comprehension (at the end of paper 2) 15 marks Flashcards
Read the following passage.
Plants require phosphate ions that they get from soil. These ions are often in poor supply and this results in poor growth of the plants. Most plants have mycorrhizae that help the plants to obtain nitrates. Mycorrhizal networks can connect the roots of plants growing next to each other. The use of fertilisers containing phosphate and nitrates in farming inhibits the growth of mycorrhizae. As a result, intensively farmed crop plants do not have mycorrhizae. Plants can defend themselves by producing defensive enzymes that destroy pathogens such as bacteria. Some plants express the genes for defensive enzymes in response to signal proteins secreted by other plants
that are being attacked by a pathogen. These signal proteins can be released into the air. Scientists have discovered that tomato plants increase production of defensive enzymes if plants next to them become infected with a pathogen. These tomato plants were connected by a mycorrhizal network that can carry signal proteins between them. The largest increase in defensive enzyme secretion that the scientists found in a tomato plant in response to the signal protein was by 122.6 per cent.
Use the information in the passage and your own knowledge to answer the following questions.
(a) Suggest and explain two reasons why a poor supply of phosphate ions results in poor growth of plants (lines 1–2). ) (2)
b)Suggest how defensive enzymes produced by plants destroy bacteria (lines 8–9).2
(c) The signal proteins secreted into the air by a plant being attacked by a pathogen act as stimuli leading to the expression of genes for defensive enzymes in other plants (lines 9–12). Suggest how they lead to the expression of these genes. 3
d) Suggest and explain the advantage to tomato plants of transmitting signal proteins through mycorrhizal networks, rather than releasing them into the air 2
(e) The largest increase in defensive enzyme secretion that the scientists found in a tomato plant in response to the signal protein was by 122.6 percent (lines 16–18). The rate of secretion of the defensive enzymes before the signal protein was produced was 450 µmol dm−3 g−1 hour−1. Calculate the rate of secretion per second after the response to the signal protein. 2
f) A student who read this passage concluded that farmers should not use fertilisers to increase yields when growing tomato plants. Evaluate his conclusion. 4
- (Required to) make ATP / glucose phosphate, so less respiration / less energy for growth;
- (Required to) make nucleotides, so less DNA / mRNA / tRNA for cell division / production of protein (for growth);
- (Required to) make RuBP / NADP, so less CO2 fixed / reduced into sugar;
- (Required to) make phospholipids for membranes;
b)Hydrolyse;
Accept digest
2. murein / glycoprotein (in cell wall);
(c) Bind to receptor (on target plant); 2. Acts as / leads to production of a transcription factor; 3. (Which) binds to promoter OR stimulates transcription of genes OR production of mRNA (for defensive enzymes); 3
(d) Direct plant-to-plant transmission; 2. (So) localised response OR
faster response OR no dilution of signal protein; 2
(e)0.278; Accept 1 mark for 1001.7 or × 100 = 122.6 2
(f) Should not use:
Fertilisers prevent development of mycorrhizae;
2. Mycorrhizae help plants to defend themselves (causing an increase in crop yield);
3Mycorrhizae help plants to take up nitrates / phosphates (causing an increase in crop yield);
Should use:
4. Fertilisers containing phosphate and nitrate increase gross primary production so increase yield;
5.Most soil is poor in phosphate so without fertiliser (tomato) plant might not get enough phosphate;
Read the following passage.
Lake Malawi in East Africa has more species of fish than any other lake in the
world. Many of these species have evolved from a common ancestor. Lake Malawi is one of the largest lakes in the world and was formed several million
years ago. Since then, the water level has fluctuated greatly. As a result, 5 what is now a large lake was at one time many smaller, separate lakes.
The country of Malawi has a total area of 118 000 km2. The actual land area is only 94 080 km2, because approximately one-fifth of the country is Lake Malawi.In December 1990, forests covered 41.4% of the actual land area of Malawi. 10 In December 2016, forests covered 26.4% of the actual land area of Malawi. Deforestation and farming along the shores of Lake Malawi have caused increased soil erosion and loss of nutrients into the lake. This has resulted in a decrease in some fish populations. The mark-release-recapture method can be used to estimate the size of a fish population. However, this method can 15 produce unreliable results in very large lake.Use the information in the passage and your own knowledge to answer the
following questions.
(a) Lake Malawi in East Africa has more species of fish than any other lake in the world (line 1).
Suggest and explain how this speciation may have occurred. 4
b)The percentage of forest cover in Malawi decreased between December 1990 and December 2016
Calculate the mean loss of forest cover in km2 per week during this time period.
2
c)Loss of nutrients into Lake Malawi has resulted in a decrease in some fish populations Explain why. 4
d)The mark-release-recapture method can be used to estimate the size of a fish population (lines 13–14). 4
e)Suggest why the mark-release-recapture method can produce unreliable results in very large lakes (lines 14–15).
Explain how.
a)1. Variation/differences due to mutation/s;
2. (Reference to) allopatric (speciation);
Ignore sympatric speciation.
3. Smaller/different lakes have different environmental conditions
OR
Smaller/different lakes have different selection pressures;
Accept different populations for different lakes.
4. Reproductive separation/isolation
OR
No gene flow
OR
Gene pools remain separate;
5. Different alleles passed on/selected
OR
Change in frequency of allele/s;
6. Eventually different species/populations cannot breed to produce fertile offspring
b)Correct answer of 10/10.4 = 2 marks;;
Ignore any numbers after 10.4
2. Working shows 14,112 = 1 mark
OR
13.09/13.1 = 1 mark;
(c) 1. (Growth/increase of) algae/surface plants/algal bloom blocks light;
2. Reduced/no photosynthesis so (submerged) plants die;
3. Saprobiotic (microorganisms) aerobically respire
OR
Saprobiotic (microorganisms) use oxygen in respiration;
Accept: Saprobiont/saprophyte/ saprotroph
Neutral: decomposer
4. Less oxygen for fish to respire;
d)Capture/collect sample, mark and release;
2. Ensure marking is not harmful (to fish)
OR
Ensure marking does not affect survival (of fish);
Accept examples e.g., marking should not be toxic.
- Allow (time for) fish to (randomly) distribute before collecting a second sample;
- (Population =) number in first sample × number in second sample divided by number of marked fish in second sample/number recaptured;
(e) 1. Less chance of recapturing fish
OR
Unlikely fish distribute randomly/evenly;
Read the following passage. Sizes of populations of normal intestinal bacteria are usually controlled by T cells that are produced slowly and in small numbers by the immune system. These T cells do not normally survive for very long. As a result, they do not release large amounts of cytokines. Cytokines are chemicals that can cause swelling of the lining of the intestines. 5 Crohn’s disease is a long-lasting disease that causes swelling of the lining of the intestines. It is believed that Crohn’s disease can be caused by a loss of tolerance to normal intestinal bacteria, as shown by an unusually large response by T cells. This response can be triggered by pathogenic bacteria in the intestines of people with a genetic tendency to Crohn’s disease. 10 Some people’s Crohn’s disease can be controlled by a drug called 5-aminosalicylic acid (5-ASA) that reduces swelling. Another drug called 6-mercaptopurine (6-MP) may also be used. 6-MP inhibits an enzyme required to make adenine and guanine. This is effective because most cells can recycle nucleotides, but T cells are not able to do so. 15 Use information from the passage and your own knowledge to answer the questions.
(a) The Crohn’s disease symptom of swelling of the lining of the intestines could be triggered by pathogenic bacteria in the intestines (lines 6–10). Suggest how 3
b)Suggest the meaning of ‘a genetic tendency to Crohn’s disease’ (line 10) 2
c)Suggest why 5-ASA is only effective in controlling the swelling of the lining of the intestines in some people with Crohn’s disease (lines 11–12). 2
d)Suggest why 6-MP can be used to control the symptoms of Crohn’s disease (lines 13–15). Do not include details of enzyme inhibition or protein synthesis in your answer 3
- A) (Presence of) antigen of the (pathogenic) bacteria; Assume bacteria are pathogenic unless otherwise stated 2. (Causes) more T cells produced / faster T cell production; 3. Against (the pathogen and) normal bacteria; 4. (Long lasting as) cells do not die / live for longer; 5. (More) cytokines / chemicals causing swelling are produced; 3 max
(b)1. (Some people) have a mutation / allele / gene; 2. (That) increases the chances / risk / makes it more likely for / causes them to have an unusually large T cell response; OR (That) lowers / removes tolerance to (normal) intestinal bacteria; 2
(c) 1. (Some people might) produce (very) large amounts of cytokine / have large amounts of swelling; 2. (That) 5-ASA drugs cannot control / reduce; OR 3. Some people may be allergic to / cannot tolerate 5-ASA; 4. So cannot take it
(d)1.(Lack of adenine and guanine) will slow / stop DNA synthesis / replication (in T cells); 2. Affects T cells more as they cannot recycle nucleotides; Needs idea of more / greater effect. Accept converse idea that ‘other’ cells not as affected as they can recycle nucleotides. 3. (6-MP therefore) suppresses / slows the (unusually large) T cell / immune response OR (6-MP causes) fewer / no T cells (to be) produced; Accept (6-MP) acts as an immunosuppressant drug 4. (So) less cytokine is produced (and therefore less swelling);
Read the following passage.
Complete achromatopsia is a form of complete colour blindness. It is caused by having only rods and no functional cone cells. People with complete achromatopsia have difficulty in seeing detail. Complete achromatopsia is caused by an autosomal recessive allele and is usually very rare in populations with only one in 40 000 being affected. However on the Pacific island of Pingelap ten percent of the population are affected. One form of red-green colour blindness is caused by a sex-linked recessive allele which affects more men than women. People with this red-green colour blindness are unable to distinguish between red and green, and also between other colours. They have green-sensitive cones but the photoreceptive pigment they contain does not function. Scientists investigated the use of gene therapy to correct red-green colour blindness in monkeys. They injected viruses containing the gene for the green-sensitive pigment directly into the eyes of the monkeys. Although the monkeys maintained two years of colour vision, there is debate on whether this form of gene therapy is worthwhile. No clinical trials of this procedure have been carried out on humans. Current research into the treatment of red-green colour blindness involves the use of induced pluripotent stem cells (iPS cells). The use of iPS cells could have advantages over the use of gene therapy. Use the information in the passage and your own knowledge to answer the following questions.
a)People with complete achromatopsia have difficulty in seeing detail Explain why. 3
b)Ten percent of the population on the Pacific island of Pingelap are affected by complete achromatopsia (lines 3–6). Use the Hardy-Weinberg equation to calculate the percentage of this population who are heterozygous for this disorder. Show your working. 2
c)Red-green colour blindness affects more men than women . Explain why 2
d)People with red-green colour blindness are unable to distinguish between red and green, and also between other colours (lines 8–10). Explain why 3
e)Current research into the treatment of red-green colour blindness involves the use of induced pluripotent stem cells (iPS cells) . Suggest how iPS cells could correct red-green colour blindness 2
f)The use of iPS cells could have advantages over the use of gene therapy to correct red-green colour blindness Using the information from the passage, suggest and explain reasons why 3
(a)1. No (functional) cones OR Only rods; 2. Cones are connected to a single neurone OR Several rods connected to a single neurone; Accept correct reference to retinal convergence Accept ‘bipolar/nerve cell’ for neurone Accept ‘many’ 2 or more for ‘several’ 3.(Cones) Separate (sets of) impulses to brain OR (Rods) Single (set of) impulse/s to brain; Accept ‘optic nerve’ for brain Reject ‘signals’, ‘messages’ for ‘impulses’ Accept ‘action potential’ 3
(b) 1 Correct answer in range 42 – 44% = 2 marks;; 2. Incorrect answer but shows that understanding that 2pq = heterozygous/carriers = 1 mark; Accept 1 – (p2 + q2) Accept understanding of 2pq by using calculation involving 2 × two different numbers 2
(c)1(Gene/allele) is on the X chromosome; 2. Females require two alleles/females can be heterozygous/carriers and males require one allele; Reference to allele is essential but only required once Reference to females and males required Reject dominant allele
d)d)1. Green sensitive pigment/cones non-functional OR Cones that detect green light non-functional; 2. Three different types of pigment/cone; 3.Other/different colours (‘seen’) due to stimulation of more than one cone/pigment; 1, 2 and 3. Reject reference to ‘green cones’/ ‘blue cones / ‘red cones’ but once only 1, 2 and 3. Reject reference to ‘green pigment’/ ‘blue pigment/ ‘red pigment but once only and only if ‘green cones’ etc, (see above) has not been rejected. 3
(e) 1.(iPS cells) divide;
2. (iPS cells) develop/differentiate into (green sensitive) cones; Accept ‘produce’/’specialise’ ‘turn in to’ / ‘genes switched on’ / ’turned on’ for ‘develop’ but ignore ‘grow’ Reject develop into ‘green cones’/blue’ cones’/’red cones’ Ignore develop/differentiate into (blue/red sensitive) cones; Reject reference to develop in to ‘green pigment’/ ‘blue pigment/ ‘red pigment 2
(f) (Use of iPS cells) long-term; Accept ‘gene therapy short-term’ or ‘only two years’ Accept ‘permanent’ 2. (Use of iPS cells) less chance of rejection/immune response; 3. (Use of iPS cells) single treatment; Accept ‘gene therapy ‘regular/frequent treatment’’ 4. Harm/side effects from using viruses (in gene therapy);
Read the following passage. Azidothymidine (AZT) is a drug used to treat people infected with human immunodeficiency virus (HIV). It inhibits the enzyme that synthesises DNA from HIV RNA. This does not destroy HIV in the body but stops or slows the development of AIDS n the past, some people who took AZT on its own eventually developed AIDS. Some of the HIV in their bodies had become resistant to AZT. To prevent this from happening, people infected with HIV are now treated with highly active antiretroviral therapy (HAART). This involves taking AZT with other anti-HIV drugs at the same time. AZT is taken in low doses. This is because people who took high doses over long periods of time suffered muscle wastage. It was found that high doses of AZT inhibit replication of mitochondria. Use information from the passage and your own knowledge to answer the questions.
(a)Suggest and explain why AZT does not destroy HIV in the body but stops or slows the development of AIDS 4
b)Suggest and explain two advantages of using HAART 4
c) Suggest why high doses of AZT lead to muscle wastage 2
(a) 1. Person (infected with HIV) has HIV DNA (in their DNA); 2. New HIV (particles) still made; 3. (AZT) inhibits reverse transcriptase; 4. (AZT) stops these (new HIV particles) from forming new HIV DNA; OR Slows / stops replication of HIV; 5. Stops destruction of more / newly infected T cells; 6. So immune system continues to work (and AIDS does not develop); 4. Context is important 4. Allow slows / stops (re)production of HIV 4. Reject (AZT) prevents DNA replication
(b) Slows / stops the development of AIDS; 2. Because HIV resistant to AZT is damaged / destroyed / prevented from replicating (by other drugs); OR 3. AZT continues to work as a drug; 4. Because HAART prevents the spread of AZT-resistant HIV to rest of the human population; OR 5. No new HIV particles made; 6. Because HAART might interfere with viral protein synthesis; Mark in pairs. Do not mix and match
(c) (Fewer mitochondria so) less (aerobic) respiration; 2. (Muscles receive) less ATP (so waste); 1. Ignore no respiration 2. Reject less energy produced 2. Ignore no ATP is made
Some insect species feed on the leaves of plants. These leaf-chewers bite off pieces of leaves. Other insect species feed on sap from phloem or xylem. These sap-feeders have sharp, piercing mouthparts that they insert directly into either xylem or phloem. Leaf-chewers and insects that feed on xylem sap are active feeders; this means they use their jaw muscles to obtain their food. In contrast, insects that feed on phloem sap are passive feeders; this means they do not use their jaw muscles to take up sap from phloem. Feeding on phloem sap presents two problems. Firstly, phloem sap has sa high sugar concentration. This could lead to a high pressure of liquid in the insect’s gut because of water entering the gut from the insect’s body tissues. A phloem-sap-feeder polymerises some of these sugars into polysaccharides which are passed out of its anus as ‘honey dew’. The second problem is that phloem sap has a low concentration of amino acids. Phloem-sap-feeding insects rely on bacteria in their guts to produce amino acids. Each phloem-sap-feeding insect receives a few of these bacteria from its parent. This has resulted in a reduction in the genetic diversity of the bacteria found within these insects. A scientist investigated the effect of three different insects on the growth of a plant called the goldenrod. He found that leaf-chewing insects and xylem-sap-feeding insects caused a much greater reduction in total leaf area than did phloem-sap-feeding insects. Use the information from the passage and you(c) Each phloem-sap-feeding insect receives a few of these bacteria from its parent. (lines 16–17). Suggest how this has caused a reduction in genetic diversity of the bacteria.r own knowledge to answer the following questions.
(a) Phloem-sap-feeders are passive feeders (lines 6–7). Phloem-sap-feeders do not use their jaw muscles to take up sap from phloem. Explain why they can take up sap without using their jaw muscles. 3
(b) A phloem-sap-feeder polymerises some of these sugars into polysaccharides Suggest the advantage of this. 2
(c) Each phloem-sap-feeding insect receives a few of these bacteria from its parent. (lines 16–17). Suggest how this has caused a reduction in genetic diversity of the bacteria. 2
d) A scientist found that leaf-chewers and xylem-sap-feeders had a greater effect on plant growth than phloem-sap-feeders Other than environmental factors, give two features the scientist would have controlled in his experiment to ensure this conclusion was valid. 2
(e) The scientist used the reduction in total leaf area of the experimental plants as an indicator of plant growth. Outline a method by which you could find the area of a plant leaf
(a) Contents of phloem vessel pushed into insect’s mouth by high pressure; 2. (High pressure in phloem vessel) caused by loading of sugars into phloem in leaf; 3. And (resulting) entry of water by osmosis. 3
(b) Polysaccharides are insoluble; 2 So do not affect water potential of gut. 2
(c) (Only few bacteria passed from parent, so) only a few (copies of) genes passed on (in bacteria); 2. May not / does not include all alleles (of genes, so diversity reduced) OR Small number of bacteria transmitted means unrepresentative sample. 2
(d)Number / mass / density of insects per plant;
2. Stage of development / size of plants / insects; Ignore any abiotic factor 2
(e) Draw around leaf on graph paper and count squares;
Alzheimer’s disease (AD) is a non-reversible brain disorder that develops over a
number of years. At the start of 2014 the number of Americans with AD was estimated to be 5.4 million. Every 30 seconds another person in America develops AD. In the brain of a person with AD there is a lower concentration of acetylcholine.
This affects communication between nerve cells and initially results in memory loss and confusion. Some of the symptoms of AD that are associated with communication between nerve cells are reduced by taking the drug donepezil.
Donepezil inhibits the enzyme acetylcholinesterase.
A gene mutation called E280A found on chromosome 14 causes early-onset AD at a mean age of 49 years. The age at which the mutation is expressed
to cause AD varies.
Yaramul is a town in a historically isolated region of the Andes Mountains. The
population of this town has the highest frequency of the mutation in the world. The origin of the mutation in this population has been traced back to a common ancestor in the 17th century. Natural selection has not reduced the frequency of the mutation in the population.
This autosomal dominant mutation involves a change in triplet 280 from GAA to
GCA. Scientists analysed chromosome 14 from 102 individuals from Yaramul.
They recorded a sample size of 204 and detected 75 E280A mutations but only
74 potential AD cases. The scientists identified individuals with the mutation by
whole genome sequencing. They had decided that a DNA probe would not be a
suitable method to detect the E280A mutation.
(a) Assuming no one with AD died in 2014, calculate the annual percentage increase in AD cases in America for 2014 (lines 2–4).
(b) Explain how donepezil could improve communication between nerve cells (lines 7–9). 3
C)(c) Suggest and explain two reasons why there is a high frequency of the E280A mutation in Yaramul (lines 13–15). 2
D(d) Explain why natural selection has not reduced the frequency of the E280A mutation in the population (lines 16–17).
(e) The age at which the E280A mutation is expressed to cause AD can vary (lines 11–12). 2
(f) One scientific study which analysed chromosome 14 involved 102 individuals. The scientists recorded a sample size of 204. In this sample they detected 75 E280A mutations but only 74 potential AD cases (lines 19–21).
(g) Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22–23). 2
Suggest explanations for the figures the scientists recorded. 2
Suggest and explain one reason for this.
A) Correct answer of 19.4 / 19.41%
19.47 / 19.5% = 2 marks;
2. Incorrect answer but shows increase of
1,048,320 OR 1,051,200 = one mark;
Accept: 19.46% for one mark.
(b) 1. Less / no acetylcholine broken down;
2. Acetylcholine attaches to receptors;
3. (More) Na+ enter to reach threshold / for depolarisation / action potential / impulse;
1. Accept: more acetylcholine present / remains.
1 and 2. Accept: remains attached for longer = 2 marks.
3. Must be sodium ions.
(c) 1. Isolated so inbreeding / low genetic diversity / small gene pool;
2. Allele inherited (through generations) from (common) ancestor;
1. Ignore: Founder effect.
1. Accept: no interbreeding with other populations.
1. Reject: interbreeding within the population.
(d) 1. AD / symptoms develops late / at 49;
2. Have already reproduced;
Note: ‘It’ is not equivalent to AD / symptom as the question stem relates to the mutation.
(e) 1. Epigenetics / environment / named factor e.g. stress, alcohol, toxins, diet, exercise, smoking;
2. methylation (of genes)
OR
acetylation (of histones);
1. Ignore: gender and lifestyle.
2. If further details are provided the context must be correct e.g. increased methylation or decreased acetylation inhibit gene expression / transcription.
(f) 1. One person was homozygous dominant / has two dominant alleles = 2 marks;
2. For one mark has two alleles / chromosomes;
1. Accept; homozygous dominant genotype e.g. ‘one person has AA’ for 2 marks.
2. Accept: is diploid or has two copies of the gene.
(g) 1. (GCA / triplet) is common / found in other places;
2. Would not know if it was the mutation / allele / gene
OR
Produces ‘false positives’
1. Accept: Probe will bind elsewhere.
