A student removed a single carpel from a flower and drew it as shown in Fig. 6.1. The scale bar on the drawing represents 5 mm. She then investigated when t...
Question 1 Report
A student removed a single carpel from a flower and drew it as shown in Fig. 6.1. The scale bar on the drawing represents 5 mm. She then investigated when the stigma was best able to receive pollen. On each of six days after the flowers opened she hand-pollinated 20 fresh flowers and later recorded how many of them formed seeds. Her results are shown in Table 6.1.
| Day after opening | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Flowers pollinated | 20 | 20 | 20 | 20 | 20 | 20 |
| Number forming seeds | 2 | 11 | 15 | 12 | 9 | 4 |
| Percentage forming seeds | 10 | 55 | ? | 60 | ? | 20 |
(a) Name the part labelled X on Fig. 6.1. [1]
(b) Measure, using the scale bar, the length of the style (the stalk between the stigma and the ovary) on Fig. 6.1, in mm. [2]
(c) Complete Table 6.1 by calculating the percentage forming seeds on day 2 and on day 4. [2]
(d) State the day on which the stigma was most able to receive pollen. [1]
(e) Describe how the percentage forming seeds changed over the six days. [3]
(f) Suggest why the student pollinated 20 flowers on each day rather than one. [2]
Answer Details
This question tests naming a flower part, a scale-bar measurement, percentage calculations, and interpreting a trend in reliability.
(a) The part labelled X is the stigma [1] - the sticky top of the carpel where pollen lands.
(b) Measure the drawn length of the style and the drawn length of the scale bar with a ruler, then scale up using \( \text{actual style} = \dfrac{\text{style on drawing}}{\text{scale bar on drawing}} \times 5\ \text{mm} \); this gives a value of about 2 to 3 mm (accept your own honest reading with working shown) [1 working, 1 answer].
(c) The percentage forming seeds is the number forming seeds out of 20, as a percentage. Day 2: \( \dfrac{15}{20} \times 100 = 75\% \) [1]. Day 4: \( \dfrac{9}{20} \times 100 = 45\% \) [1]. The completed row is:
| Day after opening | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Percentage forming seeds | 10 | 55 | 75 | 60 | 45 | 20 |
(d) The stigma was most able to receive pollen on day 2 [1], because that is when the highest percentage of flowers formed seeds.
(e) Over the six days the percentage forming seeds was low on day 0 [1], rose to a peak around day 2 [1], and then fell to a low value by day 5 [1]. This rise-then-fall shows the stigma is receptive only for a short window.
(f) Pollinating 20 flowers each day rather than one gives a more reliable, representative result [1] and reduces the effect of any one unusual (anomalous) flower [1].
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