No questions found
(a) The International Union for Conservation of Nature (IUCN) is the world’s largest global environmental organisation. The IUCN Red List of Threatened Species™ evaluates the conservation status of plant and animal species. The Sumatran tiger is categorised as critically endangered on the IUCN Red List. There are approximately 400 Sumatran tigers left in the wild.
Suggest reasons why the Sumatran tiger has become critically endangered.
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(b) Some zoos in Europe have captive breeding programmes for the Sumatran tiger, which breeds well in captivity. Once cubs have reached maturity they may be moved to other zoos. For example, as part of a coordinated breeding programme carried out in 2013, one tiger born in Chester Zoo in England was moved to France and another tiger born in the same zoo was moved to Germany. This type of coordinated breeding programme is essential for the survival of the Sumatran tiger.
Explain why it is important to the gene pool of Sumatran tigers to move tigers from one zoo to another.
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(c) Some animals do not breed well in captivity.
List the methods of assisted reproduction that may be used with animals that do not breed well in captivity.
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(a) The International Union for Conservation of Nature (IUCN) is the world’s largest global environmental organisation. The IUCN Red List of Threatened Species™ evaluates the conservation status of plant and animal species. The Sumatran tiger is categorised as critically endangered on the IUCN Red List. There are approximately 400 Sumatran tigers left in the wild.
Suggest reasons why the Sumatran tiger has become critically endangered.
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...........................................................[3]
(b) Some zoos in Europe have captive breeding programmes for the Sumatran tiger, which breeds well in captivity. Once cubs have reached maturity they may be moved to other zoos. For example, as part of a coordinated breeding programme carried out in 2013, one tiger born in Chester Zoo in England was moved to France and another tiger born in the same zoo was moved to Germany. This type of coordinated breeding programme is essential for the survival of the Sumatran tiger.
Explain why it is important to the gene pool of Sumatran tigers to move tigers from one zoo to another.
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(c) Some animals do not breed well in captivity.
List the methods of assisted reproduction that may be used with animals that do not breed well in captivity.
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531
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558
(a) In the sweet pea plant, Lathyrus odoratus, one gene codes for flower colour and one gene codes for pollen grain shape. Flower colour is either purple or red. Pollen grain shape is either long or round. The inheritance of these genes is an example of autosomal linkage.
• The allele F for purple flowers is dominant over the allele f for red flowers.
• The allele G for long pollen grains is dominant over allele g for round pollen grains.
Explain the meaning of the term autosomal linkage.
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(b) A dihybrid cross was carried out between homozygous dominant and homozygous recessive sweet pea plant parents to produce the F₁ generation.
The offspring from the F₁ generation were crossed to produce the F₂ generation.
(i) Draw a genetic diagram to show a dihybrid cross between two offspring from the F₁ generation.
Assume that these genes are closely linked and that there are no crossing over events. [4]
(ii) The actual results of the dihybrid cross are shown in Table 2.1.
Table 2.1
| phenotypes of F₂ offspring | number of individuals |
|---|---|
| purple flowers, long pollen grains | 284 |
| purple flowers, round pollen grains | 21 |
| red flowers, long pollen grains | 21 |
| red flowers, round pollen grains | 55 |
State how the results support the fact that this is an example of autosomal linkage.
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(c) (i) In a test cross, an individual of known genotype is crossed with an individual that has a dominant phenotype but unknown genotype.
State the genotype of the known individual in a test cross.
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(ii) A test cross was carried out with sweet pea plants known to be heterozygous for both flower colour and pollen grain shape.
The results of the test cross are shown in Table 2.2.
Table 2.2
| phenotypes of offspring of test cross | number of individuals |
|---|---|
| purple flowers, long pollen grains | 215 |
| purple flowers, round pollen grains | 30 |
| red flowers, long pollen grains | 32 |
| red flowers, round pollen grains | 210 |
The result of a test cross can be used to determine a crossover value (COV). A crossover value is the percentage of the total number of offspring showing recombination.
The crossover value (COV) can be calculated using the formula shown in Fig. 2.1.
$$\text{COV} = \frac{\text{number of recombinants}}{\text{total number of individuals}} \times 100$$
Fig. 2.1
Calculate the COV from the results shown in Table 2.2.
COV = ................................. % [1]
(iii) Suggest what information about the relative distance between the linked genes can be gained from crossover values.
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521
549
590
(a) In the sweet pea plant, Lathyrus odoratus, one gene codes for flower colour and one gene codes for pollen grain shape. Flower colour is either purple or red. Pollen grain shape is either long or round. The inheritance of these genes is an example of autosomal linkage.
• The allele F for purple flowers is dominant over the allele f for red flowers.
• The allele G for long pollen grains is dominant over allele g for round pollen grains.
Explain the meaning of the term autosomal linkage.
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...................................................................................................................................................... [2]
(b) A dihybrid cross was carried out between homozygous dominant and homozygous recessive sweet pea plant parents to produce the F₁ generation.
The offspring from the F₁ generation were crossed to produce the F₂ generation.
(i) Draw a genetic diagram to show a dihybrid cross between two offspring from the F₁ generation.
Assume that these genes are closely linked and that there are no crossing over events. [4]
(ii) The actual results of the dihybrid cross are shown in Table 2.1.
Table 2.1
| phenotypes of F₂ offspring | number of individuals |
|---|---|
| purple flowers, long pollen grains | 284 |
| purple flowers, round pollen grains | 21 |
| red flowers, long pollen grains | 21 |
| red flowers, round pollen grains | 55 |
State how the results support the fact that this is an example of autosomal linkage.
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......................................................................................................................................................
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(c) (i) In a test cross, an individual of known genotype is crossed with an individual that has a dominant phenotype but unknown genotype.
State the genotype of the known individual in a test cross.
...................................................................................................................................................... [1]
(ii) A test cross was carried out with sweet pea plants known to be heterozygous for both flower colour and pollen grain shape.
The results of the test cross are shown in Table 2.2.
Table 2.2
| phenotypes of offspring of test cross | number of individuals |
|---|---|
| purple flowers, long pollen grains | 215 |
| purple flowers, round pollen grains | 30 |
| red flowers, long pollen grains | 32 |
| red flowers, round pollen grains | 210 |
The result of a test cross can be used to determine a crossover value (COV). A crossover value is the percentage of the total number of offspring showing recombination.
The crossover value (COV) can be calculated using the formula shown in Fig. 2.1.
$$\text{COV} = \frac{\text{number of recombinants}}{\text{total number of individuals}} \times 100$$
Fig. 2.1
Calculate the COV from the results shown in Table 2.2.
COV = ................................. % [1]
(iii) Suggest what information about the relative distance between the linked genes can be gained from crossover values.
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523
536
516
Within a population, the variation for one characteristic is usually the result of genetic and environmental causes.
(a) Meiosis is one source of genetic variation.
(i) Describe the events that take place during prophase I of meiosis in an animal cell.
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(ii) Explain how independent assortment of homologous chromosomes leads to genetic variation during meiosis I.
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(b) Both genetic and environmental factors contribute to variation in body mass in a bird population.
One cause of variation in body mass in chickens is infection with the gut protocist Eimeria. When fed with the same diet, infected chickens have a lower gain in body mass than healthy chickens in the same population.
(i) Suggest reasons why infected chickens have a lower gain in body mass than healthy chickens.
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(ii) Suggest one environmental factor that affects the growth and reproduction of Eimeria.
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(c) A study was carried out to investigate the effect of treating infected chickens with extracts from a plant, Bidens pilosa. B. pilosa is used in traditional medicine for the treatment of some infectious diseases.
• Three populations, each containing 25 chickens infected with Eimeria were observed.
• The body mass of each chicken was measured at the start of the study.
• Each population was given a different treatment for 56 days:
◦ standard diet with no B. pilosa extract
◦ standard diet with low dose of B. pilosa extract
◦ standard diet with high dose of B. pilosa extract
• The body mass of each chicken was measured at the end of the study and the gain in body mass over the 56 days was calculated.
• The mean gain in body mass was calculated for each population.
The results are shown in Table 3.1.
Table 3.1
| treatment | mean gain in body mass / g |
|---|---|
| no B. pilosa | 1773.0 ± 23.9 |
| low dose of B. pilosa | 2093.1 ± 34.2 |
| high dose of B. pilosa | 2033.3 ± 29.9 |
A statistical analysis of the results of the study of these three populations confirmed that there was a significant difference in mean gain in body mass between low dose B. pilosa and high dose B. pilosa extract treatments.
(i) Name a statistical test that could be used to analyse the results of this study.
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(ii) Sketch on Fig. 3.1 curves to show the pattern of variation for gain in body mass in chickens treated with low dose B. pilosa extract and in chickens treated with high dose B. pilosa extract.
Label one curve ‘low dose’ and the other curve ‘high dose’.
Fig. 3.1..........................................................................................................[3]
598
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600
Within a population, the variation for one characteristic is usually the result of genetic and environmental causes.
(a) Meiosis is one source of genetic variation.
(i) Describe the events that take place during prophase I of meiosis in an animal cell.
........................................................................................................................................
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........................................................................................................................................
........................................................................................................................................
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........................................................................................................................................
........................................................................................................................................[4]
(ii) Explain how independent assortment of homologous chromosomes leads to genetic variation during meiosis I.
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................[2]
(b) Both genetic and environmental factors contribute to variation in body mass in a bird population.
One cause of variation in body mass in chickens is infection with the gut protocist Eimeria. When fed with the same diet, infected chickens have a lower gain in body mass than healthy chickens in the same population.
(i) Suggest reasons why infected chickens have a lower gain in body mass than healthy chickens.
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........................................................................................................................................
........................................................................................................................................[2]
(ii) Suggest one environmental factor that affects the growth and reproduction of Eimeria.
........................................................................................................................................
........................................................................................................................................[1]
(c) A study was carried out to investigate the effect of treating infected chickens with extracts from a plant, Bidens pilosa. B. pilosa is used in traditional medicine for the treatment of some infectious diseases.
• Three populations, each containing 25 chickens infected with Eimeria were observed.
• The body mass of each chicken was measured at the start of the study.
• Each population was given a different treatment for 56 days:
◦ standard diet with no B. pilosa extract
◦ standard diet with low dose of B. pilosa extract
◦ standard diet with high dose of B. pilosa extract
• The body mass of each chicken was measured at the end of the study and the gain in body mass over the 56 days was calculated.
• The mean gain in body mass was calculated for each population.
The results are shown in Table 3.1.
Table 3.1
| treatment | mean gain in body mass / g |
|---|---|
| no B. pilosa | 1773.0 ± 23.9 |
| low dose of B. pilosa | 2093.1 ± 34.2 |
| high dose of B. pilosa | 2033.3 ± 29.9 |
A statistical analysis of the results of the study of these three populations confirmed that there was a significant difference in mean gain in body mass between low dose B. pilosa and high dose B. pilosa extract treatments.
(i) Name a statistical test that could be used to analyse the results of this study.
........................................................................................................................................[1]
(ii) Sketch on Fig. 3.1 curves to show the pattern of variation for gain in body mass in chickens treated with low dose B. pilosa extract and in chickens treated with high dose B. pilosa extract.
Label one curve ‘low dose’ and the other curve ‘high dose’.
Fig. 3.1..........................................................................................................[3]
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Adult Drosophila fruit flies feed on yeasts on the surface of ripe fruits. Female flies lay their eggs inside the fruit and the larvae feed inside the fruit before developing into adults.
The tree Morinda citrifolia produces fruits called noni fruits. Noni fruits contain a toxin and are avoided by nearly all Drosophila species, including Drosophila yakuba in mainland Africa.
One population of D. yakuba lives on the island of Mayotte off the coast of Africa. The population of D. yakuba on Mayotte has specialised to feed on noni fruit. Flies of this population produce enzymes that give them resistance to the toxin in noni fruits. Adults are able to feed on yeasts on the fruit surface and their eggs and larvae can develop and grow inside the fruit.
Scientists have used bioinformatics to estimate that the unique D. yakuba population on Mayotte originated in the last 30 000 years.
(a) Describe how a population of Drosophila can develop resistance to noni fruit toxin.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[4]
(b) Explain why living on an island increases the likelihood of the noni toxin-resistant population of D. yakuba on Mayotte becoming a separate species to the mainland population of D. yakuba.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3]
(c) Suggest how scientists can use bioinformatics to estimate the length of time that the population of D. yakuba has been living on Mayotte.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3]
538
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Adult Drosophila fruit flies feed on yeasts on the surface of ripe fruits. Female flies lay their eggs inside the fruit and the larvae feed inside the fruit before developing into adults.
The tree Morinda citrifolia produces fruits called noni fruits. Noni fruits contain a toxin and are avoided by nearly all Drosophila species, including Drosophila yakuba in mainland Africa.
One population of D. yakuba lives on the island of Mayotte off the coast of Africa. The population of D. yakuba on Mayotte has specialised to feed on noni fruit. Flies of this population produce enzymes that give them resistance to the toxin in noni fruits. Adults are able to feed on yeasts on the fruit surface and their eggs and larvae can develop and grow inside the fruit.
Scientists have used bioinformatics to estimate that the unique D. yakuba population on Mayotte originated in the last 30 000 years.
(a) Describe how a population of Drosophila can develop resistance to noni fruit toxin.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[4]
(b) Explain why living on an island increases the likelihood of the noni toxin-resistant population of D. yakuba on Mayotte becoming a separate species to the mainland population of D. yakuba.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3]
(c) Suggest how scientists can use bioinformatics to estimate the length of time that the population of D. yakuba has been living on Mayotte.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................[3]
537
508
571
(a) A severe reduction of blood flow to the brain causes cells to die. This is called a stroke. The after-effects of a stroke can range from recovery to permanent brain damage and death.
A new emergency gene therapy treatment for people who are at risk of brain damage from a stroke was tested in mice.
• The human granulocyte colony-stimulating factor, hG-CSF, is a protein that stimulates the production of stem cells in bone marrow.
• mRNA coding for hG-CSF was obtained and used to make cDNA.
• This cDNA was inserted into an adeno-associated virus (AAV) vector and given in eye drops to mice just after they experienced a stroke.
(i) Explain what is meant by gene therapy.
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(ii) Describe the roles of reverse transcriptase and DNA polymerase in making cDNA for hG-CSF.
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(iii) The AAV vector used was unable to replicate itself within the target cells.
Suggest why the researchers chose a vector that could not replicate.
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(b) A study was carried out to investigate the effect of the gene therapy described in (a). Four groups of mice were used.
• Group A mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF once only.
• Group B mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF four times.
• Group C mice had a stroke. They received eye drops containing AAV vector carrying the $GFP$ gene coding for green fluorescent protein, instead of the cDNA for hG-CSF, once only.
• Group D mice did not have a stroke. They were not given any eye drops.
(i) Explain why the mice in group C were used in the study.
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(ii) Explain why the mice in group D were used in the study.
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(c) Table 5.1 summarises some results from the study and shows:
• the percentage of mice surviving
• the percentage of brain volume occupied by fluid-filled space
• the score on a behavioural test in which normal mice score 0.5 and brain-damaged mice score nearer to 1.0.
Table 5.1
[Table_1]
Use the results in Table 5.1 to evaluate the benefits of gene therapy treatment, with AAV vector carrying the gene for hG-CSF, for people who have a stroke.
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506
(a) A severe reduction of blood flow to the brain causes cells to die. This is called a stroke. The after-effects of a stroke can range from recovery to permanent brain damage and death.
A new emergency gene therapy treatment for people who are at risk of brain damage from a stroke was tested in mice.
• The human granulocyte colony-stimulating factor, hG-CSF, is a protein that stimulates the production of stem cells in bone marrow.
• mRNA coding for hG-CSF was obtained and used to make cDNA.
• This cDNA was inserted into an adeno-associated virus (AAV) vector and given in eye drops to mice just after they experienced a stroke.
(i) Explain what is meant by gene therapy.
......................................................................................................................................................
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......................................................................................................................................................
......................................................................................................................................................[2]
(ii) Describe the roles of reverse transcriptase and DNA polymerase in making cDNA for hG-CSF.
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(iii) The AAV vector used was unable to replicate itself within the target cells.
Suggest why the researchers chose a vector that could not replicate.
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......................................................................................................................................................[2]
(b) A study was carried out to investigate the effect of the gene therapy described in (a). Four groups of mice were used.
• Group A mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF once only.
• Group B mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF four times.
• Group C mice had a stroke. They received eye drops containing AAV vector carrying the $GFP$ gene coding for green fluorescent protein, instead of the cDNA for hG-CSF, once only.
• Group D mice did not have a stroke. They were not given any eye drops.
(i) Explain why the mice in group C were used in the study.
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(ii) Explain why the mice in group D were used in the study.
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(c) Table 5.1 summarises some results from the study and shows:
• the percentage of mice surviving
• the percentage of brain volume occupied by fluid-filled space
• the score on a behavioural test in which normal mice score 0.5 and brain-damaged mice score nearer to 1.0.
Table 5.1
[Table_1]
Use the results in Table 5.1 to evaluate the benefits of gene therapy treatment, with AAV vector carrying the gene for hG-CSF, for people who have a stroke.
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556
579
594
(a) With reference to Fig. 6.1:
name stage X
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state the precise location of stage Y
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state the numbered steps in stage Y in which decarboxylation occurs
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name the type of reaction by which ATP is made during \textbf{step 3}
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name substances Q and R.
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(b) Uncontrolled cell division can lead to a cancerous tumour. Many cancer cells break down the amino acid glutamine and convert it to a 5-carbon intermediate compound, which is shown in Fig. 6.1.
Suggest how the breakdown of glutamine can lead to the production of ATP in a cancer cell, other than that directly produced during \textbf{step 3}.
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536
582
500
(a) With reference to Fig. 6.1:
name stage X
..........................................................................................................................
state the precise location of stage Y
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state the numbered steps in stage Y in which decarboxylation occurs
..........................................................................................................................
name the type of reaction by which ATP is made during \textbf{step 3}
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name substances Q and R.
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(b) Uncontrolled cell division can lead to a cancerous tumour. Many cancer cells break down the amino acid glutamine and convert it to a 5-carbon intermediate compound, which is shown in Fig. 6.1.
Suggest how the breakdown of glutamine can lead to the production of ATP in a cancer cell, other than that directly produced during \textbf{step 3}.
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530
516
586
Fig. 7.1 shows part of a section through the leaf of a C4 plant such as maize. The letters A, B and C show three types of cell found in the leaf.
(a) Complete Table 7.1 by using the letters A, B or C from Fig. 7.1 to show the location of several compounds associated with photosynthesis in C4 plants.
You may use A, B and C once, more than once, or not at all.
Table 7.1
| compound | location |
|------------------|--------------|
| PEP carboxylase | ............. |
| RuBP | ............. |
| rubisco | ............. |
[3]
(b) Explain why the cells in C form a tight ring around the cells in B.
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(c) Several enzymes have a role in the light independent stage of photosynthesis.
Fig. 7.2 shows the activity of one of these enzymes at different temperatures for a C3 plant, Pisum sativum, and a C4 plant, Amaranthus hypochondriacus.
Describe the differences between the two curves and suggest explanations for these differences.
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Fig. 7.1 shows part of a section through the leaf of a C4 plant such as maize. The letters A, B and C show three types of cell found in the leaf.
(a) Complete Table 7.1 by using the letters A, B or C from Fig. 7.1 to show the location of several compounds associated with photosynthesis in C4 plants.
You may use A, B and C once, more than once, or not at all.
Table 7.1
| compound | location |
|------------------|--------------|
| PEP carboxylase | ............. |
| RuBP | ............. |
| rubisco | ............. |
[3]
(b) Explain why the cells in C form a tight ring around the cells in B.
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................................................................................................................................................
................................................................................................................................................[2]
(c) Several enzymes have a role in the light independent stage of photosynthesis.
Fig. 7.2 shows the activity of one of these enzymes at different temperatures for a C3 plant, Pisum sativum, and a C4 plant, Amaranthus hypochondriacus.
Describe the differences between the two curves and suggest explanations for these differences.
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567
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(a) Describe how an action potential arriving at a presynaptic neurone can result in the release of acetylcholine into the synaptic cleft.
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(b) Lidocaine is a local anaesthetic drug which blocks voltage-gated Na$^+$ channels in the postsynaptic membrane of a synapse.
Explain the effect of lidocaine action on the postsynaptic neurone.
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(c) Fig. 8.1 is a diagram of a neuromuscular junction.
(i) With reference to Fig. 8.1, name structures A, B and C.
A ........................................................
B ........................................................
C ........................................................ [3]
(ii) Outline the importance of structure A in the transmission of nerve impulses.
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(iii) Explain why the structures labelled C are needed in a neuromuscular junction.
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(a) Describe how an action potential arriving at a presynaptic neurone can result in the release of acetylcholine into the synaptic cleft.
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(b) Lidocaine is a local anaesthetic drug which blocks voltage-gated Na$^+$ channels in the postsynaptic membrane of a synapse.
Explain the effect of lidocaine action on the postsynaptic neurone.
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.................................................................................................................................
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(c) Fig. 8.1 is a diagram of a neuromuscular junction.
(i) With reference to Fig. 8.1, name structures A, B and C.
A ........................................................
B ........................................................
C ........................................................ [3]
(ii) Outline the importance of structure A in the transmission of nerve impulses.
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(iii) Explain why the structures labelled C are needed in a neuromuscular junction.
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519
(a) Outline the characteristic features of organisms in the domain Archaea. [8]
(b) Describe the role of seed banks in the protection of endangered species. [7]
509
545
505
(a) Outline the characteristic features of organisms in the domain Archaea. [8]
(b) Describe the role of seed banks in the protection of endangered species. [7]
517
581
554
(a) Describe the process of selective reabsorption and explain how the cells of the proximal convoluted tubule are adapted to carry out selective reabsorption. [9]
(b) Describe the role of the hypothalamus and the role of the posterior pituitary in osmoregulation. [6]
508
535
509
(a) Describe the process of selective reabsorption and explain how the cells of the proximal convoluted tubule are adapted to carry out selective reabsorption. [9]
(b) Describe the role of the hypothalamus and the role of the posterior pituitary in osmoregulation. [6]
525
583
575
