Biology 9700 · AS & A Level · Passage of information from parents to offspring

Passage of information from parents to offspring — practice question

Moth populations of Biston betularia occur in both Europe and North America. On each continent, the phenotype seen most often is a pale wing colour with light-grey shading, known as the typical form. A dark-winged phenotype, called the melanic form, is also present on both continents. Figure 10.1 shows the moth’s typical form, while Fig. 10.2 shows its melanic form.
(a)[3]

Two melanic European moths were crossed with each other. Their offspring showed 15 typical and 41 melanic wing colours. Draw a genetic diagram to account for these results. You may use $A$ and $a$ for the alleles.

(b)[1]

In a comparable experiment, two melanic North American moths were crossed with each other. The offspring were 10 typical and 31 melanic. What conclusion can be drawn about the allele responsible for the melanic form in moth populations on both continents?

(c(i))[1]

Researchers were unsure whether the allele producing the melanic form in European moths was at the same locus as the allele producing the melanic form in North American moths. To investigate, they made these crosses: Cross 1: European moths that were heterozygous only at the European melanic locus were crossed with North American moths that were heterozygous only at the North American melanic locus. Cross 2: The melanic offspring and the typical offspring from cross 1 were mated with each other. Explain why cross 2 is a test cross.

(c(ii))[3]

Complete Table 10.1 so that it shows the predicted outcomes when: • the European and North American melanic alleles lie at the same locus $(A/a)$ • the European and North American melanic alleles are found at two different loci $(A/a$ and $B/b)$.

(d)[2]

A light trap was employed to estimate the overall population size of $\textit{B. betularia}$ in a woodland. On the first night, 24 moths were caught. A small spot of harmless paint was then applied to mark them. On the second night, 29 moths were caught, and 8 of these had paint spots. Use the Lincoln index formula given to calculate the population size. Show your working. The formula is $N = \frac{n_1 \times n_2}{m_2}$, where $N$ is the estimate of population size, $n_1$ is the number of individuals captured in the first sample, $n_2$ is the number of individuals (both marked and unmarked) captured in the second sample, and $m_2$ is the number of marked individuals recaptured in the second sample.

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This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Cross: $Aa \times Aa$

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