Question

Consider the given branched pathway. - E- A - B - F - G - → Z The first common step, A → B, is partly inhibited by both final products, Yand Z, which act independently of each other. Suppose that a high level of Y alone decreased the rate of the A + B step from 100 s-to 60 s- and that a high level of Z alone decreased the rate from 100 s-to 40 s-7. What would be the rate of the A + B step in the presence of high levels of both Y and Z? rate:

What is the rate, please?

Answer 1

The rate of the A + B step in the presence of high levels of both Y and Z would be 24s-1. The calculation is shown in two different ways.

Calculation I:

A high level of Y alone decreased the rate of the A + B step from 100s-1 to 60s-1. So the rate of inhibition due to Y alone is 40%.

A high level of Z alone decreased the rate from 100s-1 to 40s-1. So the rate of inhibition due to Z alone is 60%.

As the inhibition rates are 40% and 60%
Total inhibition is = ( y + z) – (yz / 100 ) %
x = 40% and y = 60%

Total inhibition = (40 + 60) – [( 40 x 60) / 100] % = 100 – (2400 /100 ) % = (100 – 24)% = 76%
Inhibition = 76% of 100s-1 = ( 76 / 100 ) x 100s-1 = 76s-1
Rate = Reaction rate – inhibition= 100s-1 - 76s-1 = 24s-1

Calculation II:

A high level of Y alone decreased the rate of the A + B step from 100s-1 to 60s-1. So the rate of inhibition due to Y alone is 40%.

So, in the 1st step, due to Y the rate will be 60s-1

A high level of Z alone decreased the rate from 100s-1 to 40s-1. So the rate of inhibition due to Z alone is 60%.

So, in the 2nd step, due to Z the inhibition would be 60% of 60s-1

The inhibition = 60s-1* (60/100) = 60s-1*0.6 =36s-1

Rate = Reaction rate – inhibition= 60s-1 - 36s-1 = 24s-1

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