1). Prokaryotes and eukaryotes use several methods to regulate gene expression, but the most common method is
a). translational control.
b). transcriptional control.
c). posttranscriptional control.
d). control of mRNA passage from the nucleus.
2). The two protein subunits of the leucine zipper are held together
a). in the shape of a Y.
b). by the interaction of leucine amino acids.
c). by hydrophobic interactions.
d). All of these are correct.
3). The helix-turn-helix motif contains two helical segments, and in order for the motif to bind DNA, the ______________ fits into the major groove of the DNA.
b). recognition helix
c). zinc finger
d). leucine zipper
4). A(n) _______________ is a piece of DNA with a group of genes that are transcribed together as a unit.
5). What effect would the addition of lactose have on a repressed lac operon?
a). The operator site on the operon would move.
b). It would reinforce the repression of that gene.
c). The lac operon would be transcribed.
d). It would have no effect whatsoever.
6). A type of DNA sequence that is located far from a gene but can promote its expression is a(n)
d). TATA box.
7). Which of the following is not found in a eukaryotic transcription complex?
d). TATA-binding protein
8). DNA methylation of genes
a). inhibits transcription by blocking the base-pairing between methylated cytosine and guanine.
b). inhibits transcription by blocking the base-pairing between uracil and adenine.
c). prevents transcription by blocking the TATA sequence.
d). makes sure that genes that are turned off remain turned off.
9). Which of the following are not matched correctly?
a). RNA splicing-occurs in the nucleus
b). snRNP-splicing out exons from the transcript
c). poly-A tail-increased transcript stability
d). All are matched correctly.
10). Which of the following is not a method of posttranscriptional control in eukaryotic cells?
a). processing the transcript
b). selecting the mRNA molecules that are translated
c). digesting the DNA immediately after translation
d). selectively degrading the mRNA transcripts
Test Your Visual Understanding
1). Match the following descriptions with the appropriate lettered panels in the figure, and explain what would be needed to activate the operon if it is not activated.
i). Operon is OFF because lac repressor is bound.
ii). Operon is OFF because CAP is not bound.
iii). Operon is ON because CAP is bound and lac repressor is not.
iv). Operon is OFF both because lac repressor is bound and CAP is not.
(a)--ii). Operon is OFF because CAP is not bound.
The repressor is not blocking the operator, so to activate the operon the CAP must be bound to the CAP-binding site
(b)--iv). Operon is OFF both because lac repressor is bound and CAP is not.
In order for the operon to be active the repressor cannot be bound to the operator because it blocks the RNA-polymerase binding site. Also, the CAP must be bound to help in exposing the RNA-polymerase binding site. Therefore, to activate this operon, the repressor must be removed with lactose and the CAP must be bound due to low glucose levels.
(c)--i). Operon is OFF because lac repressor is bound.
The CAP is bound to its binding site but the repressor is also bound at the operator site. Therefore, to activate this operon, the repressor must be removed with lactose.
(d)--iii). Operon is ON because CAP is bound and lac repressor is not.
This operon is activated because low glucose levels allows the CAP to bind to its site, helping expose the RNA-polymerase binding site and also the presence of lactose keeps the repressor from binding to the operator site.
Apply Your Knowledge
1). A method of posttranscriptional control is the selective degradation of mRNA transcripts. A transcript encoding a growth factor contains the terminal sequence of AAGCUUGAAU and has a half-life of 40 minutes. Another transcript that encodes an immunoglobin has a terminal sequence of GGAUCGCCAGG and has a half-life of about 2 hours. The half-life is related to the rate of degradation by the equation: t1/2 = 0.693/K, where t1/2 is the half-life, and K is the rate of degradation. Compare the degradation rates of the two transcripts.
The degradation rate of the growth factor transcript is:
40 min. = 0.693/K or K = 0.693/40
K = 0.0173 per minute
The degradation rate of the immunoglobin transcript is:
120 min. = 0.693.K or K = 0.693/120
K = 0.00578 per minute
2). All human beings have a rich growth of E. coli bacteria in their large intestine. Will the lac operon in the bacteria present in a lactose-intolerant individual who is careful never to consume anything with lactose (milk sugar) be activated or repressed? Explain.
Answer: The lac operon will not be needed if there is no lactose and so it will be repressed. The E. coli bacteria don't need lactose to survive. They use lactose as a source of glucose. If there is no lactose present in the intestine but an alternate source of glucose, the bacteria will survive on the alternate glucose source.