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Inquiry Questions

None for this chapter.

Self Test

1). The bases of RNA are the same as those of DNA with the exception that RNA contains
    a). cysteine instead of cytosine.
    b). uracil instead of thymine.
    c). cytosine instead of guanine.
    d). uracil instead of adenine.
Answer: b

2). Which one of the following is not a type of RNA?
    a). nRNA (nuclear RNA)
    b). mRNA (messenger RNA)
    c). rRNA (ribosomal RNA)
    d). tRNA (transfer RNA)
Answer: a

3). Each amino acid in a protein is specified by
    a). several genes.
    b). a promoter.
    c). an mRNA molecule.
    d). a codon.
Answer: d

4). The three-nucleotide codon system can be arranged into ______________ combinations.
    a). 16
    b). 20
    c). 64
    d). 128
Answer: c

5). The TATA box in eukaryotes is a
    a). core promoter.
    b). – 35 sequence.
    c). – 10 sequence.
    d). 5' cap.
Answer: a

6). The site where RNA polymerase attaches to the DNA molecule to start the formation of RNA is called a(n)
    a). promoter.
    b). exon.
    c). intron.
    d). GC hairpin.
Answer: a

7). When mRNA leaves the cell's nucleus, it next becomes associated with
    a). proteins.
    b). a ribosome.
    c). tRNA.
    d). RNA polymerase.
Answer: b

8). If an mRNA codon reads UAC, its complementary anticodon will be
    a). TUC.
    b). ATG.
    c). AUG.
    d). CAG.
Answer: c

9). The nucleotide sequences on DNA that actually have information encoding a sequence of amino acids are
    a). introns.
    b). exons.
    c). UAA.
    d). UGA.
Answer: b

10). Which of the following statements is correct about prokaryotic gene expression?
    a). Prokaryotic mRNAs must have introns spliced out.
    b). Prokaryotic mRNAs are often translated before transcription is complete.
    c). Prokaryotic mRNAs contain the transcript of only one gene.
    d). All of these statements are correct.
Answer: b

Test Your Visual Understanding

1). Match the correct labels from the following list with the lettered boxes in the figure (not all labels will be used).
replication
RNA processing
transcription
translation
    a). What is this figure illustrating?
    b). Where does process A occur in the eukaryotic cell?
    c). Where does process B occur in the eukaryotic cell?
Answer:
A?Transcription
B?Translation
1a). This figure is illustrating the Central Dogma of gene expression.
1b). Transcription occurs in the nucleus of eukaryotic cells.
1c). Translation occurs in the cytoplasm of eukaryotic cells.

Apply Your Knowledge

1). Assume you had four copies per cell of mature mRNA for insulin, a protein made of two polypeptide chains with a total of 51 amino acids. Assume one ribosome can attach onto an mRNA every 20 nucleotides. One amino acid can be translated in 60 milliseconds. How many copies of insulin could be made in three minutes?
Answer: First, figure out how long it takes one ribosome to translate one molecule of insulin:
51 amino acids x 0.06 seconds (60 milliseconds) = 3.06 seconds
Next, how many insulin molecules can be translated by one ribosome in three minutes (or 180 seconds):
180/3.06 seconds = 58.8 insulin molecule
Then, figure out how many ribosomes can attach and be working at any given time:
51 amino acids = 51 codons and 51 x 3 = 153 nucleotides such that:
4 copies x 153 nucleotides/20 binding sites for ribosomes = 30.6 working ribosomes at any given time.
Therefore, one ribosome can build 58.8 molecules of insulin in 3 minutes, so 30.6 ribosomes can build 1,799 molecule of insulin in 3 minutes (58.8 x 30.6 = 1799).

2). The nucleotide sequence of a hypothetical eukaryotic gene is:
TACATACTAGTTACGTCGCCCGGAAATATC
If a mutation in this gene were to change the fifteenth nucleotide (underlined) from guanine to thymine, what effect do you think it might have on the expression of this gene?
Answer: The mRNA sequence of this gene is:
AUG UAU GAU CAA UGC AGC GGG CCU UUA UAG
The amino acid sequence for this sequence of codons is:
Met-Tyr-Asp-Glu-Cys-Ser-Gly-Pro-Leu-Stop
If the underlined guanine was changed to thymine the mRNA sequence for this codon would be UGU and the amino acid encoded by UGU is Cysteine, the same amino acid encoded by UGC, therefore there would be no effect of this mutation on the production of the protein.








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