*15.An RNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, and G = 14%.

*16.The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and identify the 5′ and 3′ ends of the RNA.

17.For the RNA molecule shown in Figure 10.1a, write out the sequence of bases on the template and nontemplate strands of DNA from which this RNA is transcribed. Label the 5′ and 3′ ends of each strand.

18.The following sequence of nucleotides is found in a single-stranded DNA template:

ATTGCCAGATCATCCCAATAGAT

Assume that RNA polymerase proceeds along this template from left to right.

19.List at least five properties that DNA polymerases and RNA polymerases have in common. List at least three differences.

20.Most RNA molecules have three phosphate groups at the 5′ end, but DNA molecules never do. Explain this difference.

21.A strain of bacteria possesses a temperature-sensitive mutation in the gene that encodes the sigma factor. The mutant bacteria produce a sigma factor that is unable to bind to RNA polymerase at elevated temperatures. What effect will this mutation have on the process of transcription when the bacteria are raised at elevated temperatures?

22.On Figure 10.5, indicate the locations of the promoters and terminators for genes a, b, and c.

23.The following diagram represents a transcription unit on a DNA molecule.

Assume that this DNA molecule is from a bacterial cell. Label the approximate locations of the promoter and terminator for this transcription unit.

24.The following diagram represents one of the Christmas-tree-like structures shown in Figure 10.3. On the diagram, identify parts a through i:

25.Provide the consensus sequence for the first three actual sequences shown in Figure 10.9.

*26.Write the consensus sequence for the following set of nucleotide sequences.

AGGAGTT

AGCTATT

TGCAATA

ACGAAAA

TCCTAAT

TGCAATT

*27.What would be the most likely effect of a mutation at the following locations in an E. coli gene?

*28.Duchenne muscular dystrophy is caused by a mutation in a gene that encompasses more than 2 million nucleotides and specifies a protein called dystrophin. However, less than 1% of the gene actually encodes the amino acids in the dystrophin protein. On the basis of what you now know about gene structure and RNA processing in eukaryotic cells, provide a possible explanation for the large size of the dystrophin gene.

29.What would be the most likely effect of moving the AAUAAA consensus sequence shown in Figure 10.19 ten nucleotides upstream?

30.Suppose that a mutation occurs in the middle of a large intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.

31.A geneticist isolates a gene that contains eight exons. He then isolates the mature mRNA produced by this gene. After making the DNA single stranded, he mixes the single-stranded DNA and RNA. Some of the single-stranded DNA hybridizes (pairs) with the complementary mRNA. Draw a picture of what the DNA–RNA hybrids will look like under the electron microscope.

*32.Draw a typical eukaryotic gene and the pre-mRNA and mRNA derived from it. Assume that the gene contains three exons. Identify the following items and, for each item, give a brief description of its function:

33.How would the deletion of the Shine–Dalgarno sequence affect a bacterial mRNA?

34.A geneticist discovers that two different proteins are encoded by the same gene. One protein has 56 amino acids, and the other has 82 amino acids. Provide a possible explanation for how the same gene can encode both of these proteins.

35. In the early 1990s, Carolyn Napoli and her colleagues were attempting to genetically engineer a variety of petunias with dark purple petals by introducing numerous copies of a gene that encodes purple petals (C. Napoli, C. Lemieux, and R. Jorgensen. 1990. Plant Cell 2:279–289). Their thinking was that extra copies of the gene would cause more purple pigment to be produced and would result in a petunia with an even darker hue of purple. However, much to their surprise, many of the plants carrying extra copies of the purple gene were completely white or had only patches of color. Molecular analysis revealed that the level of the mRNA produced by the purple gene was reduced 50-fold in the engineered plants compared with levels of mRNA in wild-type plants. Somehow, the introduction of extra copies of the purple gene silenced both the introduced copies and the plant’s own purple genes. Provide a possible explanation for how the introduction of numerous copies of the purple gene silenced all copies of the purple gene.