AQA A LEVEL BIOLOGY JUNE 2021 PAPER 1 (QUESTION PAPER)

AQA A LEVEL BIOLOGY PAPER 1 7402-1 2021 QUESTION PAPER QP
















 

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A-level

BIOLOGY

Paper 1




Time allowed: 2 hours

Materials

For this paper you must have:

a ruler with millimetre measurements

a scientific calculator.


Instructions

Use black ink or black ball-point pen.

Fill in the boxes at the top of this page.

Answer all questions.

You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages.

If you need extra space for your answer(s), use the lined pages at the end of this book. Write the question number against your answer(s).

Show all your working.

Do all rough work in this book. Cross through any work you do not want to be marked.


Information

The marks for the questions are shown in brackets.

The maximum mark for this paper is 91.






 


 


Answer all questions in the spaces provided.

 

Do not write outside the box

 



. Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.

[3 marks]





















. Scientists investigated the action of the enzyme ATP synthase. They made reaction mixtures each containing:


ATP synthase

buffer (to control pH)

substrates.


One of the substrates required in these reaction mixtures is inorganic phosphate (Pi).


Tick () one box to show which other substrate the scientists must add to the reaction mixtures to produce ATP.

[1 mark]



Adenine



Adenosine diphosphate



Glucose



Ribose

 


 


. The scientists investigated the effect of concentration of inorganic phosphate (Pi) on ATP synthase activity.


After 2 minutes, they stopped each reaction and then measured the concentration of ATP.

 

Do not write outside the box

 


Figure 1 shows the scientists’ results.

Figure 1





Suggest and explain a procedure the scientists could have used to stop each reaction.

[2 marks]











. Explain the change in ATP concentration with increasing inorganic phosphate concentration.

[2 marks]












Turn over ►

 


 


. Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.

 




[2 marks]

 

Do not write outside the box

 



 











 


 


Figure 2 shows two models of oxygen uptake found in animals.


Figure 2

 

Do not write outside the box

 




 



. Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B.

[2 marks]



















Question 2 continues on the next page












Turn over ►

 


 


. Figure 3 shows changes in concentration of oxygen in two gas exchange systems.


Figure 3


 


A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system.

 

Do not write outside the box

 


Use Figure 3 to justify this conclusion.

[2 marks]












. Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system.

[2 marks]













 


 


. Table 1 shows features of two mammals.


Bats are flying mammals; shrews are ground-living mammals.


Table 1


Mammal Mean body mass / kg 3

Mean lung volume / cm

Bat 0.096 12.48

Shrew 0.024 0.72



Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew.


Give your answer to an appropriate number of significant figures.

 

Do not write outside the box

 


 

Give one suggestion to explain this difference.

 


[3 marks]

























Answer 

 


Explanation 




 


*07*

 




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*0*

 



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. Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation.

 

Do not write outside the box

 

[3 marks]




























Question 3 continues on the next page



























Turn over ►

 


 


Table 2 shows:


mRNA codons and the amino acid coded for by each codon

the type of bond formed by the R group of some of the amino acids.


Table 2






. Crystallin is a structural protein found in the human eye. An inherited disease that leads to blindness is caused by changes in properties of crystallin. The replacement of the amino acid Arg with the amino acid Gly causes these changes.


Use information in Table 2 to suggest why this amino acid replacement changes the properties of crystallin.

 

Do not write outside the box

 

[2 marks]













 


 


. The amino acid replacement of Arg with Gly is caused by a single base substitution mutation in the DNA.  The non-mutant DNA triplet is TCC.


Complete Table 3. Give:

the mRNA codon complementary to the non-mutant DNA triplet

the mutated mRNA codon that could cause the change from Arg to Gly in the crystallin protein

the DNA triplet complementary to this mutated mRNA codon.

 

Do not write outside the box

 

[2 marks]


Table 3




Turn over for the next question






























Turn over ►

 


 


A student dissected an organ from a mammal to observe blood vessels. He dissected a slice of the organ and identified two blood vessels.

Figure 4 shows a photograph of his dissection.


Figure 4


 

Figure 5 shows a drawing of the blood vessels from his dissection.


Figure 5

 

Do not write outside the box

 


 

 


 


. Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection.

 

Do not write outside the box

 

[2 marks]


1

 




2

 





. Identify the type of blood vessel labelled as X and the type of blood vessel labelled as

Y in Figure 4.


Describe one feature that allowed you to identify the blood vessels.

[2 marks]


Blood vessel X 


Blood vessel Y 


Feature 






. Describe two precautions the student should take when clearing away after the dissection.

[2 marks]






















Turn over ►

 


 


. Describe how a sample of chloroplasts could be isolated from leaves.

 



[4 marks]

 

Do not write outside the box

 



 























 


 


. Scientists grew two groups of plants:


control plants with all the inorganic ions needed

iron-deficient plants with all the inorganic ions needed but without iron ions.

After 1 week, the scientists measured the mass of protein and the mass of chlorophyll in the chloroplasts isolated from samples of leaves of these two groups of plants.


Table 4 shows the scientists’ results.


Table 4


Mass of protein / percentage of control Mass of chlorophyll / percentage of control

40 10



Some proteins found inside the chloroplast are synthesised inside the chloroplast.


Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell.

 

Do not write outside the box

 

[2 marks]



Feature Structural difference 




. The ratio of protein to chlorophyll in control plants is 9:1


Use the information in Table 4 to calculate the ratio of protein to chlorophyll in iron-deficient plants.

[1 mark]










Ratio 


Question 5 continues on the next page




Turn over ►

 


 


. The scientists also observed the chloroplasts from the samples of leaves using an electron microscope.


Figure 6 shows a chloroplast from a control plant (image A) and a chloroplast from an iron-deficient plant (image B).


Figure 6




Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate.

[3 marks]

 

Do not write outside the box

 



 














 


 


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*17*

 




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. Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis.


Figure 7

 

Do not write outside the box

 



 


 

Calculate the rate of movement of the centromeres during phase E.


Give your answer in μm minute–1 and to 3 decimal places.

 




[2 marks]

 














    μm minute–1





















 

*1*

 



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. Name the three phases of mitosis shown by C, D and E on Figure 7.


Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases.

 

Do not write outside the box

 

[5 marks] C

























Turn over for the next question

















Turn over ►

 


 


. ADCs are molecules made of a monoclonal antibody linked to a cancer drug.


Figure 8 shows how an ADC enters and kills a tumour cell.


The process of entering the cell and the breakdown of the antibody to release the drug is very similar to phagocytosis.


Figure 8



Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell.

 

Do not write outside the box

 

[3 marks]



















 


 


. Some of the antigens found on the surface of tumour cells are also found on the surface of healthy human cells.


Use this information to explain why treatment with an ADC often causes side effects.

[2 marks]

 

Do not write outside the box

 



 














Question 7 continues on the next page






































Turn over ►

 


 


Scientists investigated whether one type of ADC could be used to treat human breast cancer.


This ADC is a monoclonal antibody combined with a drug to inhibit mitosis. The monoclonal antibody binds to a protein found on human breast cancer cells.


The scientists placed small pieces of human breast cancer tissue under the skin of mice.


The scientists then randomly divided the mice into three groups. They treated the groups as follows on day 0.


Group G – control

Group H – injected with monoclonal antibody only

Group J – injected with ADC (monoclonal antibody + drug).


Every few days, the scientists measured the volume of the tumours formed from the human breast cancer tissue.


Figure 9 shows the scientists’ results.


Figure 9

 

Do not write outside the box

 


 

 


 


. Mice in Group H were injected with 2 mg kg–1 of monoclonal antibody.

The monoclonal antibody was in a solution of concentration 500 mg dm–3

Calculate the volume of antibody solution that the scientists would have injected into a 23 g mouse. Give your answer in dm3 and in standard form.

 

Do not write outside the box

 

[2 marks]







  dm3



. Suggest one reason why there are no data for Group G and Group H after day 8

[1 mark]









. Suggest and explain two further investigations that should be done before this ADC is tested on human breast cancer patients.

[2 marks]










Turn over for the next question















Turn over ►

 


 


. Describe how a triglyceride molecule is formed.

 



[3 marks]

 

Do not write outside the box

 



 

















 


 


. Table 5 shows some properties of four fatty acids.


Table 5



Fatty acid Number of carbon atoms in the R group Number of double bonds in the R group

Caprylic acid 8 0

Palmitoleic acid 16 1

Stearic acid 18 0

Linoleic acid 18 2


Figure 10 shows diagrams of these fatty acids.


Figure 10



Put a tick () in one box that contains correct information about one of these fatty acids.

 

Do not write outside the box

 

[1 mark]




Caprylic acid is an unsaturated fatty acid represented by diagram L.



Linoleic acid is a saturated fatty acid represented by diagram N.



Palmitoleic acid is an unsaturated fatty acid represented by diagram K.



Stearic acid is a saturated fatty acid represented by diagram M.


Turn over ►

 


 


The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations.


Scientists investigated two populations of the plant, Helianthus annuus.


The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible.


They grew seeds from each population at two temperatures:


warm temperature of 24 °C

cool temperature of 10 °C


After 10 days, the scientists measured the length of each root.


Table 6 shows some of the properties of the two populations and the scientists’ results.


Table 6




Population


Temperature in natural environment In the seed – Mean percentage of fatty acids that

are saturated Mean length of root after 10 days at 24 °C / mm

(± 2 x standard deviation) Mean length of root after 10 days at 10 °C / mm

(± 2 x standard deviation)

1 Warm 10.9 8.2 (±1.0) 3.1 (±0.3)

2 Cool 6.1 5.5 (±0.9) 4.3 (±0.2)



The mean ±2 × standard deviation includes 95% of the data.

 

Do not write outside the box

 


. The scientists used a data logger to measure the length of the root rather than a ruler.


Suggest one reason why they used a data logger and explain why this was important in this investigation.

[1 mark]







 


 


. It is known that:


during respiration saturated fatty acids yield more energy than unsaturated fatty acids

saturated fatty acids have higher melting points than unsaturated fatty acids

lipases in seeds act more rapidly on liquid substrates.


Use this information and Table 6 to show how each population is better adapted for its natural environment when compared with the other population.

 

Do not write outside the box

 

[4 marks]



























. Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus.


Explain why they are both given this name.

[1 mark]















 


*27*

 




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*2*

 



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. Complete Table 7 with ticks () to show which elements are found in the following biological molecules.

 

Do not write outside the box

 

[2 marks]


Table 7



Biological molecules Element

Carbon Nitrogen Oxygen Phosphorus

Galactose

Phospholipid

RNA

Sucrose






Question 9 continues on the next page

































Turn over ►

 


 


After Watson and Crick proposed the model of DNA structure, scientists investigated the possible mechanisms for DNA replication.


Two scientists grew a bacterial population, providing them with a nitrogen source containing only the heavy isotope of nitrogen, 15 N. As soon as all the DNA in this population contained 15 N, the scientists changed the nitrogen source to one containing only the lighter isotope of nitrogen, 14 N. They changed the nitrogen source at 0 hours.


During the investigation, the scientists measured the size of the population of bacterial cells.


Figure 11 shows the scientists’ results.


Figure 11

 

Do not write outside the box

 


 

 


 


. The generation time for a population of bacteria is the time taken for all the bacteria to divide once by binary fission.


Use Figure 11 and the following equation to calculate the generation time for this population of bacteria. Give your answer in hours.

 

Do not write outside the box

 




Number of generations =

 


size of population at time +4 hours log10 �size of population at time – 4 hours�

 

log10 2

 

[2 marks]

 



















Generation time hours




Question 9 continues on the next page


























Turn over ►

 


 


At intervals during this investigation, the scientists removed samples of the bacterial population, isolated the DNA and measured the density of the DNA.


DNA made using 15 N has a higher density than DNA made using 14 N.


Figure 12 shows the scientists’ results.


Figure 12

 

Do not write outside the box

 



 

 


 


There are three possible models of DNA replication. These models are shown in Figure 13.

Figure 13



. Which of these models, P, Q or R, is supported by the results shown in Figure 12? Give the letter and name of the model supported and explain why the results do not

support the other models.

 

Do not write outside the box

 

[3 marks]


Model 


Name 


Explanation for first

unsupported model 




Explanation for second

unsupported model 















Turn over ►

 


 


. Describe the structure of DNA.

 



[5 marks]

 

Do not write outside the box

 



 





































 


 


. Name and describe five ways substances can move across the cell-surface membrane into a cell.

 




[5 marks]

 

Do not write outside the box

 



 









































Question 10 continues on the next page










Turn over ►

 


 


Figure 14 shows transmission electron micrographs of two cells, one animal cell and one prokaryotic cell.


Figure 14

 

Do not write outside the box

 



 

 


 


. Contrast the structure of the two cells visible in the electron micrographs shown in

Figure 14.

 

Do not write outside the box

 

[5 marks]











































END OF QUESTIONS












 

*37*

 



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*3*

 



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Question number Additional page, if required.

Write the question numbers in the left-hand margin.


 



 



 



 



 



 



 



 



 



 



 



 



 



 



 



 



 



 








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Version 2021
Category AQA Questions and Marking Scheme
Included files pdf
Authors vaites
Pages 55
Language English
Tags AQA A LEVEL BIOLOGY JUNE 2021 PAPER 1 (QUESTION PAPER)
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