A binding protein binds to a ligand L with a Kd of 400 nM. What is the concentration of ligand when Y is (a) 0.25, (b) 0.6, (c) 0.95? Answers: (a) 133.33nm (b) 600nm (c) 7,600nm

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(a) 133.33nm

(b) 600nm

(c) 7,600nm

Explanation:

The concentration of Y can be determined by using the formula:

where;

[L] = concentration of the binding ligand.

kd = 400 nm

Thus:

When Y = 0.25; we get :

0.25 (400 + [L]) = [L]

100 + 0.25[L] = [L]

100 = [L] - 0.25 [L]

100 = 0.75 [L]

[L] = 100/0.75

[L] = 133.33 nm

At, Y = 0.6

0.6 (400 + [L]) = [L]

240 + 0.6[L] = [L]

240 = [L] - 0.6 [L]

240 = 0.4 [L]

[L] = 240/0.4

[L] = 600 nm

At, Y = 0.95

0.95 (400 + [L]) = [L]

380 + 0.95[L] = [L]

380 = [L] - 0.95 [L]

380 = 0.05 [L]

[L] = 380/0.05

[L] = 7600 nm

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g