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Modern Genetic Analysis Chapter 14. Regulation of Gene Transcription
Solved Problems

This set of four solved problems, which are similar to Problem 4 at the end of this chapter, is designed to test understanding of the operon model. Here we are given several diploids and are asked to determine whether Z and Y gene products are made in the presence or absence of an inducer. Use a table similar to the one in Problem 4 (see page 464) as a basis for your answers, except that the column headings in order will be: Z geneinducer absent; Z geneinducer present; Y geneinducer absent; Y geneinducer present.

1. | See answer

| Solution One way to approach these problems is first to consider each chromosome separately and then to construct a diagram. The following illustration diagrams this diploid: The first chromosome is P, so no Lac enzyme can be synthesized from it. The second chromosome (O+) can be transcribed, and this transcription is repressible. However, the structural genes linked to the good promoter are defective; thus, no active Z product or Y product can be produced. The symbols to add to your table are ", , , ." See question

2. | See answer

| Solution The first chromosome is P, so no enzyme can be synthesized from it: The second chromosome is O+, so transcription will be repressed by the repressor supplied from the first chromosome, which can act in trans through the cytoplasm. However, only the Z gene from this chromosome is intact. Therefore, in the absence of an inducer, no enzyme will be made; in the presence of an inducer, only the Z gene product, b -galactosidase, will be produced. The symbols to add to the table are ", +, , ." See question

3. | See answer

| Solution Because the second chromosome is P, we need consider only the first chromosome: This chromosome is Oc, so enzyme is made in the absence of an inducer, although only the Y gene is active. The entries in the table should be ", , +, +." See question

4. | See answer

| Solution In the presence of an Is repressor, all wild-type operators will be shut off, both with and without an inducer: Therefore, the first chromosome will be unable to produce any enzyme. However, the second chromosome has an altered (Oc) operator and can produce enzyme in both the absence and the presence of an inducer. Only the Y gene is active on this chromosome, so the entries in the table should be ", , +, +." See question

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