A student investigated whether pollen grains would germinate in sugar solution. She placed a drop of sucrose solution on a cavity slide, added pollen grains...
Question 1 Report
A student investigated whether pollen grains would germinate in sugar solution. She placed a drop of sucrose solution on a cavity slide, added pollen grains from a lily flower and kept the slide at 25 °C for two hours. She then examined the slide under a microscope. Fig. 4.1 shows the field of view she saw: some grains had grown a pollen tube and some had not. The grains are magnified ×400.
(a) State what a pollen grain has produced when it has germinated. [1]
(b) Record the number of pollen grains in Fig. 4.1 that have germinated. [1]
(c) Record the total number of pollen grains shown in Fig. 4.1. [1]
(d) Calculate the percentage of the grains that have germinated. [2]
(e) Measure the image diameter, in mm, of one pollen grain and use the magnification to calculate its actual diameter in µm. [3]
(f) Suggest why the student looked at several grains rather than just one. [2]
Answer Details
This tests counting from a field of view, a percentage calculation and a magnification calculation.
(a) What a germinated grain has produced [1] It has grown a pollen tube. [1]
(b) Number germinated [1] Counting the grains that have grown a tube in Fig. 4.1 gives 4 (accept the number actually drawn with a tube). [1]
(c) Total number of grains [1] Counting every grain, with and without a tube, gives 9 (accept the number drawn). [1]
(d) Percentage germinated [2] \[\frac{\text{germinated}}{\text{total}}\times 100=\frac{4}{9}\times 100=44\%\] working [1], answer about \(44\%\) [1] (accept the value from the candidate's own counts).
(e) Actual diameter of one grain [3] Measure the image diameter of one grain with a ruler; a typical reading is \(8\ \text{mm}\) (accept 6 to 10 mm) [1]. Convert to micrometres: \(8\ \text{mm}=8000\ \mu\text{m}\) [1]. Then divide by the magnification: \[\text{actual}=\frac{\text{image size}}{\text{magnification}}=\frac{8000}{400}=20\ \mu\text{m}\] answer \(20\ \mu\text{m}\) [1].
(f) Why look at several grains [2] Looking at many grains gives a more reliable and representative result [1] and reduces the effect of any single anomalous grain [1].
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