If a disease is sex linked recessive, the disease resides on the x-chromosome (unless otherwise stated). The question asks about males, who genetically have one x-chromosome and one y-chromosome. The easiest way to solve this problem is to create a punnet square (shown below).  

As you can see, there is a 100% chance that the males will be affected. Similarly, there is a 100% chance that the daughters will be carriers of the disease. As a side note, the daughters will not be affected because they have one "good" x-chromosome to compensate for the "bad" x-chromosome that carries the mutation. 

Trisomy is a condition of an extra chromosome copy, such as in Patau (trisomy 13), Edwards (trisomy 18), Kleinfelters (XXY), and Downs (trisomy 21). Turners syndrome is a condition of monosomy (X).

Each homozygous pair of alleles will result in 100% identical dominant or recessive alleles, therefore cross multiple "1" for each homozygous pair and multiple "2" for each heterozygous pair. 

Therefore, the appropriate multiplication for determining the number of possible gene combinations is 

 possible combinations. 

All of the following are results of a nondisjunction event. Trisomy and monosomy disorders may occur as a result of improper separation in diploid human meiosis events.

The correct answer is 25% because each parent has a 50% chance of giving up their recessive allele to the offspring. There are two parents so the probability will be:

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.