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Test Prep for AP® Courses

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41 .
Which of the following is found in both prokaryotes and eukaryotes?
  1. 3’ poly-A tails
  2. 5’ caps
  3. promoters
  4. introns
42 .
The enzyme ployadenylate polymerase catalyzes the addition of adenosine monophosphate to the 3’ ends of mRNAs to form a poly-A tail. If the enzyme were blocked so that it could not function, the result would be:
  1. increased mRNA stability in eukaryotes, and decreased mRNA stability in prokaryotes
  2. decreased mRNA stability in eukaryotes, and no effect in prokaryotes
  3. no effect in eukaryotes, and increased mRNA stability in prokaryotes
  4. no effect in eukaryotes, and decreased mRNA stability in prokaryotes
43 .
Describe two ways in which gene regulation differs and two ways in which it is similar in prokaryotes and eukaryotes.
  1. Prokaryotes show co-transcriptional translation whereas eukaryotes perform transcription prior to translation; in both cell types, regulation occurs through the binding of transcription factors, activators, and repressors.
  2. Prokaryotes perform transcription prior to translation whereas eukaryotes show co-transcriptional translation (the processes occur in the same organelle).
  3. Prokaryotes show co-transcriptional translation that is regulated prior to translation whereas eukaryotes perform transcription prior to translation that is regulated only at the level of transcription. In both domains, transcription factors, activators, and repressors provide regulation.
  4. Prokaryotes show co-transcriptional translation that occurs in the nucleus whereas eukaryotes show transcription prior to translation. In both cell types, regulation occurs using transcription factors, activators, and repressors.
44 .
Lactose digestion in E. coli begins with its hydrolysis by the enzyme β -galactosidase. The gene encoding β -galactosidase, lacZ, is part of a coordinately regulated operon containing other genes required for lactose utilization. Which of the following figures correctly depicts the interactions at the lac operon when lactose is not being utilized?
  1. A short rectangular piece is divided into five segments with the left-hand two segments being longer than the three segments to their right. From left to right, the segments are labeled promoter, operator, and italicized lac Z, lac Y, and lac A. An oval labeled R N A polymerase is attached to the bottom of the promoter and has an arrow pointing to the right. A rectangular repressor has a small circle labeled lactose attached.
  2. A short rectangular piece is divided into five segments with the left-hand two segments being longer than the three segments to their right. From left to right, the segments are labeled promoter, operator, and italicized lac Z, lac Y, and lac A. A rectangular repressor is bound to the bottom of the operator and has a small circle labeled lactose attached.
  3. A short rectangular piece is divided into five segments with the left-hand two segments being longer than the three segments to their right. From left to right, the segments are labeled promoter, operator, and italicized lac Z, lac Y, and lac A. An oval labeled R N A polymerase is attached to the bottom of the operator and has an arrow pointing to the right.
  4. A short rectangular piece is divided into five segments with the left-hand two segments being longer than the three segments to their right. From left to right, the segments are labeled promoter, operator, and italicized lac Z, lac Y, and lac A. A rectangular repressor is attached to the bottom of the operator and has a circular cutout on its bottom center.
45 .
What would be the result of a mutation in the repressor protein that prevented it from binding lactose?
  1. The repressor will bind to lactose when it is removed from the operator.
  2. The repressor will bind the operator in the presence of lactose.
  3. The repressor will not bind the operator in the presence of lactose.
  4. The repressor will not bind the operator in the absence of lactose.
46 .
What type of modification might be observed in the GR gene in all newborn rats?
  1. The DNA will have many methyl molecules.
  2. The DNA will have many acetyl molecules.
  3. The DNA will have few methyl groups.
  4. The histones will have many acetyl groups.
47 .
What type of modification will be observed in the GR gene in the highly nurtured rats?
  1. The DNA will have many methyl molecules.
  2. The DNA will have many acetyl molecules.
  3. The DNA will have few methyl groups.
  4. The histones will have few acetyl groups.
48 .

The figure shows a normal gene with segments 1 through 4, a 5' UTR, and a 3'UTR. This gene has ++ expression. Below, deletion 1 is in the 5' UTR. This mutant has +++ expression. Below that, deletion 2 is in the very 5' end of the 5' UTR. This mutant has ++ expression. Below that, deletion 3 is in segment 2. This mutant has + expression. Below that, deletion 4 is in the 3' UTR. This mutant has ++ expression

The level of transcription of a gene is tested by creating deletions in the gene as shown in the illustration. These modified genes are tested for their level of transcription: (++) normal transcription levels; (+) low transcription levels; (+++) high transcription levels. Which deletion is in an enhancer involved in regulating the gene?

  1. deletion 1
  2. deletion 2
  3. deletion 3
  4. deletion 4
49 .

Five horizontal lines represent pieces of D N A with a normal gene and four genes with deletions. Each line has a thicker region with pieces labeled 1 through 4. Pieces 1 and 3 are very short and pieces 2 and 4 are approximately four times as long. The top horizontal line is labeled normal gene and has two plus symbols to the right. The second line is labeled deletion 1, has an “X” symbol just to the left of piece 1, and has three plus symbols to the right. The third line is labeled deletion 2, has an “X” symbol near its left end, and has two plus symbols. The fourth line is labeled deletion 3, has an “X” symbol over piece 2, and has a single plus symbol to the right. The fifth line is labeled deletion 4, has an “X” symbol to the right of piece 4, and has two plus symbols to the right.

The level of transcription of a gene is tested by creating deletions in the gene as shown in the illustration. These modified genes are tested for their level of transcription: (++) normal transcription levels; (+) low transcription levels; (+++) high transcription levels. Which deletion is in a repressor involved in regulating the gene?

  1. deletion 1
  2. deletion 2
  3. deletion 3
  4. deletion 4
50 .

Figure showing pre-mRNA containing sections 1 through 5, mature mRNA containing sections 1, 2, 4, and 5, and protein containing sections 2 and 4

The diagram provided shows different regions (1-5) of a pre-mRNA molecule, a mature-mRNA molecule, and the protein corresponding to the mRNA. A mutation in which region is most likely to be damaging to the cell?

  1. 1
  2. 2
  3. 3
  4. 5
51 .

Figure showing pre-mRNA containing sections 1 through 5, mature mRNA containing sections 1, 2, 4, and 5, and protein containing sections 2 and 4

What do regions 1 and 5 correspond to?

  1. exons
  2. introns
  3. promoters
  4. untranslated regions
52 .

Figure showing pre-mRNA containing a 5'UTR, segments 1 through 5, and a 3'UTR. Below, mature mRNA showing a 5' cap, 5'UTR, segments 1, 3, and 5, a 3'UTR, and a poly-A tail. Third, a mutated mature mRNA showing a 5' cap, a 5'UTR, segments 1 and 5, a 3'UTR, and a poly-A tail.

What are regions 1 through 5 in the diagram?

  1. 1, 3, and 5 are exons; 2 and 4 are introns.
  2. 2 and 4 are exons; 1,3, and 5 are introns.
  3. 1 and 5 are exons; 2, 3, and 4 are introns.
  4. 2, 3, and 4 are exons; 1 and 5 are introns.
53 .
(credit: modification of work by Mizoguchi B, Valenzuela N./PeerJ)

A research study showed a gene was alternatively spliced in two ways during embryonic development of the turtle based on environmental temperature. The image shows the two transcript variants.

What kind of alternative splicing is in this example?

  1. Exon skipping.
  2. Mutually exclusive exons.
  3. Alternative donor sites.
  4. Intron retention.
54 .

Diagram showing arrows from UV light to skin exposure to damage to DNA to p53 activated to p21 activated to apoptosis to skin peels.


The diagram illustrates the role of p53 in response to UV exposure. What would be the result of a mutation in the p53 gene that inactivates it?

  1. Skin will peel in response to UV exposure.
  2. Apoptosis will occur in response to UV exposure.
  3. No DNA damage will occur in response to UV exposure.
  4. No peeling of skin will occur in response to UV exposure.
55 .
Which of the following will not occur in response to UV exposure if a p53 mutation inactivates the p53 protein?
  1. Damage to DNA
  2. p53 activation
  3. p21 activation
  4. Apoptosis
  1. 1, 2, and 3
  2. 3 and 4
  3. 3
  4. 2, 3, and 4
56 .

Operon showing in a line: promoter, operator, trpE, trpD, trpC, trpB, and trpA

What happens when tryptophan is present?

  1. The repressor binds to the operator, and RNA synthesis is blocked.
  2. RNA polymerase binds to the operator, and RNA synthesis is blocked.
  3. Tryptophan binds to the repressor, and RNA synthesis proceeds.
  4. Tryptophan binds to RNA polymerase, and RNA synthesis proceeds.
57 .

Operon showing in a line: promoter, operator, trpE, trpD, trpC, trpB, and trpA

What happens in the absence of tryptophan?

  1. RNA polymerase binds to the repressor
  2. the repressor binds to the promoter
  3. the repressor dissociates from the operator
  4. RNA polymerase dissociates from the promoter
58 .

The figure shows Low Fixed Nitrogen with an arrow pointing to the right to a box containing Increased free calcium and -oxglutarate. Another arrow points to the right to a box containing NtcA, HetR. From this box, an arrow points to heterocyst development and PatS. PatS points to Inhibit additional heterocyst development.

Anabaena is a simple multicellular photosynthetic cyanobacterium. In the absence of fixed nitrogen, certain newly developing cells along a filament express genes that code for nitrogen-fixing enzymes and become non-photosynthetic heterocysts. The specialization is advantageous because some nitrogen-fixing enzymes function best in the absence of oxygen. Heterocysts do not carry out photosynthesis but instead provides adjacent cells with fixed nitrogen and receives fixed carbon and reduced energy carriers in return. As shown in the diagram above, when there is low fixed nitrogen in the environment, an increase in the concentration of free calcium ions and 2-oxyglutarate stimulates the expression of genes that produce two transcription factors (NtcA and HetR) that promote the expression of genes responsible for heterocyst development. HetR also causes production of a signal, PatS, that prevents adjacent cells from developing as heterocysts. Based on your understanding of the ways in which signal transmission mediates cell function, which of the following predictions is most consistent with the information given above?

  1. In an environment with low fixed nitrogen, treating the Anabaena cells with a calcium-binding compound should prevent heterocyst differentiation.
  2. A strain that overexpresses the patS gene should develop many more heterocysts in a low nitrogen environment.
  3. In an environment with abundant fixed nitrogen, free calcium levels should be high in all cells, preventing heterocysts from developing.
  4. In environments with abundant fixed nitrogen, loss of the hetR gene should induce heterocyst development.
59 .

The figure shows Low Fixed Nitrogen with an arrow pointing to the right to a box containing Increased free calcium and -oxglutarate. Another arrow points to the right to a box containing NtcA, HetR. From this box, an arrow points to heterocyst development and PatS. PatS points to Inhibit additional heterocyst development

Which of the following statements about Anabaena is false?

  1. Decreasing the concentration of free calcium ions will prevent heterocyst development.
  2. In the presence of fixed nitrogen, NtcA will not be expressed.
  3. Low fixed nitrogen levels result in increased PatS levels.
  4. A mutation in HetR that makes it nonfunctional will also allow adjacent cells to develop as heterocysts.
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