WORKED PROBLEMS

Problem 1

The fox operon, which has sequences A, B, C, and D (which may represent either structural genes or regulatory sequences), encodes enzymes 1 and 2. Mutations in sequences A, B, C, and D have the following effects, where a plus sign (+) indicates that the enzyme is synthesized and a minus sign (−) indicates that the enzyme is not synthesized.

• a. Is the fox operon inducible or repressible?
• b. Indicate which sequence (A, B, C, or D) is part of the following components of the operon. Each sequence should be used only once.

Solution Strategy

What information is required in your answer to the problem?

• a. Whether the fox operon is inducible or repressive.
• b. Which sequence represents each part of the operon.

What information is provided to solve the problem?

For each mutation, whether enzyme 1 and enzyme 2 are produced in the presence and absence of Fox.

For help with this problem, review:

Section 16.2.

Solution Steps

• a. When no mutations are present, enzymes 1 and 2 are produced in the presence of Fox but not in its absence, indicating that the operon is inducible and Fox is the inducer.

  Hint: Review Figure 16.3 for a summary of operon structure.

• b. The mutation in A allows the production of enzyme 2 in the presence of Fox, but enzyme 1 is not produced in the presence or absence of Fox, and so A must have a mutation in the structural gene for enzyme 1. With the mutation in B, neither enzyme is produced under any conditions, and so this mutation most likely occurs in the promoter and prevents RNA polymerase from binding. The mutation in C affects only enzyme 2, which is not produced in the presence or absence of Fox; enzyme 1 is produced normally (only in the presence of Fox), and so the mutation in C most likely occurs in the structural gene for enzyme 2. The mutation in D is constitutive, allowing the production of enzymes 1 and 2 whether or not Fox is present. This mutation most likely occurs in the regulator gene, producing a defective repressor that is unable to bind to the operator under any conditions.

Problem 2

A mutation occurs in the 5′ UTR of the trp operon that reduces the ability of region 2 to pair with region 3. What will the effect of this mutation be when the tryptophan level is high? When the tryptophan level is low?.

Solution Strategy

What information is required in your answer to the problem?

The effect of the mutation when tryptophan is high and when it is low.

What information is provided to solve the problem?

• The mutation occurs in the 5′ UTR of the trp operon.
• mutation reduces the ability of region 2 to pair with region 3.

For help with this problem, review:

Section 16.3.

Solution Steps

When the tryptophan level is high, regions 2 and 3 do not normally pair, and therefore the mutation will have no effect. When the tryptophan level is low, however, the ribosome normally stalls at the tryptophan codons in region 1 and does not cover region 2, and so regions 2 and 3 are free to pair, which prevents regions 3 and 4 from pairing and forming a terminator, which would end transcription. If regions 2 and 3 cannot pair, then regions 3 and 4 will pair even when tryptophan is low and attenuation will always take place. Therefore, no more tryptophan will be synthesized even in the absence of tryptophan.