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5.1.2
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A2 Biology OCR
Unit F215: Control, genomes and
environment
Module 1.2 Meiosis and variation
Notes & Questions
Andy Todd
1
5.1.2
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.
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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
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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)
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5.1.2
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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
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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
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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.
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5.1.2
o 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.
Meiosis II
o Prophase II
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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
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5.1.2
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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
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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 in their own
right.
o Telophase II
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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.
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5.1.2
Explain how meiosis and fertilisation can lead to variation through the independent
assortment of alleles.
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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.
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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
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5.1.2
Explain the terms linkage and crossing over.
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Linkage – Genes for different characteristics that are present at different loci on the
same chromosome are linked and inherited together.
o Example: Linked alleles
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
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.
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Two pure bred aliens are crossed as seen below.
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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.
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5.1.2
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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.
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F1 test cross analysis
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How did these results come about?
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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.
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5.1.2
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5.1.2
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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.
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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
Andy Todd
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A Testcross/Backcross is always a cross with a homozgous recessive
genotype for both genes.
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This will allow all alleles to be expressed in the phenotypes.
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It will allow homozygous recessive genotypes to be expressed for
both genes in the offspring if the parent is heterozygous for the two
genes.
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5.1.2
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
Offspring genotypes
Ratio
Andy Todd
AaBb,
1
ab,
Aabb,
:
1
aaBb,
:
1
aabb
:
1
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5.1.2
Genes are linked
Gametes
AB,
ab
Offspring Genotypes
Ratio
ab,
AaBb
aabb
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
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Crossing Over – When non-sister chromatids exchange alleles during prophase of
meiosis I.
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5.1.2
Use genetic diagrams to solve problems involving sex-linkage and co-dominance.
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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.
Andy Todd
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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.
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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.
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5.1.2
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These expected ratios are closely matched by our observed F 2 results
of 38 normal females : 33 short females : 41 normal males : 35 short
males.
o Conclusion
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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
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5.1.2
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5.1.2
o F2 Analysis
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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.
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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.
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These expected ratios are closely matched by our observed F 2 results
of 22 no horn : 47 one horn : 24 two horns.
o Conclusion
The horns trait shows co-dominance.
Andy Todd
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The HH genotype gives the no-horn phenotype, Hh gives one horn,
and hh gives two horns.
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Interbreeding one-horned Hh F1 individuals produces F2 offspring with
a 1:2:1 ratio
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5.1.2
Describe the interactions between loci (epistasis).
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The interaction of genes concerned with the expression of one characteristic. One
gene may mask the expression of another.
o Example: Rat coat colours
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Rat coat colour is controlled by a gene B/b, where B = Agouti (Black),
and b = Brown.
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A cross between two heterozygous agouti rats was conducted and the
results were as follows.
Parental generation
x
Bb
Bb
F1 generation
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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
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5.1.2
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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.
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Therefore the following genotypes and phenotypes are possible.
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This means that the rats in the parental cross must have actually been
BbCc x BbCc
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The following cross outlines how this has been possible.
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5.1.2
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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.
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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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5.1.2
Use the chi-squared (χ2) test to test the significance of the difference between
observed and expected results.
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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
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R – allele for round seeds
r – allele for wrinkled seeds
G – allele for yellow seeds
g – allele for green seeds
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A true heterozygous plant for round yellow seeds was crossed with
another true heterozygous plant for round yellow seeds.
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RrGg
Yellow round
x
RrGg
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.
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5.1.2
First, we must write up a Null Hypothesis.
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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’
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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)
Yellow round
169
162
Green round
54
54
Yellow wrinkled
51
54
Green wrinkled
14
18
0-E
(0-E)2
(0-E)2/E
Σ=
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5.1.2
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The completed table looks like this.
Chi-squared test
Observed (o) Expected (E)
0-E
(0-E)2
(0-E)2/E
Yellow round
169
162
7
49 0.302
Green round
54
54
0
0
0.000
Yellow wrinkled
51
54
-3
9
0.167
Green wrinkled
14
18
-4
16
0.889
Σ=
1.358
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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.
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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
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The formula looks like this;
df = n – 1
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In this example the Chi squared value is 1.358 and the df = 3 (4 - 1 = 3)
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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.
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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.
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5.1.2
Probability value (P-Value) of 0.995 = 99.5% chance that the difference is due to chance
Probability value of 0.05 = 5% chance that the difference is due to chance
DF
2
1
0.995 0.975 0.2
0.1
0.05 0.025 0.02 0.01 0.005 0.002 0.001
3.93E- 0.0009
05
82
1.642 2.706 3.841 5.024 5.412 6.635 7.879 9.55 10.828
3
2
4
3
0.0717 0.216 4.642 6.251 7.815 9.348 9.837 11.345 12.838 14.796 16.266
5
4
0.207 0.484 5.989 7.779 9.488 11.143 11.668 13.277 14.86 16.924 18.467
6
5
0.412 0.831 7.289 9.236 11.07 12.833 13.388 15.086 16.75 18.907 20.515
7
6
0.676 1.237 8.558 10.645 12.592 14.449 15.033 16.812 18.548 20.791 22.458
8
7
0.989
1.69
9.803 12.017 14.067 16.013 16.622 18.475 20.278 22.601 24.322
9
8
1.344
2.18
11.03 13.362 15.507 17.535 18.168 20.09 21.955 24.352 26.124
10
9
1.735
2.7
12.242 14.684 16.919 19.023 19.679 21.666 23.589 26.056 27.877
11
10
2.156 3.247 13.442 15.987 18.307 20.483 21.161 23.209 25.188 27.722 29.588
12
11
2.603 3.816 14.631 17.275 19.675 21.92 22.618 24.725 26.757 29.354 31.264
13
12
3.074 4.404 15.812 18.549 21.026 23.337 24.054 26.217 28.3 30.957 32.909
14
13
3.565 5.009 16.985 19.812 22.362 24.736 25.472 27.688 29.819 32.535 34.528
15
14
4.075 5.629 18.151 21.064 23.685 26.119 26.873 29.141 31.319 34.091 36.123
0.01 0.0506 3.219 4.605 5.991 7.378 7.824
Accept Null
9.21 10.597 12.429 13.816
Reject Null
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The 0.05 (5%) significance level is the cut off point.
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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’
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If the calculated chi squared value is lower than the critical value at the 0.05
significance level then we accept the null hypothesis,
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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’
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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 0.70.
Andy Todd
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5.1.2
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.
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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
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5.1.2
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
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Allele frequency

Genotype frequency
Andy Todd
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5.1.2
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 = 0.25.
2. A school of seahorses with p = 0.65.
3. A school of redhorse suckers (a type of fish) with q = 0.65.
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 = 0.25.
p = 0.25, q = 0.75 (0.25 + 0.75 = 1.0)
p2 = 0.0625
2pq = 0.375
q2 = 0.5625 (0.0625 + 0.375 + 0.5625 = 1.0)
A phenotype = 0.0625 + 0.375 = 0.4375
a phenotype = 0.5625 (0.4375 + 0.5625 = 1.0)
2. A school of seahorses with p = 0.65.
p = 0.65, q = 0.35 (0.65 + 0.35 = 1.0)
p2 = 0.4225
2pq = 0.455
q2 = 0.1225 (0.4225 + 0.455 + 0.1225 = 1.0)
A phenotype = 0.4225 + 0.455 = 0.8775
a phenotype = 0.1225 (0.8775 + 0.1225 = 1.0)
3. A school of redhorse suckers (a type of fish) with q = 0.65.
p=0.35, q= 0.65 (0.35 + 0.65 = 1.0)
p2 = 0.1225
2pq = 0.455
q2 = 0.4225 (0.1225 + 0.455 + 0.4225 = 1.0)
A phenotype = 0.1225 + 0.455 = 0.5775
a phenotype = 0.4225 (0.5775 + 0.4225 = 1.0)
Andy Todd
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5.1.2
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
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5.1.2

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
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5.1.2
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.
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5.1.2
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
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5.1.2

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
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5.1.2

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.
Artificial Selection on dogs:
Example of a breed where curly
coat is desired:
Natural Selection for antibiotic resistance in
strep bacteria.
Source of
variation in
trait.
Mutation gives rise to a curly coat Mutation gives rise to a small proportion of
in a certain dog.
bacteria resistant to the antibiotic.
Agent of
Selection
Intentional selection by breeder
Presence of antibiotic in the bacteria's
desiring more dogs showing curly environment - your throat.
coats.
Immediate
Result of
selection
Individuals with curly coats are
Greater proportion of bacteria that survive are
more likely to survive and be used resistant to the antibiotic.
for breeding.
Result after a
number of
generations
assuming a
constant
environment!
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.
Andy Todd
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.
32
5.1.2
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
Milky
Shorthorn
Holstein Friesian
Jersey
Brown
Swiss
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
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5.1.2

Modern bread Wheat (Triticum aestivum)
Wild Einkorn
AuAu
2n = 14
Einkorn
AuAu
2n = 14
Wild Grass
BB
2n = 14
STERILE AuB
Emmer Wheat
AuAuBB
4n = 28
Goat Grass
DD
2n = 14
STERILE AuBD
Polyploidy
Andy Todd
Triticum aestivum
Common Wheat
AuAuBBDD
6n = 42
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5.1.2
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]
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5.1.2
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 F1 plants all with pink flowers. The F1 were then interbred to
produce F2 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:
Cr = red, Cw = white
Parental phenotypes:
red flowers
x
white flowers
Parental genotypes:
....................................
........................................
Gametes:
....................................
........................................
F1 genotypes: ..................................................................................................
F1 phenotypes: ................................................................................................
Gametes: ........................................................................................................
F2 genotypes: ..................................................................................................
F2 phenotypes: ................................................................................................
Expected F2 phenotypic ratio:.........................................................................
[6]
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5.1.2
(c)
A chi-squared (χ2) 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.
F2 phenotype
observed numbers
red
62
pink
131
white
67
total
260
expected numbers
260
[3]
The χ2 statistic is calculated in the following way:
χ2 
(ii)

observed– expected2
expected
  “sum of ...”
Calculate the value of χ2 for the above data. Show your working.
χ2 value = ................................................................................................
[2]
(iii)
The critical value of χ2 for this type of investigation with two degrees of
freedom is 5.991.
Explain whether your answer to (b) (ii) supports the hypothesis.
................................................................................................................
................................................................................................................
[1]
[Total 16 marks]
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5.1.2
2.
Phenotype is influenced by genetic and environmental factors.
Describe one example of how the environment influences phenotype.
..................................................................................................................................
..................................................................................................................................
..................................................................................................................................
[Total 2 marks]
3
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]
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5.1.2
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]
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5.1.2
(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 F1 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
23
4
3
20
B
21
3
3
23
C
16
4
5
25
D
22
6
4
18
total
82
17
15
86
(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 = ..................................................... %
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[2]
5.1.2
(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]
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5.1.2
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.
1 ......................................................................................................................
2 ......................................................................................................................
[2]
[Total 8 marks]
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