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HW 7 Solutions

HW 7 Solutions - HW7Solutions 1 a)/nMn b s= c A a1 a 2 A a1...

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HW 7 Solutions 1. a) Both a 1 and a 2 should have the unit of 1/nM n b) ) 1 )( 1 ( ) ( ) 1 ( ) ( ) 1 /( ] ) 1 ( ) 1 [( ln ln 2 1 2 1 2 2 2 1 2 2 2 1 1 2 A a A a a a A A a F a a A A a a A a a A a F A A F F A A F s + + = + = + + + = = = c)

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2. a) 12 1 1 2 1 [] [ ] [ ] and ,or[ ] and[ ] [ ] AO BO BOK KK A O O AO BO K B == = = 2 2 10 1 202 01 0 2 1 02 0 20 2 00 2 S i n c e [ ][ ][ ]a n d [ ][ ][ ] , [ ]([ ] [ ]) ([ ] [ ]) a n d [ ] [ ] Thus, [ ] and [ ] [ ] [ ] Let [ ] , OO A O B OOO AO A O O OK AO O KB KO AO O KA rkA OO k KAKB kkOK r =− −− ++ ⎛⎞ ⎜⎟ ⎝⎠ = 2 0 . ( [ ])( [ ]) kA = b) Then ; so ( [ ])( [ ]) AB r c) A , wherem is mRNAof A protein and isdegradation rateconstant of the mRNA [ ] [ ], where is translation rate constant Using QSSA: 0 [ ], Thus [ ] Therefore, A mA m pA A p A m p m dm rdm dt d dA km dA k dt r m d kr d dt d = ( [ ])( [ ]) p AA m kkA Ad A dK AK A d) In Figure 4, one can see that there are two steady states (one for each intersection of the blue and red lines). The steady state at (0,0) is unstable because a slight increase from (0,0) gives rise to a greater increase in A. In contrast, the other steady state is stable because a small decrease from the steady state gives rise to a corresponding increase in A (synthesis > decay). Likewise, a small increase from the
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HW 7 Solutions - HW7Solutions 1 a)/nMn b s= c A a1 a 2 A a1...

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