An Introduction To Genetic Analysis 11th Edition By Griffiths – Test Bank

Chapter 11
Regulation of Gene Expression in Bacteria and Their Viruses

MULTIPLE-CHOICE QUESTIONS

Sections 11.1, 11.2, and 11.3 (Gene regulation, Discovery of the lac system: negative control, Catabolite repression of the lac operon: positive control)

1. The product of the regulator gene of the lac operon is
A) the operator.
B) the inducer.
C) the repressor.
D) the corepressor.
E) -galactosidase.
Answer: C

2. A prokaryotic operon is composed of a series of adjacent genes under the control of:
A) the same promoter.
B) the same operator.
C) an inducible promoter.
D) the same operator and promoter.
E) None of the answer options are correct.
Answer: D

3. The enzyme -galactosidase can convert the disaccharide lactose into:
A) allolactose.
B) glucose.
C) allolactose and glucose.
D) allolactose, glucose, and galactose.
E) sucrose and glucose.
Answer: D

4. The lac repressor protein controls expression of the lac operon by binding to the:
A) lac structural genes to repress expression.
B) lacZ and lacY genes only to repress expression.
C) lac operator site to repress expression.
D) lac promoter site to repress expression.
E) All of the answer options are correct.
Answer: C

5. The lac repressor (LacI) binds to:
A) lactose and DNA.
B) RNA polymerase.
C) RNA polymerase and DNA.
D) -galactosidase, permease, and transacetylase.
E) RNA and DNA.
Answer: A

6. DNA-dependent RNA polymerase binds to the:
A) repressor gene.
B) promoter.
C) operator.
D) permease gene.
E) sugar lactose.
Answer: B

7. E. coli bacteria are placed into a medium containing both glucose and lactose. Which of the gene products (i.e., enzymes) listed below do you predict will be “turned on”?
A) -galactosidase
B) lacI
C) lacP
D) permease
E) None of the answer options are correct.
Answer: E

8. A null repressor mutation (I–) results in:
A) no transcription.
B) inducible transcription.
C) transcription but no translation.
D) no translation.
E) constitutive transcription.
Answer: E

9. A constitutive operator mutation (Oc) results in:
A) no transcription.
B) inducible transcription.
C) transcription but no translation.
D) no translation.
E) constitutive transcription.
Answer: E

10. A promoter mutation (P–) results in:
A) no transcription.
B) inducible transcription.
C) transcription but no translation.
D) no translation.
E) constitutive transcription.
Answer: A

11. A super repressor mutation (IS) results in:
A) no transcription.
B) inducible transcription.
C) transcription but no translation.
D) no translation.
E) constitutive transcription.
Answer: A