Thebiotutor.com A2 Biology OCR Unit F215: Control, genomes and environment Module 1.2 Meiosis and variation Notes & Questions

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1 Thebiotutor.com A2 Biology OCR Unit F215: Control, genomes and environment Module 1.2 Meiosis and variation Notes & Questions Andy Todd 1

2 Describe, with the aid of diagrams and photographs, the behaviour of chromosomes during meiosis, and the associated behaviour of the nuclear envelope, cell membrane and centrioles. Meiosis is a type of cell division known as a reduction division. Produces gametes (Sex cells such as sperm, egg, pollen and plant egg cells) Produces 4 haploid (n) cells from one parent diploid (2n) cell. o This is important as fusion of gametes nuclei at fertilization will need to result in a diploid (2n) cell. Meiosis has two main stages: Meiosis I and Meiosis II, each with several substages. Meiosis I o Prophase I Chromatin condenses and undergoes super-coiling, resulting in it becoming shorter, thicker and visible under a light microscope. The chromosomes now come together to form homologous pairs (bivalents) At this point, each homologous chromosome pair is visible as a bivalent, a tight grouping of two chromosomes, each consisting of two sister chromatids. A Bivalent (picture below) Andy Todd 2

3 The sites of crossing-over are seen as crisscrossed non-sister chromatids and are called chiasmata (singular: chiasma). These allow swapping of paternal and maternal alleles. The nuclear envelope and nucleolus break down. Centrioles and spindle fibres form (no centrioles in plant cells) o Metaphase I Centrioles (not in plant cells) migrate to the poles of the cell Bivalents are arranged on a plane equidistant from the poles called the metaphase plate or equator Independent random assortment of Bivalents, where the orientation of the bivalent on the metaphase plate is random and independent of the orientation of the bivalents either side of it. Spindle fibres attach to the centromeres of sister chromatids. o Anaphase I Two chromosomes of each bivalent separate and start moving toward opposite poles of the cell as a result of the action of the spindle. The sister chromatids remain attached at their centromeres and move together toward the poles. A key difference between mitosis and meiosis is that sister chromatids remain joined after metaphase in meiosis I, whereas in mitosis they separate (centromeres break) At this point recombinants are formed from the crossing over events of prophase I. Andy Todd 3

4 o Telophase I (Not in plant cells) The homologous chromosome pairs complete their migration to the two poles as a result of the action of the spindle. A nuclear envelope reforms around each chromosome set, the spindle disappears, and cytokinesis follows. In animal cells, cytokinesis involves the formation of a cleavage furrow, resulting in the pinching of the cell into two cells. Meiosis II o Prophase II The centrioles duplicate. This occurs by separation of the two members of the pair, and then the formation of a daughter centriole perpendicular to each original centriole. The nuclear envelope breaks down, and the spindle apparatus forms. o Metaphase II Andy Todd 4

5 Each of the daughter cells completes the formation of a spindle apparatus. Single chromosomes align on the metaphase plate, much as chromosomes do in mitosis. This is in contrast to metaphase I, in which homologous pairs of chromosomes align on the metaphase plate. Independent random assortment of chromatids, where the orientation of the recombinant chromatids on the metaphase plate is random and independent of the orientation of the recombinant chromatids either side of it. Spindle fibres attach to the centromeres of recombinant chromatids o Anaphase II The centromeres separate, and the two chromatids of each chromosome move to opposite poles on the spindle. The separated chromatids are now called chromosomes in their own right. o Telophase II A nuclear envelope forms around each set of chromosomes. Cytokinesis takes place, producing four daughter cells (gametes, in animals), each with a haploid set of chromosomes. Because of crossing-over, some chromosomes are seen to have recombined segments of the original parental chromosomes. Andy Todd 5

6 Explain how meiosis and fertilisation can lead to variation through the independent assortment of alleles. Meiosis and fertilisation create huge levels of variation in individuals of the next generation. This results from 6 major events o Crossing Over Prophase I Homologous pairs of Chromosomes (Bivalents) form Chiasma(ta) form between homologous chromosomes Alleles swap between paternal and maternal chromatids. Maybe 3 or 4 crossing over events per Bivalent o Independent random assortment of Bivalents Metaphase I Bivalents line up along the cell equator randomly and independent of each other o Independent random assortment of (recombinant) chromatids Metaphase II Chromatids line up along the cell equator randomly and independent of each other. o Random mating Choices of mates is random o Random fusion of gametes Chances of gamete fertilising another specific gamete o Chromosome mutations Any changes to DNA further increases the variation Explain the terms allele, locus, phenotype, genotype, dominant, codominant and recessive. Allele An alternative version of a gene Locus Specific position on a chromosome, occupied by a specific gene. Genotype Combination of alleles in cells of an organism for a particular trait/characteristic. Phenotype Physical characteristic expressed due to the influence of the genotype and the environment Dominant If a dominant allele is present then the trait is expressed in the phenotype. Co-Dominant A characteristic where both alleles contribute to the phenotype. Recessive Characteristic in which the allele responsible is only expressed in the phenotype if there is no dominant allele present. Homozygous Eukaryotic cell or organism that has two identical alleles for a specific gene Heterozygous Eukaryotic cell or organism that has two different alleles for a specific gene Andy Todd 6

7 Explain the terms linkage and crossing over. Linkage Genes for different characteristics that are present at different loci on the same chromosome are linked and inherited together. o Example: Linked alleles We will use the symbol v for the vertical-eyes allele and V for the round-eyes (normal) allele, and t for the tooth allele and T for the (normal) no-tooth allele. Two pure bred aliens are crossed as seen below. F1 Analysis The F1 Aliens have the normal round-eyes and normal no-tooth phenotypes because the mutant alleles are recessive to the normal alleles. Andy Todd 7

8 A test cross is carried out where the F1 aliens are crossed with a pure bred homozygous recessive alien for vertical eyes and a tooth. This is shown below. F1 test cross analysis How did these results come about? There are obviously more of the parental phenotypes as the alleles are linked and therefore inherited together. There are still however two other phenotypes but these occur in the F1 testcross generation in lower frequencies. These individuals are a result of crossing over events that occur between the two alleles on the chromosome. Andy Todd 8

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10 Recombination frequency data are used to assign gene positions on a map. One map unit is the equivalent of a 1% recombination frequency. Thus, the v and t genes are 14.3 map units apart. When genes are linked, they do not assort independently. That is, from a testcross involving an individual heterozygous for each of two genes, the offspring will not exhibit a 1:1:1:1 phenotypic ratio expected for independently assorting genes. Instead, there will be an excess of the parental phenotypes. Results of such testcrosses can be used to calculate the map distance between the two genes involved. Test Cross o A testcross or Backcross is always used to see if genes are linked or unlinked A Testcross/Backcross is always a cross with a homozgous recessive genotype for both genes. This will allow all alleles to be expressed in the phenotypes. It will allow homozygous recessive genotypes to be expressed for both genes in the offspring if the parent is heterozygous for the two genes. Andy Todd 10

11 Genes are not Linked o This looks like a normal di-hybrid heterozygous cross o If a heterozygous AaBb was crossed in a backcross aabb Gametes AB, Ab, ab, ab ab, Offspring genotypes AaBb, Aabb, aabb, aabb Ratio 1 : 1 : 1 : 1 Andy Todd 11

12 Genes are linked Gametes AB, ab ab, Offspring Genotypes AaBb aabb Ratio 1 : 1 These are the parental genotypes They occur more frequently There occur because the genes are linked The alleles AB and ab are inherited together The ratio is a large 1: 1 Crossing Over Gametes Ab, ab, ab These are the recombinant genotypes They occur less frequently There occur because the alleles are crossed over in Prophase 1 The alleles Ab and Ab are inherited together as recombinants The ratio is a small 1: 1 Crossing Over When non-sister chromatids exchange alleles during prophase of meiosis I. Andy Todd 12

13 Use genetic diagrams to solve problems involving sex-linkage and co-dominance. Sex-Lined (X-linked) Gene with its locus on one of the sex chromosomes, X or Y. As there are few genes on the Y chromosome, in humans, most sex-linked genes are on the X chromosome o Example: Sex-Linked alleles o F1 Analysis The F1 male has the mutant phenotype because the gene is on the X chromosome and there is no allele on the Y chromosome; that is, for X-linked genes, males are hemizygous. This means that there is a mutant s allele on each of the female's X chromosomes, while the male has the normal S allele on the X as well as the Y chromosome. If all male offspring of a mutant female and true-breeding normal male have the mutant phenotype and all female offspring have the normal phenotype, the trait is inherited as an X-linked recessive. Andy Todd 13

14 These expected ratios are closely matched by our observed F2 results of 38 normal females : 33 short females : 41 normal males : 35 short males. o Conclusion Short legs can be explained by a single gene on the X chromosome with the short-legs allele recessive to the normal-legs allele. This cross illustrates the inheritance of an X-linked recessive allele. Co-Dominance - A characteristic where both alleles contribute to the phenotype in a heterozygous. The phenotype from the heterozygous individual expresses the characteristic of both alleles in the genotype. o Example: Co-Dominance Andy Todd 14

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16 o F2 Analysis How is the horns trait inherited? The total number of progeny in the F2 generation is 93. The F1 progeny of a cross of an individual with two horns and an individual with no horns all had one horn, that is, a phenotype intermediate between the two parental phenotypes. The simplest hypothesis is that the horn trait exhibits co-dominance with a single gene involved. The homozygous genotypes give the twohorns and no-horn phenotypes, while the heterozygous genotype gives the one- horn phenotype. If two homozygotes are crossed that differ in a trait and the offspring all have a phenotype intermediate between those of the two parents, then the trait exhibits co-dominance. These expected ratios are closely matched by our observed F2 results of 22 no horn : 47 one horn : 24 two horns. o Conclusion The horns trait shows co-dominance. The HH genotype gives the no-horn phenotype, Hh gives one horn, and hh gives two horns. Interbreeding one-horned Hh F1 individuals produces F2 offspring with a 1:2:1 ratio Andy Todd 16

17 Describe the interactions between loci (epistasis). The interaction of genes concerned with the expression of one characteristic. One gene may mask the expression of another. o Example: Rat coat colours Rat coat colour is controlled by a gene B/b, where B = Agouti (Black), and b = Brown. A cross between two heterozygous agouti rats was conducted and the results were as follows. Parental generation x Bb Bb F1 generation The results came out as 9:3:4!!! 9 Agouti : 3 Brown : 4 Albino This is more like a dihybrid cross (i.e two genes involved). There must be another gene at work. There is another gene at work. A gene for pigment colour and it is outlined below Brown Andy Todd 17

18 The dominant form of the C/c gene provides the pigment for which the B/b gene can colour. If there is no pigment then the result will be an albino rat regardless of what the B/b gene is instructing. In this way we say that the C/c gene has an epistatic effect over the B/b gene. Therefore the following genotypes and phenotypes are possible. This means that the rats in the parental cross must have actually been BbCc x BbCc The following cross outlines how this has been possible. Andy Todd 18

19 It is important to note how this epistasis phenomenon could actually occur. o In some cases a gene may code for an enzyme with a complementary site for another protein/gene. o It may code for a transcription factor that regulates the expression of another gene o It may code for a competitive or non-competitive inhibitor for an enzyme synthesised from another gene. Predict phenotypic ratios in problems involving epistasis. As epistasis involves two genes the epistatic crosses between two heterozygotic individuals for both genes results in all the phenotypic ratios will be variants of the 9:3:3:1 Dominant Epistasis 12:3:4 Grouping of the dominant phenotypes, genotypes (9:3 = 9+3 = 12) Recessive Epistasis 9:3:4 Grouping of the recessive phenotypes, genotypes (3:1 = 3+1 = 4) Complementary Epistasis 9:7 Grouping of the intermediate and recessive phenotypes, genotypes (3:3:1 = = 7 Andy Todd 19

20 Use the chi-squared (χ 2 ) test to test the significance of the difference between observed and expected results. Chi squared is a statistical test which allows us to compare our actual results with what we would expect if the mendelian 9:3:3:1 ratio held true. It allows us to objectively say whether are results are as we expected. Chi squared is a measure of the combined error of our results from the expected. It will always be conducted from a dihybrid cross at A-level. o Example: Dihybrid genetic cross and Chi Squared analysis R allele for round seeds r allele for wrinkled seeds G allele for yellow seeds g allele for green seeds A true heterozygous plant for round yellow seeds was crossed with another true heterozygous plant for round yellow seeds. RrGg x RrGg Yellow round Yellow round F1 results of 288 pea plants. were as follows. These are the Observed results. 9/16 Yellow round = 169 3/16 Green round = 54 3/16 Yellow wrinkled = 51 1/16 Green wrinkled = 14 What we would have expected is the following out of 288 plants if the ratio held exactly true. 9/16 Yellow round = 162 3/16 Green round = 54 3/16 Yellow wrinkled = 54 1/16 Green wrinkled = 18 Are, our observed results following the 9:3:3:1 mendelian ratio? The answer is going to be subjective so we must conduct a chi squared analysis to be sure that our conclusion is totally objective. Andy Todd 20

21 First, we must write up a Null Hypothesis. A null hypothesis always states that o There is NO significance difference between what we observed and what we expected and that any variation is totally due to chance events Below is the formula for Chi squared. You do not need to memorise this, but you will need to be able to complete the following table so you can generate your chi squared value. Chi-squared test Observed (o) Expected (E) 0 - E (0-E)2 (0-E)2/E Yellow round Green round Yellow wrinkled Green wrinkled Σ = Andy Todd 21

22 The completed table looks like this. Chi-squared test Observed (o) Expected (E) 0 - E (0-E) 2 (0-E) 2 /E Yellow round Green round Yellow wrinkled Green wrinkled Σ = Once you have calculated the chi squared value you must work out the degrees of freedom that there are and then compare the value against critical values in a confidence table like the one on the next page. Calculating the degrees of freedom (df) is straight forward. o It is the number of actual different phenotypic groups for the cross (n) minus one The formula looks like this; df = n 1 In this example the Chi squared value is and the df = 3 (4-1 = 3) Firstly you must select the correct row of critical values to compare you chi squared value against. Each row indicates the level of degrees of freedom. In this case it is rown 3 as we have 3 degrees of freedom. Then we move along the critical values and check that our chi squared value is higher or lower than the critical value at 0.05 (5%) significance level. Andy Todd 22

23 Probability value (P-Value) of = 99.5% chance that the difference is due to chance 2 DF Probability value of 0.05 = 5% chance that the difference is due to chance 3.93E Accept Null Reject Null The 0.05 (5%) significance level is the cut off point. If the calculated chi squared value is higher than the critical value at the 0.05 significance level then we reject the null hypothesis, Stating that o There is a significance difference between our observed results and those which we were expecting If the calculated chi squared value is lower than the critical value at the 0.05 significance level then we accept the null hypothesis, Stating that o There is no significance difference between our observed results and those which we were expecting, and any variation is due to chance In an answer you must say that the chi squared value which you have calculated is either higher or lower than the critical value at the 0.05 significance level and therefore you reject or accept your null hypothesis respectively. If it is lower than critical values at other significance levels to the left of the 0.05 significance level (i.e 0.70) then you also want to say that it is also lower than the critical values at significance level Andy Todd 23

24 Helpful hints when completing Genetic cross Diagrams If you are not told what the genotypes of the parents are, you can work them out by looking at the offspring phenotypes. o If offspring have recessive phenotypes then the parent must both be heterozygous. o i.e. Offspring have aabb then parents must both be AaBb If there is a small 1:1 ratio is a di-hybrid cross then the genes are likely to be linked. If there is a large 1:1 or 3:1 in a di-hybrid cross then the genes are likely to be linked. Describe the differences between continuous and discontinuous variation. & Explain the basis of continuous and discontinuous variation by reference to the number of genes which influence the variation. Variation is the differences between individuals of the same species or it can refer to the differences between individuals of different species. There are two types; Continuous o Features have a range of variation (quantitative) o Polygenic many genes contribute to the final phenotype (2 or more) o Each gene provides an additive effect on the phenotype o Genes are not linked, they are on separate chromosomes o Example height, weight etc Discontinuous o Distinct categories o Monogenic one gene contributes to the final phenotype - Epistasis if there is more than 1 gene at work o Different alleles have a large effect on the phenotype o Different genes at different loci have large effects on the phenotype. o Example blood groups, ear lobe shape etc Andy Todd 24

25 Explain that both genotype and environment contribute to phenotypic variation. The phenotype is influenced by a combination of both the environment and the individuals genotype. It is said that The genes load the gun and the environment pulls the trigger Polygenic traits are more easily influenced by the environement that monogenic trtaits Explain why variation is essential in selection. If there is no variation then all individuals have an equal chance of surviving. This would mean if the environment changed then all would live and pass on their genes to the next generation or all would die and the species would become extinct. Variation ensures evolution as some individuals of the species are more suitably adapted to the environment than others. Use the Hardy Weinberg principle to calculate allele frequencies in populations. Gene pool o Total number of different alleles within a population o Populations have gene pools o Individuals have genomes The Hardy-Weinberg model o The Hardy-Weinberg model assumes that all allele ad genotype frequencies will remain stable from generation to generation unless The population is small Mating is not random There are mutations There is migration in or out of the population There is genetic drift There is natural selection o The frequencies will not be stable in nature so it allows us to work out what is causing the changes Allele frequency Genotype frequency Andy Todd 25

26 Problems For the first three problems calculate the frequency of the three genotypes and then calculate the frequency of the two phenotypes. 1. A herd of horses with p = A school of seahorses with p = A school of redhorse suckers (a type of fish) with q = Answers For the first three problems calculate the frequency of the three genotypes and then calculate the frequency of the two phenotypes. 1. A herd of horses with p = p = 0.25, q = 0.75 ( = 1.0) p 2 = pq = q 2 = ( = 1.0) A phenotype = = a phenotype = ( = 1.0) 2. A school of seahorses with p = p = 0.65, q = 0.35 ( = 1.0) p 2 = pq = q 2 = ( = 1.0) A phenotype = = a phenotype = ( = 1.0) 3. A school of redhorse suckers (a type of fish) with q = p=0.35, q= 0.65 ( = 1.0) p 2 = pq = q 2 = ( = 1.0) A phenotype = = a phenotype = ( = 1.0) Andy Todd 26

27 Explain, with examples, how environmental factors can act as stabilising or evolutionary forces of natural selection. Selection pressure Environmental factor that confers greater chances of surviving and reproducing on some members of the population than on others. - The environmental factor could be natural or artificial (human) Environmental factors o Biotic - Living Disease Food (prey) Competition (Intraspecific or interspecific) o Abiotic Non-living Water availability (soil or pools) Mineral availability Space Light Temperature Human Factors Size of yield Quality of yield Progeny Disease resistance Temperament Stabilising pressures Stabilising selection o Favours individuals with a general phenotype. o Maintains the populations mean phenotypes by selecting against the phenotypic extremes o Example: Calf birth weights Too small = difficulty carrying out thermoregulation (homeostasis) Too large = They cannot pass through the pelvic girdle Evolutionary pressures Directional selection o Favours individuals with phenotypes at one extreme and selects against individuals with phenotypes at the other extreme o Allele frequency is shifting in 1 direction o Example: Giraffe neck lengths Andy Todd 27

28 Disruptive selection o Favours individuals at both extremes (specialists) and selects against those with a general phenotype o Supports speciation. o Example: Galapagos Finches beak size Explain how genetic drift can cause large changes in small populations. The change in allele frequency in a population, as some alleles pass to the next generation while others disappear. Chance or random selection of individuals and hence their alleles. Individuals are selected against by natural disasters and chance events as opposed to their suitability to the environment. Accelerates evolution Most exaggerated in small populations Example: Earthquake etc Andy Todd 28

29 Explain the role of isolating mechanisms in the evolution of new species, with reference to ecological (geographic), seasonal (temporal) and reproductive mechanisms. Isolating mechanism mechanism that divides populations of organisms into smaller sub groups Isolating a population will; o Decrease the gene pool o Increase inbreeding whi9ch could lead to an increase in homozygous recessive genotypes within the population o Increase the effects of genetic drift Several types of isolating mechanisms o Reproductive (mechanical) When the reproductive structures are no longer physically compatible. Example: Great Dane will not mate with a Chihuahua. o Ecological (Geographical) When two species or populations occupy different habitats within the same environment Can be separated by a physical barrier, such as a mountain range, river, road systems, islands etc. Example: Galapagos finches o Seasonal (Temporal) When two species or populations live within the same area but are reproductively active at different times. Example: American frog species. Wood frog fertile in April Tree frog fertile in June Bull frog fertile in July o Behavioural When two species or populations evolve different courtship displays which are essential for successful mating Example: Birds of paradise. Andy Todd 29

30 Explain the significance of the various concepts of the species, with reference to the biological species concept and the phylogenetic (cladistic/evolutionary) species concept. Species definition o Can interbreed to produce fertile offspring o Reproductively isolated o Have the same Morphology Physiology Anatomy Embryology Behaviour Occupy the same ecological niche Phylogeny o All living things have RNA / DNA / Proteins o Comparisons to relatedness of these molecules can determine relatedness of organisms o Haplotypes can also be used for coparisons Haplotypes are particular base sequences Can be 1 locus, several loci or an entire chromosome They are a combination of alleles at different loci transmitted together. A Phylogenic tree Andy Todd 30

31 A Monophyletic group / Clade o Group of organisms with similar haplotypes o Shows the common ancestor and all descendants o A Monophyletic group can be 1 or more clades A Paraphyletic group o Only linear more like traditional classification Andy Todd 31

32 Cladistic Classification o Based on evolutionary ancestry o Focuses on evolution rather than visible similarities o Uses DNA or RNA sequences o No distinction made between extinct and extant o Evidence is quantifiable o There are no set amount of taxonomic levels Compare and contrast natural selection and artificial selection. Natural Selection o Mechanism for evolution. Organisms that are well adapted to their environment are more likely to survive and reproduce, passing on the alleles for favourable characteristics. Artificial selection o Organisms that have characteristics that are desirable to humans are selected to survive and reproduce, passing on the alleles for favourable characteristics. Source of variation in trait. Agent of Selection Immediate Result of selection Result after a number of generations assuming a constant environment! Artificial Selection on dogs: Example of a breed where curly coat is desired: Mutation gives rise to a curly coat in a certain dog. Intentional selection by breeder desiring more dogs showing curly coats. Individuals with curly coats are more likely to survive and be used for breeding. Genes causing curlier coats desired by the breeder become wide spread. Average coat curliness in the breeding population increases. Greater proportion of curly coated dogs in the population. Natural Selection for antibiotic resistance in strep bacteria. Mutation gives rise to a small proportion of bacteria resistant to the antibiotic. Presence of antibiotic in the bacteria's environment - your throat. Greater proportion of bacteria that survive are resistant to the antibiotic. Genes causing antibiotic resistance become wide spread as long as individuals with those genes leave more offspring on average compared to those individuals that lack the resistance genes. Average level of antibiotic resistance in the bacteria population increases. Greater difficulty in treating strep throat. Andy Todd 32

33 Describe how artificial selection has been used to produce the modern dairy cow and to produce bread wheat (Triticum aestivum). Modern Dairy Cow Dairy Cows Holstein - Friesian Milky Shorthorn Brown Swiss Jersey Guernsey Ayrshire Measure Milk Yields of individuals Checks progeny of Bulls Only several bulls kept as semen can be stored and artificial inseminated Elite Cows are given hormones so they produce many eggs Eggs are fertilised in-vitro an implanted into surrogate mothers Embryos could also be cloned and divided into many more identical embryos Andy Todd 33

34 Modern bread Wheat (Triticum aestivum) Wild Einkorn A u A u 2n = 14 Einkorn A u A u 2n = 14 Wild Grass BB 2n = 14 STERILE A u B Polyploidy Emmer Wheat A u A u BB 4n = 28 STERILE A u BD Goat Grass DD 2n = 14 Triticum aestivum Common Wheat A u A u BBDD 6n = 42 Andy Todd 34

35 Questions 1. (a) (i) A gene controlling coat colour in cats is sex linked. The two alleles of this gene are black and orange. When both are present the coat colour is called tortoiseshell. Define the following terms: gene allele [2] (ii) Explain why there are no male tortoiseshell cats [2] Andy Todd 35

36 Two pure breeding strains of snapdragon, a garden plant, were obtained. One strain had red flowers and the other had white flowers. The two strains were crossed yielding F 1 plants all with pink flowers. The F 1 were then interbred to produce F 2 plants with the following colours: red 62 pink 131 white 67 The following hypothesis was proposed: Flower colour is controlled by a single gene with two codominant alleles. (b) Complete the genetic diagram to explain this cross. Use the following symbols to represent the alleles: C r = red, C w = white Parental phenotypes: red flowers x white flowers Parental genotypes: Gametes: F 1 genotypes:... F 1 phenotypes:... Gametes:... F 2 genotypes:... F 2 phenotypes:... Expected F 2 phenotypic ratio:... [6] Andy Todd 36

37 (c) A chi-squared (χ 2 ) test is carried out on the experimental data to determine whether the hypothesis is supported. (i) Complete the table below by calculating the expected numbers. F 2 phenotype observed numbers expected numbers red 62 pink 131 white 67 total [3] The χ 2 statistic is calculated in the following way: 2 observed expected χ sum of... expected 2 (ii) Calculate the value of χ 2 for the above data. Show your working. χ 2 value =... [2] (iii) The critical value of χ 2 for this type of investigation with two degrees of freedom is Explain whether your answer to (b) (ii) supports the hypothesis [1] [Total 16 marks] Andy Todd 37

38 2. Phenotype is influenced by genetic and environmental factors. 3 Describe one example of how the environment influences phenotype [Total 2 marks] During interphase preceding meiosis, each chromosome replicates itself and becomes two chromatids joined at the centromere. These identical chromatids are known as sister chromatids. During the first division of meiosis, pairing of homologous chromosomes takes place. The structure formed is called a bivalent. When paired in this way nonsister chromatids from the two chromosomes exchange segments of genetic material by breaking and rejoining. (i) State the name given to the exchange of segments of chromatids by breaking and rejoining.... [1] (ii) Name the stage of the first division of meiosis when this exchange of segments occurs.... [1] (iii) Describe the genetic difference between sister and non-sister chromatids [1] [Total 3 marks] Andy Todd 38

39 4 The following figure represents a pair of homologous chromosomes at the beginning of the first division of meiosis. The loci of two genes are shown, and both genes have two alleles. Q Q q q r r R R Complete the diagram below to show the four possible gametes formed at the end of meiosis. Use the same letters as in the figure above. [Total: 2 marks] 5 (a) Explain the meaning of the terms linkage and crossing over. linkage crossing over [3] Andy Todd 39

40 (b) In an investigation into the genes on chromosome 2 of the tomato genome, pollen from a pure-bred plant with green leaves and smooth-surfaced fruit was transferred to flowers of a plant with mottled green and yellow leaves and hairy (so-called peach ) fruit. All the F 1 generation had green leaves and smooth fruit. Describe briefly how a plant breeder ensures that the offspring produced are only from the desired cross [3] (c) Four different test crosses, A to D, were then made between F1 plants and purebred plants with mottled leaves and peach fruit. The phenotypes of 50 offspring of each of the crosses were recorded and are shown in the table below. phenotypes of offspring of test crosses cross green leaves and smooth fruit green leaves and peach fruit mottled leaves and smooth fruit mottled leaves and peach fruit A B C D total (i) Suggest one reason why, in the table above, the numbers of plants with green leaves and smooth fruit is not the same in each of the crosses A to D [1] (ii) The percentage cross over value is calculated as number of recombinan t offspring 100 total number of offspring Using the information in the table above, calculate the percentage cross over value between the loci for leaf colour and fruit surface texture. Show your working. Answer =... % [2] Andy Todd 40

41 (iii) Use annotated diagrams of tomato chromosome 2 to explain the results of the test crosses shown in the table. Use the symbols A/a for the leaf colour alleles and B/b for the fruit surface texture alleles. [6] [Total: 15 marks] 6 In this question, one mark is available for the quality of use and organisation of scientific terms. Describe the behaviour of chromosomes during meiosis which results in genetic variation among Daphnia populations. [7] Quality of Written Communication [1] [Total 8 marks] Andy Todd 41

42 7 The following are different stages in meiosis. Each stage has been given a letter. anaphase II metaphase II anaphase I prophase I telophase II metaphase I M N P Q R S (i) Using only the letters, arrange these stages in the correct sequence [1] (ii) State the letter of the stage when each of the following processes occur. pairing of chromosomes... centromeres divide... crossing over... bivalents align on equator... nuclear membrane reforms... [5] (iii) State two processes that occur in a cell during interphase to prepare for a meiotic division [2] [Total 8 marks] Andy Todd 42