Y13 AS RECAP XMAS WORK LIST
Time:
317 minutes
Marks:
234 marks
Name:
________________________
Class:
________________________
Date:
________________________
Comments:
Page 1 of 79
1
(a)
Describe how the structures of starch and cellulose molecules are related to their functions. (5)
(b)
Describe the processes involved in the transport of sugars in plant stems. (5) (Total 10 mark s)
2
(a)
What two measurements are needed to calculate an index of diversity? 1
.....................................................................................................................
2
..................................................................................................................... (2)
(b)
A herbicide is a chemical used to kill weeds. Ecologists investigated the effect of a herbicide on crop yield and the diversity of insects. They sprayed different fields with the same volume of different concentrations of the herbicide. At harvest, the ecologists determined the mean crop yield and the mean index of diversity of insects for fields that had received the same concentration of the herbicide. The figure below shows their results.
Concentration of herbicide sprayed on field / mg dm −3 (i)
Some fields acted as controls. They were sprayed with a solution that did not contain the herbicide. Explain the purpose of these control fields. ............................................................................................................... ............................................................................................................... ............................................................................................................... (1)
Page 2 of 79
(ii)
Suggest an explanation for the relationship between the concentration of herbicide and the mean crop yield. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (2)
(iii)
Explain theof relationship of diversity insects. between the concentration of herbicide and the mean index ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ...............................................................................................................
(Extra space) ........................................................................................ ............................................................................................................... ............................................................................................................... (3) (Total 8 marks )
Page 3 of 79
3
There are nine subspecies of giraffe. These subspecies evolved when populations of giraffe were separated for long time periods. Each subspecies has distinct coloured skin markings. Some biologists have suggested that up to six of these subspecies should be classified as different species. (a)
Explain how different subspecies of giraffe may have evolved from a common ancestor. Use information from the passage in your answer. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (5)
(b)
Biologists compared the mitochondrial DNA of the different subspecies of giraffe. They used the results from comparing this DNA to conclude that six of the nine subspecies are separate species. Suggest how they came to this conclusion. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (2) (Total 7 marks )
Page 4 of 79
4
Lake Malawi in East Africa contains around 400 different species of cichlids which are small, brightly coloured fish. All these species have evolved from a common ancestor. (a)
Describe one way in which scientists could find out whether cichlids from two different populations belong to the same species. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ......................................................................................................................
(b)
(2)
During the last 700 000 years there have been long periods when the water level was much lower and Lake Malawi split up into many smaller lakes. Explain how speciation of the cichlids may have occurred following the formation of separate, smaller lakes. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (4)
(c)
Many species of cichlids are similar in size and, apart from their colour, in appearance. Suggest how the variety of colour patterns displayed by these cichlids may help to maintain the fish as separate species. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2) (Total 8 marks )
Page 5 of 79
5
(a)
(i)
Why is the genetic code described as being universal? ............................................................................................................... ............................................................................................................... (1)
(ii)
The genetic code uses four different DNA bases. What is the maximum number of different DNA triplets that can be made using these four bases?
(1)
Transcription of a gene produces pre-mRNA. (b)
Name the process that removes base sequences from pre-mRNA to form mRNA. ........................................................................................................................ (1)
Page 6 of 79
(c)
The figure below shows part of a pre-mRNA molecule. Geneticists identified two mutations that can affect this pre-mRNA, as shown in the figure. Base sequence coding for amino acids
Mutation 1, single base deletion (i)
Base sequence removed from pre-mRNA
Base sequence coding for amino acids
Mutation 2, single base substitution
Mutation 1 leads to the production of a non-functional protein. Explain why. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ...............................................................................................................
(Extra space) ........................................................................................ ............................................................................................................... ............................................................................................................... (3)
(ii)
What effect might mutation 2 have on the protein produced? Explain your answer. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (2) (Total 8 marks )
Page 7 of 79
6
Scientists investigated the presence of bacteria resistant to the antibiotic neomycin in turkeys, chickens and the farmers who kept the turkeys and chickens. They looked for Escherichia coli (E. coli) resistant to neomycin. At 46 farms, the scientists obtained samples of bacteria from faeces of turkeys, turkey farmers, chickens and chicken farmers. The turkey farmers very often used turkey food containing neomycin. The chicken farmers did not use chicken food containing neomycin very often. The bacteria were grown on nutrient agar in cultures. The nutrient agar contained neomycin. Any resistant bacteria grew and divided to form visible colonies. The results are shown in the table
Percentage of samples of faeces containingE. Samples taken from coli resistant to neomycin Turkeys Turkeysfarmers Chickens Chickenfarmers
(a)
81 57 24 8
Suggest two hypotheses the scientists were testing in this investigation. Hypothesis 1........................................................................................................ ............................................................................................................................. ............................................................................................................................. Hypothesis 2........................................................................................................ ............................................................................................................................. ............................................................................................................................. (2)
(b)
(i)
Describe what the results in the table show. ................................................................................................................... ................................................................................................................... ................................................................................................................... ................................................................................................................... ................................................................................................................... (2)
Page 8 of 79
(ii)
Suggest and explain one reason for the observed differences in percentage of neomycin-resistant E. coli in turkeys and chickens. ................................................................................................................... ................................................................................................................... ................................................................................................................... ................................................................................................................... ................................................................................................................... (2)
(c)
The scientists followed strict safety guidelines when collecting samples of faeces. Apart from the risk of contamination from E. coli this was especially important when collecting samples from humans. Explain why. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (1)
(d)
Use the information provided to identify and explain one way in which the scientists increased the reliability of their method. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (2)
(e)
Suggest how the scientists could use DNA to investigate whether the neomycin-resistant bacteria in farmers were identical to the strain of bacteria in the birds they kept. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (2)
Page 9 of 79
(f)
At one time, most animal feeds contained antibiotics that increased the rate of animal growth. In the UK, fewer animal feeds now contain antibiotics. Suggest reasons why. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. ............................................................................................................................. (4) (Total 15 mark s)
7
(a)
Describe how you would test a piece of food for the presence of lipid. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2)
Page 10 of 79
The figure below shows a phospholipid.
X (b)
Y
The part of the phospholipid labelled A is formed from a particular molecule. Name this molecule. ........................................................................................................................ (1)
(c)
Name the type of bond between A and fatty acid X. ........................................................................................................................ (1)
Page 11 of 79
(d)
Which of the fatty acids, X or Y, in the figure above is unsaturated? Explain your answer. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (1)
Scientists investigated the percentages of different types of lipid in plasma membranes from different types of cell. The table shows some of their results.
Type of lipid
(e)
Percentage of lipid in plasma membrane by mass Cell lining ileum of mammal
Red blood cell of mammal
The bacterium Escherichia coli
Cholesterol
17
23
0
Glycolipid
7
3
0
Phospholipid
54
60
70
Others
22
14
30
The scientists expressed their results as Percentage of lipid in plasma membrane by mass. Explain how they would find these values. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2)
Cholesterol increases the stability of plasma membranes. Cholesterol does this by making membranes less flexible. (f)
Suggest one advantage of the different percentage of cholesterol in red blood cells compared with cells lining the ileum. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (1)
Page 12 of 79
(g)
E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell maintains a constant shape. Explain why. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (Total 10 mark s)
Page 13 of 79
8
Malaria is a disease that is spread by insects called mosquitoes. In Africa, DDT is a pesticide used to kill mosquitoes, to try to control the spread of malaria. Mosquitoes have a gene called KDR. Today, some mosquitoes have an allele of this gene,KDR minus, that gives them resistance to DDT. The other allele, KDR plus, does not give resistance. Scientists investigated the frequency of the KDR minus allele in a population of mosquitoes in an African country over a period of 10 years. The figure below shows the scientists’ results.
Year (a)
Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygous for the KDR gene in this population in 2003. Show your working.
Frequency of heterozygotes in population in 2003 ................................... (2)
Page 14 of 79
(b)
Suggest an explanation for the results in the figure above. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4)
The KDR plus allele codes for the sodium ion channels found in neurones. (c)
When DDT binds to a sodium ion channel, the channel remains open all the time. Use this information to suggest how DDT kills insects. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2)
(d)
Suggest how the KDR minus allele gives resistance to DDT. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (Total 10 mark s)
Page 15 of 79
9
The body markings of cheetahs vary, in particular the pattern of bands on their tails. Cheetahs are solitary animals but the young stay with their mother until they are between 14 and 18 months old. Scientists investigated the banding pattern on the tails of cheetahs living in the wild. • •
They drove a car alongside a walking cheetah and used binoculars to study the tail pattern. They gave each cheetah a banding pattern score based on the width of the dark and light bands on the end of the tail. They scored the width of the bands on the right and left side of the tail using a 5 point scale of width.
•
A typical pattern on the right side of one cheetah’s tail is shown inFigure 1.
Figure 1 Band number
1
Bandwidthscore
3
23
4
5
6
7
11
4
3
3
3
The scientists collected data from each cheetah on four separate occasions. Figure 2 shows the data for one of the cheetahs.
Figure 2
Side of tail
Mean band width score (± standard deviation) Band1
Band2
Band3
Right
3.00 (± 0.82)
1.00 (± 0.00)
1.00 (± 0.00)
3.75 (± 0.50)
2.75 (± 0.50)
3.00 (± 0.00)
3.00 (± 0.00)
Left
3.75 (± 0.50)
3.25 (± 0.50)
2.00 (± 0.50)
3.00 (± 0.00)
2.00 (± 0.00)
2.50 (± 0.50)
3.00 (± 0.50)
(a)
Band4
Band5
Band6
Band7
The scientists only used data from cheetahs which were fully grown. Suggest why. ........................................................................................................................ ........................................................................................................................ (1)
Page 16 of 79
(b)
The scientists estimated the width of the bands on the same cheetah on four separate occasions. They did not always get the same score. (i)
Give two pieces of evidence from Figure 2 which show that the scientists sometimes obtained different scores for the same band. 1 ............................................................................................................ ............................................................................................................... 2 ............................................................................................................ ...............................................................................................................
(ii)
(2)
The method the scientists used resulted in them getting different scores for the same band. Suggest why. ............................................................................................................... ............................................................................................................... (1)
(c)
What is the evidence from Figure 2 that the dark and light bands do not form rings of equal width around the tail? ........................................................................................................................ ........................................................................................................................ (1)
Page 17 of 79
(d)
The scientists found the difference in banding pattern between • •
offspring in the same family cheetahs chosen randomly.
Explain how scientists could use this information to show that some variation in tail banding was genetic. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ (3) (Total 8 marks )
10
(a)
Messenger RNA (mRNA) is used during translation to form polypeptides. Describe how mRNA is produced in the nucleus of a cell. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (6) Page 18 of 79
(b)
Describe the structure of proteins. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (5)
(c)
Describe how proteins are digested in the human gut. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4) (Total 15 mark s)
Page 19 of 79
11
Scientists used fossil leaves from one species of pine tree to investigate whether changes in the concentration of carbon dioxide in the air over long periods of time had led to changes in the number of stomata in the leaves. Their method is outlined below. • •
They selected sites of different ages. They collected between 11 and 24 fossil leaves from each site.
• •
They found the mean number of stomata per mm 2 on the leaves from each site. They estimated the age of each sample by dating organic remains around the leaves at each site.
They compared results from the fossil leaves with leaves from the same species of pine tree growing today. They knew the concentration of carbon dioxide in the air at different times in the past. Their results are shown in the table.
Age of sample / years present day 5000
(a)
Concentration of carbon dioxide in the air / % 0.0350 0.0270
Mean number of stomata per mm2 (± standard deviation) 92 (±2) 87 (±4)
10 000
0.0250
95 (±2)
15 000
0.0205
108 (±6)
20 000
0.0195
115 (±4)
25 000
0.0188
118 (±6)
30 000
0.0190
130 (±6)
The concentration of carbon dioxide in the air has changed with time. Use the data to describe how. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2)
Page 20 of 79
(b)
The scientists calculated the mean number of stomata per mm2 and the standard deviation. What does the standard deviation show? ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(c)
(2)
The scientists found the age of the fossil leaves by dating the organic remains around them. Would this have affected the accuracy of their data? Explain your answer. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (1)
(d)
30 000 years ago the mean number of stomata per mm2 on the lower epidermis of pine tree leaves was muchdioxide higher than it is today.ofThis when the carbon concentration the would air washave low.enabled the plant to grow faster Explain why. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (1)
Page 21 of 79
(e)
A student who saw these results concluded that as the carbon dioxide concentration of the air had increased the number of stomata per mm2 in leaves had decreased. Do the results support this conclusion? ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ (3)
(f)
The leaves of plants that grow in dry areas usually have a low number of stomata per mm2. Use your knowledge of leaf structure to suggest three other adaptations that the leaves might have that enable the plants to grow well in dry conditions. 1
.....................................................................................................................
2
.....................................................................................................................
3
..................................................................................................................... (3) (Total 12 mark s)
Page 22 of 79
12
The figure below represents a capillary surrounded by tissue fluid. The values of the hydrostatic pressure are shown.
Arteriole end
direction of blood flow
Hydrostatic pressure = 4.3 kPa
Venule end
Hydrostatic pressure = 1.6 kPa
Tissue fluid Hydrostatic pressure = 1.1 kPa (a)
Use the information in the figure above to explain how tissue fluid is formed. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2)
(b)
The hydrostatic pressure falls from the arteriole end of the capillary to the venule end of the capillary. Explain why. ........................................................................................................................ ........................................................................................................................ (1)
(c)
High blood pressure leads to an accumulation of tissue fluid. Explain how. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ (3)
Page 23 of 79
(d)
The water potential of the blood plasma is more negative at the venule end of the capillary than at the arteriole end of the capillary. Explain why. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ (3) (Total 9 marks )
13
Read the following passage. During the course of a day, we come into contact with many poisonous substances. These include industrial and household chemicals. The skin acts as a barrier and prevents many of these substances entering and harming the body.
5
The skin is one of the largest organs in the body. It is composed of several layers of tissue. The outer layer consists of dead cells packed with keratins. Keratins are a group of proteins that differ from each other in their primary structure. Each keratin molecule consists of several polypeptide chains, each individual chain wound into a spiral or helix. The polypeptide chains include many sulphur-containing amino acids and these help to give the keratin molecules their characteristic strength.
Use information from the passage and your own knowledge to answer the questions. (a)
What is the evidence from the passage that keratin molecules have a quaternary structure? ...................................................................................................................... ......................................................................................................................
(1)
Page 24 of 79
(b)
Explain how sulphur-containing amino acids help to give keratin molecules their characteristic strength (lines 8–9). ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2)
(c)
Explain (line 6). why differences in primary structure result in keratins with different properties ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2)
(d)
The skin prevents poisonous substances entering and harming the body (line 3). Explain why these substances are unable to pass through the outer layer of skin cells by active transport. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (3)
Page 25 of 79
(e)
Skin cells may be studied with a transmission electron microscope or an optical microscope. Explain the advantages and limitations of using a transmission electron microscope to study cells. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (6) (Total 14 mark s)
Page 26 of 79
14
Nicotine is the addictive substance in tobacco. When nicotine reaches the brain, it binds to a specific protein. This causes the release of chemicals that give a feeling of reward to the smoker. This reward is part of the reason why people find it difficult to stop smoking. Scientists have developed a vaccine against nicotine to help people stop smoking. They set up an investigation, which involved a large number of volunteers. Once a month for 5 months, one group of volunteers was given the vaccine and the other group was given a placebo. At regular intervals, the scientists measured the concentration of antibodies to nicotine in the blood of each group of volunteers. They also calculated the percentage of volunteers who had stopped smoking from months 2 to 6 of the investigation. (a)
(i)
In this investigation, neither the volunteers nor the scientists knew if a particular volunteer was receiving the vaccine or a placebo. Suggest two reasons why this made the scientists’ results more reliable. 1 ............................................................................................................ ............................................................................................................... 2 ............................................................................................................ ............................................................................................................... (2)
(ii)
The scientists measured the concentration of nicotine in the blood of two volunteers who smoked the same number of cigarettes per day. Suggest two reasons why the concentration of nicotine in the blood of these smokers might be different. 1 ............................................................................................................ ............................................................................................................... 2 ............................................................................................................ ............................................................................................................... (2)
Page 27 of 79
(b)
(i)
Suggest how this vaccine could help people to stop smoking. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ...............................................................................................................
(Extra space) ........................................................................................ ............................................................................................................... ............................................................................................................... (3)
(ii)
Some people have suggested that this vaccine should not be given free to smokers on the National Health Service (NHS). Evaluate this suggestion. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ...............................................................................................................
(Extra space) ........................................................................................ ............................................................................................................... ............................................................................................................... (3)
The scientists measured the concentration of antibodies to nicotine in the blood of the volunteers for 12 months after the first vaccination. As a result of these measurements, they divided the volunteers who received the nicotine vaccine into three groups: • • •
high antibody responders medium antibody responders low antibody responders.
The figure below shows their results. The scientists also recorded the number of volunteers who had stopped smoking from months 2 to 6 of the investigation. Page 28 of 79
The table below shows these results.
Month when vaccine or placebo was given
Page 29 of 79
Group
Percentage of volunteers who had stopped smoking from months 2 to 6 of the investigation
Highantibodyresponders
56.6
Lowantibodyresponders
38.1
Mediumantibodyresponders Placebo (c)
32.1 31.3
A journalist reported that this vaccine is a major breakthrough in helping people to stop smoking. Do these data support this statement? Explain your answer. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (5) (Total 15 mark s)
Page 30 of 79
15
Many sports drinks contain water, sodium chloride and carbohydrates. The manufacturers of the sports drinks claim that carbohydrates provide an energy boost. The sodium chloride is used to increase absorption of glucose in the small intestine. Scientists investigated the effect of a sports drink on the performance of runners in 5 km races. They recruited 100 runners who had previously run a 5 km race in similar times. During this race, Race 1, they had water they could drink. The scientists divided the runners into two equal groups, P and Q. Both groups ran a second 5 km race, Race 2. During this race: •
group P had water available
•
group Q had the sports drink available.
The scientists recorded the mean time for each group to complete this race. The following figure shows their results.
(a)
Use the figure to calculate the percentage decrease in the mean time taken for group Q to complete Race 2 compared with Race 1. Show your working.
............................. % (2)
Page 31 of 79
(b)
One of the runners concluded that the sports drink improved performance. Do these data support his conclusion? ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3)
(c)
The runners were matched for the time taken to run the first race. Give three other factors for which they should have been matched. Factor 1 .......................................................................................................... ........................................................................................................................ Factor 2 .......................................................................................................... ........................................................................................................................ Factor 3 .......................................................................................................... ........................................................................................................................ (3)
Page 32 of 79
(d)
The sports drink contains sodium chloride. Sodium chloride increases uptake of glucose in the small intestine. Explain how. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4) (Total 12 mark s)
16
(a)
Describe how DNA is replicated. (6)
(b)
The graph shows information about the movement of chromatids in a cell that has just started metaphase of mitosis.
Page 33 of 79
(i)
What was the duration of metaphase in this cell? minutes (1)
(ii)
Use line X to calculate the duration of anaphase in this cell. minutes (1)
(iii)
Complete line Y on the graph. (2)
(c)
A doctor investigated the number of cells in different stages of the cell cycle in two tissue samples, C and D. One tissue sample was taken from a cancerous tumour. The other was taken from non-cancerous tissue. The table shows his results.
Percentage of cells in each stage of the cell cycle Stage of the cell cycle
(i)
Tissue sample C
Tissue sample D
Interphase
82
45
Prophase
4
16
Metaphase
5
18
Anaphase
5
12
Telophase
4
9
In tissue sample C, one cell cycle took 24 hours. Use the data in the table to calculate the time in which these cells were in interphase during one cell cycle. Show your working.
Time cells in interphase ...................................... hours (2)
Page 34 of 79
(ii)
Explain how the doctor could have recognised which cells were in interphase when looking at the tissue samples. ............................................................................................................... ............................................................................................................... ............................................................................................................... (1)
(iii)
Which tissue sample, C or D, was taken from a cancerous tumour? Use information in the table to explain your answer. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (2) (Total 15 mark s)
Page 35 of 79
17
(a)
Explain how the structure of DNA is related to its functions. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) .................................................................................................. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (6)
Page 36 of 79
Scientists investigated three genes, C, D and E, involved in controlling cell division. They studied the effect of mutations in these genes on the risk of developing lung cancer. The scientists analysed genes C, D and E from healthy people and people with lung cancer. • •
If a person had a normal allele for a gene, they used the symbol N. If a person had two mutant alleles for a gene, they used the symbol M.
They used their data to calculate the risk of developing lung cancer for people with different combinations of N and M alleles of the genes. A risk value of 1.00 indicates no increased risk. The following table shows the scientists’ results.
GeneC
GeneD
GeneE
Risk of developing lung cancer
N
N
N
1.00
M
N
N
1.30
N
N
M
1.78
N
M
N
1.45
N = at least one copy of the normal allele is present M = two copies of the mutant allele are present (b)
What do these data suggest about the relative importance of the mutant alleles of genes C, D and E on increasing the risk of developing lung cancer? Explain your answer. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3)
Page 37 of 79
Chemotherapy is the use of a drug to treat cancer. The drug kills dividing cells. The figure below shows the number of healthy cells and cancer cells in the blood of a patient receiving chemotherapy. The arrows labelled F to I show when the drug was given to the patient.
Time / days (c)
Calculate the rate at which healthy cells were killed between days 42 and 46.
.............. cells killed per unit volume of blood per day (1)
Page 38 of 79
(d)
Describe similarities and differences in the response of healthy cells and cancer cells to the drug between times F and G. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ (3)
(e)
More cancer cells could be destroyed if the drug was given more frequently. Suggest why the drug was not given more frequently. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (Total 15 mark s)
Page 39 of 79
18
Read the following passage. In a human, there are over 200 different types of cell clearly distinguishable from each other. What is more, many of these types include a number of different varieties. White blood cells, for example, include lymphocytes and granulocytes.
5
10
Although different animal cells have many features in common, each type has adaptations. associated with its function in the organism. As an example, most cells contain the same organelles, but the number may differ from one type of cell to another. Muscle cells contain many mitochondria, while enzyme-secreting cells from salivary glands have particularly large amounts of rough endoplasmic reticulum. The number of a particular kind of organelle may change during the life of the cell. An example of this change is provided by cells in the tail of a tadpole. As a tadpole matures into a frog, its tail is gradually absorbed until it disappears completely. Absorption is associated with an increase in the number of lysosomes in the cells of the tail.
Use information from the passage and your own knowledge to answer the following questions. (a)
Explain the link between. (i)
mitochondria and muscle cells (lines 6 - 7); ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (3)
(ii)
rough endoplasmic reticulum and enzyme-secreting cells from salivary glands (lines 7 - 8). ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2)
Page 40 of 79
(b)
Use information in the passage to explain how a tadpole’s tail is absorbed as a tadpole changes into a frog. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2)
(c)
Starting with some lettuce leaves, describe how you would obtain a sample oftoundamaged chloroplasts. Use your knowledge of cell fractionation and ultracentrifugation answer this question. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (6) (Total 13 mark s)
Page 41 of 79
19
(a)
ATP is useful in many biological processes. Explain why. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4)
Page 42 of 79
(b)
Describe how ATP is made in mitochondria. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (6)
Page 43 of 79
(c)
Plants produce ATP in their chloroplasts during photosynthesis. They also produce ATP during respiration. Explain why it is important for plants to produce ATP during respiration in addition to during photosynthesis. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................
(Extra space) ................................................................................................. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (5) (Total 15 mark s)
Page 44 of 79
20
Read the following passage.
5
Straw consists of three main organic substances – cellulose, hemicellulose and lignin. Cellulose molecules form chains which pack together into fibres. Hemicellulose is a small molecule formed mainly from five-carbon (pentose) sugar monomers. It acts as a cement holding cellulose fibres together. Like hemicellulose, lignin is a polymer, but it is not a carbohydrate. It covers the cellulose in the cell wall and supplies additional strength. In addition to these three substances, there are small amounts of other biologically important polymers present.
The other main component of straw is water. Water content is variable but may be determined by heating a known mass of straw at between 80 and 90°C until it reaches a constant mass. 10 The loss in mass is the water content. Since straw is plentiful, it is possible that it could be used for the production of a range of organic substances. The first step is the conversion of cellulose to glucose. It has been suggested that an enzyme could be used for this process. There is a difficulty here, however. The lignin which covers the cellulose protects the cellulose from enzyme attack. Use information from the passage and your own knowledge to answer the following questions. (a)
(i)
Give one way in which the structure of a hemicellulose molecule is similar to the structure of a cellulose molecule. ............................................................................................................. ............................................................................................................. (1)
(ii)
Complete the table to show two ways in which the structure of a hemicellulose molecule differs from the structure of a cellulose molecule.
Hemicellulose
Cellulose
..........................................................
..........................................................
..........................................................
..........................................................
..........................................................
..........................................................
..........................................................
.......................................................... (2)
(b)
Name one biologically important polymer, other than those mentioned in the passage, which would be found in straw. ...................................................................................................................... (1) Page 45 of 79
(c)
Explain why the following steps were necessary in finding the water content of straw: (i)
heating the straw until it reaches constant mass(line 9); ............................................................................................................. ............................................................................................................. .............................................................................................. (1)
(ii)
not heating the straw above 90°C (line 9). ............................................................................................................. ............................................................................................................. ............................................................................................................. ............................................................................................................. (2)
(d)
A covering of lignin protects cellulose from enzyme attack (line 14). Use your knowledge of the way in which enzymes work to explain why cellulose-digesting enzymes do not digest lignin. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (2)
Page 46 of 79
(e)
Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function in cells. ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... ...................................................................................................................... (6) (Total 15 mark s)
Page 47 of 79
Mark schemes 1
(a)
Starch (max 3) 1. Helical/ spiral shape so compact; 2. Large (molecule)/insoluble so osmotically inactive; 3. 4.
Accept: does not affect water potential/ψ. Branched so glucose is (easily) released for respiration;
Ignore: unbranched. Large (molecule) so cannot leave cell/cross cell-surface membrane;
Cellulose (max 3) 5. Long, straight/unbranched chains of 6. Joined by hydrogen bonding;
7. 8.
glucose;
Note: references to ‘strong hydrogen bonds’ disqualifies this mark point. To form (micro/macro)fibrils; Provides rigidity/strength; 5 max
(b)
1.
(At source) sucrose is actively (transported) into the phloem/sieve element/tube;
Accept: ‘sugar/s’ for sucrose but reject other named sugars e.g. glucose. 2.
Accept: co-transport (with H+ ions). By companion/transfer cells;
4.
Lowers water potential in phloem/sieve element/tube and water enters by osmosis; (Produces) high (hydrostatic) pressure;
5.
Accept: pressure gradient. Mass flow/transport towards sink/roots/storage tissue;
6.
Accept: sieve element/tube. At sink/roots sugars are removed/unloaded;
3.
Accept: at sink/roots sugars are used in respiration/stored. 5 max
[10]
2
(a)
1.
Number of (individuals of) each species;
Accept: ‘population’ for ‘number’ 2.
Total number of individuals / number of species; Accept: ‘species richness’
MP2 allows for other types of diversity index 2
Page 48 of 79
(b)
(i)
(Shows) results are due to the herbicide / are not due to another factor / (to) compare the effect of using and not using the herbicide / shows the effect of adding the herbicide;
Neutral: allows a comparison Neutral: ensures results are due to the independent variable Reject: ‘insecticide’ Accept: ‘pesticide’ 1
(ii)
1.
(More) weeds killed so more crops / plants survive / higher yield / less competition;
2.
High concentrations (of herbicide) harm / damage / kill / are toxic to crops / plants;
Accept: ‘pesticide’ Neutral: ‘insecticide’ Accept: use of figures (eg 400+) 2
(iii)
1.
Reduced plant diversity / fewer plant species / fewer varieties of plant;
Accept: ‘weed’ for ‘plant’ Neutral: fewer plants Accept: only one crop species remains 2.
Fewer habitats / niches; Q Neutral: fewer homes / shelters
3.
Fewer food sources / varieties of food;
Neutral: less food 3
[8]
Page 49 of 79
3
(a)
1.
No interbreeding / gene pools are separate / geographic(al) isolation;
2. 3. 4.
Accept: reproductive isolation as an alternative to no interbreeding. Mutation linked to (different) markings/colours; Selection/survival linked to (different) markings/colours; Adapted organisms breed / differential reproductive success;
5.
Note: ’passed on to offspring’ on its own isnot sufficient for reproduction. Change/increase in allele frequency/frequencies; 5
(b)
1.
(Compare DNA) base sequence / base pairing / (DNA) hybridisation; Ignore: compare chromosomes / ‘genetic make-up’.
Accept: (compare) genes / introns / exons. 2.
Note: reference to only comparing alleles is 1 max. Different in six (species) /different in different species / similar in three (subspecies) /similar in same species/subspecies; Ignore: compare chromosomes / ‘genetic make-up’. Reject: ‘same alleles/ same DNA bases in three species/subspecies’. Note: mark point 2 can be awarded without mark point 1. 2 [7]
(a)
breed together; if fertile offspring, then same species;
4
2
(b)
isolation of two populations; variation already present due to mutations; different environmental conditions / selection pressures leading to selection of different features and hence different alleles; different frequency of alleles; separate gene pools / no interbreeding; 4
(c)
selection of mate dependent on colour pattern; prevents interbreeding / keeps gene pools separate; 2
[8]
5
(a)
(i)
(In all organisms / DNA,) the same triplet codes for the same amino acid;
Accept codon / same three bases / nucleotides Accept plurals if both triplets and amino acids Reject triplets code for an amino acid Reject reference to producing amino acid 1
(ii)
64; 1
Page 50 of 79
(b)
Splicing;
Ignore deletion references Accept RNA splicing 1
(c)
(i)
1.
(Mutation) changes triplets / codons after that point / causes frame shift;
Accept changes splicing site Ignore changes in sequence of nucleotides / bases 2.
Changes amino acid sequence (after this) / codes for different amino acids (after this);
Accept changes primary structure Reject changes amino acid formed / one amino acid changed 3.
Affects hydrogen / ionic / sulfur bond (not peptide bond);
4.
Changes tertiary structure of protein (so non-functional);
Neutral 3-D structure 3 max
(ii)
1.
Intron non-coding (DNA) / only exons coding;
Context is the intron Do not mix and match from alternatives Neutral references to introns removed during splicing 1.and 2. Ignore ref. to code degenerate and get same / different amino acid in sequence 2.
(So) not translated / no change in mRNA produced / no effect (on protein) / no effect on amino acid sequence;
Accept does not code for amino acids OR 3.
Prevents / changes splicing;
4.
(So) faulty mRNA formed;
Accept exons not joined together / introns not removed 5.
Get different amino acid sequence; 2 max
[8]
Page 51 of 79
6
(a)
1.
Type of feed affects (antibiotic) resistant bacteria (in animals);
Accept: null hypotheses Accept predictions, for example
2.
More antibiotic resistant bacteria form in animals fed with antibiotics in their food (Antibiotic) resistant resistant infect /are passed on to animals/farmer / resistant resistant are passed between animals; Accept: bird to bird/bird to human/human to human Accept: a link (exists) between (antibiotic) resistance in animals and their keepers/farmers – as lowest level QWC
3.
4.
Incidence turkeys; of (antibiotic) resistant resistant differs in chickens and
Accept: a comparison, eg ‘more resistant bacteria in chickens than turkeys' Incidence of (antibiotic) resistant resistant differs in chicken farmers and turkey farmers; Accept: a comparison, eg ‘more resistant bacteria in chickens than turkeys' Max 2
(b)
(i)
1.
Large(r) percentage of resistant bacteria in turkeys/low(er) percentage of resistant bacteria in chickens;
Accept: E coli for bacteria 2.
Ignore: number, eg. ignore ‘more’/’fewer’ turkeys/chickens Large(r) percentage of resistant bacteria in turkey farmers/low(er) percentage of resistant bacteria in chicken farmers; 2
(ii)
1.
(More) antibiotic in turkey feed kills (more) non-resistant bacteria / resistant bacteria survive;
Accept: antibiotic creates selection pressure 2.
Survive must be explicit, notimplied by ‘reproduce’ (Resistant bacteria) reproduce / pass on gene for resistance; 2
(c)
(Human) faeces contain pathogens;
Accept: harmful organisms (d)
1.
Large number of farms / farmers (surveyed) / 46;
2.
‘Reliable’ is used in the question stem So results are (likely to be) representative / can identify anomalous results;
1
Ignore: reproducible / accurate / valid / reliable Accept valid explanation of replicates minimising effects of chance 2
Page 52 of 79
(e)
1. 2.
(DNA) hybridisation (of gene for resistance in bacteria taken from bird and farmer); (Identical) strands separate at high(est) temperature; OR
3. 4.
Compare base/nucleotide sequence (of gene for resistance in bacteria taken from bird and farmer); (Identical strains) have identical/same base sequences
Mark in pairs, do not mix and match. Accept: bacteria in bird and farmer/both types of bacteria have identical base sequences = 2 marks (f)
1.
(Antibiotic use has) increased cases of bacterial resistance;
2.
Accept: number Transfer/horizontal transmission of (resistance) gene to pathogens/harmful bacteria;
3.
Accept: conjugation (Antibiotic) resistant bacteria cause harm / medical treatments less effective;
4. 5. 6.
Accept: superbug Avoids side effects on animals; Increased demand for organic food; Antibiotic/resistant bacteria could be present in human food;
7. 8.
High cost of antibiotics; Legislation has controlled antibiotic use;
2
Accept: EU/government guidelines 4 max [15]
7
(a)
1. 2.
Dissolve in alcohol, then add water; White emulsion shows presence of lipid. 2
(b)
Glycerol. 1
(c)
Ester. 1
(d)
Y (no mark) Contains double bond between (adjacent) carbon atoms in hydrocarbon chain. 1
(e)
1. 2.
Divide mass of each lipid by total mass of all lipids (in that type of cell); Multiply answer by 100. 2
(f)
Red blood cells free in blood / not supported by other cells so cholesterol helps to maintain shape;
Allow converse for cell from ileum – cell supported by others in endothelium so cholesterol has less effect on maintaining shape. 1
Page 53 of 79
(g)
1. 2. 3.
Cell unable to change shape; (Because) cell has a cell wall; (Wall is) rigid / made of peptidoglycan / murein. 2 max
[10]
8
(a)
0.32.
Correct answer = 2 marks Accept 32% for 1 mark max Incorrect answer but identifying 2pq as heterozygous = 1 mark 2
(b)
1. 2. 3. 4.
Mutation produced KDR minus / resistance allele; DDT use provides selection pressure; Mosquitoes with KDR minus allele more likely (to survive) to reproduce; Leading to increase in KDR minus allele in population. 4
(c)
1. 2.
Neurones remain depolarised; So no action potentials / no impulse transmission. 2
(d)
1. 2.
(Mutation) changes shape of sodium ion channel (protein) / of receptor (protein); DDT no longer complementary / no longer able to bind. 2
[10]
9
(a)
Banding pattern changes as cheetah gets older / difficult to judge as tail is short / fluffy; 1
(b)
(i)
Mean not (always) a whole number; Standard deviation not (always) zero; 2
(ii)
Movement of tail / angle of sight / confused it with another band / subjective estimation;
Accept reference to Figure 1 E.g. Bands 2 and 3 have same thickness but look different 1
(c)
Band width not the same on both sides of tail; 1
(d)
Offspring of the same family will be more similar genetically; As have same mother (and father) / parent; Expect to see more differences in randomly chosen cheetahs; 3
[8]
Page 54 of 79
10
(a)
1. 2. 3. 4. 5. 6. 7.
Helicase; Breaks hydrogen bonds; Only one DNA strand acts as template; RNA nucleotides attracted to exposed bases; (Attraction) according to base pairing rule; RNA polymerase joins (RNA) nucleotides together; Pre-mRNA spliced to remove introns. 6 max
(b)
1. 2. 3.
Polymer of amino acids; Joined by peptide bonds; Formed by condensation;
4. 5.
Primary structure is order of amino acids; Secondary structure is folding of polypeptide chain due to hydrogen bonding;
Accept alpha helix / pleated sheet 6. 7.
Tertiary structure is 3-D folding due to hydrogen bonding and ionic / disulfide bonds; Quaternary structure is two or more polypeptide chains. 5 max
(c)
1. 2. 3. 4.
Hydrolysis of peptide bonds; Endopeptidases break polypeptides into smaller peptide chains; Exopeptidases remove terminal amino acids; Dipeptidases hydrolyse / break down dipeptides into amino acids. 4
[15]
11
(a)
1.
The more recent the sample the greater the concentration;
Accept converse This could be expressed by reference to time e.g. ‘concentration has increased since 25 000 years ago 2.
Increases most in last 5000 years / more or less constant / slight increase between 30 000 and 15 000 years ago; 2
(b)
1.
Variation in data / spread of data;
Reject references to range e.g. ‘range of data’ 2.
Around the mean;
Both marks are possible in the context of using the data 2
(c)
1.
Yes as pine leaves not in organic matter of the same age;
2.
No as organic matter would be the same age as the pine leaves;
Accept either approach 1 max
Page 55 of 79
(d)
Can get more CO2 for photosynthesis;
More CO2 enters leaf is insufficient. Accept light-independent (reaction) as equivalent 1
(e)
Any three from: 1.
(Overall data show) negative correlation;
Do not allow description of correlation because in question stem 2.
Little change in number of stomata in last 10 000 years;
3.
Small sample size;
4.
Only one species studied;
5.
Other factors / named factor may have affected number of stomata;
6.
Evidence does not support the conclusion between 30 000 and 25 000 years ago / between 5000 years ago and present day;
Accept reference to either one of these age ranges 7.
Appropriate reference to standard deviations (in comparing means);
E.g. no overlap between 15 000 and 10 000 years ago 3 max
(f)
Any three from : 1.
Thick cuticle;
2.
Small leaves / low surface area;
Accept other ways of describing ‘small’, e.g. ‘needle-like’ 3.
Hairy leaves;
4.
Sunken stomata;
5.
Rolled leaves; 3 max
[12]
(a)
12
1. 2.
(Overall) outward pressure of 3.2 kPa; Forces small molecules out of capillary. 2
(b)
Loss of water / loss of fluid / friction (against capillary lining). 1
(c)
1. 2. 3.
High blood pressure = high hydrostatic pressure; Increases outward pressure from (arterial) end of capillary / reduces inward pressure at (venule) end of capillary; (So) more tissue fluid formed / less tissue fluid is reabsorbed.
Allow lymph system not able to drain tissues fast enough 3
Page 56 of 79
(d)
1. 2. 3.
Water has left the capillary; Proteins (in blood) too large to leave capillary; Increasing / giving higher concentration of blood proteins (and thus wp). 3
[9]
13
(a)
Several / more than one polypeptide chain in molecule;
Evidence must only relate to 4º structure 1
(b)
Chemical bonds formed between sulphur-containing groups / R-groups / form stronger disulphide bonds; Bind chain(s) to each other; 2
(c)
Different number / sequences of amino acids; Bonds in different places which gives different shape; 2
(d)
Outer layer of skin cells are dead / do not respire Do not contain mitochondria / do not produce ATP / release energy; Cells do not have required proteins / carriers; 3
(e)
Advantages: 1
Small objects can be seen;
2
TEM has high resolution as wavelength of electrons shorter;
Accept better Limitations: 3
Cannot look at living cells as cells must be in a vacuum;
4
must cut section / thin specimen;
5
Preparation may create artefact
6
Does not produce colour image; 6
[14]
14
(a)
(i)
1.
(Scientists) can‘t show bias / influence / may have a vested interest / work for the company developing the vaccine;
Relates to the scientists 2.
(Volunteers) can’t show psychological / mental effects / ‘placebo effect’ / expectations;
Relates to the volunteers Accept: reduces the ‘Hawthorne effect’ /demand characteristics Neutral: so they have no idea what they are taking 2
Page 57 of 79
(ii)
Any two suitable suggestions, eg
Neutral: refs. to age and health 1.
Amount of nicotine in cigarettes;
Neutral: different types of cigarette / different ways / frequency of smoking 2.
Amount inhaled / absorbed / time since last cigarette;
Neutral: absorption by gut / digestion Accept: absorption by mouth 3.
(Different) amounts excreted / metabolism / rate of binding (of nicotine) to protein;
Accept: broken down (differently) 4.
(Different) blood volumes;
Neutral: different body masses 5.
Nicotine from passive smoking / other smokers / other sources;
6.
Some volunteers received the vaccine / placebo;
Accept: some volunteers would have / would not have the antibodies 2 max
(b)
(i)
1.
Antibodies to nicotine produced / antibodies bind to nicotine; Q Reject: vaccine contains / produces antibodies Q Neutral: antibodies digest / kill / fight nicotine
2.
(So) nicotine does not bind to protein / does not reach the brain; Q Reject: any reference to ‘active site’
Neutral: idea that the antibodies bind to the protein 3.
(So) cigarettes / smoking does not satisfy addiction / reward smokers / release (reward) chemicals; 3
(ii)
(Agree): 1. 2.
People choose to smoke / know the risks; Should spend this money on education / preventing people from starting to smoke / treating other health problems / vaccines are expensive;
(Disagree): 3.
Unethical not to treat;
4.
Less money needed to treat the effects of smoking / cancer / smokers pay taxes so are entitled to treatment; 3 max
Page 58 of 79
(c)
1.
High antibody responders have a high % to stop smoking / are more likely to stop smoking;
‘People producing a high concentration of antibodies’ is equivalent to ‘high antibody responders’ Accept: reference to values from the table 2.
Only a few may be high antibody responders / no numbers on how many are high / medium / low antibody responders;
Neutral: not all people are high antibody responders 3.
Percentage who stopped smoking is similar for placebo group and low / medium responders / some / % of placebo group (still) stopped smoking / placebo has the lowest value / % to stop smoking;
Accept: reference to values from the table 4.
Large sample size / double blind so reliable / representative;
5.
Antibody levels peak at / drop after 5 months / boosters may be needed at / after 5 months;
6.
May start smoking again after 5 / 6 months / do not know the percentage who stopped smoking after 5 / 6 months;
7.
Nicotine is not the only factor responsible for making people smoke;
Must mention nicotine Do not accept: correlation does not mean causation / could be due to other factors 5 max
[15]
15
(a)
Answer of 9.09 / 9.1;; = 2 marks Calculation of the difference in mean time (2) divided by srcinal time (22); = 1 mark
Ignore number of decimal places as long as they are correct 2
Page 59 of 79
(b)
(Yes)
Can mix and match yes or no approach, all 5 responses are available (No) 1.
Faster running time after sports drink;
‘Faster running time in group Q’ is insufficient but accept ‘faster running time in group Q in Race 2’ 2.
Mean times given so there will be variation in the group;
3.
No standard deviations to know the spread of the data (about the mean) / whether they overlap;
Accept ‘no stats analysis’ 4.
Improvement in running time only small in both groups / both groups improved in Race 2;
5.
Did not drink the same volumes; 3 max
(c)
1.
Age;
2.
Gender / sex;
3.
Ethnicity;
4.
Food / fluid intake before the race;
Any fluid / food is included here eg coffee, alcohol 5.
Amount of sleep / rest / exercise before the race;
6.
Reference to one named health factor eg diabetic or non-diabetic, smoker or non-smoker;
Reference to medication is included here 3 max
Page 60 of 79
(d)
1.
Sodium ions and glucose absorbed by co-transport;
Only penalise omission of ‘ions’ once in marking points 1,3, 4 and 5 2.
(Co-transport) via carrier / channel protein;
Accept via symport Only reward reference to carrier / channel proteins in the context of co-transport 3.
Sodium ions removed (from epithelial cell) by active transport into blood;
4.
Maintains low concentration of sodium ions (in epithelial cell) / maintains sodium ion concentration gradient (between small intestine and epithelial cell);
Principle: marking points 3, 5, and 6 require consideration of ‘what moves’, ‘where it moves to’ and ‘how it moves’ to achieve credit 5.
Sodium ions enter epithelial cells by facilitated diffusion taking glucose with them (from small intestine);
Reference to diffuse / diffusion for movement is required. Accept facilitated diffusion 6.
Glucose moved by facilitated diffusion into blood (from epithelial cells); 4 max
[12]
(a)
1.
16
Strands separate / H-bonds break;
1. Q Neutral: strands split 1. Accept: strands unzip 2.
DNA helicase (involved);
3.
Both strands / each strand act(s) as (a) template(s);
4.
(Free) nucleotides attach;
4. Neutral: bases attach 4. Accept: nucleotides attracted 5.
Complementary / specific base pairing / AT and GC;
6.
DNA polymerase joins nucleotides (on new strand);
6. Reject: if wrong function of DNA polymerase 7.
H-bonds reform;
8.
Semi-conservative replication / new DNA molecules contain one old strand and one new strand;
8. Reject: if wrong context e.g. new DNA molecules contain half of each srcinal strand 6 max
Page 61 of 79
(b)
(i)
18;
Do not accept 17.5 1
(ii)
10; 1
(iii)
1.
Horizontal until 18 minutes;
Allow + / - one small box 2.
(Then) decreases as straight line to 0
m at 28 minutes;
2. Allow lines that start from the wrong place, ending at 0 at 28 minutes 2
(c)
(i)
Two marks for correct answer of 19.68 or 19.7;;
Accept 19hrs 41mins One mark for incorrect answers in which candidate clearly multiplies by 0.82;
Allow one mark for incorrect answers that clearly show 82% of 24 (hours) 2
(ii)
1.
No visible chromosomes / chromatids / visible nucleus; 1
(iii)
D (no mark) 1.
Lower % (of cells) in interphase / higher % (of cells) in mitosis / named stage of mitosis;
1. Accept: ‘less’ or ‘more’ instead of ‘%’ 1. Do not accept: higher % (of cells) in each / all stage(s) 2.
(So) more cells dividing / cells are dividing quicker;
2. Accept: uncontrolled cell division 2. Do not award if TissueC is chosen 2
[15]
Page 62 of 79
17
(a)
1.
Sugar-phosphate (backbone) / double stranded / helix so provides strength / stability / protects bases / protects hydrogen bonds;
Must be a direct link / obvious to get the mark Neutral: reference to histones 2.
Long / large molecule so can store lots of information;
3.
Helix / coiled so compact;
Accept: can store in a small amount of space for ‘compact’ 4.
Base sequence allows information to be stored / base sequence codes for amino acids / protein; Accept: base sequence allows transcription
5.
Double stranded so replication can occur semi-conservatively / strands can act as templates / complementary base pairing / A-T and G-C so accurate replication / identical copies can be made;
6.
(Weak) hydrogen bonds for replication / unzipping / strand separation / many hydrogen bonds so stable / strong;
Accept: 'H-bonds' for ‘hydrogen bonds’ 6
(b)
1.
(Mutation) in E produces highest risk / 1.78;
2.
(Mutation) in D produces next highest risk / 1.45;
3.
(Mutation) in C produces least risk / 1.30;
Must be stated directly and not implied E > D > C = 3 marks
Accept: values of 0.78, 0.45 and 0.30 for MP1, MP2 and MP3 respectively If no mark is awarded, a principle mark can be given for the idea that all mutant alleles increase the risk 3
(c)
180; 1
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(d)
(Similarities): 1.
Same / similar pattern / both decrease, stay the same then increase;
2.
Number of cells stays the same for same length of time;
Ignore: wrong days stated (Differences): (Per unit volume of blood) 3.
Greater / faster decrease in number of healthy cells / more healthy cells killed / healthy cells killed faster; Accept: converse for cancer cells
Accept: greater percentage decrease in number of cancer cells / greater proportion of cancer cells killed 4.
Greater / faster increase in number of healthy cells / more healthy cells replaced / divide / healthy cells replaced / divide faster;
Accept: converse for cancer cells For differences, statements made must be comparative 3 max
(e)
1.
More / too many healthy cells killed;
2.
(So) will take time to replace / increase in number;
Neutral: will take time to ‘repair’ 3.
Person may die / have side effects; 2 max
[15]
18
(a)
(i)
Mitochondria site of respiration; Production of ATP / release of energy; For contraction;
Do not award credit for making or producing energy. 3
(ii)
Enzymes are proteins; Proteins synthesised / made on ribosomes; 2
(b)
Lysosomes produce / contain enzymes; Which break down / hydrolyse proteins / substances / cells of tail; 2
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(c)
1. Chop up (accept any reference to crude breaking up); 2. Cold; 3. Buffer solution; 4. Isotonic / same water potential; 5. Filter and centrifuge filtrate; 6. Centrifuge supernatant; 7. At higher speed; 8. Chloroplasts in (second) pellet; max 6
[13]
(a)
1.
Releases energy in small / manageable amounts;
1. Accept less than glucose
19 2.
(Broken down) in a one step / single bond broken immediate energy compound / makes energy available rapidly;
2. Accept easily broken down 3.
Phosphorylates / adds phosphate makes (phosphorylated substances) more reactive / lowers activation energy;
3. Do not accept phosphorus or P on its own 4.
Reformed / made again;
4. Must relate to regeneration 4
(b)
1.
Substrate level phosphorylation / ATP produced in Krebs cycle; Accept alternatives for reduced NAD
2.
Krebs cycle / link reaction produces reduced coenzyme / reduced NAD / reduced FAD;
2. Accept description of either Krebs cycle or link reaction 3.
Electrons released from reduced / coenzymes / NAD / FAD;
4.
(Electrons) pass along carriers / through electron transport chain / through series of redox reactions;
5.
Energy released;
5. Allow this mark in context of electron transport or chemiosmosis 6.
ADP / ADP + Pi;
6. Accept H+ or hydrogen ions and cristae 7.
Protons move into intermembrane space;
7. Allow description of movement through membrane 8.
ATP synthase;
8. Accept ATPase. Reject stalked particles 6 max
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(c)
1.
In the dark no ATP production in photosynthesis;
1. In context of in photosynthetic tissue / leaves 2.
Some tissues unable to photosynthesise / produce ATP;
3.
ATP cannot be moved from cell to cell / stored;
4.
Plant uses more ATP than produced in photosynthesis;
5.
ATP for active transport / synthesis (of named substance); 5
[15]
20
(a)
(i)
both are polymers / polysaccharides / built up from many sugar units / both contain glycosidic bonds / contain (C)arbon, (H)ydrogen and (O)xygen; 1
(ii)
hemicellulose shorter / smaller than cellulose / fewer carbons; hemicellulose from pentose / five-carbon sugars and cellulose from hexose / glucose / six-carbon sugars;
(only credit answers which compare like with like.) 2
(b)
protein / nucleic acid / enzyme / RNA / DNA / starch / amylose / amylopectin polypeptide; 1
(c)
(i)
to make sure that all the water has been lost;
(ii)
only water given off below 90 °C; (above 90°C) other substances straw burnt / oxidised / broken down; and lost as gas / produce loss in mass;
1
2
(d)
enzymes are specific; shape of lignin molecules will not fit active site (of enzyme); OR shape of active site (of enzyme); will not fit molecule; 2 max
(e)
1. made from -glucose; 2. joined by condensation / removing molecule of water / glycosidic bond; 3. 1 : 4 link specified or described; 4. “flipping over” of alternate molecules; 5. hydrogen bonds linking chains / long straight chains; 6. cellulose makes cell walls strong / cellulose fibres are strong; 7. can resist turgor pressure / osmotic pressure / pulling forces; 8. bond difficult to break; 9. resists digestion / action of microorganisms / enzymes;
(allow maximum of 4 marks for structural features) 6 max
[15] Page 66 of 79
Examiner reports 1
(a)
There was a considerable range in the quality of answers for this question. Some students provided detailed explanations of how the structures of starch and cellulose are related to their functions. At the other end of the spectrum, students referred to these molecules as polypeptides and provided details on protein structure. 10% of students gained full marks and almost 15% gained zero. Although many students appreciated the importance of starch being compact they did not always relate this to its helical/spiral structure. Most students stated that starch did not affect the water potential of cells due to its insolubility. Fewer students related the branching of starch to the faster release of glucose for respiration or discussed its inability to leave a cell. Generally, more students gained marks when describing the structure and function cellulose. Most appreciated theHowever, strength a of cellulose but a common error was to relate of this to ‘strong hydrogen bonds’. significant number of students gained credit for describing (micro/macro) fibrils.
(b)
As in question 10.1, 10% of students obtained maximum marks. However, almost 30% of students scored zero in this question. The responses from students obtaining low marks included a number of factual errors; sugars being produced by sinks, phloem consisting of dead cells, the transport of sugars in the xylem, transport of glucose rather than sucrose, and transport of sugars using cohesion tension. Conversely, there were also some superb answers which included detailed explanations that went beyond the requirements of the mark scheme. This was particularly evident in the details provided on the loading of sugars from the source into the phloem via companion cells. Better answers contained extensive explanations of how co-transport is involved in this process. A significant number of students gained credit for explaining how the pressure gradient between the source and sink was established including the osmotic movement of water into the phloem. Surprisingly, few students obtained the mark for referring to the mass flow of sugars towards sinks. However, most students did describe the unloading of sugars at sinks.
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2
3
Parts (a), (b)(ii) and (b)(iii) proved to be good discriminators. (a)
70% of students scored full marks. Those who scored one mark often gave both alternatives of the second mark point. Weaker responses often lacked clarity; for example, ‘number of individuals’ and ‘different species withina population’. Students who failedto score often thought that the ‘size of the area’ and ‘standard deviation values’ are needed to calculate an index of diversity. It should be noted that although the specification requires students to be able to calculate one specific index of diversity, the mark scheme was amended so that other types of index of diversity could be credited.
(b)
(i)
Most students were aware that the purpose of the control fields was to ensure that the results are due to the herbicide, or not due to another factor. Those who failed to score typically gave stock How Science Works responses, which could apply to any investigation. These usually referred to comparing groups or results, ensuring that the results were due to the independent variable, or simply that these fields acted as controls. Students should be reminded of the need to relate their answers to the specific investigation or context outlined.
(ii)
Half of students scored one mark and this was usually for appreciating that the herbicide killed more weeds, which led to less competition. However, the ability to explain the effect of high concentrations of herbicide, in terms of damage to the crop, proved to be a good discriminator. Unfortunately, many students did not read information in the introduction carefully enough. They thought that the herbicide killed insects, which meant that fewer crops were eaten. The weakest responses usually went no further than to describe the graph.
(iii)
Just under half of students scored at least two marks. This was usually for ‘fewer habitats’ and ‘fewer food sources’. It was only the best responses that referred to ‘fewer plant species’ being present. Similarly, the ability to express these ideas discriminated well. Weaker responses often referred to ‘less food’ and ‘less plants’, which were not credited. As mentioned in part (i), some students wrongly thought that the herbicide killed insects, which directly led to a decrease in their index of diversity.
(a)
This question proved to be a very effective discriminator. 60% of students were able to obtain at least three marks often by reciting the basic principles of speciation. The marks most frequently credited referred to geographical isolation, differential reproductive success and change in allele frequency. Many of these students failed to gain further credit as they did not use the specific information provided on the coloured skin markings. Better answers applied this information, outlining that mutations could result in particular skin colourations and selection for these giraffes could eventually lead to speciation. Consequently, only 10% of students achieved maximum marks.
(b)
One in every three students obtained both marks. These students provided clear explanations on how the mitochondrial DNA of different giraffes could be compared to determine if they were the same species. The most frequently used method was to compare DNA base sequences. Students obtaining a single mark often failed to outline specifically how they would compare mitochondrial DNA. However, these students gained credit for stating that the DNA of different species of giraffes would have more differences than would subspecies.
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4
5
Most candidates had little difficulty obtaining at least half the available marks for this question. (a)
The vast majority of candidates gained both marks, almost invariably for indicating that organisms of the same species would breed together to produce fertile offspring. A few weaker candidates referred to DNA but these answers were only credited when a specific method of comparing the DNA, e.g. DNA sequencing, was mentioned.
(b)
Most candidates were able to gain at least one or two marks, often for referring to variation being present in each population and the different selection pressures in the different environments. Better candidates had little difficulty obtaining maximum marks by explaining that organisms with favourable alleles would survive and pass these alleles on to future generations, resulting in a change in the frequency of alleles. However, some weaker candidates provided descriptions akin to Lamarckism, although these were not as prevalent as in previous years.
(c)
Unfortunately, a significant number of candidates considered colour and camouflage rather than colour and mate selection. However, candidates making the correct link usually obtained both marking points.
(a)
(i)
This part asked students why the genetic code is described as universal. Universal in this context means found in all organisms. A large percentage of students wrote that it is universal because it is found everywhere. Only a quarter of students made correct references to the triplet code used in DNA. Some had the correct ideatobut wrote things such as, ‘The same triplet codes for all amino acids’ and failed score.
(ii)
50% of students gave the correct answer.
(b)
This part discriminated well, but with over 40% getting all three marks. Most stated or described the idea of a frame shift. However, some wrote that this changed the sequence of bases afterwards, rather than the sequence of codons. Another fairly common misconception was that mRNA leads to the synthesis, or formation, of amino acids.
(c)
This part proved more challenging and only about a third obtained both marks. Most correct answers revolved around the idea of introns being non-coding and thus not affecting an amino acid sequence. Students who failed to score often ignored the fact that the mutation was in an intron and wrote about possible effects of a substitution on amino acid sequences. In the figure, it clearly states that the intron is removed from pre-mRNA.
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6
(a)
Most students demonstrated a sound understanding of the principle that a hypothesis is a statement that is tested by an investigation. Predictions were most often given and many gave null hypotheses. Marks were missed when students failed to make any reference to ‘resistance’ in the context of bacteria, and it was not uncommon for the resistance to be linked to the animal or human rather than to Escherichia coli.
(b)
(i)
Many students showed good skill and precision by selecting the right information from the table and made careful references to ‘percentages’ and ‘resistance’ to gain full marks.
(ii)
The principle of natural selection is well understood, but often marks were missed because of a misconception that antibiotics cause mutations. Imprecise expressions were common, such as ‘bacteria thrive’ or they ‘grow’ rather than they ‘reproduce’. Occasionally, students wrote that ‘resistant alleles survived’, with no role implied for the bacteria that contained them.
(c)
Approximately a quarter of students achieved the mark on this question because they made a clear link between human faeces and contaminating pathogens. Many mentioned Escherichia coli, which the question explicitly told them not to do, and poor written communication was evident in answers suggesting that what is spread is a ‘disease’ rather than an ‘agent of disease’.
(d)
The principle of measuring the reliability of the design of an investigation is very well understood, with the majority of answers gaining two marks. Some students misread the question because they suggested changes in the design to increase reliability, and others did not notice ‘reliability’ was given in the question’s stem so it should not have been included as an answer.
(e)
This question differentiated well. Students made appropriate suggestions using either DNA hybridisation or nucleotide sequencing to address the context of this question; some wrote at length about both of these techniques. A minority of responses discussed using antibodies to compare the similarities, possibly they think DNA is a kind of protein.
(f)
Few students achieved more than three marks on this question because many tended to dwell far too long on giving a full explanation of how natural selection increased cases of resistance; they did not look to give broader reasons relating to society’s view. Contamination of the human food chain, avoiding adverse side effects, the generation of superbugs, and high costs were common answers, although students giving more than any two of these was rare. Very few answers contained suggestions about changes in legislation and, although prevention of the horizontal and vertical transmission of genes conferring resistance was often cited, it was rare for students to make clear the principle of avoiding transmission of resistance into other harmful bacteria, so this marking point was not often awarded.
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9
11
(a)
There was widespread recognition that tail band width would be likely to change with age.
(b)
In part (a), many candidates lacked the mathematical understanding to appreciate that a mean which had a value with decimal places suggested that measurements of the same band must differ. Likewise, they did not appreciate that a standard deviation with a value other than zero indicated variation in the measurements of the same band. However in part (b), having read the description of the procedure, most recognised that viewing an animal's tail through binoculars from a moving vehicle was likely to give rise to inconsistent data.
(c)
Most candidates correctly used the data about the width of bands from the left and right sides of the tail as evidence that rings of equal width were not found.
(d)
The most frequently awarded mark was for showing an understanding that unrelated animals would be expected to show more variation than animals from the same family. It was less usual to find a link to the idea that members of one family are genetically closely related, or a reference to the animals’ parentage.
(a)
In general, students made good use of the data, as required, answering this question. Weaker answers only identified the overall trend.
(b)
There was some misinterpretation of this question by students. All that was required was a description of what standard deviation shows, namely, the variation in, or spread of data about, the mean value. Some attempted to discuss standard deviation values related to the data obtained but there was no specific direction to do so in this case.
(c)
Given that two possible approaches to this question were possible, the majority of students were able to make one acceptable line of reasoning.
(d)
The link between carbon dioxide and photosynthesis was not apparent to a large number of students. Thus, they failed to gain the mark for this question.
(e)
Although all responses were seen within the work that was moderated, most students could not make three relevant points – the mark allocation was the key for this – to justify whether the results supported the conclusion. It was rare to see recognition of a negative correlation, although some described such but were only repeating the question stem and, for doing so, there is no credit. This type of question reflects a weakness with the assimilation of resource material. The likelihood of another factor being responsible was the most common point made, but this supports the idea that many students produce rehearsed answers without showing a comprehension of what is in the resources.
(f)
Many students finished strongly with a question relying on recall. The Marking Guidelines specifically referred to “thick” cuticle but some assessors credited reference to ‘waxy’ as an alternative.
13
(a)
Difficulties were experienced with this question where answers were frequently unselective, relating not only to quaternary structure but to aspects of secondary and tertiary structure as well. To gain credit here, candidates needed to confine their answers to the fact that keratin molecules consisted of several polypeptide chains.
(b)
Most candidates clearly appreciated that the bonds formed between sulphur- containing amino acids were strong and helped to bind the individual polypeptide chains. Less able candidates often confused these bonds with peptide bonds or did little more than paraphrase the wording of the question. Page 71 of 79
(c)
As was not infrequently the case with the answers to many of the questions in this paper, less able candidates gave the impression of relying on the recall of mark schemes from broadly similar past questions. In this case they either simply described the primary structure of a protein, which gained little credit, or described how the primary structure of a protein affected its tertiary structure which was potentially, at least, a better option. Those who read the question carefully were usually able to comment on differences in the amino acid sequence leading to differences in bonding and in molecular shape. There was some confusion, presumably among candidates who had also completed Module 2 or 3, between amino acids, proteins and bases.
(d)
As in part (c), the principal requirement here was to answer the question as written. Unfortunately, the response offered by many was no more than a description of active transport. In this question candidates were expected to use this knowledge along with information available in the passage to explain why substances were unable to pass through the outer layer of skin cells. Those who approached the question in the right way generally pointed out that the cells were dead and progressed to make an appropriate comment about respiration and the release of energy or generation of ATP. A not infrequent misconception was that since movement against a concentration gradient involves active transport, active transport cannot be involved in movement down a gradient.
(e)
The many good answers to this part of the question suggested that most candidates had a clear understanding of the principles of electron microscopy and were able to offer a lucid account of its advantages and limitations. Less able candidates were usually able to explain the advantages associated with high resolution but the limitations they suggested concerning expense, size, the production of black and white images and the need for technical support were of a more anecdotal nature and seldom gained significant credit.
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14
Parts (a)(ii), (b)(i), (b)(ii) and (c) proved to be good discriminators. (a)
(b)
(i)
Nearly all students scored one mark and this was usually for suggesting that this method prevented the scientists from being biased. Unfortunately, the second suggestion provided by some also related to the scientists, rather than the volunteers; for example, ‘they may have a vested interest’. Students who scored a second mark often referred to reducing the placebo effect or psychological effects.
(ii)
One-third of students scored full marks. The most common mark points awarded were for suggesting that different types of cigarette contain different amounts of nicotine, different amounts may be absorbed, different amounts may be excreted and that the volunteers may have different blood volumes. Unfortunately, a lack of precision let down some students; for example, ‘they smoked different cigarettes’ and ‘they had different body masses’.
(i)
Just over half of students scored at least two marks. This was usually for mentioning that nicotine will not bind to the protein, so the smoker will not feel rewarded. Surprisingly, few students clearly expressed the idea that the vaccine stimulates the production of antibodies to nicotine, or that these antibodies bind to nicotine. A common misconception seen in weaker responses was that the vaccine contains antibodies to nicotine. Similarly, some students thought that this vaccine contained a weakened strain of bacteria. Generally, there were three incorrect approaches to this question, which were all due to not reading the introduction carefully enough. The first was that the vaccine causes the release of reward chemicals, meaning that a person would not need to smoke. The second was that the vaccine stops smokers from feeling addicted, rather than stopping them from feeling rewarded. The third was that the antibodies to nicotine bind to the protein in the brain, instead of to nicotine. Unfortunately, some students had the second mark disqualified for referring to the protein receptor in the brain as an enzyme. A minority also thought that the ability of the vaccine to stop people smoking could be spread within the population to other smokers by herd immunity.
(ii)
Just under half of students scored at least two marks. The most common mark points awarded were for appreciating that people choose to smoke, the vaccine would be expensive and less money would be needed to treat the effects of smoking. Relatively few referred to it being unethical not to treat smokers, or that money would be better spent in preventing people from smoking. Students who failed to score often gave vague responses; for example, ‘if it is free, more people will stop smoking’ and ‘it will prevent people from dying of cancer’. A minority suggested that the vaccine should not be used at all, due to the Government losing millions of pounds each yearset. in tax on cigarettes. Some of the weakest responses did not answer theas, ‘it question These typically contained stock How Science Works phrases such is only one study’ and ‘we do not know the sample size’.
(c)
It was disappointing that only one-fifth of students scored at least three marks. Again, weaker responses often contained stock How Science Works phrases, which did not apply specifically to this investigation. The question clearly asked students to use the data to evaluate the statement made by the journalist. This said, many students did note that high antibody responders are more likely to stop smoking. Many also realised that the placebo group and low antibody responders had a similar percentage of volunteers who stopped smoking. The next most accessible mark point was that the volunteers may start smoking again after five or six months. Better responses also noted the peak, or drop, in the concentration of antibodies. However, some failed to mention when this occurred, or Page 73 of 79
quoted an incorrect time from the graph. Relatively few students suggested that only a small proportion of the population may be high antibody responders, or that the large sample size produced more reliable or representative results. Overall, it was evident that many students did not analyse the data in the graph and table in enough detail, particularly in relation to the timing of events.
15
16
(a)
Calculating percentage change still remains a problem for a number of students.
(b)
Students generally did not score more than two marks for this question. Most were able to gain the ‘Yes’ mark, identifying that running times were faster after the sports drink, and recognise that both groups had shown improvement. Few considered that the runners might have drunk different volumes or that, as mean times rather than individual times were given, there could be variation in performances. It was rare to see students comment on the lack of standard error bars and the implications of such.
(c)
Most students scored at least two marks in this question but some struggled to identify relevant factors beyond age and sex. There was often a lenient interpretation of marking point 6 despite the requirement that a health factor should be named.
(d)
There were some excellent answers to this question but, equally, there were some very poor ones as well. In some cases, assessors gave credit when the required answer for a particular marking point was incomplete. This was particularly noticeable with marking points 3, 5 and 6. Many students omitted a reference to sodium ions and this omission was not always recognised by assessors.
(a)
This proved to be an excellent discriminator. Just over 70% of students scored at least half marks. Many were aware of the breakingit of hydrogen bonds, the rolethat of DNA helicase complementary base pairing. However, was only better responses referred to theand attachment of free nucleotides (as opposed to free bases) and both strands acting as templates. DNA polymerase was frequently mentioned but its role was often confused in weaker responses. This enzyme joins nucleotides on the newly formed strand, it does not cause complementary base pairing. Some students negated the mark for semi-conservative replication through poor expression. The most common examples of this included ‘each new DNA molecule contains half of the srcinal strand’ and ‘new strands contain half of the srcinal strand’. Very few students wrote about hydrogen bonds reforming.
(b)
(c)
(i)
Two-thirds of students correctly gave the duration of metaphase as 18 minutes.
(ii)
80% of students correctly calculated the duration of anaphase as 10 minutes.
(iii)
This proved to be a good discriminator. Most students gained one mark for extending the horizontal line to 18 minutes, or decreasing this line to 0 m at 28 minutes. Weaker responses often showed the horizontal line increasing.
(i)
70% of students correctly calculated the time the cells were in interphase as 19.7 hours. Very few students gained the principle mark for multiplying by 0.82.
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(ii)
Just under half of students were aware that cells in interphase could be detected by a visible nucleus or the inability to see chromosomes. Weaker responses typically referred to the inability to see DNA or that the cells in interphase would contain twice the amount of chromosomes.
(iii)
This proved to be a good discriminator. Most students were aware that cancer cells divide more rapidly than healthy cells. However, it was only better responses that referred to data in the table and correctly linked this to tissue D. Some students wrongly thought that more cells in interphase meant more rapid cell division due to increased DNA replication.
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17
Parts (a), (b) and (d) proved to be good discriminators. (a)
It was disappointing that only just below 40% of students scored at least half marks. This was mainly due to simply describing the structure of DNA, without explaining how these features relate to its functions. Some students wrote about DNA structure and function in different paragraphs. This made it unclear which feature went with which function, as no direct links had been made. In contrast, there were some truly excellent responses, which had clearly been well planned before putting pen to paper. The most common mark points awarded were for the sugar-phosphate backbone providing strength or protecting bases, the helix allowing the molecule to be compact, weak hydrogen bonds allowing strand separation or replication and the two strands acting as templates or allowing semi-conservative replication. Relatively few students linked complementary base pairing with accurate replication or the production of identical copies of DNA. Similarly, few students referred to DNA as a large molecule that can store lots of information, or the base sequence coding for amino acids. Weaker responses often mentioned this in the context of the genetic code being degenerate. Indeed, some students thought that the base sequence causes amino acids to be produced. The ability to convey that many hydrogen bonds provide stability was rarely seen. It was also unfortunate that a number of students wasted their time by writing about irrelevant topics such as the differences between prokaryotic and eukaryotic DNA and the role of histones. There were also some lengthy accounts of DNA replication, enzyme structure and the different levels of protein structure.
(b)
Many students scored at least two marks for stating that a mutation in gene E produces the highest risk and a mutation in gene C produces the lowest risk. However, only the best responses also referred to gene D. Students who did not mention any of the genes usually picked up one mark for noting that all of the mutant alleles increase the risk of lung cancer. Surprisingly, some thought that a mutation in gene D produces the highest risk.
18
(c)
Just fewer than 40% of students gave the correct answer of 180.
(d)
Two-thirds of students scored at least two marks. Many were able to identify the decrease, plateau and increase for healthy cells and cancer cells. However, relatively few made reference to the plateau occurring for the same length of time. Students who failed to gain a mark for a similarity usually ignored the plateau. Most students spotted that a greater number of healthy cells were killed or that they experienced a faster decrease in number. Similarly, it was impressive to see that some used data from the graph to calculate that a greater proportion of cancer cells were killed. Many students also noted the faster increase in the number of healthy cells.
(e)
Half of students scored full marks. This was usually for mentioning that too many healthy cells would be killed, which could kill the patient or cause side effects. However, relatively few appreciated that it would take time to replace the healthy cells that had been killed.
(a)
Superficial answers along the lines that ‘muscles have many mitochondria because we use our muscles in everyday life’ failed to gain credit but, even where candidates had some understanding of the function of these organelles, errors crept into the accounts they gave. There were still many thermodynamically incorrect statements about ‘making’ energy and a significant number were reluctant to associate mitochondria with respiration, or having done so, expressed their ideas incorrectly in terms of the ATP produced by mitochondria being used for respiration or to ‘allow the muscle to respire’. Many candidates were clearly of the opinion that rough endoplasmic reticulum synthesises protein so did not refer to ribosomes in their answers to part (ii). Others were somewhat equivocal about the status of enzymes and proteins. Page 76 of 79
19
(b)
There was considerable confusion between lymphocytes and phagocytes with many candidates describing reabsorption of the tail in terms of it being engulfed by lysosomes. Most, however, realised that lysosomes contained enzymes, although there were some who were of the opinion that they were enzymes. There was less clarity about the function of these enzymes. Their digestive role was seldom recognised and most responses concluded with a vague statement about ‘destroying’ the tail.
(c)
Responses to this part of the question were frequently marred by misuse of terminology. Homogenate and supernatant, fractionation and centrifugation were often confused. Where possible, examiners ignored such incorrect usage, but where this obscured the underlying meaning, marks were withheld. Most candidates clearly appreciated the significance of adding a cold, isotonic buffer although this was generally added at a rather late stage in the procedure. From this point on, many accounts became less convincing. Homogenisation as a process which breaks open cells was perhaps not understood, with many answers giving the impression that whole leaves, or small pieces of leaf were centrifuged; few referred to filtering before centrifugation, while others appeared to lose track of precisely what they were centrifuging, supernatant or pellet.
(a)
Some good answers were given to this question, with candiates being confident in their understanding of the way in which ATP rapidly releases small, manageable amounts of energy in a single hydrolytic reaction. Marking points 5 and 6 were the least often seen, and the use of ATP to lower activation energy was very rarely seen, although answers frequently referred to activation of glucose in glycolysis.
(b)
Many excellent answers were given in this section that included six or more of the marking points and showed excellent understanding of the processes involved in ATP formation, including chemiosmosis. A significant number gave an account of the whole process of respiration, including glycolysis, using up the space provided and indicating that the answer continued on a separate sheet. One or two included the digestion and absorption of carbohydrates. Weaker students often gained marking points 1, 2 and 6. There was confusion over protons and electrons and hydrogen ions/atoms and molecules. Some students confused the processes of respiration and the light-independent reaction of photosynthesis. Glycerate 3-phosphate (GP) and triose phosphate (TP) were sometimes said to be involved in the Krebs cycle, as was NADP. The movement of protons through the inner mitochondrial membrane into the intermembrane space was often only loosely described, with protons passing into the membrane, along the membrane, or out of the mitochondrion.
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20
(c)
Many students did not appear to have any real understanding of the relationship between photosynthesis and respiration. Statements such as ‘plants have to respire so they can make the carbon dioxide so they can photosynthesise’ were not atypical. The weakest students completely reversed the roles of the two processes. Most commonly, students gained two marks, for referring to the uses of ATP in active transport and synthesis. Marking points 1 and 4 were seen rather less often and marking points 2 and 3 were fairly rarely made. Some students demonstrated good knowledge but not the ability to be selective, giving accounts in some detail of both photosynthesis and respiration which failed to address the question fully.
(a)
(i)
Answers to parts of this question were not infrequently marred by lack of knowledge of the basic structure of cellulose as a polymer of -glucose. Thus, although all that was required here was to note that both molecules were polymers, many disqualified their answers by referring to cellulose as also being a pentose.
(ii)
Limited question technique frequently restricted the credit available. Many candidates concentrated on functional rather than structural differences. As a consequence, the answer boxes were often so full that they rarely compared like with like and offered a valid comparison. Among the better, more focused, answers were some which unfortunately were a little too concise, referring to hemicellulose as a pentose and cellulose as a hexose. Questions requiring structural similarities are likely to remain a feature of BYA1. Candidates clearly need an effective strategy for answering them.
(b)
Starch and protein were correctly identified by many, but a range of incorrect responses included glycogen, phospholipid and various monosaccharides.
(c)
(i)
Answers suggested that, although candidates were clearly familiar with the term “constant mass”, they were by no means all conversant with the idea that it represented the point at which all water had been lost.
(ii)
There were many correct answers. Answers to this second part, such as “Going over 90 °C would start to boil the water so that we would be unable to calculate the water content” were frequent and suggested that candidates had failed to focus on the information provided in the second paragraph of the passage. The better candidates at whom this question was directed were generally able to point out, however, that high temperatures might lead to other substances being broken down and a consequent loss in mass.
(d)
Although most candidates were aware of the specific nature of enzyme action, they experienced varying degrees of difficulty in relating the general concepts involved to the context of this question. Those candidates who gained least credit were inclined to reword the question and offer an explanation in terms of the lignin covering. Others offered responses centred around lignin acting as an enzyme inhibitor. Better candidates clearly understood the concepts of molecular shape and fit and were able to apply them to this situation.
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(e)
Answers to this part of the question ranged from those of the more able candidates who wrote clearly and logically about cellulose structure and function, often with a pleasing level of accuracy and detail, to those which did not gain credit. Among the latter were many who failed to attempt this part of the question and others who confused cellulose with other molecular components of plant cells such as starch and plasma membranes. There was much confusion between hydrogen bonds and glycosidic bonds, and between -glucose and -pleated sheets. Other incorrect assertions which frequently arose were that cellulose is formed from alternating - and -glucose residues, and that it contains both 1-4 and 1-6 linkages. Many candidates correctly identified strength as one of the molecule’s properties and went further in discussing the importance of this in withstanding pressures resulting from osmosis. A frequent error, however, was to assign the function of energy storage to cellulose.
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