A double-reciprocal plot of 1/V0 versus 1/[S] for an enzyme in the presence of increasing concentrations of an uncompetitive inhibitor will have lines corresponding to the different inhibitor concentrations that are BEST described by which statement?a) The lines will intersect at the x-axis to the left of the y-axis. b) The lines will cross the y-axis and will be parallel to each other. c) The lines will intersect to the left of the y-axis but above the x-axis. d) The lines will intersect below the x-axis, to the right of the y-axis. e) The lines will be parallel to each other but will not cross the y-axis.

The answer is b) The lines will cross the y-axis and will be parallel to each other.

Explanation:

In the attached image you can see the double reciprocal graph where non-competitive inhibition is described. This non-competitive inhibition is characterized in that the inhibitor can only bind to the enzyme substrate complex.

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