Gradients, intercepts and logarithmic scales

Overview

Recall questions on calculating rates from gradients, y = mx + c, drawing tangents to curves, intercepts and the compensation point, extrapolation, and why logarithmic scales are used for microbial growth.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   How is a rate calculated from a straight-line graph?

   Model answer

   Choose two points far apart on the line and divide the change in y by the change in x between them; this gradient is the rate.

2. 2

   Calculate the rate of water uptake when a straight line on a potometer graph passes through (1.0 min, 12 mm) and (6.0 min, 87 mm).

   Model answer

   Rate = (87 − 12) ÷ (6.0 − 1.0) = 75 ÷ 5.0 = 15 mm/min.

3. 3

   What do m and c represent in the equation y = mx + c?

   Model answer

   m is the gradient of the straight line and c is the intercept, the value of y where the line crosses the y-axis.

4. 4

   What does a straight-line graph passing through the origin show about two variables?

   Model answer

   The y variable is directly proportional to the x variable (y ∝ x), so c = 0 and doubling x doubles y.

5. 5

   What shape of graph is expected for rate of reaction against substrate concentration when enzyme is in excess, and why?

   Model answer

   A straight line through the origin, because spare active sites are always available, so the rate rises in proportion to substrate concentration.

6. 6

   How is the rate at one point on a curved graph found?

   Model answer

   Draw a tangent that just touches the curve at that point using a ruler, then calculate the gradient of the tangent as change in y ÷ change in x.

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