17. Selection and evolution
A section of Biology, 9700
Listing 10 of 104 questions
The following passage is a summary of the main principles of natural selection. Some of the words have been omitted. Write the most appropriate term in each space. Individuals in a population have great potential and yet the numbers in a population remain roughly . This is because many die due to environmental factors and therefore do not reproduce. There is amongst members of a population and those with the features best adapted to the environment survive. They reproduce and pass on their to their offspring. This may lead to a change in the pool of the population and over time may lead to evolutionary change.
9700_s11_qp_42
THEORY
2011
Paper 4, Variant 2
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The following passage is a summary of the main principles of natural selection. Some of the words have been omitted. Write the most appropriate term in each space. Individuals in a population have great potential and yet the numbers in a population remain roughly . This is because many die due to environmental factors and therefore do not reproduce. There is amongst members of a population and those with the features best adapted to the environment survive. They reproduce and pass on their to their offspring. This may lead to a change in the pool of the population and over time may lead to evolutionary change.
9700_s11_qp_43
THEORY
2011
Paper 4, Variant 3
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Within a population, the variation for one characteristic is usually the result of genetic and environmental causes. Meiosis is one source of genetic variation. Describe the events that take place during prophase I of meiosis in an animal cell. Explain how independent assortment of homologous chromosomes leads to genetic variation during meiosis I. 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 protoctist Eimeria. When fed with the same diet, infected chickens have a lower gain in body mass than healthy chickens in the same population. Suggest reasons why infected chickens have a lower gain in body mass than healthy chickens. Suggest one environmental factor that affects the growth and reproduction of Eimeria. 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. Name a statistical test that could be used to analyse the results of this study. Sketch on 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’. number of chickens body mass gain / g
9700_s18_qp_42
THEORY
2018
Paper 4, Variant 2
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In continuous variation, a population shows a range of phenotypes between two extremes with no distinct groups. Height and mass are examples of phenotypic traits that show continuous variation. Describe the genetic basis for continuous variation. Environmental factors can contribute to continuous variation. Suggest two environmental factors that may affect the body mass of an animal. Humans have used selective breeding (artificial selection) for thousands of years to improve the quality of livestock. Outline the principles of selective breeding in livestock.
9700_s21_qp_42
THEORY
2021
Paper 4, Variant 2
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Explain what is meant by stabilising selection. shows the distribution of fur colour in a population of mice. The colour of the soil where the mice lived was mid‑brown. white numbers in population light brown mid- brown fur colour dark brown black Suggest why mice with white fur and mice with black fur were present in the lowest numbers in . An environmental event caused the colour of the soil to change to dark brown. On , sketch a curve to show the distribution of fur colour in the mouse population after many generations. Explain what is meant by genetic drift.
9700_s21_qp_43
THEORY
2021
Paper 4, Variant 3
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Phenotypic variation exists in natural populations. There are many causes of variation. Natural selection determines which phenotypes are advantageous. Variation in a particular characteristic can be described as either discontinuous or continuous. Table 5.1 contains a list of statements that apply to discontinuous variation, continuous variation or both. Complete both columns of Table 5.1. Put a tick (✓) in the box if the statement applies and leave the box empty if the statement does not apply. Table 5.1 statement discontinuous variation continuous variation often involves one gene only environmental factors may affect gene expression there is an additive effect of genes that contributes to the phenotype there are distinct differences between the various forms of a characteristic There is variation in the quantity of vitamin D stored in the body. Vitamin D has an important role in keeping bones healthy. The main storage form of vitamin D in the body is serum 25‑hydroxyvitamin D (serum 25‑OHD). A study was carried out on 262 healthy women to investigate if the concentration of serum 25‑OHD varied between summer and winter. The women had taken no vitamin D supplements. The age range of the women in the sample was 40 to 72 years old. Table 5.2 shows the results of the study. Table 5.2 group mean concentration of serum 25‑OHD / ng cm–3 standard deviation sampled during summer n = 138 32.7 7.6 sampled during winter n = 124 28.5 8.3 whole sample n = 262 30.7 8.2 Additional analysis showed that there was no significant correlation between age and serum 25‑OHD concentration. Explain what is meant by standard deviation. The t‑test was used to compare the mean concentration of serum 25‑OHD when sampled during the summer with the mean concentration of serum 25‑OHD when sampled during the winter, as shown in Table 5.2. Calculate the value of t using the formula provided. key to symbols: ¯x = mean s = sample standard deviation n = sample size (number of observations) Give your answer to four significant figures. There is space for your working. t test value t n s n s x x = + - f p The critical value at the 0.0001 probability level is 3.773. State the conclusion that can be made about the results of the study shown in Table 5.2 and explain how the result of your calculation in can be used to support this conclusion. Suggest the likely causes of variation in quantity of vitamin D stored in the body in this sample of women.
9700_w24_qp_42
THEORY
2024
Paper 4, Variant 2
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Salmon can be genetically modified (GM) to produce increased quantities of growth hormone, which is a protein. GM salmon modified in this way have a faster growth rate and reach their maximum body mass at a younger age than non-GM salmon. Within any population of salmon there is variation in body mass. This is an example of continuous variation. Explain what is meant by continuous variation and how it can be caused. Scientists investigated whether injection of very young non-GM salmon with recombinant growth hormone could cause an increase in the growth rate of the salmon. The scientists used two groups of non-GM salmon: • a control group of salmon that were not injected with recombinant growth hormone • an experimental group of salmon that were injected with 1.0 µg of recombinant growth hormone at the start of the experiment and once a week for the next six weeks. The mean body mass of the salmon in the two groups at the start of the experiment was the same (5.3 g). After six weeks, the body mass of every salmon was measured again. The results are summarised in Table 3.1. Table 3.1 no injection with recombinant growth hormone injected with recombinant growth hormone number of non-GM salmon 28 body mass / g range 6.5–8.6 7.2–12.7 mean (xr ) 7.7 9.4 standard deviation 0.4 1.1 A student decided that a t-test should be performed on the results shown in Table 3.1. Calculate the value of t for the results shown in Table 3.1 using the formula for the t-test: t = n s n s x x + - r r f p Give your answer to two decimal places. Show your working. t = The critical value at p = 0.05 for these data is 2.01. The student used the results in Table 3.1 and the t-test to conclude that the injections of recombinant growth hormone cause an increase in the growth rate of the non-GM salmon. Comment on the extent to which the conclusion made by the student can be supported. Suggest one advantage, other than cost, of farming GM salmon that produce increased quantities of growth hormone instead of farming non-GM salmon that are injected with recombinant growth hormone each week.
9700_m23_qp_42
THEORY
2023
Paper 4, Variant 2
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The fruitfly, Drosophila, has many different species. Three of these species, Drosophila pseudoobscura, D. persimilis and D. miranda, are thought to be closely related. Samples of these three species were collected from the western United States of America. shows where these species naturally occur. Key D. pseudoobscura D. persimilis and D. miranda The base sequences of four regions of DNA of each species were sequenced. The divergence of these base sequences in D. pseudoobscura and D. persimilis from the sequences in D. miranda was calculated. The results are shown in Table 5.1. Table 5.1 DNA region Drosophila species percentage divergence of base sequence from that of D. miranda pseudoobscura 2.5 persimilis 2.4 pseudoobscura 8.1 persimilis 7.3 pseudoobscura 2.1 persimilis 1.7 pseudoobscura 1.9 persimilis 1.7 With reference to Table 5.1, describe the evidence that D. miranda may be more closely related to D. persimilis than to D. pseudoobscura. Suggest why there is more divergence in some regions of DNA than in others. The area where D. pseudoobscura is found is separated from the areas where the other two species are found by a high range of mountains. Explain how the species D. pseudoobscura could have evolved from a population of D. miranda.
9700_s10_qp_42
THEORY
2010
Paper 4, Variant 2
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Questions Discovered
104