Cellular Control Unit 1 Communication, Homeostasis and Energy

Cellular Control
Unit 1
Communication, Homeostasis and Energy
Meiosis
Module 1: Cellular Control
Learning outcomes

describe, with the aid of diagrams and
photographs,
 the behaviour of chromosomes during
meiosis,
 the associated behaviour of the nuclear
envelope, cell membrane and centrioles.
 (Names of the main stages are expected,
but not the subdivisions of prophase);
Reproduction and variation

Asexual reproduction
 Single organism divides by mitosis
 New organism is genetically identical to
the parent

Sexual reproduction
 Meiosis produces haploid gametes
 Which fuse at fertilisation to form a diploid
zygote
 This produces genetic variation amongst
offspring
Human Life Cycle
Diploid
Zygote
46
fertilisation
Mitosis
Haploid
Sperm
23
Haploid
Egg
23
Meiosis
Adult
46
Self assessment questions
The fruit fly Drosophila melangaster has
eight chromosomes in its body cells. How
many chromosomes will there be in a
Drosophila sperm?
 The symbol n is used to indicate the number
of chromosomes in one set – the haploid
number of chromosomes. For example in
humans n = 23, in a horse n = 32.

 How many chromosomes are there in a gamete
of a horse?
 What is the diploid number of chromosomes (2n)
of a horse?
Meiosis

Meiosis is a reduction division
 Resulting daughter cells have half the original
number of chromosomes
 Daughter cells are haploid
 Can be used for sexual reproduction
 Source of genetic variation

Meiosis has two divisions
 meiosis I and meiosis II

Each division has 4 stages
 Prophase, metaphase, anaphase, telophase
Meiosis

You can view an animation of Meiosis at
http://www.cellsalive.com/meiosis.htm
Meiosis I
Prophase I
Chromatin condenses
Homologous pairs form
a bivalent
Nucleolus disappears
Spindle forms
Anaphase I
Homologous
chromosome in each
bivalent are pulled to
opposite poles
Metaphase I
Bivalents
line up on
equator of cell
Telophase I
Two
new nuclear
envelopes form
Cell divides by
cytokinesis
Early Prophase 1
Late Prophase 1
Metaphase 1
Anaphase 1
Telophase 1
Cytokinesis 1
Meiosis II
Prophase II
Metaphase II
Anaphase II
Telophase II
Nucleolus disappears
Chromosomes condense
Spindle forms
Centromeres divide
Chromatids pulled apart
to opposite poles
Chromosomes
arrange
themselves on equator
Attach by centromere to
spindle fibres
nuclear
envelopes reform
around haploid nuclei
Cell divides by cytokinesis
Prophase II
Metaphase II
Anaphase II
Telophase II
Cytokinesis II
Learning outcomes

explain how meiosis and fertilisation
can lead to variation through the
independent assortment of alleles
Key words





Allele
Locus
Crossing over
Maternal chromosome
Paternal chromosome
Alleles, locus and homologous
chromosomes
Meiosis and variation



Meiosis enables sexual reproduction to
occur by the production of haploid
gametes.
Sexual reproduction increases genetic
variation
Genetic variation increases the
chances of evolution through natural
selection
Meiosis and Variation
Crossing over – prophase I
Independent assortment of chromosomes –
metaphase I
 Random assortment of chromatids –
metaphase II
 Random fertilisation
 Chromosome mutations


 Number of chromosomes
▪ Non-disjunction - polysomy or polyploidy
 Structure of chromosomes
▪ Inversion, deletion, translocation
Crossing over
During metaphase I
During metaphase I
No crossing over
Crossing over – new
combinations of alleles
Independent Assortment
Learning Outcomes


explain the terms allele, locus,
phenotype, genotype, dominant,
codominant and recessive;
explain the terms linkage and crossingover;
Glossary





Gene
Locus
Allele
Genotype
Phenotype





Heterozygous
Homozygous
Monohybrid cross
Dominant allele
Recessive allele
Genetics


Genetics is the study of inheritance
Allele
 different varieties of the same gene

Locus
 position of a gene on a chromosome
Genetics

Dominant
 An allele whose effect is expressed in the
phenotype if one copy present

Recessive
 An allele which only expresses as a
homozygote

Co-dominant
 Both alleles have an effect on the
phenotype

Genotype
 genetic constitution of the organism

Phenotype
 appearance of character resulting from
inherited information

Homozygous
 Individual is true breeding
 Possesses two alleles of a gene e.g. RR or
rr

Heterozygous
 Two different alleles for a gene e.g. Rr
Monohybrid inheritance

Mendel’s First Law
 principle of segregation
“The alleles of a gene exist in pairs but
when gametes are formed, the
members of each pair pass into
different gametes, thus each gamete
contains only one of each allele.”
Inheritance of height in pea
plants
gene
Height of
pea
plants

Allele relationship
Symbol
Tall
Dominant
T
dwarf
recessive
t
Follow out the following cross to the F2 generation
 Homozygous tall pea plant with a homozygous dwarf pea
plant

Write out the genotypic and phenotypic ratios from
the F2 generation
Inheritance of height in pea
plants

Laying out the cross










P phenotype
P genotype
Gametes
F1 genotype
F1 phenotype
F1 self-fertilised
Gametes
Random fertilisation
F2 genotypic ratio
F2 phenotypic ratio
Pupil Activity

Answer the questions on monohybrid
inheritance
 Remember to write out each cross in full.
Cystic Fibrosis


Cystic Fibrosis is caused by a mutation
to a gene on one of the autosomes.
Mutation
 Changes the shape of the
transmembrane chloride ion channels
(CFTR protein)
 The CFTR gene is found on Chromosome 7
 The faulty gene is recessive
Genetic Cross conventions


Use symbols to represent two alleles
Alleles of the same gene should be
given the same letter
 Capital letter represents the dominant
allele
 Small letter represents the recessive allele

Choose letters where the capital and
small letter look different
 The examiner needs to be in no doubt
about what you have written
Inheritance of cystic fibrosis

Three possible genotypes
 FF
 Ff
 ff
unaffected
unaffected
cystic fibrosis
Remember gametes can only contain one
allele for the CFTR gene
 At fertilisation, any gamete from the father
can fertilise any gamete from the mother

 This can be shown in a genetic diagram
Genetic diagram showing the chances of a
heterozygous man and a heterozygous woman having
a child with cystic fibrosis.
Phenotype ratio of offspring

Genotype ratio 1FF:2Ff:1ff


Phenotype ratio 3 unaffected:1cystic fibrosis
Can also be expressed as
 25% chance of the child having cystic fibrosis
 Probability of 0.25 that a child will inherit the
disease
 Probability that 1 in 4 that a child from these
parents will have this disease.
Learning Outcome

Use genetic diagrams to solve
problems involving sex-linkage and
codominance.
Sex-Linkage



Sex-linked genes are genes whose loci
are on the X or Y chromosomes
The sex chromosomes are not
homologous, as many genes present
on the X are not present on the Y.
Examples
 Haemophilia
 Fragile X syndrome
 Red green colour blindness
Sex Chromosomes
Factor VIII and Haemophilia

Haemophilia is caused by a recessive
allele of a gene that codes for a faulty
version of the protein factor VIII
 XH
 Xh
normal allele
haemophilia allele
possible genotypes and
phenotypes
Inheritance of Haemophilia
Pedigree for a sex linked
recessive disease
Codominance


Codominance describes a pair of
alleles, neither of which is dominant
over the other.
This means both have an effect on the
phenotype when present together in
the genotype
Codominance example

Flower colour in plants
 CR
 Cw


red
white
Genotypes
 CRCR
 CRCW
 CWCW
red flowers
pink flowers
white flowers

Write out a genetic
cross between a pure
breeding red plant
and a pure breeding
white plant.
Carry out the cross to
the F2 generation.
 Write out the genotype
and phenotype ratio for
the F2 generation
Revision Question



Coat colour in Galloway cattle is controlled by a
gene with two alleles. The CR allele produces red
hairs and therefore a red coat colour. The Cw allele
produces white hairs.
A farmer crossed a true-breeding, red-coated cow
with a true-breeding white-coated bull. The calf
produced had roan coat colouring (made up of
an equal number of red and white hairs).
Explain the result and draw a genetic diagram to
predict the outcome of crossing two roan coloured
animals.
Inheritance of A, B, AB and O
blood groups

Human blood groups give an example
of codominance and multiple alleles
 There are 3 alleles present
▪ IA
▪ IB
▪ Io
 IA


and IB are codominant
Io is recessive
Remember each human will only have
two alleles
Blood Groups
Genotype
Phenotype
IAIA
IA Io
IAIB
IBIB
IB Io
Io Io
Blood Group A
Blood Group A
Blood Group AB
Blood Group B
Blood Group B
Blood Group o
Inheritance of blood groups

Carry out genetic crosses for the
following examples
 Two parents have blood groups A and B,
the father is IAIo and the mother is IBIo
 Father has blood group AB and the
mother has blood group O
 Mother is homozygous blood group A and
the father is heterozygous B.
Learning Outcome


Describe the interactions between loci
(epistasis).
Predict phenotypic ratios in problems
involving epistasis.
Dihybrid Inheritance

Monohybrid cross
 Inheritance of one gene

Dihybrid cross
 Inheritance of two genes
Example – dihybrid cross

Tomato plants
 Stem colour
A
purple stem
a
green stem
d
potato leaves
 Leaf shape
D

cut leaves
NOTE
 In the heterozygote AaDd due to independent
assortment in meiosis there are 4 possible gamete
combinations
AD
Ad
aD
ad
Crosses

Cross a heterozygous plant with a
plant with a green stem and potato
leaves

Cross two heterozygous tomato plants
Dihybrid Inheritance
A woman with cystic fibrosis has blood
group A (genotype IAIo). Her partner does
not have cystic fibrosis and is not a carrier
for it. He has blood group O.
 Write down the genotypes of these two
people.
 With the help of a full and correctly laid out
genetic diagram, determine the possible
genotypes and phenotypes of any children
that they may have.

Autosomal linkage



Each Chromosome carries a large
number of linked genes
If two genes are on the same
chromosome then independent
assortment can not take place.
The genes are transmitted together
and are said to be linked.
Linked Genes

Where linked genes are involved the
offspring of a dihybrid cross will result in
a 3:1 ratio instead of the 9:3:3:1 ratio.

Example:
 In peas, the genes for plant height and
seed colour are on the same
chromosome (i.e. linked)
Learning Outcome


Describe the interactions between loci
(epistasis).
Predict phenotypic ratios in problems
involving epistasis.
Flower colour in sweet pea

Flower colour
 Colourless precursor of a pigment
C
 Gene that controls conversion of this pigment to
purple
P
 Both dominant alleles need to be present for the
purple colour to develop

Cross
 Cross two white flowered plants with the
genotypes CCpp and ccPP
 Follow this cross through to the F2 generation
Interactions of unlinked genes


A single character maybe influenced
by two or more unlinked genes.
E.g. determination of comb shape in
domestic poultry
 Dominant allele P
pea comb
 Dominant allele R
rose comb
 Two dominant alleles walnut comb
 No dominant alleles single comb
Genetic Crosses


Carry out a genetic cross between a
true-breeding pea comb and a true
breeding rose comb
Follow this cross through to the F2
generation
Inheritance of coat colour in
mice

Wild mice have a coat colour that is
referred to as “agouti”.
 Agouti (A) is dominant to black (a)
 C is a dominant gene required for coat
colour to develop
 A homozygous recessive cc means that
no pigment can be formed and the
individual is albino
Inheritance of coat colour in
mice


Carry out a cross between a purebreeding black mouse (aaCC) and an
albino (AAcc)
Follow this cross through to the F2
generation.
Epistasis


This is the interaction of different gene
loci so that one gene locus masks or
suppresses the expression of another
gene locus.
Genes can
 Work antagonistically resulting in masking
 Work complementary
Epistasis ratios

9 : 3 : 4 ratio
 Suggests recessive epistasis

9 : 7 ratio
 Suggests epistasis by complementary
action

12 : 3 : 1 ratio or 13 : 3 ratio
 Suggests dominant epistasis
Predicting phenotypic ratios

Read through pages 132 and 133 of
your textbook
 Answer questions 1 – 7
 Complete the stretch and challenge
question on “eye colour in humans”

Read through and complete the
worksheet provided for you on
epistasis
Learning outcome

Use the chi-squared (χ2) test to test the
significance of the difference
between observed and expected
results.
χ2 (chi-squared) test

Allows us to compare observed and
expected results and decide if there is
a significant difference between
them.
χ2 (chi-squared) test

Where
 Σ = the sum of
 O = observed value
 E = expected value
χ2 (chi-squared) test

Compare the χ2 value to a table of
probabilities
 The probability that the differences between our
expected and observed values are due to
chance.
If the χ2 value represents a probability of
0.05 or larger, the differences are not
significant
 If the χ2 value represents a probability of less
than 0.05, it is likely that the results are not
due to chance and there is a significant
difference.

Degrees of freedom

The degrees of freedom takes into
account the number of comparisons
made.
 Degrees of freedom
= number of classes of data - 1
Table of χ2 values
Degrees
of
freedom
1
2
3
4
Probability greater than
0.1
0.05
0.01
0.001
2.71
4.60
6.25
7.78
10.83
13.82
16.27
18.46
3.84
5.99
7.82
9.49
6.64
9.21
11.34
13.28
Critical value
95% certain that the results are
not due to chance
Table of χ2 values
Degrees
of
freedom
1
2
3
4
Probability greater than
0.1
0.05
0.01
0.001
2.71
4.60
6.25
7.78
10.83
13.82
16.27
18.46
3.84
5.99
7.82
9.49
Accept null hypothesis
There is no significant difference, results
have occurred due to chance
6.64
9.21
11.34
13.28
Table of χ2 values
Degrees
of
freedom
1
2
3
4
Probability greater than
0.1
0.05
0.01
0.001
2.71
4.60
6.25
7.78
10.83
13.82
16.27
18.46
3.84
5.99
7.82
9.49
6.64
9.21
11.34
13.28
Reject null hypothesis: accept experimental hypothesis
Difference is significant, not due to chance
Mammal question
 χ2 value = 51.8
 Degrees of freedom = 3
 Critical value (p=0.05) =



7.82
Reject the null hypothesis
There is a significant difference
between observed and expected
results
Suggestions?
 The two genes are linked
Variation
What did you learn at AS
level?
Learning Outcomes




Define the term variation.
Discuss the fact that variation occurs
within as well as between species.
Describe the differences between
continuous and discontinuous
variation, using examples of a range
of characteristics found in plants,
animals and microorganisms.
Explain both genetic and
environmental causes of variation.
Variation

Variation is the differences that exist
between individual organisms.
 Interspecific variation (between species)
▪ Differences that are used to assign
individuals to different species
 Intraspecific variation (within a species)
▪ Individuals of the same species show variation

Variation can be inherited or
influenced by the environment.
Types of variation

There are two main types of variation
 Continuous variation
 Discontinuous variation

There are two main causes of variation
 Genetic variation
 Environmental variation
Continuous variation





Existence of a range of types between two
extremes
Most individuals are close to a mean value
Low numbers of individuals at the extremes
Both genes and the environment interact in
controlling the features
Examples
 Height in humans
 Length of leaves on a bay tree
 Length of stalk of a toad stool
Continuous variation

Use a tally chart and plot results in a
histogram
Discontinuous variation
2 or more distinct categories with no
intermediate values
 Examples







Earlobes
Blood groups
Bacteria
Flowers
attached or unattached
A, B, AB or o
flagella or no flagella
colour of petals
Genetically determined
The environment has little or no effect on
discontinuous variation
Discontinuous variation
Causes of variation

Genetic Variation
 Genes inherited from parents provide information
used to define our characteristics

Environmental Variation
 Gives differences in phenotype (appearance)
but not passed on by parents to offspring
 Examples
▪ Skin colour tans with exposure to sunlight
▪ Plant height determined by where the seed lands
Variation
What you need to know for
A2!!
Learning outcomes
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.
 Explain that both genotype and
environment contribute to phenotypic
variation.
 Explain why variation is essential in selection.

variation

Variation can be:
 Discontinuous
▪ Each organism falls into one of a few clear-cut
categories, no intermediate values
▪ Qualitative differences between phenotypes
 Continuous
▪ No definite categories
▪ A continuous range of values between two
extremes
▪ Quantitative differences between phenotypes
Genes and variation

Discontinuous (qualitative) variation
 Monogenic inheritance
 Different alleles at same gene locus
 Different gene loci have different effects
 Epistasis, codominance, dominance and
recessive patterns of inheritance
Genes and Variation

Continuous (quantitative) variation
 Polygenic inheritance
 Two or more genes
 Each gene has an additive effect
 Unlinked genes
Polygenic Inheritance

Example –length of corn cobs
 Three genes – A/a, B/b and C/c
 Each dominant allele adds 2 cm length
 Each recessive allele adds 1 cm length

So
 AABBCC
 aabbcc

Hmmm!!
= 12 cm long
= 6 cm long
 How long would AaBBCc be?
 How long would aaBbCc be?
Genotype, environment and
phenotype


The environment can affect the
expression of the genotype
examples
 AABBCC has the genetic potential to
produce cobs 12cm long
▪ This could be affected by
▪ Lack of water, light or minerals
 Obesity in humans
▪ Affected by diet and exercise
Genotype, environment and
phenotype

The environment influences the
expression of polygenic traits more
than monogenic traits.
Learning Outcomes

Use the Hardy–Weinberg principle to
calculate allele frequencies in
populations.
Population genetics

What is a population?
 Group of individuals of the same species
that can interbreed
 Populations are dynamic

The set of genetic information carried
by a population is the gene pool.
Allele Frequency

To measure the frequency of an allele
you need to know
 Mechanism of inheritance of that trait
 How many different alleles of the gene for
that trait are in the population
Hardy-Weinberg principle


The Hardy-Weinberg principle is a
fundamental concept of population
genetics
It makes the following assumptions
 Population is very large
 Random mating
 No selective advantage
 No mutation, migration or genetic drift
The equations



p frequency of the dominant allele
q frequency of the recessive allele
The frequency of the allele will be in
the range 0 – 1.
 0 – no one has the allele
 0.5 – half the population has the allele
 1 – only allele for that gene in the
population
Ok – the equations


Equation 1
p+q=1
Equation 2
p2 + 2pq + q2 = 1
 Where
▪ p2
frequency of genotype DD
▪ 2pq frequency of genotype Dd
▪ q2
frequency of genotype dd
Calculating the frequency of
cystic fibrosis in the population







1 in 3300 babies are born with cystic fibrosis
All babies with cystic fibrosis have genotype
nn
Calculate q2
Calculate q
Calculate p
Calculate frequency of genotype Nn
If we have 30,000 people in our population
how many will be carriers of the cystic
fibrosis allele
Question



Phenylketonuria, PKU, is a genetic
disease caused by a recessive allele.
About one in 15 000 people in a
population are born with PKU.
Use the hardy-Weinberg equations to
calculate the frequency of the PKU
allele in the population.
State the meaning of the symbols that
you use, and show all your working.
The Answer


Calculate q2 = 1 / 15000 = 0.000067
Calculate q = 0.0082
Another question

Explain why the Hardy-Weinberg
principle does not need to be used to
calculate the frequency of
codominant alleles.
Pupil Activity


Answer the Hardy-Weinberg practice
question.
You have 10 minutes
 Starting NOW!!
The Answers





q2 = 0.52 / q = 0.72
p = 1 – 0.72 = 0.28
p+q=1
p2 + 2pq + q2 = 1
Answer = 2pq / use of appropriate
numbers;
Answer = 40%;
The other answers

Any three from:
 Small founder population / common ancestor;
 Genetic isolation / small gene pool / no





immigration /
no migration / in-breeding;
High probability of mating with person having Hallele;
Reproduction occurs before symptoms of disease
are apparent;
Genetic argument – Hh x hh = 50% / Hh x Hh =
75% affected offspring;
No survival / selective disadvantage;
Learning Outcomes


Explain, with examples, how
environmental factors can act as
stabilising or evolutionary forces of
natural selection.
Explain how genetic drift can cause
large changes in small populations.
Variation and Natural Selection
The set of alleles in a population is it’s gene
pool
 Each individual can have any combination
of alleles in the gene pool

 producing variation
 Some individuals more likely to survive
 They reproduce and pass genes on to offspring
 Advantageous alleles become more frequent in
the population
Environmental Resistance


Environmental factors that limit the
growth of a population offer
environmental resistance
These factors can be biotic or abiotic
Selection pressures


An environmental factor that “selects”
for some members of a population
over others
Confers an advantage onto certain
individuals
Stabilising Selection



If the environment stays stable
The same alleles will be selected for in
successive generations
Nothing changes, this is called
stabilising selection
Stabilising Selection
Stabilising Selection
Directional Selection




Change in the environment resulting in
a change in the selection pressures on
the population
Previously disadvantageous alleles
maybe selected for
Change in the genetically determined
characteristics of subsequent
generations of the species
A.k.a. evolution
Directional Selection
Directional Selection
Genetic Drift


A change in the gene pool and
characteristics within the population.
This change has occurred by chance
rather than as the result of natural
selection.
Genetic Drift and Islands

Genetic drift is thought to happen
relatively frequently in populations on
islands.
 Small populations
 Geographically separated from other
members of their species

Evidence
 Many isolated islands have their own
endemic species of plants and animals
Genetic Drift




Reduces genetic variation
Reduce the ability of the population
to survive in a new environment
May contribute to the extinction of a
population or species
Could lead to the production of a
new species
Genetic Drift – Frog Hoppers
The colours of the
common frog-hopper
are determined by
seven different alleles
of a single gene.
 The range of colours
and their frequencies,
on different islands in
the Isles of Scilly, are
very variable,
 There are different
selection pressures on
the different islands

Genetic Drift – Frog Hoppers
The answers
Learning Outcomes

Explain the role of isolating
mechanisms in the evolution of new
species, with reference to ecological
(geographic), seasonal (temporal)
and reproductive mechanisms.
Speciation


Speciation is the formation of a new
species.
Species
 Group of organisms, with similar
morphology and physiology, which can
interbreed with one another to produce
fertile offspring.
Speciation

In the production of a new species,
some individuals must
 Becomes morphologically or
physiologically different from members of
the original species
 No longer be able to breed with the
members of the original species to
produce fertile offspring.
Isolation


Splitting apart of a “splinter group”
Geographical isolation
 Organisms are separated by a physical
barrier

Reproductive isolation
 Two groups have become so different
that they are no longer able to interbreed
 They are now a different species
Isolating Mechanisms

Large populations may be split into
sub-groups by
 Geographic barriers
 Ecological barriers
 Temporal barriers
 Reproductive barriers
Geographical Barriers
(AS recap)


Geographical barrier separates two
populations of a species
Two groups evolve along different lines
 Different selection pressures
 Genetic drift


If barrier breaks down and two
populations come together again,
they may have changed so much
that they can no longer interbreed
They are now two different species
Isolating Mechanisms


Speciation occurs when organisms live
in the same place
The barriers which can prevent two
closely related species from
interbreeding include
 Ecological
 Temporal
 Reproductive
Ecological Barriers


Ecological barriers exist where two
species live in the same area at the
same time, but rarely meet.
Example
 Two different species of crayfish,
Orconectes virilis and orconectes immunis,
both live in freshwater habitats in North
America
Meet the Crayfish

Orconectes virilis
 Not good at digging, can’t
survive summer drying
 Lives in streams and lake
margins

Orconectes immunis
 Lives in ponds and swamps,
 Can easily burrow into the
mud when the pond dries
up

In streams and lake margins
O. virillis is more aggressive
and will drive O. immunis
out of crevices where it tries
to shelter
Temporal Barriers



Two species live in the same place,
and may even share the same habitat
Do not interbreed as they are active
at different times of the day, or
reproduce at different times of year
Example – flowering shrubs in Western
Australia
Meet the shrubs



Banksia
attenuata
flowers in the
summer
Banksia
menziesii
flowers in the
winter
They can not
interbreed
Reproductive barriers

Even if species share the same habitat
and are reproductively active at the
same time, they may not be able to
interbreed





Different courtship behaviours
Mechanical problems with mating
Gamete incompatibility
Zygote inviability
Hybrid sterility
Meet the Mallards

Different courtship
behaviours
 A male mallard duck will
only mate with a female
who displays the correct
courtship behaviour
 Although the pintail
female looks similar to
the Mallard female, her
courtship behaviour will
only attract a pintail
male.
Learning Outcomes

Explain the significance of the various
concepts of the species, with
reference to the biological species
concept and the phylogenetic
(cladistic/evolutionary) species
concept.
The Species Concept

In AS biology you defined a species as
 “a group of organisms, with similar
morphological, physiological, biochemical
and behavioural features, which can
interbreed to produce fertile offspring,
and are reproductively isolated from other
species”
The two species concept

Group of organisms

 Capable of
interbreeding
 Capable of
producing fertile
offspring
 Reproductively
isolated from other
groups

This is the Biological
Species concept
Group of organisms
showing similarities
in characteristics






Morphological
Physiological
biochemical
Ecological
Behavioural
This is the
phylogenetic
species concept
Biological Species concept



Group of organisms that can
interbreed and produce fertile
offspring.
Clear cut definition
Limitation
 Can only be used for organisms that
reproduce sexually
Phylogenetic species concept

Also known as the
 Evolutionary species concept
 Cladistic species concept


Different morphology between the
two groups and certain that they
evolved from a common ancestor
Not rigorous but allows decisions to be
made
Comparing the genetics


Closely related organisms have similar
molecular structures for DNA, RNA and
proteins.
Biologists can compare specific base
sequences (haplotypes)
 The number of differences caused by
base substitutions can be expressed as the
% divergence
Cladistics

Clade
 Group of organisms with similar haplotypes

In cladistic classification systems is
assumes that the taxa are
monophyletic, this means that it
includes an ancestral organism and all
it’s descendents.
Cladistic classification





Focuses on evolution
Places importance on using molecular
analysis
Uses DNA and RNA sequencing
Uses computer programmes
Makes no distinction between extinct
and still existing species
Learning Outcomes


Compare and contrast natural
selection and artificial selection.
Describe how artificial selection has
been used to produce the modern
dairy cow and to produce bread
wheat (Triticum aestivum).
Selection

Natural Selection
 Mechanism for evolution
 Organisms best adapted to their
environments more likely to survive to
reproductive age
 Favourable characteristics are passed on
 Produces organisms that are well adapted
to their environment
Artificial Selection



Humans select the favourable
characteristics
Humans allow those organisms to
breed
Produces populations that show one
characteristic to an extreme
 Other characteristics retained may be
disadvantageous
Artificial Selection and the
modern dairy cow

Breeds of cows with higher milk production
have been artificially selected for
 Milk yield from each cow is measured and





recorded
Test progeny of bulls
Elite cows given hormones to produce many
eggs
Eggs fertilised in vitro
Embryos implanted into surrogate mothers
A few elite cows produce more offspring
than they would naturally
Disadvantage to high milk
yields

Health costs for artificially selected
cows is higher due to
 Mastitis
 Ketosis and milk fever
 Lameness
 Respiratory problems
Artificial selection and bread
wheat (Triticum aestivum)

Polyploidy
 Nuclei contain more than one diploid set
of chromosomes


Wild species of wheat have a diploid
number (2n) of 14
Modern bread wheat is hexaploid
(6n), It has 42 chromosomes in the
nucleus of every cell
Getting from the ancestors to
modern bread wheat
Wild einkorn
AUAU
2n = 14
Domestication and
artificial selection
Einkorn
AUAU
2n = 14
x
Wild Grass
BB
2n = 14
Einkorn
AUAU
2n = 14
x
Wild Grass
BB
2n = 14
Sterile hybrid P
AuB
Mutation that double
chromosome number
Emmer Wheat
AUAUBB
4n = 28
x
Wild Grass
DD
2n = 14
Emmer Wheat
AUAUBB
4n = 28
x
Wild Grass
DD
2n = 14
Sterile hybrid Q
AuBD
Mutation that double
chromosome number
Common Wheat
AUAUBBDD
6n = 42
Continuing selection in wheat

Breeders are continuing to try and
improve wheat varieties
 Resistance to fungal infections
 High protein content
 Straw stiffness
 Resistance to lodging
 Increased yield