AP Biology · Topic 6.8

Biotechnology Practice

Part of Gene Expression and Regulation.(IST-2.D)

Practice questions

36

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Sample questions

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  1. Sample 1difficulty 1/5

    EcoRI recognizes the palindromic sequence GAATTC and cuts both strands between G and A, generating four-nucleotide 5' overhangs.

    EcoRI recognition site 5' G A A T T C 3' 3' C T T A A G 5' Cuts produce 5' overhangs (sticky ends)

    What feature of EcoRI's cleavage pattern enables ligation between two DNA molecules cut by the same enzyme?

    • A

      EcoRI methylates the overhangs to enhance ligation.

    • B

      EcoRI cuts both strands at the same position, creating blunt ends that randomly join.

    • C

      The ends are phosphorylated by EcoRI.

    • D

      Both fragments have complementary single-stranded overhangs that can anneal.

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    Why

    EcoRI cleaves between G and A on both strands, leaving 5' AATT overhangs. Identical sticky ends from any source can base-pair and be sealed by DNA ligase, the basis of recombinant DNA cloning.

  2. Sample 2difficulty 1/5

    The plasmid map shows an origin of replication (ori), an ampicillin-resistance gene (amp^R), a multiple cloning site (MCS), and an EcoRI site.

    EcoRI amp^R ori MCS pUC-like

    Why is the amp^R gene important when cloning a foreign gene into this plasmid?

    • A

      It is the origin of replication for the plasmid.

    • B

      It cuts the foreign gene to insert it.

    • C

      It encodes the foreign protein of interest.

    • D

      It serves as a selectable marker so only transformed bacteria grow on ampicillin plates.

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    Why

    Selectable markers like amp^R allow researchers to grow only cells that took up the plasmid by plating on antibiotic-containing media. The ori drives replication; the MCS provides cloning sites.

  3. Sample 3difficulty 2/5

    Southern blot of three patients Lane 1 Lane 2 Lane 3

    Lane 2 shows two bands, while Lanes 1 and 3 each show one. What does the Southern blot most likely indicate about Patient 2?

    • A

      Patient 2 is heterozygous for two restriction-fragment alleles at this locus.

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    • B

      Patient 2 has no DNA in the lane.

    • C

      Patient 2's DNA was not digested.

    • D

      Patient 2 is homozygous and has a deletion.

    Why

    Southern blots detect specific DNA fragments by hybridization to a labeled probe. Two bands of different sizes in one lane suggest a heterozygote inheriting two distinct alleles distinguished by a restriction-site polymorphism.

  4. Sample 4difficulty 2/5

    A DNA sample run alongside a size ladder produces two visible bands.

    - (cathode, top) + (anode, bottom) 10 kb 5 kb 1 kb 0.2 kb sample

    Approximately what fragment sizes does the sample contain?

    • A

      Both bands are below 0.2 kb

    • B

      About 6 kb and about 0.5 kb

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    • C

      Cannot estimate without sequence data

    • D

      Both bands are 10 kb

    Why

    DNA migrates toward the positive electrode; smaller fragments travel farther. Comparing to the ladder, the sample bands fall between 5-10 kb (<del>6 kb) and between 1-0.2 kb (</del>0.5 kb).

  5. Sample 5difficulty 2/5

    A genetic test for sickle-cell disease uses restriction enzymes that cut the normal beta-globin allele but not the sickle (HbS) allele. The gel below shows results for four family members along with a size ladder.

    L F M C1 C2 1.3 1.0 0.7 0.3 L=ladder, F=father, M=mother, C1, C2=children (kb)

    Which children are heterozygous (carriers)?

    • A

      Both children are heterozygous.

    • B

      Only C1 is a carrier.

    • C

      Both parents are heterozygous; C1 is HbS/HbS and C2 is HbA/HbA.

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    • D

      C1 and C2 are both unaffected and homozygous normal.

    Why

    Both parents show two bands (1.3 uncut and 0.7 cut), indicating heterozygotes. C1 shows only the 1.3 uncut band (HbS/HbS); C2 shows only the 0.7 cut band (HbA/HbA).