PROBLEM SET

1 Experimental Background

E. coli produces beta-galactosidase when lactose is present. X-gal is a colorless, lactose-like
molecule that can be split into two fragments by beta-galactosidase. One of these product
molecules is blue. We identified two genes, A and B, that might be involved in the lactose
processing in E. coli. Using CRISPR/Cas9 techniques, we sufficiently produced different cell
lines that had double-stranded breaks (DSBs) in genes A and B. After recovery, we plated the
modified colonies on agar, containing glucose and corresponding treatment.

Table 1. Results. White and blue demonstrate the quantification of colored colonies.

TreatmentControlGene A DSBGene B DSBGene A/B DSB
WhiteBlueWhiteBlueWhiteBlueWhiteBlue
No Lactose+++– –+++++++++
Lactose+++– –+++++++++
No Lactose + X-gal+++– –++++++++++
Lactose + X-gal++++++++++++++

Table 2. List of possible genes by mass spec.

Lac-AThiogalactocide transcetylase
Lac-IRegulator gene
Lac-OOperator gene
Lac-PPromoter gene
Lac-YLactose permease
Lac-ZBeta-galactosidase

Furthermore, studies have found out that there might be a competing interaction between the
usage of glucose and galactose. We identified an important gene in this regulatory pathway, and
we performed a DSB within this gene.

Table 2. Results. Blue colonies on agar plates were counted and quantified under the Control and Gene C DSB column.

cAMP levelControlGene C DSB
Glucose +/ Lactose –
Glucose -/ Lactose +++++++
Glucose +/ Lactose ++

Figure 1. Experimental Setup.

2 Problem Set

Question 1: Based on the Table 1 and Figure 1 provided, speculate the functions of gene A and B and give the name of the gene. If possible, draw the corresponding construct, i.e., the gene and the location of the DSB.

Question 2: Why are there fewer blue colonies in the Glucose+/Lactose+ group than in the Glucose-/Lactose+ group?

Question 3: Based on the different results in Experiment One and Experiment Two, what is the fundamental difference between Gene A/B and Gene C? (Hint, think from the gene regulation, rather than the outcomes)

Solution

  1. Gene A: Lac-Z, Beta-galactosidase
    Gene B: Lac-I, regulatory control of the expression of Lac-Z
  1. Glucose is preferred over lactose. The agar plates contains X-gal, which will turn blue if
    bacteria are actively transcribing beta-galactosidase. Thus, more blue colonies means more
    lactose usage. Glucose+/Lactose+ shows fewer blue colonies, meaning that less lactose is
    used in the presence of glucose. One might speculate that the reason behind this is that
    bacteria use glucose more readily. Lactose needs to be converted to glucose and galactose
    before it can be used by bacteria.
  1. Gene A/B is a negative regulator (repressor), whereas Gene C is a positive regulator (ac-
    tivator).
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