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Biology for AP® Courses

Test Prep for AP® Courses

Biology for AP® CoursesTest Prep for AP® Courses
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42 .
What characteristic of the genetic code points to a common ancestry for all organisms?
  1. The code is degenerate
  2. The code contains 64 codons.
  3. The genetic code is almost universal.
  4. The code contains stop codons
43 .
What process transfers heritable material to the next generation?
  1. replication
  2. splicing
  3. transcription
  4. translation
44 .
When comparing transcription of heritable information in prokaryotes and eukaryotes, which events are the same?
  1. Transcription by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved.
  2. Translation by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved.
  3. Transcription by polymerase, recognition of a highly variable sequence in the promoter, and termination by a hairpin loop are conserved.
  4. Transcription by polymerase, recognition of a consensus sequence in the promoter, and elongation by a hairpin loop are conserved.
45 .
Which of the following cell structures does not contain heritable information?
  1. chloroplast
  2. cytoplasmic membrane
  3. mitochondria
  4. nucleus
46 .
How does the enzyme reverse transcriptase violate the central dogma of molecular biology in HIV?
  1. The enzyme reverse transcriptase reverse transcribes the RNA in the genome of HIV to DNA.
  2. The enzyme reverse transcriptase translates the RNA of the HIV into protein and then back to DNA.
  3. The enzyme reverse transcriptase transcribes the DNA straight into the protein molecules.
  4. The enzyme reverse transcriptase transcribes DNA to RNA, then again to DNA. There is no protein synthesis.
47 .
Radioactive deoxythymidine triphosphate is supplied to the protist Euglena to use for replication. After an interval of time, the cells are homogenized and different fractions are analyzed for radioactivity content in large nucleic acid molecules. Which fraction of the cells will not be radioactively labeled?
  1. nucleus
  2. mitochondrion
  3. chloroplast
  4. plasma membrane
48 .
You sequence a gene of interest and isolate the matching mRNA. You find that the mRNA is considerably shorter than the DNA sequence. Why is that?
  1. There was an experimental mistake. The mRNA should have the same length as the gene.
  2. The mRNA should be longer than the DNA sequence because the promoter is also transcribed.
  3. The processed mRNA is shorter because introns were removed.
  4. The mRNA is shorter because the signal sequence to cross the nuclear membrane was removed.
49 .
A mutation in the promoter region of the gene for the beta-globin can cause beta-thalassemia, a hereditary condition which causes anemia. Why would mutations in the promoter region lead to low levels of hemoglobin?
  1. The globin chains produced are too long to form functional hemoglobin.
  2. The globin chains are too short to form functional hemoglobin.
  3. Fewer globin chains are synthesized because less mRNA is transcribed.
  4. Globin chains do not fold properly and are non-functional.
50 .

A table of with the codon and what amino acid it codes for, which correlate as follows: GCA - alanine, AAG - lysine, GUU - valine, AAU -asparagine, UGC - cysteine, UCG and UCU - serine, UUA - Leucine, and UAA - stop.

You are given three mRNA sequences:

  1. 5’-UCG-GCA- AAU-UUA -GUU-3’
  2. 5’-UCU-GCA- AAU-UUA -GUU-3’
  3. 5’-UCU-GCA- AAU-UAA -GUU-3’

Using the table, write the peptide encoded by each of the mRNA sequences.

    1. Serine-alanine-asparagine-leucine-valine
    2. Serine-alanine-asparagine-leucine-valine
    3. Serine-alanine-asparagine(-stop)
    1. Serine-phenylalanine-asparagine-leucine-valine
    2. Serine-alanine-asparagine-leucine-valine
    3. Serine-alanine-asparagine (-stop)
    1. Serine-alanine-asparagine-leucine-valine
    2. Serine-alanine-asparagine (-stop)
    3. Serine-alanine-asparagine-leucine-valine
    1. Serine-alanine-asparagine-leucine-valine
    2. Serine-arginine-asparagine-leucine-valine
    3. Serine-alanine-asparagine(-stop)
51 .

A table of with the codon and what amino acid it codes for, which correlate as follows: GCA - alanine, AAG - lysine, GUU - valine, AAU -asparagine, UGC - cysteine, UCG and UCU - serine, UUA - Leucine, and UAA - stop.

You are given three mRNA sequences:

  1. 5’-UCG-GCA- AAU-UUA -GUU-3’
  2. 5’-UCU-GCA- AAU-UUA -GUU-3’
  3. 5’-UCU-GCA- AAU-UAA -GUU-3’

Using the peptide encoded by each of the above, compare the three peptides obtained. How are peptides 2 and 3 different from 1? What would be the consequence for the cell in each case?

  1. There is a silent mutation in peptide 2 and peptide 3 has a stop codon due to mutation.
  2. There is a silent mutation in peptide 3 and peptide 2 has a stop codon due to mutation.
  3. There is a different amino acid in peptide 2 and peptide 3 has a stop codon due to mutation.
  4. There isn’t a mutation in peptide 2 and peptide 3 has a stop codon due to mutation.
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