Melanism (dark-coloured wings) in a species of moth is controlled by a gene with two alleles. The dominant allele codes for the melanic form and the recessive allele codes for the non-melanic form (pale, speckled wings). Birds are a predator of this species. The melanic form is better camouflaged in areas polluted by carbon deposits from smoke. The non-melanic form is better camouflaged in areas unpolluted by smoke. The frequencies of these two forms of moth were investigated in two populations living in different areas, X and Y, by using moth traps and counting the numbers of each type of moth. The investigation was carried out over a period of time corresponding to twenty-four generations of this species. During this time period, human activity increased in both areas X and Y as towns and roads were developed. During generation 1, a large factory started releasing lots of smoke in area X. Fig. 2.1 shows the results of the investigation. [Figure 2.1]

1. (a)Suggest two variables in this investigation that should be standardised.[2]
2. (b)Describe and explain the changes in the two populations, as shown in Fig. 2.1.[3]
3. (c)If the factory in area X closed during generation 10, sketch a line on Fig. 2.1 to predict the likely effect on the frequency of the melanic form of the moth in area X for the next 14 generations.[1]
4. (d)The investigators assumed that the changes in the frequencies of the different types of moth in both areas X and Y were the result of predation by birds. Suggest two other possible reasons for the increase in frequency of the melanic form.[2]
5. (e)(i)The investigators carried out test crosses by breeding a heterozygous male melanic moth with several female non-melanic moths. The number of moths of each form in the next generation was recorded. melanic 56 non-melanic 48 Complete Table 2.1 to calculate the chi-squared (χ²) value for determining whether or not these results were significantly different from those expected for crosses of this type. The formula for calculating χ² is: χ² = Σ (O-E)² E v = degrees of freedom v=c-1 c = number of classes Table 2.1 category O E (O-E)² E melanic 56 non-melanic 48 χ² =[3]
6. (e)(ii)Table 2.2 shows some critical values for χ² at different probabilities. [Table 2.2] Use Table 2.2 to explain whether or not the χ² value calculated in Table 2.1 is significant.[1]

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