Biology 0610/61 October/November 2020: question paper and mark scheme
Cambridge IGCSE Biology (0610), Paper 6 Alternative to Practical, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The enzyme lipase catalyses the break-down of fats into fatty acids and glycerol.
A student investigated how the concentration of lipase affected the break-down of fat in milk.
An increase in the concentration of fatty acids in the milk will change the pH of the milk.
The student used a pH indicator, bromothymol blue, and observed the colour changes of the indicator in the milk and lipase solution.
Table 1.1 shows the colour of bromothymol blue indicator at different pH values.
Table 1.1
pH 6 7 8 colour yellow green blue
(a) Step 1 Four test-tubes were labelled L1, L2, L3 and L4.
Step 2 Solutions containing different concentrations of lipase enzyme were made using the volumes of 2% lipase solution and distilled water shown in Table 1.2.
Table 1.2
volume of 2% lipase volume of distilled percentage concentration test-tube solution / cm3 water / cm3 of lipase solution
L1 3.00 0.00 2.0
L2 1.50 1.50 1.0
L3 0.75 2.25
L4 0.00 3.00 0.0
(i) Calculate the percentage concentration of lipase solution in test-tube L3 using the information in Table 1.2.
Space for working.
……% [1]
Step 3 Another four test-tubes were labelled M1, M2, M3 and M4.
Step 4 Three different substances were added to each of test-tubes M1, M2, M3 and M4:
• 5 drops of bromothymol blue indicator
• 2 cm3 of sodium carbonate solution
• 2 cm3 of milk.
The student observed that the contents of test-tubes M1, M2, M3 and M4 were all blue.
Step 5 All of the test-tubes were put into a water-bath containing warm water and left for five minutes.
Step 6 After five minutes test-tubes M1 and L1 were removed from the water-bath.
Step 7 The contents of test-tube M1 were poured into L1 and a stop-clock started.
Step 8 The colour of the bromothymol blue indicator in test-tube L1 was observed.
The time taken for the bromothymol blue indicator to become yellow was recorded.
If the colour had not changed to yellow in five minutes the result was recorded as
>300.
Step 9 Steps 6, 7 and 8 were repeated for test-tubes M2 and L2, M3 and L3, M4 and L4.
The times from step 8 and step 9 are shown in Fig. 1.1.
no change
00 : 25 00 : 38 01 : 17 at 5 minutes
L1 L2 L3 L4
Fig. 1.1
(ii) Prepare a table for the results. Convert the times shown in Fig. 1.1 to seconds and record them in your table.
[3]
(iii) State a conclusion for these results.
…… [1]
(b) (i) Identify the control in this investigation and explain why a control was used.
control ……
explanation ……
[2]
(ii) Using the information in Table 1.1 and the results, estimate the pH values in test-tube L1 and test-tube L4 at the end of the investigation.
L1 ……
L4 ……
[1]
(iii) State the variable that was deliberately changed (the independent variable) in this investigation.
…… [1]
(iv) State two variables that were kept constant in this investigation.
1 ……
2 ……
[2]
(v) Suggest why all of the test-tubes were placed into a water-bath, in step 5, for five minutes before mixing their contents.
…… [1]
(vi) State the potential source of error in step 8.
…… [1]
(c) Describe how you would safely test lipase for the presence of protein and state the result of a positive test.
method ……
positive result ……
safety precaution ……
[3]
(d) The average temperature of the human body is 37 °C. Humans produce lipase for fat digestion. A student thought that lipase would work best at human body temperature.
Plan an investigation to find out if 37 °C is the optimum (best) temperature for lipase activity.
…… [6]
[Total: 22]
Mark scheme
1(a)(i) 0.5(%) ; 1
1(a)(ii) table drawn with a minimum of two columns and a header line ; 3
column and row headings with units ; percentage concentration of lipase and time / s
correct time recorded in seconds for each concentration ; 25, 38, 77, >300
1(a)(iii) as the, enzyme / lipase, concentration / percentage, increases the time 1
to change colour, decreases / is faster ; ora
the higher the enzyme concentration the faster the, fat / milk, is broken
down /AW ;
1(b)(i) control: distilled water / L4 / 0% lipase ; 2
explanation: to show that lipase / enzyme is causing the change (of
colour in the indicator / of pH) ;
1(b)(ii) L1 = 6 and L4 = 8 ; 1
1(b)(iii) lipase / enzyme, concentration ; 1
1(b)(iv) any two from: 2
volume of milk ;
total volume, of (enzyme) solution ;
volume of sodium carbonate solution ;
number of drops of, indicator / bromothymol blue, used ;
1(b)(v) to give enough time for the (contents of the) test-tubes to, reach the 1
temperature of the water-bath / equilibrate AW ;
1(b)(vi) idea of: colour change is subjective / end-point is hard to judge ; 1
1(c) add biuret, reagent / solution (to lipase solution) ; 3
colour change (from blue) to purple / lilac / violet / mauve ;
wear gloves / safety glasses / goggles ;
1(d) independent variable 6
1 at least three temperatures ;
2 one below 37°C, one above 37°C, and one at 37°C ;
new method
3 method of maintaining temperature described ;
4 named alternative sources of fat ;
5 correct reference to emulsion test
6 correct reference to use of a different indicator, e.g. UI
7 and 8 max 2 from given method ;;
• same, volume of sodium carbonate
• same number of drops of indicator
• same volume of milk
• measure time taken for indicator to change colour
• idea of equilibrate enzyme and milk before mixing / AW
9 and 10 max 2 from variables kept constant ;;
• same volume of indicator
• same type of milk / fat
• same volume of fat ;
• same amount / volume / concentration, of lipase
11 ref. to two or more repeats ;
(a) Fig. 2.1 is a labelled diagram of the parts of a flower.
petal
stigma anther stamen style carpel filament ovary
sepal
Fig. 2.1
Fig. 2.2 is a photograph showing the parts of a flower that have been separated.
A
B
C
Fig. 2.2
Complete Table 2.1 using the information in Fig. 2.1 and Fig. 2.2 by stating the:
• names of flower parts A, B and C
• number of each of the flower parts A, B and C visible in Fig. 2.2.
Table 2.1
letter on Fig. 2.2 name of flower part number visible
A
B
C
[3]
(b) Fig. 2.3 shows a photograph of a germinating pollen grain.
pollen tube
X
pollen grain
Y
magnification ×350
Fig. 2.3
(i) Make a large drawing of the germinating pollen grain shown in Fig. 2.3.
Label the pollen tube.
[5]
(ii) Measure the length of line XY on Fig. 2.3.
length of line XY …… mm
Calculate the actual length of the pollen grain in Fig. 2.3 using the formula.
length of XY on Fig. 2.3 magnification = actual length of XY
Include the unit.
Space for working.
[3]
(c) Some students collected pollen from the anthers of flowers to investigate the effect of two different solutions, S1 and S2, on the germination of pollen.
Two microscope slides were prepared.
Slide one had 210 pollen grains and two drops of solution S1.
Slide two had 250 pollen grains and two drops of solution S2.
Every 10 minutes the students counted and recorded the number of pollen grains that had germinated.
The percentage of pollen grains that had germinated was calculated.
Fig. 2.4 shows a drawing of the pollen grains as seen with a light microscope.
germinated pollen grain
ungerminated pollen grain
Fig. 2.4
The results of the investigation are shown in Table 2.2.
Table 2.2
percentage germination
time / minutes 10 20 30 40 50 60
solution S1 5 18 26 38 51 51
solution S2 3 8 18 28 36 51
(i) State two conclusions for these results.
1 ……
2 ……
[2]
(ii) The results in Table 2.2 are shown as percentages rather than as the actual number of germinated pollen grains.
Explain why this enables a valid comparison to be made between the results for S1 and S2.
…… [1]
(iii) Describe how the percentage germination in Table 2.2 was calculated.
…… [2]
(d) The students prepared another three slides using solutions A, B and C and left them for 60 minutes. They measured the length of the pollen tubes in 20 germinated pollen grains.
The results are shown in Table 2.3.
Table 2.3
solution average length of pollen tube / µm
A 190
B 220
C 265
Fig. 2.5 shows the grid that the students used to plot a graph of their results.
average length of pollen tube / μm
A
solution
Fig. 2.5
Use the information in Table 2.3 to complete the graph in Fig. 2.5 by:
• adding the scale for the y-axis
• plotting the bars for solutions B and C.
[2]
[Total: 18]
Mark scheme
2(a) 3 one mark per row
letter on Fig. 2.2 name of flower part number visible
A stigma 1
B filament 8
C sepal 4
;;;
2(b)(i) lines: clear single outer line without shading ; 5
size: X-Y bigger than 35 mm ;
rounded structures inside grain ;
pollen tube top edge with one indentation ;
label to pollen tube ;
2(b)(ii) line XY 35 ±1 (mm) ; 3 MP1 correct measurement
0.1 mm / 100 µm / 0.01 cm ;; MP2 correct calculation
A ecf measurement ÷ 350
MP3 correct unit
2(c)(i) any two from: 2
pollen germinated faster / germination rate is higher, in S1 (than S2)
(at the start) / AW ; ora
by 60 minutes no difference between (percentage germination) in S1
and S2 ;
(pollen in) S1 stops germinating after 50 minutes / AW ;
only 51% / not all pollen grains, germinated ;
2(c)(ii) idea that different numbers of pollen were counted on each slide ; 1
2(c)(iii) divide the number germinated by the total ; 2
multiply by 100 ;
2(d) y-axis scaled correctly with suitable numbers ; 2
plot for B and C correct and labelled ;
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Biology 0610/61 for October/November 2020 is out of 40 marks, with 1 hour to answer it, across 2 questions.
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