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.

Autosomal recessive inheritance means that in order for the Wilson's Disease phenotype to display, the person must have two copies of the mutated gene. If we represent the dominant, wild type allele as , and the mutant, recessive allele as , we can calculate the chance that a child of two carriers (each ) will have Wilson's Disease ().

Using a Punnett square, we can predict the offspring of the two carrier parents.

Child 1: ; wild type

Child 2: ; wild type

Child 3: ; wild type

Child 4: ; Wilson's disease

The chance of a child being affected is one in four. The 1:2:1 genotypic ratio should be familiar to students for heterozygous crosses.

A Mendelian gene generally has two types of alleles: dominant and recessive. An individual always contains two sets of chromosomes (homologous chromosomes) that contain the gene. If both chromosomes contain a dominant allele, then the dominant trait is expressed. If both contain the recessive allele, then the recessive trait is expressed.

If an individual carries both alleles (a heterozygous individual), then only the dominant trait is observed. This occurs because the recessive allele is silenced and is not expressed in the presence of a dominant allele. If the recessive allele is expressed in conjunction with the dominant allele, then it is called incomplete dominance. In normal heterozygous genes, however, the recessive allele is not expressed.

Each chromosome carries only one copy of each gene, and cannot accommodate two alleles for a single trait.

The passage states that the F1 generation only had round, yellow seeds. Test crossing an F1 offspring lead to an equal ratio of four different phenotypes. When you test cross, you are crossing the F1 offspring with a homozygous recessive individual; therefore, the test cross individual had recessive seed shape and seed color.

???? x aabb

If round and yellow were recessive, then the F2 offspring would all be round and yellow (because you would cross round/yellow with round/yellow).

aabb x aabb (all offspring round and yellow)

If only round was recessive then you wouldn’t get any wrinkled seeds in F2 generation; similarly, if only yellow was recessive then you wouldn’t get any green seeds.

Aabb x aabb (all offspring have one recessive trait, bb)

If round and yellow are both dominant, then means that wrinkled and green are recessive. If the F1 offspring is heterozygous for both traits, then we can see the observed ratios from the test cross.

AaBb x aabb

1 AaBb (round/yellow), 1 aaBb (wrinkled/yellow), 1 Aabb (round/green), 1 aabb (wrinkled/green)

We can conclude that green and wrinkled must be recessive to yellow and round.