The man's genotype can be written as , where  represents the allele with the disorder. The woman, who has a normal genotype, is therefore .

Their possible children are represented by the Punnett square shown here:

Any sons will inherit the chromosome from the father and an unaffected from their moth. Any daughters will inherit chromosomes from both parents. By necessity, all daughters will inherit an normal from the mother and an affected  from the father. All of the couple's daughters will be carriers, while all of their sons will be phenotypically normal; therefore, there is a 50% chance that their child will be a carrier.

For two children, the probability is .

Since B is completely dominant over b, and you obtained a 3:1 ratio of phenotypes, the parents must have been heterozygous, therefore you can represent their genotypes as Bb.

If either parent were BB, you would obtain only brown offspring. If either parent were bb, they would have been white (rather than bronw).

Because the plants are homozygous dominant for color, you need only consider seed shape. Crossing a heterozygous plant with another heterozygous plant will always lead to 75% of offspring in the F1 generation displaying the dominant trait.

All of the offsrping will also be yellow, though this does not directly relate to the answer.

When examining two autosomal dominant traits, a cross between two heterozygous individuals will always produce F1 offspring in the same 9:3:3:1 ratio. Nine of the sixteen are individuals who are dominant for both traits, six are individuals who are dominant for one trait and recessive for the other, and the single individual is recessive for both traits. Three individuals will be yellow with wrinkled seeds, and three will be green with round seeds.

Since one of the parent has blue antennae, we know for that they have a homozygous recessive genotype (bb). The information given in the question does not tell us whether or not the parent with green antennae is homozygous dominant (BB) or heterozygous (Bb). The cross could be set up as either BB x bb or Bb x bb. These two crosses would give you a 0% chance and a 50% chance of having a child with blue antennae, respectively.

BB x bb: Offspring will all be Bb, and cannot show the blue phenotype.

Bb x bb: Half of the offspring will be Bb and half will be bb. Half of the offspring will show the blue phenotype.

If the dog with the unknown genotype is heterozygous, our cross would be: Aa x aa. In this model, A represents the dominant yellow-fur allele and a represents the recessive black-fur allele. We the black furred dog has to be genotype aa because he displays the recessive phenotype. The offspring produced from this cross would be 2 Aa and 2 aa genotypes, based on a punnet square.

This cross would generate offspring that are 50% Aa and 50% aa. This corresponds to 50% yellow and 50% black.

First, we need to find the genotype of the man. We know that his father has the disease, and that his mother is homozygous dominant. For his father to display the autosomal recessive trait, he must be homozygous recessive. We can thus write the man's parents' genotypes, using A as the dominant allele and a as the recessive allele.

Parents: AA (mother) x aa (father)

Offspring: All offspring will be Aa.

The man must inherit one allele from each parent, meaning he must be heterozygous.

This man, we are told, marries a heterozygous woman and they have a child. Again, we can write out this cross.

Man and woman: Aa (man) x Aa (woman)

Offspring: 1 AA (dominant), 2 Aa (dominant), 1 aa (recessive)

The possible offspring from this cross show a three-to-one ratio for displaying the dominant phenotype. Only one of the four possible offspring will be homozygous recessive and display the recessive phenotype. This one-in-four chance correspond to a 25% chance.

We know that the allele is autosomal, the father is heterozygous and black, and the mother is homozygous and white. From this information, we can determine that black must be dominant to white since the heterozygote shows the black phenotype. We will use B to represent the dominant black allele and b to represent the recessive white allele.

Cross: Bb x bb

Offspring: Half Bb (black) and half bb (white)

We can see that roughly half of the offspring will show the white phenotype. If there are twelve offspring, six of them will be white.

The first cross will produce a generation of double heterozygous flowers. Self-crossing this generation becomes a dihybrid cross problem.

In order to solve this problem, you can use either a punnett square and count the squares that display this phenotype, or you can use probability. Since purple is dominant, three out of four flowers will display the color purple. Since wrinkled is recessive to round, only one out of four flowers will display this phenotype. Since the flowers in question must express both of these traits, we can simply multiply these two fractions together.

You should be familiar with the 9:3:3:1 ratio of dihybrid cross phenotypes. A phenotype of one dominant trait and one recessive trait will always be observed with a  frequency.

Alleles are simply different versions of the same gene that encode for variations of the same characteristic. For example, different eye colors are encoded by different alleles of the same gene.