AMPK Methods and Protocols

21. Compare the signal from the different IPs to see how much of
    the different subunit isoforms are coprecipitated with the
    others.

200 μl total volume

• 20 μl Protein G agarose beads


• 10 μl antibody


• 40 μl lysate___
  = 130 μl IP-buffer

Mix the reagents for the IP

Incubate with rotation over night at 4°C

If the total volume is

190 μl excl. beads,

then transfer 170 μl.

Leave 20 μl to avoid

transferring beads

A

Protein G agarose. 20 μl (50:50 slurry) solution

(10 μl beads : 10 μl buffer)

Antibody against target protein (2 μg in 10 μl)

Lysate. 200 μg = 40 μl if conc. is 5 mg/ml

Spin down the beads.

Transfer supernatant (SN) to new

tube with 6xSB. This is the PPost-IP

The beads left in the

tube is the IIP

Wash beads 3x

Protein conc. in the

Post-IP / Pre-IP

After wash, add

Sample-buffer the IIP

200 mg protein in 190 ml

170 ml (SN) of 190 ml = 179 mg

179 mg in 170 ml + 34 ml 6xSB

= 0.88 mg/ml

Add 20 ml 2xSB to beads.

Precipitated protein from

200 mg lysate in 20 ml

≈10 mg/ml

B

C

Fig. 1Schematic representation of the IP procedure. (A) Example of the constituents of an IP sample with
indicated volumes of Protein G agarose, antibody, lysate, and IP buffer. (B) Separation of IP and Post-IP,
transferring the supernatant from the IP to a new tube with sample buffer. (C) Resulting protein concentration
of the Post-IP and the equivalent of the IP for the precipitated protein

Analysis of Heterotrimeric Complex Stoichiometry 209