The bacterium, Escherichia coli, can use glucose or disaccharides, such as lactose, in its metabolism. Lactose needs to be hydrolysed by the enzyme β-galactosidase to form glucose and galactose, which can then be used by E. coli. The production of β-galactosidase is controlled by a length of DNA called the lac operon.

1. (a)[Figure 7.1] shows the lac operon when lactose is absent. On [Figure 7.2], draw the positions of RNA polymerase and the repressor molecule when lactose is present.[2]
2. (b)The protein coded for by lacY is not involved in the control of gene expression.
3. (b)(i)Name the type of gene represented by lacY.[1]
4. (b)(ii)Name the protein product coded for by lacY and state the precise role of this protein.[2]
5. (c)In an investigation into the growth of E. coli, a sample of the bacterium was grown in a medium that contained limited concentrations of glucose and lactose. The population size of E. coli was measured at regular intervals. [Figure 7.3] shows the population growth curve obtained for this investigation. Describe and suggest explanations for the population growth curve shown in [Figure 7.3].[4]

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