Download Problems 11

BS 50—Genetics and Genomics
Week of Nov 7
Additional Practice Problems for Section
Answer Key
1.
Genotype
β-galactosidase (lacZ)
Not
Induced
induced
lactose permease (lacY)
Not
Induced
induced
I + P + O+ Z + Y +
0
+
0
+
I + P – O+ Z + Y +
0
0
0
0
I s P + O+ Z + Y –
0
0
0
0
I – P – O+ Z + Y –
0
0
0
0
I + P + Oc Z + Y –
+
+
0
0
I – P + Oc Z + Y +
+
+
+
+
I + P + O+ Z – Y + / I + P + O+ Z + Y –
0
+
0
+
I + P + O+ Z – Y + / I + P + Oc Z + Y –
+
+
0
+
I + P – Oc Z – Y + / I + P + O+ Z + Y –
0
+
0
0
I s P + O+ Z – Y + / I + P + O+ Z + Y –
0
0
0
0
I + P – O+ Z + Y – / I – P + O+ Z – Y +
0
0
0
+
I + P – Oc Z + Y – / I – P + O+ Z – Y +
0
0
0
+
Now consider any difference in the above partial diploids depending upon which of the
lac operons is located on the bacterial chromosome and which on the F’ plasmid. Think
of this over the course of time. Does the expression and activity of the structural genes
change over time? Why or why not?
If a wild-type repressor gene (I+) is contained on an F’ plasmid which is
used to transfer DNA to an I– cell, there will be a delay in repression until
there is enough time for the repressor to be transcribed and translated. So
for instance, transfecting an I– P+ O+ Z– Y+ cell with an F’ plasmid with
genotype I+ P+ O+ Z+ Y–, you will get production of β-gal and permease
until enough repressor is made to block transcription.
2. The activity of the enzyme β-galactosidase produced by cells containing certain
mutations was measured (in relative units) when the cells were grown in media
supplemented with different carbon sources. (Glycerol can be metabolized by these cells
but is not a sugar.)
wildtype
phenotype 1
phenotype 2
phenotype 3
phenotype 4
glycerol
0
0
0
1000
0
lactose
1000
10
0
1000
1000
lactose + glucose
10
10
0
10
1000
Match the mutant phenotype with the mutation(s) below. (Different mutations may have
the same phenotype.) BRIEFLY explain your reasoning for each answer. (4 points each)
a. Is:
A) #2 — since the repressor does not bind lactose, levels of β -galactosidase should be
the same with and without lactose and the same as wildtype in the absence of
lactose.
b. Oc:
B) #3 — leads to absence of negative control by repressor binding but still under the
positive control of CAP-cAMP binding
c. defective CAP-cAMP DNA binding site:
C) #1 — leads to absence of positive control by CAP-cAMP binding but still under
the negative control of repressor binding
d. nonsense mutation in the lacZ gene:
D) #2 — no β -galactosidase under any conditions since the gene cannot encode a
full-length protein
e. nonsense mutation in the crp gene (encoding CAP protein)
E) #1 — same as (C); leads to absence of positive control by CAP-cAMP binding but
still under the negative control of repressor binding
3)
(15pts)
i)
ii)
iii)
(5pts) An allosteric repressor protein binds DNA in the presence of ethanol
and does not in its absence…. The binding of ethanol to repressor alters the
protein’s shape to its active form.
(5pts) Mutations in two loci (c and d) affect the control of this operon. Strain
2 indicates d- is recessive, strain 3 indicates c- is dominant. Mutations in the
operator would be cis-dominant, so c must be the operator.
(5pts) NO. Both a- and b- are recessive (strain 1). A promoter mutation
would prevent transcription and also be cis-dominant. For example, in strain
1, if a was the promoter then b would not be transcribed on the achromosome; since the other chromosome is b- no b product would be made
in this diploid and the strain would be unable to synthesize ethanol.
4.
Genotype
Lactose
ß-gal
lac I+ O+ Z+ Y+
lac I+ Oc Z+ Y–
+
+
lac I– O+ Z– Y+
F´ lac I+O+Z–Y– / lac I–O+Z+Y+
F´ lac I+OcZ–Y+ / lac I–O+Z+Y–
F´ lac I–O+Z–Y+ / lac I+OcZ+Y–
+
+
+
Glycerol
permease
ß-gal
Glucose + IPTG
permease
ß-gal
permease
+
+
+
+
+
+
+
+
+
Even though IPTG is an inducer, the presence of glucose prevents the binding of CAP so
high levels of transcription (and product) will not occur with any of these genotypes.