The question states "based SOLELY on this information." The only information you have is that the round phenotype occurs naturally in the environment.

By definition, a wild type phenotype is something that occurs naturally in the environment (naturally in the "wild"). This observation makes statement III true.

By examining the given passage, we can also conclude that the yellow phenotype is dominant; however, this conclusion cannot be reached solely from the information provided in the question. The only conclusion you can make from this information is that the round phenotype is wild type. Also, remember that a wild type allele isn’t always dominant; there are recessive wild type alleles. For example, the wild type allele for Drosophila wings is recessive, whereas the mutant wrinkled wings are dominant.

Carriers are heterozygotes. In the case of recessive inheritance, a carrier will not know that they are carrying a single allele for the disease. In the Punnett Square, crossing two heterozygotes will result in 25% homozygous dominant and 50% heterozygous carriers, with 25% homozygous recessive who will have cystic fibrosis

There is a 25% chance that a child, regardless of sex, will have the disease. Sickle cell anemia is an autosomal recessive trait, so female and male offsprings are affected equally. Two carriers that mate will have a 25% chance of having a child with the disease, 50% chance of having a child who is also a carrier, and 25% chance of having a child who is normal, meaning he/she has two normal alleles. This can be seen by drawing a Punnett square for two heterozygous parents.

First, we know that muscular dystrophy must be recessive, as the grandmother in the question is a carrier.

Below is the Punnett Square for the grandparents, where the mother is a carrier and father is unaffected. X represents an unaffected allele, while X' represents the allele for dystrophy. The probability their daughter is a carrier is 50%, and the probability she is unaffected is 50%.

If the daughter is a carrier, the Punnett square between her and an unaffected male (the father) would be identical to that of the grandparents. In this case, the chance of the son being affected is 50%.

If the daughter is not a carrier, then her son cannot be affected.

In order for the son to be affected, two events must take place: the daughter must be a carrier (50% chance), and the son must inherit the affected allele (50% chance).

There is a 25% chance that the son will be affected; thus, there is a 75% chance that he will be unaffected.

A test cross takes place when an unknown flower is crossed with a flower that is homozygous recessive for the trait. This is used to determine if the unknown flower is heterozygous or homozygous dominant. In this case, a flower recessive for the color trait will be white.

Crossing with a known homozygous recessive flower allows us to determine if the purple flower is homozygous dominant or heterozygous. If all offspring are purple, then the purple parent is homozygous dominant. If any white offspring are produced, then the purple parent must carry the white allele and be heterozygous.

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.