Question

Please elaborate on the calculations

Question 24 Based on the observed effects, what is the minimum dose of Compound X required to reduce the HCN current to half the normal current? A. 5 x 10-4 umol X B. 1 x 102 umol C. 5 x 10-2 umol D. 5 x 10-2 mol 1.0 0.8 0.6 0.4 0) -8 -7 -6 -5 -4 log(XD Figure 3 Inhibition of HCN currents by addition of compound X Key terms: test HCN blocking, concentration effect Cause and effect: Figure 3 shows that as concentration of X increases, the Ih is inhibited more; the y-axis is (change in current)/(max current) C is correct. A closer look at Figure 3 shows that the y-axis represents the inhibition of lh as a proportion of lhmax. The question is asking us to find the X dosage that would cut the lh in half. On Figure 3 we can see that Δ1Mmax = 0.5 at approximately 10-5 M on the x-axis. The passage tells us that solutions of 5 mL were prepared. To find the number of moles, we simply multiply volume by concentration: 105 M x 5 x 10-3 L- 5 x 10-8 mol 5 x 108 mol x (1 umol/ 10-6 mol)-5 x 102 umol

Answer 1

According to the given figure:

5 6 8 6420 08 1000 (mtiplIV

The inhibition of the HCN current as a function of concetration of compound X is shown in the plot. Notice that the values for change in current respect to the maximum current (ΔI/I_h) are increasing when [X] is increasing too.

To reduce the HCN current to half the normal current, we must take into account the middle point on the curve given in the plot, as you can see:

Normal current 1.0 0.8 0.6 Half of normal current 0.4 0.2 -8 .7 -6 -5 .4 log(XD

The half of normal current corresponds to a value of Log([X]) = -5. Solving it for the concentration of X:

$[X]=10^{-5}M$

This is the minimum concentratof of X compound to reduce to half the value of normal current of HCN. For 5 mL of sample, we can calculate the number of moles accoring to the definition of molarity:

n(sto) V(sln)

Where

M is the molarity (mmol/mL)

n_(sto) is the number of moles of the solute (mmol)

V_(sln) is the volume of the solution (mL)

Solving for the number of moles of solute, we have:

M * 2 (stoj (sin)

Substituting known values for compound X:

10-5M * 5mL = n(sto)

510-mmol -

Converting this result into μmol units from the conversion factor.

1.0m mol-1000μ mol

We have the minimum amount of compound X is:

1000μ 77201 n(x) = 5 * 10-5mmol * 1.0mmol

っ* 10-11.7720

Finally, correct option is C.