A haploid cell has only one set of chromosomes as compared to the two sets normally found in diploid cells. In a haploid cell there are 22 autosomes and 1 sex chromosome for a total of 23 chromosomes.   

Haploid cells have 23 chromosomes each (22 autosomes and 1 sex chromosome) so when they fuse the resulting cell will have 46 chromosomes (44 autosomes and 2 sex chromosomes). This idea is demonstrated during fertilization, where two haploid cells (gametes) fuse to form the zygote, which is diploid.

A somatic cell is a diploid cell so it will have 22 pairs of autosomes and one pair of sex chromosomes for a total of 46 chromosomes.  

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).