E.F.ACADEMY
JWh
9/9/2014
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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.
 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 sub-stages.
 Meiosis I
 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) the process of synapsis.

 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
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 Crossing over may occur, this is a random event in which genetic material is exchanged between sister chromatids. The site at which crossing over occurs is called a chiasma. (pl chiasmata)
 These allow exchange of alleles.
 The nuclear envelope and nucleolus disappear.
 Centrioles and spindle fibres form (no centrioles in plant cells)

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Metaphase 1
 Centrioles (not in plant cells) migrate to the poles of the cell
 Bivalents are arranged at the equator
 Independent random assortment of Bivalents: The separation of each pair is random. For each pair there is a maternal and a paternal chromosome which came from the mother and father. There is no mechanism to cause maternal or paternal chromosomes to either be attracted by or repel one another. So the movement of each chromosome pair is random with respect to other chromosome pairs.
 Spindle fibres attach to the centromeres of sister chromatids.
Anaphase 1

 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.
Telophase I (Not in plant cells
)
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 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.

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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.
Metaphase II
 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
Anaphase II
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 The centromeres separate, and the two chromatids of each chromosome move to opposite poles on the spindle.
 The separated chromatids are now called chromosomes.
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.

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 Because of crossing-over, some chromosomes are seen to have recombined segments of the original parental chromosomes.

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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:
1.
Crossing Over
 Prophase I
 Homologous pairs of Chromosomes (Bivalents) form
 Chiasma(ta) form between homologous chromosomes
 Alleles swap between paternal and maternal chromatids.
 Several crossings over events occur per Bivalent
2.
Independent random assortment of Bivalents
 Metaphase I
 Bivalents line up along the cell equator randomly and independent of each other
3.
Independent random assortment of (recombinant) chromatids
 Metaphase II
 Chromatids line up along the cell equator randomly and independent of each other.
4.
Random mating
 Choices of mates is random
5.
Random fusion of gametes
 Chances of gamete fertilising another specific gamete
6.
Chromosome mutations
 Any changes to DNA further increases the variation

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Explain the terms allele, locus, phenotype, genotype, dominant, co-dominant and recessive.
 Allele – An alternative form 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/feature.
 Phenotype – Physical characteristic shown 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 homozygous.
 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.

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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.
Example: Linked alleles
 In Drosophila; The symbol R for the red eyes allele and r for the white eyes (normal) allele, and N for the normal wing allele and n for the curly wing allele.
Phenotype
Genotype
Gametes
F1
Red Normal x White curly
RRNN
RN x x rrnn rn
RrNn all red eyed normal wing.

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If the F1 are allowed to breed amongst themselves, then:
RrNn x RrNn Genotype
Gametes RN rn x RN rn
(remember; the genes are linked)
F2 genotypes: RRNN: rRnN: RrNn: rrnn
3 red eyed normal winged: 1 white eyed curly winged
There may be a small number of red eyed curly wing and white eye normal wing: these are due to crossing over having taken place.
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 F2 cross generation in lower frequencies.
 These individuals are a result of crossing over events that occur between the two alleles on the chromosome.

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To find the genotype of a dominant looking individual the unknown must be crossed with a homozygous recessive fly. Any hidden recessive genes will make themselves shown. This is called a TEST-CROSS.
Phenotype red eye, normal wings x
Genotype RrNn x rrnn white eye, curly wing
Gametes RN rn
Offspring genotypes
Offspring Phenotypes
Ratio of offspring: x rn
So the dominant looking fly was heterozygous
RrNn; rrnn
Red eyed, normal wing; white eye, curly wing
1 : 1
If the parent had been homozygous then all the offspring would have been Red eyed, normal wing.
Genes are not Linked
 This looks like a normal di-hybrid heterozygous cross
 If a heterozygous AaBb was crossed in a backcross aabb
Gametes: AB, Ab, aB, ab x ab,
Offspring genotypes:
Ratio
AaBb, Aabb, aaBb, aabb
1 : 1 : 1 : 1

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Use genetic diagrams to solve problems involving sex-linkage and co-dominance.
Sex-Lined (X-linked) – Gene with its locus on the sex chromosomes X . As there are few genes on the Y chromosome, most sex-linked genes are on the X chromosome.
Example: Sex-Linked alleles

F1 Analysis
 Half of the F1 males have the mutant phenotype because the gene is on
15 the X chromosome and there is no allele on the Y chromosome.
 The allele is on the non- homologous region of the X chromosome.
 This means that there is a mutant allele on one of the X chromosomes of the female, while the male has the normal allele on the X as well as the
Y chromosome.
 The female is a carrier of the recessive gene.
 Half of the female offspring will also be carriers.
Examples: Haemophilia.
Haemophilia in the Descendants of Queen Victoria (God Bless Her)

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Co-dominance
A characteristic where both alleles contribute to the phenotype.
The phenotype from the heterozygous individual shows the characteristic of both alleles in the genotype. o Example: Co-Dominance
In Snapdragons the gene for red flowers is co-dominant with the gene for white flowers:

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F1 Analysis
 How is the colour inherited?
 The F1 progeny of a cross between one white and one red individual all had pink flowers, that is, a phenotype intermediate between the two parental phenotypes.
 The simplest hypothesis is that the colour exhibits co-dominance with a single gene involved.
 The homozygous genotypes give the red and white phenotypes, while the heterozygous genotype gives the pink 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.
F2 Analysis
 Two pink individuals produced red, pink and white offspring in the ratio
1:2:1
 The RR genotype gives the red phenotype, Rr gives pink, and rr gives white
.

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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 predicted 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.
Eg:
A plant with pink flowers was crossed with another plant with white flowers.
200 seeds were grown and the expected ratio was 1:1
The observed numbers of plants was 120/80
First, we must write up a Null Hypothesis.
 A null hypothesis always states that:
‘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.

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Observed
120
80
Expected
100
100
(O-E)
20
-20
(O-E) 2
400
400
(O-E) 2
E
4
4
4 + 4 = 8
  
 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.
 Calculating the degrees of freedom (df) is straight forward. It is the number of actual different phenotypic groups for the cross (n) minus one

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 The formula looks like this; df = n – 1
 In this example the Chi squared value is 8 and the df = 1 (2 - 1 = 1)
 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.
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,

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Stating that
‘There is a significance difference between our observed results and those expected’
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
‘There is no significance difference between our observed results and those expected’
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.
In the example above 8 is greater than the  2 value of 3.84. Therefore the

Null Hypothesis is rejected.

'There is a statistically significant difference between our observed and expected results'.


Describe the differences between continuous and discontinuous variation.
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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;
1.
Continuous
 Features have a range of variation (quantitative)
 Polygenic – many genes contribute to the final phenotype (2 or more)
 Each gene provides an additive effect on the phenotype
 Genes are not linked, they are on separate chromosomes
 Examples – height, weight
2.
Discontinuous
 Distinct categories
 Monogenic – one gene contributes to the final phenotype
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Epistasis if there is more than 1 gene at work
 Different alleles have a large effect on the phenotype
 Different genes at different loci have large effects on the phenotype.
 Example – blood groups, ear lobe shape
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: Mouse coat colours
 Mouse coat colour is controlled by a gene B/b, where B = Agouti (Black), and b = Black.
 A second gene D allows coat colour to develop, dd results in albino
(white) mice.
A cross between two heterozygous agouti mice was conducted and the results were as follows.
Parental generation

DB
Db dB db
Parental Genotypes DdBb
Gametes: DB, Db, dB, db x
DB Db x DdBb
DB, Db, dB, db dB
DDBB
DDBb
DdBB
DdBb
DDBb
DDbb
DdBb
Ddbb
DdBB
DdBb ddBB ddbb db
DdBb
Ddbb ddBb ddbb
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D.B. = Agouti = 9
D.bb = Black = 3 dd.. = White = 4

The dominant form of the D/d gene provides the pigment for which the B/b
25 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 D/d gene has an epistatic effect over the B/b gene
.
It is important to note how this epistasis phenomenon could actually occur.
 In some cases a gene may code for an enzyme with a complementary site for another protein/gene.
 It may code for a transcription factor that regulates the expression of another gene
 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 = 3+3+1 = 7

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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.
 Polygenic traits are more easily influenced by the environment that monogenic traits
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.
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 Gene pool
 Total number of different alleles within a population
 Populations have gene pools
 Individuals have genomes
The Hardy-Weinberg model
 The Hardy-Weinberg model assumes that all allele ad genotype frequencies will remain stable from generation to generation and:-
1.
The population is large
2.
Mating is random
3.
There are no mutations
4.
There is no migration in or out of the population
5.
There is no genetic drift
6.
There is no selection
 The frequencies will not be stable in nature so it allows us to work out what is causing the changes
Hardy – Weinberg
Allele frequency:
Genotype frequency p + q = 1 p 2 + 2pq + q 2 = 1

28 e.g. In ladybirds colour is controlled by two alleles. bb gives a red ladybird and
BB or Bb gives black. In a population there were 296 red and 50 black ladybirds. Use Hardy Weinberg to calculate the frequency of the dominant allele p and the recessive allele q. p + q = 1 p 2 + 2pq + q 2 = 1
There were (296 + 50) = 346 ladybirds.
296 = red and therefore bb; bb = q 2
So 296/346 = q 2 q 2 = 0.86
So q = /0.86 q = 0.92 p + q = 1 p + 0.92 = 1 p = 0.08 p = 0.08 q = 0.92

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Explain, with examples, how environmental factors can act as stabilising or evolutionary forces of natural selection. 
1.
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)
2. Environmental factors a. Biotic - Living
 Disease
 Food (prey)
 Competition (Intraspecific or interspecific) b. Abiotic – Non-living
 Water availability (soil or pools)
 Mineral availability
 Space
 Light
 Temperature c. Human Factors
 Size of yield
 Quality of yield
 Progeny
 Disease resistance
 Temperament
3.
Stabilising pressures a.
Stabilising selection
 Favours individuals well adapted to unchanging environment
 Maintains the populations mean phenotypes by selecting against the phenotypic extremes
 Example: Coelacanth – unchanged for millions of years.

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4.
Evolutionary pressures a.
Directional selection
 Favours individuals with phenotypes at one extreme and selects against individuals with phenotypes at the other extreme
 Allele frequency is shifting in 1 direction
 Example: Giraffe neck lengths
5.
Disruptive selection
 Favours individuals at both extremes (specialists) and selects against those with a general phenotype
 Supports speciation.
 Example: Bonobo /chimpanzee

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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

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Explain the role of isolating mechanisms in the evolution of new species, with reference to ecological (geographic), seasonal (temporal) and reproductive mechanisms.
1.
Isolating mechanism – mechanism that divides populations of organisms into smaller sub groups a.
Isolating a population will;
 Decrease the gene pool
 Increase inbreeding which could lead to an increase in homozygous recessive genotypes within the population
 Increase the effects of genetic drift
2.
Several types of isolating mechanisms a.
Reproductive (mechanical)
 When the reproductive structures are no longer physically compatible.
 Example: Great Dane will not mate with a Chihuahua. b.
Ecological (Geographical) Allopatric Speciation
 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 or islands.
 Example: Galapagos finches c.
Seasonal (Temporal) Sympatric Speciation
 When two species or populations live within the same area but are reproductively active at different times.
 Example: American frog species. i.
Wood frog – fertile in April ii.
Tree frog – fertile in June iii.
Bull frog – fertile in July

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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 i.
Can interbreed to produce fertile offspring ii.
Reproductively isolated iii.
Have the same: a.
Morphology b.
Physiology c.
Anatomy d.
Embryology e.
Behaviour f.
Occupy the same ecological niche d.
Behavioural Sympatric Speciation
 When two species or populations evolve different courtship displays which are essential for successful mating
 Example: Birds of paradise.
 Phylogeny i.
All living things have RNA / DNA / Proteins ii.
Comparisons to relatedness of these molecules can determine how closely related organisms are. iii.
Haplotypes can also be used for comparisons a.
Haplotypes are particular base sequences b.
Can be 1 locus, several loci or an entire chromosome c.
They are a combination of alleles at different loci transmitted together.

A Phylogenic tree
A Monophyletic group / Clade

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Group of organisms with similar haplotypes
 Shows the common ancestor and all descendants
 A Monophyletic group can be 1 or more clades 


A Paraphyletic group
Only linear – more like traditional classification
 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 i. 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 i. Organisms that have characteristics that are desirable to humans are selected to survive and reproduce, passing on the alleles for favourable characteristics.
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Selective breeding of domesticated animals shows that artificial selection can cause evolution.
Zebu
Low productivity
High disease resistance
Aberdeen Angus
High productivity
Low disease resistance
Droughtmaster
High productivity and disease resistance

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Selective breeding is: the process by which humans breed other animals and plants for particular traits.
Typically, strains that are selectively bred are domesticated, and the breeding is normally done by a professional breeder.
Bred animals are known as breeds, while bred plants are known as varieties, cultigens, or cultivars.
Selectively bred Maize cf. original wild type
The offspring of two pure bred animals but of different breeds is called a crossbreed, and crossbred plants are called hybrids.
The aims of selective breeding (in food organisms) is to increase yield, increase disease/pest resistance, decrease growing time, decrease non-food growth (Stalks on plants).