To answer this question we need to set up punnett squares for each potential cross. The unknown purple flower can be represented as A_ because we know it must contain at least one dominant allele to show purple coloration.

An A_ x AA would yield only purple flowers, which would not be useful.

An A_ x A_ could yield either all purple offspring or a 3:1 purple to white ratio; however, this cross would not answer our question because we wouldn't know which unknown purple flower was homozygous and which was heterozygous if the first result is achieved.

The only option that gives a definitive result is A_ x aa. If the unknown is homozygous, all offspring will be purple; if it is heterozygous, we will see a 1:1 purple to white ratio.

Let’s assign the letters A and B for the genes that code for seed shape and seed color, respectively. A dominant allele is assigned the capital letter and the recessive allele is assigned the lower case letter.

In order to produce the observed F2 generation, all F1 plants must be heterozygous for both traits. This means that the F1 generation has a genotype of AaBb and the test cross has a genotype of aabb. Remember that the test cross will always be against a homozygous recessive individual. The results of the F1 test cross would be:

F1: AaBb x aabb

F2: AaBb, Aabb, aaBb, aabb

If the yellow phenotype is dominant and the wrinkled phenotype is recessive, then our target F2 plant has a genotype of aaBb. Using this genotype in a test cross, we can find our answer.

F2: aaBb x aabb

Offspring: half aaBb (wrinkled/yellow) and half aabb (wrinkled/green)

The probability of a wrinkled green offspring is 0.5, or 50%.

Finding probabilities of an offspring from a complex cross (polyhybrid crosses) can also be done by multiplying the probability of each individual trait. In this case, the probability of wrinkled is 1 (100%) and the probability of green is 0.5 (50%), resulting in the 0.5 probability. If each gene independently assorts, then you can find individual probabilities of each trait in a genotype and multiply them together.

The main purpose of a test cross is to determine unknown genotypes of previous generations. A specimen with unknown genotype is crossed with a specimen of known genotype, and the phenotypes of the offspring are observed to determine the heritability ratios of the cross. Statement I is true.

When a researcher employs a test cross, he is crossing an unknown individual with a homozygous recessive individual, not homozygous dominant. This allows all traits from the unknown individual to be expressed in the offspring. If a homozygous dominant individual were used, then all offspring would show the dominant phenotypes and the cross would be useless. Statement II is false.

A test cross can be used for any cross; it doesn’t matter if it is monohybrid, dihybrid, or polyhybrid. If you can obtain an individual that is recessive for all the traits you are analyzing, then you can employ a test cross. Statement III is false.

In order for the woman to be a carrier, the disorder must be X-linked recessive, and the woman must be heterozygous for the allele. The information about the man's parents becomes irrelevant as soon as we know that he does not have the disorder. Males only carry one copy of the X-chromosome; thus, for the man to be unaffected, he must carry the dominant, unaffected allele.  Now we know that the woman's genotype is  and the man's is , where represented the affected allele. The Punnett square for this couple is drawn below.

Males cannot be carriers for X-linked traits, as they only carry one copy of the X-chromosome; thus, we are looking only at daughters (50%). Of the daughters, 50% will be heterozygous for the trait (). There is a one in four chance the child will be a female who is a carrier, or 25%.

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.