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AQA MERGED QUESTIONS AND MARK SCHEME BIOLOGY PAPER 2 FOR MAY 2024
Typology: Exams
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Question Mark 1 2 3 4 5 6 7 8 9
TOTAL
Answer all questions in the spaces provided.
(^0 1). 1 Water has a high heat capacity and a large latent heat of vaporisation.
Describe the importance of each of these properties to living organisms. [2 marks]
High heat capacity
Large latent heat of vaporisation
(^0 1). 2 Figure 1 shows that water loss from a porous pot can cause the upward movement of water.
Figure 1
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Do not write outside the (^0 2). 1 Figure 2 shows an image of a red blood cell at a magnification of Ă 5500^ box
Figure 2
Calculate the actual diameter in ÎŒm of the red blood cell between points P and Q.
Show your working. [2 marks]
Answer ÎŒm
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Do not write outside the (^0 2). 2 A haemocytometer is a special microscope slide that can be used to determine the^ box mean number of red blood cells in 0.004 mm 3 of blood.
The volume of blood in the body of the adult was 4.8 dm 3
Calculate the total number of red blood cells in the body of this adult.
Show your working. [2 marks]
Answer
(^0 2). 3 The solution used to dilute the blood had to have the same water potential as the blood.
Explain why. [2 marks]
Question 2 continues on the next page
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Do not write outside the (^0 3). 1 Describe how you would use cell fractionation and ultracentrifugation to obtain a^ box sample of mitochondria from muscle tissue. [4 marks]
Question 3 continues on the next page
Do not write outside the (^0 3). 2 Mitochondrial respiratory capacity is a measure of maximum ATP production in a^ box mitochondrion. Scientists investigated the effect of a resistance exercise training (RET) programme on the respiratory capacity of mitochondria in skeletal muscle tissue. RET develops muscle strength.
The scientists:
Figure 3 shows some of the scientistsâ results.
The error bars represent ± 2 standard deviations from the mean, which includes over 95% of the data.
Figure 3
Do not write outside the (^0 4). 1 Where does transcription occur in a eukaryotic cell?^ box [1 mark]
0 4. 2 The genome of an adenovirus is a single, linear molecule of double-stranded DNA. Adenoviruses use eukaryotic host cells to transcribe their genes in protein synthesis. The process of transcription of adenovirus genes is similar to the process of transcription of genes in eukaryotes.
Scientists investigated the process of transcription of this viral DNA.
Figure 4 shows one of the experiments carried out by these scientists.
Figure 4
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Do not write outside the Looking at the results of the experiment in Figure 4, the scientists concluded that^ box splicing had taken place.
Use Figure 4 to describe and explain why the scientistsâ conclusion was justified. [3 marks]
0 4. 3 Describe and explain how the results of the experiment in (^) Figure 4 would differ if the scientists had used prokaryotic DNA. [2 marks]
0 4. 4 Errors in the precise location of splicing in the DNA molecule can lead to mutations.
Explain why. [ 1 mark]
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Do not write outside the (^0 5). 4 A student used an optical microscope to observe a piece of tissue from the lower^ box surface of an orchid leaf.
The piece of leaf tissue observed was very thin.
Explain why this was important. [2 marks]
0 5. 5 The student produced a biological drawing of the leaf tissue they viewed through an optical microscope.
Give three ways the student could ensure they produce a correct biological drawing of the leaf tissue.
Assume the student uses a sharp pencil. [3 marks]
1
Do not write outside the (^0 6). 1 Disinfectants are used to kill microorganisms on non-living surfaces. A student^ box investigated the effect of different concentrations of disinfectant X on the growth of Bacillus subtilis.
The student:
Table 2 shows the studentâs results.
Table 2
Percentage concentration of disinfectant X
Percentage light absorbance 100 100 87 52 10 10
The student prepared the different concentrations of disinfectant X.
Describe how the student made 5 cm^3 of the 60% concentration using distilled water and undiluted disinfectant. [1 mark]
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Do not write outside the (^0 7). 1 Edwardsâ syndrome is a condition resulting from an extra chromosome 18.^ box A chromosome mutation in the second meiotic division is the most frequent cause of Edwardsâ syndrome.
Explain how a chromosome mutation in the second meiotic division could result in an extra chromosome 18.
In your answer, name the type of chromosome mutation which would result in the extra chromosome. [2 marks]
0 7. 2 Complete trisomy 18 is the most common type of Edwardsâ syndrome. This occurs when all the cells of the body have an extra chromosome 18.
Explain why all the cells of the body have an extra chromosome 18. [2 marks]
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Do not write outside the (^0 7). 3 Mosaic trisomy 18 is another type of Edwardsâ syndrome. This occurs due to a^ box chromosome mutation after fertilisation.
In mosaic trisomy, the body has cells with an extra chromosome 18 and cells with the correct number of chromosomes.
Explain how cells with different numbers of chromosomes are produced in mosaic trisomy. [1 mark]
(^0 7). 4 The age of the female parent is a factor linked to the risk of a child having Edwardsâ syndrome.
Which statistical test should be used to test whether this link is statistically significant?
Tick (ïŒ) one box. [1 mark]
Correlation coefficient
Chi-squared
Studentâs t-test
Question 7 continues on the next page
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Do not write outside the (^0 8) Lyme disease is most frequently caused by the bacterium Borrelia burgdorferi.^ box Lyme disease can be difficult to diagnose.
Figure 5 shows an ELISA test that is used to find out if a person has antibodies to B. burgdorferi.
Figure 5
A false positive in this test is a result which incorrectly indicates that antibodies to B. burgdorferi are present.
0 8. 1 Failure to thoroughly wash the well in Step 4 can result in a false positive.
Explain why. [2 marks]
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Do not write outside the (^0 8). 2 A false positive can be produced if a person has been infected by another bacterium^ box that causes a disease called syphilis.
Suggest why. [1 mark]
(^0 8). 3 A false negative in this test is often produced if a person is tested within 2 weeks of being infected with B. burgdorferi.
Explain why. [2 marks]
(^0 8). 4 Sometimes, symptoms of Lyme disease can persist for 6 months following antibiotic treatment. This condition is known as Post-Treatment Lyme Disease Syndrome (PTLDS).
Scientists investigated the symptoms experienced by a large number of PTLDS patients and a control group. During a 2-week period, they asked all the participants:
The scientists used a statistical test to determine if there was a difference in the intensity of symptoms of PTLDS between these two groups.