The primary RNA contains introns and exons because it has not been processed yet, and therefore the introns have not been spliced out. Mature mRNA contains only exons, which are the coding sequences that ultimately get translated. Intron regions are non-coding and are not included in mature transcripts. Note that post-translational modifications such as splicing only occurs in eukaryotes.

This question requires you to know that preRNA contains both intronic and exonic regions, but the introns get spliced out to produce the mRNA. Therefore, you had to subtract the total intron basepairs (250) from the total length of the preRNA (1200), which gives an mRNA length of 950 basepairs. 

The promoter is a specific segment of DNA that signals the starting point of transcription. RNA polymerase attaches to the promoter and proceeds to create the mRNA primary transcript.

DNA polymerase binds to the RNA primer to begin DNA replication. Ribosomes bind to the 5' cap on eukaryotic mRNA.

The important thing to remember about the lac operon is that it is transcribed when glucose is absent from the cell, but lactose is present and can be utilized. As a result, the operon's transcription would be high if there are both high levels of lactose available, and very little amounts of glucose.

The lac operon is set up in a way so that the lac repressor is able to be transcribed, regardless of glucose and lactose levels. The lac repressor will then attach to the operator, which inhibits transcription. If lactose is present, it will bind to the lac repressor, and make it detach from the operator.

This process allows the operon to be transcribed in the event that glucose is absent. If glucose is absent, but lactose is not present, then the repressor will remain in place and transcription will not take place.

The question informs us that the mutational defect in the gene involves the enzyme's ability to load copper onto apoceruloplasmin. Healthy individuals are able to load copper to apoceruloplasmin, creating serum-soluble ceruloplasmin. With this process disrupted in an individual with Wilson's Disease, we would expect that less ceruloplasmin would be produced because copper could not be transported. We would expect to see reduced serum levels of the complete protein, and high levels of copper building up in hepatocytes and circulatory serum.

The addition of a number of nucleotides that is not a multiple of three shifts the reading frame of the codons in the gene. One base pair was inserted early in the strand, thus shifting the codon reading frame +1 to the right.

Missense and nonsense mutations imply base pair substitutions, which did not occur in the diagram. Similarly, nothing was duplicated, and repeat expansion would require multiple repetitions of a short DNA sequence.

This question requires the knowledge that a dominant negative is expected to work in the same direction as a genetic mutant; it is a loss-of-function of the gene, by expressing a version of the gene that outcompetes the actual gene but cannot perform the proper biological process. Because we are assuming the dominant negative works perfectly, it should act just like the mutant and reduce cell number by 50%, giving us 50,000 cells. 

Transcription factors can be activated or deactivated by any number of processes occurring within the cell and nucleus, and this it not limited to phosphatases (which remove phosphate groups from proteins). All of the other answers accurately describe possible activity and function of transcription factors.

Only the first statement in this question is false. Transcription factors typically (in fact, almost always) bind upstream of the gene to enhancer or promoter regions, and are rarely found interacting with the gene's coding region itself.