AS & A Level Biology - 9700 Paper 5 2011 Winter Zone 1

(a) (i) State the two dependent variables in this investigation.
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(ii) Apart from temperature and pH, which have little effect, state two variables that should be standardised during this investigation.
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(b) (i) Water with no suspended algae transmits 100% of the light. State how the data to plot the absorption spectrum was obtained.
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(ii) State the data which would be used to plot the action spectrum.
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(c) Using the information in Fig.1.3, suggest why using two different solvents gives a better separation of these pigments than just using one solvent.
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(d) Outline a procedure that a student could use to extract the photosynthetic pigments and obtain these chromatograms.
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(e) Different photosynthetic pigments absorb different wavelengths of light.
Table 1.1 shows some information about the pigments, P, Q, R, S, and T, found in these unicellular algae, including the wavelength of light at which maximum light absorption occurs.

[Table_1]

One of the strains of algae lacks one of the pigments.

Using the information in Table 1.1, Fig. 1.2 and Fig. 1.3:
(i) identify the strain of alga that lacks one of these pigments and state the letter of the missing pigment
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(ii) state the evidence that supports your answer to (i).
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(iii) In water, the shorter the wavelength of light, the deeper it travels. Suggest why it is an advantage to have the pigment that you identified in (i).
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A student carried out some investigations into the inheritance of body colour and wing length in the fruit fly, Drosophila melanogaster, to test the hypothesis: The inheritance of body colour and wing length in fruit flies is controlled by two genes on separate chromosomes. The student carried out three genetic crosses. To carry out each cross the following procedure was used:

• male and virgin female adult fruit flies were placed into a breeding unit containing a culture medium for their larvae
• after mating and egg laying, the adult fruit flies were removed
• newly emerged adult fruit flies were sexed by observing the shape of the last abdominal segment.

1. Suggest one factor that might affect the rate of development of the fruit flies from egg to adult. State one method by which it might be controlled.
   • factor ...............................................................................................................................................
   • method of control ............................................................................................................................
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2. Adult fruit flies are about 2.5 mm long. Suggest how the student might have observed the last abdominal segment in order to sex them. ..................................................................................................................................................... [1] The three crosses the student carried out were:
   • cross 1: pure breeding fruit flies with grey bodies and long wings × pure breeding flies with ebony bodies and short wings
   • cross 2: the offspring of cross 1 (offspring 1) were crossed with each other
   • cross 3: offspring 1 × pure breeding flies with ebony bodies and short wings
   • male and female pupae were transferred to separate breeding units
   • the number of newly emerged adult flies in each phenotype was counted
   Table 2.1 shows the results of the three crosses.

   cross	parental phenotypes	total number of offspring	number of offspring of each phenotype
   1	pure breeding grey body long wings × pure breeding ebony body short wings	62	62 grey body long wings
   2	offspring 1 grey body long wings × offspring 1 grey body long wings	75	42 grey body long wings, 15 grey body short wings, 13 ebony body long wings, 5 ebony body short wings
   3	offspring 1 grey body long wings × pure breeding ebony body short wings	64	15 grey body long wings, 19 grey body short wings, 13 ebony body long wings, 17 ebony body short wings

3. The student concluded that the results of cross 2 showed that the two genes were on separate chromosomes. State the evidence for this conclusion. ..................................................................................................................................................... [1]
4. The student used the chi-squared test ($\chi^2$ test) to analyse the results for cross 3. The student predicted that the numbers of fruit flies with each phenotype in this cross should be in the ratio 1 : 1 : 1 : 1.
   1. State the null hypothesis for this test. ..................................................................................................................................................... [1]
   2. Complete Table 2.2 to calculate the value of $\chi^2$ for the results of cross 3. The equation for the calculation of $\chi^2$ is: $$ \chi^2 = \sum \frac{(O - E)^2}{E} \\\\ O = \text{Observed result} \\\\ E = \text{Expected result} $$

      offspring phenotype	O	E	\(\frac{(O-E)^2}{E}\)
      grey bodies long wings	15
      grey bodies short wings	19
      ebony bodies long wings	13
      ebony bodies short wings	17

      $\chi^2 =$ ................................................................................................................... [3] Table 2.3 shows some critical values for chi-squared at four different probability levels.

      degrees of freedom	probability (p)
      1	2.71	3.84	6.64	10.83
      2	4.61	5.99	9.21	13.82
      3	6.25	7.82	11.34	16.27
      4	7.78	9.49	13.28	18.46

   3. State why the student should look for the critical value at 3 degrees of freedom in this investigation. .................................................................................................................................................... [1]
   4. State the conclusion from the $\chi^2$ value calculated in part (ii). .................................................................................................................................................... [1]