Question

While performing the assays for the LDH kinetics experiment, you will pipet 25ul of 250 ug/ml LDH into 975 ul of assay buffer (Tris, Lactate, and NAD). What is the final concentration of LDH in the assay?

I am having trouble if the V₂ in M₁V₁=M₂V₂ is either 975 ul or if its 1000 ul

Answer 1

The molar concentration of LDH = 250 _$\mu$ g/mL (M₁) and the volume V₁ = 25 _$\mu$ L.

The volume V₂ of assay buffer solution = 975 _$\mu$ L

We use the normality equation, N₁V₁ = N₂V₂

or n₁M₁V₁ = n₂M₂V₂

or M₁V₁ = M₂V₂ (n₁ and n₂ = 1, considering that the LDH and buffer solution ratio is 1:1)

250 X 25 = M₂ X 975

M2 = 250X25 975

or M₂ = 6.41 _$\mu$ g/mL

The final concentration (M₂) of LDH in the assay is 6.41 _$\mu$ g/mL.

The final volume V₂ is taken as 975 _$\mu$ L.