, and her father is affected. She will receive one affected paternal X-chromosome (100%) and has a 50% chance of received an affected maternal copy. The probability of her being affected is given by the calculation.

We can tell from the pedigree that the trait is X-linked recessive. Individual 18 is female; her mother is a carrier, and her father is affected. She will receive one affected paternal X-chromosome (100%) and has a 50% chance of received an affected maternal copy. The probability of her being a carrier is given by the calculation.

We can tell from the pedigree that the trait is X-linked recessive. Individual 18 is female; her mother is a carrier, and her father is affected. She will receive one affected paternal X-chromosome (100%) and has a 50% chance of received an affected maternal copy. Because we know there is a 100% probability of her receiving at least one affected chromosome (from the father), there is a 0% chance that she will be unaffected. She must be either affected or a carrier.

We can tell from the pedigree that the trait is X-linked recessive. Individual 12 is a female carrier, meaning she must have the genotype , where is the affected allele. Individual 17 is an unaffected male, meaning he must have genotype . If this couple has a son, there is a 100% chance he will inherit the Y-chromosome from his father, and a 50% chance of receiving the affected allele from the mother; thus, there is a 50% chance that the son will be affected.

We can tell from the pedigree that the trait is X-linked recessive. The mother, individual 20, will have the genotype , where represents the recessive trait. If individual 20 is unaffected, then he has the genotype . The daughter, individual 21, will receive the unaffected paternal X-chromosome and an affected maternal X-chromosome; thus, there is a 100% probability that she will be a carrier.

When a disease is X-linked dominant, a heterozygous mother crossing with a healthy father will result in four different progeny possibilities: a healthy or sick daughter, or a healthy or sick son. In this scenario, sons and daughters both have a 50% probability of having the disease.

Suppose X is a healthy allele and X^A is an affected allele.

Heterozygous mother crossed with healthy father: XX^A x XY

Child 1: XX (daughter, healthy)

Child 2: XX^A (daughter, affected)

Child 3: XY (son, healthy)

Child 4: X^AY (son, affected)

In order for the offspring to be hemophilic, they must not possess the dominant allele, either from the father or the mother. The cross can be given as X^HX x XY, where X^H is the affected allele and X is the healthy allele. Sons necessarily inherit the Y chromosome from the father, giving the sons possible genotypes of X^HY or XY; the sons have a 50% chance of being affected. Daughters necessarily inherit the unaffected X chromosome from the father, giving them the possible genotypes of X^HX or XX; the daughters cannot be affected, but have a 50% chance of being carriers.