Atp is a (+) allosteric effector, and ctp is a (−) allosteric effector of the enzyme atcase. both of these heterotropic effectors bind to the regulatory subunits on atcase. the substrates of atcase, aspartate and carbamoyl phosphate, bind the enzyme active site with positive cooperativity (i. e., they exert a “+” homotropic effect on activity). as the concentrations of the substrates change from values where [s]≪km to values where [s] is saturating ([s]≫km), how will the binding constants for each of the two allosteric effectors change? in other words, does atp bind atcase with higher affinity when [s] is low or high? does ctp bind atcase with higher affinity when [s] is low or high? match the letters in the left column to the appropriate blanks in the sentences on the right.

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8