Download unit 5h.1 5b.4 genetics evolution genes alleles

Unit 5: Genetics,
evolution & biodiversity
5H.1 & 5B.4 Genetics & evolution;
genes & alleles
By Mr. Wilson
Gene expression
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The process by which a gene’s DNA sequence is
converted into the structures and functions of a
cell, thus producing a particular characteristic in
the phenotype.
It is highly regulated within cells (genes are
‘switched on and off ’ depending on various
factors).
There are complex environmental influences on
gene expression creating numerous phenotypes.
Phenotype variation
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Both genes and the environment can interact to
produce phenotype variation.
Examples – Genes for being tall, but poor
nutrition (environmental factor) means you don’t
get that tall.
Pea plants – Dwarf & tall varieties (genes), but a
range of heights in each group due to nutrition,
sunlight, competition etc. (environment).
Characteristics which show continuous variation
are affected more strongly by the environment.
Monohybrid Inheritance
The inheritance of just 1 characteristic.
 If a parent (P) which is homozygous
dominant for a characteristic/trait is crossed
with a parent which is homozygous recessive
for the same characteristic, all of the
offspring (F1) will be heterozygous for the
characteristic.
 They will all show the dominant characteristic
in the phenotype.
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Monohybrid Inheritance
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Crossing of these F1
offspring (they are all
heterozygous) produces
a 1:2:1 ratio of
homozygous dominant
to heterozygous to
homozygous recessive in
the genotype of the F2
generation.
There is a 3:1 ratio in the
phenotype.
Monohybrid Inheritance
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Knock up a quick genetic cross diagram to show
the phenotype ratio produced when crossing a
parent which is heterozygous for a characteristic
with a parent which is homozygous recessive for
the same characteristic. Use W and w.
What do these terms mean: LOCUS, GENE,
ALLELE, GENOTYPE, PHENOTYPE,
HOMOZYGOUS, HETEROZYGOUS,
HAPLOID, DIPLOID, DOMINANT &
RECESSIVE. Ask now if unsure!
Dihybrid Inheritance
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Dihybrid inheritance occurs because genes on
different chromosomes show
INDEPENDENT ASSORTMENT.
In other words: The emergence of one
characteristic or trait will not affect the
emergence of another different characteristic.
Crossing a parent which is double homozygous
dominant with one which is double homozygous
recessive results in F1 offspring which are all
double heterozygous.
Dihybrid Inheritance
The male and
female of the F1
offspring make 4
types of gamete RY, Ry, ry and rY.
 If these offspring
are crossed there is
a 9:3:3:1 ratio in
the offspring
phenotypes.
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Dihybrid Inheritance
Dihybrid Inheritance
Knock up a genetic cross diagram to show the
phenotype ratios when crossing a parent which
is double heterozygous for a characteristic
(NnGg) with a double homozygous recessive
parent for that characteristic (nngg).
 Could you explain, without the use of a genetic
cross diagram, why the phenotypes of all the
offspring of this cross: NnGg and NNGG,
would express both dominant characteristics?
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Dihybrid Inheritance
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Drosophila melanogaster are an insect species
commonly known as fruit flies.
Drosophila means ‘dew-loving’ from the Greek.
They are often used to study dihybrid inheritance
as they are easy to breed and their many
characteristics are easy to spot.
We used to kill them to note the phenotypes, but
now we just knock them out for a bit.
The most commonly observed phenotype is
known as the wild-type.
Review
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Mendel’s first law – ‘The law of segregation’
This is basically the concept of alleles.
Which of the four parts of this law is now
known to be incorrect today?
Mendel’s second law – ‘The law of independent
assortment’.
One genes inheritance is not affected by that of
another (unlinked; on different chromosomes).
Which was worked out by doing monohybrid
crosses and which by doing dihybrid crosses?
Test crosses
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If you want to know a genotype for a characteristic
when the dominant phenotype is in evidence
(remember it could be homozygous dominant or
heterozygous) you need to perform a test cross!
This is always done by crossing the unknown
phenotype with a homozygous recessive genotype
for the same characteristic and observing the
PHENOTYPE RATIOS obtained.
If you get a phenotype ratio of 1:1 in the offspring
the organism MUST have been heterozygous! If
the offspring all show the dominant phenotype over
and over again the organism is (or is extremely,
extremely likely to be) homozygous dominant!
Test crosses
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A pea is yellow, but we don’t know if it is
homozygous dominant (YY) or heterozygous
(Yy). Do a genetic cross diagram to show how
you would find it’s genotype.
A pea is round and yellow, but we don’t know if
it is double homozygous dominant (RRYY) or
double heterozygous (RrYy).
How would you find out?
What ratios would you look for in the offspring
phenotypes?
A little bit of statistics…..
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The Chi-squared (X2) test estimates the
probability that differences between observed
and expected data is due to chance.
In other words we can find out if what we
observe is too different from the expected to be
valid or if the difference is too slight to matter.
This is commonly applied to patterns of dihybrd
inheritance as we can cross & observe
phenotype ratios and compare them to what we
would expect based on Mendelian rules.
Chi-squared test
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If we expect a ratio of 1:1:1:1 from crossing a
double heterozygous genotype (RrYy) with a
double homozygous recessive genotype (rryy)
and we got the following numbers of offspring
we could use Chi-squared to test the results:
Round yellow = 57
Round green = 51
Wrinkled yellow = 49
Wrinkled green = 53
Chi-squared test
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First we calculate our expected values for each
genotype: (total offspring) = 210/4 = 52.5.
Then we work out the difference between each
observed value and the expected value and
square the result to eliminate negative values.
RYRY = 57 – 52.5 = 4.5 squared = 20.25
Rryy = 51 – 52.5 = -1.5 squared = 2.25
rrYy = 49 – 52.5 = -3.5 squared = 12.25
rryy = 53 – 52.5 = 0.5
squared = 0.25
Chi-squared test
Next we work out the value of Chi-squared
using this formula:
 Basically we divide
each of our 4 values
from the last section
by the expected & add
all the new values together
giving us one final value: Chi-squared!
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Chi-squared test
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This would be:
(RrYy); 20.25/52.5 = 0.38
(Rryy); 2.25/52.5 = 0.043
(rrYy); 12.25/52.5 = 0.233
(rryy); 0.25/52.5 = 0.00048
These 4 values added together = 0.66
This is our magic number – Chi-squared!!
Chi-squared test
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Now we work out something called our degrees
of freedom (easy – it’s the number that is one
less than the total number of classes of data).
We had four classes of data (4 genotypes in our
examples), so we have 3 degrees of freedom.
This is important for finding the correct row to
use in a probability table.
We find the closest match to our value in the
table along this row and we can see the
probability that our results are due to chance.
Chi-squared test
Degrees
of
freedom
Probability (p)
0.90 0.50
0.20
0.10
0.05
0.02
0.01
1
0.02 0.46
1.64
2.71
3.84
5.41
6.64
2
0.21 1.39
3.22
4.61
5.99
7.82
9.21
3
0.58 2.37
4.64
6.25
7.82
9.84
11.34
Chi-squared test
Can you see that our value for Chi-squared is
just under 0.90 probability so there’s about a
90% probability our differences are due to
chance.
 We can therefore think of our results as
reliable.
 In exams you would not be expected to carry
out a full test, but calculate some values and
perhaps work out degrees of freedom and a
probability from a given value.
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Chi-squared test
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A dihybrid cross between two double
heterozygous pea plants gave the following
phenotypes in the offspring:
Round yellow – 39 seeds
Round green – 10 seeds
Wrinkled yellow – 13 seeds
Wrinkled green – 4 seeds
Find the probability that the differences between
observed and expected values are due to chance.
Incomplete Dominance
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INCOMPLETE DOMINANCE occurs when a
heterozygous genotype creates an intermediate
phenotype. Examples are flower colour in
carnations & snap dragons and SICKLE CELL
ANAEMIA in humans.
Crossing a carnation which is homozygous
dominant for flower colour (red flower) with one
which is homozygous recessive for flower colour
(white flower) results in all offspring being
heterozygous with the intermediate phenotype
(pink flower).
Crossing these heterozygotes (with the intermediate
pink flower phenotype) will result in the reemergence of the parental phenotypes.
Incomplete Dominance; flower
colour in carnations
Incomplete Dominance; flower
colour in carnations
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You may sometimes see this written:
Parents – CRCR and CWCW
F1 offspring – all CRCW
F2 offspring – CRCR, CRCW, CWCR, CWCW.
C = the allele for flower colour
R denotes red and W denotes white.
CRCW offspring are intermediate types. They
have PINK flowers.
Co-dominance & multiple alleles;
ABO blood groups
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3 alleles exist for determination of blood group.
I is used to denote the allele in genetic cross
diagrams or there would be 3 different letters to
use.
IA is the allele for the synthesis of a protein marker
(antigen A) on red blood cells.
IB is the allele for the synthesis of a protein marker
(antigen B) on red blood cells.
IO is an allele that does not code for an antigen on
red blood cells. (Recessive to A and B). A and B are
both dominant (co-dominant).
ABO blood groups
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Remember – there may be 3 alleles, but only 1
will be present in any gamete and only 2 will be
present in a zygote or somatic cell (only 2 alleles
maximum, at set loci, per homologous or paired
chromosome).
Write down the 6 possible blood group
genotypes.
How many different phenotypes could there be?
Complete the genetic cross diagrams sheet.
ABO
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If all four blood groups are present in the F1
offspring of a cross what are the genotypes of
the parents?
Several offspring show a 3:1 phenotype ratio of
group A to group O. What are the genotypes of
the parents?
Don’t be confused – we use a given letter to
denote the alleles with co-dominance as we
shouldn’t use completely different letters to
denote two alleles of the same gene!
Sickle cell anaemia
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The sickle cell allele is created by a point
mutation and is thought of as recessive.
There are 3 phenotypes: Normal, Sickle cell trait
and Sickle cell anaemia.
This is an example of INCOMLPETE
DOMINANCE because Sickle cell trait
haemoglobin is an intermediate phenotype (only
slightly less ‘normal’ than normal Hb)
H is often used to denote the allele with N added
for normal Hb and S added for abnormal Hb,
although single letters an also be used.
Sickle cell anaemia
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Therefore:
HNHN is a normal individual.
HNHS is someone with sickle cell trait.
HSHS is someone with sickle cell anaemia.
Why are the parents of someone with sickle cell
anaemia generally heterozygous (with sickle cell trait)?
Sickle cell anaemia offers protection against the malaria
parasite; Plasmodium as it alters the shape of
erythrocytes hindering the protozoan’s entry. It is
SELECTED FOR within populations in, or who have
ancestry in, hot countries where malaria is prevalent.
Sickle cell anaemia
Inheriting Sickle Cell Anaemia
Autosomal linkage
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When two genes are linked on the same autosomal
(not X or Y) chromosome.
In this case genes are often inherited together as
opposed to separately.
Linked genes tend to assort together in meiosis.
The Mendelian rules of independent assortment
therefore do not apply and ratios in dihybrid crosses
may be altered.
E.g. Fruit flies – gene for eye colour and wing
length are on the same chromosome and often
inherited together.
Autosomal linkage
In this example
genes A, B, C, D
and E are linked.
 They are most
likely inherited
together as they
appear on the same
chromosome.
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Autosomal linkage
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For example – if gene A and B were linked and
a parent had the genotype AaBb they would
only be able to make 2 types of gamete AB and
ab in meiosis, instead of the usual 4 (AB, Ab, aB,
ab).
This is because the two genes are on the same
chromosome and can’t assort independently of
each other to make the 4 different types.
Work this out with some quick chromosome
diagrams and check the change in ratios from a
cross between 2 double heterozygotes for genes
A & B.
Autosomal linkage
An expected 9:3:3:1 phenotype ratio ends up
being a 3:1 ratio.
 Crossing a double heterozygous individual
with a double homozygous recessive
individual should give a 1:1:1:1 phenotype
ratio. What would happen if the genes were
linked? Do a quick test cross.
 Generally autosomal linkage sways offspring
towards the parental genotypes in form.
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Autosomal linkage
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Oh…..the exception to the rule….
Because of CROSSING OVER in meiosis some
genes that would ordinarily be linked may not
be. You could get 4 types of gamete after all!
AB & ab because they are linked. Ab & aB only
from crossing over.
The gametes you don’t expect, however, occur
much less frequently than they would if the
genes were unlinked. Ratios can be ‘odd’.
Autosomal linkage
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The further apart linked genes are on a
chromosome the more likely they are to become
unlinked by crossing over.
Genes A & B are linked, but are fairly far apart
on a chromosome. A cross between AaBb and
AaBb is performed. Show with diagrams what
phenotype ratios might be expected.
It is believed that the genes that contribute to
red hair colour in humans are strongly linked to
the genes that contribute to fair skin in humans.
Gene interaction
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Interactions between different genes may
sometimes provide completely new phenotypes.
The classic example is comb shape in poultry.
There are many phenotypes, but we will focus
on 4. Rose, pea, walnut & single.
Rose is caused by gene R being present, but
gene P not being present (absent).
Pea is caused by gene P being present, but gene
R not being present (absent).
Gene interaction
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If both R and P are present a new phenotype
called ‘walnut’ appears.
If neither R or P are present a new phenotype
called ‘single’ (comb) appears.
Draw a genetic cross diagram to show the
offspring from crossing a true breeding rose
combed individual with a true breeding pea
combed individual. How would you represent
the alleles?
Now cross 2 of these F1 offspring. What
happens in the F2 generation with regard to
comb shape?
Gene interaction
 Clockwise:
 Rose,
pea, walnut
& single.
 Extend
yourself!
 What’s Epistasis?
Sex determination in humans
X
(egg)
X
(egg)
X
(sperm)
XX
Female
XX
Female
Y
(sperm)
XY
Male
XY
Male
Sex determination in humans
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These are the sex chromosomes (pair 23)
The Y chromosome is smaller & has less genes
than the X chromosome (the Y has more-or-less
only genes to make you MALE).
The X chromosome, however, has the genes
that would make you FEMALE (if you inherit
both) and many other functional genes, which
are said to be X-linked or sex-linked.
The male/female balance is maintained by this
pattern of inheritance.
X (or sex) -linked inheritance
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Both males & females can inherit genes carried
on the X chromosome, but females may inherit
2 alleles whereas males can only inherit 1.
The male will always be homozygous, but the
female can be homozygous (recessive or
dominant), or heterozygous.
This gives females an advantage when they
inherit a faulty (recessive) allele for a condition
because the condition will not be expressed in
the phenotype of heterozygotes.
X-linked inheritance
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If a male inherits the faulty allele he WILL have
the condition.
Males are far more likely than females to show
recessive phenotypes for X-linked conditions.
Classic examples in humans are:
HAEMOPHILIA and RED/GREEN
COLOUR BLINDNESS.
To be affected females need 2 copies of the
recessive allele that causes these conditions, but
males only need 1.
X-linked inheritance
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What parental
genotypes must be in
evidence to produce
an affected daughter
in this way?
Why is this situation
unlikely or suspected
before the birth of
the daughter?
X-linked inheritance; haemophilia
X-linked inheritance;
haemophilia pedigree chart
X-linked inheritance; red/green
colour blindness
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Clockwise:
Information, genotypes
of Elizabeth & John, 4
possible gametes
produced by Elizabeth
& John, F1 offspring.
Note – the C allele is
for normal colour
vision & c is colour
blindness; both located
on the X chromosome.