Answer:

With a single-suited hand of 22 HCP, West opens 2C (as taught in beginner's lessons with 1NT = 15-17, 2NT = 20-21).  East has 10 HCP and knows West has no more than 4 losers.  East has one winner (AD) but needs to know more about West's hand. From here on East drives the contract.

If East bids 3D, West will bid 3S to show a 5-card suit and game-force.  East has only 2 spades.  If she bids 3NT the contract will end but she needs to call for Aces, so she will bid 4C.  Assuming standard Gerber, the response by West of 4S shows an Ace is missing. 

If East bids 2D (waiting bid), West will then bid 2S to show a 5-card suit and game-force. East then bids 3D to show long diamonds.  This tells nothing about her HCP.  At this stage West bids 3NT and East then bids 4C as above. Note that West's bid of 3NT denies 6 spades, so West has a balanced 5332 shape hand and therefore has 22+ HCP at least.  So East can determine from the bidding that slam in 6NT is feasible with a combined holding of 32-33 unless the missing AK happen to be in the same suit with NS. This means that hearts and clubs are at risk from East's perspective.

So how can East deduce in which suit are the missing A and K? If the AC is given to South and AH to West, the point count is the same but NS can make 2 tricks in clubs.

... and I don't know how to go on from here Nerelle. Then into control cue bidding ...