Download Mendelian or qualitative genetics

Mendelian or qualitative genetics
 Traits that are controlled by one to a few
genes and show distinct phenotypic classes.
Mendel developed two laws that dealt with
how units of heredity (genes) were inherited.
The two laws were:
The Law of Segregation
The Law of Independent Assortment
Law of Segregation - A pair of alleles for a
given gene (trait) separate or segregate in the
gametes equally.
In meiosis this relates to the separation of the
homologous chromosome pairs in anaphase I.
1
Law of independent assortment - Allelic pairs
of genes for two traits will behave
independently of each other (unless they are
close to each other on the same chromosome).
In meiosis this relates to the sorting of nonhomologous chromosomes in the reductional
division.
2
Terminology
gene - region of DNA that codes for either a
protein, tRNA or rRNA.
allele - one of a series of possible alternative
forms of a given gene. The difference
in the genes relates to differences in
the DNA sequence that affect the
functioning of the gene product.
genotype - the genetic make-up of an
organism.
phenotype - physical appearance of an
organism.
homozygous - having identical alleles at a
given gene locus on
homologous chromosomes.
heterozygous - having dissimilar alleles at a
given gene locus on
homologous chromosomes.
3
dominant - alleles that are phenotypically
expressed in the heterozygous
state.
recessive - alleles that require the
homozygous state to be
phenotypically expressed.
Abbreviations for alleles
dominant allele  capital letter or ‘+’
recessive allele  lower case letter
example : RR - homozygous dominant
Rr - heterozygous
rr - homozygous recessive
4
Mendel’s laws
Law of segregation - A pair of alleles for a
given gene (trait) separate or segregate in the
gametes equally.
example: seed shape in peas
Two phenotypes - smooth and wrinkled
smooth
X
wrinkled

F1 smooth
(filial)
X (means self-pollinated)

F2 - 3 smooth
1 wrinkled
5
gametes
phenotypic ratio
3 - S_  since S is dominant it does not
matter what allele is here.
1 - ss
genotypic ratio
3 genotypes possible
1 - SS smooth homozygous
2 - Ss smooth heterozygous
1 - ss wrinkled homozygous
How to test for the genotype of an individual.
6
1. testcross - cross F2 individuals to a
homozygous recessive individual. Because
one parent is homozygous recessive any
variation in the phenotype of the progeny
will be due to the other, unknown genotype,
parent.
SS x ss

100 % - Ss
Ss x ss

50% - Ss
50% - ss
ss x ss

100% - ss
So if you observe segregation in a testcross
then the unknown genotype is heterozygous.
7
2. If possible, self the individual.
SS
Ss
ss
X

all SS
smooth
X

3 smooth
1 wrinkled
X

all ss
wrinkled
So if the progeny segregate then the unknown
genotype was heterozygous.
8
Pedigree analysis
If a species has few progeny per year, few
progeny in a lifetime, and/or long durations
between generations, it can be difficult to get
enough progeny to do genetic analysis of a
trait.
A way around this problem is to do pedigree
analysis of the family, looking back several
generations.
Symbols
male 
female 
mating 
example:
I
parents
II
children
9
Law of independent assortment - Allelic pairs
of genes for two traits will behave
independently of each other (unless they are
close to each other on the same chromosome).
Phenotypic ratios
9 - smooth, yellow
3 - wrinkled, yellow
3 - smooth, green
1 - wrinkled, green
Genotypic ratios
smooth, yellow: 1 SSYY
2 SSYy
2 SsYY
4 SsYy
wrinkled, yellow 1 ssYY
2 ssYy
smooth, green: 1 SSyy
2 Ssyy
wrinkled, green: 1 ssyy
10
Another method to determine phenotypic and
genotypic ratios is to use the split fork
method.
If you self a F1 with smooth-yellow (SsYy)
seed with red flowers (Rr) the phenotypic
ratios in the F2 can be calculated in the
following way.
3 red
27 S_Y_R_
3 yellow
1 white
9 S_Y_rr
3 red
9 S_yyR_
1 white
3 S_yyrr
3 red
9 ssY_R_
1 white
3 ssY_rr
3 red
3 ssyyR_
1 white
1 ssyyrr
3 smooth
1 green
3 yellow
1 wrinkled
1 green
11
Sometimes in an experiment you do not get a
perfect segregation ratio.
If you have a ratio of 2.23 - 1, is it close
enough to a 3 - 1 ratio to say you are
observing a single gene trait under control of
a completely dominant allele?
To aid in you decision you can use probability
and statistical analysis to determine if the
observed ratio is close enough to the expected
ratio.
12
Probability and Statistical Testing
Probability - Two basic rules, the product
rule and the sum rule
product rule - multiply the probabilities
together of two independent events to
determine the probability of two independent
events will occur together.
sum rule - add the probability of two events
together when the events are mutually
exclusive events.
Genetic ratios can be used as examples of
both the product rule and the sum rule.
13
Punnett square analysis
A (p= .5)
a (p= .5)
A
(p= .5)
AA
(p= .25)
Aa
(p= .25)
a
(p= .5)
Aa
(p= .25)
aa
(p= .25)
 To get the probability of gametes coming
together you use the product rule. For
example the probability of an AA genotype
is .5 x .5 = .25.
 Each zygote is a mutually exclusive event so
the probability of a heterozygote (Aa)
occurring uses the sum rule, .25 + .25 = .5.
14
From these probabilities genotypic ratios can
be determined.
AA = .25 = 1
Aa = .50 = 2
aa = .25 = 1
Probabilities can also be used to determine
phenotypic ratios
Probability of at least one dominant allele
being present:
AA (.25) + Aa (.25) + Aa (.25) = .75
Probability of homozygous recessive (aa) is
.25.
A_ = .75 = 3
aa = .25 = 1
15
So probabilities will give the same genotypic
and phenotypic ratios as found by Mendel.
When dealing with 3, 4, or 5 gene models you
can use probability to calculate the
probability of a specific genotype or
phenotype occurring. You can also calculate
the phenotypic or genotypic ratios. For
example:
F2 phenotypes from the split fork example
smooth-yellow-red = .75 x .75 x .75 = .43
smooth-yellow-white = .75 x .75 x .25 = .14
smooth-green-red = .75 x .25 x .75 = .14
smooth-green-white = .75 x .25 x .25 = .05
wrinkled-yellow-red = .25 x .75 x .75 = .14
wrinkled-yellow-white = .25 x .75 x .25 = .05
wrinkled-green-red = .25 x .25 x .75 = .05
wrinkled-green-white = .25 x .25 x .25 = .02
To get the ratios just multiply the probability
by the total number of expected progeny.
In this case with three genes, 4 x 4 x 4 = 64
16
smooth-yellow-red = 64 x .43  27
For genotypes:
Example: What is the probability of
recovering the following genotypes from this
cross:
Aa Bb Cc Dd Ee x Aa Bb Cc Dd Ee
a) AA BB CC DD EE
b) Aa Bb Cc Dd Ee
c) Aa BB CC DD Ee
Since each gene behaves independently, you
can calculate the probability of a given
genotype by multiplying the probabilities for
each gene.
a) .25 x .25 x .25 x .25 x .25 = .001
b) .5 x .5 x .5 x .5 x .5 = .0313
c) .5 x .25 x .25 x .25 x .5 = .0039
17
To determine the ratio again just multiply by
the total number of individuals expected.
5 genes = 45 = 1,024
a) .001 x 1,024 = 1
b) .0313 x 1,024 = 32
c) .0039 x 1,024 = 4
The numbers you obtain may differ slightly
depending on the level of rounding you are
doing.
18
Statistical Testing
Chi square analysis
2 =  (observed - expected)2/ expected
This analysis can be used to determine if the
phenotypic of genotypic ratio observed is
close enough to the expected ratio to be
accepted.
Example: ratio 2.23 : 1, is this a 3 : 1 ratio
one gene model
need to look at the actual numbers of each
genotype observed.
Out of 149 plants
103 showed the dominant phenotype
46 showed the recessive phenotype
These are your observed values
19
If the progeny segregated in the expected 3:1
ratio the expected number of dominant and
recessive phenotypes would be:
149/4 x 3 = 111.75 expected progeny with the
dominant phenotype
149/4 x 1 = 37.25 expected progeny with the
recessive phenotype
These are your expected values
Now you plug the values into the formula
2 = (103 - 112)2/112 + (46 - 37)2/ 37
= (-9)2/112 + (9)2/37
= 81/112 + 81/37
= .723 + 2.189
= 2.912
To determine if you should accept the 2.23 : 1
20
ratio as an acceptable deviation from 3 : 1
one gene model, check your Chi square value
in the Chi square table.
degrees of
freedom
probabilities
0.10
0.05
1
2.71
3.84
2
4.61
5.99
3
6.25
7.82
The degrees of freedom can be determined by
taking the number of phenotypic classes (n)
and subtracting 1.
df = n - 1
In our example we have 2 phenotypic classes
so the degrees of freedom is (2 - 1) = 1
21
If our Chi square value is less than the value
at 0.05 for the proper degrees of freedom we
can accept 2.23 : 1 as an acceptable deviation
from 3 : 1.
So at 1 degree of freedom the 0.05 value is
3.84 and our value is 2.912 so we can accept
the hypothesis that this is a 3 : 1 ratio.
If our value was greater than 3.84 we would
reject the hypothesis that it is a 3 : 1 one gene
model
22
Probability can also be used to determine the
chances of a combination of events occurring.
example: In a testcross (Aa x aa) you get 2
progeny.
What are the chances you will get:
a) 2 heterozygotes
b) 2 homozygous recessive
c) one heterozygote and one homozygote
The genotype of one off-spring is independent
of the genotype of the other off-spring.
a) 2 heterozygotes = .5 x .5 = .25
b) 2 homozygous recessives = .5 x .5 = .25
23
c) 1 heterozygote and 1 homozygote
2 ways this could occur:
1st progeny Aa second progeny aa
1st progeny aa second progeny Aa
example of mutually exclusive events so you
take the probability of the first combination
plus the probability of the second
combination.
Aa (.5) x aa (.5) + aa (.5) x Aa (.5) = .5
The distribution pattern is 1 : 2 : 1 which are
the coefficients for a binomial equation.
1p2 + 2pq + 1q2 = (p + q)2 = 1
24
So in our example:
if Aa = p = .5 and aa = q = .5
1p2 + 2pq + 1q2 = (p + q)2 = 1
1 (Aa)(Aa) + 2 (Aa)(aa) + 1 (aa)(aa) = 1
.25
+
.5
+
.25
=1
The advantage of a binomial equation is that
it can be expanded to determine the
probabilities for more than 2 off-spring.
(p + q)N
Where N is the total number of progeny
example: if N = 3
(p + q)3 = 1p3 + 3p2q + 3pq2 + q3
25
So if you were interested in the probability of
having a combination of 2 boys and 1 girl:
boy = p = .5
girl = q = .5
2 boys and one girl would be p2q so from the
formula the probability of this combination
occurring is 3p2q = 3 x .5 x .5 x .5 = .3750
Why multiply by three?
Three possible combinations:
girl-boy-boy, boy-girl-boy, and boy-boy-girl
The more progeny involved the more complex
the binomial expansion.
26
There is a way to determine the probability
for one combination occurring using the
following formula:
(n!/w!x!) pwqx
p = probability of an event
q = probability of a different event
n = total number of progeny
w = number of progeny with p event
x = number of progeny with q event
The factorials (n!/w!x!) will give the
coefficient from the binomial expansion for
the combination of events in question.
27
Variation in Mendelian ratios
1. incomplete dominance
2. over-dominance
3. co-dominance
4. multiple alleles
5. environment
6. epistasis (gene interactions)
7. lethal genes
8. gene linkage
28
1. Incomplete dominance - when the
heterozygote expresses a unique phenotype
that is intermediate between the two
homozygous phenotypes.
Example: herbicide resistant lettuce
While the plants segregated 3 : 1 for
resistance or susceptibility to a herbicide, it
was observed that the resistant group could
be broken into two levels of resistance,
intermediate and complete.
Resistant: 61 plants - 20 complete resistance
41 intermediate resist.
Susceptible: 22 plants
So in reality we do not have complete
dominance but instead incomplete
dominance.
29
In the case of incomplete dominance, the
phenotypic ratio is the same as the genotypic
ratio.
ratio genotype
complete resist. - 20
1
RR
intermed. resist. - 41
2
Rr
susceptible
1
rr
- 22
Biochemical explanation
The herbicide binds to an essential enzyme in
the plant preventing the enzyme from acting.
The resistance gene codes for the same
enzyme but because of a difference in the
amino acid sequence of the enzyme the
herbicide can no longer bind and disable the
enzyme.
30
Complete resistance - RR - 100% of the
enzyme is resistant
Intermed. resist.
- Rr - 50% of the enzyme
is resistant
Susceptible
- rr - none of the enzyme
is the resistant form
2. Over-dominance - when the heterozygote
phenotype exceeds one or both of the
homozygous phenotypes.
Example: head shape in club wheats
The F1 club head is longer than the club
parent and wider than the common wheat
parent. Head shape is controlled by a single
gene.
31
3. Co-dominance - when the phenotypes of
both alleles are expressed in the
heterozygote.
examples:
 sickle cell anemia
HbAHbA x HbSHbS

HbAHbS
Both types of hemoglobin are expressed in the
heterozygote.
32
4. Multiple alleles - when there are more than
two alleles for a given gene. Can result in
combinations of complete dominance and
co-dominance expression.
examples:
 eye color in Drosophila
 self-incompatibility in plants
 blood type in humans
 coat color in animals
33
Self-incompatibility in plants
example: Brassica - four allele system
S1, S2, S3, and S4
If the same allele is present in the male and
female gamete the pollen tube will stop
growing before it reaches the ovary.
Cross S1S2 x S1S2

sterile
Cross S1S2 x S2S3

S2 S3
S1 S3
34
Cross S1S2 x S3S4

S1 S3
S1 S4
S2 S3
S2 S4
Blood type in humans
Three alleles, A, B, and O
 A and B are co-dominant
 Both A and B are dominant to O
blood type
A
B
AB
O
genotype
IAIA or IAIO
IBIB or IBIO
IAIB
IOIO
AB - universal recipient
O - universal donor
35
Coat color in rabbits
c+ - wild type
ch - himalayan
cch - chinchilla
c - albino
There is a gradation in dominance for coat
color
c+  cch  ch  c
36
With multiple alleles the number of possible
genotypes increases greatly.
You can calculate the number of different
genotypes with the following formula:
n(n+1)/2
where n is the number of alleles
examples:
Rabbit coat color: n = 4
number of genotypes = 4(4+1)/2
= 20/2
= 10
blood type: n = 3
number of genotypes = 3(3+1)/2
= 12/2
=6
37
With the presence of multiple alleles it can be
difficult at times to determine if the observed
variation for a trait is due to two genes or
allelic variation at one gene locus.
The way to determine if the variation you are
observing is allelic is to do a complementation
test.
example:
You have two individuals that are both white
variants from the normal red flower color.
If you cross them and the progeny are red
then the trait is controlled by more than one
gene and the two white variants have a
mutation in different genes.
If you cross them and the progeny are all
white then the two variants have a mutation
in the same gene and the trait may be
controlled by only one gene.
38
Biochemically it would work like this:
substrate  intermediate  product
(white) A (white)
B (red)
‘A’ is an enzyme that converts the substrate
to an intermediate and is controlled by gene
A.
‘B’ is an enzyme that converts the
intermediate to the final product and is
controlled by gene B.
1) 2 gene model
plant 1 white aaBB x plant 2 white AAbb

F1 AaBb all red
The flowers are red because the F1
individuals have one functional gene/allele at
39
each gene loci. Hence the genes compliment
each other.
2) 1 gene model
plant 1 white a1a1 x plant 2 white a2a2

F1 a1a2 all white
The flowers in the F1 individuals are white
because they do not have a dominant allele at
the A locus. The shades of white may vary
depending on the mutation in each parent.
So it is possible to have multiple alleles (a1, a2,
a3, etc.) at a single gene locus that give
various shades of white depending on the
location of the mutation in the gene.
40
5. Environment - the environment can
influence the expression and level of
expression of a gene. Factors such as
temperature, light, and nutrition can
reduce or prevent expression of a gene or
genes.
Example: temperature response in fur color.
Rabbits - Himalayan
As temperature decreases the fur on the
extremities darkens.


How to measure environmental effects:
41
 plants - grow large populations of a single
genotype in several environments then
compare the differences in expression of a
trait. Any differences observed have to be
due to the different environments.
 animals - multiple matings to produce
individuals with similar genotypes and
study the F1 progeny in several
environments.
 humans - work with twins
1) Separate identical twins at birth and
place them in separate environments.
2) Study difference between identical and
fraternal twins for expression of various
traits.
42
 If the similarity (concordance) within the
sets of twins for a trait is the same between
identical and fraternal twins than the
expression of that trait is under more
environmental than genetic control.
 if the level of concordance differs
significantly between identical and
fraternal twins with a higher level of
concordance in the sets of identical twins
then the trait is under more genetic control
than environmental control.
43
Level of concordance among identical and
fraternal twins.
trait
percent concordance
identical
fraternal
hair color
89
22
diabetes mellitus
65
18
measles
95
87
schizophrenia
80
13
manic-depressive
77
19
alcohol drinking
100
86
coffee drinking
94
79
44
You can also see the effect of environment by
looking at just the concordance of identical
twins.
example: diabetes mellitus
A concordance of 65% in identical twins
means that out of 100 sets of identical twins
65 sets were the same for the trait while 35
sets had only one of the two showing the
genetic disorder.
The difference observed is due to the
environment. Differences in the environment
(diet?) must be causing the differences
because with identical twins there are no
genetic differences.
45
Two terms are used to describe the
phenotypic expression of a genotype.
 penetrance - Percent of individuals with a
given genotype that express the phenotype
associated with that genotype in a
population.
 expressivity - The degree or extent that a
given genotype is expressed phenotypically
in an individual.
46
6. Epistasis - when the action of one gene pair
interferes with or masks the expression of
all allelic alternatives of another gene. This
genetic interaction results in modified two
gene ratios:
9:3:4
9:6:1
9:7
12 : 3 : 1
13 : 3
15 : 1
Note that all the ratios still total 16. So
independent assortment (9 : 3 : 3 : 1) still is
occurring.
47
example:
blood type in humans
AB
x O

O phenotype
How is this possible?
 adoption
 babies switched at birth
 parents visited Chernobyl
 one gene masking the expression of another
gene.
How would this work?
48
1
2
precursor  intermediary  product
steps 1 and 2 are protein (enzyme) mediated
reactions.
If enzyme 1 does not function then it does not
matter if there is a functional enzyme 2.
blood type:
A antigen
A allele
precursor  H substance
B allele
B antigen
49
So if the true genotypes of the parents were:
ABHh
x
OOHh

Both AOhh or BOhh would give the
O phenotype
So the H gene interferes with the expression
of the blood antigen gene when H is in the
homozygous recessive (hh) condition. This is
called recessive epistasis.
Recessive epistasis - when the homozygous
condition is required to mask the expression
of a second gene. Because of the requirement
of the homozygous condition the largest class
in a F2 segregation is not affected.
50
1) single recessive epistasis - 9 : 3 : 4 ratio
Where the homozygous recessive condition
at one gene masks the expression of
another gene.
example: coat color in mammals
color expression
B - black
b - brown
hair color
C - color
c - no color
Mating:
BBcc x bbCC
white
brown

F1 BbCc x BbCc
black
black

9 - B_C_ - black  9
3 - bbC_ - brown  3
3 - B_cc - white  4
51
1 - bbcc - white 
2) double recessive epistasis - 9 : 6 : 1
Where the homozygous condition at either
gene masks expression of the second gene
and the double homozygous recessive has
its own unique phenotype.
example: fruit shape in squash
disk shape
AABB
x
long shape
aabb

F1 disk shape
AaBb
X

9 - A_B_
3 - A_bb
3 - aaB_
1 - aabb
- disk
- sphere
- sphere
- long
 9
 6

1
52
3) double recessive epistasis - 9 : 7
Where the homozygous recessive condition
at either gene masks expression of the
other gene and the double homozygous
recessive does not have a unique
phenotype.
example: pea flower color
white x white
AAbb
aaBB

AaBb
F1 all purple
X

9 A_B_ - purple  9
3 A_bb - white
3 aaB_ - white
7
1 aabb - white
53
Biochemical explanation:
A
B
precursor  intermediary  product
(white)
(white)
(purple)
If either gene product is non-functional the
flower color will be white.
54
4) single dominant epistasis - 12 : 3 : 1 or 13 : 3
Where the presence of one dominant allele for
one gene will mask the expression of a second
gene.
example: summer squash color
AAbb
white
x
aaBB
yellow

F1 AaBb - all white
X

9 - A_B_ - white  12
3 - A_bb - white
3 - aaB_ - yellow  3
1 - aabb - green  1
So the presence of the dominant ‘A’ allele
masks the expression of both the dominant
and recessive B alleles.
55
To get the 13 : 3 ratio the double homozygous
phenotype can not have its own unique
phenotype.
5) double dominant epistasis - 15 : 1
Where the dominant allele of both genes will
mask the expression of the homozygous
recessive of the other gene.
example: fruit shape in shepherd’s purse
round (AABB) x narrow (aabb)

F1 round AaBb
X

9 - A_B_ - round
3 - A_bb - round  15
3 - aaB_ - round
1 - aabb - narrow  1
56
Summary:
Epistasis - the masking of the expression of
one gene by another gene. Can be either
recessive or dominant.
 recessive epistasis is when the homozygous
recessive condition at one gene will mask
the expression at another gene locus. Can
be either single or double recessive
epistasis.
 dominant epistasis is when either the
homozygous dominant or heterozygous
condition at one gene will mask the
expression at another gene locus. Can be
either single or dominant epistasis.
 Gene interaction can also involve other
types of gene action such as incomplete
dominance or lethal genes.
57

7. Lethal genes - when certain genotypes do
not survive so the phenotypic and
genotypic ratios are altered.
example: coat color in mice
Two phenotypes - dark and yellow
mate: dark x yellow

1 : 1 dark to yellow
mate: yellow x yellow

2 : 1 yellow to dark
58
Why a 2 : 1 ratio instead of a 3: 1 ratio?
If: AY is the allele for yellow
A is the allele for dark
AYA x AYA

1: AYAY
homozygous
lethal
2:AYA
yellow
1:AA
dark
So really have a 3 : 1 but the homozygous
dominant genotype does not survive.
How can you confirm this?
If all yellows tested are heterozygous that
means the homozygous dominant genotype is
missing.
This is an example of a recessive lethal allele.
59
Recessive lethal - requires the homozygous
condition to express lethality. In the
heterozygous condition may make
modifications in the organism’s phenotype,
example - Manx cats (little or no tail)
Dominant lethal - requires only the
heterozygous condition to express lethality.
How could a dominant lethal allele stay in a
population?
example: Huntington’s disease
60
8. Gene linkage - when Mendel’s second law
does not apply because two gene are
located close together on the same
chromosome. Genes can be linked on the
sex chromosomes and on the autosomes.
Sex linkage - when genes for traits are
carried on the sex chromosomes you can get
differential expression of the trait depending
on the sex of the individual.
In sex linkage you can get expression of a
trait when only one copy of the gene is
present. This is called a hemizygous gene or
hemizygous expression.
example: a gene on the X chromosome in
humans. In the male (XY) the allele on the X
is the only one present so it is expressed.
61
Expression of a sex linked trait can be as a
recessive or dominant allele.
Recessive expression
1) female expresses the trait
x

males
females
expresses trait
does not express trait
2) male expresses the trait
x

male
does not express trait
female
does not express trait
62
Key to identifying a sex linked trait is that
expression of the trait in the progeny differs
between reciprocal crosses.
3) female is a carrier for the trait
x

male
50% express
50% do not express
female
50% are carriers and do
not express
50% do not express
With a female carrier you get expression in
50% of the male progeny but in none of the
female progeny. So you have a difference in
expression based on the sex of the progeny.
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Dominant expression
1) female heterozygous for the trait
x

male
50% express
50 % do not express
female
50% express
50% do not express
2) female homozygous for the trait
x

male
express
female
express
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3) male carries the trait
x

male
does not express
female
express the trait
So you can again compare difference in
expression in the progeny of reciprocal
crosses to determine if a trait is sex linked.
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Other forms of sex linkage:
1) Gene on the Y chromosome
 few genes are on the Y chromosome but if it
occurs it will never be observed in females
and will be observed in all male progeny.
2) Gene associated with mitochondrial DNA
- if a gene is on the mitochondrial DNA the
trait will be only passed through the maternal
side. So if the mother expresses the trait all
her children will express it, while if the father
expresses the trait none of the children will
express the trait.
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Autosomal gene linkage
When you do not get independent assortment
of two genes because they are on the same
chromosome.
example:
AAbb x aaBB

AaBb
X

expect: 9 - A_B_
3 - A_bb
3 - aaB_
1 - aabb
observe: 1 - AAbb
2 - AaBb
1 - aaBB
Why? If genes are very close to each other on
a chromosome they will behave like a single
gene giving a 3 : 1 or a 1 : 2 : 1 ratio instead
of a 9: 3 : 3 : 1 ratio
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The difference between the 3 : 1 ratio and 1 :
2: 1 ratio is the relationship of the dominant
alleles for the two genes. If the dominant
alleles are linked on the same chromosome
then they are said to be coupled and the ratio
will be 3 : 1. If the dominant allele for one
gene is linked to the recessive of the second
gene then they are said to be in repulsion and
a 1 : 2 : 1 ratio will occur instead of a 9 : 3 : 3
: 1 ratio.
Is it possible to break the linkage between two
genes on a chromosome?
Yes, by crossing-over between non-sister
chromatids.
How often does recombination occur between
two linked genes?
Depends on the distance between the two
genes.
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In a plant where the alleles are coupled, if
crossing-over is occurring between two linked
genes you would expect to see most of the
progeny showing the parental phenotypes,
A_B_ or aabb. The recombinants would be
few but have the phenotypes of A_bb and
aaB_.
The higher the number of recombinants the
further apart the genes are on the
chromosome.
Since the frequency of recombination is a
function of the distance between genes, the
recombination frequencies can be used to
map genes on a chromosome.
Gene mapping - 3 point testcross
A way to determine the gene order and the
distance between three genes that are on the
same chromosome.
example: HhBbDd x hhbbdd
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a completely homozygous recessive line is
used in the cross so the recombinants
observed come from only one parent.
progeny
phenotype
number
HBD
hbd
495
505
hBD
Hbd
51
56
HbD
hBd
36
39
HBd
hbD
2
1
1185
Note: Because one parent is homozygous
recessive all the alleles donated by that parent
to the progeny would be ‘hbd’ so they can be
dropped from the progeny phenotypes.
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The first thing to do is identify the parental
phenotypes and the recombinant phenotypes.
parental phenotypes
The phenotypes with the highest number of
progeny are the parental phenotypes.
HBD
hbd
495
505
recombinant phenotypes
To determine the distance between genes it is
important to identify the phenotypes that
have had recombination between only two of
the three genes.
The way to identify which genes have been
involved in a recombination event is to
compare the relationship of the genes in the
recombinant to the relationship of the same
genes in the parent.
71
If they differ then a cross-over has occurred
between those two genes.
example:
parental phenotypes - HB
hb
So the dominant alleles are together and the
recessive alleles are together in the parents.
Recombinants for these two genes would then
have the following phenotypes:
Hb
hB
H-B recombinants
hBD
Hbd
HbD
hBd
51
56
36
39
182
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You can do the same for the other two gene
combinations, HD and BD.
H-D recombinants
hBD
Hbd
HBd
hbD
51
56
2
1
110
B-D recombinants
HbD
hBd
HBd
hbD
36
39
2
1
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To map gene order and distance the order of
the genes needs to be determined first.
73
There are two ways to determine the gene
order.
1) Look at the two gene recombinants and the
combination with the highest number of
recombinants is the pair of genes that are
furthest apart.
HB = 182 recombinants
HD = 110 recombinants
BD = 78 recombinants
So the genes that are furthest apart are H and
B and the gene order is H - D - B.
2) Identify the gene involved in a doublecross-over because it will be the gene in the
middle. Since a double cross-over require
two recombination events to occur together
the probability of it occurring is low so the
3 gene recombinant group with the lowest
number of progeny is the result of a double
cross-over.
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Compare the double cross-over recombinant
phenotypes to the parental
phenotypes. The gene that has had an allelic
change between the recombinants and the
parentals is the gene in the middle.
parentals
low recombinants
HBD
hbd
HBd - 2
hbD - 1
Note that the relationship of H to B remained
the same but the relationship of H to D and B
to D changed. So D is the gene in the middle.
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Calculating distances between genes
If the middle gene is identified then you
calculate the distance from the middle gene to
the two outer genes.
H  D  B
formula:
(no. of single cross-overs + no. of double cross-overs) x 100
total no. of progeny
HD
hBD
Hbd
51
56
(107+3)/1185 x 100 = 9.3 mu
HBd
hbD
2
1
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BD
HbD
hBd
36
39
(75+3)/1185 x 100 = 6.5 mu
HBd
hbD
2
1
The linkage map would look like this:
H  9.3 mu  D  6.5mu  B
Why count double cross-overs in both
calculations?
In a double cross-over there is a single crossover between H and D and a single cross-over
between D and B.
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Sometimes you do not get as many double
cross-overs as you would expect off the
probabilities/frequencies of single cross-overs.
The reduction in double-cross-over events is
known as interference.
Interference occurs when the first cross-over
prevents the second cross-over from
occurring.
An interference value can be calculated using
the observed double cross-overs (ob. dco) and
the expected double cross-overs exp. dco).
Expected double cross-overs is calculated by
multiplying the probability/frequency of the
first cross-over by the probability/frequency
of the second cross-over.
exp. dco = (freq. 1st sco x freq. 2nd sco) x total progeny
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so in our example:
exp. dco = (.093 x .065) x 1185 = 7.16
expected double cross-overs  7
ob. dco/exp. dco = coefficient of coincidence
interference = 1 - coefficient of coincidence
If ob. dco = 3 and exp. dco = 7
coefficient of coincidence = 3/7 = 0.428
interference = 1 - 0.428 = 0.57
So approximately 57% of the time the double
cross-over does not occur.
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How can you use linkage data?
You can select for a difficult trait to observe if
it is linked to a more easily identified trait.
You can also estimate the number of progeny
you would have to screen to recover
recombinants that break an undesired
combination of linked alleles.
example:
3 genes with the following linkage map:
A  20mu  B  10mu  C
and an interference value of 60%.
Out of 1000 progeny, how many
recombinants would expect between A-B and
B-C.
To get the answer you need to first calculate
the number of double cross-overs you would
expect.
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freq. of single cross-overs (sco) A-B is .20
freq. of single cross-overs (sco) B-C is .10
exp. dco = (sco A-B x sco B-c) x total progeny
exp. dco = (.20 x .10) = .020 x 1000 = 20
So you would expect 20 double cross-overs
but there is an interference value of 60%
obs. dco = exp. dco - (exp. dco x interf. value)
obs. dco = 20 - (20 x .6) = 20 - 12 = 8
So only 8 double cross-overs would be
observed.
Use your observed double cross-overs to
calculate the number of single cross-overs
expected for each two gene combination.
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formula:
(total progeny x freq sco) - obs. dco
For recombinants between A-B out of 1000
progeny.
(1000 x .20) - 8 = 200 - 8 = 192
So there would be 192 A_B recombinants out
of 1000 progeny.
For recombinants between B-C out of 1000
progeny.
(1000 x .10) - 8 = 100 - 8 = 92
So there would be 92 B-C recombinants out
of 1000 progeny.
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