Download Variation with standard deviation

Lesson objectives
the different types of variation
To include intraspecific and interspecific variation AND
the differences between continuous and discontinuous
variation, using examples of a range of characteristics
found in plants, animals and microorganisms AND both
genetic and environmental causes of variation.
An opportunity to use standard deviation to measure
the spread of a set of data
and/or
Student’s t-test to compare means of data values of
two populations
and/or
Spearman’s rank correlation coefficient to consider
the relationship of the data.
Variation
The presence of variety
(of differences between individuals)
Variation within species
Variation between species
• Usually obvious.
• Variation used to classify one species from
another.
Continuous & Discontinuous
Continuous variation:
A full range of
intermediate
phenotypes between
two extremes.
Discontinuous variation:
Discrete groups of
phenotypes with no or very
few individuals in between
What causes variation?
Inherited / genetic variation
• Genes (inherited from parents)
• Alleles (versions of these genes)
Sexual reproduction – random shuffling of
alleles – new combinations of parental
alleles in offspring
Mutations – “mistakes” in the DNA change
base sequence and may bring about a
new version of the gene (i.e. a new allele)
Mutations
• When cells divide they need
new chromosomes to fill the
nucleus.
• The chromosomes replicate
and sometimes a mistake is
made, causing a change in a
gene on that chromosome.
• This is called a mutation.
• We all inherit some mutations
but usually we aren’t aware of
them as we have 2 copies of
each gene.
Can mutations have other
Causes?
• Radiation can increase the number of mutations.
This includes any high energy rays such as
– ultra violet
– X rays
– Ionising radiation
• Some chemicals can also lead to mutations such
as Mustard Gas which has been used in
chemical warfare.
Are Mutations Always Bad?
Mutations
provide
variation
For the individual
mutations can be
bad, good or
neutral.
Mutations create variation which leads
differences in success at surviving which
leads to natural selection and therefore
the possibility of evolution.
Without them there would be
no differences between
individuals, and this would
mean everyone would be
equally likely to survive or
die.
Attached earlobes
(recessive
Free earlobes
(dominant)
Environmental causes of variation
A combination of both?
• Environmental and genetic variation are
linked. E.g. Height
• Not all genes are active at any one time.
Changes in the environment affect which
genes are active.
Question to try:
For each of these examples of variation
between sunflower plants, suggest whether
they are caused by genes alone, environment
alone, or an interaction between both.
•The height of the plant
•The colour of the plant petals
•The diameter of the mature flower
•The percentage of seeds that
develop after fertilisation
Look at the following data
50 petals for the flowers of a rush (Luzula sylvatica)
3.1 3.2 2.7 3.1 3.0 3.2 3.3 3.3 3.2 3.2
3.3 3.2 2.9 3.4 3.2 3.1 3.2 3.1 2.9 3.0
3.1 3.3 2.8 3.1 2.9 3.2 3.0 3.0 3.0 3.0
3.5 3.1 3.0 3.2 3.1 3.1 3.3 3.0 2.9 2.8
3.1 2.8 3.3 3.4 3.1 2.9 3.4 3.0 3.3 2.9
1. Calculate the mean petal length of this sample
2. Count up the number of petals of each length. Draw a
histogram to display these results.
3. What is the mode for these results?
4. What is the median petal length?
Standard Deviation
• This measures the spread of the data from
the mean.
There is more variation in leaf
length so leaves in this group
varies a lot from the mean.
There will be a large standard
deviation
There is less variation in leaf
length so overall each leaf is
closer to the mean
There will be a small standard
deviation
Displaying the data
• These leaf lengths can be drawn as histograms
and the spread of the data compared
Number
of leaves
Number
of leaves
Leaf length (mm)
Leaf length (mm)
Calculating the Standard
deviation
This means the
sum of
This the symbol
for mean
n is the
number of
values
Worked Example
Tree
Height
(m)
A
22
B
27
C
26
D
29
Adding and
subtracting the
standard deviation to
the mean will include
68% of the data.
Error Bars
• These can be drawn onto a graph.
• They’re drawn by adding 1SD to the value and subtracting 1SD from
the value.
• They can be added to line or bar graphs.
30
28
Mean
Height (m)
From our example on tree height
Mean = 26
SD = 2.9
26
24
22
20
18
26+ 2.9 = 28.9
26-2.9 = 23.1
Produce a list of human characteristics to fill the
following table.
For each state if it is environmental / inherited or both.
State whether it is continuous or discontinuous.
Inherited
Environmental
Both
Students T Test
This is a statistical test to determine whether 2
sets of data are statistically different.
A continuous characteristic such as length, height,
width, mass can be measured and compared
between 2 different groups
eg males and females
lower shore and upper shore
sun and shade leaves
Null Hypothesis
• This is a negative statement.
• Usually it will state that there is no link
between 2 factors.
• eg There is no similarity between 2 sets of
data
To be able to use the T Test, the data must be normally
distributed.
The mean height is
likely to be close to
the peak of the curve
frequency
Height of men (cm)
When comparing the distributions for male and
female height there is a difference in the position of
the mean but there is also a lot of overlap.
Are they statistically different?
Collate and display the
data as histograms to
determine whether each
set is normally
distributed
Numbers
We need to know whether the
data is significantly different
from each other by comparing
the means and the spread of
the data, so we will need to
calculate:
• The means of each set
• The standard deviations of
each set
Length (mm)
Numbers
Length (mm)
The Student T Formula
This compared
the means
This compares
the standard
deviations
squared (known
as the variances
Understanding What The value
of t means
The value of t which is
calculated must be compared
to the table of values.
Work out the number of
degrees of freedom
(n1+ n2 )- 2
If the value of t is equal to or more
than the critical value there is
significant difference between the
two sets of data
We can reject the Null Hypothesis
which states that there is no
difference between the 2 sets of
data
Probability 
Degrees of
freedom
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.05
0.025
0.01
6.3138
2.9200
2.3534
2.1319
2.0150
1.9432
1.8946
1.8595
1.8331
1.8124
1.7959
1.7823
1.7709
1.7613
1.7530
1.7459
1.7396
1.7341
1.7291
1.7247
12.7065
4.3026
3.1824
2.7764
2.5706
2.4469
2.3646
2.3060
2.2621
2.2282
2.2010
2.1788
2.1604
2.1448
2.1314
2.1199
2.1098
2.1009
2.0930
2.0860
31.8193
6.9646
4.5407
3.7470
3.3650
3.1426
2.9980
2.8965
2.8214
2.7638
2.7181
2.6810
2.6503
2.6245
2.6025
2.5835
2.5669
2.5524
2.5395
2.5280
T table